2

CHAPTER 0

..

Preliminaries

0-2

y 35 30 25 20 15 10 5 0

x 1

2

3

4

5

6

7

8

FIGURE 0.1 The Fibonacci sequence

these is the attempt to find patterns to help us better describe the world. The other theme is the interplay between graphs and functions. By connecting the powerful equation-solving techniques of algebra with the visual images provided by graphs, you will significantly improve your ability to make use of your mathematical skills in solving real-world problems.

0.1

POLYNOMIALS AND RATIONAL FUNCTIONS

The Real Number System and Inequalities Although mathematics is far more than just a study of numbers, our journey into calculus begins with the real number system. While this may seem to be a fairly mundane starting place, we want to give you the opportunity to brush up on those properties that are of particular interest for calculus. The most familiar set of numbers is the set of integers, consisting of the whole numbers and their additive inverses: 0, ±1, ±2, ±3, . . . . A rational number is any number of the p 27 form q , where p and q are integers and q = 0. For example, 23 , − 73 and 125 are all rational numbers. Notice that every integer n is also a rational number, since we can write it as the n quotient of two integers: n = . 1 p The irrational numbers are all those real numbers that cannot be written in the form q , where p and q are integers. Recall that rational numbers have decimal expansions that either ¯ 1 = 0.125 and 1 = 0.166666¯ are terminate or repeat. For instance, 12 = 0.5, 13 = 0.33333, 8 6 all rational numbers. By contrast, irrational numbers have decimal expansions that do not repeat or terminate. For instance, three familiar irrational numbers and their decimal expansions are √ 2 = 1.41421 35623 . . . , π = 3.14159 26535 . . . and

e = 2.71828 18284 . . . .

We picture the real numbers arranged along the number line displayed in Figure 0.2 (the real line). The set of real numbers is denoted by the symbol R.

0-3

SECTION 0.1

..

2 5 4 3 2 1

Polynomials and Rational Functions

3 0

1

3

p

2

3

4

5

e

FIGURE 0.2 The real line

For real numbers a and b, where a < b, we define the closed interval [a, b] to be the set of numbers between a and b, including a and b (the endpoints), that is, a

FIGURE 0.3 A closed interval

a

[a, b] = {x ∈ R | a ≤ x ≤ b},

b

as illustrated in Figure 0.3, where the solid circles indicate that a and b are included in [a, b]. Similarly, the open interval (a, b) is the set of numbers between a and b, but not including the endpoints a and b, that is,

b

FIGURE 0.4 An open interval

(a, b) = {x ∈ R | a < x < b}, as illustrated in Figure 0.4, where the open circles indicate that a and b are not included in (a, b). You should already be very familiar with the following properties of real numbers.

THEOREM 1.1 If a and b are real numbers and a < b, then (i) (ii) (iii) (iv)

For any real number c, a + c < b + c. For real numbers c and d, if c < d, then a + c < b + d. For any real number c > 0, a · c < b · c. For any real number c < 0, a · c > b · c.

REMARK 1.1 We need the properties given in Theorem 1.1 to solve inequalities. Notice that (i) says that you can add the same quantity to both sides of an inequality. Part (iii) says that you can multiply both sides of an inequality by a positive number. Finally, (iv) says that if you multiply both sides of an inequality by a negative number, the inequality is reversed.

We illustrate the use of Theorem 1.1 by solving a simple inequality.

EXAMPLE 1.1

Solving a Linear Inequality

Solve the linear inequality 2x + 5 < 13. Solution We can use the properties in Theorem 1.1 to isolate the x. First, subtract 5 from both sides to obtain (2x + 5) − 5 < 13 − 5

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0-4

2x < 8.

or

Finally, divide both sides by 2 (since 2 > 0, the inequality is not reversed) to obtain x < 4. We often write the solution of an inequality in interval notation. In this case, we get the interval (−∞, 4). You can deal with more complicated inequalities in the same way.

EXAMPLE 1.2

Solving a Two-Sided Inequality

Solve the two-sided inequality 6 < 1 − 3x ≤ 10. Solution First, recognize that this problem requires that we find values of x such that 6 < 1 − 3x

and

1 − 3x ≤ 10.

Here, we can use the properties in Theorem 1.1 to isolate the x by working on both inequalities simultaneously. First, subtract 1 from each term, to get 6 − 1 < (1 − 3x) − 1 ≤ 10 − 1 5 < −3x ≤ 9.

or

Now, divide by −3, but be careful. Since −3 < 0, the inequalities are reversed. We have 5 −3x 9 > ≥ −3 −3 −3 5 > x ≥ −3. 3 5 −3 ≤ x < − , 3 −

or We usually write this as

or in interval notation as [−3, − 53 ). y

You will often need to solve inequalities involving fractions. We present a typical example in the following.

8 4

4

EXAMPLE 1.3 x

2

1

2

4 8

FIGURE 0.5 y=

x −1 x +2

4

Solving an Inequality Involving a Fraction

x −1 ≥ 0. x +2 Solution In Figure 0.5, we show a graph of the function, which appears to indicate that the solution includes all x < −2 and x ≥ 1. Carefully read the inequality and observe that there are only three ways to satisfy this: either both numerator and denominator are positive, both are negative or the numerator is zero. To visualize this, we draw number lines for each of the individual terms, indicating where each is positive, negative or zero and use these to draw a third number line indicating the value of the quotient, as shown in the margin. In the third number line, we have placed an “ ”

Solve the inequality

0-5

SECTION 0.1

0

0

x2

2

0

2

5

For inequalities involving a polynomial of degree 2 or higher, factoring the polynomial and determining where the individual factors are positive and negative, as in example 1.4, will lead to a solution.

x1 x2

1

Polynomials and Rational Functions

above the −2 to indicate that the quotient is undefined at x = −2. From this last number line, you can see that the quotient is nonnegative whenever x < −2 or x ≥ 1. We write the solution in interval notation as (−∞, −2) ∪ [1, ∞).

x1

1

..

y

EXAMPLE 1.4

20

Solving a Quadratic Inequality

Solve the quadratic inequality

x

6 4 2

2

4

FIGURE 0.6

0

(1.2)

x3

3

(x + 3)(x − 2) > 0.

This can happen in only two ways: when both factors are positive or when both factors are negative. As in example 1.3, we draw number lines for both of the individual factors, indicating where each is positive, negative or zero and use these to draw a number line representing the product. We show these in the margin. Notice that the third number line indicates that the product is positive whenever x < −3 or x > 2. We write this in interval notation as (−∞, −3) ∪ (2, ∞).

y = x2 + x − 6 0

Solution In Figure 0.6, we show a graph of the polynomial on the left side of the inequality. Since this polynomial factors, (1.1) is equivalent to

6

10

(1.1)

x 2 + x − 6 > 0.

10

No doubt, you will recall the following standard definition. x2

2

0

3

0

(x 3)(x 2)

2

DEFINITION 1.1

The absolute value of a real number x is |x| =

x, −x,

if x ≥ 0. if x < 0

Make certain that you read Definition 1.1 correctly. If x is negative, then −x is positive. This says that |x| ≥ 0 for all real numbers x. For instance, using the definition, |− 4 | = −(−4) = 4, notice that for any real numbers a and b, |a · b| = |a| · |b|.

NOTES

However,

For any two real numbers a and b, |a − b| gives the distance between a and b. (See Figure 0.7.)

|a + b| = |a| + |b|,

in general. (To verify this, simply take a = 5 and b = −2 and compute both quantities.) However, it is always true that |a + b| ≤ |a| + |b|.

a b a

b

FIGURE 0.7 The distance between a and b

This is referred to as the triangle inequality. The interpretation of |a − b| as the distance between a and b (see the note in the margin) is particularly useful for solving inequalities involving absolute values. Wherever possible, we suggest that you use this interpretation to read what the inequality means, rather than merely following a procedure to produce a solution.

6

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EXAMPLE 1.5

0-6

Solving an Inequality Containing an Absolute Value

Solve the inequality |x − 2| < 5. 5

5

2 5 3

257

2

FIGURE 0.8 |x − 2| < 5

(1.3)

Solution Before you start trying to solve this, take a few moments to read what it says. Since |x − 2| gives the distance from x to 2, (1.3) says that the distance from x to 2 must be less than 5. So, find all numbers x whose distance from 2 is less than 5. We indicate the set of all numbers within a distance 5 of 2 in Figure 0.8. You can now read the solution directly from the figure: −3 < x < 7 or in interval notation: (−3, 7). Many inequalities involving absolute values can be solved simply by reading the inequality correctly, as in example 1.6.

EXAMPLE 1.6

Solving an Inequality with a Sum Inside an Absolute Value

Solve the inequality |x + 4| ≤ 7. 7

7

4 7 11 4

(1.4)

Solution To use our distance interpretation, we must first rewrite (1.4) as

4 7 3

|x − (−4)| ≤ 7. This now says that the distance from x to −4 is less than or equal to 7. We illustrate the solution in Figure 0.9, from which it follows that −11 ≤ x ≤ 3 or [−11, 3].

FIGURE 0.9 | x + 4 |≤ 7

Recall that for any real number r > 0, |x| < r is equivalent to the following inequality not involving absolute values: −r < x < r. In example 1.7, we use this to revisit the inequality from example 1.5.

EXAMPLE 1.7

An Alternative Method for Solving Inequalities

Solve the inequality |x − 2| < 5. y

Solution This is equivalent to the two-sided inequality (x2, y2)

y2

−5 < x − 2 < 5. Adding 2 to each term, we get the solution

Distance

y2 y1

−3 < x < 7, or in interval notation (−3, 7), as before.

y1

(x1, y1)

x2 x1

x1

x2

FIGURE 0.10 Distance

x

Recall that the distance between two points (x1 , y1 ) and (x2 , y2 ) is a simple consequence of the Pythagorean Theorem and is given by d{(x1 , y1 ), (x2 , y2 )} = (x2 − x1 )2 + (y2 − y1 )2 . We illustrate this in Figure 0.10.

0-7

SECTION 0.1

EXAMPLE 1.8

..

Polynomials and Rational Functions

7

Using the Distance Formula

Find the distance between the points (1, 2) and (3, 4). Solution The distance between (1, 2) and (3, 4) is d{(1, 2), (3, 4)} =

(3 − 1)2 + (4 − 2)2 =

√

4+4=

√

8.

Equations of Lines Year 1960 1970 1980 1990

x

The federal government conducts a nationwide census every 10 years to determine the population. Population data for the last several decades are shown in the accompanying table. One difficulty with analyzing these data is that the numbers are so large. This problem is remedied by transforming the data. We can simplify the year data by defining x to be the number of years since 1960. Then, 1960 corresponds to x = 0, 1970 corresponds to x = 10 and so on. The population data can be simplified by rounding the numbers to the nearest million. The transformed data are shown in the accompanying table and a scatter plot of these data points is shown in Figure 0.11. Most people would say that the points in Figure 0.11 appear to form a straight line. (Use a ruler and see if you agree.) To determine whether the points are, in fact, on the same line (such points are called colinear), we might consider the population growth in each of the indicated decades. From 1960 to 1970, the growth was 24 million. (That is, to move from the first point to the second, you increase x by 10 and increase y by 24.) Likewise, from 1970 to 1980, the growth was 24 million. However, from 1980 to 1990, the growth was only 22 million. Since the rate of growth is not constant, the data points do not fall on a line. Notice that to stay on the same line, y would have had to increase by 24 again. The preceding argument involves the familiar concept of slope.

U.S. Population 179,323,175 203,302,031 226,542,203 248,709,873

y

0

179

10

203

20

227

30

249

Transformed data y 250 200

DEFINITION 1.2 For x1 = x2 , the slope of the straight line through the points (x1 , y1 ) and (x2 , y2 ) is the number y2 − y1 m= . (1.5) x2 − x1

150 100 50 x 10

20

FIGURE 0.11 Population data

30

When x1 = x2 , the line through (x1 , y1 ) and (x2 , y2 ) is vertical and the slope is undefined. y , We often describe slope as “the change in y divided by the change in x,” written x Rise or more simply as (see Figure 0.12a on the following page). Run The slope of a straight line is the same no matter which two points on the line you select. Referring to Figure 0.12b (where the line has positive slope), notice that for any four points A, B, D and E on the line, the two right triangles ABC and DEF are similar. Recall that for similar triangles, the ratios of corresponding sides must be the same. In this case, this says that y y = x x

8

CHAPTER 0

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Preliminaries

0-8

y

y

B (x2, y2)

y2

y A

y y2 y1 Rise y1

x

E

C

y

(x1, y1) D

x x2 x1

x1

F

x

Run x

x2

x

FIGURE 0.12a

FIGURE 0.12b

Slope

Similar triangles and slope

and so, the slope is the same no matter which two points on the line are selected. Furthermore, a line is the only curve with constant slope. Notice that a line is horizontal if and only if its slope is zero.

EXAMPLE 1.9

Finding the Slope of a Line

Find the slope of the line through the points (4, 3) and (2, 5). Solution From (1.5), we get m=

EXAMPLE 1.10

y2 − y1 5−3 2 = = = −1. x2 − x1 2−4 −2

Using Slope to Determine if Points Are Colinear

Use slope to determine whether the points (1, 2), (3, 10) and (4, 14) are colinear. Solution First, notice that the slope of the line joining (1, 2) and (3, 10) is m1 =

y2 − y1 10 − 2 8 = = = 4. x2 − x1 3−1 2

Similarly, the slope through the line joining (3, 10) and (4, 14) is m2 =

y2 − y1 14 − 10 = = 4. x2 − x1 4−3

Since the slopes are the same, the points must be colinear. Recall that if you know the slope and a point through which the line must pass, you have enough information to graph the line. The easiest way to graph a line is to plot two points and then draw the line through them. In this case, you need only to find a second point.

0-9

SECTION 0.1

..

Polynomials and Rational Functions

9

y

EXAMPLE 1.11

Graphing a Line

If a line passes through the point (2, 1) with slope 23 , find a second point on the line and then graph the line. y2 − y1 Solution Since slope is given by m = , we take m = 23 , y1 = 1 and x1 = 2, x2 − x1 to obtain

4 3 2 1

2 y2 − 1 = . 3 x2 − 2

x 1

2

3

4

5

You are free to choose the x-coordinate of the second point. For instance, to find the point at x2 = 5, substitute this in and solve. From

1

2 y2 − 1 y2 − 1 = = , 3 5−2 3

FIGURE 0.13a Graph of straight line

we get 2 = y2 − 1 or y2 = 3. A second point is then (5, 3). The graph of the line is shown in Figure 0.13a. An alternative method for finding a second point is to use the slope

y

m=

4

y 2

3 2 1 x 3 1

2

3

4

x 5

1

FIGURE 0.13b

2 y = . 3 x

The slope of 23 says that if we move three units to the right, we must move two units up to stay on the line, as illustrated in Figure 0.13b. In example 1.11, the choice of x = 5 was entirely arbitrary; you can choose any x-value you want to find a second point. Further, since x can be any real number, you can leave x as a variable and write out an equation satisfied by any point (x, y) on the line. In the general case of the line through the point (x0 , y0 ) with slope m, we have from (1.5) that y − y0 m= . (1.6) x − x0 Multiplying both sides of (1.6) by (x − x0 ), we get

Using slope to find a second point

y − y0 = m(x − x0 ) or

POINT-SLOPE FORM OF A LINE y = m(x − x0 ) + y0 .

y 7

Equation (1.7) is called the point-slope form of the line.

6 5

EXAMPLE 1.12

4

Finding the Equation of a Line Given Two Points

Find an equation of the line through the points (3, 1) and (4, −1), and graph the line.

3 2

−1 − 1 −2 = = −2. Using (1.7) with slope 4−3 1 m = −2, x-coordinate x0 = 3 and y-coordinate y0 = 1, we get the equation of the line:

Solution From (1.5), the slope is m =

1 x 1

(1.7)

1

2

3

4

y = −2(x − 3) + 1. FIGURE 0.14 y = −2(x − 3) + 1

(1.8)

To graph the line, plot the points (3, 1) and (4, −1), and you can easily draw the line seen in Figure 0.14.

10

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Preliminaries

0-10

In example 1.12, you may be tempted to simplify the expression for y given in (1.8). As it turns out, the point-slope form of the equation is often the most convenient to work with. So, we will typically not ask you to rewrite this expression in other forms. At times, a form of the equation called the slope-intercept form is more convenient. This has the form y = mx + b, where m is the slope and b is the y-intercept (i.e., the place where the graph crosses the y-axis). In example 1.12, you simply multiply out (1.8) to get y = −2x + 6 + 1 or y = −2x + 7. As you can see from Figure 0.14, the graph crosses the y-axis at y = 7. Theorem 1.2 presents a familiar result on parallel and perpendicular lines.

THEOREM 1.2 Two (nonvertical) lines are parallel if they have the same slope. Further, any two vertical lines are parallel. Two (nonvertical) lines of slope m 1 and m 2 are perpendicular whenever the product of their slopes is −1 (i.e., m 1 · m 2 = −1). Also, any vertical line and any horizontal line are perpendicular. y

Since we can read the slope from the equation of a line, it’s a simple matter to determine when two lines are parallel or perpendicular. We illustrate this in examples 1.13 and 1.14.

20 10

4

x

2

2

EXAMPLE 1.13

4

Find an equation of the line parallel to y = 3x − 2 and through the point (−1, 3).

10

Solution It’s easy to read the slope of the line from the equation: m = 3. The equation of the parallel line is then

20

y = 3[x − (−1)] + 3

FIGURE 0.15

or simply y = 3(x + 1) + 3. We show a graph of both lines in Figure 0.15.

Parallel lines y

EXAMPLE 1.14

Finding the Equation of a Perpendicular Line

Find an equation of the line perpendicular to y = −2x + 4 and intersecting the line at the point (1, 2).

4 2 x

2

Finding the Equation of a Parallel Line

2

4

2 4

FIGURE 0.16 Perpendicular lines

Solution The slope of y = −2x + 4 is −2. The slope of the perpendicular line is then −1/(−2) = 12 . Since the line must pass through the point (1, 2), the equation of the perpendicular line is 1 (x − 1) + 2. 2 We show a graph of the two lines in Figure 0.16. y=

We now return to this subsection’s introductory example and use the equation of a line to estimate the population in the year 2000.

0-11

SECTION 0.1

EXAMPLE 1.15

..

Polynomials and Rational Functions

11

Using a Line to Estimate Population

Given the population data for the census years 1960, 1970, 1980 and 1990, estimate the population for the year 2000. Solution We began this subsection by showing that the points in the corresponding table are not colinear. Nonetheless, they are nearly colinear. So, why not use the straight line connecting the last two points (20, 227) and (30, 249) (corresponding to the populations in the years 1980 and 1990) to estimate the population in 2000? (This is a simple example of a more general procedure called extrapolation.) The slope of the line joining the two data points is

y 300

200

m= 100

249 − 227 22 11 = = . 30 − 20 10 5

The equation of the line is then y=

x 10

20

30

40

50

11 (x − 30) + 249. 5

See Figure 0.17 for a graph of the line. If we follow this line to the point corresponding to x = 40 (the year 2000), we have the estimated population

FIGURE 0.17 Population

11 (40 − 30) + 249 = 271. 5 That is, the estimated population is 271 million people. The actual census figure for 2000 was 281 million, which indicates that the U.S. population has grown at a rate that is faster than linear.

Functions A

B f

x

For any two subsets A and B of the real line, we make the following familiar definition.

y

DEFINITION 1.3

REMARK 1.2 Functions can be defined by simple formulas, such as f (x) = 3x + 2, but in general, any correspondence meeting the requirement of matching exactly one y to each x defines a function.

A function f is a rule that assigns exactly one element y in a set B to each element x in a set A. In this case, we write y = f (x). We call the set A the domain of f. The set of all values f (x) in B is called the range of f . That is, the range of f is { f (x) | x ∈ A}. Unless explicitly stated otherwise, the domain of a function f is the largest set of real numbers for which the function is defined. We refer to x as the independent variable and to y as the dependent variable.

By the graph of a function f, we mean the graph of the equation y = f (x). That is, the graph consists of all points (x, y), where x is in the domain of f and where y = f (x). Notice that not every curve is the graph of a function, since for a function, only one y-value corresponds to a given value of x. You can graphically determine whether a curve is the graph of a function by using the vertical line test: if any vertical line intersects the graph in more than one point, the curve is not the graph of a function.

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CHAPTER 0

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0-12

EXAMPLE 1.16

Using the Vertical Line Test

Determine which of the curves in Figures 0.18a and 0.18b correspond to functions. y

y 1 x

1

1

x 0.5

1

2

1

FIGURE 0.18a y

Solution Notice that the circle in Figure 0.18a is not the graph of a function, since a vertical line at x = 0.5 intersects the circle twice (see Figure 0.19a). The graph in Figure 0.18b is the graph of a function, even though it swings up and down repeatedly. Although horizontal lines intersect the graph repeatedly, vertical lines, such as the one at x = 1.2, intersect only once (see Figure 0.19b). 0.5

1

FIGURE 0.18b

x 1

You are already familiar with a number of different types of functions, and we will only briefly review these here and in sections 0.4 and 0.5. The functions that you are probably most familiar with are polynomials. These are the simplest functions to work with because they are defined entirely in terms of arithmetic.

FIGURE 0.19a Curve fails vertical line test

DEFINITION 1.4

y

A polynomial is any function that can be written in the form f (x) = an x n + an−1 x n−1 + · · · + a1 x + a0 ,

1

x 0.5

1

2

1

FIGURE 0.19b

where a0 , a1 , a2 , . . . , an are real numbers (the coefficients of the polynomial) with an = 0 and n ≥ 0 is an integer (the degree of the polynomial). Note that the domain of every polynomial function is the entire real line. Further, recognize that the graph of the linear (degree 1) polynomial f (x) = ax + b is a straight line.

Curve passes vertical line test

EXAMPLE 1.17

Sample Polynomials

The following are all examples of polynomials: f (x) = 2 (polynomial of degree 0 or constant), f (x) = 3x + 2 (polynomial of degree 1 or linear polynomial), f (x) = 5x 2 − 2x + 1 (polynomial of degree 2 or quadratic polynomial), f (x) = x 3 − 2x + 1 (polynomial of degree 3 or cubic polynomial), f (x) = −6x 4 + 12x 2 − 3x + 13 (polynomial of degree 4 or quartic polynomial),

0-13

SECTION 0.1

..

Polynomials and Rational Functions

13

and f (x) = 2x 5 + 6x 4 − 8x 2 + x − 3 (polynomial of degree 5 or quintic polynomial). We show graphs of these six functions in Figures 0.20a–0.20f. y

y 120

15 y

10

4 1

x

2

2

4 40

5 10 15

x

3 2

80

5

3

x

6 4 2

2

4

6

FIGURE 0.20a

FIGURE 0.20b

FIGURE 0.20c

f (x) = 2

f (x) = 3x + 2

f (x) = 5x 2 − 2x + 1

y

y

10

20

5

10

1

x 1

2

2

3

y 20 10

x

1

1

5

10

10

20

2 3

2

x

1

1 10

FIGURE 0.20d

FIGURE 0.20e

FIGURE 0.20f

f (x) = x 3 − 2x + 1

f (x) = −6x 4 + 12x 2 − 3x + 13

f (x) = 2x 5 + 6x 4 − 8x 2 + x − 3

DEFINITION 1.5 Any function that can be written in the form f (x) =

p(x) , q(x)

where p and q are polynomials, is called a rational function.

Notice that since p(x) and q(x) are polynomials, they are both defined for all x, and p(x) so, the rational function f (x) = is defined for all x for which q(x) = 0. q(x)

14

CHAPTER 0

..

Preliminaries

0-14

y

EXAMPLE 1.18 10

A Sample Rational Function

Find the domain of the function

5

3

f (x) =

x 2 + 7x − 11 . x2 − 4

x

1

1

3

Solution Here, f (x) is a rational function. We show a graph in Figure 0.21. Its domain consists of those values of x for which the denominator is nonzero. Notice that

5

x 2 − 4 = (x − 2)(x + 2)

10

and so, the denominator is zero if and only if x = ±2. This says that the domain of f is

FIGURE 0.21

{x ∈ R | x = ±2} = (−∞, −2) ∪ (−2, 2) ∪ (2, ∞).

x 2 + 7x − 11 f (x) = x2 − 4

√ The square root function is defined in the usual way. When we√ write y = x, we mean that y is that number for which y 2 =√x and y ≥ 0. In particular, 4 = 2. Be careful not to write erroneous statements such as 4 = ±2. In particular, be careful to write √ x 2 = |x|. √ Since x 2 is asking for the nonnegative number whose square is x 2 , we are looking for |x| and not x. We can say √ x 2 = x, only for x ≥ 0. √ Similarly, for any integer n ≥ 2, y = n x whenever y n = x, where for n even, x ≥ 0 and y ≥ 0.

EXAMPLE 1.19

Finding the Domain of a Function Involving a Square Root or a Cube Root

Find the domains of f (x) =

√

x 2 − 4 and g(x) =

√ 3

x 2 − 4.

Solution Since even roots are defined only for nonnegative values, f (x) is defined only for x 2 − 4 ≥ 0. Notice that this is equivalent to having x 2 ≥ 4, which occurs when x ≥ 2 or x ≤ −2. The domain of f is then (−∞, −2] ∪ [2, ∞). On the other hand, odd roots are defined for both positive and negative values. Consequently, the domain of g(x) is the entire real line, (−∞, ∞). We often find it useful to label intercepts and other significant points on a graph. Finding these points typically involves solving equations. A solution of the equation f (x) = 0 is called a zero of the function f or a root of the equation f (x) = 0. Notice that a zero of the function f corresponds to an x-intercept of the graph of y = f (x).

y 10 8

2

EXAMPLE 1.20

Finding Zeros by Factoring

6

Find all x- and y-intercepts of f (x) = x 2 − 4x + 3.

4

Solution To find the y-intercept, set x = 0 to obtain

2

y = 0 − 0 + 3 = 3. x 1

2

3

4

FIGURE 0.22 y = x 2 − 4x + 3

6

To find the x-intercepts, solve the equation f (x) = 0. In this case, we can factor to get f (x) = x 2 − 4x + 3 = (x − 1)(x − 3) = 0. You can now read off the zeros: x = 1 and x = 3, as indicated in Figure 0.22.

0-15

SECTION 0.1

..

Polynomials and Rational Functions

15

Unfortunately, factoring is not always so easy. Of course, for the quadratic equation ax 2 + bx + c = 0 (for a = 0), the solution(s) are given by the familiar quadratic formula: x=

EXAMPLE 1.21

−b ±

√

b2 − 4ac . 2a

Finding Zeros Using the Quadratic Formula

Find the zeros of f (x) = x 2 − 5x − 12. Solution You probably won’t have much luck trying to factor this. However, from the quadratic formula, we have √ √ −(−5) ± (−5)2 − 4 · 1 · (−12) 5 ± 25 + 48 5 ± 73 x= = = . 2·1 2 2 √

So, the two solutions are given by x = 52 + 273 ≈ 6.772 and x = (No wonder you couldn’t factor the polynomial!)

5 2

−

√

73 2

≈ −1.772.

Finding zeros of polynomials of degree higher than 2 and other functions is usually trickier and is sometimes impossible. At the least, you can always find an approximation of any zero(s) by using a graph to zoom in closer to the point(s) where the graph crosses the x-axis, as we’ll illustrate shortly. A more basic question, though, is to determine how many zeros a given function has. In general, there is no way to answer this question without the use of calculus. For the case of polynomials, however, Theorem 1.3 (a consequence of the Fundamental Theorem of Algebra) provides a clue.

THEOREM 1.3 A polynomial of degree n has at most n distinct zeros.

REMARK 1.3 Polynomials may also have complex zeros. For instance, f (x) = x 2 + 1 has only the complex zeros x = ±i, where i is the imaginary number defined √ by i = −1.

Notice that Theorem 1.3 does not say how many zeros a given polynomial has, but rather, that the maximum number of distinct (i.e., different) zeros is the same as the degree. A polynomial of degree n may have anywhere from 0 to n distinct real zeros. However, polynomials of odd degree must have at least one real zero. For instance, for the case of a cubic polynomial, we have one of the three possibilities illustrated in Figures 0.23a, 0.23b and 0.23c on the following page. In these three figures, we show the graphs of cubic polynomials with 1, 2 and 3 distinct, real zeros, respectively. These are the graphs of the functions f (x) = x 3 − 2x 2 + 3 = (x + 1)(x 2 − 3x + 3), g(x) = x 3 − x 2 − x + 1 = (x + 1)(x − 1)2 and

h(x) = x 3 − 3x 2 − x + 3 = (x + 1)(x − 1)(x − 3),

respectively. Note that you can see from the factored form where the zeros are (and how many there are).

16

CHAPTER 0

..

Preliminaries

0-16

y

y

x1

y

x2

x

x1

x2

x3

x

x

x1

FIGURE 0.23a

FIGURE 0.23b

FIGURE 0.23c

One zero

Two zeros

Three zeros

y 4

Theorem 1.4 provides an important connection between factors and zeros of polynomials.

2

THEOREM 1.4 (Factor Theorem) 2

x

1

1

2

For any polynomial f, f (a) = 0 if and only if (x − a) is a factor of f (x).

3

2

EXAMPLE 1.22

Find the zeros of f (x) = x 3 − x 2 − 2x + 2.

FIGURE 0.24a y = x 3 − x 2 − 2x + 2

0.2

1.41

1.39

x

0.2

FIGURE 0.24b Zoomed in on zero near x = −1.4 0.02 x 1.40

1.42

0.02

Solution By calculating f (1), you can see that one zero of this function is x = 1, but how many other zeros are there? A graph of the function (see Figure 0.24a) shows that there are two other zeros of f , one near x = −1.5 and one near x = 1.5. You can find these zeros more precisely by using your graphing calculator or computer algebra system to zoom in on the locations of these zeros (as shown in Figures 0.24b and 0.24c). From these zoomed graphs it is clear that the two remaining zeros of f are near x = 1.41 and x = −1.41. You can make these estimates more precise by zooming in even more closely. Most graphing calculators and computer algebra systems can also find approximate zeros, using a built-in “solve” program. In Chapter 3, we present a versatile method (called Newton’s method) for obtaining accurate approximations to zeros. The only way to find the exact solutions is to factor the expression (using either long division or synthetic division). Here, we have √ √ f (x) = x 3 − x 2 − 2x + 2 = (x − 1)(x 2 − 2) = (x − 1)(x − 2)(x + 2), √ √ from which you can see that the zeros are x = 1, x = 2 and x = − 2. Recall that to find the points of intersection of two curves defined by y = f (x) and y = g(x), we set f (x) = g(x) to find the x-coordinates of any points of intersection.

EXAMPLE 1.23

0.04

FIGURE 0.24c Zoomed in on zero near x = 1.4

Finding the Zeros of a Cubic Polynomial

Finding the Intersections of a Line and a Parabola

Find the points of intersection of the parabola y = x 2 − x − 5 and the line y = x + 3. Solution A sketch of the two curves (see Figure 0.25) shows that there are two intersections, one near x = −2 and the other near x = 4. To determine these precisely,

0-17

SECTION 0.1

Polynomials and Rational Functions

17

we set the two functions equal and solve for x:

y

x 2 − x − 5 = x + 3.

20

Subtracting (x + 3) from both sides leaves us with 0 = x 2 − 2x − 8 = (x − 4)(x + 2).

10

4 2

..

This says that the solutions are exactly x = −2 and x = 4. We compute the corresponding y-values from the equation of the line y = x + 3 (or the equation of the parabola). The points of intersection are then (−2, 1) and (4, 7). Notice that these are consistent with the intersections seen in Figure 0.25.

x 4

6

10

Unfortunately, you won’t always be able to solve equations exactly, as we did in examples 1.20–1.23. We explore some options for dealing with more difficult equations in section 0.2.

FIGURE 0.25 y = x + 3 and y = x 2 − x − 5

EXERCISES 0.1 WRITING EXERCISES 1. If the slope of the line passing through points A and B equals the slope of the line passing through points B and C, explain why the points A, B and C are colinear.

In exercises 11–16, find a second point on the line with slope m and point P, graph the line and find an equation of the line.

2. If a graph fails the vertical line test, it is not the graph of a function. Explain this result in terms of the definition of a function.

11. m = 2, P = (1, 3)

3. You should not automatically write the equation of a line in slope-intercept form. Compare the following forms of the same line: y = 2.4(x − 1.8) + 0.4 and y = 2.4x − 3.92. Given x = 1.8, which equation would you rather use to compute y? How about if you are given x = 0? For x = 8, is there any advantage to one equation over the other? Can you quickly read off the slope from either equation? Explain why neither form of the equation is “better.” 4. To understand Definition 1.1, you must believe that | x |= −x for negative x’s. Using x = −3 as an example, explain in words why multiplying x by −1 produces the same result as taking the absolute value of x.

12. m = −2, P = (1, 4) 13. m = 0, P = (−1, 1) 14. m = 12 , P = (2, 1) 15. m = 1.2, P = (2.3, 1.1) 16. m = − 14 , P = (−2, 1) In exercises 17–22, determine if the lines are parallel, perpendicular, or neither. 17. y = 3(x − 1) + 2 and y = 3(x + 4) − 1 18. y = 2(x − 3) + 1 and y = 4(x − 3) + 1 19. y = −2(x + 1) − 1 and y = 12 (x − 2) + 3

In exercises 1–4, determine if the points are colinear.

20. y = 2x − 1 and y = −2x + 2

1. (2, 1), (0, 2), (4, 0)

2. (3, 1), (4, 4), (5, 8)

21. y = 3x + 1 and y = − 13 x + 2

3. (4, 1), (3, 2), (1, 3)

4. (1, 2), (2, 5), (4, 8)

22. x + 2y = 1 and 2x + 4y = 3

In exercises 5–10, find the slope of the line through the given points. 5. (1, 2), (3, 6)

6. (1, 2), (3, 3)

In exercises 23–26, find an equation of a line through the given point and (a) parallel to and (b) perpendicular to the given line.

7. (3, −6), (1, −1)

8. (1, −2), (−1, −3)

23. y = 2(x + 1) − 2 at (2, 1)

24. y = 3(x − 2) + 1 at (0, 3)

25. y = 2x + 1 at (3, 1)

26. y = 1 at (0, −1)

9. (0.3, −1.4), (−1.1, −0.4)

10. (1.2, 2.1), (3.1, 2.4)

18

..

CHAPTER 0

Preliminaries

In exercises 27–30, find an equation of the line through the given points and compute the y-coordinate of the point on the line corresponding to x 4.

0-18

In exercises 31–34, use the vertical line test to determine whether the curve is the graph of a function. y

31.

y

27.

10 5

5

4 3

x

3 2

2

2

3

5

1

10 x 1

2

3

4

5

y

28.

y

32. 6

10

5

5

4

4

3

2

x 2

2

4

5

1 x 1

2

3

4

5

10

6

y

29. 4

y

33.

3

6

2

4

1

2 x 0.5

1.0

1.5

x

3 2 1

2.0

1

2

y

30.

1

2.0

0.5

1.0

2

1

y

34.

3.0

x

x 1

0.5

1

1.5

2

3

0-19

SECTION 0.1

In exercises 35–40, identify the given function as polynomial, rational, both or neither.

Speed

35. f (x) = x 3 − 4x + 1 x 2 + 2x − 1 x +1 √ 39. f (x) = x 2 + 1 37. f (x) =

..

4 x2 − 1

38. f (x) =

x 3 + 4x − 1 x4 − 1

40. f (x) = 2x − x 2/3 − 6

46. f (x) =

4x x 2 + 2x − 6

In exercises 47–50, find the indicated function values. 47. f (x) = x 2 − x − 1;

f (0), f (2), f (−3), f (1/2)

x +1 ; f (0), f (2), f (−2), f (1/2) x −1 √ 49. f (x) = x + 1; f (0), f (3), f (−1), f (1/2) 48. f (x) =

50. f (x) =

3 ; x

19

36. f (x) = 3 − 2x + x 4

In exercises 41–46, find the domain of the function. √ √ 41. f (x) = x + 2 42. f (x) = 2x + 1 √ √ 44. f (x) = x 2 − 4 43. f (x) = 3 x − 1 45. f (x) =

Polynomials and Rational Functions

Time

FIGURE A Bicycle speed

60. Figure B shows the population of a small country as a function of time. During the time period shown, the country experienced two influxes of immigrants, a war and a plague. Identify these important events. Population

f (1), f (10), f (100), f (1/3)

In exercises 51–54, a brief description is given of a physical situation. For the indicated variable, state a reasonable domain.

Time

51. A parking deck is to be built; x = width of deck (in feet).

FIGURE B

52. A parking deck is to be built on a 200 -by-200 lot; x = width of deck (in feet). 53. A new candy bar is to be sold; x = number of candy bars sold in the first month. 54. A new candy bar is to be sold; x = cost of candy bar (in cents). In exercises 55–58, discuss whether you think y would be a function of x. 55. y = grade you get on an exam, x = number of hours you study

Population In exercises 61–66, find all intercepts of the given graph. 61. y = x 2 − 2x − 8

62. y = x 2 + 4x + 4

63. y = x 3 − 8

64. y = x 3 − 3x 2 + 3x − 1

65. y =

x2 − 4 x +1

66. y =

2x − 1 x2 − 4

56. y = probability of getting lung cancer, x = number of cigarettes smoked per day

In exercises 67–74, factor and/or use the quadratic formula to find all zeros of the given function.

57. y = a person’s weight, x = number of minutes exercising per day

67. f (x) = x 2 − 4x + 3

68. f (x) = x 2 + x − 12

69. f (x) = x 2 − 4x + 2

70. f (x) = 2x 2 + 4x − 1

71. f (x) = x 3 − 3x 2 + 2x

72. f (x) = x 3 − 2x 2 − x + 2

73. f (x) = x 6 + x 3 − 2

74. f (x) = x 3 + x 2 − 4x − 4

58. y = speed at which an object falls, x = weight of object 59. Figure A shows the speed of a bicyclist as a function of time. For the portions of this graph that are flat, what is happening to the bicyclist’s speed? What is happening to the bicyclist’s speed when the graph goes up? down? Identify the portions of the graph that correspond to the bicyclist going uphill; downhill.

75. The boiling point of water (in degrees Fahrenheit) at elevation h (in thousands of feet above sea level) is given by B(h) = −1.8h + 212. Find h such that water boils at 98.6◦ . Why would this altitude be dangerous to humans?

20

..

CHAPTER 0

Preliminaries

0-20

76. The spin rate of a golf ball hit with a 9 iron has been measured at 9100 rpm for a 120-compression ball and at 10,000 rpm for a 60-compression ball. Most golfers use 90-compression balls. If the spin rate is a linear function of compression, find the spin rate for a 90-compression ball. Professional golfers often use 100-compression balls. Estimate the spin rate of a 100-compression ball. 77. The chirping rate of a cricket depends on the temperature. A species of tree cricket chirps 160 times per minute at 79◦ F and 100 times per minute at 64◦ F. Find a linear function relating temperature to chirping rate. 78. When describing how to measure temperature by counting cricket chirps, most guides suggest that you count the number of chirps in a 15-second time period. Use exercise 77 to explain why this is a convenient period of time. 79. A person has played a computer game many times. The statistics show that she has won 415 times and lost 120 times, and the winning percentage is listed as 78%. How many times in a

0.2

40 20 x

2

2

2. Solve the equation | x − 2 | + | x − 3 | = 1. (Hint: It’s an unusual solution, in that it’s more than just a couple of numbers.) Then, solve the equation √ √ x + 3 − 4 x − 1 + x + 8 − 6 x − 1 = 1. (Hint: If you make the correct substitution, you can use your solution to the previous equation.)

4

EXAMPLE 2.1

FIGURE 0.26a y

8 4 x 1

FIGURE 0.26b y = 3x 2 − 1

Generating a Calculator Graph

Use your calculator or computer to sketch a graph of f (x) = 3x 2 − 1.

y = 3x 2 − 1

1

1. Suppose you have a machine that will proportionally enlarge a photograph. For example, it could enlarge a 4 × 6 photograph to 8 × 12 by doubling the width and height. You could make an 8 × 10 picture by cropping 1 inch off each side. Explain how you would enlarge a 3 12 × 5 picture to an 8 × 10. A friend returns from Scotland with a 3 12 × 5 picture showing the Loch Ness monster in the outer 14 on the right. If you use your procedure to make an 8 × 10 enlargement, does Nessie make the cut?

The relationships between functions and their graphs are central topics in calculus. Graphing calculators and user-friendly computer software allow you to explore these relationships for a much wider variety of functions than you could with pencil and paper alone. This section presents a general framework for using technology to explore the graphs of functions. Recall that the graphs of linear functions are straight lines and the graphs of quadratic polynomials are parabolas. One of the goals of this section is for you to become more familiar with the graphs of other functions. The best way to become familiar is through experience, by working example after example.

60

2

EXPLORATORY EXERCISES

GRAPHING CALCULATORS AND COMPUTER ALGEBRA SYSTEMS

y

4

row must she win to raise the reported winning percentage to 80%?

2

Solution You should get an initial graph that looks something like that in Figure 0.26a. This is simply a parabola opening upward. A graph is often used to search for important points, such as x-intercepts, y-intercepts or peaks and troughs. In this case, we could see these points better if we zoom in, that is, display a smaller range of x- and y-values than the technology has initially chosen for us. The graph in Figure 0.26b shows x-values from x = −2 to x = 2 and y-values from y = −2 to y = 10. You can see more clearly in Figure 0.26b that the parabola bottoms out roughly at the point (0, −1) and crosses the x-axis at approximately x = −0.5 and x = 0.5. You can make this more precise by doing some algebra. Recall that an x-intercept is a point where y = 0 or f (x) = 0. Solving 3x 2 − 1 = 0 gives 3x 2 = 1 or x 2 = 13 , so that x = ± 13 ≈ ±0.57735.

20

..

CHAPTER 0

Preliminaries

0-20

76. The spin rate of a golf ball hit with a 9 iron has been measured at 9100 rpm for a 120-compression ball and at 10,000 rpm for a 60-compression ball. Most golfers use 90-compression balls. If the spin rate is a linear function of compression, find the spin rate for a 90-compression ball. Professional golfers often use 100-compression balls. Estimate the spin rate of a 100-compression ball. 77. The chirping rate of a cricket depends on the temperature. A species of tree cricket chirps 160 times per minute at 79◦ F and 100 times per minute at 64◦ F. Find a linear function relating temperature to chirping rate. 78. When describing how to measure temperature by counting cricket chirps, most guides suggest that you count the number of chirps in a 15-second time period. Use exercise 77 to explain why this is a convenient period of time. 79. A person has played a computer game many times. The statistics show that she has won 415 times and lost 120 times, and the winning percentage is listed as 78%. How many times in a

0.2

40 20 x

2

2

2. Solve the equation | x − 2 | + | x − 3 | = 1. (Hint: It’s an unusual solution, in that it’s more than just a couple of numbers.) Then, solve the equation √ √ x + 3 − 4 x − 1 + x + 8 − 6 x − 1 = 1. (Hint: If you make the correct substitution, you can use your solution to the previous equation.)

4

EXAMPLE 2.1

FIGURE 0.26a y

8 4 x 1

FIGURE 0.26b y = 3x 2 − 1

Generating a Calculator Graph

Use your calculator or computer to sketch a graph of f (x) = 3x 2 − 1.

y = 3x 2 − 1

1

1. Suppose you have a machine that will proportionally enlarge a photograph. For example, it could enlarge a 4 × 6 photograph to 8 × 12 by doubling the width and height. You could make an 8 × 10 picture by cropping 1 inch off each side. Explain how you would enlarge a 3 12 × 5 picture to an 8 × 10. A friend returns from Scotland with a 3 12 × 5 picture showing the Loch Ness monster in the outer 14 on the right. If you use your procedure to make an 8 × 10 enlargement, does Nessie make the cut?

The relationships between functions and their graphs are central topics in calculus. Graphing calculators and user-friendly computer software allow you to explore these relationships for a much wider variety of functions than you could with pencil and paper alone. This section presents a general framework for using technology to explore the graphs of functions. Recall that the graphs of linear functions are straight lines and the graphs of quadratic polynomials are parabolas. One of the goals of this section is for you to become more familiar with the graphs of other functions. The best way to become familiar is through experience, by working example after example.

60

2

EXPLORATORY EXERCISES

GRAPHING CALCULATORS AND COMPUTER ALGEBRA SYSTEMS

y

4

row must she win to raise the reported winning percentage to 80%?

2

Solution You should get an initial graph that looks something like that in Figure 0.26a. This is simply a parabola opening upward. A graph is often used to search for important points, such as x-intercepts, y-intercepts or peaks and troughs. In this case, we could see these points better if we zoom in, that is, display a smaller range of x- and y-values than the technology has initially chosen for us. The graph in Figure 0.26b shows x-values from x = −2 to x = 2 and y-values from y = −2 to y = 10. You can see more clearly in Figure 0.26b that the parabola bottoms out roughly at the point (0, −1) and crosses the x-axis at approximately x = −0.5 and x = 0.5. You can make this more precise by doing some algebra. Recall that an x-intercept is a point where y = 0 or f (x) = 0. Solving 3x 2 − 1 = 0 gives 3x 2 = 1 or x 2 = 13 , so that x = ± 13 ≈ ±0.57735.

0-21

SECTION 0.2

..

Graphing Calculators and Computer Algebra Systems

21

Notice that in example 2.1, the graph suggested approximate values for the two x-intercepts, but we needed the algebra to find the values exactly. We then used those values to obtain a view of the graph that highlighted the features that we wanted. Before investigating other graphs, we should say a few words about what a computeror calculator-generated graph really is. Although we call them graphs, what the computer actually does is light up some tiny screen elements called pixels. If the pixels are small enough, the image appears to be a continuous curve or graph. By graphing window, we mean the rectangle defined by the range of x- and y-values displayed. The graphing window can dramatically affect the look of a graph. For example, suppose the x’s run from x = −2 to x = 2. If the computer or calculator screen is wide enough for 400 columns of pixels from left to right, then points will be displayed for x = −2, x = −1.99, x = −1.98, . . . . If there is an interesting feature of this function located between x = −1.99 and x = −1.98, you will not see it unless you zoom in some. In this case, zooming in would reduce the difference between adjacent x’s. Similarly, suppose that the y’s run from y = 0 to y = 3 and that there are 600 rows of pixels from top to bottom. Then, there will be pixels corresponding to y = 0, y = 0.005, y = 0.01, . . . . Now, suppose that f (−2) = 0.0049 and f (−1.99) = 0.0051. Before points are plotted, function values are rounded to the nearest y-value, in this case 0.005. You won’t be able to see any difference in the y-values of these points. If the actual difference is important, you will have to zoom in some to see it.

REMARK 2.1 Most calculators and computer drawing packages use one of the following two schemes for defining the graphing window for a given function. r Fixed graphing window: Most calculators follow this method. Graphs are plotted

in a preselected range of x- and y-values, unless you specify otherwise. For example, the Texas Instruments graphing calculators’ default graphing window plots points in the rectangle defined by −10 ≤ x ≤ 10 and −10 ≤ y ≤ 10. r Automatic graphing window: Most computer drawing packages and some calculators use this method. Graphs are plotted for a preselected range of x-values and the computer calculates the range of y-values so that all of the calculated points will fit in the window. Get to know how your calculator or computer software operates, and use it routinely as you progress through this course. You should always be able to reproduce the computer-generated graphs used in this text by adjusting your graphing window appropriately.

REMARK 2.2 To be precise, f (M) is a local maximum of f if there exist numbers a and b with a < M < b such that f (M) ≥ f (x) for all x such that a < x < b.

Graphs are drawn to provide visual displays of the significant features of a function. What qualifies as significant will vary from problem to problem, but often the x- and y-intercepts and points known as extrema are of interest. The function value f (M) is called a local maximum of the function f if f (M) ≥ f (x) for all x’s “nearby” x = M. Similarly, the function value f (m) is a local minimum of the function f if f (m) ≤ f (x) for all x’s “nearby” x = m. A local extremum is a function value that is either a local maximum or local minimum. Whenever possible, you should produce graphs that show all intercepts and extrema.

EXAMPLE 2.2

Sketching a Graph

Sketch a graph of f (x) = x 3 + 4x 2 − 5x − 1 showing all intercepts and extrema.

22

CHAPTER 0

..

Preliminaries

0-22

Solution Depending on your calculator or computer software, you may initially get a graph that looks like one of those in Figure 0.27a or 0.27b. y

y 10

200

100

4 y 30 20 10 4

2

x 2

4

FIGURE 0.28 y = x 3 + 4x 2 − 5x − 1 y

y a1x a0

x

FIGURE 0.29a Line, a1 < 0 y

x

2

x

10

2

4

5

10

10

FIGURE 0.27a

FIGURE 0.27b

y = x 3 + 4x 2 − 5x − 1

y = x 3 + 4x 2 − 5x − 1

Neither graph is completely satisfactory, although both should give you the idea of a graph that (reading left to right) rises to a local maximum near x = −3, drops to a local minimum near x = 1, and then rises again. To get a better graph, notice the scales on the x- and y-axes. The graphing window for Figure 0.27a is the rectangle defined by −5 ≤ x ≤ 5 and −6 ≤ y ≤ 203. The graphing window for Figure 0.27b is defined by the rectangle −10 ≤ x ≤ 10 and −10 ≤ y ≤ 10. From either graph, it appears that we need to show y-values larger than 10, but not nearly as large as 203, to see the local maximum. Since all of the significant features appear to lie between x = −6 and x = 6, one choice for a better window is −5 ≤ x ≤ 5 and −6 ≤ y ≤ 30, as seen in Figure 0.28. There, you can clearly see the three x-intercepts, the local maximum and the local minimum. The graph in example 2.2 was produced by a process of trial and error with thoughtful corrections. You are unlikely to get a perfect picture on the first try. However, you can enlarge the graphing window (i.e., zoom out) if you need to see more, or shrink the graphing window (i.e., zoom in) if the details are hard to see. You should get comfortable enough with your technology that this revision process is routine (and even fun!). In the exercises, you will be asked to graph a variety of functions and discuss the shapes of the graphs of polynomials of different degrees. Having some knowledge of the general shapes will help you decide whether you have found an acceptable graph. To get you started, we now summarize the different shapes of linear, quadratic and cubic polynomials. Of course, the graphs of linear functions [ f (x) = a1 x + a0 ] are simply straight lines of slope a1 . Two possibilities are shown in Figures 0.29a and 0.29b. The graphs of quadratic polynomials [ f (x) = a2 x 2 + a1 x + a0 ] are parabolas. The parabola opens upward if a2 > 0 and opens downward if a2 < 0. We show typical parabolas in Figures 0.30a and 0.30b. y

y y a2x2 a1x a0

y a2x 2 a1x a0

y a1x a0

x x

x

FIGURE 0.29b Line, a1 > 0

FIGURE 0.30a

FIGURE 0.30b

Parabola, a2 > 0

Parabola, a2 < 0

0-23

..

SECTION 0.2

Graphing Calculators and Computer Algebra Systems

23

The graphs of cubic functions [ f (x) = a3 x 3 + a2 x 2 + a1 x + a0 ] are somewhat S-shaped. Reading from left to right, the function begins negative and ends positive if a3 > 0, and begins positive and ends negative if a3 < 0, as indicated in Figures 0.31a and 0.31b, respectively. y y y a3x3 a2x 2 a1x a0

y a3x3 a2x 2 a1x a0

x

x

y y a3x3 a2x 2 a1x a0

Inflection point

FIGURE 0.31a

FIGURE 0.31b

Cubic: one max, min, a3 > 0

Cubic: one max, min, a3 < 0

Some cubics have one local maximum and one local minimum, as do those in Figures 0.31a and 0.31b. Many curves (including all cubics) have what’s called an inflection point, where the curve changes its shape (from being bent upward, to being bent downward, or vice versa), as indicated in Figures 0.32a and 0.32b. You can already use your knowledge of the general shapes of certain functions to see how to adjust the graphing window, as in example 2.3.

x

FIGURE 0.32a Cubic: no max or min, a3 > 0

EXAMPLE 2.3

y

Sketching the Graph of a Cubic Polynomial

Sketch a graph of the cubic polynomial f (x) = x 3 − 20x 2 − x + 20. y

a3x3 a2x 2

a1x a0

Solution Your initial graph probably looks like Figure 0.33a or 0.33b.

Inflection point y

y 10

x

x

4

FIGURE 0.32b Cubic: no max or min, a3 < 0

4 200 400 600

10

x

5

5

10

10

FIGURE 0.33a

FIGURE 0.33b

f (x) = x 3 − 20x 2 − x + 20

f (x) = x 3 − 20x 2 − x + 20

However, you should recognize that neither of these graphs looks like a cubic; they look more like parabolas. To see the S-shape behavior in the graph, we need to consider a larger range of x-values. To determine how much larger, we need some of the concepts

24

CHAPTER 0

..

Preliminaries

y 20 10

x

10 400 800

0-24

of calculus. For the moment, we use trial and error, until the graph resembles the shape of a cubic. You should recognize the characteristic shape of a cubic in Figure 0.33c. Although we now see more of the big picture (often referred to as the global behavior of the function), we have lost some of the details (such as the x-intercepts), which we could clearly see in Figures 0.33a and 0.33b (often referred to as the local behavior of the function).

1200

Rational functions have some properties not found in polynomials, as we see in examples 2.4, 2.5 and 2.6.

FIGURE 0.33c f (x) = x 3 − 20x 2 − x + 20

EXAMPLE 2.4

Sketching the Graph of a Rational Function

x −1 and describe the behavior of the graph near x = 2. x −2 Solution Your initial graph should look something like Figure 0.34a or 0.34b. From either graph, it should be clear that something unusual is happening near x = 2. Zooming in closer to x = 2 should yield a graph like that in Figure 0.35.

Sketch a graph of f (x) =

y

y

4

1e08

10

5e07

5 x

20

5e07

10

1e08 x 2

10

10

x −1 x −2

y 10 5 x 4

x

5

5

5

10

10

FIGURE 0.34a

FIGURE 0.34b y=

x −1 x −2

In Figure 0.35, it appears that as x increases up to 2, the function values get more and more negative, while as x decreases down to 2, the function values get more and more positive. This is also observed in the following table of function values.

FIGURE 0.35

4

4

x −1 y= x −2

20

y=

y

x

f(x)

x

f(x)

1.8

−4

2.2

6

1.9

−9

2.1

11

1.99

−99

2.01

101

1.999

−999

2.001

1001

1.9999

−9999

2.0001

10,001

8

5 10

FIGURE 0.36 Vertical asymptote

Note that at x = 2, f (x) is undefined. However, as x approaches 2 from the left, the graph veers down sharply. In this case, we say that f (x) tends to −∞. Likewise, as x approaches 2 from the right, the graph rises sharply. Here, we say that f (x) tends to ∞ and there is a vertical asymptote at x = 2. (We’ll define this more carefully in Chapter 1.) It is common to draw a vertical dashed line at x = 2 to indicate this (see Figure 0.36). Since f (2) is undefined, there is no point plotted at x = 2.

0-25

SECTION 0.2

..

Graphing Calculators and Computer Algebra Systems

25

Many rational functions have vertical asymptotes. Notice that there is no point plotted on the vertical asymptote since the function is undefined at such an x-value (due to the division by zero when that value of x is substituted in). Given a rational function, you can locate possible vertical asymptotes by finding where the denominator is zero. It turns out that if the numerator is not zero at that point, there is a vertical asymptote at that point.

EXAMPLE 2.5

A Graph with Several Vertical Asymptotes

x −1 . − 5x + 6 Solution Note that the denominator factors as

Find all vertical asymptotes for f (x) =

x2

x 2 − 5x + 6 = (x − 2)(x − 3), so that the only possible locations for vertical asymptotes are x = 2 and x = 3. Since neither x-value makes the numerator (x − 1) equal to zero, there are vertical asymptotes at both x = 2 and x = 3. A computer-generated graph gives little indication of how the function behaves near the asymptotes. (See Figure 0.37a and note the scale on the y-axis.) y y 2e08 5

1e08 x 1

x

4

1e08

1

4

5

5

2e08 10

3e08

FIGURE 0.37a x −1 y= 2 x − 5x + 6

FIGURE 0.37b y=

x −1 x 2 − 5x + 6

y 0.2

x

10

10

We can improve the graph by zooming in in both the x- and y-directions. Figure 0.37b shows a graph of the same function using the graphing window defined by the rectangle −1 ≤ x ≤ 5 and −13 ≤ y ≤ 7. This graph clearly shows the vertical asymptotes at x = 2 and x = 3.

20

As we see in example 2.6, not all rational functions have vertical asymptotes. 0.2

EXAMPLE 2.6 0.4

FIGURE 0.38 y=

x −1 x2 + 4

A Rational Function with No Vertical Asymptotes

Find all vertical asymptotes of

x −1 . x2 + 4

Solution Notice that x 2 + 4 = 0 has no (real) solutions, since x 2 + 4 > 0 for all real numbers, x. So, there are no vertical asymptotes. The graph in Figure 0.38 is consistent with this observation.

26

..

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Preliminaries

0-26

Graphs are useful for finding approximate solutions of difficult equations, as we see in examples 2.7 and 2.8. y

EXAMPLE 2.7 12

Find approximate solutions of the equation x 2 =

8

4

4

Finding Zeros Approximately

x

2

2

4

FIGURE 0.39a y = x2 −

√

x +3

y 10 8

√

x + 3. √ Solution You could rewrite√this equation as x 2 − x + 3 = 0 and then look for zeros in the graph of f (x) = x 2 − x + 3, seen in Figure 0.39a. Note that two zeros are clearly indicated: one near −1, the√other near 1.5. However, since you know very little of the nature of the function x 2 − x + 3, you cannot say whether or not there are any other zeros, ones that don’t show up in the window seen in Figure 0.39a. On the other hand, if you graph the two functions on either side of the equation on the same set of axes, as in Figure 0.39b, you can clearly see two points where the graphs intersect (corresponding to the two zeros seen in Figure 0.39a). Further, since you know the general shapes of both of the graphs, you can infer from Figure 0.39b that there are no other intersections (i.e., there are no other zeros of f ). This is important information that you cannot obtain from Figure 0.39a. Now that you know how many solutions there are, you need to estimate their values. One method is to zoom in on the zeros graphically. We leave it as an exercise to verify that the zeros are approximately x = 1.4 and x = −1.2. If your calculator or computer algebra system has a solve command, you can use it to quickly obtain an accurate approximation. In this case, we get x ≈ 1.452626878 and x ≈ −1.164035140.

6

When using the solve command on your calculator or computer algebra system, be sure to check that the solutions make sense. If the results don’t match what you’ve seen in your preliminary sketches and zooms, beware! Even high-tech equation solvers make mistakes occasionally.

4 2 4

x

2

2

4

EXAMPLE 2.8

FIGURE 0.39b √

y = x and y =

x +3

2

Find all points of intersection of the graphs of y = 2 cos x and y = 2 − x.

y 4

1

x 1

3

5

2

Finding Intersections by Calculator: An Oversight

Solution Notice that the intersections correspond to solutions of the equation 2 cos x = 2 − x. Using the solve command on the Tl-92 graphing calculator, we found intersections at x ≈ 3.69815 and x = 0. So, what’s the problem? A sketch of the graphs of y = 2 − x and y = 2 cos x (we discuss this function in the next section) clearly shows three intersections (see Figure 0.40). The middle solution, x ≈ 1.10914, was somehow passed over by the calculator’s solve routine. The lesson here is to use graphical evidence to support your solutions, especially when using software and/or functions with which you are less than completely familiar.

FIGURE 0.40 y = 2 cos x and y = 2 − x

You need to look skeptically at the answers provided by your calculator’s solver program. While such solvers provide a quick means of approximating solutions of equations, these programs will sometimes return incorrect answers, as we illustrate with example 2.9. So, how do you know if your solver is giving you an accurate answer or one that’s incorrect? The only answer to this is that you must carefully test your calculator’s solution, by separately calculating both sides of the equation (by hand) at the calculated solution.

0-27

SECTION 0.2

EXAMPLE 2.9

..

Graphing Calculators and Computer Algebra Systems

27

Solving an Equation by Calculator: An Erroneous Answer

1 1 = . x x Solution Certainly, you don’t need a calculator to solve this equation, but consider what happens when you use one. Most calculators report a solution that is very close to zero, while others report that the solution is x = 0. Not only are these answers incorrect, but the given equation has no solution, as follows. First, notice that the 1 equation makes sense only when x = 0. Subtracting from both sides of the equation x leaves us with x = 0, which can’t possibly be a solution, since it does not satisfy the original equation. Notice further that, if your calculator returns the approximate solution x = 1 × 10−7 and you use your calculator to compute the values on both sides of the equation, the calculator will compute 1 x + = 1 × 10−7 + 1 × 107 , x 1 which it approximates as 1 × 107 = , since calculators carry only a finite number of x digits. In other words, although

Use your calculator’s solver program to solve the equation x +

1 × 10−7 + 1 × 107 = 1 × 107 , your calculator treats these numbers as the same and so incorrectly reports that the equation is satisfied. The moral of this story is to be an intelligent user of technology and don’t blindly accept everything a calculator tells you. We want to emphasize again that graphing should be the first step in the equation-solving process. A good graph will show you how many solutions to expect, as well as give their approximate locations. Whenever possible, you should factor or use the quadratic formula to get exact solutions. When this is impossible, approximate the solutions by zooming in on them graphically or by using your calculator’s solve command. Always compare your results to a graph to see if there’s anything you’ve missed.

EXERCISES 0.2 WRITING EXERCISES 1. Explain why there is a significant difference among Figures 0.33a, 0.33b and 0.33c. 2. In Figure 0.36, the graph approaches the lower portion of the vertical asymptote from the left, whereas the graph approaches the upper portion of the vertical asymptote from the right. Use the table of function values found in example 2.4 to explain how to determine whether a graph approaches a vertical asymptote by dropping down or rising up. 3. In the text, we discussed the difference between graphing with a fixed window versus an automatic window. Discuss the advantages and disadvantages of each. (Hint: Consider the case of a first graph of a function you know nothing about and the case of hoping to see the important details of a graph for which you know the general shape.)

x3 + 1 with each of the following x graphing windows: (a) −10 ≤ x ≤ 10, (b) − 1000 ≤ x ≤ 1000. Explain why the graph in (b) doesn’t show the details that the graph in (a) does.

4. Examine the graph of y =

In exercises 1–30, sketch a graph of the function showing all extrema, intercepts and asymptotes. 1. f (x) = x 2 − 1

2. f (x) = 3 − x 2

3. f (x) = x 2 + 2x + 8

4. f (x) = x 2 − 20x + 11

5. f (x) = x 3 + 1

6. f (x) = 10 − x 3

7. f (x) = x 3 + 2x − 1

8. f (x) = x 3 − 3x + 1

28

CHAPTER 0

..

Preliminaries

9. f (x) = x 4 − 1

0-28

10. f (x) = 2 − x 4

11. f (x) = x 4 + 2x − 1

12. f (x) = x 4 − 6x 2 + 3

13. f (x) = x 5 + 2

14. f (x) = 12 − x 5

15. f (x) = x 5 − 8x 3 + 20x − 1 3 17. f (x) = x −1

16. f (x) = x 5 + 5x 4 + 2x 3 + 1 4 18. f (x) = x +2

3x 19. f (x) = x −1

4x 20. f (x) = x +2

3x 2 21. f (x) = x −1

4x 2 22. f (x) = x +2

23. f (x) =

2 x2 − 4

24. f (x) =

6 x2 − 9

25. f (x) =

3 x2 + 4

26. f (x) =

6 x2 + 9

27. f (x) =

x2

29. f (x) = √

x +2 +x −6 3x

x2 + 4

28. f (x) =

x2

30. f (x) = √

55. cos x = x 2 − 1

56. sin x = x 2 + 1

60. f (x) = x 4 − 2x + 1

2x

61. f (x) = x 4 − 7x 3 − 15x 2 − 10x − 1410

x2 + 1

62. f (x) = x 6 − 4x 4 + 2x 3 − 8x − 2

x +4 x2 − 9

33. f (x) =

4x x 2 + 3x − 10

34. f (x) =

x +2 x 2 − 2x − 15

35. f (x) =

4x x2 + 4

36. f (x) = √

3x

63. Graph y = x 2 in the graphing window −10 ≤ x ≤ 10, −10 ≤ y ≤ 10, without drawing the x- and y-axes. Adjust the graphing window for y = 2(x − 1)2 + 3 so that (without the axes showing) the graph looks identical to that of y = x 2. 64. Graph y = x 2 in the graphing window −10 ≤ x ≤ 10, −10 ≤ y ≤ 10. Separately graph y = x 4 with the same graphing window. Compare and contrast the graphs. Then graph the two functions on the same axes and carefully examine the differences in the intervals −1 < x < 1 and x > 1.

x2 − 9

3x x4 − 1

In exercises 39–42, a standard graphing window will not reveal all of the important details of the graph. Adjust the graphing window to find the missing details. 39. f (x) = 13 x 3 −

54. (x + 1)2/3 = 2 − x

59. f (x) = x 4 − 3x 3 − x + 1

x −1 + 4x + 3

32. f (x) =

38. f (x) =

53. (x 2 − 1)2/3 = 2x + 1

58. f (x) = x 3 − 4x 2 + 2

3x x2 − 4

x2 + 1 + 3x 2 + 2x

52. x 3 + 1 = −3x 2 − 3x

57. f (x) = x 3 − 3x + 1

31. f (x) =

x3

51. x 3 − 3x 2 = 1 − 3x

In exercises 57–62, use a graphing calculator or computer graphing utility to estimate all zeros.

In exercises 31–38, find all vertical asymptotes.

37. f (x) =

In exercises 49–56, determine the number of (real) solutions. Solve for the intersection points exactly if possible and estimate the points if necessary. √ √ 49. x − 1 = x 2 − 1 50. x 2 + 4 = x 2 + 2

1 x 400

40. f (x) = x 4 − 11x 3 + 5x − 2 √ 41. f (x) = x 144 − x 2

65. In this exercise, you will find an equation describing all points equidistant from the x-axis and the point (0, 2). First, see if you can sketch a picture of what this curve ought to look like. For a point (x, y) that is on the curve, explain why y 2 = x 2 + (y − 2)2 . Square both sides of this equation and solve for y. Identify the curve. 66. Find an equation describing all points equidistant from the x-axis and (1, 4) (see exercise 65).

42. f (x) = 15 x 5 − 78 x 4 + 13 x 3 + 72 x 2 − 6x In exercises 43–48, adjust the graphing window to identify all vertical asympotes. 4x x2 − 1

43. f (x) =

3 x −1

44. f (x) =

46. f (x) =

2x x +4

x2 − 1 47. f (x) = √ x4 + x

45. f (x) =

2

3x x2 − 1

48. f (x) = √

2x x2 + x

EXPLORATORY EXERCISES 1. Suppose that a graphing calculator is set up with pixels corresponding to x = 0, 0.1, 0.2, 0.3, . . . , 2.0 and y = 0, 0.1, 0.2, 0.3, . . . , 4.0. For the function f (x) = x 2 , compute the indicated function values and round off to give pixel coordinates [e.g., the point (1.19, 1.4161) has pixel coordinates (1.2, 1.4)].

0-29

SECTION 0.3

(a) f (0.4), (b) f (0.39), (c) f (1.17), (d) f (1.20), (e) f (1.8), (f) f (1.81). Repeat (c)–(d) if the graphing window is zoomed in so that x = 1.00, 1.01, . . . , 1.20 and y = 1.30, 1.31, . . . , 1.50. Repeat (e)–(f ) if the graphing window is zoomed in so that x = 1.800, 1.801, . . . , 1.820 and y = 3.200, 3.205, . . . , 3.300. 2. Graph y = x 2 − 1, y = x 2 + x − 1, y = x 2 + 2x − 1, y = x 2 − x − 1, y = x 2 − 2x − 1 and other functions of the form y = x 2 + cx − 1. Describe the effect(s) a change in c has on the graph.

0.3

Inverse Functions

29

3. Figures 0.31 and 0.32 provide a catalog of the possible types of graphs of cubic polynomials. In this exercise, you will compile a catalog of graphs of fourth-order polynomials (i.e., y = ax 4 + bx 3 + cx 2 + d x + e). Start by using your calculator or computer to sketch graphs with different values of a, b, c, d and e. Try y = x 4 , y = 2x 4 , y = −2x 4 , y = x 4 + x 3 , y = x 4 + 2x 3 , y = x 4 − 2x 3 , y = x 4 + x 2 , y = x 4 − x 2 , y = x 4 − 2x 2 , y = x 4 + x, y = x 4 − x and so on. Try to determine what effect each constant has.

INVERSE FUNCTIONS f

x

y

Domain { f}

Range { f } g

FIGURE 0.41 g(x) = f −1 (x) y 8 y x3

6 4 2 2

..

x 1

The notion of an inverse relationship is basic to many areas of science. The number of common inverse problems is immense. As only one example, take the case of the electrocardiogram (EKG). In an EKG, technicians connect a series of electrodes to a patient’s chest and use measurements of electrical activity on the surface of the body to infer something about the electrical activity on the surface of the heart. This is referred to as an inverse problem, since physicians are attempting to determine what inputs (i.e., the electrical activity on the surface of the heart) cause an observed output (the measured electrical activity on the surface of the chest). The mathematical notion of inverse is much the same as that just described. Given an output (in this case, a value in the range of a given function), we wish to find the input (the value in the domain) that produced that output. That is, given a y ∈ Range{ f }, find the x ∈ Domain{ f } for which y = f (x). (See the illustration of the inverse function g shown in Figure 0.41.) For instance, suppose that f (x) = x 3 and y = 8. Can you find an x such that x 3 = 8? That is, can you find the x-value corresponding the √ to y = 8? (See Figure 0.42.) Of course, √ solution of this particular equation is x = 3 8 = 2. In general, if x 3 = y, then x = 3 y. In light of this, we say that the cube root function is the inverse of f (x) = x 3 .

2

2 4

EXAMPLE 3.1

If f (x) = x 3 and g(x) = x 1/3 , show that

FIGURE 0.42 Finding the x-value corresponding to y = 8

Two Functions That Reverse the Action of Each Other

f (g(x)) = x

and

g( f (x)) = x,

for all x. Solution For all real numbers x, we have f (g(x)) = f (x 1/3 ) = (x 1/3 )3 = x and

g( f (x)) = g(x 3 ) = (x 3 )1/3 = x.

Notice in example 3.1 that the action of f undoes the action of g and vice versa. We take this as the definition of an inverse function. (Again, think of Figure 0.41.)

0-29

SECTION 0.3

(a) f (0.4), (b) f (0.39), (c) f (1.17), (d) f (1.20), (e) f (1.8), (f) f (1.81). Repeat (c)–(d) if the graphing window is zoomed in so that x = 1.00, 1.01, . . . , 1.20 and y = 1.30, 1.31, . . . , 1.50. Repeat (e)–(f ) if the graphing window is zoomed in so that x = 1.800, 1.801, . . . , 1.820 and y = 3.200, 3.205, . . . , 3.300. 2. Graph y = x 2 − 1, y = x 2 + x − 1, y = x 2 + 2x − 1, y = x 2 − x − 1, y = x 2 − 2x − 1 and other functions of the form y = x 2 + cx − 1. Describe the effect(s) a change in c has on the graph.

0.3

Inverse Functions

29

3. Figures 0.31 and 0.32 provide a catalog of the possible types of graphs of cubic polynomials. In this exercise, you will compile a catalog of graphs of fourth-order polynomials (i.e., y = ax 4 + bx 3 + cx 2 + d x + e). Start by using your calculator or computer to sketch graphs with different values of a, b, c, d and e. Try y = x 4 , y = 2x 4 , y = −2x 4 , y = x 4 + x 3 , y = x 4 + 2x 3 , y = x 4 − 2x 3 , y = x 4 + x 2 , y = x 4 − x 2 , y = x 4 − 2x 2 , y = x 4 + x, y = x 4 − x and so on. Try to determine what effect each constant has.

INVERSE FUNCTIONS f

x

y

Domain { f}

Range { f } g

FIGURE 0.41 g(x) = f −1 (x) y 8 y x3

6 4 2 2

..

x 1

The notion of an inverse relationship is basic to many areas of science. The number of common inverse problems is immense. As only one example, take the case of the electrocardiogram (EKG). In an EKG, technicians connect a series of electrodes to a patient’s chest and use measurements of electrical activity on the surface of the body to infer something about the electrical activity on the surface of the heart. This is referred to as an inverse problem, since physicians are attempting to determine what inputs (i.e., the electrical activity on the surface of the heart) cause an observed output (the measured electrical activity on the surface of the chest). The mathematical notion of inverse is much the same as that just described. Given an output (in this case, a value in the range of a given function), we wish to find the input (the value in the domain) that produced that output. That is, given a y ∈ Range{ f }, find the x ∈ Domain{ f } for which y = f (x). (See the illustration of the inverse function g shown in Figure 0.41.) For instance, suppose that f (x) = x 3 and y = 8. Can you find an x such that x 3 = 8? That is, can you find the x-value corresponding the √ to y = 8? (See Figure 0.42.) Of course, √ solution of this particular equation is x = 3 8 = 2. In general, if x 3 = y, then x = 3 y. In light of this, we say that the cube root function is the inverse of f (x) = x 3 .

2

2 4

EXAMPLE 3.1

If f (x) = x 3 and g(x) = x 1/3 , show that

FIGURE 0.42 Finding the x-value corresponding to y = 8

Two Functions That Reverse the Action of Each Other

f (g(x)) = x

and

g( f (x)) = x,

for all x. Solution For all real numbers x, we have f (g(x)) = f (x 1/3 ) = (x 1/3 )3 = x and

g( f (x)) = g(x 3 ) = (x 3 )1/3 = x.

Notice in example 3.1 that the action of f undoes the action of g and vice versa. We take this as the definition of an inverse function. (Again, think of Figure 0.41.)

30

..

CHAPTER 0

Preliminaries

REMARK 3.1

0-30

DEFINITION 3.1

Pay close attention to the notation. Notice that f −1 (x) 1 . We write does not mean f (x) the reciprocal of f (x) as

Assume that f and g have domains A and B, respectively, and that f (g(x)) is defined for all x ∈ B and g( f (x)) is defined for all x ∈ A. If

1 = [ f (x)]−1 . f (x)

f (g(x)) = x,

for all x ∈ B

g( f (x)) = x,

for all x ∈ A,

and

we say that g is the inverse of f, written g = f −1 . Equivalently, f is the inverse of g, f = g −1 .

Observe that many familiar functions have no inverse. y

EXAMPLE 3.2 20

Show that f (x) = x 2 has no inverse on the interval (−∞, ∞). Solution Notice that f (4) = 16 and f (−4) = 16. That is, there are two x-values that produce the same y-value. So, if we were to try to define an inverse of f, how would we define f −1 (16)? Look at the graph of y = x 2 (see Figure 0.43) to see what the problem is. For each y > 0, there are two x-values for which y = x 2 . Because of this, the function does not have an inverse.

12 8 4

4

A Function with No Inverse

x

2

2

4

FIGURE 0.43 y = x2

√ For f (x) = x 2 , it is tempting to jump √ to2the conclusion that g(x) = x is the inverse of f (x). Notice that although f (g(x)) = ( x) = x√for all x ≥ 0 (i.e., for all x in the domain of g(x)), it is not generally true that g( f (x)) = x 2 = x. In fact, this last equality holds only for x √ ≥ 0. However, for f (x) = x 2 restricted to the domain x ≥ 0, we do have that −1 f (x) = x.

DEFINITION 3.2 A function f is called one-to-one when for every y ∈ Range{ f }, there is exactly one x ∈ Domain{ f } for which y = f (x).

y

REMARK 3.2

y f (x)

x a

Observe that an equivalent definition of one-to-one is the following. A function f (x) is one-to-one if and only if the equality f (a) = f (b) implies a = b. This version of the definition is often useful for proofs involving one-to-one functions.

b

FIGURE 0.44a f (a) = f (b), for a = b So, f does not pass the horizontal line test and is not one-to-one.

It is helpful to think of the concept of one-to-one in graphical terms. Notice that a function f is one-to-one if and only if every horizontal line intersects the graph in at most one point. This is referred to as the horizontal line test. We illustrate this in Figures 0.44a and 0.44b. The following result should now make sense.

0-31

SECTION 0.3

..

Inverse Functions

31

y

THEOREM 3.1

y f (x)

A function f has an inverse if and only if it is one-to-one.

This theorem simply says that every one-to-one function has an inverse and every function that has an inverse is one-to-one. However, it says nothing about how to find an inverse. For very simple functions, we can find inverses by solving equations. x

a

EXAMPLE 3.3

Find the inverse of f (x) = x 3 − 5.

FIGURE 0.44b

Solution Note that it is not entirely clear from the graph (see Figure 0.45) whether f passes the horizontal line test. To find the inverse function, write y = f (x) and solve for x (i.e., solve for the input x that produced the observed output y). We have

Every horizontal line intersects the curve in at most one point. So, f passes the horizontal line test and is one-to-one.

y = x 3 − 5.

y

Adding 5 to both sides and taking the cube root gives us (y + 5)1/3 = (x 3 )1/3 = x.

40

So, x = f −1 (y) = (y + 5)1/3 . Reversing the variables x and y gives us

20

4

Finding an Inverse Function

f −1 (x) = (x + 5)1/3 .

x

2

2

4

20

EXAMPLE 3.4

A Function That Is Not One-to-One

Show that f (x) = 10 − x 4 does not have an inverse.

40

Solution You can see from a graph (see Figure 0.46) that f is not one-to-one; for instance, f (1) = f (−1) = 9. Consequently, f does not have an inverse.

FIGURE 0.45 y = x3 − 5

Most often, we cannot find a formula for an inverse function and must be satisfied with simply knowing that the inverse function exists. Example 3.5 is typical of this situation.

y 20 4

2

2 20 40 60 80 100

FIGURE 0.46 y = 10 − x 4

4

x

EXAMPLE 3.5

Finding Values of an Inverse Function

Given that f (x) = x 5 + 8x 3 + x + 1 has an inverse, find f −1 (1) and f −1 (11). Solution First, notice that from the graph shown in Figure 0.47 (on the following page), the function looks like it might be one-to-one, but how can we be certain of this? (Remember that graphs can be deceptive!) Until we develop some calculus, we will be unable to verify this. Ideally, we would show that f has an inverse by finding a formula for f −1 , as in example 3.3. However, in this case, we must solve the equation y = x 5 + 8x 3 + x + 1 for x. Think about this for a moment: you should realize that we can’t solve for x in terms of y here. We need to assume that the inverse exists, as indicated in the instructions.

32

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..

Preliminaries

0-32

Turning to the problem of finding f −1 (1) and f −1 (11), you might wonder if this is possible, since we were unable to find a formula for f −1 (x). While it’s certainly true that we have no such formula, you might observe that f (0) = 1, so that f −1 (1) = 0. By trial and error, you might also discover that f (1) = 11 and so, f −1 (11) = 1.

y 3 2 1 3 2

1 1

x 1

2

3

2

In example 3.5, we examined a function that has an inverse, although we could not find that inverse algebraically. Even when we can’t find an inverse function explicitly, we can say something graphically. Notice that if (a, b) is a point on the graph of y = f (x) and f has an inverse, f −1 , then since

3

b = f (a),

FIGURE 0.47

That is, (b, a) is a point on the graph of y = f −1 (x). This tells us a great deal about the inverse function. In particular, we can immediately obtain any number of points on the graph of y = f −1 (x), simply by inspection. Further, notice that the point (b, a) is the reflection of the point (a, b) through the line y = x (see Figure 0.48). It now follows that given the graph of any one-to-one function, you can draw the graph of its inverse simply by reflecting the entire graph through the line y = x. In example 3.6, we illustrate the symmetry of a function and its inverse.

y yx a

f −1 (b) = f −1 ( f (a)) = a.

we have that

y = x 5 + 8x 3 + x + 1

(b, a)

(a, b)

b

EXAMPLE 3.6 x b

a

FIGURE 0.48 Reflection through y = x

The Graph of a Function and Its Inverse

Draw a graph of f (x) = x 3 and its inverse. Solution From example 3.1, the inverse of f (x) = x 3 is f −1 (x) = x 1/3 . Notice the symmetry of their graphs shown in Figure 0.49.

y y

yx

yx 1

y x 1/3

y f (x)

yx

3

x

1

1

y f 1(x)

1

FIGURE 0.49 y = x 3 and y = x 1/3

x

FIGURE 0.50

Graphs of f and f −1

Observe that we can use this symmetry principle to draw the graph of an inverse function, even when we don’t have a formula for that function (see Figure 0.50).

0-33

SECTION 0.3

..

Inverse Functions

33

y 3

y f (x)

EXAMPLE 3.7

yx

Draw a graph of f (x) = x 5 + 8x 3 + x + 1 and its inverse.

2 y f 1(x)

1 3 2

1 1

Drawing the Graph of an Unknown Inverse Function

x 1

2

3

2 3

Solution In example 3.5, we were unable to find a formula for the inverse function. Despite this, we can draw a graph of f −1 with ease. We simply take the graph of y = f (x) seen in Figure 0.47 and reflect it across the line y = x, as shown in Figure 0.51. (When we introduce parametric equations in section 9.1, we will see a clever way to draw this graph with a graphing calculator.) In example 3.8, we apply our theoretical knowledge of inverse functions in a medical setting.

FIGURE 0.51

y = f (x) and y = f −1 (x)

EXAMPLE 3.8

Determining the Proper Dosage of a Drug

Suppose that the injection of a certain drug raises the level of a key hormone in the body. Physicians want to determine the dosage that produces a healthy hormone level. Dosages of 1, 2, 3 and 4 mg produce hormone levels of 12, 20, 40 and 76, respectively. If the desired hormone level is 30, what is the proper dosage?

y

y

TODAY IN MATHEMATICS Kim Rossmo (1955– ) A Canadian criminologist who developed the Criminal Geographic Targeting algorithm that indicates the most probable area of residence for serial murderers, rapists and other criminals. Rossmo served 21 years with the Vancouver Police Department. His mentors were Professors Paul and Patricia Brantingham of Simon Fraser University. The Brantinghams developed Crime Pattern Theory, which predicts crime locations from where criminals live, work and play. Rossmo inverted their model and used the crime sites to determine where the criminal most likely lives. The premiere episode of the television drama Numb3rs was based on Rossmo’s work.

80

80

60

60

40

40

20

20 x 1

2

3

4

x 1

2

3

FIGURE 0.52a

FIGURE 0.52b

Hormone data

Approximate curve

4

Solution A plot of the points (1, 12), (2, 20), (3, 40) and (4, 76) summarizes the data (see Figure 0.52a). The problem is an inverse problem: given y = 30, what is x? It is tempting to argue the following: since 30 is halfway between 20 and 40, the x-value should be halfway between 2 and 3: x = 2.5. This method of solution is called linear interpolation, since the point (x, y) = (2.5, 30) lies on the line through the points (2, 20) and (3, 40). However, this estimate does not take into account all of the information we have. The points in Figure 0.52a suggest a graph that is curving up. If this is the case, x = 2.6 or x = 2.7 may be a better estimate of the required dosage. In Figure 0.52b, we have sketched a smooth curve through the data points and indicated a graphical solution of the problem. More advanced techniques (e.g., polynomial interpolation) have been developed by mathematicians to make the estimate of such quantities as accurate as possible.

34

CHAPTER 0

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Preliminaries

0-34

EXERCISES 0.3 WRITING EXERCISES 1. Explain in words (and a picture) why the following is true: if f (x) is increasing for all x, then f has an inverse. 2. Suppose the graph of a function passes the horizontal line test. Explain why you know that the function has an inverse (defined on the range of the function). 3. Radar works by bouncing a high-frequency electromagnetic pulse off of a moving object, then measuring the disturbance in the pulse as it is bounced back. Explain why this is an inverse problem by identifying the input and output.

17. f (x) = 18. f (x) =

√ √

(b) f −1 (1)

y 4 2

4

x

2

2

4

2

4

2

4

2 4

y

20. 4 2

1 − 2x 1 and g(x) = (x = 0, x = −2) x +2 x

4

5. f (x) = x 3 − 2

6. f (x) = x 3 + 4

7. f (x) = x 5 − 1

8. f (x) = x 5 + 4 10. f (x) = x 4 − 2x − 1 √ 12. f (x) = x 2 + 1

2 4

y

21. 4

In exercises 13–18, assume that the function has an inverse. Without solving for the inverse, find the indicated function values. 13. f (x) = x 3 + 4x − 1,

(a) f −1 (−1),

(b) f −1 (4)

14. f (x) = x 3 + 2x + 1,

(a) f −1 (1),

(b) f −1 (13)

15. f (x) = x 5 + 3x 3 + x,

(a) f −1 (−5),

(b) f −1 (5)

16. f (x) = x 5 + 4x − 2,

(a) f −1 (38),

(b) f −1 (3)

x

2

In exercises 5–12, determine whether the function is one-to-one. If it is, find the inverse and graph both the function and its inverse.

9. f (x) = x 4 + 2 √ 11. f (x) = x 3 + 1

x 5 + 4x 3 + 3x + 1, (a) f −1 (3),

19.

In exercises 1–4, show that f (g(x)) x and g( f (x)) x for all x:

4. f (x) =

(b) f −1 (2)

In exercises 19–22, use the given graph to graph the inverse function.

4. Each human disease has a set of symptoms associated with it. Physicians attempt to solve an inverse problem: given the symptoms, they try to identify the disease causing the symptoms. Explain why this is not a well-defined inverse problem (i.e., logically it is not always possible to correctly identify diseases from symptoms alone).

1. f (x) = x 5 and g(x) = x 1/5 1/3 2. f (x) = 4x 3 and g(x) = 14 x x −1 3 3. f (x) = 2x + 1 or g(x) = 3 2

(a) f −1 (4),

x 3 + 2x + 4,

2

4

x

2 2 4

0-35

SECTION 0.3

4

35

41. Graph f (x) = (x − 2)2 and find an interval on which it is oneto-one. Find the inverse of the function restricted to that interval. Graph both functions.

2

4

Inverse Functions

40. Graph f (x) = x 2 + 2 for x ≤ 0 and verify that it is one-to-one. Find its inverse. Graph both functions.

y

22.

..

x

2

2

4

2 4

In exercises 23–26, use linear interpolation (see example 3.8) to estimate f − 1 (b). Use the apparent curving of the graph to conjecture whether the estimate is too high or too low. 23. (1, 12), (2, 20), (3, 26), (4, 30), b = 23 24. (1, 12), (2, 10), (3, 6), (4, 0), b = 8

42. Graph f (x) = (x + 1)4 and find an interval on which it is oneto-one. Find the inverse of the function restricted to that interval. Graph both functions. √ 43. Graph f (x) = x 2 − 2x and find an interval on which it is one-to-one. Find the inverse of the function restricted to that interval. Graph both functions. x 44. Graph f (x) = 2 and find an interval on which it is one-tox −4 one. Find the inverse of the function restricted to that interval. Graph both functions. 45. Graph f (x) = sin x and find an interval on which it is one-toone. Find the inverse of the function restricted to that interval. Graph both functions.

26. (1, 12), (2, 20), (3, 36), (4, 50), b = 32

46. Graph f (x) = cos x and find an interval on which it is one-toone. Find the inverse of the function restricted to that interval. Graph both functions.

In exercises 27–36, use a graph to determine whether the function is one-to-one. If it is, graph the inverse function.

In exercises 47–52, discuss whether the function described has an inverse.

27. f (x) = x 3 − 5

47. The income of a company varies with time.

28. f (x) = x − 3

48. The height of a person varies with time.

29. f (x) = x + 2x − 1

49. For a dropped ball, its height varies with time.

30. f (x) = x − 2x − 1

50. For a ball thrown upward, its height varies with time.

31. f (x) = x − 3x − 1

51. The shadow made by an object depends on its threedimensional shape.

25. (1, 12), (2, 6), (3, 2), (4, 0), b = 5

2 3 3 5

3

32. f (x) = x + 4x − 2 5

33. f (x) =

3

1 x +1

4 x2 + 1 x 35. f (x) = x +4 x 36. f (x) = √ 2 x +4 34. f (x) =

52. The number of calories burned depends on how fast a person runs. 53. Suppose that your boss informs you that you have been awarded a 10% raise. The next week, your boss announces that due to circumstances beyond her control, all employees will have their salaries cut by 10%. Are you as well off now as you were two weeks ago? Show that increasing by 10% and decreasing by 10% are not inverse processes. Find the inverse for adding 10%. (Hint: To add 10% to a quantity you can multiply the quantity by 1.10.)

Exercises 37–46 involve inverse functions on restricted domains. √ 37. Show that f (x) = x 2 (x ≥ 0) and g(x) = x (x ≥ 0) are inverse functions. Graph both functions. √ 38. Show that f (x) = x 2 − 1 (x ≥ 0) and g(x) = x + 1 (x ≥ −1) are inverse functions. Graph both functions.

1. Find all values of k such that f (x) = x 3 + kx + 1 is one-toone.

39. Graph f (x) = x 2 for x ≤ 0 and verify that it is one-to-one. Find its inverse. Graph both functions.

2. Find all values of k such that f (x) = x 3 + 2x 2 + kx − 1 is one-to-one.

EXPLORATORY EXERCISES

36

CHAPTER 0

0.4

..

Preliminaries

0-36

TRIGONOMETRIC AND INVERSE TRIGONOMETRIC FUNCTIONS Many phenomena encountered in your daily life involve waves. For instance, music is transmitted from radio stations in the form of electromagnetic waves. Your radio receiver decodes these electromagnetic waves and causes a thin membrane inside the speakers to vibrate, which, in turn, creates pressure waves in the air. When these waves reach your ears, you hear the music from your radio (see Figure 0.53). Each of these waves is periodic, meaning that the basic shape of the wave is repeated over and over again. The mathematical description of such phenomena involves periodic functions, the most familiar of which are the trigonometric functions. First, we remind you of a basic definition.

FIGURE 0.53 Radio and sound waves

NOTES DEFINITION 4.1

When we discuss the period of a function, we most often focus on the fundamental period.

A function f is periodic of period T if f (x + T ) = f (x)

y

for all x such that x and x + T are in the domain of f. The smallest such number T > 0 is called the fundamental period.

(cos u, sin u)

u

1 u

sin u

cos u

FIGURE 0.54 Definition of sin θ and cos θ : cos θ = x and sin θ = y

x

There are several equivalent ways of defining the sine and cosine functions. We want to emphasize a simple definition from which you can easily reproduce many of the basic properties of these functions. Referring to Figure 0.54, begin by drawing the unit circle x 2 + y 2 = 1. Let θ be the angle measured (counterclockwise) from the positive x-axis to the line segment connecting the origin to the point (x, y) on the circle. Here, we measure θ in radians, where the radian measure of the angle θ is the length of the arc indicated in the figure. Again referring to Figure 0.54, we define sin θ to be the y-coordinate of the point on the circle and cos θ to be the x-coordinate of the point. From this definition, it follows that sin θ and cos θ are defined for all values of θ , so that each has domain −∞ < θ < ∞, while the range for each of these functions is the interval [−1, 1].

0-37

SECTION 0.4

..

Trigonometric and Inverse Trigonometric Functions

37

REMARK 4.1 Unless otherwise noted, we always measure angles in radians.

Note that since the circumference of a circle (C = 2πr ) of radius 1 is 2π , we have that 360◦ corresponds to 2π radians. Similarly, 180◦ corresponds to π radians, 90◦ corresponds to π/2 radians, and so on. In the accompanying table, we list some common angles as measured in degrees, together with the corresponding radian measures.

Angle in degrees

0◦

30◦

45◦

60◦

90◦

135◦

180◦

270◦

360◦

Angle in radians

0

π 6

π 4

π 3

π 2

3π 4

π

3π 2

2π

THEOREM 4.1 The functions f (θ ) = sin θ and g(θ ) = cos θ are periodic, of period 2π .

PROOF Referring to Figure 0.54, since a complete circle is 2π radians, adding 2π to any angle takes you all the way around the circle and back to the same point (x, y). This says that sin(θ + 2π) = sin θ cos(θ + 2π ) = cos θ,

and

for all values of θ. Furthermore, 2π is the smallest angle for which this is true. You are likely already familiar with the graphs of f (x) = sin x and g(x) = cos x shown in Figures 0.55a and 0.55b, respectively. y

y

1

1

x r

w

q

q

1

w

r

2p

x

p

p 1

FIGURE 0.55a

FIGURE 0.55b

y = sin x

y = cos x

2p

38

CHAPTER 0

x

sin x

cos x

0

0

1

π 6

π 3

1 2 √ 2 2 √ 3 2

π 2

1

π 4

Preliminaries

0-38

Notice that you could slide the graph of y = sin x slightly to the left or right and get an exact copy of the graph of y = cos x. Specifically, we have the relationship

√

3 2 √ 2 2

π sin x + = cos x. 2

1 2

The accompanying table lists some common values of sine and cosine. Notice that many of these can be read directly from Figure 0.54.

0

3π 4

√ 3 2 √ 2 2

5π 6

1 2

√ 2 2 √ − 23

π

0

−1

3π 2

−1

0

2π

0

1

2π 3

..

− 12

EXAMPLE 4.1

−

Solving Equations Involving Sines and Cosines

Find all solutions of the equations (a) 2 sin x − 1 = 0 and (b) cos2 x − 3 cos x + 2 = 0. Solution For (a), notice that 2 sin x − 1 = 0 if 2 sin x = 1 or sin x = 12 . From the unit circle, we find that sin x = 12 if x = π6 or x = 5π . Since sin x has period 2π, additional 6 π 5π π solutions are 6 + 2π, 6 + 2π, 6 + 4π and so on. A convenient way of indicating that any integer multiple of 2π can be added to either solution is to write x = π6 + 2nπ or x = 5π + 2nπ, for any integer n. Part (b) may look rather difficult at first. However, 6 notice that it looks like a quadratic equation using cos x instead of x. With this clue, you can factor the left-hand side to get

REMARK 4.2 Instead of writing (sin θ)2 or (cos θ )2 , we usually use the notation sin2 θ and cos2 θ , respectively.

0 = cos2 x − 3 cos x + 2 = (cos x − 1)(cos x − 2), from which it follows that either cos x = 1 or cos x = 2. Since −1 ≤ cos x ≤ 1 for all x, the equation cos x = 2 has no solution. However, we get cos x = 1 if x = 0, 2π or any integer multiple of 2π . We can summarize all the solutions by writing x = 2nπ , for any integer n. We now give definitions of the remaining four trigonometric functions.

DEFINITION 4.2 sin x . cos x cos x The cotangent function is defined by cot x = . sin x 1 The secant function is defined by sec x = . cos x

The tangent function is defined by tan x =

REMARK 4.3 Most calculators have keys for the functions sin x, cos x and tan x, but not for the other three trigonometric functions. This reflects the central role that sin x, cos x and tan x play in applications. To calculate function values for the other three trigonometric functions, you can simply use the identities cot x = and

1 , tan x

sec x =

csc x =

1 . sin x

1 cos x

The cosecant function is defined by csc x =

1 . sin x

We show graphs of these functions in Figures 0.56a, 0.56b, 0.56c and 0.56d. Notice in each graph the locations of the vertical asymptotes. For the “co” functions cot x and csc x, the division by sin x causes vertical asymptotes at 0, ±π, ±2π and so on (where sin x = 0). For tan x and sec x, the division by cos x produces vertical asymptotes at ±π/2, ±3π/2, ±5π/2 and so on (where cos x = 0). Once you have determined the vertical asymptotes, the graphs are relatively easy to draw. Notice that tan x and cot x are periodic, of period π , while sec x and csc x are periodic, of period 2π.

0-39

SECTION 0.4

..

Trigonometric and Inverse Trigonometric Functions

39

y y

x 2p w

q

p

q

p

w

x 2p w

2p

p

q

q

FIGURE 0.56a

FIGURE 0.56b

y = tan x

y = cot x

1

q

p

q 1

2p w

2p

y

y

p

w

p

q

x w

2p w

2p

w

1

x

1

p

q

2p

p

FIGURE 0.56c

FIGURE 0.56d

y = sec x

y = csc x

It is important to learn the effect of slight modifications of these functions. We present a few ideas here and in the exercises.

EXAMPLE 4.2

Altering Amplitude and Period

Graph y = 2 sin x and y = sin 2x, and describe how each differs from the graph of y = sin x (see Figure 0.57a). y

y

2

y

2

1

2

1 w

q

1 w

q x

w

q

x w

q

1

1

2

2

2p

x

p

p 1 2

FIGURE 0.57a

FIGURE 0.57b

FIGURE 0.57c

y = sin x

y = 2 sin x

y = sin (2x)

2p

40

CHAPTER 0

..

Preliminaries

0-40

Solution The graph of y = 2 sin x is given in Figure 0.57b. Notice that this graph is similar to the graph of y = sin x, except that the y-values oscillate between −2 and 2 instead of −1 and 1. Next, the graph of y = sin 2x is given in Figure 0.57c. In this case, the graph is similar to the graph of y = sin x except that the period is π instead of 2π (so that the oscillations occur twice as fast). The results in example 4.2 can be generalized. For A > 0, the graph of y = A sin x oscillates between y = −A and y = A. In this case, we call A the amplitude of the sine curve. Notice that for any positive constant c, the period of y = sin cx is 2π/c. Similarly, for the function A cos cx, the amplitude is A and the period is 2π/c. The sine and cosine functions can be used to model sound waves. A pure tone (think of a single flute note) is a pressure wave described by the sinusoidal function A sin ct. (Here, we are using the variable t, since the air pressure is a function of time.) The amplitude A determines how loud the tone is perceived to be and the period determines the pitch of the note. In this setting, it is convenient to talk about the frequency f = c/2π. The higher the frequency is, the higher the pitch of the note will be. (Frequency is measured in hertz, where 1 hertz equals 1 cycle per second.) Note that the frequency is simply the reciprocal of the period.

EXAMPLE 4.3

Finding Amplitude, Period and Frequency

Find the amplitude, period and frequency of (a) f (x) = 4 cos 3x and (b) g(x) = 2 sin(x/3). Solution (a) For f (x), the amplitude is 4, the period is 2π/3 and the frequency is 3/(2π) (see Figure 0.58a). (b) For g(x), the amplitude is 2, the period is 2π/(1/3) = 6π and the frequency is 1/(6π ) (see Figure 0.58b). y

y 2

4

x 2p

o

i

i

o

2p

3p 2p p

x p

2p 3p

2

4

FIGURE 0.58a

FIGURE 0.58b

y = 4 cos 3x

y = 2 sin (x/3)

There are numerous formulas or identities that are helpful in manipulating the trigonometric functions. You should observe that, from the definition of sin θ and cos θ (see Figure 0.54), the Pythagorean Theorem gives us the familiar identity sin2 θ + cos2 θ = 1, since the hypotenuse of the indicated triangle is 1. This is true for any angle θ . In addition, sin(−θ ) = − sin θ

and

cos(−θ) = cos θ

0-41

SECTION 0.4

..

Trigonometric and Inverse Trigonometric Functions

41

We list several important identities in Theorem 4.2.

THEOREM 4.2 For any real numbers α and β, the following identities hold: sin (α + β) = sin α cos β + sin β cos α cos (α + β) = cos α cos β − sin α sin β sin2 α = 12 (1 − cos 2α)

(4.1) (4.2) (4.3)

cos2 α = 12 (1 + cos 2α).

(4.4)

From the basic identities summarized in Theorem 4.2, numerous other useful identities can be derived. We derive two of these in example 4.4.

EXAMPLE 4.4

Deriving New Trigonometric Identities

Derive the identities sin 2θ = 2 sin θ cos θ and cos 2θ = cos2 θ − sin2 θ . Solution These can be obtained from formulas (4.1) and (4.2), respectively, by substituting α = θ and β = θ . Alternatively, the identity for cos 2θ can be obtained by subtracting equation (4.3) from equation (4.4). y

The Inverse Trigonometric Functions

1 x q

q 1

FIGURE 0.59

y = sin x on − π2 , π2

We now expand the set of functions available to you by defining inverses to the trigonometric functions. To get started, look at a graph of y = sin x (see Figure 0.57a). Notice that we cannot define an inverse function, since sin x is not one-to-one. Although the sine function does not have an inverse function, we can define one by modifying the domain of the sine. We do this by choosing a portion of the sine curve that passes the horizontal line test.

If we restrict the domain to the interval − π2 , π2 , then y = sin x is one-to-one there (see Figure 0.59) and, hence, has an inverse. We thus define the inverse sine function by y = sin−1 x

if and only if

sin y = x and − π2 ≤ y ≤

π . 2

(4.5)

Think of this definition as follows: if y = sin−1 x, then y is the angle between − π2 and π2 for which sin y = x. Note that we could have selected any interval on which sin x is one-toone, but − π2 , π2 is the most convenient. To verify that these are inverse functions, observe that sin (sin−1 x) = x,

REMARK 4.4 and Mathematicians often use the notation arcsin x in place of sin−1 x. People read sin−1 x interchangeably as “inverse sine of x” or “arcsine of x.”

sin−1 (sin x) = x,

for all x ∈ [−1, 1] π π for all x ∈ − , . 2 2

(4.6)

−1 Read equation (4.6) very carefully. It does not say

that sin (sin x) = x for all x, but rather, only for those in the restricted domain, − π2 , π2 . For instance, sin−1 (sin π ) = π , since

sin−1 (sin π ) = sin−1 (0) = 0.

42

..

CHAPTER 0

Preliminaries

0-42

EXAMPLE 4.5 Evaluate (a) sin−1

Evaluating the Inverse Sine Function

√ 3 2

and (b) sin−1 − 12 .

Solution For (a), we look for the angle θ in the interval − π2 , π2 for which √ √

sin θ = 23 . Note that since sin π3 = 23 and π3 ∈ − π2 , π2 , we have that √

sin−1 23 = π3 . For (b), note that sin − π6 = − 12 and − π6 ∈ − π2 , π2 . Thus, sin−1 − 12 = − π6 . Judging by example 4.5, you might think that (4.5) is a roundabout way of defining a function. If so, you’ve got the idea exactly. In fact, we want to emphasize that what we know about the inverse sine function is principally through reference to the sine function. Recall from our discussion in section 0.3 that we can draw of y = sin−1 x π aπ graph

simply by reflecting the graph of y = sin x on the interval − 2 , 2 (from Figure 0.59) through the line y = x (see Figure 0.60).

Turning to y = cos x, observe that restricting the domain to the interval − π2 , π2 , as we did for the inverse sine function, will not work here. (Why not?) The simplest way to make cos x one-to-one is to restrict its domain to the interval [0, π ] (see Figure 0.61). Consequently, we define the inverse cosine function by

y q

x

1

1

q

FIGURE 0.60 y = sin−1 x

y = cos−1 x

y

if and only if

cos y = x and 0 ≤ y ≤ π.

Note that here, we have

1

cos (cos−1 x) = x,

for all x ∈ [−1, 1]

cos−1 (cos x) = x,

for all x ∈ [0, π ].

x q

p

and

1

As with the definition of arcsine, it is helpful to think of cos−1 x as that angle θ in [0, π ] for which cos θ = x. As with sin−1 x, it is common to use cos−1 x and arccos x interchangeably.

FIGURE 0.61 y = cos x on [0, π]

EXAMPLE 4.6

√ Evaluate (a) cos−1 (0) and (b) cos−1 − 22 .

y p

q

1

Evaluating the Inverse Cosine Function

x 1

FIGURE 0.62 y = cos−1 x

Solution For (a), you will need to find that angle θ in [0, π ] for which cos θ = 0. It’s not hard to see that cos−1 (0) = π2 . (If you calculate this on your calculator and get 90, your calculator is in degrees mode. In this event, you should immediately √ change it to radians mode.) For (b), look for the angle θ ∈ [0, π ] for which cos θ = − 22 . Notice √ = − 22 and 3π that cos 3π ∈ [0, π ]. Consequently, 4 4 √ cos−1 − 22 = 3π . 4 Once again, we obtain the graph of this inverse function by reflecting the graph of y = cos x on the interval [0, π ] (seen in Figure 0.61) through the line y = x (see Figure 0.62).

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Trigonometric and Inverse Trigonometric Functions

43

We can define inverses for each of the four remaining trigonometric functions in similar ways. For y = tan x, we restrict the domain to the interval − π2 , π2 . Think about why the endpoints of this interval are not included (see Figure 0.63). Having done this, you should readily see that we define the inverse tangent function by y = tan−1 x

tan y = x and − π2 < y <

if and only if

π . 2

The graph of y = tan−1 x is then as seen in Figure 0.64, found by reflecting the graph in Figure 0.63 through the line y = x. y 6 4 y

2 q

x q

q 2 4

y 10

4

2

4

6

q

FIGURE 0.63 y = tan x on − π2 , π2

p

1 1

x

2

6

5

FIGURE 0.64 y = tan−1 x

x

q

EXAMPLE 4.7

5

Evaluating an Inverse Tangent

Evaluate tan−1 (1).

Solution You must look for the angle θ on the interval − π2 , π2 for which tan θ = 1. This is easy enough. Since tan π4 = 1 and π4 ∈ − π2 , π2 , we have that tan−1 (1) = π4 .

10

FIGURE 0.65 y = sec x on [0, π ]

We now turn to defining an inverse for sec x. First, we must issue a disclaimer. There are several reasonable ways in which to suitably restrict the domain and different authors restrict We have (somewhat arbitrarily) chosen to restrict the domain to be π itdifferently.

0, 2 ∪ π2 , π . Why not use all of [0, π ]? You need only think about the definition of sec x to see why we needed to exclude the value x = π2 . See Figure 0.65 for a graph of sec x on this domain. Note the vertical asymptote at x = π2 . Consequently, we define the inverse secant function by

y p

q

10

6

5

1 1

y = sec−1 x

x 5

FIGURE 0.66 y = sec−1 x

if and only if

sec y = x and y ∈ 0, π2 ∪ π2 , π .

A graph of sec−1 x is shown in Figure 0.66.

10

EXAMPLE 4.8

√

Evaluating an Inverse Secant

Evaluate sec−1 (− 2).

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REMARK 4.5 We can likewise define inverses to cot x and csc x. As these functions are used only infrequently, we will omit them here and examine them in the exercises.

Function Domain sin−1 x [−1, 1] cos−1 x −1

tan

x

Range π π

−2, 2

[0, π] (−∞, ∞) − π2 , π2 [−1, 1]

0-44

Solution You must look for the angle θ with θ ∈ 0, π2 ∪ π2 , π , for which √ √ √ sec θ = − 2. Notice that if sec θ = − 2, then cos θ = − √12 = − 22 . Recall from √

example 4.6 that cos 3π = − 22 . Further, the angle 3π is in the interval π2 , π and so, 4 4 √ sec−1 (− 2) = 3π . 4 Calculators do not usually have built-in functions for sec x or sec−1 x. In this case, you must convert the desired secant value to a cosine value and use the inverse cosine function, as we did in example 4.8. We summarize the domains and ranges of the three main inverse trigonometric functions in the margin. In many applications, we need to calculate the length of one side of a right triangle using the length of another side and an acute angle (i.e., an angle between 0 and π2 radians). We can do this rather easily, as in example 4.9.

EXAMPLE 4.9

Finding the Height of a Tower

A person 100 feet from the base of a tower measures an angle of 60◦ from the ground to the top of the tower (see Figure 0.67). (a) Find the height of the tower. (b) What angle is measured if the person is 200 feet from the base? h

sin θ 100 ft

l θ cos θ

Solution For (a), we first convert 60◦ to radians: π π 60◦ = 60 = radians. 180 3 We are given that the base of the triangle in Figure 0.67 is 100 feet. We must now compute the height h of the tower. Using the similar triangles indicated in Figure 0.67, we have sin θ h = , cos θ 100

FIGURE 0.67 Height of a tower

so that the height of the tower is √ sin θ π = 100 tan θ = 100 tan = 100 3 ≈ 173 feet. cos θ 3 For part (b), the similar triangles in Figure 0.67 give us √ √ h 3 100 3 tan θ = = = . 200 200 2 π Since 0 < θ < , we have 2

√ 3 −1 θ = tan ≈ 0.7137 radians (about 41 degrees). 2 h = 100

In example 4.10, we simplify expressions involving both trigonometric and inverse trigonometric functions.

EXAMPLE 4.10

Simplifying Expressions Involving Inverse Trigonometric Functions

Simplify (a) sin (cos−1 x) and (b) tan (cos−1 x). Solution Do not look for some arcane formula to help you out. Think first: cos−1 x is the angle (call it θ ) for which x = cos θ . First, consider the case where x > 0. Looking

0-45

SECTION 0.4

..

Trigonometric and Inverse Trigonometric Functions

45

at Figure 0.68, we have drawn a right triangle, with hypotenuse 1 and adjacent angle θ . From the definition of the sine and cosine, then, we have that the base of the triangle is cos θ = x and the altitude is sin θ, which by the Pythagorean Theorem is sin (cos−1 x) = sin θ = 1 − x 2 . sin u 1 x2 1

u cos1x

Wait! We have not yet finished part (a). Figure 0.68 shows 0 < θ < π2 , but by definition, θ = cos−1 x could range from 0 to π . Does our answer change if π2 < θ < π? To see that it doesn’t change, note that if 0 ≤ θ ≤ π, then sin θ ≥ 0. From the Pythagorean identity sin2 θ + cos2 θ = 1, we get sin θ = ± 1 − cos2 θ = ± 1 − x 2 . Since sin θ ≥ 0, we must have

cos u x

FIGURE 0.68 θ = cos−1 x

sin θ =

1 − x 2,

for all values of x. For part (b), you can read from Figure 0.68 that

√ 1 − x2 sin θ tan (cos x) = tan θ = = . cos θ x Note that this last identity is valid, regardless of whether x = cos θ is positive or negative. −1

EXERCISES 0.4 WRITING EXERCISES 1. Many students are comfortable using degrees to measure angles and don’t understand why they must learn radian measures. As discussed in the text, radians directly measure distance along the unit circle. Distance is an important aspect of many applications. In addition, we will see later that many calculus formulas are simpler in radians form than in degrees. Aside from familiarity, discuss any and all advantages of degrees over radians. On balance, which is better? 2. A student graphs f (x) = cos x on a graphing calculator and gets what appears to be a straight line at height y = 1 instead of the usual cosine curve. Upon investigation, you discover that the calculator has graphing window −10 ≤ x ≤ 10, −10 ≤ y ≤ 10 and is in degrees mode. Explain what went wrong and how to correct it.

5. In example 4.3, f (x) = 4 cos 3x has period 2π/3 and g(x) = 2 sin (x/3) has period 6π. Explain why the sum h(x) = 4 cos 3x + 2 sin (x/3) has period 6π. 6. Give a different range for sec−1 x than that given in the text. For which x’s would the value of sec−1 x change? Using the calculator discussion in exercise 4, give one reason why we might have chosen the range that we did. In exercises 1 and 2, convert the given radians measure to degrees. 1. (a) 2. (a)

π 4 3π 5

π 3 (b) π7

(b)

(c)

π 6

(c) 2

(d)

4π 3

(d) 3

3. Inverse functions are necessary for solving equations. The restricted range we had to use to define inverses of the trigonometric functions also restricts their usefulness in equation solving. Explain how to use sin−1 x to find all solutions of the equation sin u = x.

In exercises 3 and 4, convert the given degrees measure to radians. 3. (a) 180◦ (b) 270◦ (c) 120◦ (d) 30◦

4. Discuss how to compute sec−1 x, csc−1 x and cot−1 x on a calculator that has built-in functions only for sin−1 x, cos−1 x and tan−1 x.

In exercises 5–14, find all solutions of the given equation.

4. (a) 40◦

(b) 80◦

5. 2 cos x − 1 = 0 √ 7. 2 cos x − 1 = 0

(c) 450◦

(d) 390◦

6. 2 sin x + 1 = 0 √ 3=0

8. 2 sin x −

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0-46

10. sin2 x − 2 sin x − 3 = 0

51. f (x) = sin 2x − cos 5x

11. sin2 x + cos x − 1 = 0

12. sin 2x − cos x = 0

52. f (x) = cos 3x − sin 7x

13. cos x + cos x = 0

14. sin x − sin x = 0

9. sin2 x − 4 sin x + 3 = 0

2

2

In exercises 15–24, sketch a graph of the function.

In exercises 53–56, use the range for θ to determine the indicated function value.

15. f (x) = sin 2x

16. f (x) = cos 3x

53. sin θ = 13 , 0 ≤ θ ≤

π ; 2

find cos θ.

17. f (x) = tan 2x

18. f (x) = sec 3x 20. f (x) = 4 cos (x + π)

π ; 2

find sin θ.

19. f (x) = 3 cos (x − π/2)

54. cos θ = 45 , 0 ≤ θ ≤

21. f (x) = sin 2x − 2 cos 2x

22. f (x) = cos 3x − sin 3x

23. f (x) = sin x sin 12x

24. f (x) = sin x cos 12x

55. sin θ = 12 ,

π 2

≤ θ ≤ π;

find cos θ.

56. sin θ = 12 ,

π 2

≤ θ ≤ π;

find tan θ.

25. f (x) = 3 sin 2x

26. f (x) = 2 cos 3x

In exercises 57–64, use a triangle to simplify each expression. Where applicable, state the range of x’s for which the simplification holds.

27. f (x) = 5 cos 3x

28. f (x) = 3 sin 5x

57. cos (sin−1 x)

58. cos (tan−1 x)

29. f (x) = 3 cos (2x − π/2)

30. f (x) = 4 sin (3x + π)

31. f (x) = −4 sin x

32. f (x) = −2 cos 3x

59. tan (sec−1 x)

60. cot (cos−1 x)

61. sin cos−1 12

62. cos sin−1 12

63. tan cos−1 35

64. csc sin−1 23

In exercises 25–32, identify the amplitude, period and frequency.

In exercises 33–36, prove that the given trigonometric identity is true. 33. sin (α − β) = sin α cos β − sin β cos α 34. cos (α − β) = cos α cos β + sin α sin β 35. (a) cos (2θ ) = 2 cos2 θ − 1

(b) cos (2θ) = 1 − 2 sin2 θ

36. (a) sec2 θ = tan2 θ + 1

(b) csc2 θ = cot2 θ + 1

In exercises 37–46, evaluate the inverse function by sketching a unit circle and locating the correct angle on the circle. 37. cos−1 0

38. tan−1 0

39. sin−1 (−1)

40. cos−1 (1)

41. sec−1 1

42. tan−1 (−1)

43. sec−1 2

44. csc−1 2 √ 46. tan−1 3

45. cot−1 1 47. Prove that, for some constant β,

4 cos x − 3 sin x = 5 cos (x + β). Then, estimate the value of β. 48. Prove that, for some constant β, √ 2 sin x + cos x = 5 sin (x + β). Then, estimate the value of β. In exercises 49–52, determine whether the function is periodic. If it is periodic, find the smallest (fundamental) period. 49. f (x) = cos 2x + 3 sin π x √ 50. f (x) = sin x − cos 2x

In exercises 65–68, use a graphing calculator or computer to determine the number of solutions of each equation, and numerically estimate the solutions (x is in radians). 65. 2 cos x = 2 − x

66. 3 sin x = x

67. cos x = x − 2

68. sin x = x 2

2

69. A person sitting 2 miles from a rocket launch site measures 20◦ up to the current location of the rocket. How high up is the rocket? 70. A person who is 6 feet tall stands 4 feet from the base of a light pole and casts a 2-foot-long shadow. How tall is the light pole? 71. A surveyor stands 80 feet from the base of a building and measures an angle of 50◦ to the top of the steeple on top of the building. The surveyor figures that the center of the steeple lies 20 feet inside the front of the structure. Find the distance from the ground to the top of the steeple. 72. Suppose that the surveyor of exercise 71 estimates that the center of the steeple lies between 20 and 21 inside the front of the structure. Determine how much the extra foot would change the calculation of the height of the building. 73. A picture hanging in an art gallery has a frame 20 inches high, and the bottom of the frame is 6 feet above the floor. A person whose eyes are 6 feet above the floor stands x feet from the wall. Let A be the angle formed by the ray from the person’s eye to the bottom of the frame and the ray from the person’s

0-47

SECTION 0.4

eye to the top of the frame. Write A as a function of x and graph y = A(x).

20"

..

Trigonometric and Inverse Trigonometric Functions

47

The player can catch the ball by running to keep the angle ψ constant (this makes it appear that the ball is moving in a straight line). Assuming that all triangles shown are right tan α triangles, show that tan ψ = and then solve for ψ. tan β

A

Ball

6'

c

x

74. In golf, the goal is to hit a ball into a hole of diameter 4.5 inches. Suppose a golfer stands x feet from the hole trying to putt the ball into the hole. A first approximation of the margin of error in a putt is to measure the angle A formed by the ray from the ball to the right edge of the hole and the ray from the ball to the left edge of the hole. Find A as a function of x. 75. In an AC circuit, the voltage is given by v(t) = v p sin 2π ft, where v p is the peak voltage and f is the frequency in Hz. A voltmeter actually measures an√average (called the root-meansquare) voltage, equal to v p / 2. If the voltage has amplitude 170 and period π/30, find the frequency and meter voltage. 76. An old-style LP record player rotates records at 33 13 rpm (revolutions per minute). What is the period (in minutes) of the rotation? What is the period for a 45-rpm record? 77. Suppose that the ticket sales of an airline (in of thousands dollars) is given by s(t) = 110 + 2t + 15 sin 16 πt , where t is measured in months. What real-world phenomenon might cause the fluctuation in ticket sales modeled by the sine term? Based on your answer, what month corresponds to t = 0? Disregarding seasonal fluctuations, by what amount is the airline’s sales increasing annually? 78. Piano tuners sometimes start by striking a tuning fork and then the corresponding piano key. If the tuning fork and piano note each have frequency 8, then the resulting sound is sin 8t + sin 8t. Graph this. If the piano is slightly out-of-tune at frequency 8.1, the resulting sound is sin 8t + sin 8.1t. Graph this and explain how the piano tuner can hear the small difference in frequency.

Home plate

a b Outfielder

EXPLORATORY EXERCISES 1. In his book and video series The Ring of Truth, physicist Philip Morrison performed an experiment to estimate the circumference of the earth. In Nebraska, he measured the angle to a bright star in the sky, then drove 370 miles due south into Kansas and measured the new angle to the star. Some geometry shows that the difference in angles, about 5.02◦ , equals the angle from the center of the earth to the two locations in Nebraska and Kansas. If the earth is perfectly spherical (it’s not) and the circumference of the portion of the circle measured out by 5.02◦ is 370 miles, estimate the circumference of the earth. This experiment was based on a similar experiment by the ancient Greek scientist Eratosthenes. The ancient Greeks and the Spaniards of Columbus’ day knew that the earth was round, they just disagreed about the circumference. Columbus argued for a figure about half of the actual value, since a ship couldn’t survive on the water long enough to navigate the true distance. 2. An oil tank with circular cross sections lies on its side. A stick is inserted in a hole at the top and used to measure the depth d of oil in the tank. Based on this measurement, the goal is to compute the percentage of oil left in the tank.

79. Give precise definitions of csc−1 x and cot−1 x. 80. In baseball, outfielders are able to easily track down and catch fly balls that have very long and high trajectories. Physicists have argued for years about how this is done. A recent explanation involves the following geometry.

d

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0-48

To simplify calculations, suppose the circle is a unit circle with center at (0, 0). Sketch radii extending from the origin to the top of the oil. The area of oil at the bottom equals the area of the portion of the circle bounded by the radii minus the area of the triangle formed above the oil in the figure.

1

1 u d

Start with the triangle, which has area one-half base times height. Explain why the height is 1 − d. Find a right triangle in the figure (there are two of them) with hypotenuse 1 (the radius of the circle) and one vertical side of length 1 − d. The horizontal side has length equal to one-half the base of the larger triangle. Show that this equals 1 − (1 − d)2 . The area of the portion of the circle equals π θ/2π = θ/2, where

0.5

θ is the angle at the top of the triangle. Find this angle as a function of d. (Hint: Go back to the right triangle used above with upper angle θ/2.) Then find the area filled with oil and divide by π to get the portion of the tank filled with oil. 3. Computer graphics can be misleading. This exercise works best using a “disconnected” graph (individual dots, not connected). Graph y = sin x 2 using a graphing window for which each pixel represents a step of 0.1 in the x- or y-direction. You should get the impression of a sine wave that oscillates more and more rapidly as you move to the left and right. Next, change the graphing window so that the middle of the original screen (probably x = 0) is at the far left of the new screen. You will likely see what appears to be a random jumble of dots. Continue to change the graphing window by increasing the x-values. Describe the patterns or lack of patterns that you see. You should find one pattern that looks like two rows of dots across the top and bottom of the screen; another pattern looks like the original sine wave. For each pattern that you find, pick adjacent points with x-coordinates a and b. Then change the graphing window so that a ≤ x ≤ b and find the portion of the graph that is missing. Remember that, whether the points are connected or not, computer graphs always leave out part of the graph; it is part of your job to know whether or not the missing part is important.

EXPONENTIAL AND LOGARITHMIC FUNCTIONS Some bacteria reproduce very quickly, as you may have discovered if you have ever had an infected cut or strep throat. Under the right circumstances, the number of bacteria in certain cultures will double in as little as an hour. In this section, we discuss some functions that can be used to model such rapid growth. Suppose that initially there are 100 bacteria at a given site and the population doubles every hour. Call the population function P(t), where t represents time (in hours) and start the clock running at time t = 0. Since the initial population is 100, we have P(0) = 100. After 1 hour, the population will double to 200, so that P(1) = 200. After another hour, the population will have doubled again to 400, making P(2) = 400 and so on. To compute the bacterial population after 10 hours, you could calculate the population at 4 hours, 5 hours and so on, or you could use the following shortcut. To find P(1), double the initial population, so that P(1) = 2 · 100. To find P(2), double the population at time t = 1, so that P(2) = 2 · 2 · 100 = 22 · 100. Similarly, P(3) = 23 · 100. This pattern leads us to P(10) = 210 · 100 = 102,400. Observe that the population can be modeled by the function P(t) = 2t · 100. We call P(t) an exponential function because the variable t is in the exponent. There is a subtle question here: what is the domain of this function? We have so far used only integer

48

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0-48

To simplify calculations, suppose the circle is a unit circle with center at (0, 0). Sketch radii extending from the origin to the top of the oil. The area of oil at the bottom equals the area of the portion of the circle bounded by the radii minus the area of the triangle formed above the oil in the figure.

1

1 u d

Start with the triangle, which has area one-half base times height. Explain why the height is 1 − d. Find a right triangle in the figure (there are two of them) with hypotenuse 1 (the radius of the circle) and one vertical side of length 1 − d. The horizontal side has length equal to one-half the base of the larger triangle. Show that this equals 1 − (1 − d)2 . The area of the portion of the circle equals π θ/2π = θ/2, where

0.5

θ is the angle at the top of the triangle. Find this angle as a function of d. (Hint: Go back to the right triangle used above with upper angle θ/2.) Then find the area filled with oil and divide by π to get the portion of the tank filled with oil. 3. Computer graphics can be misleading. This exercise works best using a “disconnected” graph (individual dots, not connected). Graph y = sin x 2 using a graphing window for which each pixel represents a step of 0.1 in the x- or y-direction. You should get the impression of a sine wave that oscillates more and more rapidly as you move to the left and right. Next, change the graphing window so that the middle of the original screen (probably x = 0) is at the far left of the new screen. You will likely see what appears to be a random jumble of dots. Continue to change the graphing window by increasing the x-values. Describe the patterns or lack of patterns that you see. You should find one pattern that looks like two rows of dots across the top and bottom of the screen; another pattern looks like the original sine wave. For each pattern that you find, pick adjacent points with x-coordinates a and b. Then change the graphing window so that a ≤ x ≤ b and find the portion of the graph that is missing. Remember that, whether the points are connected or not, computer graphs always leave out part of the graph; it is part of your job to know whether or not the missing part is important.

EXPONENTIAL AND LOGARITHMIC FUNCTIONS Some bacteria reproduce very quickly, as you may have discovered if you have ever had an infected cut or strep throat. Under the right circumstances, the number of bacteria in certain cultures will double in as little as an hour. In this section, we discuss some functions that can be used to model such rapid growth. Suppose that initially there are 100 bacteria at a given site and the population doubles every hour. Call the population function P(t), where t represents time (in hours) and start the clock running at time t = 0. Since the initial population is 100, we have P(0) = 100. After 1 hour, the population will double to 200, so that P(1) = 200. After another hour, the population will have doubled again to 400, making P(2) = 400 and so on. To compute the bacterial population after 10 hours, you could calculate the population at 4 hours, 5 hours and so on, or you could use the following shortcut. To find P(1), double the initial population, so that P(1) = 2 · 100. To find P(2), double the population at time t = 1, so that P(2) = 2 · 2 · 100 = 22 · 100. Similarly, P(3) = 23 · 100. This pattern leads us to P(10) = 210 · 100 = 102,400. Observe that the population can be modeled by the function P(t) = 2t · 100. We call P(t) an exponential function because the variable t is in the exponent. There is a subtle question here: what is the domain of this function? We have so far used only integer

0-49

SECTION 0.5

..

Exponential and Logarithmic Functions

49

values of t, but for what other values of t does P(t) make √sense? Certainly, rational powers make sense, as in P(1/2) = 21/2 · 100, where 21/2 = 2. This says that the number of bacteria in the culture after a half hour is approximately √ P(1/2) = 21/2 · 100 = 2 · 100 ≈ 141. It’s a simple matter to interpret fractional powers as roots. For instance, √ x 1/2 = x, √ x 1/3 = 3 x, √ x 1/4 = 4 x, √ √ 3 x 2/3 = x 2 = ( 3 x)2 , √ 10 x 3.1 = x 31/10 = x 31 and so on. But, what about irrational powers? They are harder to define, but they work exactly the way you would want them to. For instance, since π is between 3.14 and 3.15, 2π is between 23.14 and 23.15 . In this way, we define 2x for x irrational to fill in the gaps in the graph of y = 2x for x rational. That is, if x is irrational and a < x < b, for rational numbers a and b, then 2a < 2x < 2b . This is the logic behind the definition of irrational powers. If for some reason you wanted to find the bacterial population after π hours, you can use your calculator or computer to obtain the approximate population: P(π ) = 2π · 100 ≈ 882. For your convenience, we now summarize the usual rules of exponents.

RULES OF EXPONENTS r For any integers m and n,

x m/n =

√ n

√ x m = ( n x)m .

r For any real number p,

x−p = r For any real numbers p and q,

1 . xp

(x p )q = x p·q . r For any real numbers p and q,

x p · x q = x p+q .

Throughout your calculus course, you will need to be able to quickly convert back and forth between exponential form and fractional or root form.

EXAMPLE 5.1

Converting Expressions to Exponential Form

√ 5 3x 2 Convert each to exponential form: (a) 3 x 5 , (b) √ , (c) √ and (d) (2x · 23+x )2 . 3 x 2 x

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0-50

Solution For (a), simply leave the 3 alone and convert the power: √ 3 x 5 = 3x 5/2 . For (b), use a negative exponent to write x in the numerator: 5 = 5x −1/3 . √ 3 x For (c), first separate the constants from the variables and then simplify: 3x 2 3 3 3 x2 = x 2−1/2 = x 3/2 . √ = 1/2 2x 2 2 2 x For (d), first work inside the parentheses and then square: (2x · 23+x )2 = (2x+3+x )2 = (22x+3 )2 = 24x+6 . The function in part (d) of example 5.1 is called an exponential function with a base of 2.

DEFINITION 5.1 For any constant b > 0, the function f (x) = b x is called an exponential function. Here, b is called the base and x is the exponent. Be careful to distinguish between algebraic functions such as f (x) = x 3 and g(x) = x 2/3 and exponential functions. For exponential functions such as h(x) = 2x , the variable is in the exponent (hence the name), instead of in the base. Also, notice that the domain of an exponential function is the entire real line, (−∞, ∞), while the range is the open interval (0, ∞), since b x > 0 for all x. While any positive real number can be used as a base for an exponential function, three bases are the most commonly used in practice. Base 2 arises naturally when analyzing processes that double at regular intervals (such as the bacteria at the beginning of this section). Our standard counting system is base 10, so this base is commonly used. However, far and away the most useful base is the irrational number e. Like π , the number e has a surprising tendency to occur in important calculations. We define e by e = lim

n→∞

1+

1 n

n .

(5.1)

Note that equation (5.1) has at least two serious shortcomings. First, we have not yet said what the notation lim means. (In fact, we won’t define this until Chapter 1.) Second, it’s n→∞ unclear why anyone would ever define a number in such a strange way. We will not be in a position to answer the second question until Chapter 4 (but the answer is worth the wait). It suffices for the moment to say that equation (5.1) means that e can be approximated by calculating values of (1 + 1/n)n for large values of n and that the larger the value of n, the closer the approximation will be to the actual value of e. In particular, if you look at the sequence of numbers (1 + 1/2)2 , (1 + 1/3)3 , (1 + 1/4)4 and so on, they will get progressively closer and closer to (i.e., home in on) the irrational number e.

0-51

SECTION 0.5

..

Exponential and Logarithmic Functions

51

To get an idea of the value of e, compute several of these numbers: 1 10 1+ = 2.5937 . . . , 10 1000 1 1+ = 2.7169 . . . , 1000 10,000 1 1+ = 2.7181 . . . 10,000 and so on. You should compute enough of these values to convince yourself that the first few digits of the decimal representation of e (e ≈ 2.718281828459 . . .) are correct.

EXAMPLE 5.2

Computing Values of Exponentials

Approximate e4 , e−1/5 and e0 . Solution From a calculator, we find that e4 = e · e · e · e ≈ 54.598. From the usual rules of exponents, e−1/5 =

1 e1/5

1 =√ ≈ 0.81873. 5 e

(On a calculator, it is convenient to replace −1/5 with −0.2.) Finally, e0 = 1. The graphs of the exponential functions summarize many of their important properties.

EXAMPLE 5.3

Sketching Graphs of Exponentials

Sketch the graphs of the exponential functions y = 2x , y = e x , y = e2x , y = e x/2 , y = (1/2)x and y = e−x . Solution Using a calculator or computer, you should get graphs similar to those that follow. y

4

y

30

30

20

20

10

10 x

2

2

4

4

x

2

2

FIGURE 0.69a

FIGURE 0.69b

y = 2x

y = ex

4

Notice that each of the graphs in Figures 0.69a, 0.69b, 0.70a and 0.70b starts very near the x-axis (reading left to right), passes through the point (0, 1) and then rises steeply. This is true for all exponentials with base greater than 1 and with a positive coefficient in the exponent. Note that the larger the base (e > 2) or the larger the coefficient in the

52

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0-52

y

4

y

30

30

20

20

10

10 x

2

2

4

4

x

2

2

FIGURE 0.70a

FIGURE 0.70b

y = e2x

y = e x/2

y

4

4

y

30

30

20

20

10

10 x

2

2

4

4

FIGURE 0.71a

x

2

2

4

FIGURE 0.71b y = e−x

y = (1/2)x

exponent (2 > 1 > 1/2), the more quickly the graph rises to the right (and drops to the left). Note that the graphs in Figures 0.71a and 0.71b are the mirror images in the y-axis of Figures 0.69a and 0.69b, respectively. The graphs rise as you move to the left and drop toward the x-axis as you move to the right. It’s worth noting that by the rules of exponents, (1/2)x = 2−x and (1/e)x = e−x . In Figures 0.69–0.71, each exponential function is one-to-one and, hence, has an inverse function. We define the logarithmic functions to be inverses of the exponential functions.

DEFINITION 5.2 For any positive number b = 1, the logarithm function with base b, written logb x, is defined by y = logb x

if and only if

x = by .

That is, the logarithm logb x gives the exponent to which you must raise the base b to get the given number x. For example, log10 10 = 1 (since 101 = 10), log10 100 = 2 (since 102 = 100), log10 1000 = 3 (since 103 = 1000) and so on. The value of log10 45 is less clear than the preceding three values, but the idea is the same: you need to find the number y such that 10 y = 45. The answer is between

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SECTION 0.5

..

Exponential and Logarithmic Functions

53

1 and 2, but to be more precise, you will need to employ trial and error. You should get log10 45 ≈ 1.6532. Observe from Definition 5.2 that for any base b > 0 (b = 1), if y = logb x, then x = b y > 0. That is, the domain of f (x) = logb x is the interval (0, ∞). Likewise, the range of f is the entire real line, (−∞, ∞). As with exponential functions, the most useful bases turn out to be 2, 10, and e. We usually abbreviate log10 x by log x. Similarly, loge x is usually abbreviated ln x (for natural logarithm).

EXAMPLE 5.4

Evaluating Logarithms

Without using your calculator, determine log(1/10), log(0.001), ln e and ln e3 . Solution Since 1/10 = 10−1 , log(1/10) = −1. Similarly, since 0.001 = 10−3 , we have that log(0.001) = −3. Since ln e = loge e1 , ln e = 1. Similarly, ln e3 = 3. We want to emphasize the inverse relationship defined by Definition 5.2. That is, b x and logb x are inverse functions for any b > 0 (b = 1). In particular, for the base e, we have eln x = x

for any x > 0

and

ln (e x ) = x

for any x.

(5.2)

We demonstrate this as follows. Let y = ln x = loge x. By Definition 5.2, we have that x = e y = eln x . We can use this relationship between natural logarithms and exponentials to solve equations involving logarithms and exponentials, as in examples 5.5 and 5.6.

EXAMPLE 5.5

Solving a Logarithmic Equation

Solve the equation ln(x + 5) = 3 for x. Solution Taking the exponential of both sides of the equation and writing things backward (for convenience), we have e3 = eln(x+5) = x + 5, from (5.2). Subtracting 5 from both sides gives us e3 − 5 = x.

EXAMPLE 5.6

Solving an Exponential Equation

Solve the equation e x+4 = 7 for x. Solution Taking the natural logarithm of both sides and writing things backward (for simplicity), we have from (5.2) that ln 7 = ln (e x+4 ) = x + 4. Subtracting 4 from both sides yields ln 7 − 4 = x.

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y

0-54

As always, graphs provide excellent visual summaries of the important properties of a function.

2 1 x 1

1

2

3

4

5

EXAMPLE 5.7

Sketching Graphs of Logarithms

Sketch graphs of y = log x and y = ln x, and briefly discuss the properties of each.

2

Solution From a calculator or computer, you should obtain graphs resembling those in Figures 0.72a and 0.72b. Notice that both graphs appear to have a vertical asymptote at x = 0 (why would that be?), cross the x-axis at x = 1 and very gradually increase as x increases. Neither graph has any points to the left of the y-axis, since log x and ln x are defined only for x > 0. The two graphs are very similar, although not identical.

3

FIGURE 0.72a y = log x

The properties just described graphically are summarized in Theorem 5.1.

y 2

THEOREM 5.1

1 x 1

1

2

3

4

5

For any positive base b = 1, (i) logb x is defined only for x > 0, (ii) logb 1 = 0 and (iii) if b > 1, then logb x < 0 for 0 < x < 1 and logb x > 0 for x > 1.

2 3

FIGURE 0.72b y = ln x

PROOF (i) Note that since b > 0, b y > 0 for any y. So, if logb x = y, then x = b y > 0. (ii) Since b0 = 1 for any number b = 0, logb 1 = 0 (i.e., the exponent to which you raise the base b to get the number 1 is 0). (iii) We leave this as an exercise. All logarithms share a set of defining properties, as stated in Theorem 5.2.

THEOREM 5.2 For any positive base b = 1 and positive numbers x and y, we have (i) logb (x y) = logb x + logb y, (ii) logb (x/y) = logb x − logb y and (iii) logb (x y ) = y logb x.

As with most algebraic rules, each one of these properties can dramatically simplify calculations when it applies.

EXAMPLE 5.8

Simplifying Logarithmic Expressions

Write each as a single logarithm: (a) log2 27x − log2 3x and (b) ln 8 − 3 ln (1/2). Solution First, note that there is more than one order in which to work each problem. For part (a), we have 27 = 33 and so, 27x = (33 )x = 33x . This gives us log2 27x − log2 3x = log2 33x − log2 3x = 3x log2 3 − x log2 3 = 2x log2 3 = log2 32x .

0-55

..

SECTION 0.5

Exponential and Logarithmic Functions

55

For part (b), note that 8 = 23 and 1/2 = 2−1 . Then, ln 8 − 3 ln (1/2) = 3 ln 2 − 3(− ln 2) = 3 ln 2 + 3 ln 2 = 6 ln 2 = ln 26 = ln 64. In some circumstances, it is beneficial to use the rules of logarithms to expand a given expression, as in example 5.9.

EXAMPLE 5.9

Expanding a Logarithmic Expression

Use the rules of logarithms to expand the expression ln

x 3 y4 . z5

Solution From Theorem 5.2, we have that 3 4 x y ln = ln (x 3 y 4 ) − ln (z 5 ) = ln (x 3 ) + ln (y 4 ) − ln (z 5 ) z5 = 3 ln x + 4 ln y − 5 ln z. Using the rules of exponents and logarithms, we can rewrite any exponential as an exponential with base e, as follows. For any base a > 0, we have x

a x = eln (a ) = e x ln a .

(5.3)

This follows from Theorem 5.2 (iii) and the fact that eln y = y, for all y > 0.

EXAMPLE 5.10

Rewriting Exponentials as Exponentials with Base e

Rewrite the exponentials 2x , 5x and (2/5)x as exponentials with base e. Solution From (5.3), we have x

2x = eln (2 ) = e x ln 2 , x

and

5x = eln (5 ) = e x ln 5 x 2 x = eln [(2/5) ] = e x ln (2/5) . 5

Just as we can rewrite an exponential with any positive base in terms of an exponential with base e, we can rewrite any logarithm in terms of natural logarithms, as follows. For any positive base b (b = 1), we will show that logb x =

ln x . ln b

(5.4)

Let y = logb x. Then by Definition 5.2, we have that x = b y . Taking the natural logarithm of both sides of this equation, we get by Theorem 5.2 (iii) that ln x = ln(b y ) = y ln b. Dividing both sides by ln b (since b = 1, ln b = 0) gives us y= establishing (5.4).

ln x , ln b

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0-56

Equation (5.4) is useful for computing logarithms with bases other than e or 10. This is important since, more than likely, your calculator has keys only for ln x and log x. We illustrate this idea in example 5.11.

EXAMPLE 5.11

Approximating the Value of Logarithms

Approximate the value of log7 12. Solution From (5.4), we have log7 12 =

ln 12 ≈ 1.2769894. ln 7

Hyperbolic Functions There are two special combinations of exponential functions, called the hyperbolic sine and hyperbolic cosine functions, that have important applications. For instance, the Gateway Arch in Saint Louis was built in the shape of a hyperbolic cosine graph. (See the photograph in the margin.) The hyperbolic sine function [denoted by sinh (x)] and the hyperbolic cosine function [denoted by cosh (x)] are defined by sinh x =

Saint Louis Gateway Arch

e x − e−x 2

and

cosh x =

e x + e−x . 2

Graphs of these functions are shown in Figures 0.73a and 0.73b. The hyperbolic functions (including the hyperbolic tangent, tanh x, defined in the expected way) are often convenient to use when solving equations. For now, we verify several basic properties that the hyperbolic functions satisfy in parallel with their trigonometric counterparts. y

4

y

10

10

5

5 x

2

2

4

4

x

2

2

5

5

10

10

FIGURE 0.73a

FIGURE 0.73b

y = sinh x

y = cosh x

EXAMPLE 5.12

4

Computing Values of Hyperbolic Functions

Compute f (0), f (1) and f (−1), and determine how f (x) and f (−x) compare for each function: (a) f (x) = sinh x and (b) f (x) = cosh x.

0-57

SECTION 0.5

..

Exponential and Logarithmic Functions

57

e0 − e−0 1−1 = = 0. Note that this 2 2 1 −1 e −e means that sinh 0 = sin 0 = 0. Also, we have sinh 1 = ≈ 1.18, while 2 e−1 − e1 sinh (−1) = ≈ −1.18. Notice that sinh (−1) = − sinh 1. In fact, for any x, 2 e−x − e x −(e x − e−x ) sinh (−x) = = = − sinh x. 2 2

Solution For part (a), we have sinh 0 =

[The same rule holds for the sine function: sin (−x) = −sin x.] For part (b), we have e0 + e−0 1+1 cosh 0 = = = 1. Note that this means that cosh 0 = cos 0 = 1. Also, 2 2 1 −1 e +e e−1 + e1 we have cosh 1 = ≈ 1.54, while cosh (−1) = ≈ 1.54. Notice that 2 2 cosh (−1) = cosh 1. In fact, for any x, cosh (−x) =

e−x + e x e x + e−x = = cosh x. 2 2

[The same rule holds for the cosine function: cos (−x) = cos x.]

Fitting a Curve to Data You are familiar with the idea that two points determine a straight line. As we see in example 5.13, two points will also determine an exponential function.

EXAMPLE 5.13

Matching Data to an Exponential Curve

Find the exponential function of the form f (x) = aebx that passes through the points (0, 5) and (3, 9). Solution We must solve for a and b, using the properties of logarithms and exponentials. First, for the graph to pass through the point (0, 5), this means that 5 = f (0) = aeb · 0 = a, so that a = 5. Next, for the graph to pass through the point (3, 9), we must have 9 = f (3) = ae3b = 5e3b . To solve for b, we divide both sides of the equation by 5 and take the natural logarithm of both sides, which yields 9 ln = ln e3b = 3b, 5 Year

U.S. Population

1790

3,929,214

1800

5,308,483

1810

7,239,881

1820

9,638,453

1830

12,866,020

1840

17,069,453

1850

23,191,876

1860

31,443,321

from (5.2). Finally, dividing by 3 gives us the value for b: 1 9 b = ln . 3 5 1

Thus, f (x) = 5e 3 ln (9/5) x. Consider the population of the United States from 1790 to 1860, found in the accompanying table. A plot of these data points can be seen in Figure 0.74 (where the vertical scale represents the population in millions). This shows that the population was increasing, with larger and larger increases each decade. If you sketch an imaginary curve through these points, you will probably get the impression of a parabola or perhaps the right half of a

U.S. population (in millions)

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cubic or exponential. And that’s the question: are these data best modeled by a quadratic function, a cubic function, an exponential function or what? We can use the properties of logarithms from Theorem 5.2 to help determine whether a given set of data is modeled better by a polynomial or an exponential function, as follows. Suppose that the data actually come from an exponential, say, y = aebx (i.e., the data points lie on the graph of this exponential). Then,

35 30 25 20 15 10

ln y = ln (aebx ) = ln a + ln ebx = ln a + bx.

5 1 2 3 4 5 6 7 8 Number of decades since 1780

FIGURE 0.74

If you draw a new graph, where the horizontal axis shows values of x and the vertical axis corresponds to values of ln y, then the graph will be the line ln y = bx + c (where the constant c = ln a). On the other hand, suppose the data actually came from a polynomial. If y = bx n (for any n), then observe that

U.S. Population 1790–1860

Natural logarithm of U.S. population (millions)

0-58

ln y = ln (bx n ) = ln b + ln x n = ln b + n ln x.

3

In this case, a graph with horizontal and vertical axes corresponding to x and ln y, respectively, will look like the graph of a logarithm, ln y = n ln x + c. Such semi-log graphs (i.e., graphs of ln y versus x) let us distinguish the graph of an exponential from that of a polynomial: graphs of exponentials become straight lines, while graphs of polynomials (of degree ≥ 1) become logarithmic curves. Scientists and engineers frequently use semi-log graphs to help them gain an understanding of physical phenomena represented by some collection of data.

2

EXAMPLE 5.14

1

Determine whether the population of the United States from 1790 to 1860 was increasing exponentially or as a polynomial.

4

1 2 3 4 5 6 7 8 Number of decades since 1780

FIGURE 0.75 Semi-log plot of U.S. population

Using a Semi-Log Graph to Identify a Type of Function

Solution As already indicated, the trick is to draw a semi-log graph. That is, instead of plotting (1, 3.9) as the first data point, plot (1, ln 3.9) and so on. A semi-log plot of this data set is seen in Figure 0.75. Although the points are not exactly colinear (how would you prove this?), the plot is very close to a straight line with ln y-intercept of 1 and slope 0.3. You should conclude that the population is well modeled by an exponential function. The exponential model would be y = P(t) = aebt , where t represents the number of decades since 1780. Here, b is the slope and ln a is the ln y-intercept of the line in the semi-log graph. That is, b ≈ 0.3 and ln a ≈ 1 (why?), so that a ≈ e. The population is then modeled by P(t) = e · e0.3t million.

EXERCISES 0.5 WRITING EXERCISES 1. Starting from a single cell, a human being is formed by 50 generations of cell division. Explain why after n divisions there are 2n cells. Guess how many cells will be present after 50 divisions, then compute 250 . Briefly discuss how rapidly exponential functions increase.

2. Explain why the graphs of f (x) = 2−x and g(x) = the same.

1 x 2

are

3. Compare f (x) = x 2 and g(x) = 2x for x = 12 , x = 1, x = 2, x = 3 and x = 4. In general, which function is bigger for large values of x? For small values of x?

0-59

SECTION 0.5

4. Compare f (x) = 2x and g(x) = 3x for x = −2, x = − 12 , x = 12 and x = 2. In general, which function is bigger for negative values of x? For positive values of x? In exercises 1–6, convert each exponential expression into fractional or root form. −3

−2

2. 4

1. 2

2/5

2/3

4. 6

5. 5

3. 3

−2/3

6. 4

In exercises 17–20, use a calculator or computer to estimate each value. 18. 4e−2/3

12 e

14 20. √ e

In exercises 21–30, sketch a graph of the given function. 21. f (x) = e2x

22. f (x) = e3x

23. f (x) = 2e x/4

24. f (x) = e−x

25. f (x) = 3e−2x

26. f (x) = 10e−x/3

27. f (x) = ln 2x

28. f (x) = ln x 2

29. f (x) = e2 ln x

30. f (x) = e−x/4 sin x

2

In exercises 31–40, solve the given equation for x. 31. e2x = 2

32. e4x = 3

33. e x (x 2 − 1) = 0

34. xe−2x + 2e−2x = 0

35. ln 2x = 4

36. 2 ln 3x = 1

37. 4 ln x = −8

38. x 2 ln x − 9 ln x = 0

39. e2 ln x = 4

40. ln (e2x ) = 6

In exercises 41 and 42, use the definition of logarithm to determine the value. 41. (a) log3 9 42. (a) log4

1 16

1 27

(b) log4 64

(c) log3

(b) log4 2

(c) log9 3

59

In exercises 43 and 44, use equation (5.4) to approximate the value. 43. (a) log3 7 44. (a) log4

1 10

1 24

(b) log4 60

(c) log3

(b) log4 3

(c) log9 8

In exercises 45–50, rewrite the expression as a single logarithm. 46. 2 ln 4 − ln 3

ln 4 − ln 2

48. 3 ln 2 − ln 12

49. ln 34 + 4 ln 2

50. ln 9 − 2 ln 3

47.

In exercises 13–16, find the integer value of the given expression without using a calculator. √ 8 2 3/2 2/3 13. 4 14. 8 15. 1/2 16. 2 (1/3)2

19.

Exponential and Logarithmic Functions

45. ln 3 − ln 4

1/2

In exercises 7–12, convert each expression into exponential form. √ 1 2 3 7. 2 8. x 2 9. 3 x x 4 1 3 10. 2 11. √ 12. √ x 2 x 2 x3

17. 2e−1/2

..

1 2

In exercises 51–54, find a function of the form f (x) aebx with the given function values. 51. f (0) = 2, f (2) = 6

52. f (0) = 3, f (3) = 4

53. f (0) = 4, f (2) = 2

54. f (0) = 5, f (1) = 2

55. A fast-food restaurant gives every customer a game ticket. With each ticket, the customer has a 1-in-10 chance of winning a free meal. If you go 10 times, estimate your chances of at winning 10 least one free meal. The exact probability is 1 − 109 . Compute this number and compare it to your guess. 56. In exercise 55, if you had 20 tickets with a 1-in-20 chance of winning, would you expect your probability of winning at least once to increase or decrease? Compute the probability 20 1 − 19 to find out. 20 57. In general, if you have n chances of winning with a 1-in-n chance on each n try, the probability of winning at least once is 1 − 1 − n1 . As n gets larger, what number does this probability approach? (Hint: There is a very good reason that this question is in this section!) 58. If y = a · x m , show that ln y = ln a + m ln x. If v = ln y, u = ln x and b = ln a, show that v = mu + b. Explain why the graph of v as a function of u would be a straight line. This graph is called the log-log plot of y and x. 59. For the given data, compute v = ln y and u = ln x, and plot points (u, v). Find constants m and b such that v = mu + b and use the results of exercise 58 to find a constant a such that y = a · xm. x y

2.2 14.52

2.4 17.28

2.6 20.28

2.8 23.52

3.0 27.0

3.2 30.72

3.6 13.66

3.8 14.81

60. Repeat exercise 59 for the given data. x y

2.8 9.37

3.0 10.39

3.2 11.45

3.4 12.54

61. Construct a log-log plot (see exercise 58) of the U.S. population data in example 5.14. Compared to the semi-log plot of the data in Figure 0.75, does the log-log plot look linear? Based on this, are the population data modeled better by an exponential function or a polynomial (power) function?

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62. Construct a semi-log plot of the data in exercise 59. Compared to the log-log plot already constructed, does this plot look linear? Based on this, are these data better modeled by an exponential or power function? 63. The concentration [H+ ] of free hydrogen ions in a chemical solution determines the solution’s pH, as defined by pH = − log [H+ ]. Find [H+ ] if the pH equals (a) 7, (b) 8 and (c) 9. For each increase in pH of 1, by what factor does [H+ ] change? 64. Gastric juice is considered an acid, with a pH of about 2.5. Blood is considered alkaline, with a pH of about 7.5. Compare the concentrations of hydrogen ions in the two substances (see exercise 63). 65. The Richter magnitude M of an earthquake is defined in terms of the energy E in joules released by the earthquake, with log10 E = 4.4 + 1.5M. Find the energy for earthquakes with magnitudes (a) 4, (b) 5 and (c) 6. For each increase in M of 1, by what factor does E change? 66. It puzzles some people who have not grown up around earthquakes that a magnitude 6 quake is considered much more severe than a magnitude 3 quake. Compare the amount of energy released in the two quakes. (See exercise 65.) 67. The decibel level of a noise is defined in terms of the intensity I of the noise, with dB = 10 log (I /I0 ). Here, I0 = 10−12 W/m2 is the intensity of a barely audible sound. Compute the intensity levels of sounds with (a) dB = 80, (b) dB = 90 and (c) dB = 100. For each increase of 10 decibels, by what factor does I change? 68. At a basketball game, a courtside decibel meter shows crowd noises ranging from 60 dB to 110 dB. Compare the intensity level of the 110-dB crowd noise with that of the 60-dB noise. (See exercise 67.) 69. Use a graphing calculator to graph y = xe−x , y = xe−2x , y = xe−3x and so on. Estimate the location of the maximum for each. In general, state a rule for the location of the maximum of y = xe−kx . 70. In golf, the task is to hit a golf ball into a small hole. If the ground near the hole is not flat, the golfer must judge how much the ball’s path will curve. Suppose the golfer is at the point (−13, 0), the hole is at the point (0, 0) and the path of the ball is, for −13 ≤ x ≤ 0, y = −1.672x + 72 ln (1 + 0.02x). Show that the ball goes in the hole and estimate the point on the y-axis at which the golfer aimed. Exercises 71–76 refer to the hyperbolic functions. 71. Show that the range of the hyperbolic cosine is cosh x ≥ 1 and the range of the hyperbolic sine is the entire real line. 72. Show that cosh2 x − sinh2 x = 1 for all x. 73. The Saint Louis Gateway Arch is both 630 feet wide and 630 feet tall. (Most people think that it looks taller than

0-60

it is wide.) One model xfor the outline of the arch is y = 757.7 − 127.7 cosh 127.7 for y ≥ 0. Use a graphing calculator to approximate the x- and y-intercepts and determine if the model has the correct horizontal and vertical measurements. 74. To model the outline of the Gateway Arch with a parabola, you can start with y = −c(x + 315)(x − 315) for some constant c. Explain why this gives the correct x-intercepts. Determine the constant c that gives a y-intercept of 630. Graph this parabola and the hyperbolic cosine in exercise 73 on the same axes. Are the graphs nearly identical or very different? 75. Find all solutions of sinh (x 2 − 1) = 0. 76. Find all solutions of cosh (3x + 2) = 0. 77. On a standard piano, the A below middle C produces a sound wave with frequency 220 Hz (cycles per second). The frequency of the A one octave higher is 440 Hz. In general, doubling the frequency produces the same note an octave higher. Find an exponential formula for the frequency f as a function of the number of octaves x above the A below middle C. 78. There are 12 notes in an octave on a standard piano. Middle C is 3 notes above A (see exercise 77). If the notes are tuned equally, this means that middle C is a quarter-octave above A. Use x = 14 in your formula from exercise 77 to estimate the frequency of middle C.

EXPLORATORY EXERCISES 1. Graph y = x 2 and y = 2x and approximate the two positive solutions of the equation x 2 = 2x . Graph y = x 3 and y = 3x , and approximate the two positive solutions of the equation x 3 = 3x . Explain why x = a will always be a solution of x a = a x , a > 0. What is different about the role of x = 2 as a solution of x 2 = 2x compared to the role of x = 3 as a solution of x 3 = 3x ? To determine the a-value at which the change occurs, graphically solve x a = a x for a = 2.1, 2.2, . . . , 2.9, and note that a = 2.7 and a = 2.8 behave differently. Continue to narrow down the interval of change by testing a = 2.71, 2.72, . . . , 2.79. Then guess the exact value of a. 2. Graph y = ln x and describe the behavior near x = 0. Then graph y = x ln x and describe the behavior near x = 0. Repeat this for y = x 2 ln x, y = x 1/2 ln x and y = x a ln x for a variety of positive constants a. Because the function “blows up” at x = 0, we say that y = ln x has a singularity at x = 0. The order of the singularity at x = 0 of a function f (x) is the smallest value of a such that y = x a f (x) doesn’t have a singularity at x = 0. Determine the order of the singularity at x = 0 for (a) f (x) = x1 , (b) f (x) = 12 and (c) f (x) = 13 . The x x higher the order of the singularity, the “worse” the singularity is. Based on your work, how bad is the singularity of y = ln x at x = 0?

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0.6

..

Transformations of Functions

61

TRANSFORMATIONS OF FUNCTIONS You are now familiar with a long list of functions: polynomials, rational functions, trigonometric functions, exponentials and logarithms. One important goal of this course is to more fully understand the properties of these functions. To a large extent, you will build your understanding by examining a few key properties of functions. We expand on our list of functions by combining them. We begin in a straight-forward fashion with Definition 6.1.

DEFINITION 6.1 Suppose that f (x) and g(x) are functions with domains D1 and D2 , respectively. The functions f + g, f − g and f · g are defined by ( f + g)(x) = f (x) + g(x), ( f − g)(x) = f (x) − g(x) ( f · g)(x) = f (x) · g(x),

and

f for all x in D1 ∩ D2 (i.e., x ∈ D1 , and x ∈ D2 ). The function is defined by g f f (x) (x) = , g g(x) for all x in D1 ∩ D2 such that g(x) = 0.

In example 6.1, we examine various combinations of several simple functions.

EXAMPLE 6.1

Combinations of Functions

If f (x) = x − 3 and g(x) =

√

x − 1, determine the functions f + g, 3 f − g and

f , g

stating the domains of each. Solution First, note that the domain of f is the entire real line and the domain of g is the set of all x ≥ 1. Now, √ ( f + g)(x) = x − 3 + x − 1 √ √ and (3 f − g)(x) = 3(x − 3) − x − 1 = 3x − 9 − x − 1. Notice that the domain of both ( f + g) and (3 f − g) is {x|x ≥ 1}. For f f (x) x −3 , (x) = =√ g g(x) x −1 the domain is {x|x > 1}, where we have added the restriction x = 1 to avoid dividing by 0.

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Definition 6.1 and example 6.1 show us how to do arithmetic with functions. An operation on functions that does not directly correspond to arithmetic is the composition of two functions.

DEFINITION 6.2

f

f (g(x))

The composition of functions f and g, written f ◦ g, is defined by ( f ◦ g)(x) = f (g(x)), for all x such that x is in the domain of g and g(x) is in the domain of f .

g(x)

x

g

The composition of two functions is a two-step process, as indicated in the margin schematic. Be careful to notice what this definition is saying. In particular, for f (g(x)) to be defined, you first need g(x) to be defined, so x must be in the domain of g. Next, f must be defined at the point g(x), so that the number g(x) will need to be in the domain of f.

( f ◦ g)(x) = f (g(x))

EXAMPLE 6.2

Finding the Composition of Two Functions

For f (x) = x 2 + 1 and g(x) = identify the domain of each.

√

x − 2, find the compositions f ◦ g and g ◦ f and

Solution First, we have

√ ( f ◦ g)(x) = f (g(x)) = f ( x − 2) √ = ( x − 2)2 + 1 = x − 2 + 1 = x − 1.

It’s tempting to write that the domain of f ◦ g is the entire real line, but look more carefully. Note that for x to be in the domain of g, we must have x ≥ 2. The domain of f is the whole real line, so this places no further restrictions on the domain of f ◦ g. Even though the final expression x − 1 is defined for all x, the domain of ( f ◦ g) is {x|x ≥ 2}. For the second composition, (g ◦ f )(x) = g( f (x)) = g(x 2 + 1) = (x 2 + 1) − 2 = x 2 − 1. The resulting square root requires x 2 − 1 ≥ 0 or |x| ≥ 1. Since the “inside” function f is defined for all x, the domain of g ◦ f is {x ∈ R|x| ≥ 1}, which we write in interval notation as (−∞, −1] ∪ [1, ∞). As you progress through the calculus, you will often find yourself needing to recognize that a given function is a composition of simpler functions. For now, it is an important skill to practice.

EXAMPLE 6.3

Identifying Compositions of Functions

Identify functions f and g√such that the given function can be written as ( f ◦ g)(x) for √ each of (a) x 2 + 1, (b) ( x + 1)2 , (c) sin x 2 and (d) cos2 x. Note that more than one answer is possible for each function. 2 Solution (a) Notice that x√ + 1 is inside the square root. So, one choice is to have 2 g(x) = x + 1 and f (x) = x.

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SECTION 0.6

Transformations of Functions

63

√ √ (b) Here, x + 1 is inside the square. So, one choice is g(x) = x + 1 and f (x) = x 2 . (c) The function can be rewritten as sin (x 2 ), with x 2 clearly inside the sine function. Then, g(x) = x 2 and f (x) = sin x is one choice. (d) The function as written is shorthand for (cos x)2 . So, one choice is g(x) = cos x and f (x) = x 2 .

y

In general, it is quite difficult to take the graphs of f (x) and g(x) and produce the graph of f (g(x)). If one of the functions f and g is linear, however, there is a simple graphical procedure for graphing the composition. Such linear transformations are explored in the remainder of this section. The first case is to take the graph of f (x) and produce the graph of f (x) + c for some constant c. You should be able to deduce the general result from example 6.4.

10 8 6 4 2 4

..

EXAMPLE 6.4 x

2

2

4

Graph y = x 2 and y = x 2 + 3; compare and contrast the graphs. Solution You can probably sketch these by hand. You should get graphs like those in Figures 0.76a and 0.76b. Both figures show parabolas opening upward. The main obvious difference is that x 2 has a y-intercept of 0 and x 2 + 3 has a y-intercept of 3. In fact, for any given value of x, the point on the graph of y = x 2 + 3 will be plotted exactly 3 units higher than the corresponding point on the graph of y = x 2 . This is shown in Figure 0.77a. In Figure 0.77b, the two graphs are shown on the same set of axes. To many people, it does not look like the top graph is the same as the bottom graph moved up 3 units.

FIGURE 0.76a y=x

Vertical Translation of a Graph

2

y 10 8

y

y

6 25 4 2 4

2

x 2

25

Move graph up 3 units

20

20

15

15

10

10

5

5

4

FIGURE 0.76b y = x2 + 3

4

2

x 2

4

4

2

x 2

FIGURE 0.77a

FIGURE 0.77b

Translate graph up

y = x 2 and y = x 2 + 3

4

This is an unfortunate optical illusion. Humans usually mentally judge distance between curves as the shortest distance between the curves. For these parabolas, the shortest distance is vertical at x = 0 but becomes increasingly horizontal as you move away from the y-axis. The distance of 3 between the parabolas is measured vertically. In general, the graph of y = f (x) + c is the same as the graph of f (x) shifted up (if c > 0) or down (if c < 0) by |c| units. We usually refer to f (x) + c as a vertical translation (up or down, by |c| units).

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In example 6.5, we explore what happens if a constant is added to x.

EXAMPLE 6.5

A Horizontal Translation

Compare and contrast the graphs of y = x 2 and y = (x − 1)2 . Solution The graphs are shown in Figures 0.78a and 0.78b, respectively. y

y

Move graph to the right one unit

4

6

2

x 2

FIGURE 0.79 Translation to the right

4

10

8

8

6

6

4

4

x

2

2

4

4

2

x 2

FIGURE 0.78a

FIGURE 0.78b

y = x2

y = (x − 1)2

4

4

10

2

10 8

y

4

Notice that the graph of y = (x − 1)2 appears to be the same as the graph of y = x 2 , except that it is shifted 1 unit to the right. This should make sense for the following reason. Pick a value of x, say, x = 13. The value of (x − 1)2 at x = 13 is 122 , the same as the value of x 2 at x = 12, 1 unit to the left. Observe that this same pattern holds for any x you choose. A simultaneous plot of the two functions (see Figure 0.79) shows this. In general, for c > 0, the graph of y = f (x − c) is the same as the graph of y = f (x) shifted c units to the right. Likewise (again, for c > 0), you get the graph of f (x + c) by moving the graph of y = f (x) to the left c units. We usually refer to f (x − c) and f (x + c) as horizontal translations (to the right and left, respectively, by c units). To avoid confusion on which way to translate the graph of y = f (x), focus on what makes the argument (the quantity inside the parentheses) zero. For f (x), this is x = 0, but for f (x − c) you must have x = c to get f (0) [i.e., the same y-value as f (x) when x = 0]. This says that the point on the graph of y = f (x) at x = 0 corresponds to the point on the graph of y = f (x − c) at x = c.

EXAMPLE 6.6

Comparing Vertical and Horizontal Translations

Given the graph of y = f (x) shown in Figure 0.80a, sketch the graphs of y = f (x) − 2 and y = f (x − 2). Solution To graph y = f (x) − 2, simply translate the original graph down 2 units, as shown in Figure 0.80b. To graph y = f (x − 2), simply translate the original graph to

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SECTION 0.6

..

Transformations of Functions

65

the right 2 units (so that the x-intercept at x = 0 in the original graph corresponds to an x-intercept at x = 2 in the translated graph), as seen in Figure 0.80c. y

3

y

y

15

15

15

10

10

10

5

5

5

x

1 5

2

x

3 2 1 5

3

10

10

15

15

1

2

3

1

x 1

2

3

4

5

15

FIGURE 0.80a

FIGURE 0.80b

FIGURE 0.80c

y = f (x)

y = f (x) − 2

y = f (x − 2)

Example 6.7 explores the effect of multiplying or dividing x or y by a constant.

EXAMPLE 6.7

Comparing Some Related Graphs

Compare and contrast the graphs of y = x 2 − 1, y = 4(x 2 − 1) and y = (4x)2 − 1. Solution The first two graphs are shown in Figures 0.81a and 0.81b, respectively. y

y

y

10

40

10

8

32

8

6

24

4

16

2

8

y 4(x 2 1)

6 4

3 2 1 2

x 1

2

3

3 2 1 8

2 x 1

2

3 2 1

y x2 1 x 1

2

3

3 4

FIGURE 0.81a

FIGURE 0.81b

FIGURE 0.81c

y = x2 − 1

y = 4(x 2 − 1)

y = x 2 − 1 and y = 4(x 2 − 1)

These graphs look identical until you compare the scales on the y-axes. The scale in Figure 0.81b is four times as large, reflecting the multiplication of the original function by 4. The effect looks different when the functions are plotted on the same scale, as in Figure 0.81c. Here, the parabola y = 4(x 2 − 1) looks thinner and has a different y-intercept. Note that the x-intercepts remain the same. (Why would that be?) The graphs of y = x 2 − 1 and y = (4x)2 − 1 are shown in Figures 0.82a and 0.82b, respectively.

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y

y

y

10

10

10

8

8

8

6

6

6

4

4

4

2

2 x

3 2 1 2

1

2

3

0.75

0.25 2

y (4x)2 1

y x2 1 x 0.25

0.75

3 2 1 2

x 1

2

3

FIGURE 0.82a

FIGURE 0.82b

FIGURE 0.82c

y = x2 − 1

y = (4x)2 − 1

y = x 2 − 1 and y = (4x)2 − 1

Can you spot the difference here? In this case, the x-scale has now changed, by the same factor of 4 as in the function. To see this, note that substituting x = 1/4 into (4x)2 − 1 produces (1)2 − 1, exactly the same as substituting x = 1 into the original function. When plotted on the same set of axes (as in Figure 0.82c), the parabola y = (4x)2 − 1 looks thinner. Here, the x-intercepts are different, but the y-intercepts are the same.

y 20

10

4

x

2

2

4

FIGURE 0.83a y = x2 y

EXAMPLE 6.8

40

2

A Translation and a Stretching

Describe how to get the graph of y = 2x 2 − 3 from the graph of y = x 2 . Solution You can get from x 2 to 2x 2 − 3 by multiplying by 2 and then subtracting 3. In terms of the graph, this has the effect of multiplying the y-scale by 2 and then shifting the graph down by 3 units (see the graphs in Figures 0.83a and 0.83b).

20

4

We can generalize the observations made in example 6.7. Before reading our explanation, try to state a general rule for yourself. How are the graphs of the functions c f (x) and f (cx) related to the graph of y = f (x)? Based on example 6.7, notice that to obtain a graph of y = c f (x) for some constant c > 0, you can take the graph of y = f (x) and multiply the scale on the y-axis by c. To obtain a graph of y = f (cx) for some constant c > 0, you can take the graph of y = f (x) and multiply the scale on the x-axis by 1/c. These basic rules can be combined to understand more complicated graphs.

x 2

FIGURE 0.83b y = 2x 2 − 3

4

EXAMPLE 6.9

A Translation in Both x - and y -Directions

Describe how to get the graph of y = x 2 + 4x + 3 from the graph of y = x 2 . Solution We can again relate this (and the graph of every quadratic) to the graph of y = x 2 . We must first complete the square. Recall that in this process, you take the coefficient of x (4), divide by 2 (4/2 = 2) and square the result (22 = 4). Add and subtract this number and then, rewrite the x-terms as a perfect square. We have y = x 2 + 4x + 3 = (x 2 + 4x + 4) − 4 + 3 = (x + 2)2 − 1.

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67

To graph this function, take the parabola y = x 2 (see Figure 0.84a) and translate the graph 2 units to the left and 1 unit down (see Figure 0.84b). y

y

4

20

20

10

10

x

2

2

4

6

4

2

FIGURE 0.84a

FIGURE 0.84b

y = x2

y = (x + 2)2 − 1

x 2

The following table summarizes our discoveries in this section. Transformations of f (x) Transformation

Form

Effect on Graph

Vertical translation

f (x) + c

|c| units up (c > 0) or down (c < 0)

Horizontal translation

f (x + c)

|c| units left (c > 0) or right (c < 0)

Vertical scale

c f (x) (c > 0)

multiply vertical scale by c

Horizontal scale

f (cx) (c > 0)

divide horizontal scale by c

You will explore additional transformations in the exercises.

EXERCISES 0.6 WRITING EXERCISES 1. The restricted domain of example 6.2 may be puzzling. Consider the following analogy. Suppose you have an airplane flight from New York to Los Angeles with a stop for refueling in Minneapolis. If bad weather has closed the airport in Minneapolis, explain why your flight will be canceled (or at least rerouted) even if the weather is great in New York and Los Angeles. 2. Explain why the graphs of y = 4(x 2 − 1) and y = (4x)2 − 1 in Figures 0.81c and 0.82c appear “thinner” than the graph of y = x 2 − 1.

3. As illustrated in example 6.9, completing the square can be used to rewrite any quadratic function in the form a(x − d)2 + e. Using the transformation rules in this section, explain why this means that all parabolas (with a > 0) will look essentially the same. 4. Explain why the graph of y = f (x + 4) is obtained by moving the graph of y = f (x) four units to the left, instead of to the right.

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In exercises 1–6, find the compositions f ◦ g and g ◦ f , and identify their respective domains. √ 1. f (x) = x + 1, g(x) = x − 3 √ 2. f (x) = x − 2, g(x) = x + 1 3. f (x) = e , g(x) = ln x √ 4. f (x) = 1 − x, g(x) = ln x

y 10 5

x

4

5. f (x) = x 2 + 1, g(x) = sin x 1 6. f (x) = 2 , g(x) = x 2 − 2 x −1

x

2

2

4

5 10

In exercises 7–16, identify functions f(x) and g(x) such that the given function equals ( f ◦g)(x). √ √ 1 7. x 4 + 1 8. 3 x + 3 9. 2 x +1 1 11. (4x + 1)2 + 3 12. 4 (x + 1)2 + 3 10. 2 + 1 x 2 13. sin3 x 14. sin x 3 15. e x +1 16. e4x−2

In exercises 39–44, complete the square and explain how to transform the graph of y x 2 into the graph of the given function. 39. f (x) = x 2 + 2x + 1

40. f (x) = x 2 − 4x + 4

41. f (x) = x 2 + 2x + 4

42. f (x) = x 2 − 4x + 2

43. f (x) = 2x + 4x + 4

44. f (x) = 3x 2 − 6x + 2

2

In exercises 17–22, identify functions f (x), g(x) and h(x) such that the given function equals [ f ◦(g◦h)] (x). √ 3 17. √ 18. e4x + 1 sin x + 2 √ 19. cos3 (4x − 2) 20. ln x 2 + 1

2 2 21. 4e x − 5 22. tan−1 (3x + 1) In exercises 23–30, use the graph of y f (x) given in the figure to graph the indicated function. 23. f (x) − 3

24. f (x + 2)

25. f (x − 3)

26. f (x) + 2

27. f (2x)

28. 3 f (x)

29. 4 f (x) − 1

30. 3 f (x + 2)

46. f (x) = −3(x 2 − 1) 47. f (x) = −3(x 2 − 1) + 2 48. f (x) = −2(x 2 − 1) − 1 In exercises 49–52, graph the given function and compare to the graph of y (x − 1)2 − 1 x 2 − 2x. 49. f (x) = (−x)2 − 2(−x) 50. f (x) = (−2x)2 − 2(−2x) 52. f (x) = (−3x)2 − 2(−3x) − 3

10

53. Based on exercises 45–48, state a rule for transforming the graph of y = f (x) into the graph of y = c f (x) for c < 0.

8

54. Based on exercises 49–52, state a rule for transforming the graph of y = f (x) into the graph of y = f (cx) for c < 0.

6 4

55. Sketch the graph of y = |x|3 . Explain why the graph of y = |x|3 is identical to that of y = x 3 to the right of the y-axis. For y = |x|3 , describe how the graph to the left of the y-axis compares to the graph to the right of the y-axis. In general, describe how to draw the graph of y = f (|x|) given the graph of y = f (x).

2 2

45. f (x) = −2(x 2 − 1)

51. f (x) = (−x)2 − 2(−x) + 1

y

4

In exercises 45–48, graph the given function and compare to the graph of y x 2 – 1.

x 2

4

2

In exercises 31–38, use the graph of y f (x) given in the figure to graph the indicated function. 31. f (x − 4)

32. f (x + 3)

33. f (2x)

34. f (2x − 4)

35. f (3x + 3)

36. 3 f (x)

37. 2 f (x) − 4

38. 3 f (x) + 3

56. For y = x 3 , describe how the graph to the left of the y-axis compares to the graph to the right of the y-axis. Show that for f (x) = x 3 , we have f (−x) = − f (x). In general, if you have the graph of y = f (x) to the right of the y-axis and f (−x) = − f (x) for all x, describe how to graph y = f (x) to the left of the y-axis.

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57. Iterations of functions are important in a variety of applications. To iterate f (x), start with an initial value x0 and compute x1 = f (x0 ), x2 = f (x1 ), x3 = f (x2 ) and so on. For example, with f (x) = cos x and x0 = 1, the iterates are x1 = cos 1 ≈ 0.54, x2 = cos x1 ≈ cos 0.54 ≈ 0.86, x3 ≈ cos 0.86 ≈ 0.65 and so on. Keep computing iterates and show that they get closer and closer to 0.739085. Then pick your own x0 (any number you like) and show that the iterates with this new x0 also converge to 0.739085. 58. Referring to exercise 57, show that the iterates of a function can be written as x1 = f (x0 ), x2 = f ( f (x0 )), x3 = f ( f ( f (x0 ))) and so on. Graph y = cos (cos x), y = cos (cos (cos x)) and y = cos (cos (cos (cos x))). The graphs should look more and more like a horizontal line. Use the result of exercise 57 to identify the limiting line. 59. Compute several iterates of f (x) = sin x (see exercise 57) with a variety of starting values. What happens to the iterates in the long run? 60. Repeat exercise 59 for f (x) = x 2 . 61. In cases where the iterates of a function (see exercise 57) repeat a single number, that number is called a fixed point. Explain why any fixed point must be a solution of the equation f (x) = x. Find all fixed points of f (x) = cos x by solving the equation cos x = x. Compare your results to that of exercise 57. 62. Find all fixed points of f (x) = sin x (see exercise 61). Compare your results to those of exercise 59.

EXPLORATORY EXERCISES 1. You have explored how completing the square can transform any quadratic function into the form y = a(x − d)2 + e. We

..

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69

concluded that all parabolas with a > 0 look alike. To see that the same statement is not true of cubic polynomials, graph y = x 3 and y = x 3 − 3x. In this exercise, you will use completing the cube to determine how many different cubic graphs there are. To see what “completing the cube” would look like, first show that (x + a)3 = x 3 + 3ax 2 + 3a 2 x + a 3 . Use this result to transform the graph of y = x 3 into the graphs of (a) y = x 3 − 3x 2 + 3x − 1 and (b) y = x 3 − 3x 2 + 3x + 2. Show that you can’t get a simple transformation to y = x 3 −3x 2 +4x −2. However, show that y = x 3 −3x 2 + 4x −2 can be obtained from y = x 3 + x by basic transformations. Show that the following statement is true: any cubic (y = ax 3 + bx 2 + cx + d) can be obtained with basic transformations from y = ax 3 + kx for some constant k. 2. In many applications, it is important to take a section of a graph (e.g., some data) and extend it for predictions or other analysis. For example, suppose you have an electronic signal equal to f (x) = 2x for 0 ≤ x ≤ 2. To predict the value of the signal at x = −1, you would want to know whether the signal was periodic. If the signal is periodic, explain why f (−1) = 2 would be a good prediction. In some applications, you would assume that the function is even. That is, f (x) = f (−x) for all x. In this case, you want f (x) = 2(−x) = −2x for −2 ≤ x ≤ 0. −2x if − 2 ≤ x ≤ 0 . Graph the even extension f (x) = 2x if 0 ≤ x ≤ 2 2 Find the even extension for (a) f (x) = x + 2x + 1, 0 ≤ x ≤ 2 and (b) f (x) = e−x , 0 ≤ x ≤ 2. 3. Similar to the even extension discussed in exploratory exercise 2, applications sometimes require a function to be odd; that is, f (−x) = − f (x). For f (x) = x 2 , 0 ≤ x ≤ 2, the odd extension requires that for −2≤ x ≤ 0, f (x) = − f (−x) = −x 2 if − 2 ≤ x ≤ 0 −(−x)2 = −x 2 so that f (x) = . Graph x2 if 0 ≤ x ≤ 2 y = f (x) and discuss how to graphically rotate the right half of the graph to get the left half of the graph. Find the odd extension for (a) f (x) = x 2 + 2x, 0 ≤ x ≤ 2 and (b) f (x) = e−x − 1, 0 ≤ x ≤ 2.

Review Exercises WRITING EXERCISES The following list includes terms that are defined and theorems that are stated in this chapter. For each term or theorem, (1) give a precise definition or statement, (2) state in general terms what it means and (3) describe the types of problems with which it is associated.

Slope of a line Domain Graphing window Inverse function Sine function e Composition

Parallel lines Intercepts Local maximum One-to-one function Cosine function Exponential function

Perpendicular lines Zeros of a function Vertical asymptote Periodic function Arcsine function Logarithm

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1-2

A BRIEF PREVIEW OF CALCULUS: TANGENT LINES AND THE LENGTH OF A CURVE In this section, we approach the boundary between precalculus mathematics and the calculus by investigating several important problems requiring the use of calculus. Recall that the slope of a straight line is the change in y divided by the change in x. This fraction is the same regardless of which two points you use to compute the slope. For example, the points (0, 1), (1, 4), and (3, 10) all lie on the line y = 3x + 1. The slope of 3 can be obtained from any two of the points. For instance, 4−1 10 − 1 =3 or m= = 3. 1−0 3−0 In the calculus, we generalize this problem to find the slope of a curve at a point. For instance, suppose we wanted to find the slope of the curve y = x 2 + 1 at the point (1, 2). You might think of picking a second point on the parabola, say (2, 5). The slope of the line through these two points (called a secant line; see Figure 1.2a) is easy enough to compute. We have m=

m sec =

5−2 = 3. 2−1

However, using the points (0, 1) and (1, 2), we get a different slope (see Figure 1.2b): m sec =

2−1 = 1. 1−0

y

y

6

6

4

4

2

2

0.5

x 0.5

1

1.5

2

2

y

2.5

0.5

x 0.5

1

1.5

2

2.5

2

FIGURE 1.2a

FIGURE 1.2b

Secant line, slope = 3

Secant line, slope = 1

2.10 2.05 2.00 1.95 1.90 x 0.96 0.98 1.00 1.02 1.04

FIGURE 1.3 y = x2 + 1

For curves other than straight lines, the slopes of secant lines joining different points are generally not the same, as seen in Figures 1.2a and 1.2b. If you get different slopes using different pairs of points, then what exactly does it mean for a curve to have a slope at a point? The answer can be visualized by graphically zooming in on the specified point. Take the graph of y = x 2 + 1 and zoom in tight on the point (1, 2). You should get a graph something like the one in Figure 1.3. The graph looks very much like a straight line. In fact, the more you zoom in, the straighter the curve appears to be and the less it matters which two points are used to compute a slope. So, here’s the strategy: pick several points on the parabola, each closer to the point (1, 2) than the previous one. Compute the slopes of the lines through (1, 2) and each of the points. The closer the second point gets to (1, 2), the closer the computed slope is to the answer you seek.

1-3

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A Brief Preview of Calculus

75

For example, the point (1.5, 3.25) is on the parabola fairly close to (1, 2). The slope of the line joining these points is m sec =

3.25 − 2 = 2.5. 1.5 − 1

The point (1.1, 2.21) is even closer to (1, 2). The slope of the secant line joining these two points is m sec =

2.21 − 2 = 2.1. 1.1 − 1

Continuing in this way, observe that the point (1.01, 2.0201) is closer yet to the point (1, 2). The slope of the secant lines through these points is m sec =

2.0201 − 2 = 2.01. 1.01 − 1

The slopes of the secant lines (2.5, 2.1, 2.01) are getting closer and closer to the slope of the parabola at the point (1, 2). Based on these calculations, it seems reasonable to say that the slope of the curve is approximately 2. Example 1.1 takes our introductory example just a bit further.

EXAMPLE 1.1

Estimating the Slope of a Curve

Estimate the slope of y = x 2 + 1 at x = 1. Solution We focus on the point whose coordinates are x = 1 and y = 12 + 1 = 2. To estimate the slope, choose a sequence of points near (1, 2) and compute the slopes of the secant lines joining those points with (1, 2). (We showed sample secant lines in Figures 1.2a and 1.2b.) Choosing points with x > 1 (x-values of 2, 1.1 and 1.01) and points with x < 1 (x-values of 0, 0.9 and 0.99), we compute the corresponding y-values using y = x 2 + 1 and get the slopes shown in the following table. Second Point (2, 5) (1.1, 2.21) (1.01, 2.0201)

msec 5−2 =3 2−1 2.21 − 2 = 2.1 1.1 − 1 2.0201 − 2 = 2.01 1.01 − 1

Second Point (0, 1) (0.9, 1.81) (0.99, 1.9801)

msec 1−2 =1 0−1 1.81 − 2 = 1.9 0.9 − 1 1.9801 − 2 = 1.99 0.99 − 1

Observe that in both columns, as the second point gets closer to (1, 2), the slope of the secant line gets closer to 2. A reasonable estimate of the slope of the curve at the point (1, 2) is then 2. In Chapter 2, we develop a powerful technique for computing such slopes exactly (and easily). Note what distinguishes the calculus problem from the corresponding algebra problem. The calculus problem involves a process we call a limit. While we presently can

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only estimate the slope of a curve using a sequence of approximations, the limit allows us to compute the slope exactly.

EXAMPLE 1.2

Estimating the Slope of a Curve

Estimate the slope of y = sin x at x = 0. Solution This turns out to be a very important problem, one that we will return to later. For now, choose a sequence of points near (0, 0) and compute the slopes of the secant lines joining those points with (0, 0). The following table shows one set of choices.

y

q

q x

Second Point

msec

Second Point

msec

(1, sin 1) (0.1, sin 0.1) (0.01, sin 0.01)

0.84147 0.99833 0.99998

(−1, sin (−1)) (−0.1, sin (−0.1)) (−0.01, sin (−0.01))

0.84147 0.99833 0.99998

Note that as the second point gets closer and closer to (0, 0), the slope of the secant line (m sec ) appears to get closer and closer to 1. A good estimate of the slope of the curve at the point (0, 0) would then appear to be 1. Although we presently have no way of computing the slope exactly, this is consistent with the graph of y = sin x in Figure 1.4. Note that near (0, 0), the graph resembles that of y = x, a straight line of slope 1.

FIGURE 1.4 y = sin x

A second problem requiring the power of calculus is that of computing distance along a curved path. While this problem is of less significance than our first example (both historically and in the development of the calculus), it provides a good indication of the need for mathematics beyond simple algebra. You should pay special attention to the similarities between the development of this problem and our earlier work with slope. Recall that the (straight-line) distance between two points (x1 , y1 ) and (x2 , y2 ) is d{(x1 , y1 ), (x2 , y2 )} =

(x2 − x1 )2 + (y2 − y1 )2 .

For instance, the distance between the points (0, 1) and (3, 4) is y

d{(0, 1), (3, 4)} =

(3, 4)

4 3 2 1

(0, 1) x 1

2

3

FIGURE 1.5a y = (x − 1)2

4

√ (3 − 0)2 + (4 − 1)2 = 3 2 ≈ 4.24264.

However, this is not the only way we might want to compute the distance between these two points. For example, suppose that you needed to drive a car from (0, 1) to (3, 4) along a road that follows the curve y = (x − 1)2 (see Figure 1.5a). In this case, you don’t care about the straight-line distance connecting the two points, but only about how far you must drive along the curve (the length of the curve or arc length). √ Notice that the distance along the curve must be greater than 3 2 (the straight-line distance). Taking a cue from the slope problem, we can formulate a strategy for obtaining a sequence of increasingly accurate approximations. Instead of using just one line segment √ to get the approximation of 3 2, we could use two line segments, as in Figure 1.5b. Notice that the sum of the lengths of the two line segments appears to be a much better approximation to the actual length of the curve than the straight-line distance used previously. This

1-5

SECTION 1.1

y

A Brief Preview of Calculus

77

y (3, 4)

4

(3, 4)

4

3

3

2

2

1

..

(0, 1)

1

(0, 1)

(2, 1)

x 1 1.5 2

3

4

x 1

2

3

FIGURE 1.5b

FIGURE 1.5c

Two line segments

Three line segments

4

distance is d2 = d{(0, 1), (1.5, 0.25)} + d{(1.5, 0.25), (3, 4)} = (1.5 − 0)2 + (0.25 − 1)2 + (3 − 1.5)2 + (4 − 0.25)2 ≈ 5.71592. You’re probably way ahead of us by now. If approximating the length of the curve with two line segments gives an improved approximation, why not use three or four or more? Using the three line segments indicated in Figure 1.5c, we get the further improved approximation

No. of Segments 1 2 3 4 5 6 7

d3 = d{(0, 1), (1, 0)} + d{(1, 0), (2, 1)} + d{(2, 1), (3, 4)} = (1 − 0)2 + (0 − 1)2 + (2 − 1)2 + (1 − 0)2 + (3 − 2)2 + (4 − 1)2 √ √ = 2 2 + 10 ≈ 5.99070.

Distance 4.24264 5.71592 5.99070 6.03562 6.06906 6.08713 6.09711

Note that the more line segments we use, the better the approximation appears to be. This process will become much less tedious with the development of the definite integral in Chapter 4. For now we list a number of these successively better approximations (produced using points on the curve with evenly spaced x-coordinates) in the table found in the margin. The table suggests that the length of the curve is approximately 6.1 (quite far from the straight-line distance of 4.2). If we continued this process using more and more line segments, the sum of their lengths would approach the actual length of the curve (about 6.126). As in the problem of computing the slope of a curve, the exact arc length is obtained as a limit.

y 1

y = sin x

EXAMPLE 1.3

Estimating the Arc Length of a Curve

Estimate the arc length of the curve y = sin x for 0 ≤ x ≤ π (see Figure 1.6a).

q

p

x

FIGURE 1.6a Approximating the curve with two line segments

Solution The endpoints of the curve on this interval are (0, 0) and (π, 0). The distance between these points is d1 = π . The point on the graph of y = sin x corresponding to the midpoint of the interval [0, π ] is (π /2, 1). The distance from (0, 0) to (π/2, 1) plus the distance from (π/2, 1) to (π , 0) (illustrated in Figure 1.6a) is π 2 π 2 d2 = +1+ + 1 ≈ 3.7242. 2 2

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Limits and Continuity

√ √ Using the five points (0, 0), (π/4, 1/ 2), (π/2, 1), (3π/4, 1/ 2), and (π, 0) (i.e., four line segments, as indicated in Figure 1.6b), the sum of the lengths of these line segments is π 2 1 2 π 2 1 d4 = 2 + +2 + 1− √ ≈ 3.7901. 4 2 4 2

y y = sin x

d

q

w

1-6

p

Using nine points (i.e., eight line segments), you need a good calculator and some patience to compute the distance of 3.8125. A table showing further approximations is given in the margin. At this stage, it would be reasonable to estimate the length of the sine curve on the interval [0, π ] as slightly more than 3.8.

x

FIGURE 1.6b Approximating the curve with four line segments

BEYOND FORMULAS

Number of Line Segments 8 16 32 64

In the process of estimating both the slope of a curve and the length of a curve, we make some reasonably obvious (straight-line) approximations and then systematically improve on those approximations. In each case, the shorter the line segments are, the closer the approximations are to the desired value. The essence of this is the concept of limit, which separates precalculus mathematics from the calculus. At first glance, this limit idea might seem of little practical importance, since in our examples we never compute the exact solution. In the chapters to come, we will find remarkably simple shortcuts to exact answers. Can you think of ways to find the exact slope in example 1.1?

Sum of Lengths 3.8125 3.8183 3.8197 3.8201

EXERCISES 1.1 WRITING EXERCISES 1. Explain why each approximation of arc length in example 1.3 is less than the actual arc length. 2. To estimate the slope of f (x) = x 2 + 1 at x = 1, you would compute the slopes of various secant lines. Note that y = x 2 + 1 curves up. Explain why the secant line connecting (1, 2) and (1.1, 2.21) will have slope greater than the slope of the tangent line. Discuss how the slope of the secant line between (1, 2) and (0.9, 1.81) compares to the slope of the tangent line.

9. f (x) = e x , a = 0 11. f (x) = ln x, a = 1

1. f (x) = x 2 + 1, a = 1

2. f (x) = x 2 + 1, a = 2

3. f (x) = cos x, a = 0

4. f (x) = cos x, a = π/2

5. f (x) = x 3 + 2, a = 1

6. f (x) = x 3 + 2, a = 2

7. f (x) =

√

x + 1, a = 0

8. f (x) =

√

x + 1, a = 3

12. f (x) = ln x, a = 2

In exercises 13–20, estimate the length of the curve y f (x) on the given interval using (a) n 4 and (b) n 8 line segments. (c) If you can program a calculator or computer, use larger n’s and conjecture the actual length of the curve. 13. f (x) = x 2 + 1, 0 ≤ x ≤ 2 14. f (x) = x 3 + 2, 0 ≤ x ≤ 1

In exercises 1–12, estimate the slope (as in example 1.1) of y f (x) at x a.

10. f (x) = e x , a = 1

15. f (x) = cos x, 0 ≤ x ≤ π/2 16. f (x) = sin x, 0 ≤ x ≤ π/2 √ 17. f (x) = x + 1, 0 ≤ x ≤ 3 18. f (x) = 1/x, 1 ≤ x ≤ 2 19. f (x) = x 2 + 1, −2 ≤ x ≤ 2 20. f (x) = x 3 + 2, −1 ≤ x ≤ 1

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79

26. Use the technique of exercises 21–23 to estimate the area below y = x 3 and above the x-axis between x = 0 and x = 2.

21. An important problem in calculus is finding the area of a region. Sketch the parabola y = 1 − x 2 and shade in the region above the x-axis between x = −1 and x = 1. Then sketch in the following rectangles: (1) height f (− 34 ) and width 12 extending from x = −1 to x = − 12 . (2) height f (− 14 ) and width 1 extending from x = − 12 to x = 0. (3) height f ( 14 ) and width 2 1 extending from x = 0 to x = 12 . (4) height f ( 34 ) and width 12 2 extending from x = 12 to x = 1. Compute the sum of the areas of the rectangles. Based on your sketch, does this give you a good approximation of the area under the parabola?

EXPLORATORY EXERCISE 1. Several central concepts of calculus have been introduced in this section. An important aspect of our future development of calculus is to derive simple techniques for computing quantities such as slope and arc length. In this exercise, you will learn how to directly compute the slope of a curve at a point. Suppose you want the slope of y = x 2 at x = 1. You could start by computing slopes of secant lines connecting the point (1, 1) with nearby points. Suppose the nearby point has x-coordinate 1 + h, where h is a small (positive or negative) number. Explain why the corresponding y-coordinate is (1 + h)2 . Show (1 + h)2 − 1 = 2 + h. As that the slope of the secant line is 1+h−1 h gets closer and closer to 0, this slope better approximates the slope of the tangent line. Letting h approach 0, show that the slope of the tangent line equals 2. In a similar way, show that the slope of y = x 2 at x = 2 is 4 and find the slope of y = x 2 at x = 3. Based on your answers, conjecture a formula for the slope of y = x 2 at x = a, for any unspecified value of a.

22. To improve the approximation of exercise 21, divide the interval [−1, 1] into 8 pieces and construct a rectangle of the appropriate height on each subinterval. Compared to the approximation in exercise 21, explain why you would expect this to be a better approximation of the actual area under the parabola. 23. Use a computer or calculator to compute an approximation of the area in exercise 21 using (a) 16 rectangles, (b) 32 rectangles, (c) 64 rectangles. Use these calculations to conjecture the exact value of the area under the parabola. 24. Use the technique of exercises 21–23 to estimate the area below y = sin x and above the x-axis between x = 0 and x = π. 25. Use the technique of exercises 21–23 to estimate the area below y = x 3 and above the x-axis between x = 0 and x = 1.

1.2

..

THE CONCEPT OF LIMIT In this section, we develop the notion of limit using some common language and illustrate the idea with some simple examples. The notion turns out to be a rather subtle one, easy to think of intuitively, but a bit harder to pin down in precise terms. We present the precise definition of limit in section 1.6. There, we carefully define limits in considerable detail. The more informal notion of limit that we introduce and work with here and in sections 1.3, 1.4 and 1.5 is adequate for most purposes. As a start, consider the functions

y f(x) 4 f(x)

f (x) =

2

x

2

x

2

FIGURE 1.7a y=

x2 − 4 x −2

x

x2 − 4 x −2

and

g(x) =

x2 − 5 . x −2

Notice that both functions are undefined at x = 2. So, what does this mean, beyond saying that you cannot substitute 2 for x? We often find important clues about the behavior of a function from a graph (see Figures 1.7a and 1.7b). Notice that the graphs of these two functions look quite different in the vicinity of x = 2. Although we can’t say anything about the value of these functions at x = 2 (since this is outside the domain of both functions), we can examine their behavior in the vicinity of

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79

26. Use the technique of exercises 21–23 to estimate the area below y = x 3 and above the x-axis between x = 0 and x = 2.

21. An important problem in calculus is finding the area of a region. Sketch the parabola y = 1 − x 2 and shade in the region above the x-axis between x = −1 and x = 1. Then sketch in the following rectangles: (1) height f (− 34 ) and width 12 extending from x = −1 to x = − 12 . (2) height f (− 14 ) and width 1 extending from x = − 12 to x = 0. (3) height f ( 14 ) and width 2 1 extending from x = 0 to x = 12 . (4) height f ( 34 ) and width 12 2 extending from x = 12 to x = 1. Compute the sum of the areas of the rectangles. Based on your sketch, does this give you a good approximation of the area under the parabola?

EXPLORATORY EXERCISE 1. Several central concepts of calculus have been introduced in this section. An important aspect of our future development of calculus is to derive simple techniques for computing quantities such as slope and arc length. In this exercise, you will learn how to directly compute the slope of a curve at a point. Suppose you want the slope of y = x 2 at x = 1. You could start by computing slopes of secant lines connecting the point (1, 1) with nearby points. Suppose the nearby point has x-coordinate 1 + h, where h is a small (positive or negative) number. Explain why the corresponding y-coordinate is (1 + h)2 . Show (1 + h)2 − 1 = 2 + h. As that the slope of the secant line is 1+h−1 h gets closer and closer to 0, this slope better approximates the slope of the tangent line. Letting h approach 0, show that the slope of the tangent line equals 2. In a similar way, show that the slope of y = x 2 at x = 2 is 4 and find the slope of y = x 2 at x = 3. Based on your answers, conjecture a formula for the slope of y = x 2 at x = a, for any unspecified value of a.

22. To improve the approximation of exercise 21, divide the interval [−1, 1] into 8 pieces and construct a rectangle of the appropriate height on each subinterval. Compared to the approximation in exercise 21, explain why you would expect this to be a better approximation of the actual area under the parabola. 23. Use a computer or calculator to compute an approximation of the area in exercise 21 using (a) 16 rectangles, (b) 32 rectangles, (c) 64 rectangles. Use these calculations to conjecture the exact value of the area under the parabola. 24. Use the technique of exercises 21–23 to estimate the area below y = sin x and above the x-axis between x = 0 and x = π. 25. Use the technique of exercises 21–23 to estimate the area below y = x 3 and above the x-axis between x = 0 and x = 1.

1.2

..

THE CONCEPT OF LIMIT In this section, we develop the notion of limit using some common language and illustrate the idea with some simple examples. The notion turns out to be a rather subtle one, easy to think of intuitively, but a bit harder to pin down in precise terms. We present the precise definition of limit in section 1.6. There, we carefully define limits in considerable detail. The more informal notion of limit that we introduce and work with here and in sections 1.3, 1.4 and 1.5 is adequate for most purposes. As a start, consider the functions

y f(x) 4 f(x)

f (x) =

2

x

2

x

2

FIGURE 1.7a y=

x2 − 4 x −2

x

x2 − 4 x −2

and

g(x) =

x2 − 5 . x −2

Notice that both functions are undefined at x = 2. So, what does this mean, beyond saying that you cannot substitute 2 for x? We often find important clues about the behavior of a function from a graph (see Figures 1.7a and 1.7b). Notice that the graphs of these two functions look quite different in the vicinity of x = 2. Although we can’t say anything about the value of these functions at x = 2 (since this is outside the domain of both functions), we can examine their behavior in the vicinity of

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x2 − 4 , we compute x −2 some values of the function for x close to 2, as in the following tables.

y

this point. We consider these functions one at a time. First, for f (x) =

10 5 10

x 5 5 10

FIGURE 1.7b y=

x2 − 5 x −2

10

x

f (x)

1.9 1.99 1.999 1.9999

3.9 3.99 3.999 3.9999

x2 − 4 x− 2

x

f (x)

2.1 2.01 2.001 2.0001

4.1 4.01 4.001 4.0001

x2 − 4 x− 2

Notice that as you move down the first column of the table, the x-values get closer to 2, but are all less than 2. We use the notation x → 2− to indicate that x approaches 2 from the left side. Notice that the table and the graph both suggest that as x gets closer and closer to 2 (with x < 2), f (x) is getting closer and closer to 4. In view of this, we say that the limit of f(x) as x approaches 2 from the left is 4, written lim f (x) = 4.

x→2−

Likewise, we need to consider what happens to the function values for x close to 2 but larger than 2. Here, we use the notation x → 2+ to indicate that x approaches 2 from the right side. We compute some of these values in the second table. Again, the table and graph both suggest that as x gets closer and closer to 2 (with x > 2), f (x) is getting closer and closer to 4. In view of this, we say that the limit of f(x) as x approaches 2 from the right is 4, written lim f (x) = 4.

x→2+

We call lim− f (x) and lim+ f (x) one-sided limits. Since the two one-sided limits of x→2

x→2

f (x) are the same, we summarize our results by saying that the limit of f(x) as x approaches 2 is 4, written lim f (x) = 4.

x→2

The notion of limit as we have described it here is intended to communicate the behavior of a function near some point of interest, but not actually at that point. We finally observe that we can also determine this limit algebraically, as follows. Notice that since the expression x2 − 4 in the numerator of f (x) = factors, we can write x −2 x2 − 4 x→2 x − 2 (x − 2)(x + 2) = lim x→2 x −2 = lim (x + 2) = 4,

lim f (x) = lim

x→2

x→2

Cancel the factors of (x − 2). As x approaches 2, (x + 2) approaches 4.

where we can cancel the factors of (x − 2) since in the limit as x → 2, x is close to 2, but x = 2, so that x − 2 = 0.

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SECTION 1.2

x 1.9 1.99 1.999 1.9999

..

The Concept of Limit

81

x2 − 5 , as x → 2. Based on the x −2 graph in Figure 1.7b and the table of approximate function values shown in the margin, observe that as x gets closer and closer to 2 (with x < 2), g(x) increases without bound. Since there is no number that g(x) is approaching, we say that the limit of g(x) as x approaches 2 from the left does not exist, written Similarly, we consider one-sided limits for g(x) =

x2 − 5 g(x) x− 2 13.9 103.99 1003.999 10,003.9999

lim g(x) does not exist.

x→2−

Similarly, the graph and the table of function values for x > 2 (shown in the margin) suggest that g(x) decreases without bound as x approaches 2 from the right. Since there is no number that g(x) is approaching, we say that

x2 − 5 x− 2

x

g(x)

2.1 2.01 2.001 2.0001

−5.9 −95.99 −995.999 −9995.9999

lim g(x) does not exist.

x→2+

Finally, since there is no common value for the one-sided limits of g(x) (in fact, neither limit exists), we say that the limit of g(x) as x approaches 2 does not exist, written lim g(x) does not exist.

x→2

Before moving on, we should summarize what we have said about limits.

A limit exists if and only if both corresponding one-sided limits exist and are equal. That is, lim f (x) = L , for some number L, if and only if lim− f (x) = lim+ f (x) = L .

x→a

x→a

x→a

In other words, we say that lim f (x) = L if we can make f (x) as close as we might like to x→a L, by making x sufficiently close to a (on either side of a), but not equal to a. Note that we can think about limits from a purely graphical viewpoint, as in example 2.1. y

EXAMPLE 2.1 2

Use the graph in Figure 1.8 to determine lim− f (x), lim+ f (x), lim f (x) and lim f (x). x→1

1 2

1

Determining Limits Graphically

x 1 1 2

2

x→1

x→1

x→−1

Solution For lim− f (x), we consider the y-values as x gets closer to 1, with x < 1. x→1

That is, we follow the graph toward x = 1 from the left (x < 1). Observe that the graph dead-ends into the open circle at the point (1, 2). Therefore, we say that lim− f (x) = 2. x→1

For lim+ f (x), we follow the graph toward x = 1 from the right (x > 1). In this case, x→1

FIGURE 1.8 y = f (x)

the graph dead-ends into the solid circle located at the point (1, −1). For this reason, we say that lim+ f (x) = −1. Because lim− f (x) = lim+ f (x), we say that lim f (x) does x→1

x→1

x→1

x→1

not exist. Finally, we have that lim f (x) = 1, since the graph approaches a y-value of x→−1

1 as x approaches −1 both from the left and from the right.

82

..

CHAPTER 1

Limits and Continuity

1-10

y

EXAMPLE 2.2 3

3

x

f(x)

x

Q

x 2

f(x)

Further, note that

3

lim −

FIGURE 1.9 lim

x→−3

x→−3

1 3x + 9 =− x2 − 9 2

−3.1 −3.01 −3.001 −3.0001

−2.9 −2.99 −2.999 −2.9999

lim

x→−3

1 3x + 9 =− . x2 − 9 2

In example 2.2, the limit exists because both one-sided limits exist and are equal. In example 2.3, neither one-sided limit exists.

y

EXAMPLE 2.3

30

Cancel factors of (x + 3).

Finally, since the function approaches the same value as x → −3 both from the right and from the left (i.e., the one-sided limits are equal), we write

3x 9 x2 − 9 −0.508475 −0.500835 −0.500083 −0.500008

x

3x + 9 3(x + 3) = lim − 2 x→−3 (x + 3)(x − 3) x −9 3 1 = lim − =− , x→−3 x − 3 2

since (x − 3) → −6 as x → −3. Again, the cancellation of the factors of (x + 3) is valid since in the limit as x → −3, x is close to −3, but x = −3, so that x + 3 = 0. Likewise, 3x + 9 1 lim =− . x→−3+ x 2 − 9 2

3x 9 x2 − 9 −0.491803 −0.499168 −0.499917 −0.499992

x

A Limit Where Two Factors Cancel

3x + 9 Evaluate lim 2 . x→−3 x − 9 Solution We examine a graph (see Figure 1.9) and compute some function values for x near −3. Based on this numerical and graphical evidence, it’s reasonable to conjecture that 3x + 9 3x + 9 1 lim + 2 = lim − 2 =− . x→−3 x − 9 x→−3 x − 9 2

f(x)

A Limit That Does Not Exist

3x + 9 exists. x2 − 9 Solution We first draw a graph (see Figure 1.10) and compute some function values for x close to 3. 3x + 9 Based on this numerical and graphical evidence, it appears that, as x → 3+ , 2 x −9 is increasing without bound. Thus,

Determine whether lim

x→3

x 3

x

f(x) 30

FIGURE 1.10 y=

3x + 9 x2 − 9

x

3x + 9 does not exist. x2 − 9 Similarly, from the graph and the table of values for x < 3, we can say that lim

x→3+

3x + 9 does not exist. x2 − 9 Since neither one-sided limit exists, we say lim

x→3−

x 3.1 3.01 3.001 3.0001

3x 9 x2 − 9 30 300 3000 30,000

3x + 9 does not exist. x2 − 9 Here, we considered both one-sided limits for the sake of completeness. Of course, you should keep in mind that if either one-sided limit fails to exist, then the limit does not exist. lim

x→3

1-11

SECTION 1.2

3x 9 x2 − 9 −30 −300 −3000 −30,000

x 2.9 2.99 2.999 2.9999

EXAMPLE 2.4

sin x . x Solution Unlike some of the limits considered previously, there is no algebra that will simplify this expression. However, we can still draw a graph (see Figure 1.11) and compute some function values.

0.1 0.01 0.001 0.0001 0.00001

f (x)

x x 2

4

lim

sin x x 0.998334 0.999983 0.99999983 0.9999999983 0.999999999983

x −0.1 −0.01 −0.001 −0.0001 −0.00001

sin x =1 x from which we conjecture that x→0+

sin x =1 x

sin x x 0.998334 0.999983 0.99999983 0.9999999983 0.999999999983

The graph and the tables of values lead us to the conjectures: lim

FIGURE 1.11 x→0

Approximating the Value of a Limit

x→0

x

2

83

Evaluate lim

1

4

The Concept of Limit

Many limits cannot be resolved using algebraic methods. In these cases, we can approximate the limit using graphical and numerical evidence, as we see in example 2.4.

y

x

..

and

lim

x→0−

sin x = 1, x

sin x = 1. x In Chapter 2, we examine these limits with greater care (and prove that these conjectures are correct). lim

x→0

REMARK 2.1 Computer or calculator computation of limits is unreliable. We use graphs and tables of values only as (strong) evidence pointing to what a plausible answer might be. To be certain, we need to obtain careful verification of our conjectures. We see how to do this in sections 1.3–1.7.

EXAMPLE 2.5 y

Evaluate lim

x→0

1

x

4

4

A Case Where One-Sided Limits Disagree

x . |x|

Solution The computer-generated graph shown in Figure 1.12a is incomplete. Since x is undefined at x = 0, there is no point at x = 0. The graph in Figure 1.12b |x| correctly shows open circles at the intersections of the two halves of the graph with the y-axis. We also have

1

lim

x→0+

FIGURE 1.12a y=

x |x|

x x = lim+ x→0 x |x| = lim+ 1 x→0

=1

Since |x| = x, when x > 0.

84

CHAPTER 1

..

Limits and Continuity

1-12

y

and

lim

x→0−

1

x

x→0

= −1.

x 2

f(x)

It now follows that

lim

x→0

1

FIGURE 1.12b lim

x→0

Since |x| = −x, when x < 0.

= lim− −1

f(x) x 2

x x = lim− x→0 −x |x|

x does not exist. |x|

x does not exist, |x|

since the one-sided limits are not the same. You should also keep in mind that this observation is entirely consistent with what we see in the graph.

EXAMPLE 2.6

A Limit Describing the Movement of a Baseball Pitch

The knuckleball is one of the most exotic pitches in baseball. Batters describe the ball as unpredictably moving left, right, up and down. For a typical knuckleball speed of 60 mph, the left/right position of the ball (in feet) as it crosses the plate is given by f (ω) =

1.7 5 − sin(2.72ω) ω 8ω2

(derived from experimental data in Watts and Bahill’s book Keeping Your Eye on the Ball), where ω is the rotational speed of the ball in radians per second and where f (ω) = 0 corresponds to the middle of home plate. Folk wisdom among baseball pitchers has it that the less spin on the ball, the better the pitch. To investigate this theory, we consider the limit of f (ω) as ω → 0+ . As always, we look at a graph (see Figure 1.13) and generate a table of function values. The graphical and numerical evidence suggests that lim+ f (ω) = 0. ω→0

y 1.5

1.0

0.5

v 2

4

6

8

10

FIGURE 1.13 5 1.7 − y= sin(2.72ω) ω 8ω2

ω

f (ω)

10 1 0.1 0.01 0.001 0.0001

0.1645 1.4442 0.2088 0.021 0.0021 0.0002

The limit indicates that a knuckleball with absolutely no spin doesn’t move at all (and therefore would be easy to hit). According to Watts and Bahill, a very slow rotation rate of about 1 to 3 radians per second produces the best pitch (i.e., the most movement). Take another look at Figure 1.13 to convince yourself that this makes sense.

1-13

SECTION 1.2

..

The Concept of Limit

85

EXERCISES 1.2 WRITING EXERCISES 1. Suppose your professor says, “You can think of the limit of f (x) as x approaches a as what f (a) should be.” Critique this statement. What does it mean? Does it provide important insight? Is there anything misleading about it? Replace the phrase in italics with your own best description of what the limit is. 2. Your friend’s professor says, “The limit is a prediction of what f (a) will be.” Compare and contrast this statement to the one in exercise 1. Does the inclusion of the word prediction make the limit idea seem more useful and important?

2. For the function graphed below, identify each limit or state that it does not exist. (a) lim f (x)

(b) lim f (x)

(c) lim f (x)

(d) lim f (x)

(e) lim f (x)

(f) lim f (x)

(g) lim f (x)

(h) lim f (x)

(i) lim f (x)

(j) lim f (x)

x→0−

x→0+ x→2−

x→0

x→1−

x→−2 x→1+

x→1

x→−1

x→3

3. We have observed that lim f (x) does not depend on the actual x→a

y

value of f (a), or even onwhether f (a) exists. In principle, x 2 if x = 2 are as “normal” as functions such as f (x) = 13 if x = 2 functions such as g(x) = x 2 . With this in mind, explain why it is important that the limit concept is independent of how (or whether) f (a) is defined. 4. The most common limit encountered in everyday life is the speed limit. Describe how this type of limit is very different from the limits discussed in this section.

4 2 6

2

x 2

4

6

2 4

1. For the function graphed below, identify each limit or state that it does not exist. (a) lim f (x)

(b) lim f (x)

(c) lim f (x)

(d) lim f (x)

(e) lim f (x)

(f) lim f (x)

(g) lim f (x)

(h) lim f (x)

x→0−

x→0+ x→1−

x→0

x→2−

x→−1 x→2+

x→2

(i) lim f (x)

(j) lim f (x)

x→−2

x→3

y

3. Sketch the graph of f (x) = (a) lim f (x)

(b) lim f (x)

(c) lim f (x)

(d) lim f (x)

x→2− x→2

2 2 4

x→2+ x→1

3 x −1 4. Sketch the graph of f (x) = 0 √ x +1−2 identify each limit. (a) lim f (x)

(b) lim f (x)

(d) lim f (x)

(e) lim f (x)

x→0−

x→−1

2 6

6

5. Sketch the graph of f (x) =

lim f (x)

x→−1−

(c) lim f (x) x→−1

x <0 x = 0 and x >0

x→0

x→3

x2 + 1 3x + 1

if if

x < −1 and idenx ≥ −1

tify each limit. (a)

if if if

(c) lim f (x)

x→0+

4

x <2 and identify x ≥2

if if

each limit.

4

x

2x x2

(b)

lim f (x)

x→−1+

(d) lim f (x) x→1

86

CHAPTER 1

..

Limits and Continuity

2x + 1 if if 6. Sketch the graph of f (x) = 3 2x + 1 if identify each limit. (a)

lim f (x)

x→−1−

(d) lim f (x) x→1

(b)

1-14

x < −1 −1 ≤ x < 1 and x >1

lim f (x)

(c) lim f (x)

x→−1+

x→−1

x→0

8. Evaluate f (−1.5), f (−1.1), f (−1.01) and f (−1.001), and x +1 conjecture a value for lim f (x) for f (x) = 2 . Evaluate x→−1− x −1 f (−0.5), f (−0.9), f (−0.99) and f (−0.999), and conjecture x +1 . Does lim f (x) a value for lim f (x) for f (x) = 2 x→−1 x→−1+ x −1 exist? In exercises 9–14, use numerical and graphical evidence to conjecture values for each limit. x2 + x 10. lim 2 x→−1 x − x − 2 sin x 12. lim x→π x − π 14. lim e−1/x

2

x→0

In exercises 15–26, use numerical and graphical evidence to conjecture whether the limit at x a exists. If not, describe what is happening at x a graphically. x2 − 1 x→1 x 2 − 2x + 1 √ 5−x −2 17. lim √ x→1 10 − x − 3 1 19. lim sin x→0 x 15. lim

x −2 x→2 |x − 2|

16. lim

x→−1

x2 − 1 x +1

x 2 + 4x 18. lim √ x→0 x3 + x2 1 20. lim x sin x→0 x |x + 1| x→−1 x 2 − 1

21. lim

22. lim

23. lim ln x

24. lim x ln (x 2 )

x→0

tan x x→0 x

25. lim

x→0

tan−1 x x→0 x

26. lim

x2 + 1 x +1 , lim and similar limits to investix→1 x − 1 x→2 x 2 − 4 gate the following. Suppose that f (x) and g(x) are polynomials with g(a) = 0 and f (a) = 0. What can you conjecture about f (x) ? lim x→a g(x)

27. Compute lim

x→−1

(e) lim f (x)

7. Evaluate f (1.5), f (1.1), f (1.01) and f (1.001), and conjecx −1 . Evaluate ture a value for lim f (x) for f (x) = √ x→1+ x −1 f (0.5), f (0.9), f (0.99) and f (0.999), and conjecture a value x −1 for lim f (x) for f (x) = √ . Does lim f (x) exist? x→1 x→1− x −1

x2 − 1 9. lim x→1 x − 1 x2 + x 11. lim x→0 sin x tan x 13. lim x→0 sin x

x +1 sin x , lim and similar limits to invesx 2 + 1 x→π x tigate the following. Suppose that f (x) and g(x) are functions with f (a) = 0 and g(a) = 0. What can you conjecture about f (x) lim ? x→a g(x)

28. Compute lim

In exercises 29–32, sketch a graph of a function with the given properties. 29. f (−1) = 2, f (0) = −1, f (1) = 3 and lim f (x) does not exist. x→1

30. f (x) = 1 for −2 ≤ x ≤ 1, lim f (x) = 3 and lim f (x) = 1. x→1+

x→−2

31. f (0) = 1, lim f (x) = 2 and lim f (x) = 3. x→0−

x→0+

32. lim f (x) = −2, f (0) = 1, f (2) = 3 and lim f (x) does not x→0

x→2

exist. 33. As we see in Chapter 2, the slope of the tangent √ line to the √ 1+h−1 . curve y = x at x = 1 is given by m = lim h→0 h √ Estimate the slope m. Graph y = x and the line with slope m through the point (1, 1). 34. As we see √ in Chapter 2, the velocity of an object that has traveled x √ miles in x hours at the x = 1 hour mark is given x −1 . Estimate this limit. by v = lim x→1 x − 1 π 35. Consider the following arguments concerning lim sin . x→0+ x π increases without bound; First, as x > 0 approaches 0, x since sin t oscillates for increasing t, the limit does not exist. Second: taking x = 1, 0.1, 0.01 and so on, we compute sin π = sin 10π = sin 100π = · · · = 0; therefore the limit equals 0. Which argument sounds better to you? Explain. Explore the limit and determine which answer is correct. 36. Consider the following argument concerning lim e−1/x . As x x→0 1 −1 approaches 0, increases without bound and decreases x x t without bound. Since e approaches 0 as t decreases without bound, the limit equals 0. Discuss all the errors made in this argument. 37. Numerically estimate lim (1 + x)1/x and lim (1 + x)1/x . x→0+

x→0−

Note that the function values for x > 0 increase as x decreases, while for x < 0 the function values decrease as x increases. Explain why this indicates that, if lim (1 + x)1/x x→0

exists, it is between function values for positive and negative x’s. Approximate this limit correct to eight digits. 38. Explain what is wrong with the following logic (you may use exercise 37 to convince yourself that the answer is wrong, but discuss the logic without referring to exercise 37): as x → 0, it is clear that (1 + x) → 1. Since 1 raised to any power is 1, lim (1 + x)1/x = lim (1)1/x = 1. x→0

x→0

1-15

SECTION 1.3

39. Numerically estimate lim x sec x . Try to numerically estimate x→0+

lim x sec x . If your computer has difficulty evaluating the func-

x→0−

tion for negative x’s, explain why. 40. Explain what is wrong with the following logic (note from exercise 39 that the answer is accidentally correct): since 0 to any power is 0, lim x sec x = lim 0sec x = 0. x→0

x→0

41. Give an example of a function f such that lim f (x) exists but x→0

f (0) does not exist. Give an example of a function g such that g(0) exists but lim g(x) does not exist. x→0

42. Give an example of a function f such that lim f (x) exists and x→0

f (0) exists, but lim f (x) = f (0). x→0

43. In the text, we described lim f (x) = L as meaning “as x gets x→a

closer and closer to a, f (x) is getting closer and closer to L.” As x gets closer and closer to 0, it is true that x 2 gets closer and closer to −0.01, but it is certainly not true that lim x 2 = −0.01. Try to modify the description of limit to x→0

make it clear that lim x 2 = −0.01. We explore a very precise x→0

definition of limit in section 1.6. 44. In Figure 1.13, the final position of the knuckleball at time t = 0.68 is shown as a function of the rotation rate ω. The batter must decide at time t = 0.4 whether to swing at the pitch. At t = 0.4, the left/right position of the ball is given 1 5 by h(ω) = − sin (1.6ω). Graph h(ω) and compare to ω 8ω2 Figure 1.13. Conjecture the limit of h(ω) as ω → 0. For ω = 0, is there any difference in ball position between what the batter sees at t = 0.4 and what he tries to hit at t = 0.68? 45. A parking lot charges $2 for each hour or portion of an hour, with a maximum charge of $12 for all day. If f (t) equals the total parking bill for t hours, sketch a graph of y = f (t) for 0 ≤ t ≤ 24. Determine the limits lim f (t) and lim f (t), if t→3.5

t→4

they exist. 46. For the parking lot in exercise 45, determine all values of a with 0 ≤ a ≤ 24 such that lim f (t) does not exist. Briefly t→a

discuss the effect this has on your parking strategy (e.g., are there times where you would be in a hurry to move your car or times where it doesn’t matter whether you move your car?).

1.3

..

Computation of Limits

87

EXPLORATORY EXERCISES 1. In a situation similar to that of example 2.6, the left/right position of a knuckleball pitch in baseball can be modeled by 5 (1 − cos 4ωt), where t is time measured in seconds P= 8ω2 (0 ≤ t ≤ 0.68) and ω is the rotation rate of the ball measured in radians per second. In example 2.6, we chose a specific t-value and evaluated the limit as ω → 0. While this gives us some information about which rotation rates produce hardto-hit pitches, a clearer picture emerges if we look at P over its entire domain. Set ω = 10 and graph the resulting func1 tion (1 − cos 40t) for 0 ≤ t ≤ 0.68. Imagine looking at a 160 pitcher from above and try to visualize a baseball starting at the pitcher’s hand at t = 0 and finally reaching the batter, at t = 0.68. Repeat this with ω = 5, ω = 1, ω = 0.1 and whatever values of ω you think would be interesting. Which values of ω produce hard-to-hit pitches? 2. In this exercise, the results you get will depend on the accuracy of your computer or calculator. Work this exercise and compare your results with your classmates’ results. We will incos x − 1 vestigate lim . Start with the calculations presented x→0 x2 in the table (your results may vary): x

f(x)

0.1 0.01 0.001

−0.499583. . . −0.49999583. . . −0.4999999583. . .

Describe as precisely as possible the pattern shown here. What would you predict for f (0.0001)? f (0.00001)? Does your computer or calculator give you this answer? If you continue trying powers of 0.1 (0.000001, 0.0000001 etc.) you should eventually be given a displayed result of −0.5. Do you think this is exactly correct or has the answer just been rounded off? Why is rounding off inescapable? It turns out that −0.5 is the exact value for the limit, so the round-off here is somewhat helpful. However, if you keep evaluating the function at smaller and smaller values of x, you will eventually see a reported function value of 0. This round-off error is not so benign; we discuss this error in section 1.7. For now, evaluate cos x at the current value of x and try to explain where the 0 came from.

COMPUTATION OF LIMITS Now that you have an idea of what a limit is, we need to develop some means of calculating limits of simple functions. In this section, we present some basic rules for dealing with common limit problems. We begin with two simple limits.

1-15

SECTION 1.3

39. Numerically estimate lim x sec x . Try to numerically estimate x→0+

lim x sec x . If your computer has difficulty evaluating the func-

x→0−

tion for negative x’s, explain why. 40. Explain what is wrong with the following logic (note from exercise 39 that the answer is accidentally correct): since 0 to any power is 0, lim x sec x = lim 0sec x = 0. x→0

x→0

41. Give an example of a function f such that lim f (x) exists but x→0

f (0) does not exist. Give an example of a function g such that g(0) exists but lim g(x) does not exist. x→0

42. Give an example of a function f such that lim f (x) exists and x→0

f (0) exists, but lim f (x) = f (0). x→0

43. In the text, we described lim f (x) = L as meaning “as x gets x→a

closer and closer to a, f (x) is getting closer and closer to L.” As x gets closer and closer to 0, it is true that x 2 gets closer and closer to −0.01, but it is certainly not true that lim x 2 = −0.01. Try to modify the description of limit to x→0

make it clear that lim x 2 = −0.01. We explore a very precise x→0

definition of limit in section 1.6. 44. In Figure 1.13, the final position of the knuckleball at time t = 0.68 is shown as a function of the rotation rate ω. The batter must decide at time t = 0.4 whether to swing at the pitch. At t = 0.4, the left/right position of the ball is given 1 5 by h(ω) = − sin (1.6ω). Graph h(ω) and compare to ω 8ω2 Figure 1.13. Conjecture the limit of h(ω) as ω → 0. For ω = 0, is there any difference in ball position between what the batter sees at t = 0.4 and what he tries to hit at t = 0.68? 45. A parking lot charges $2 for each hour or portion of an hour, with a maximum charge of $12 for all day. If f (t) equals the total parking bill for t hours, sketch a graph of y = f (t) for 0 ≤ t ≤ 24. Determine the limits lim f (t) and lim f (t), if t→3.5

t→4

they exist. 46. For the parking lot in exercise 45, determine all values of a with 0 ≤ a ≤ 24 such that lim f (t) does not exist. Briefly t→a

discuss the effect this has on your parking strategy (e.g., are there times where you would be in a hurry to move your car or times where it doesn’t matter whether you move your car?).

1.3

..

Computation of Limits

87

EXPLORATORY EXERCISES 1. In a situation similar to that of example 2.6, the left/right position of a knuckleball pitch in baseball can be modeled by 5 (1 − cos 4ωt), where t is time measured in seconds P= 8ω2 (0 ≤ t ≤ 0.68) and ω is the rotation rate of the ball measured in radians per second. In example 2.6, we chose a specific t-value and evaluated the limit as ω → 0. While this gives us some information about which rotation rates produce hardto-hit pitches, a clearer picture emerges if we look at P over its entire domain. Set ω = 10 and graph the resulting func1 tion (1 − cos 40t) for 0 ≤ t ≤ 0.68. Imagine looking at a 160 pitcher from above and try to visualize a baseball starting at the pitcher’s hand at t = 0 and finally reaching the batter, at t = 0.68. Repeat this with ω = 5, ω = 1, ω = 0.1 and whatever values of ω you think would be interesting. Which values of ω produce hard-to-hit pitches? 2. In this exercise, the results you get will depend on the accuracy of your computer or calculator. Work this exercise and compare your results with your classmates’ results. We will incos x − 1 vestigate lim . Start with the calculations presented x→0 x2 in the table (your results may vary): x

f(x)

0.1 0.01 0.001

−0.499583. . . −0.49999583. . . −0.4999999583. . .

Describe as precisely as possible the pattern shown here. What would you predict for f (0.0001)? f (0.00001)? Does your computer or calculator give you this answer? If you continue trying powers of 0.1 (0.000001, 0.0000001 etc.) you should eventually be given a displayed result of −0.5. Do you think this is exactly correct or has the answer just been rounded off? Why is rounding off inescapable? It turns out that −0.5 is the exact value for the limit, so the round-off here is somewhat helpful. However, if you keep evaluating the function at smaller and smaller values of x, you will eventually see a reported function value of 0. This round-off error is not so benign; we discuss this error in section 1.7. For now, evaluate cos x at the current value of x and try to explain where the 0 came from.

COMPUTATION OF LIMITS Now that you have an idea of what a limit is, we need to develop some means of calculating limits of simple functions. In this section, we present some basic rules for dealing with common limit problems. We begin with two simple limits.

88

..

CHAPTER 1

Limits and Continuity

1-16

y

For any constant c and any real number a,

yc

c

lim c = c.

(3.1)

x→a

x

x

x

a

FIGURE 1.14

In other words, the limit of a constant is that constant. This certainly comes as no surprise, since the function f (x) = c does not depend on x and so, stays the same as x → a (see Figure 1.14). Another simple limit is the following.

lim c = c

x→a

For any real number a, lim x = a.

(3.2)

x→a

y f(x)

Again, this is not a surprise, since as x → a, x will approach a (see Figure 1.15). Be sure that you are comfortable enough with the limit notation to recognize how obvious the limits in (3.1) and (3.2) are. As simple as they are, we use them repeatedly in finding more complex limits. We also need the basic rules contained in Theorem 3.1.

yx

a f(x)

THEOREM 3.1 x

x a

FIGURE 1.15 lim x = a

x→a

x

Suppose that lim f (x) and lim g(x) both exist and let c be any constant. The x→a x→a following then apply: (i) lim [c · f (x)] = c · lim f (x), x→a

x→a

(ii) lim [ f (x) ± g(x)] = lim f (x) ± lim g(x), x→a x→a x→a

(iii) lim [ f (x) · g(x)] = lim f (x) lim g(x) and x→a

x→a

x→a

f (x) x→a = if lim g(x) = 0 . x→a g(x) lim g(x) lim f (x)

(iv) lim

x→a

x→a

The proof of Theorem 3.1 is found in Appendix A and requires the formal definition of limit discussed in section 1.6. You should think of these rules as sensible results that you would certainly expect to be true, given your intuitive understanding of what a limit is. Read them in plain English. For instance, part (ii) says that the limit of a sum (or a difference) equals the sum (or difference) of the limits, provided the limits exist. Think of this as follows. If as x approaches a, f (x) approaches L and g(x) approaches M, then f (x) + g(x) should approach L + M. Observe that by applying part (iii) of Theorem 3.1 with g(x) = f (x), we get that, whenever lim f (x) exists, x→a

lim [ f (x)]2 = lim [ f (x) · f (x)] x→a

2 = lim f (x) lim f (x) = lim f (x) .

x→a

x→a

x→a

x→a

Likewise, for any positive integer n, we can apply part (iii) of Theorem 3.1 repeatedly, to yield

n lim [ f (x)]n = lim f (x) (3.3) x→a

(see exercises 67 and 68).

x→a

1-17

SECTION 1.3

..

Computation of Limits

89

Notice that taking f (x) = x in (3.3) gives us that for any integer n > 0 and any real number a, lim x n = a n .

(3.4)

x→a

That is, to compute the limit of any positive power of x, you simply substitute in the value of x being approached.

EXAMPLE 3.1

Finding the Limit of a Polynomial

Apply the rules of limits to evaluate lim (3x 2 − 5x + 4). x→2

Solution We have lim (3x 2 − 5x + 4) = lim (3x 2 ) − lim (5x) + lim 4

x→2

x→2

x→2

= 3 lim x 2 − 5 lim x + 4

By Theorem 3.1 (i).

= 3 · (2)2 − 5 · 2 + 4 = 6.

By (3.4).

x→2

EXAMPLE 3.2

By Theorem 3.1 (ii).

x→2

x→2

Finding the Limit of a Rational Function

Apply the rules of limits to evaluate lim

x→3

x 3 − 5x + 4 . x2 − 2

Solution We get lim (x 3 − 5x + 4) x 3 − 5x + 4 x→3 lim = x→3 x2 − 2 lim (x 2 − 2)

By Theorem 3.1 (iv).

x→3

= =

lim x 3 − 5 lim x + lim 4

x→3

x→3 x→3 2 lim x − lim 2 x→3 x→3

33 − 5 · 3 + 4 16 = . 32 − 2 7

By Theorem 3.1 (i) and (ii).

By (3.4).

You may have noticed that in examples 3.1 and 3.2, we simply ended up substituting the value for x, after taking many intermediate steps. In example 3.3, it’s not quite so simple.

EXAMPLE 3.3

Finding a Limit by Factoring

x2 − 1 . x→1 1 − x Solution Notice right away that

Evaluate lim

lim (x 2 − 1) x2 − 1 x→1 lim = , x→1 1 − x lim (1 − x) x→1

since the limit in the denominator is zero. (Recall that the limit of a quotient is the quotient of the limits only when both limits exist and the limit in the denominator is not

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1-18

zero.) We can resolve this problem by observing that x2 − 1 (x − 1)(x + 1) Factoring the numerator and = lim factoring −1 from denominator. x→1 1 − x x→1 −(x − 1) (x + 1) Simplifying and = lim = −2, substituting x = 1. x→1 −1 where the cancellation of the factors of (x − 1) is valid because in the limit as x → 1, x is close to 1, but x = 1, so that x − 1 = 0. lim

In Theorem 3.2, we show that the limit of a polynomial is simply the value of the polynomial at that point; that is, to find the limit of a polynomial, we simply substitute in the value that x is approaching.

THEOREM 3.2 For any polynomial p(x) and any real number a, lim p(x) = p(a).

x→a

PROOF Suppose that p(x) is a polynomial of degree n ≥ 0, p(x) = cn x n + cn−1 x n−1 + · · · + c1 x + c0 . Then, from Theorem 3.1 and (3.4), lim p(x) = lim (cn x n + cn−1 x n−1 + · · · + c1 x + c0 )

x→a

x→a

= cn lim x n + cn−1 lim x n−1 + · · · + c1 lim x + lim c0 x→a

x→a

x→a

x→a

= cn a n + cn−1 a n−1 + · · · + c1 a + c0 = p(a). Evaluating the limit of a polynomial is now easy. Many other limits are evaluated just as easily.

THEOREM 3.3 Suppose that lim f (x) = L and n is any positive integer. Then, x→a lim n f (x) = n lim f (x) = n L , x→a

x→a

where for n even, we assume that L > 0. The proof of Theorem 3.3 is given in Appendix A. Notice that this result says that we may (under the conditions outlined in the hypotheses) bring limits “inside” nth roots. We can then use our existing rules for computing the limit inside.

EXAMPLE 3.4 Evaluate lim

x→2

√ 5

3x 2

Evaluating the Limit of an nth Root of a Polynomial − 2x.

Solution By Theorems 3.2 and 3.3, we have √ 5 5 lim 3x 2 − 2x = 5 lim (3x 2 − 2x) = 8. x→2

x→2

1-19

REMARK 3.1 In general, in any case where the limits of both the numerator and the denominator are 0, you should try to algebraically simplify the expression, to get a cancellation, as we do in examples 3.3 and 3.5.

SECTION 1.3

EXAMPLE 3.5 √

Evaluate lim

x→0

..

Computation of Limits

91

Finding a Limit by Rationalizing

x +2− x

√

2

.

√ √ Solution First, notice that both the numerator ( x + 2 − 2) and the denominator (x) approach 0 as x approaches 0. Unlike example 3.3, we can’t factor the numerator. However, we can rationalize the numerator, as follows: √ √ √ √ √ √ x +2− 2 ( x + 2 − 2)( x + 2 + 2) x +2−2 = = √ √ √ √ x x( x + 2 + 2) x( x + 2 + 2) x 1 = √ √ =√ √ , x( x + 2 + 2) x +2+ 2 where the last equality holds if x = 0 (which is the case in the limit as x → 0). So, we have √ √ x +2− 2 1 1 1 lim = lim √ √ =√ √ = √ . x→0 x→0 x x +2+ 2 2+ 2 2 2 So that we are not restricted to discussing only the algebraic functions (i.e., those that can be constructed by using addition, subtraction, multiplication, division, exponentiation and by taking nth roots), we state the following result now, without proof.

THEOREM 3.4 For any real number a, we have (i) lim sin x = sin a,

(v) lim sin−1 x = sin−1 a, for −1 < a < 1,

(ii) lim cos x = cos a,

(vi) lim cos−1 x = cos−1 a, for −1 < a < 1,

(iii) lim e x = ea and

(vii) lim tan−1 x = tan−1 a, for −∞

x→a x→a x→a

x→a x→a x→a

(iv) lim ln x = ln a, for a > 0. (viii) if p is a polynomial and lim f (x) = L , x→a

then lim f ( p(x)) = L .

x→ p(a)

x→a

Notice that Theorem 3.4 says that limits of the sine, cosine, exponential, natural logarithm, inverse sine, inverse cosine and inverse tangent functions are found simply by substitution. A more thorough discussion of functions with this property (called continuity) is found in section 1.4.

EXAMPLE 3.6 Evaluate lim sin x→0

−1

Evaluating a Limit of an Inverse Trigonometric Function

x +1 . 2

Solution By Theorem 3.4, we have x +1 1 π lim sin−1 = sin−1 = . x→0 2 2 6 So much for limits that we can compute using elementary rules. Many limits can be found only by using more careful analysis, often by an indirect approach. For instance, consider the following problem.

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1-20

y

EXAMPLE 3.7

A Limit of a Product That Is Not the Product of the Limits

Evaluate lim (x cot x). x→0

Solution Your first reaction might be to say that this is a limit of a product and so, must be the product of the limits: lim (x cot x) = lim x lim cot x This is incorrect!

x p

q

q

x→0

p

x→0

x→0

= 0 · ? = 0,

(3.5) where we’ve written a “?” since you probably don’t know what to do with lim cot x. x→0

Since the first limit is 0, do we really need to worry about the second limit? The problem here is that we are attempting to apply the result of Theorem 3.1 in a case where the hypotheses are not satisfied. Specifically, Theorem 3.1 says that the limit of a product is the product of the respective limits when all of the limits exist. The graph in Figure 1.16 suggests that lim cot x does not exist. You should compute some function

FIGURE 1.16 y = cot x

x→0

values, as well, to convince yourself that this is in fact the case. So, equation (3.5) does not hold and we’re back to square one. Since none of our rules seem to apply here, the most reasonable step is to draw a graph (see Figure 1.17) and compute some function values. Based on these, we conjecture that

y 1

lim (x cot x) = 1,

0.99

x→0

0.98

0.97 x

0.3

0.3

FIGURE 1.17 y = x cot x

x ±0.1 ±0.01 ±0.001 ±0.0001 ±0.00001

x cot x 0.9967 0.999967 0.99999967 0.9999999967 0.999999999967

which is definitely not 0, as you might have initially suspected. You can also think about this limit as follows: cos x x lim (x cot x) = lim x = lim cos x x→0 x→0 x→0 sin x sin x x = lim lim cos x x→0 sin x x→0 lim cos x

1 = = 1, sin x 1 lim x→0 x since lim cos x = 1 and where we have used the conjecture we made in example 2.4 x→0 sin x = 1. (We verify this last conjecture in section 2.6, using the Squeeze that lim x→0 x Theorem, which follows.) =

x→0

At this point, we introduce a tool that will help us determine a number of important limits.

THEOREM 3.5 (Squeeze Theorem) Suppose that f (x) ≤ g(x) ≤ h(x) for all x in some interval (c, d), except possibly at the point a ∈ (c, d) and that lim f (x) = lim h(x) = L ,

x→a

x→a

for some number L. Then, it follows that lim g(x) = L , also.

x→a

1-21

SECTION 1.3

y

FIGURE 1.18 The Squeeze Theorem

REMARK 3.2 The Squeeze Theorem also applies to one-sided limits.

Computation of Limits

x

EXAMPLE 3.8

Using the Squeeze Theorem to Verify the Value of a Limit

1 Determine the value of lim x cos . x→0 x 2

Solution Your first reaction might be to observe that this is a limit of a product and so, might be the product of the limits: 1 1 ? 2 2 lim x cos lim cos = lim x . This is incorrect! (3.6) x→0 x→0 x→0 x x However, the graph of y = cos x1 found in Figure 1.19 suggests that cos x1 oscillates back and forth between −1 and 1. Further, the closer x gets to 0, the more rapid the oscillations become. You should compute some function values, as well, to convince yourself that lim cos x1 does not exist. Equation (3.6) then does not hold and x→0

we’re back to square one. Since none of our rules seem to apply here, the most reasonable step is to draw a graph and compute some function values in an effort to see what is going on. The graph of y = x 2 cos x1 appears in Figure 1.20 and a table of function values is shown in the margin. y

y

1

0.03

x

0.2

0.2

1

y = cos

x 2 cos (1/x) −0.008 8.6 × 10−5 5.6 × 10−7 −9.5 × 10−9 −9.99 × 10−11

x

0.3

0.3

0.03

FIGURE1.19

x ±0.1 ±0.01 ±0.001 ±0.0001 ±0.00001

93

The proof of Theorem 3.5 is given in Appendix A, since it depends on the precise definition of limit found in section 1.6. However, if you refer to Figure 1.18, you should clearly see that if g(x) lies between f (x) and h(x), except possibly at a itself and both f (x) and h(x) have the same limit as x → a, then g(x) gets squeezed between f (x) and h(x) and therefore should also have a limit of L. The challenge in using the Squeeze Theorem is in finding appropriate functions f and h that bound a given function g from below and above, respectively, and that have the same limit as x → a.

y h(x) y g(x) y f(x)

a

..

1 x

FIGURE 1.20 y = x 2 cos

1 x

The graph and the table of function values suggest the conjecture 1 2 = 0, lim x cos x→0 x which we prove using the Squeeze Theorem. First, we need to find functions f and h such that 1 2 f (x) ≤ x cos ≤ h(x), x

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for all x = 0 and where lim f (x) = lim h(x) = 0. Recall that

y 0.03

x→0

y x2

x

0.3

0.3

0.03

1-22

y x2

x→0

1 −1 ≤ cos ≤ 1, x

for all x = 0. If we multiply (3.7) through by x 2 (notice that since x 2 ≥ 0, this multiplication preserves the inequalities), we get 1 −x 2 ≤ x 2 cos ≤ x 2, x for all x = 0. We illustrate this inequality in Figure 1.21. Further, lim (−x 2 ) = 0 = lim x 2 .

FIGURE 1.21

y = x 2 cos x1 , y = x 2 and y = −x 2

(3.7)

x→0

x→0

So, from the Squeeze Theorem, it now follows that 1 lim x 2 cos = 0, x→0 x also, as we had conjectured.

BEYOND FORMULAS

TODAY IN MATHEMATICS Michael Freedman (1951– ) An American mathematician who first solved one of the most famous problems in mathematics, the four-dimensional Poincar´e conjecture. A winner of the Fields Medal, the mathematical equivalent of the Nobel Prize, Freedman says, “Much of the power of mathematics comes from combining insights from seemingly different branches of the discipline. Mathematics is not so much a collection of different subjects as a way of thinking. As such, it may be applied to any branch of knowledge.” Freedman finds mathematics to be an open field for research, saying that, “It isn’t necessary to be an old hand in an area to make a contribution.”

To resolve the limit in example 3.8, we could not apply the rules for limits contained in Theorem 3.1. So, we resorted to an indirect method of finding the limit. This tour de force of graphics plus calculation followed by analysis is sometimes referred to as the Rule of Three. (The Rule of Three presents a general strategy for attacking new problems. The basic idea is to look at problems graphically, numerically and analytically.) In the case of example 3.8, the first two elements of this “rule” (the graphics in Figure 1.20 and the accompanying table of function values) suggest a plausible conjecture, while the third element provides us with a careful mathematical verification of the conjecture. In what ways does this sound like the scientific method? Functions are often defined by different expressions on different intervals. Such piecewise-defined functions are important and we illustrate such a function in example 3.9.

EXAMPLE 3.9

A Limit for a Piecewise-Defined Function

Evaluate lim f (x), where f is defined by x→0 2 x + 2 cos x + 1, for x < 0 f (x) = . for x ≥ 0 e x − 4, Solution Since f is defined by different expressions for x < 0 and for x ≥ 0, we must consider one-sided limits. We have lim f (x) = lim− (x 2 + 2 cos x + 1) = 2 cos 0 + 1 = 3,

x→0−

x→0

by Theorem 3.4. Also, we have lim f (x) = lim+ (e x − 4) = e0 − 4 = 1 − 4 = −3.

x→0+

x→0

Since the one-sided limits are different, we have that lim f (x) does not exist. x→0

1-23

..

SECTION 1.3

Computation of Limits

95

We end this section with an example of the use of limits in computing velocity. In section 2.1, we see that for an object moving in a straight line, whose position at time t is given by the function f (t), the instantaneous velocity of that object at time t = 1 (i.e., the velocity at the instant t = 1, as opposed to the average velocity over some period of time) is given by the limit f (1 + h) − f (1) lim . h→0 h

EXAMPLE 3.10

Evaluating a Limit Describing Velocity

Suppose that the position function for an object at time t (seconds) is given by f (t) = t 2 + 2 (feet), find the instantaneous velocity of the object at time t = 1. Solution Given what we have just learned about limits, this is now an easy problem to solve. We have f (1 + h) − f (1) [(1 + h)2 + 2] − 3 lim = lim . h→0 h→0 h h While we can’t simply substitute h = 0 (why not?), we can write [(1 + h)2 + 2] − 3 (1 + 2h + h 2 ) − 1 = lim h→0 h→0 h h lim

Expanding the squared term.

2h + h 2 h(2 + h) = lim h→0 h→0 h h 2+h = lim = 2. h→0 1

= lim

Canceling factors of h.

So, the instantaneous velocity of this object at time t = 1 is 2 feet per second.

EXERCISES 1.3 WRITING EXERCISES 1. Given your knowledge of the graphs of polynomials, explain why equations (3.1) and (3.2) and Theorem 3.2 are obvious. Name five non-polynomial functions for which limits can be evaluated by substitution. 2. Suppose that you can draw the graph of y = f (x) without lifting your pencil from your paper. Explain why lim f (x) = f (a), for every value of a. x→a

3. In one or two sentences, explain the Squeeze Theorem. Use a real-world analogy (e.g., having the functions represent the locations of three people as they walk) to indicate why it is true. 4. Given the graph in Figure 1.20 and the calculations that follow, it may be unclear why we insist on using the Squeeze Theorem before concluding that lim [x 2 cos (1/x)] is indeed 0. Review x→0

section 1.2 to explain why we are being so fussy.

In exercises 1–34, evaluate the indicated limit, if it exists. Assume sin x that lim 1. x→0 x √ 3 2. lim 2x + 1 1. lim (x 2 − 3x + 1) x→0

x→2

−1

2

3. lim cos (x ) x→0

x2 − x − 6 x→3 x −3 2 x −x −2 7. lim x→2 x2 − 4 sin x 9. lim x→0 tan x xe−2x+1 11. lim 2 x→0 x + x 5. lim

x −5 x2 + 4 x2 + x − 2 6. lim 2 x→1 x − 3x + 2 x3 − 1 8. lim 2 x→1 x + 2x − 3 tan x 10. lim x→0 x 4. lim

x→2

12. lim x 2 csc2 x x→0+

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√

x +4−2 2x 14. lim √ x→0 x 3− x +9 √ x −1 x −2 lim √ 16. lim x→1 x→4 x − 4 x −1 2 2 1 2 lim − 2 18. lim − x→1 x→0 x −1 x −1 x |x| 2x 1−e −1/x 2 lim 20. lim sin(e ) x→0 1 − e x x→0 sin |x| sin2 (x 2 ) lim 22. lim x→0 x→0 x x4 2x if x < 2 lim f (x), where f (x) = x 2 if x ≥ 2 x→2 2 x + 1 if x < −1 lim f (x), where f (x) = 3x + 1 if x ≥ −1 x→−1 2 x + 2 if x < −1 lim f (x), where f (x) = 3x + 1 if x ≥ −1 x→0 2x if x < 2 lim f (x), where f (x) = x 2 if x ≥ 2 x→1 2x + 1 if x < −1 if −1 < x < 1 lim f (x), where f (x) = 3 x→−1 2x + 1 if x > 1 2x + 1 if x < −1 if −1 < x < 1 lim f (x), where f (x) = 3 x→1 2x + 1 if x > 1 (2 + h)2 − 4 (1 + h)3 − 1 lim 30. lim h→0 h→0 h √ h h2 x2 + x + 4 − 2 lim √ 32. lim √ h→0 x→0 x2 + x h2 + h + 3 − h + 3 1 1 +t tan 2x lim 2 34. lim t→−2 2 + t x→0 5x Use numerical and graphical evidence to conjecture the value of lim x 2 sin (1/x). Use the Squeeze Theorem to x→0 prove that you are correct: identify the functions f and h, show graphically that f (x) ≤ x 2 sin (1/x) ≤ h(x) and justify lim f (x) = lim h(x).

13. lim

x→0

15. 17. 19. 21. 23. 24. 25. 26. 27.

28. 29. 31. 33. 35.

x→0

x→0

36. Why can’t you use the Squeeze Theorem as in exercise 35 to prove that lim x 2 sec (1/x) = 0? Explore this limit graphically. x→0 √ 37. Use the Squeeze Theorem to prove that lim [ x cos2 (1/x)] = 0. x→0+

Identify √the functions f and h, show graphically that f (x) ≤ x cos2 (1/x) ≤ h(x) for all x > 0, and justify lim f (x) = 0 and lim h(x) = 0. x→0+

1 1 − cos x 43. Given that lim = , quickly evaluate 2 + x→0 x 2 √ 1 − cos x . lim x→0+ x sin x 1 − cos2 x 44. Given that lim = 1, quickly evaluate lim . x→0 x x→0 x2 g(x) if x < a for polynomials g(x) and 45. Suppose f (x) = h(x) if x > a h(x). Explain why lim f (x) = g(a) and determine lim f (x). x→a −

x→0

In exercises 39–42, either find the limit or explain why it does not exist. 39. lim 16 − x 2 40. lim 16 − x 2 + − x→4 x→4 41. lim x 2 + 3x + 2 42. lim x 2 + 3x + 2 x→−2+

x→a +

46. Explain how to determine lim f (x) if g and h are polynomials x→a g(x) if x < a if x = a . and f (x) = c h(x) if x > a 47. Evaluate each limit and justify each step by citing the appropriate theorem or equation. x −2 (a) lim (x 2 − 3x + 1) (b) lim 2 x→2 x→0 x + 1 48. Evaluate each limit and justify each step by citing the appropriate theorem or equation. xe x (a) lim [(x + 1) sin x] (b) lim x→−1 x→1 tan x In exercises 49–52, use the given position function f (t) to find the velocity at time t a. 49. f (t) = t 2 + 2, a = 2

50. f (t) = t 2 + 2, a = 0

51. f (t) = t , a = 0

52. f (t) = t 3 , a = 1

3

√ 53. In Chapter 2, the slope of the tangent √ line to the curve y = x 1+h−1 . Compute the at x = 1 is defined by m = lim h→0 h √ slope m. Graph y = x and the line with slope m through the point (1, 1). 54. In Chapter 2, √ an alternative form for the limit in exercise 53 is x −1 given by lim . Compute this limit. x→1 x − 1 In exercises 55–62, use numerical evidence to conjecture the value of the limit if it exists. Check your answer with your Computer Algebra System (CAS). If you disagree, which one of you is correct? 55. lim (1 + x)1/x x→0+

58. lim x ln x x→0+

x→0+

38. Suppose that f (x) is bounded: that is, there exists a constant M such that | f (x)| ≤ M for all x. Use the Squeeze Theorem to prove that lim x 2 f (x) = 0.

x→−2−

1-24

−1

61. lim tan x→0

1 x

57. lim x −x

56. lim e1/x x→0+ 1 59. lim sin x→0 x 1 62. lim ln x→0 x

2

x→0+

60. lim e1/x x→0

In exercises 63–66, use lim f (x) 2, lim g(x) − 3 and x→a

x→a

lim h(x) 0 to determine the limit, if possible.

x→a

63. lim [2 f (x) − 3g(x)] x→a f (x) + g(x) 65. lim x→a h(x)

64. lim [3 f (x)g(x)] x→a 3 f (x) + 2g(x) 66. lim x→a h(x)

1-25

SECTION 1.4

67. Assume that lim f (x) = L. Use Theorem 3.1 to prove that x→a 3

lim [ f (x)] = L . Also, show that lim [ f (x)] = L . 3

4

x→a

4

x→a

68. How did you work exercise 67? You probably used Theorem 3.1 to work from lim [ f (x)]2 = L 2 to lim [ f (x)]3 = L 3 , and x→a

x→a

then used lim [ f (x)]3 = L 3 to get lim [ f (x)]4 = L 4 . Going one x→a

x→a

step at a time, we should be able to reach lim [ f (x)]n = L n x→a

for any positive integer n. This is the idea of mathematical induction. Formally, we need to show the result is true for a specific value of n = n 0 [we show n 0 = 2 in the text], then assume the result is true for a general n = k ≥ n 0 . If we show that we can get from the result being true for n = k to the result being true for n = k + 1, we have proved that the result is true for any positive integer n. In one sentence, explain why this is true. Use this technique to prove that lim [ f (x)]n = L n

..

Continuity and Its Consequences

97

76. Suppose a state’s income tax code states that tax liability is 12% on the first $20,000 of taxable earnings and 16% on the remainder. Find constants a and b for the tax function a + 0.12x if x ≤ 20,000 T (x) = b + 0.16(x − 20,000) if x > 20,000 such that lim T (x) = 0 and lim T (x) exists. Why is it x→0+

x→20,000

important for these limits to exist? 77. The greatest integer function is denoted by f (x) = [x] and equals the greatest integer that is less than or equal to x. Thus, [2.3] = 2, [−1.2] = −2 and [3] = 3. In spite of this last fact, show that lim [x] does not exist. x→3

78. Investigate the existence of (a) lim [x], (b) (c) lim [2x], and (d) lim (x − [x]). x→1.5

x→1

lim [x],

x→1.5

x→1

x→a

for any positive integer n. 69. Find all the errors in the following incorrect string of equalities: 1 x 1 = lim 2 = lim x lim 2 = 0 · ? = 0. x→0 x x→0 x→0 x x 70. Find all the errors in the following incorrect string of equalities: lim

x→0

0 sin 2x = = 1. x 0 71. Give an example of functions f and g such that lim [ f (x) + g(x)] exists but lim f (x) and lim g(x) do not exist. lim

x→0

x→0

x→0

x→0

72. Give an example of functions f and g such that lim [ f (x) · g(x)] exists but at least one of lim f (x) and lim g(x) x→0

x→0

x→0

does not exist. 73. If lim f (x) exists and lim g(x) does not exist, is it always true x→a

x→a

that lim [ f (x) + g(x)] does not exist? Explain. x→a

74. Is the following true or false? If lim f (x) does not exist, then x→0

lim

x→0

Compute lim T (x); why is this good? Compute

1.4

sin x = 1. Using graphical x sin 2x , and numerical evidence, conjecture the value of lim x→0 x sin 3x sin π x sin x/2 , lim and lim . In general, conjeclim x→0 x→0 x→0 x x x sin cx for any constant c. Given that ture the value of lim x→0 x sin cx = 1 for any constant c = 0, prove that your conlim x→0 cx jecture is correct. x→0

75. Suppose a state’s income tax code states the tax liability on x dollars of taxable income is given by 0.14x if 0 ≤ x < 10,000 . T (x) = 1500 + 0.21x if 10,000 ≤ x x→0+

1. The value x = 0 is called a zero of multiplicity n (n ≥ 1) f (x) for the function f if lim n exists and is nonzero but x→0 x f (x) = 0. Show that x = 0 is a zero of multiplicity 2 lim x→0 x n−1 2 for x , x = 0 is a zero of multiplicity 3 for x 3 and x = 0 is a zero of multiplicity 4 for x 4 . For polynomials, what does multiplicity describe? The reason the definition is not as straightforward as we might like is so that it can apply to nonpolynomial functions, as well. Find the multiplicity of x = 0 for f (x) = sin x; f (x) = x sin x; f (x) = sin x 2 . If you know that x = 0 is a zero of multiplicity m for f (x) and multiplicity n for g(x), what can you say about the multiplicity of x = 0 for f (x) + g(x)? f (x) · g(x)? f (g(x))? 2. We have conjectured that lim

1 does not exist. Explain. f (x)

why is this bad?

EXPLORATORY EXERCISES

lim

x→10,000

T (x);

CONTINUITY AND ITS CONSEQUENCES When you describe something as continuous, just what do you have in mind? For example, if told that a machine has been in continuous operation for the past 60 hours, most of us would interpret this to mean that the machine has been in operation all of that time, without

1-25

SECTION 1.4

67. Assume that lim f (x) = L. Use Theorem 3.1 to prove that x→a 3

lim [ f (x)] = L . Also, show that lim [ f (x)] = L . 3

4

x→a

4

x→a

68. How did you work exercise 67? You probably used Theorem 3.1 to work from lim [ f (x)]2 = L 2 to lim [ f (x)]3 = L 3 , and x→a

x→a

then used lim [ f (x)]3 = L 3 to get lim [ f (x)]4 = L 4 . Going one x→a

x→a

step at a time, we should be able to reach lim [ f (x)]n = L n x→a

for any positive integer n. This is the idea of mathematical induction. Formally, we need to show the result is true for a specific value of n = n 0 [we show n 0 = 2 in the text], then assume the result is true for a general n = k ≥ n 0 . If we show that we can get from the result being true for n = k to the result being true for n = k + 1, we have proved that the result is true for any positive integer n. In one sentence, explain why this is true. Use this technique to prove that lim [ f (x)]n = L n

..

Continuity and Its Consequences

97

76. Suppose a state’s income tax code states that tax liability is 12% on the first $20,000 of taxable earnings and 16% on the remainder. Find constants a and b for the tax function a + 0.12x if x ≤ 20,000 T (x) = b + 0.16(x − 20,000) if x > 20,000 such that lim T (x) = 0 and lim T (x) exists. Why is it x→0+

x→20,000

important for these limits to exist? 77. The greatest integer function is denoted by f (x) = [x] and equals the greatest integer that is less than or equal to x. Thus, [2.3] = 2, [−1.2] = −2 and [3] = 3. In spite of this last fact, show that lim [x] does not exist. x→3

78. Investigate the existence of (a) lim [x], (b) (c) lim [2x], and (d) lim (x − [x]). x→1.5

x→1

lim [x],

x→1.5

x→1

x→a

for any positive integer n. 69. Find all the errors in the following incorrect string of equalities: 1 x 1 = lim 2 = lim x lim 2 = 0 · ? = 0. x→0 x x→0 x→0 x x 70. Find all the errors in the following incorrect string of equalities: lim

x→0

0 sin 2x = = 1. x 0 71. Give an example of functions f and g such that lim [ f (x) + g(x)] exists but lim f (x) and lim g(x) do not exist. lim

x→0

x→0

x→0

x→0

72. Give an example of functions f and g such that lim [ f (x) · g(x)] exists but at least one of lim f (x) and lim g(x) x→0

x→0

x→0

does not exist. 73. If lim f (x) exists and lim g(x) does not exist, is it always true x→a

x→a

that lim [ f (x) + g(x)] does not exist? Explain. x→a

74. Is the following true or false? If lim f (x) does not exist, then x→0

lim

x→0

Compute lim T (x); why is this good? Compute

1.4

sin x = 1. Using graphical x sin 2x , and numerical evidence, conjecture the value of lim x→0 x sin 3x sin π x sin x/2 , lim and lim . In general, conjeclim x→0 x→0 x→0 x x x sin cx for any constant c. Given that ture the value of lim x→0 x sin cx = 1 for any constant c = 0, prove that your conlim x→0 cx jecture is correct. x→0

75. Suppose a state’s income tax code states the tax liability on x dollars of taxable income is given by 0.14x if 0 ≤ x < 10,000 . T (x) = 1500 + 0.21x if 10,000 ≤ x x→0+

1. The value x = 0 is called a zero of multiplicity n (n ≥ 1) f (x) for the function f if lim n exists and is nonzero but x→0 x f (x) = 0. Show that x = 0 is a zero of multiplicity 2 lim x→0 x n−1 2 for x , x = 0 is a zero of multiplicity 3 for x 3 and x = 0 is a zero of multiplicity 4 for x 4 . For polynomials, what does multiplicity describe? The reason the definition is not as straightforward as we might like is so that it can apply to nonpolynomial functions, as well. Find the multiplicity of x = 0 for f (x) = sin x; f (x) = x sin x; f (x) = sin x 2 . If you know that x = 0 is a zero of multiplicity m for f (x) and multiplicity n for g(x), what can you say about the multiplicity of x = 0 for f (x) + g(x)? f (x) · g(x)? f (g(x))? 2. We have conjectured that lim

1 does not exist. Explain. f (x)

why is this bad?

EXPLORATORY EXERCISES

lim

x→10,000

T (x);

CONTINUITY AND ITS CONSEQUENCES When you describe something as continuous, just what do you have in mind? For example, if told that a machine has been in continuous operation for the past 60 hours, most of us would interpret this to mean that the machine has been in operation all of that time, without

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any interruption at all, even for a moment. Mathematicians mean much the same thing when we say that a function is continuous. A function is said to be continuous on an interval if its graph on that interval can be drawn without interruption, that is, without lifting your pencil from the paper. It is helpful for us to first try to see what it is about the functions whose graphs are shown in Figures 1.22a–1.22d that makes them discontinuous (i.e., not continuous) at the point x = a. y

y

x

a x

a

FIGURE 1.22b

FIGURE 1.22a

f (a) is defined, but lim f (x) does

f (a) is not defined (the graph has a hole at x = a).

not exist (the graph has a jump at x = a).

x→a

y

y f (a)

a

x a

FIGURE 1.22c lim f (x) exists and f (a) is defined,

FIGURE 1.22d

but lim f (x) = f (a) (the graph has

lim f (x) does not exist (the

x→a

x→a

REMARK 4.1 The definition of continuity all boils down to the one condition in (iii), since conditions (i) and (ii) must hold whenever (iii) is met. Further, this says that a function is continuous at a point exactly when you can compute its limit at that point by simply substituting in.

a hole at x = a).

x

x→a

function “blows up” at x = a).

This suggests the following definition of continuity at a point.

DEFINITION 4.1 A function f is continuous at x = a when (i) f (a) is defined, (ii) lim f (x) exists and x→a

(iii) lim f (x) = f (a).

Otherwise, f is said to be discontinuous at x = a.

x→a

For most purposes, it is best for you to think of the intuitive notion of continuity that we’ve outlined above. Definition 4.1 should then simply follow from your intuitive understanding of the concept.

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..

Continuity and Its Consequences

99

y

EXAMPLE 4.1

Finding Where a Rational Function Is Continuous

Determine where f (x) = 4

y

x2 2x 3 x1

x 2 + 2x − 3 is continuous. x −1

Solution Note that x 2 + 2x − 3 (x − 1)(x + 3) = x −1 x −1 = x + 3, for x = 1.

f (x) =

x 1

Factoring the numerator. Canceling common factors.

This says that the graph of f is a straight line, but with a hole in it at x = 1, as indicated in Figure 1.23. So, f is discontinuous at x = 1, but continuous elsewhere.

FIGURE 1.23 y=

x 2 + 2x − 3 x −1

EXAMPLE 4.2

Removing a Discontinuity

Make the function from example 4.1 continuous everywhere by redefining it at a single point.

REMARK 4.2 You should be careful not to confuse the continuity of a function at a point with its simply being defined there. A function can be defined at a point without being continuous there. (Look back at Figures 1.22b and 1.22c.)

Solution In example 4.1, we saw that the function is discontinuous at x = 1, since it is undefined there. So, suppose we just go ahead and define it, as follows. Let x 2 + 2x − 3 , if x = 1 g(x) = x −1 a, if x = 1, for some real number a. Notice that g(x) is defined for all x and equals f (x) for all x = 1. Here, we have x 2 + 2x − 3 x→1 x −1 = lim (x + 3) = 4.

lim g(x) = lim

x→1

y

x→1

Observe that if we choose a = 4, we now have that lim g(x) = 4 = g(1)

x→1

y g(x)

4

and so, g is continuous at x = 1. Note that the graph of g is the same as the graph of f seen in Figure 1.23, except that we now include the point (1, 4) (see Figure 1.24). Also, note that there’s a very simple way to write g(x). (Think about this.)

x 1

FIGURE 1.24 y = g(x)

When we can remove a discontinuity by redefining the function at that point, we call the discontinuity removable. Not all discontinuities are removable, however. Carefully examine Figures 1.22a–1.22d and convince yourself that the discontinuities in Figures 1.22a and 1.22c are removable, while those in Figures 1.22b and 1.22d are nonremovable. Briefly, a function f has a removable discontinuity at x = a if lim f (x) exists and either f (a) is x→a undefined or lim f (x) = f (a). x→a

EXAMPLE 4.3

Nonremovable Discontinuities

1 1 . Find all discontinuities of f (x) = 2 and g(x) = cos x x

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y

1-28

Solution You should observe from Figure 1.25a (also, construct a table of function values) that lim

x→0

4

1 does not exist. x2

Hence, f is discontinuous at x = 0. Similarly, observe that lim cos(1/x) does not exist, due to the endless oscillation of x→0

2

x

3

cos(1/x) as x approaches 0 (see Figure 1.25b). In both cases, notice that since the limits do not exist, there is no way to redefine either function at x = 0 to make it continuous there.

3

From your experience with the graphs of some common functions, the following result should come as no surprise.

FIGURE 1.25a 1 y= 2 x y

THEOREM 4.1 All polynomials are continuous everywhere. Additionally, sin x, cos x, tan−1 x and e x √ n are continuous everywhere, x is continuous for all x, when n is odd and for x > 0, when n is even. We also have ln x is continuous for x > 0 and sin−1 x and cos−1 x are continuous for −1 < x < 1.

1

x

0.2

0.2

PROOF 1

We have already established (in Theorem 3.2) that for any polynomial p(x) and any real number a,

FIGURE 1.25b

lim p(x) = p(a),

y = cos (1/x)

x→a

from which it follows that p is continuous at x = a. The rest of the theorem follows from Theorem 3.4 in a similar way. From these very basic continuous functions, we can build a large collection of continuous functions, using Theorem 4.2.

THEOREM 4.2 Suppose that f and g are continuous at x = a. Then all of the following are true: (i) ( f ± g) is continuous at x = a, (ii) ( f · g) is continuous at x = a and (iii) ( f /g) is continuous at x = a if g(a) = 0.

Simply put, Theorem 4.2 says that a sum, difference or product of continuous functions is continuous, while the quotient of two continuous functions is continuous at any point at which the denominator is nonzero.

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101

PROOF (i) If f and g are continuous at x = a, then lim [ f (x) ± g(x)] = lim f (x) ± lim g(x)

x→a

x→a

x→a

= f (a) ± g(a)

From Theorem 3.1. Since f and g are continuous at a.

= ( f ± g)(a), by the usual rules of limits. Thus, ( f ± g) is also continuous at x = a. Parts (ii) and (iii) are proved in a similar way and are left as exercises. y

EXAMPLE 4.4 150

50 5

x 4 − 3x 2 + 2 . x 2 − 3x − 4 Solution Here, f is a quotient of two polynomial (hence continuous) functions. The graph of the function indicated in Figure 1.26 suggests a vertical asymptote at around x = 4, but doesn’t indicate any other discontinuity. From Theorem 4.2, f will be continuous at all x where the denominator is not zero, that is, where Determine where f is continuous, for f (x) =

100

10

Continuity for a Rational Function

x 5 50

10

100 150

FIGURE 1.26 y=

x 4 − 3x 2 + 2 x 2 − 3x − 4

x 2 − 3x − 4 = (x + 1)(x − 4) = 0. Thus, f is continuous for x = −1, 4. (Think about why you didn’t see anything peculiar about the graph at x = −1.) With the addition of the result in Theorem 4.3, we will have all the basic tools needed to establish the continuity of most elementary functions.

THEOREM 4.3 Suppose that lim g(x) = L and f is continuous at L. Then, x→a lim f (g(x)) = f lim g(x) = f (L). x→a

x→a

A proof of Theorem 4.3 is given in Appendix A. Notice that this says that if f is continuous, then we can bring the limit “inside.” This should make sense, since as x → a, g(x) → L and so, f (g(x)) → f (L), since f is continuous at L.

COROLLARY 4.1 Suppose that g is continuous at a and f is continuous at g(a). Then, the composition f ◦ g is continuous at a.

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PROOF From Theorem 4.3, we have

lim ( f ◦ g)(x) = lim f (g(x)) = f

x→a

x→a

lim g(x)

x→a

= f (g(a)) = ( f ◦ g)(a).

EXAMPLE 4.5

Since g is continuous at a.

Continuity for a Composite Function

Determine where h(x) = cos(x 2 − 5x + 2) is continuous. Solution Note that h(x) = f (g(x)), where g(x) = x 2 − 5x + 2 and f (x) = cos x. Since both f and g are continuous for all x, h is continuous for all x, by Corollary 4.1.

y

DEFINITION 4.2 If f is continuous at every point on an open interval (a, b), we say that f is continuous on (a, b). Following Figure 1.27, we say that f is continuous on the closed interval [a, b], if f is continuous on the open interval (a, b) and lim f (x) = f (a)

x→a +

x a

b

and

lim f (x) = f (b).

x→b−

Finally, if f is continuous on all of (−∞, ∞), we simply say that f is continuous. (That is, when we don’t specify an interval, we mean continuous everywhere.)

FIGURE 1.27 f continuous on [a, b]

For many functions, it’s a simple matter to determine the intervals on which the function is continuous. We illustrate this in example 4.6. y

EXAMPLE 4.6

Continuity on a Closed Interval

Determine the interval(s) where f is continuous, for f (x) =

x

4 − x2 > 0

2

FIGURE 1.28 y=

√ 4 − x2

4 − x 2.

Solution First, observe that f is defined only for −2 ≤ x ≤ 2. Next, note that f is the composition of two continuous functions and hence, is continuous for all x for which 4 − x 2 > 0. We show a graph of the function in Figure 1.28. Since

2

2

√

for −2 < x < 2, we have that f is continuous for all x in the interval (−2, 2), by Theorem 4.1 and Corollary 4.1. Finally, √ √ we test the endpoints to see that lim− 4 − x 2 = 0 = f (2) and lim + 4 − x 2 = 0 = f (−2), so that f is continuous x→2

x→−2

on the closed interval [−2, 2].

EXAMPLE 4.7

Interval of Continuity for a Logarithm

Determine the interval(s) where f (x) = ln(x − 3) is continuous. Solution It follows from Theorem 4.1 and Corollary 4.1 that f is continuous whenever (x − 3) > 0 (i.e., for x > 3). Thus, f is continuous on the interval (3, ∞).

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103

The Internal Revenue Service presides over some of the most despised functions in existence. Look up the current Tax Rate Schedules. In 2002, the first few lines (for single taxpayers) looked like: For taxable amount over $0 $6000 $27,950

but not over $6000 $27,950 $ 67,700

your tax liability is 10% 15% 27%

minus $0 $300 $3654

Where do the numbers $300 and $3654 come from? If we write the tax liability T (x) as a function of the taxable amount x (assuming that x can be any real value and not just a whole dollar amount), we have if 0 < x ≤ 6000 0.10x if 6000 < x ≤ 27,950 T (x) = 0.15x − 300 0.27x − 3654 if 27,950 < x ≤ 67,700. Be sure you understand our translation so far. Note that it is important that this be a continuous function: think of the fairness issues that would arise if it were not!

EXAMPLE 4.8

Continuity of Federal Tax Tables

Verify that the federal tax rate function T (x) is continuous at the “joint” x = 27,950. Then, find a to complete the table. (You will find b and c as exercises.) For taxable amount over

but not over

your tax liability is

minus

$67,700 $141,250 $307,050

$141,250 $307,050 —

30% 35% 38.6%

a b c

Solution For T (x) to be continuous at x = 27,950, we must have lim

x→27,950−

T (x) =

lim

x→27,950+

T (x).

Since both functions 0.15x − 300 and 0.27x − 3654 are continuous, we can compute the one-sided limits by substituting x = 27,950. Thus, lim

x→27,950−

and

lim

x→27,950+

T (x) = 0.15(27,950) − 300 = 3892.50

T (x) = 0.27(27,950) − 3654 = 3892.50.

Since the one-sided limits agree and equal the value of the function at that point, T (x) is continuous at x = 27,950. We leave it as an exercise to establish that T (x) is also continuous at x = 6000. (It’s worth noting that the function could be written with equal signs on all of the inequalities; this would be incorrect if the function were discontinuous.) To complete the table, we choose a to get the one-sided limits at x = 67,700 to match. We have lim

x→67,700−

T (x) = 0.27(67,700) − 3654 = 14,625,

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while

lim

x→67,700+

T (x) = 0.30(67,700) − a = 20,310 − a.

So, we set the one-sided limits equal, to obtain 14,625 = 20,310 − a a = 20,310 − 14,625 = 5685.

or

Theorem 4.4 should seem an obvious consequence of our intuitive definition of continuity.

HISTORICAL NOTES

THEOREM 4.4 (Intermediate Value Theorem)

Karl Weierstrass (1815–1897) A German mathematician who proved the Intermediate Value Theorem and several other fundamental results of the calculus. Weierstrass was known as an excellent teacher whose students circulated his lecture notes throughout Europe, because of their clarity and originality. Also known as a superb fencer, Weierstrass was one of the founders of modern mathematical analysis.

Suppose that f is continuous on the closed interval [a, b] and W is any number between f (a) and f (b). Then, there is a number c ∈ [a, b] for which f (c) = W .

Theorem 4.4 says that if f is continuous on [a, b], then f must take on every value between f (a) and f (b) at least once. That is, a continuous function cannot skip over any numbers between its values at the two endpoints. To do so, the graph would need to leap across the horizontal line y = W , something that continuous functions cannot do (see Figure 1.29a). Of course, a function may take on a given value W more than once (see Figure 1.29b). We must point out that, although these graphs make this result seem reasonable, like any other result, Theorem 4.4 requires proof. The proof is more complicated than you might imagine and we must refer you to an advanced calculus text. y

y f (b)

f (b) W f (c)

yW

a

yW

x c

b

f (a)

a

x c1 c2

c3

b

f (a)

FIGURE 1.29a

FIGURE 1.29b

An illustration of the Intermediate Value Theorem

More than one value of c

In Corollary 4.2, we see an immediate and useful application of the Intermediate Value Theorem.

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105

y

COROLLARY 4.2 f(b)

Suppose that f is continuous on [a, b] and f (a) and f (b) have opposite signs [i.e., f (a) · f (b) < 0]. Then, there is at least one number c ∈ (a, b) for which f (c) = 0. (Recall that c is then a zero of f .) y f (x)

a

x c

b f(a)

Notice that Corollary 4.2 is simply the special case of the Intermediate Value Theorem where W = 0 (see Figure 1.30). The Intermediate Value Theorem and Corollary 4.2 are examples of existence theorems; they tell you that there exists a number c satisfying some condition, but they do not tell you what c is.

The Method of Bisections

FIGURE 1.30 Intermediate Value Theorem where c is a zero of f

In example 4.9, we see how Corollary 4.2 can help us locate the zeros of a function.

EXAMPLE 4.9

Finding Zeros by the Method of Bisections

Find the zeros of f (x) = x 5 + 4x 2 − 9x + 3. y 20 10

2

1

x 1

2

10 20

FIGURE 1.31 y = x 5 + 4x 2 − 9x + 3

Solution If f were a quadratic polynomial, you could certainly find its zeros. However, you don’t have any formulas for finding zeros of polynomials of degree 5. The only alternative is to approximate the zeros. A good starting place would be to draw a graph of y = f (x) like the one in Figure 1.31. There are three zeros visible on the graph. Since f is a polynomial, it is continuous everywhere and so, Corollary 4.2 says that there must be a zero on any interval on which the function changes sign. From the graph, you can see that there must be zeros between −3 and −2, between 0 and 1 and between 1 and 2. We could also conclude this by computing say, f (0) = 3 and f (1) = −1. Although we’ve now found intervals that contain zeros, the question remains as to how we can find the zeros themselves. While a rootfinding program can provide an accurate approximation, the issue here is not so much to get an answer as it is to understand how to find one. We suggest a simple yet effective method, called the method of bisections. For the zero between 0 and 1, a reasonable guess might be the midpoint, 0.5. Since f (0.5) ≈ −0.469 < 0 and f (0) = 3 > 0, there must be a zero between 0 and 0.5. Next, the midpoint of [0, 0.5] is 0.25 and f (0.25) ≈ 1.001 > 0, so that the zero is on the interval (0.25, 0.5). We continue in this way to narrow the interval on which there’s a zero until the interval is sufficiently small so that any point in the interval can serve as an adequate approximation to the actual zero. We do this in the following table. a

b

f(a)

f(b)

Midpoint

f (midpoint)

0 0 0.25 0.375 0.375 0.40625 0.40625 0.40625 0.40625

1 0.5 0.5 0.5 0.4375 0.4375 0.421875 0.4140625 0.41015625

3 3 1.001 0.195 0.195 0.015 0.015 0.015 0.015

−1 −0.469 −0.469 −0.469 −0.156 −0.156 −0.072 −0.029 −0.007

0.5 0.25 0.375 0.4375 0.40625 0.421875 0.4140625 0.41015625 0.408203125

−0.469 1.001 0.195 −0.156 0.015 −0.072 −0.029 −0.007 0.004

If you continue this process through 20 more steps, you ultimately arrive at the approximate zero x = 0.40892288, which is accurate to at least eight decimal places.

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This method of bisections is a tedious process, if you’re working it with pencil and paper. It is interesting because it’s reliable and it’s a simple, yet general method for finding approximate zeros. Computer and calculator rootfinding utilities are very useful, but our purpose here is to provide you with an understanding of how basic rootfinding works. We discuss a more powerful method for finding roots in Chapter 3.

EXERCISES 1.4 WRITING EXERCISES 1. Think about the following “real-life” functions, each of which is a function of the independent variable time: the height of a falling object, the velocity of an object, the amount of money in a bank account, the cholesterol level of a person, the heart rate of a person, the amount of a certain chemical present in a test tube and a machine’s most recent measurement of the cholesterol level of a person. Which of these are continuous functions? For each function you identify as discontinuous, what is the real-life meaning of the discontinuities? 2. Whether a process is continuous or not is not always clear-cut. When you watch television or a movie, the action seems to be continuous. This is an optical illusion, since both movies and television consist of individual “snapshots” that are played back at many frames per second. Where does the illusion of continuous motion come from? Given that the average person blinks several times per minute, is our perception of the world actually continuous? (In what cognitive psychologists call temporal binding, the human brain first decides whether a stimulus is important enough to merit conscious consideration. If so, the brain “predates” the stimulus so that the person correctly identifies when the stimulus actually occurred.) 3. When you sketch the graph of the parabola y = x 2 with pencil or pen, is your sketch (at the molecular level) actually the graph of a continuous function? Is your calculator or computer’s graph actually the graph of a continuous function? On many calculators, you have the option of a connected or disconnected graph. At the pixel level, does a connected graph show the graph of a function? Does a disconnected graph show the graph of a continuous function? Do we ever have problems correctly interpreting a graph due to these limitations? In the exercises in section 1.7, we examine one case where our perception of a computer graph depends on which choice is made.

In exercises 1–6, use the given graph to identify all discontinuities of the functions. y

1. 5

x 5

y

2. 5

x 5

y

3. 5

x 6

4. For each of the graphs in Figures 1.22a–1.22d, describe (with an example) what the formula for f (x) might look like to produce the given discontinuity.

1-35

..

SECTION 1.4

y

4.

107

In exercises 13–24, find all discontinuities of f (x). For each discontinuity that is removable, define a new function that removes the discontinuity.

5

x 5

x −1 x2 − 1 4x 15. f (x) = 2 x +4 17. f (x) = x 2 tan x

4x x2 + x − 2 3x 16. f (x) = 2 x − 2x − 4 18. f (x) = x cot x

13. f (x) =

14. f (x) =

19. f (x) = x ln x 2

20. f (x) = e−4/x sin x 22. f (x) = x 1

21. f (x) =

2x x2

2

if x < 1 if x ≥ 1

3x − 1 if x ≤ −1 23. f (x) = x 2 + 5x if −1 < x < 1 3 3x if x ≥ 1 if x < 0 2x if 0 < x ≤ π 24. f (x) = sin x x − π if x > π

y

5.

Continuity and Its Consequences

5

if x = 0 if x = 0

x 5

In exercises 25–32, determine the intervals on which f (x) is continuous. √ √ 25. f (x) = x + 3 26. f (x) = x 2 − 4 √ 28. f (x) = (x − 1)3/2 27. f (x) = 3 x + 2 1 29. f (x) = sin(x 2 + 2) 30. f (x) = cos x 31. f (x) = ln(x + 1) 32. f (x) = ln(4 − x 2 )

y

6. 5

x 5

In exercises 7–12, explain why each function is discontinuous at the given point by indicating which of the three conditions in Definition 4.1 are not met. x x2 − 1 7. f (x) = at x = 1 8. f (x) = at x = 1 x −1 x −1 1 9. f (x) = sin at x = 0 10. f (x) = e1/x at x = 0 x 2 if x < 2 x if x = 2 at x = 2 11. f (x) = 3 3x − 2 if x > 2 2 x if x < 2 12. f (x) = at x = 2 3x − 2 if x > 2

In exercises 33–35, determine values of a and b that make the given function continuous. 2 sin x if x < 0 33. f (x) = a x if x = 0 b cos x if x > 0 x if x < 0 ae +x1 −1 if 0 ≤ x ≤ 2 34. f (x) = sin 2 2 x − x + b if x > 2 a(tan−1 x + 2) if x < 0 if 0 ≤ x ≤ 3 35. f (x) = 2ebx + 1 ln(x − 2) + x 2 if x > 3 36. Prove Corollary 4.1. 37. Suppose that a state’s income tax code states that the tax liability on x dollars of taxable income is given by if x ≤ 0 0 if 0 < x < 10,000 T (x) = 0.14x c + 0.21x if 10,000 ≤ x. Determine the constant c that makes this function continuous for all x. Give a rationale why such a function should be continuous.

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38. Suppose a state’s income tax code states that tax liability is 12% on the first $20,000 of taxable earnings and 16% on the remainder. Find constants a and b for the tax function 0 if x ≤ 0 if 0 < x ≤ 20,000 T (x) = a + 0.12x b + 0.16(x − 20,000) if x > 20,000 such that T (x) is continuous for all x. 39. In example 4.8, find b and c to complete the table. 40. In example 4.8, show that T (x) is continuous for x = 6000. In exercises 41–46, use the Intermediate Value Theorem to verify that f (x) has a zero in the given interval. Then use the method of bisections to find an interval of length 1/32 that contains the zero. 41. f (x) = x 2 − 7, [2, 3] 42. f (x) = x 3 − 4x − 2, [2, 3]

44. f (x) = x 3 − 4x − 2, [−2, −1] 45. f (x) = cos x − x, [0, 1] 46. f (x) = e x + x, [−1, 0] A function is continuous from the right at x a if lim f (x) f (a). In exercises 47–50, determine whether f (x) x→a

is continuous from the right at x 2. x2 if x < 2 47. f (x) = 3x − 1 if x ≥ 2 2 if x < 2 x if x = 2 48. f (x) = 3 3x − 3 if x > 2 x2 if x ≤ 2 49. f (x) = 3x − 3 if x > 2 50. f (x) =

x2 3x − 2

for some function g(T ). Explain why it is reasonable that f (T ) be continuous. Determine a function g(T ) such that 0 ≤ g(T ) ≤ 100 for 30 ≤ T ≤ 34 and the resulting function f (T ) is continuous. [Hint: It may help to draw a graph first and make g(T ) linear.] x 2 , if x = 0 and g(x) = 2x, show that 54. If f (x) = 4, if x = 0 lim f (g(x)) = f lim g(x) . x→0

55. If you push on a large box resting on the ground, at first nothing will happen because of the static friction force that opposes motion. If you push hard enough, the box will start sliding, although there is again a friction force that opposes the motion. Suppose you are given the following description of the friction force. Up to 100 pounds, friction matches the force you apply to the box. Over 100 pounds, the box will move and the friction force will equal 80 pounds. Sketch a graph of friction as a function of your applied force based on this description. Where is this graph discontinuous? What is significant physically about this point? Do you think the friction force actually ought to be continuous? Modify the graph to make it continuous while still retaining most of the characteristics described. 400 , we have f (−1) > 0 and f (2) < 0. x Does the Intermediate Value Theorem guarantee a zero of f (x) between x = −1 and x = 2? What happens if you try the method of bisections?

56. For f (x) = 2x −

if x < 2 if x > 2

51. Define what it means for a function to be continuous from the left at x = a and determine which of the functions in exercises 47–50 are continuous from the left at x = 2. g(x) 52. Suppose that f (x) = and h(a) = 0. Determine whether h(x) each of the following statements is always true, always false, or maybe true/maybe false. Explain. (a) lim f (x) does not exist. (b) f (x) is discontinuous at x = a.

53. The sex of newborn Mississippi alligators is determined by the temperature of the eggs in the nest. The eggs fail to develop unless the temperature is between 26◦ C and 36◦ C. All eggs between 26◦ C and 30◦ C develop into females, and eggs between 34◦ C and 36◦ C develop into males. The percentage of females decreases from 100% at 30◦ C to 0% at 34◦ C. If f (T ) is the percentage of females developing from an egg at T ◦ C, then 100 if 26 ≤ T ≤ 30 f (T ) = g(T ) if 30 < T < 34 0 if 34 ≤ T ≤ 36,

x→0

43. f (x) = x 3 − 4x − 2, [−1, 0]

1-36

x→a

57. On Monday morning, a saleswoman leaves on a business trip at 7:13 A.M. and arrives at her destination at 2:03 P.M. The following morning, she leaves for home at 7:17 A.M. and arrives at 1:59 P.M. The woman notices that at a particular stoplight along the way, a nearby bank clock changes from 10:32 A.M. to 10:33 A.M. on both days. Therefore, she must have been at the same location at the same time on both days. Her boss doesn’t believe that such an unlikely coincidence could occur. Use the Intermediate Value Theorem to argue that it must be true that at some point on the trip, the saleswoman was at exactly the same place at the same time on both Monday and Tuesday. 58. Suppose you ease your car up to a stop sign at the top of a hill. Your car rolls back a couple of feet and then you drive through

1-37

SECTION 1.4

the intersection. A police officer pulls you over for not coming to a complete stop. Use the Intermediate Value Theorem to argue that there was an instant in time when your car was stopped (in fact, there were at least two). What is the difference between this stopping and the stopping that the police officer wanted to see? 59. Suppose a worker’s salary starts at $40,000 with $2000 raises every 3 months. Graph the salary function s(t); why is it discon2000 tinuous? How does the function f (t) = 40,000 + t (t in 3 months) compare? Why might it be easier to do calculations with f (t) than s(t)? 60. Prove the final two parts of Theorem 4.2. 61. Suppose that f (x) is a continuous function with consecutive zeros at x = a and x = b; that is, f (a) = f (b) = 0 and f (x) = 0 for a < x < b. Further, suppose that f (c) > 0 for some number c between a and b. Use the Intermediate Value Theorem to argue that f (x) > 0 for all a < x < b. 62. Use the method of bisections to estimate the other two zeros in example 4.9. 63. Suppose that f (x) is continuous at x = 0. Prove that lim x f (x) = 0. x→0

64. The converse of exercise 63 is not true. That is, the fact lim x f (x) = 0 does not guarantee that f (x) is continuous at x→0

x = 0. Find a counterexample; that is, find a function f such that lim x f (x) = 0 and f (x) is not continuous at x = 0. x→0

65. If f (x) is continuous at x = a, prove that g(x) = | f (x)| is continuous at x = a.

..

Continuity and Its Consequences

109

x = 1. For the method of bisections, we guess the midpoint, x = 0.5. Is there any reason to suspect that the solution is actually closer to x = 0 than to x = 1? Using the function values f (0) = −1 and f (1) = 5, devise your own method of guessing the location of the solution. Generalize your method to using f (a) and f (b), where one function value is positive and one is negative. Compare your method to the method of bisections on the problem x 3 + 5x − 1 = 0; for both methods, stop when you are within 0.001 of the solution, x ≈ 0.198437. Which method performed better? Before you get overconfident in your method, compare the two methods again on x 3 + 5x 2 − 1 = 0. Does your method get close on the first try? See if you can determine graphically why your method works better on the first problem. 2. You have probably seen the turntables on which luggage rotates at the airport. Suppose that such a turntable has two long straight parts with a semicircle on each end (see the figure). We will model the left/right movement of the luggage. Suppose the straight part is 40 ft long, extending from x = −20 to x = 20. Assume that our luggage starts at time t = 0 at location x = −20, and that it takes 60 s for the luggage to reach x = 20. Suppose the radius of the circular portion is 5 ft and it takes the luggage 30 s to complete the half-circle. We model the straight-line motion with a linear function x(t) = at + b. Find constants a and b so that x(0) = −20 and x(60) = 20. For the circular motion, we use a cosine (Why is this a good choice?) x(t) = 20 + d · cos (et + f ) for constants d, e and f . The requirements are x(60) = 20 (since the motion is continuous), x(75) = 25 and x(90) = 20. Find values of d, e and f to make this work. Find equations for the position of the luggage along the backstretch and the other semicircle. What would the motion be from then on?

66. Determine whether the converse of exercise 65 is true. That is, if | f (x)| is continuous at x = a, is it necessarily true that f (x) must be continuous at x = a? 67. Let f (x) be a continuous function for x ≥ a and define h(x) = max f (t). Prove that h(x) is continuous for x ≥ a. a≤t≤x

Would this still be true without the assumption that f (x) is continuous? 68. Graph f (x) = tinuities.

sin |x 3 − 3x 2 + 2x| and determine all disconx 3 − 3x 2 + 2x Luggage carousel

EXPLORATORY EXERCISES 1. In the text, we discussed the use of the method of bisections to find an approximate solution of equations such as f(x) = x 3 + 5x −1 = 0. We can start by noticing that f (0) = −1 and f (1) = 5. Since f (x) is continuous, the Intermediate Value Theorem tells us that there is a solution between x = 0 and

3. Determine all x’s for which each function is continuous. 0 if x is irrational , f (x) = x if x is rational 2 x + 3 if x is irrational g(x) = 4x if x is rational and cos 4x if x is irrational . h(x) = sin 4x if x is rational

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1-38

LIMITS INVOLVING INFINITY; ASYMPTOTES y

10

f(x)

In this section, we revisit some old limit problems to give more informative answers and examine some related questions.

x x

3

x

3

EXAMPLE 5.1 Examine lim

x→0

f(x) 10

FIGURE 1.32 1 1 = ∞ and lim = −∞ lim x→0+ x x→0− x x

1 x

0.1 0.01 0.001 0.0001 0.00001

10 100 1000 10,000 100,000

x −0.1 −0.01 −0.001 −0.0001 −0.00001

A Simple Limit Revisited

1 . x

Solution Of course, we can draw a graph (see Figure 1.32) and compute a table of function values easily, by hand. (See the tables in the margin.) 1 1 While we say that the limits lim+ and lim− do not exist, the behavior of the x→0 x x→0 x 1 function is clearly quite different for x > 0 than for x < 0. Specifically, as x → 0+ , x 1 increases without bound, while as x → 0− , decreases without bound. To x communicate more about the behavior of the function near x = 0, we write 1 =∞ x

(5.1)

1 = −∞. x

(5.2)

lim+

x→0

and

1 x

lim

x→0−

1 approaches the vertical line x = 0, as x x → 0, as seen in Figure 1.32. When this occurs, we say that the line x = 0 is a vertical asymptote. It is important to note that while the limits (5.1) and (5.2) do not exist, we say that they “equal” ∞ and −∞, respectively, only to be specific as to why they do not exist. Finally, in view of the one-sided limits (5.1) and (5.2), we say that

Graphically, this says that the graph of y =

−10 −100 −1000 −10,000 −100,000

lim

x→0

REMARK 5.1 It may at first seem 1 x does not exist and then to write 1 lim = ∞. Note that since x→0+ x ∞ is not a real number, there is no contradiction here. (When we say that a limit “does not exist,” we are saying that there is no real number L that the function values are approaching.) We say that 1 lim = ∞ to indicate that as x→0+ x + x → 0 , the function values are increasing without bound. contradictory to say that lim

x→0+

EXAMPLE 5.2

1 does not exist. x

A Function Whose One-Sided Limits Are Both Infinite

1 Evaluate lim 2 . x→0 x Solution The graph (in Figure 1.33) seems to indicate a vertical asymptote at x = 0. A table of values is easily constructed by hand (see the accompanying tables). x

1 x2

x

1 x2

0.1 0.01 0.001 0.0001 0.00001

100 10,000 1 × 106 1 × 108 1 × 1010

−0.1 −0.01 −0.001 −0.0001 −0.00001

100 10,000 1 × 106 1 × 108 1 × 1010

1-39

..

SECTION 1.5

y f (x)

lim+

1 =∞ x2

lim

1 = ∞. x2

x→0

4

and

x→0−

Since both one-sided limits agree (i.e., both tend to ∞), we say that

2

lim

x

x→0

x

x

3

3

FIGURE 1.33 lim

111

From this, we can see that

f(x)

x→0

Limits Involving Infinity; Asymptotes

1 =∞ x2

1 = ∞. x2

This one concise statement says that the limit does not exist, but also that f (x) has a vertical asymptote at x = 0 where f (x) → ∞ as x → 0 from either side.

REMARK 5.2 Mathematicians try to convey as much information as possible with as few symbols as 1 1 possible. For instance, we prefer to say lim 2 = ∞ rather than lim 2 does not x→0 x x→0 x exist, since the first statement not only says that the limit does not exist, but also says 1 that 2 increases without bound as x approaches 0, with x > 0 or x < 0. x

y

EXAMPLE 5.3

10

f(x)

Evaluate lim

x→5

5 x 5

x

x 10

5 f(x)

10

1 = ∞ and (x − 5)3 1 lim = −∞ x→5− (x − 5)3

lim

1 . (x − 5)3

Solution In Figure 1.34, we show a graph of the function. From the graph, you should get a pretty clear idea that there’s a vertical asymptote at x = 5 and just how the function is blowing up there (to ∞ from the right side and to −∞ from the left). You can verify this behavior algebraically, by noticing that as x → 5, the denominator approaches 0, while the numerator approaches 1. This says that the fraction grows large in absolute value, without bound as x → 5. Specifically, as x → 5+ ,

FIGURE 1.34 x→5+

A Case Where Infinite One-Sided Limits Disagree

(x − 5)3 → 0

and

(x − 5)3 > 0.

We indicate the sign of each factor by printing a small “+” or “−” sign above or below each one. This enables you to see the signs of the various terms at a glance. In this case, we have +

1 lim = ∞. x→5+ (x − 5)3

Since (x − 5)3 > 0, for x > 5.

+

Likewise, as

x → 5− ,

(x − 5)3 → 0

and

(x − 5)3 < 0.

In this case, we have +

1 lim− = −∞. x→5 (x − 5)3

Since (x − 5)3 < 0, for x < 5.

−

Finally, we say that

1 does not exist, x→5 (x − 5)3 lim

since the one-sided limits are different.

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1-40

Learning from the lessons of examples 5.1, 5.2 and 5.3, you should recognize that if the denominator tends to 0 and the numerator does not, then the limit in question does not exist. In this event, we can determine whether the limit tends to ∞ or −∞ by carefully examining the signs of the various factors.

EXAMPLE 5.4 y

Evaluate lim

x→−2

10 f (x) 5 x

x

4

x

Another Case Where Infinite One-Sided Limits Disagree

x +1 . (x − 3)(x + 2)

Solution First, notice from the graph of the function shown in Figure 1.35 that there appears to be a vertical asymptote at x = −2. Further, the function appears to tend to ∞ as x → −2+ , and to −∞ as x → −2− . You can verify this behavior, by observing that −

4

x +1 lim =∞ x→−2+ (x − 3) (x + 2)

5

−

f (x)

10

+

Since (x + 1) < 0, (x − 3) < 0 and (x + 2) > 0, for −2 < x < −1.

−

and

FIGURE 1.35 x +1 does not exist. lim x→−2 (x − 3)(x + 2)

x +1 lim = −∞. x→−2− (x − 3) (x + 2) −

−

Since (x + 1) < 0, (x − 3) < 0 and (x + 2) < 0, for x < −2.

So, we can see that x = −2 is indeed a vertical asymptote and that x +1 does not exist. x→−2 (x − 3)(x + 2) lim

EXAMPLE 5.5 y

A Limit Involving a Trigonometric Function

Evaluate limπ tan x. x→ 2

Solution Notice from the graph of the function shown in Figure 1.36 that there appears π to be a vertical asymptote at x = . 2 You can verify this behavior by observing that +

q

q

p

sin x limπ − tan x = limπ − =∞ cos x x→ 2 x→ 2

x

+

w

+

and

limπ + tan x = limπ +

x→ 2

FIGURE 1.36 y = tan x

x→ 2

sin x = −∞. cos x −

Since sin x > 0 and cos x > 0 π for 0 < x < . 2 Since sin x > 0 and cos x < 0 π < x < π. for 2

π So, we can see that x = is indeed a vertical asymptote and that 2 limπ tan x does not exist. x→ 2

Limits at Infinity We are also interested in examining the limiting behavior of functions as x increases without bound (written x → ∞) or as x decreases without bound (written x → −∞).

1-41

SECTION 1.5

y

Returning to f (x) =

10 f (x)

3

x

3

f(x)

1 1 = 0 and lim =0 x→−∞ x x

lim

x→−∞

1 = 0. x

1 . x

1 Solution We show a graph of y = f (x) in Figure 1.38. Since as x → ±∞, → 0, x we get that 1 lim 2 − =2 x→∞ x 1 and lim 2 − = 2. x→−∞ x

8 6 4 x

Thus, the line y = 2 is a horizontal asymptote.

x f (x) x

2

Finding Horizontal Asymptotes

Look for any horizontal asymptotes of f (x) = 2 −

y

f(x)

Similarly,

EXAMPLE 5.6

FIGURE 1.37 lim

113

Notice that in Figure 1.37, the graph appears to approach the horizontal line y = 0, as x → ∞ and as x → −∞. In this case, we call y = 0 a horizontal asymptote.

10

x→∞

Limits Involving Infinity; Asymptotes

1 1 , we can see that as x → ∞, → 0. In view of this, we write x x 1 lim = 0. x→∞ x

x x

..

2 2

1 , for any positive rational power t, xt 1 as x → ±∞, is largely the same as we observed for f (x) = . x As you can see in Theorem 5.1, the behavior of

FIGURE 1.38

1 = 2 and lim 2 − x→∞ x 1 lim 2 − =2 x→−∞ x

THEOREM 5.1 For any rational number t > 0, lim

x→±∞

1 = 0, xt

where for the case where x → −∞, we assume that t =

REMARK 5.3 All of the usual rules for limits stated in Theorem 3.1 also hold for limits as x → ±∞.

p where q is odd. q

A proof of Theorem 5.1 is given in Appendix A. Be sure that the following argument 1 makes sense to you: for t > 0, as x → ∞, we also have x t → ∞, so that t → 0. x In Theorem 5.2, we see that the behavior of a polynomial at infinity is easy to determine.

THEOREM 5.2 For a polynomial of degree n > 0, pn (x) = an x n + an−1 x n−1 + · · · + a0 , we have ∞, if an > 0 lim pn (x) = . −∞, if an < 0 x→∞

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1-42

PROOF lim pn (x) = lim (an x n + an−1 x n−1 + · · · + a0 ) x→∞ an−1 a0 = lim x n an + + ··· + n x→∞ x x = ∞, an−1 a0 if an > 0, since lim an + + · · · + n = an x→∞ x x and lim x n = ∞. The result is proved similarly for an < 0. We have

x→∞

x→∞

Observe that you can make similar statements regarding the value of lim pn (x), but x→−∞

be careful: the answer will change depending on whether n is even or odd. (We leave this as an exercise.) In example 5.7, we again see the need for caution when applying our basic rules for limits (Theorem 3.1), which also apply to limits as x → ∞ or as x → −∞.

y

EXAMPLE 5.7

4 x x

10

10 f (x) 4

A Limit of a Quotient That Is Not the Quotient of the Limits

5x − 7 . 4x + 3 Solution You might be tempted to write lim (5x − 7) 5x − 7 x→∞ lim = x→∞ 4x + 3 lim (4x + 3)

Evaluate lim

x→∞

x→∞

FIGURE 1.39 lim

x→∞

5 5x − 7 = 4x + 3 4

x

5x − 7 4x 3

10 100 1000 10,000 100,000

1 1.223325 1.247315 1.249731 1.249973

This is an incorrect use of Theorem 3.1 (iv), since the limits in the numerator and the denominator do not exist.

∞ = This is incorrect! (5.3) = 1. ∞ The graph in Figure 1.39 and some function values (see the accompanying table) suggest that the conjectured value of 1 is incorrect. Recall that the limit of a quotient is the quotient of the limits only when both limits exist (and the limit in the denominator is nonzero). Since both the limit in the denominator and that in the numerator tend to ∞, the limits do not exist. Further, when a limit looks like ∞ , the actual value of the limit can be anything at ∞ all. For this reason, we call ∞ an indeterminate form, meaning that the value of the ∞ expression cannot be determined solely by noticing that both numerator and denominator tend to ∞. Rule of Thumb: When faced with the indeterminate form ∞ in calculating the ∞ limit of a rational function, divide numerator and denominator by the highest power of x appearing in the denominator. Here, we have Multiply numerator and 5x − 7 5x − 7 (1/x) lim = lim · 1 denominator by . x→∞ 4x + 3 x→∞ 4x + 3 (1/x) x 5 − 7/x 4 + 3/x lim (5 − 7/x)

= lim

x→∞

=

x→∞

lim (4 + 3/x)

Multiply through by

1 . x

By Theorem 3.1 (iv).

x→∞

5 = 1.25, 4 which is consistent with what we observed both graphically and numerically earlier. =

1-43

SECTION 1.5

..

Limits Involving Infinity; Asymptotes

115

In example 5.8, we apply our rule of thumb to a common limit problem.

EXAMPLE 5.8

Finding Slant Asymptotes

4x + 5 and find any slant asymptotes. x→∞ −6x 2 − 7x Solution As usual, we first examine a graph (see Figure 1.40a). Note that here, the graph appears to tend to −∞ as x → ∞. Further, observe that outside of the interval [−2, 2], the graph looks very much like a straight line. If we look at the graph in a somewhat larger window, this linearity is even more apparent (see Figure 1.40b). 3

Evaluate lim

y

y

20

6

x

6

6

x

20

20

20

6

FIGURE 1.40a 4x 3 + 5 y= −6x 2 − 7x

FIGURE 1.40b y=

Using our rule of thumb, we have 4x 3 + 5 (1/x 2 ) 4x 3 + 5 lim = lim · x→∞ −6x 2 − 7x x→∞ −6x 2 − 7x (1/x 2 ) 4x + 5/x 2 x→∞ −6 − 7/x = −∞, = lim

4x 3 + 5 −6x 2 − 7x

Multiply numerator and 1 denominator by 2 . x Multiply through by

1 . x2

since as x → ∞, the numerator tends to ∞ and the denominator tends to −6. To further explain the behavior seen in Figure 1.40b, we perform a long division. We have 4x 3 + 5 2 7 5 + 49/9x =− x+ + . −6x 2 − 7x 3 9 −6x 2 − 7x Since the third term in this expansion tends to 0 as x → ∞, the function values approach those of the linear function 2 7 − x+ , 3 9 as x → ∞. For this reason, we say that the function has a slant (or oblique) asymptote. That is, instead of approaching a vertical or horizontal line, as happens with vertical or horizontal asymptotes, the graph is approaching the slanted straight line 2 7 y = − x + . (This is the behavior we’re seeing in Figure 1.40b.) 3 9 Limits involving exponential functions are very important in many applications.

116

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1-44

y

EXAMPLE 5.9 10

Two Limits of an Exponential Function

Evaluate lim− e1/x and lim+ e1/x . x→0

5

10

x

5

5

10

5

x→0

Solution A computer-generated graph is shown in Figure 1.41a. Although it is an unusual looking graph, it appears that the function values are approaching 0, as x approaches 0 from the left and tend to infinity as x approaches 0 from the right. To 1 verify this, recall that lim− = −∞ and lim e x = 0. (See Figure 1.41b for a graph x→−∞ x→0 x of y = e x .) Combining these results, we get

FIGURE 1.41a

lim e1/x = 0.

y = e1/x

x→0−

y

Similarly, lim+ x→0

30

we have

1 = ∞ and lim e x = ∞. (Again, see Figure 1.41b.) In this case, x→∞ x lim e1/x = ∞.

20

x→0+

10 4

As we see in example 5.10, inverse trigonometric functions may have horizontal asymptotes.

x

2

2

4

FIGURE 1.41b

EXAMPLE 5.10

y = ex

Evaluate lim tan−1 x and lim tan−1 x.

y

x→∞

1

4

Two Limits of an Inverse Trigonometric Function

x

2

2

4

1

FIGURE 1.42a y = tan−1 x y

x→−∞

Solution The graph of y = tan−1 x (shown in Figure 1.42a) suggests a horizontal asymptote of about y = −1.5 as x → −∞ and about y = 1.5 as x → ∞. We can be more precise with this, as follows. For lim tan−1 x, we are looking for the angle that θ x→∞ π π must approach, with − < θ < , such that tan θ tends to ∞. Referring to the graph 2 2 π− of y = tan x in Figure 1.42b, we see that tan x tends to ∞ as x approaches . 2 π+ Likewise, tan x tends to −∞ as x approaches − , so that 2 π π lim tan−1 x = and lim tan−1 x = − . x→∞ x→−∞ 2 2 In example 5.11, we consider a model of the size of an animal’s pupils. Recall that in bright light, pupils shrink to reduce the amount of light entering the eye, while in dim light, pupils dilate to allow in more light. (See the chapter introduction.)

−π 2

π 2

x

EXAMPLE 5.11

Finding the Size of an Animal’s Pupils

Suppose that the diameter of an animal’s pupils is given by f (x) mm, where x is the 160x −0.4 + 90 intensity of light on the pupils. If f (x) = , find the diameter of the pupils 4x −0.4 + 15 with (a) minimum light and (b) maximum light. FIGURE 1.42b y = tan x

Solution For part (a), notice that f (0) is undefined, since 0−0.4 indicates a division by 0. We therefore consider the limit of f (x) as x approaches 0, but we compute a

1-45

SECTION 1.5

y

..

Limits Involving Infinity; Asymptotes

117

one-sided limit since x cannot be negative. A computer-generated graph of y = f (x) with 0 ≤ x ≤ 10 is shown in Figure 1.43a. It appears that the y-values approach 20 as x approaches 0. To compute the limit, we multiply numerator and denominator by x 0.4 (to eliminate the negative exponents). We then have

20 15

lim+

10

x→0

160x −0.4 + 90 160x −0.4 + 90 x 0.4 = lim+ · −0.4 x→0 4x + 15 4x −0.4 + 15 x 0.4 = lim+

5

x→0

x 2

4

6

8

10

FIGURE 1.43a y = f (x) y

160 + 90x 0.4 160 = = 40 mm. 4 + 15x 0.4 4

This limit does not seem to match our graph, but notice that Figure 1.43a shows a gap near x = 0. In Figure 1.43b, we have zoomed in so that 0 ≤ x ≤ 0.1. Here, a limit of 40 looks more reasonable. For part (b), we consider the limit as x tends to ∞. From Figure 1.43a, it appears that the graph has a horizontal asymptote at a value close to y = 10. We compute the limit

40

160x −0.4 + 90 90 = = 6 mm. x→∞ 4x −0.4 + 15 15 lim

30

So, the pupils have a limiting size of 6 mm, as the intensity of light tends to ∞.

20

In our final example, we consider the velocity of a falling object.

10 x 0.02 0.04 0.06 0.08 0.1

FIGURE 1.43b

EXAMPLE 5.12

Finding the Limiting Velocity of a Falling Object

The velocity in ft/s of a falling object is modeled by

y = f (x)

√

32 1 − e−2t 32k √ v(t) = − , k 1 + e−2t 32k where k is a constant that depends upon the size and shape of the object and the density of the air. Find the limiting velocity of the object, that is, find lim v(t) and compare limiting t→∞ velocities for skydivers with k = 0.00016 (head first) and k = 0.001 (spread eagle). Solution Observe that the only place√that t appears in the expression for v(t) is in the √ −2t 32k −2t 32k two identical exponential terms: e . Also notice that lim e = 0, since lim e x = 0. We then have x→−∞

t→∞

√ 32 1 − e−2t 32k √ lim v(t) = lim − t→∞ t→∞ k 1 + e−2t 32k 1 − lim e−2t √32k 32 32 32 1 − 0 t→∞ =− √ =− =− ft/s, k k 1+0 k 1 + lim e−2t 32k

t→∞

where the negative sign indicates a downward direction. So, with k = 0.00016, the 32 limiting velocity is − 0.00016 ≈ −447 ft/s (about 300 mph!), and with k = 0.001, the 32 limiting velocity is − 0.001 ≈ −179 ft/s (about 122 mph).

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1-46

EXERCISES 1.5 WRITING EXERCISES 1. It may seem odd that we use ∞ in describing limits but do not count ∞ as a real number. Discuss the existence of ∞: is it a number or a concept? 2. In example 5.7, we dealt with the “indeterminate form” ∞ . ∞ Thinking of a limit of ∞ as meaning “getting very large” and a limit of 0 as meaning “getting very close to 0,” explain why the following are indeterminate forms: ∞ , 0 , ∞ − ∞, and ∞ 0 ∞ · 0. Determine what the following non-indeterminate forms represent: ∞ + ∞, −∞ − ∞, ∞ + 0 and 0/∞. 3. On your computer or calculator, graph y = 1/(x − 2) and look for the horizontal asymptote y = 0 and the vertical asymptote x = 2. Most computers will draw a vertical line at x = 2 and will show the graph completely flattening out at y = 0 for large x’s. Is this accurate? misleading? Most computers will compute the locations of points for adjacent x’s and try to connect the points with a line segment. Why might this result in a vertical line at the location of a vertical asymptote?

In exercises 5–24, determine each limit (answer as appropriate, with a number, ∞ , − ∞ or does not exist). −x −2/3 5. lim √ 6. lim (x 2 − 2x − 3) x→2− x→−1− 4 − x2 −x 2x 2 − x + 1 7. lim √ 8. lim x→−∞ x→∞ 4x 2 − 3x − 1 4 + x2 x3 − 2 2x 2 − 1 9. lim 10. lim x→∞ 3x 2 + 4x − 1 x→∞ 4x 3 − 5x − 1 12. lim ln 2x 11. lim ln 2x x→0+

x→∞

13. lim e

−2/x

14. lim e−2/x

x→0+

x→∞

−1

15. lim cot x→∞

17. lim e

16. lim sec−1 x

x

x→∞

2x−1

18. lim e1/x

x→∞

20. lim (e−3x cos 2x)

19. lim sin 2x x→∞

x→∞

ln(2 + e ) ln(1 + e x ) 23. lim e− tan x 3x

21. lim

x→∞

4. Many students learn that asymptotes are lines that the graph gets closer and closer to without ever reaching. This is true for many asymptotes, but not all. Explain why vertical asymptotes are never reached or crossed. Explain why horizontal or slant asymptotes may, in fact, be crossed any number of times; draw one example.

In exercises 1–4, determine each limit (answer as appropriate, with a number, ∞ , − ∞ or does not exist). 1. (a) lim

x→1−

(c) lim

x→1

2. (a)

lim

x→−1−

x→−1

3. (a) lim

x→2−

(c) lim

x→2

x→1+

1 − 2x x2 − 1

1 − 2x x2 − 1

x −4 x 2 − 4x + 4

(b)

lim

x→−1+

1 − 2x x2 − 1

(b) lim

x→2+

x −4 x 2 − 4x + 4

x −4 x 2 − 4x + 4

lim

x→−1

1−x (x + 1)2

1−x (x + 1)2

x→π/2

lim

x→−1+

1−x (x + 1)2

x→∞

24. lim tan−1 (ln x) x→0+

34. f (x) = 3e−1/x

In exercises 35–38, determine all vertical and slant asymptotes. x3 4 − x2 x3 37. y = 2 x +x −4 35. y =

x2 + 1 x −2 x4 38. y = 3 x +2 36. y =

In exercises 39–48, use graphical and numerical evidence to conjecture a value for the indicated limit. x2 x→∞ 2x x 2 − 4x + 7 41. lim x→∞ 2x 2 + x cos x x 3 + 4x + 5 43. lim x→∞ e x/2 39. lim

(b)

22. lim sin(tan−1 x)

In exercises 25–34, determine all horizontal and vertical asymptotes. For each vertical asymptote, determine whether f (x) → ∞ or f (x) → − ∞ on either side of the asymptote. x x 26. f (x) = √ 25. f (x) = √ 4 + x2 4 − x2 x x2 27. f (x) = 28. f (x) = 2 4−x 4 − x2 2 1−x 3x + 1 30. f (x) = 2 29. f (x) = 2 x − 2x − 3 x +x −2 ln(x + 2) 31. f (x) = ln(1 − cos x) 32. f (x) = ln(x 2 + 3x + 3) 33. f (x) = 4 tan−1 x − 1

1 − 2x x2 − 1

x→−1−

(c) lim

(b) lim

1 − 2x x2 − 1

(c) lim

4. (a)

1 − 2x x2 − 1

2

x→0

2x x→∞ x 2 2x 3 + 7x 2 + 1 42. lim x→−∞ x 3 − x sin x

40. lim

44. lim (e x/3 − x 4 ) x→∞

1-47

SECTION 1.5

x − cos (π x) x→−1 x +1 x 47. lim x→0+ cos x − 1

ex − 1 x→0 x ln x 2 48. lim 2 x→0 x

45. lim

46. lim

In exercises 49–52, use graphical and numerical evidence to conjecture the value of the limit. Then, verify your conjecture by finding the limit exactly. 49. lim ( 4x 2 − 2x + 1 − 2x) (Hint: Multiply and divide by the x→∞ √ conjugate expression: 4x 2 − 2x + 1 + 2x and simplify.) 50. lim ( x 2 + 3 − x) (See the hint for exercise 49.) x→∞ 51. lim ( 5x 2 + 4x + 7 − 5x 2 + x + 3) (See the hint for x→∞

exercise 49.) 3 2x 1 x . Hint : e = lim 1 + 52. lim 1 + x→−∞ x→∞ x x 53. Sketch the graph of f (x) = e−x cos x. Identify the horizontal asymptote. Is this asymptote approached in both directions (as x → ∞ and x → −∞)? How many times does the graph cross the horizontal asymptote? 54. Explain why it is reasonable that lim f (x) = lim f (1/x) and x→∞

lim f (x) = lim f (1/x).

x→−∞

x→0+

x→0−

55. In the exercises of section 1.2, we found that lim (1 + x)1/x = x→0+

lim (1 + x)1/x . (It turns out that both limits equal the irra-

x→0−

tional number e.) Use this result and exercise 54 to argue that lim (1 + 1/x)x = lim (1 + 1/x)x . x→∞

x→−∞

56. One of the reasons for saying that infinite limits do not exist is that we would otherwise invalidate Theorem 3.1 in section 1.3. Find examples of functions with infinite limits such that parts (ii) and (iv) of Theorem 3.1 do not hold. 57. Suppose that the length of a small animal t days after birth 300 is h(t) = mm. What is the length of the animal at 1 + 9(0.8)t birth? What is the eventual length of the animal (i.e., the length as t → ∞)? 58. Suppose that the length of a small animal t days after birth 100 mm. What is the length of the animal at is h(t) = 2 + 3(0.4)t birth? What is the eventual length of the animal (i.e., the length as t → ∞)? 59. Suppose that the size of the pupil of a certain animal is given by f (x) (mm), where x is the intensity of the light on the pupil. 80x −0.3 + 60 If f (x) = , find the size of the pupil with no light 2x −0.3 + 5 and the size of the pupil with an infinite amount of light. 60. Repeat exercise 59 with f (x) =

80x −0.3 + 60 . 8x −0.3 + 15

61. Modify the functions in exercises 59 and 60 to find a function f such that lim f (x) = 8 and lim f (x) = 2. x→0+

x→∞

..

Limits Involving Infinity; Asymptotes

119

62. After an injection, the concentration of a drug in a muscle varies according to a function of time f (t). Suppose that t is measured in hours and f (t) = e−0.02t − e−0.42t . Find the limit of f (t) both as t → 0 and t → ∞, and interpret both limits in terms of the concentration of the drug. 63. Suppose an object with initial velocity v0 = 0 ft/s and (constant) mass m slugs is accelerated by a constant force F pounds for t seconds. According to Newton’s laws of motion, the object’s speed will be v N = Ft/m. According to Einstein’s √ theory of relativity, the object’s speed will be v E = Fct/ m 2 c2 + F 2 t 2 , where c is the speed of light. Compute lim v N and lim v E . t→∞

t→∞

64. According to Einstein’s theory of relativity, themass of an object traveling at speed v is given by m = m 0 / 1 − v 2 /c2 , where c is the speed of light (about 9.8 × 108 ft/s). Compute lim m and explain why m 0 is called the “rest mass.” Compute v→0

lim m and discuss the implications. (What would happen if

v→c−

you were traveling in a spaceship approaching the speed of light?) How much does the mass of a 192-pound man (m 0 = 6) increase at the speed of 9000 ft/s (about 4 times the speed of sound)? 65. In example 5.12, the velocity t seconds after jump√ of a skydiver 32 1 − e−2t 32k √ . Find the limiting ing is given by v(t) = − k 1 + e−2t 32k velocity with k = 0.00064 and k = 0.00128. By what factor does a skydiver have to change the value of k to cut the limiting velocity in half? 66. Graph the velocity function in exercise 65 with k = 0.00016 (representing a headfirst dive) and estimate how long it takes for the diver to reach a speed equal to 90% of the limiting velocity. Repeat with k = 0.001 (representing a spread-eagle position). 67. Ignoring air resistance, the maximum height reached by a v02 R rocket launched with initial velocity v0 is h = m/s, 19.6R − v02 where R is the radius of the earth. In this exercise, we interpret this as a function of v0 . Explain why the domain of this function must be restricted to v0 ≥ 0. There is an additional restriction. Find the (positive) value ve such that h is undefined. Sketch a possible graph of h with 0 ≤ v0 < ve and discuss the significance of the vertical asymptote at ve . (Explain what would happen to the rocket if it is launched with initial velocity ve .) Explain why ve is called the escape velocity. p(x) with the q(x) degree (largest exponent) of p(x) less than the degree of q(x). Determine the horizontal asymptote of y = f (x).

68. Suppose that f (x) is a rational function f (x) =

p(x) with q(x) the degree of p(x) greater than the degree of q(x). Determine whether y = f (x) has a horizontal asymptote.

69. Suppose that f (x) is a rational function f (x) =

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Limits and Continuity

1-48

p(x) . If q(x) y = f (x) has a horizontal asymptote y = 2, how does the degree of p(x) compare to the degree of q(x)? p(x) Suppose that f (x) is a rational function f (x) = . If q(x) y = f (x) has a slant asymptote y = x + 2, how does the degree of p(x) compare to the degree of q(x)? x2 − 4 Find a quadratic function q(x) such that f (x) = has q(x) one horizontal asymptote y = 2 and two vertical asymptotes x = ±3. x2 − 4 Find a quadratic function q(x) such that f (x) = has q(x) 1 one horizontal asymptote y = − 2 and exactly one vertical asymptote x = 3. x −4 has two horizonFind a function g(x) such that f (x) = g(x) tal asymptotes y = ±1 and no vertical asymptotes.

70. Suppose that f (x) is a rational function f (x) =

71.

72.

73.

74.

In exercises 75–80, label the statement as true or false (not always true) for real numbers a and b. 75. If lim f (x) = a and lim g(x) = b, then x→∞

x→∞

lim [ f (x) + g(x)] = a + b.

x→∞

76. If lim f (x) = a and lim g(x) = b, then lim x→∞

x→∞

x→∞

f (x) a = . g(x) b

77. If lim f (x) = ∞ and lim g(x) = ∞, then x→∞

x→∞

lim [ f (x) − g(x)] = 0.

x→∞

78. If lim f (x) = ∞ and lim g(x) = ∞, then x→∞

x→∞

lim [ f (x) + g(x)] = ∞.

f (x) = 0. x→∞ x→∞ x→∞ g(x) f (x) = 1. 80. If lim f (x) = ∞ and lim g(x) = ∞, then lim x→∞ x→∞ x→∞ g(x) x→∞

79. If lim f (x) = a and lim g(x) = ∞, then lim

In exercises 81 and 82, determine all vertical and horizontal asymptotes. 4x if x < 0 x −4 2 x 81. f (x) = if 0 ≤ x < 4 x −2 −x e if x ≥ 4 x +1 x +3 if x < 0 x 2 − 4x x if 0 ≤ x < 2 82. f (x) = e + 1 2 x − 1 if x ≥ 2 x 2 − 7x + 10 83. Explain why lim (e−at sin t) = 0 for any positive constant a. t→∞

Although this is theoretically true, it is not necessarily useful in practice. The function e−at sin t is a simple model for a

spring-mass system, such as the suspension system on a car. Suppose t is measured in seconds and the car passengers cannot feel any vibrations less than 0.01 (inches). If suspension system A has the vibration function e−t sin t and suspension system B has the vibration function e−t/4 sin t, determine graphically how long it will take before the vibrations damp out, that is, | f (t)| < 0.01. Is the result lim (e−at sin t) = 0 much consolat→∞

tion to the owner of car B? 84. (a) State and prove a result analogous to Theorem 5.2 for lim pn (x), for n odd. x→−∞

(b) State and prove a result analogous to Theorem 5.2 for lim pn (x), for n even. x→−∞

85. It is very difficult to find simple statements in calculus that are always true; this is one reason that a careful development of the theory is so important. You may have heard the simple rule: g(x) , simply set the to find the vertical asymptotes of f (x) = h(x) denominator equal to 0 [i.e., solve h(x) = 0]. Give an example where h(a) = 0 but there is not a vertical asymptote at x = a. 86. In exercise 85, you needed to find an example indicating that the following statement is not (necessarily) true: if h(a) = 0, g(x) has a vertical asymptote at x = a. This is then f (x) = h(x) g(x) has not true, but perhaps its converse is true: if f (x) = h(x) a vertical asymptote at x = a, then h(a) = 0. Is this statement true? What if g and h are polynomials? In exercises 87–90, use numerical evidence to conjecture a decimal representation for the limit. Check your answer with your computer algebra system (CAS); if your answers disagree, which one is correct? 88. lim (ln x)x x→1+ ln x 90. lim x→−∞ x 2

87. lim x 1/ ln x x→0+

89. lim x 1/x x→∞

2 −1

EXPLORATORY EXERCISES 1. Suppose you are shooting a basketball from a (horizontal) distance of L feet, releasing the ball from a location h feet below the basket. To get a perfect swish, it is necessary that the initial velocity v0 and initial release angle θ0 satisfy the equation h

u0 10

√

L

v0 = gL/ 2 cos2 θ0 (tan θ0 − h/L). For a free throw, take L = 15, h = 2 and g = 32 and graph v0 as a function of θ0 .

1-49

SECTION 1.6

What is the significance of the two vertical asymptotes? Explain in physical terms what type of shot corresponds to each vertical asymptote. Estimate the minimum value of v0 (call it vmin ). Explain why it is easier to shoot a ball with a small initial velocity. There is another advantage to this initial velocity. Assume that the basket is 2 ft in diameter and the ball is 1 ft in diameter. For a free throw, L = 15 ft is perfect. What is the maximum horizontal distance the ball could travel and still go in the basket (without bouncing off the backboard)? What is the minimum horizontal distance? Call these numbers L max and L min . Find the angle θ1 corresponding to vmin and L min and the angle θ2 corresponding to vmin and L max . The difference |θ2 − θ1 | is the angular margin of error. Brancazio has shown that the angular margin of error for vmin is larger than for any other initial velocity.

1.6

..

Formal Definition of the Limit

121

2. In applications, it is common to compute lim f (x) to dex→∞

termine the “stability” of the function f (x). Consider the function f (x) = xe−x . As x → ∞, the first factor in f (x) goes to ∞, but the second factor goes to 0. What does the product do when one term is getting smaller and the other term is getting larger? It depends on which one is changing faster. What we want to know is which term “dominates.” Use graphical and numerical evidence to conjecture the value of lim (xe−x ). Which term dominates? In the limit x→∞

lim (x 2 e−x ), which term dominates? Also, try lim (x 5 e−x ).

x→∞

x→∞

Based on your investigation, is it always true that exponentials dominate polynomials? Are you positive? Try to determine which type of function, polynomials or logarithms, dominates.

FORMAL DEFINITION OF THE LIMIT We have now spent many pages discussing various aspects of the computation of limits. This may seem a bit odd, when you realize that we have never actually defined what a limit is. Oh, sure, we have given you an idea of what a limit is, but that’s about all. Once again, we have said that lim f (x) = L ,

x→a

HISTORICAL NOTES Augustin Louis Cauchy (1789–1857) A French mathematician who developed the ε-δ definitions of limit and continuity. Cauchy was one of the most prolific mathematicians in history, making important contributions to number theory, linear algebra, differential equations, astronomy, optics and complex variables. A difficult man to get along with, a colleague wrote, “Cauchy is mad and there is nothing that can be done about him, although right now, he is the only one who knows how mathematics should be done.”

if f (x) gets closer and closer to L as x gets closer and closer to a. So far, we have been quite happy with this somewhat vague, although intuitive, description. In this section, however, we will make this more precise, and you will begin to see how mathematical analysis (that branch of mathematics of which the calculus is the most elementary study) works. Studying more advanced mathematics without an understanding of the precise definition of limit is somewhat akin to studying brain surgery without bothering with all that background work in chemistry and biology. In medicine, it has only been through a careful examination of the microscopic world that a deeper understanding of our own macroscopic world has developed, and good surgeons need to understand what they are doing and why they are doing it. Likewise, in mathematical analysis, it is through an understanding of the microscopic behavior of functions (such as the precise definition of limit) that a deeper understanding of the mathematics will come about. We begin with the careful examination of an elementary example. You should certainly believe that lim (3x + 4) = 10.

x→2

Suppose that you were asked to explain the meaning of this particular limit to a fellow student. You would probably repeat the intuitive explanation we have used so far: that as x gets closer and closer to 2, (3x + 4) gets arbitrarily close to 10. But, exactly what do we mean by close? One answer is that if lim (3x + 4) = 10, we should be able to make x→2

(3x + 4) as close as we like to 10, just by making x sufficiently close to 2. But can we

1-49

SECTION 1.6

What is the significance of the two vertical asymptotes? Explain in physical terms what type of shot corresponds to each vertical asymptote. Estimate the minimum value of v0 (call it vmin ). Explain why it is easier to shoot a ball with a small initial velocity. There is another advantage to this initial velocity. Assume that the basket is 2 ft in diameter and the ball is 1 ft in diameter. For a free throw, L = 15 ft is perfect. What is the maximum horizontal distance the ball could travel and still go in the basket (without bouncing off the backboard)? What is the minimum horizontal distance? Call these numbers L max and L min . Find the angle θ1 corresponding to vmin and L min and the angle θ2 corresponding to vmin and L max . The difference |θ2 − θ1 | is the angular margin of error. Brancazio has shown that the angular margin of error for vmin is larger than for any other initial velocity.

1.6

..

Formal Definition of the Limit

121

2. In applications, it is common to compute lim f (x) to dex→∞

termine the “stability” of the function f (x). Consider the function f (x) = xe−x . As x → ∞, the first factor in f (x) goes to ∞, but the second factor goes to 0. What does the product do when one term is getting smaller and the other term is getting larger? It depends on which one is changing faster. What we want to know is which term “dominates.” Use graphical and numerical evidence to conjecture the value of lim (xe−x ). Which term dominates? In the limit x→∞

lim (x 2 e−x ), which term dominates? Also, try lim (x 5 e−x ).

x→∞

x→∞

Based on your investigation, is it always true that exponentials dominate polynomials? Are you positive? Try to determine which type of function, polynomials or logarithms, dominates.

FORMAL DEFINITION OF THE LIMIT We have now spent many pages discussing various aspects of the computation of limits. This may seem a bit odd, when you realize that we have never actually defined what a limit is. Oh, sure, we have given you an idea of what a limit is, but that’s about all. Once again, we have said that lim f (x) = L ,

x→a

HISTORICAL NOTES Augustin Louis Cauchy (1789–1857) A French mathematician who developed the ε-δ definitions of limit and continuity. Cauchy was one of the most prolific mathematicians in history, making important contributions to number theory, linear algebra, differential equations, astronomy, optics and complex variables. A difficult man to get along with, a colleague wrote, “Cauchy is mad and there is nothing that can be done about him, although right now, he is the only one who knows how mathematics should be done.”

if f (x) gets closer and closer to L as x gets closer and closer to a. So far, we have been quite happy with this somewhat vague, although intuitive, description. In this section, however, we will make this more precise, and you will begin to see how mathematical analysis (that branch of mathematics of which the calculus is the most elementary study) works. Studying more advanced mathematics without an understanding of the precise definition of limit is somewhat akin to studying brain surgery without bothering with all that background work in chemistry and biology. In medicine, it has only been through a careful examination of the microscopic world that a deeper understanding of our own macroscopic world has developed, and good surgeons need to understand what they are doing and why they are doing it. Likewise, in mathematical analysis, it is through an understanding of the microscopic behavior of functions (such as the precise definition of limit) that a deeper understanding of the mathematics will come about. We begin with the careful examination of an elementary example. You should certainly believe that lim (3x + 4) = 10.

x→2

Suppose that you were asked to explain the meaning of this particular limit to a fellow student. You would probably repeat the intuitive explanation we have used so far: that as x gets closer and closer to 2, (3x + 4) gets arbitrarily close to 10. But, exactly what do we mean by close? One answer is that if lim (3x + 4) = 10, we should be able to make x→2

(3x + 4) as close as we like to 10, just by making x sufficiently close to 2. But can we

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1-50

actually do this? For instance, can we force (3x + 4) to be within distance 1 of 10? To see what values of x will guarantee this, we write an inequality that says that (3x + 4) is within 1 unit of 10: |(3x + 4) − 10| < 1. Eliminating the absolute values, we see that this is equivalent to −1 < (3x + 4) − 10 < 1

y y 3x 4

− 1 < 3x − 6 < 1.

or

11 10 9

Since we need to determine how close x must be to 2, we want to isolate x − 2, instead of x. So, dividing by 3, we get − x 2W

2

2W

FIGURE 1.44 1 1 2 − < x < 2 + guarantees 3 3 that |(3x + 4) − 10| < 1

1 1 < x −2< 3 3 |x − 2| <

or

1 . 3

(6.1)

Reversing the steps that lead to inequality (6.1), we see that if x is within distance 13 of 2, then (3x + 4) will be within the specified distance (1) of 10. (See Figure 1.44 for a graphical interpretation of this.) So, does this convince you that you can make (3x + 4) as close as you want to 10? Probably not, but if you used a smaller distance, perhaps you’d be more convinced.

EXAMPLE 6.1

Exploring a Simple Limit

Find the values of x for which (3x + 4) is within distance

1 of 10. 100

Solution We want |(3x + 4) − 10| <

1 . 100

Eliminating the absolute values, we get −

1 1 < (3x + 4) − 10 < 100 100

or

−

1 1 < 3x − 6 < . 100 100

Dividing by 3 yields

−

1 1 < x −2< , 300 300

which is equivalent to

|x − 2| <

1 . 300

So, based on example 6.1, are you now convinced that we can make (3x + 4) as close as desired to 10? All we’ve been able to show is that we can make (3x + 4) pretty close to 10. So, how close do we need to be able to make it? The answer is arbitrarily close, as close as anyone would ever demand. We can show that this is possible by repeating the arguments in example 6.1, this time for an unspecified distance, call it ε (epsilon, where ε > 0).

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..

Formal Definition of the Limit

123

y 10 ´

EXAMPLE 6.2

y 3x 4

Verifying a Limit

Show that we can make (3x + 4) within any specified distance ε of 10 (no matter how small ε is), just by making x sufficiently close to 2.

10 10 ´

Solution The objective is to determine the range of x-values that will guarantee that (3x + 4) stays within ε of 10 (see Figure 1.45 for a sketch of this range). We have |(3x + 4) − 10| < ε. ´

23

2

´

23

x

This is equivalent to or

− ε < 3x − 6 < ε.

Dividing by 3, we get

−

FIGURE 1.45 The range of x-values that keep |(3x + 4) − 10| < ε

− ε < (3x + 4) − 10 < ε

or

ε ε < x −2< 3 3 ε |x − 2| < . 3

ε also implies 3 ε that |(3x + 4) − 10| < ε. This says that as long as x is within distance of 2, (3x + 4) 3 will be within the required distance ε of 10. That is, ε |(3x + 4) − 10| < ε whenever |x − 2| < . 3

Notice that each of the preceding steps is reversible, so that |x − 2| <

Take a moment or two to recognize what we’ve done in example 6.2. By using an unspecified distance, ε, we have verified that we can indeed make (3x + 4) as close to 10 as might be demanded (i.e., arbitrarily close; just name whatever ε > 0 you would like), simply by making x sufficiently close to 2. Further, we have explicitly spelled out what “sufficiently close to 2” means in the context of the present problem. Thus, no matter how close we are asked to make (3x + 4) to 10, we can accomplish this simply by taking x to be in the specified interval. Next, we examine this more precise notion of limit in the case of a function that is not defined at the point in question.

EXAMPLE 6.3

Proving That a Limit Is Correct

2x + 2x − 4 = 6. x→1 x −1 Solution It is easy to use the usual rules of limits to establish this result. It is yet another matter to verify that this is correct using our new and more precise notion of limit. In this case, we want to know how close x must be to 1 to ensure that 2

Prove that lim

f (x) =

2x 2 + 2x − 4 x −1

is within an unspecified distance ε > 0 of 6. First, notice that f is undefined at x = 1. So, we seek a distance δ (delta, δ > 0), such that if x is within distance δ of 1, but x = 1 (i.e., 0 < |x − 1| < δ), then this guarantees that | f (x) − 6| < ε.

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Notice that we have specified that 0 < |x − 1| to ensure that x = 1. Further, | f (x) − 6| < ε is equivalent to −ε <

2x 2 + 2x − 4 − 6 < ε. x −1

Finding a common denominator and subtracting in the middle term, we get −ε <

2x 2 + 2x − 4 − 6(x − 1) <ε x −1

or

−ε <

2x 2 − 4x + 2 < ε. x −1

Since the numerator factors, this is equivalent to y

−ε < y f (x)

6 ´

2(x − 1)2 < ε. x −1

Since x = 1, we can cancel two of the factors of (x − 1) to yield

6

−ε < 2(x − 1) < ε

6 ´

−

or

ε ε < x−1 < , 2 2

Dividing by 2.

which is equivalent to |x − 1| < ε/2. So, taking δ = ε/2 and working backward, we see that requiring x to satisfy x ´ 12

1

0 < |x − 1| < δ =

´ 12

FIGURE 1.46

ε 0 < |x − 1| < guarantees that 2 2x 2 + 2x − 4 6−ε < < 6 + ε. x −1

will guarantee that

ε 2

2 2x + 2x − 4 − 6 < ε. x −1

We illustrate this graphically in Figure 1.46. What we have seen so far motivates us to make the following general definition, illustrated in Figure 1.47.

y

DEFINITION 6.1 (Precise Definition of Limit) y f (x)

For a function f defined in some open interval containing a (but not necessarily at a itself), we say

L ´ L L ´

lim f (x) = L ,

x→a

x

a ad

if given any number ε > 0, there is another number δ > 0, such that 0 < |x − a| < δ guarantees that | f (x) − L| < ε.

ad

FIGURE 1.47 a − δ < x < a + δ guarantees that L − ε < f (x) < L + ε.

Notice that example 6.2 amounts to an illustration of Definition 6.1 for lim (3x + 4). There, we found that δ = ε/3 satisfies the definition.

x→2

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..

Formal Definition of the Limit

125

REMARK 6.1 TODAY IN MATHEMATICS

We want to emphasize that this formal definition of limit is not a new idea. Rather, it is a more precise mathematical statement of the same intuitive notion of limit that we have been using since the beginning of the chapter. Also, we must in all honesty point out that it is rather difficult to explicitly find δ as a function of ε, for all but a few simple examples. Despite this, learning how to work through the definition, even for a small number of problems, will shed considerable light on a deep concept.

Paul Halmos (1916– ) A Hungarian-born mathematician who earned a reputation as one of the best mathematical writers ever. For Halmos, calculus did not come easily, with understanding coming in a flash of inspiration only after a long period of hard work. “I remember standing at the blackboard in Room 213 of the mathematics building with Warren Ambrose and suddenly I understood epsilons. I understood what limits were, and all of that stuff that people had been drilling into me became clear. . . . I could prove the theorems. That afternoon I became a mathematician.’’

Example 6.4, although only slightly more complex than the last several problems, provides an unexpected challenge.

EXAMPLE 6.4

Using the Precise Definition of Limit

Use Definiton 6.1 to prove that lim (x 2 + 1) = 5. x→2

Solution If this limit is correct, then given any ε > 0, there must be a δ > 0 for which 0 < |x − 2| < δ guarantees that |(x 2 + 1) − 5| < ε. Notice that |(x 2 + 1) − 5| = |x 2 − 4|

Factoring the difference

= |x + 2||x − 2|.

of two squares.

(6.2)

Our strategy is to isolate |x − 2| and so, we’ll need to do something with the term |x + 2|. Since we’re interested only in what happens near x = 2, anyway, we will only consider x’s within a distance of 1 from 2, that is, x’s that lie in the interval [1, 3] (so that |x − 2| < 1). Notice that this will be true if we require δ ≤ 1 and |x − 2| < δ. In this case, we have |x + 2| ≤ 5,

Since x ∈ [1, 3].

and so, from (6.2), |(x 2 + 1) − 5| = |x + 2||x − 2| ≤ 5|x − 2|.

(6.3)

Finally, if we require that

y

5|x − 2| < ε, 5 ´

(6.4)

then we will also have from (6.3) that |(x 2 + 1) − 5| ≤ 5|x − 2| < ε.

5

Of course, (6.4) is equivalent to 5 ´

|x − 2| <

y x2 1

x 2d

2

2d

FIGURE 1.48 0 < |x − 2| < δ guarantees that |(x 2 + 1) − 5| < ε.

ε . 5

ε So, in view of this, we now have two restrictions: that |x − 2| < 1 and that |x − 2| < . 5 ε To ensure that both restrictions are met, we choose δ = min 1, i.e., the minimum 5 ε of 1 and . Working backward, we get that for this choice of δ, 5 0 < |x − 2| < δ will guarantee that |(x 2 + 1) − 5| < ε, as desired. We illustrate this in Figure 1.48.

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Exploring the Definition of Limit Graphically As you can see from example 6.4, this business of finding δ’s for a given ε is not easily accomplished. There, we found that even for the comparatively simple case of a quadratic polynomial, the job can be quite a challenge. Unfortunately, there is no procedure that will work for all problems. However, we can explore the definition graphically in the case of more complex functions. First, we reexamine example 6.4 graphically.

EXAMPLE 6.5

Exploring the Precise Definition of Limit Graphically

Explore the precise definition of limit graphically, for lim (x 2 + 1) = 5.

5.5

x→2

ε Solution In example 6.4, we discovered that for δ = min 1, , 5 2 0 < |x − 2| < δ implies that |(x + 1) − 5| < ε.

5.3 y

5.1 4.9 4.7 4.5 1.9

1.95

2 x

2.05

2.1

FIGURE 1.49 y = x2 + 1

This says that (for ε ≤ 5 ) if we draw a graph of y = x 2 + 1 and restrict the x-values to ε ε lie in the interval 2 − , 2 + , then the y-values will lie in the interval (5 − ε, 5 + ε). 5 5 1 Take ε = , for instance. If we draw the graph in the window defined by 2 1 1 2− ≤ x ≤2+ and 4.5 ≤ y ≤ 5.5, then the graph will not run off the top or 10 10 bottom of the screen (see Figure 1.49). Of course, we can draw virtually the same picture for any given value of ε, since we have an explicit formula for finding δ given ε. For most limit problems, we are not so fortunate.

EXAMPLE 6.6

Exploring the Definition of Limit for a Trigonometric Function

1 Graphically find a δ > 0 corresponding to (a) ε = and (b) ε = 0.1 for 2 πx lim sin = 0. x→2 2 2π Solution This limit seems plausible enough. After all, sin = 0 and f (x) = sin x 2 is a continuous function. However, the point is to verify this carefully. Given any ε > 0, we want to find a δ > 0, for which πx 0 < |x − 2| < δ guarantees that sin − 0 < ε. 2 πx Note that since we have no algebra for simplifying sin , we cannot accomplish this 2 symbolically. Instead, we’ll try to graphically find δ’s corresponding to the specific ε’s 1 given. First, for ε = , we would like to find a δ > 0 for which if 0 < |x − 2| < δ, then 2

y 0.5

x 1

1.5

2

2.5

0.5

FIGURE 1.50a y = sin

πx 2

3

−

1 πx 1 < sin −0< . 2 2 2

πx 1 1 Drawing the graph of y = sin with 1 ≤ x ≤ 3 and − ≤ y ≤ , we get 2 2 2 Figure 1.50a. If you trace along a calculator or computer graph, you will notice that the graph stays on the screen (i.e., the y-values stay in the interval [−0.5, 0.5]) for

1-55

SECTION 1.6

y

..

Formal Definition of the Limit

x ∈ [1.666667, 2.333333]. Thus, we have determined experimentally that for ε =

0.5

127

1 , 2

δ = 2.333333 − 2 = 2 − 1.666667 = 0.333333 2–δ

2+δ

2

0.5

FIGURE 1.50b y = sin

πx 2

x

will work. (Of course, any value of δ smaller than 0.333333 will also work.) To illustrate this, we redraw the last graph, but restrict x to lie in the interval [1.67, 2.33] (see Figure 1.50b). In this case, the graph stays in the window over the entire range of displayed x-values. Taking ε = 0.1, we look for an interval of x-values that will πx guarantee that sin stays between −0.1 and 0.1. We redraw the graph from Figure 2 1.50a, with the y-range restricted to the interval [−0.1, 0.1] (see Figure 1.51a). Again, tracing along the graph tells us that the y-values will stay in the desired range for x ∈ [1.936508, 2.063492]. Thus, we have experimentally determined that δ = 2.063492 − 2 = 2 − 1.936508 = 0.063492 will work here. We redraw the graph using the new range of x-values (see Figure 1.51b), since the graph remains in the window for all values of x in the indicated interval. y

y

0.1

0.1

x 1.7

2

2–δ

2.3

0.1

2+δ

2

x

0.1

FIGURE 1.51a πx y = sin 2

FIGURE 1.51b y = sin

πx 2

It is important to recognize that we are not proving that the above limit is correct. To prove this requires us to symbolically find a δ for every ε > 0. The idea here is to use these graphical illustrations to become more familiar with the definition and with what δ and ε represent. x 0.1 0.01 0.001 0.0001

x 2 2x √ x 3 4x 2 1.03711608 1.0037461 1.00037496 1.0000375

EXAMPLE 6.7

Exploring the Definition of Limit Where the Limit Does Not Exist

x 2 + 2x Determine whether or not lim √ = 1. x→0 x 3 + 4x 2 Solution We first construct a table of function values. From the table alone, we might be tempted to conjecture that the limit is 1. However, we would be making a huge error, as we have not considered negative values of x or drawn a graph. This kind of carelessness is dangerous. Figure 1.52a (on the following page) shows the default graph drawn by our computer algebra system. In this graph, the function values do not quite look like they are approaching 1 as x → 0 (at least as x → 0− ). We now investigate the limit graphically for ε = 12 . We need to find a δ > 0 for which 0 < |x| < δ guarantees that 1−

1 1 x 2 + 2x <1+ <√ 2 2 x 3 + 4x 2

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1-56

y 16

y

12

1.5

8 1 4

4

x

2

2

0.5

4

4

x

0.1

0.1

FIGURE 1.52a

FIGURE 1.52b

x 2 + 2x y= √ x 3 + 4x 2

x 2 + 2x y= √ x 3 + 4x 2

x 2 + 2x 1 3 <√ < . 3 2 2 2 x + 4x

or

We try δ = 0.1 to see if this is sufficiently small. So, we set the x-range to the interval [−0.1, 0.1] and the y-range to the interval [0.5, 1.5] and redraw the graph in this window (see Figure 1.52b). Notice that no points are plotted in the window for any x < 0. According to the definition, the y-values must lie in the interval (0.5, 1.5) for all x in the interval (−δ, δ). Further, you can see that δ = 0.1 clearly does not work since x = −0.05 lies in the interval (−δ, δ), but f (−0.05) ≈ −0.981 is not in the interval (0.5, 1.5). You should convince yourself that no matter how small you make δ, there is an x in the interval (−δ, δ) such that f (x) ∈ / (0.5, 1.5). (In fact, notice that for all x’s in the interval (−1, 0), f (x) < 0.) That is, there is no choice of δ that makes the defining inequality true for ε = 12 . Thus, the conjectured limit of 1 is incorrect. You should note here that, while we’ve only shown that the limit is not 1, it’s somewhat more complicated to show that the limit does not exist.

Limits Involving Infinity Recall that we had observed that lim

x→0

written

1 does not exist, but to be more descriptive, we had x2 lim

x→0

1 = ∞. x2

By this statement, we mean that the function increases without bound as x → 0. Just as with our initial intuitive notion of lim f (x) = L, this description is imprecise and needs to be x→a 1 more carefully defined. When we say that 2 increases without bound as x → 0, we mean x 1 that we can make 2 as large as we like, simply by making x sufficiently close to 0. So, x 1 given any large positive number, M, we must be able to make 2 > M, for x sufficiently x close to 0. We measure closeness here the same way as we did before and arrive at the following definition.

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..

Formal Definition of the Limit

129

y

DEFINITION 6.2 For a function f defined in some open interval containing a (but not necessarily at a itself), we say lim f (x) = ∞,

M

x→a

if given any number M > 0, there is another number δ > 0, such that 0 < |x − a| < δ guarantees that f (x) > M. (See Figure 1.53 for a graphical interpretation of this.) x ad

a

ad

FIGURE 1.53 lim f (x) = ∞

Similarly, we had said that if f decreases without bound as x → a, then lim f (x) = −∞. Think of how you would make this more precise and then consider the x→a following definition.

x→a

DEFINITION 6.3

y ad

a

ad x

For a function f defined in some open interval containing a (but not necessarily at a itself), we say lim f (x) = −∞, x→a

if given any number N < 0, there is another number δ > 0, such that 0 < |x − a| < δ guarantees that f (x) < N . (See Figure 1.54 for a graphical interpretation of this.)

N

It’s easy to keep these definitions straight if you think of their meaning. Don’t simply memorize them. FIGURE 1.54 lim f (x) = −∞

x→a

EXAMPLE 6.8

Using the Definition of Limit Where the Limit Is Infinite

1 = ∞. x2 Solution Given any (large) number M > 0, we need to find a distance δ > 0 such that if x is within δ of 0 (but not equal to 0) then

Prove that lim

x→0

1 > M. x2

(6.5)

Since both M and x 2 are positive, (6.5) is equivalent to x2 <

1 . M

√ Taking the square root of both sides and recalling that x 2 = |x|, we get 1 |x| < . M 1 So, for any M > 0, if we take δ = and work backward, we have that 0 < |x − 0| < δ M guarantees that 1 > M, x2 1 as desired. Note that this says, for instance, that for M = 100, 2 > 100, whenever x 1 1 0 < |x| < = . (Verify that this works, as an exercise.) 100 10

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There are two remaining limits that we have yet to place on a careful footing. Before reading on, try to figure out for yourself what appropriate definitions would look like. If we write lim f (x) = L, we mean that as x increases without bound, f (x) gets x→∞ closer and closer to L. That is, we can make f (x) as close to L as we like, by choosing x sufficiently large. More precisely, we have the following definition. y

DEFINITION 6.4 For a function f defined on an interval (a, ∞), for some a > 0, we say

L ´ L L´

lim f (x) = L ,

x→∞

if given any ε > 0, there is a number M > 0 such that x > M guarantees that | f (x) − L| < ε.

x M

(See Figure 1.55 for a graphical interpretation of this.)

FIGURE 1.55 lim f (x) = L

Similarly, we have said that lim f (x) = L means that as x decreases without bound,

x→∞

x→−∞

f (x) gets closer and closer to L. So, we should be able to make f (x) as close to L as desired, just by making x sufficiently large in absolute value and negative. We have the following definition. y

DEFINITION 6.5 L ´ L L´

For a function f defined on an interval (−∞, a), for some a < 0, we say lim f (x) = L ,

x→−∞

if given any ε > 0, there is a number N < 0 such that x < N guarantees that N

| f (x) − L| < ε.

x

(See Figure 1.56 for a graphical interpretation of this.) FIGURE 1.56 lim f (x) = L

x→−∞

We use Definitions 6.4 and 6.5 essentially the same as we do Definitions 6.1–6.3, as we see in example 6.9.

EXAMPLE 6.9 Prove that lim

x→−∞

Using the Definition of Limit Where x Is Becoming Infinite

1 = 0. x

1 Solution Here, we must show that given any ε > 0, we can make within ε of 0, x simply by making x sufficiently large in absolute value and negative. So, we need to determine those x’s for which 1 − 0 < ε x 1 < ε. or (6.6) x

1-59

SECTION 1.6

..

Formal Definition of the Limit

131

Since x < 0, |x| = −x, and so (6.6) becomes

REMARK 6.2 You should take care to note the commonality among the definitions of the five limits we have given. All five deal with a precise description of what it means to be “close.” It is of considerable benefit to work through these definitions until you can provide your own words for each. Don’t just memorize the formal definitions as stated here. Rather, work toward understanding what they mean and come to appreciate the exacting language mathematicians use.

1 < ε. −x Dividing both sides by ε and multiplying by x (remember that x < 0 and ε > 0, so that this will change the direction of the inequality), we get −

1 > x. ε

1 So, if we take N = − and work backward, we have satisfied the definition and thereby ε proved that the limit is correct. We don’t use the limit definitions to prove each and every limit that comes along. Actually, we use them to prove only a few basic limits and to prove the limit theorems that we’ve been using for some time without proof. Further use of these theorems then provides solid justification of new limits. As an illustration, we now prove the rule for a limit of a sum.

THEOREM 6.1 Suppose that for a real number a, lim f (x) = L 1 and lim g(x) = L 2 . Then, x→a

x→a

lim [ f (x) + g(x)] = lim f (x) + lim g(x) = L 1 + L 2 .

x→a

x→a

x→a

PROOF Since lim f (x) = L 1 , we know that given any number ε1 > 0, there is a number δ1 > 0 for x→a which 0 < |x − a| < δ1 guarantees that | f (x) − L 1 | < ε1 .

(6.7)

Likewise, since lim g(x) = L 2 , we know that given any number ε2 > 0, there is a number x→a δ2 > 0 for which 0 < |x − a| < δ2 guarantees that |g(x) − L 2 | < ε2 .

(6.8)

Now, in order to get lim [ f (x) + g(x)] = (L 1 + L 2 ),

x→a

we must show that, given any number ε > 0, there is a number δ > 0 such that 0 < |x − a| < δ guarantees that |[ f (x) + g(x)] − (L 1 + L 2 )| < ε. Notice that |[ f (x) + g(x)] − (L 1 + L 2 )| = |[ f (x) − L 1 ] + [g( x) − L 2 ]| ≤ | f (x) − L 1 | + |g(x) − L 2 |,

(6.9)

by the triangle inequality. Of course, both terms on the right-hand side of (6.9) can be made ε arbitrarily small, from (6.7) and (6.8). In particular, if we take ε1 = ε2 = , then as long as 2 0 < |x − a| < δ1 and 0 < |x − a| < δ2 ,

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1-60

we get from (6.7), (6.8) and (6.9) that |[ f (x) + g(x)] − (L 1 + L 2 )| ≤ | f (x) − L 1 | + |g(x) − L 2 | ε ε < + = ε, 2 2 as desired. Of course, this will happen if we take 0 < |x − a| < δ = min{δ1 , δ2 }. The other rules for limits are proven similarly, using the definition of limit. We show these in Appendix A.

EXERCISES 1.6 WRITING EXERCISES 1. In his 1726 masterpiece Mathematical Principles of Natural Philosophy, which introduces many of the fundamentals of calculus, Sir Isaac Newton described the important limit f (a + h) − f (a) (which we study at length in Chapter 2) lim h→0 h as “the limit to which the ratios of quantities decreasing without limit do always converge, and to which they approach nearer than by any given difference, but never go beyond, nor ever reach until the quantities vanish.” If you ever get weary of all the notation that we use in calculus, think of what it would look like in words! Critique Newton’s definition of limit, addressing the following questions in the process. What restrictions do the phrases “never go beyond” and “never reach” put on the limit process? Give an example of a simple limit, not necessarily f (a + h) − f (a) of the form lim , that violates these restrich→0 h tions. Give your own (English language) description of the limit, avoiding restrictions such as Newton’s. Why do mathematicians consider the ε−δ definition simple and elegant? 2. You have computed numerous limits before seeing the definition of limit. Explain how this definition changes and/or improves your understanding of the limit process. 3. Each word in the ε−δ definition is carefully chosen and precisely placed. Describe what is wrong with each of the following slightly incorrect “definitions” (use examples!): (a) There exists ε > 0 such that there exists a δ > 0 such that if 0 < |x − a| < δ, then | f (x) − L| < ε. (b) For all ε > 0 and for all δ > 0, if 0 < |x − a| < δ, then | f (x) − L| < ε. (c) For all δ > 0 there exists ε > 0 such that 0 < |x − a| < δ and | f (x) − L| < ε. 4. In order for the limit to exist, given every ε > 0, we must be able to find a δ > 0 such that the if/then inequalities are true. To prove that the limit does not exist, we must find a particular ε > 0 such that the if/then inequalities are not true for any choice of δ > 0. To understand the logic behind the swapping of the “for every” and “there exists” roles, draw an

analogy with the following situation. Suppose the statement, “Everybody loves somebody” is true. If you wanted to verify the statement, why would you have to talk to every person on earth? But, suppose that the statement is not true. What would you have to do to disprove it? In exercises 1–8, numerically and graphically determine a δ corresponding to (a) ε 0.1 and (b) ε 0.05. Graph the function in the ε − δ window [x-range is (a − δ, a δ) and y-range is (L − ε, L ε)] to verify that your choice works. 1. lim (x 2 + 1) = 1

2. lim (x 2 + 1) = 5

3. lim cos x = 1

4. lim cos x = 0

x→0

x→0

5. lim

x→1

7. lim

x→1

√

x→2

x→π/2

x +3=2

√

6. lim

x→−2

x +2 =3 x2

8. lim

x→2

x +3=1

x +2 =1 x2

In exercises 9–20, symbolically find δ in terms of ε. 9. lim 3x = 0

10. lim 3x = 3

x→0

x→1

11. lim (3x + 2) = 8

12. lim (3x + 2) = 5

13. lim (3 − 4x) = −1

14. lim (3 − 4x) = 7

x +x −2 =3 15. lim x→1 x −1

16. lim

17. lim (x − 1) = 0

18. lim (x 2 − x + 1) = 1

19. lim (x − 1) = 3

20. lim (x 3 + 1) = 1

x→2

x→1

x→1

x→−1

2

2

x→1

x→−1

x2 − 1 = −2 x +1

x→1

2

x→2

x→0

21. Determine a formula for δ in terms of ε for lim (mx + b). (Hint: x→a

Use exercises 9–14.) Does the formula depend on the value of a? Try to explain this answer graphically. 22. Based on exercises 17 and 19, does the value of δ depend on the value of a for lim (x 2 + b)? Try to explain this graphically. x→a

1-61

SECTION 1.6

23. Modify the ε − δ definition to define the one-sided limits lim f (x) and lim f (x). x→a −

x→a +

24. Symbolically find the largest δ corresponding to ε = 0.1 in the definition of lim 1/x = 1. Symbolically find the largest x→1−

δ corresponding to ε = 0.1 in the definition of lim 1/x = 1.

49. f (x) = 50. f (x) =

x→1+

Which δ could be used in the definition of lim 1/x = 1? Briefly x→1

explain. Then prove that lim 1/x = 1. x→1

In exercises 25–30, find a δ corresponding to M 100 or N − 100 (as appropriate) for each limit. 25. lim

x→1+

2 =∞ x −1

27. lim cot x = ∞ x→0+

29. lim √ − x→2

2 4 − x2

=∞

26. lim

x→1−

2 = −∞ x −1

28. lim cot x = −∞ x→π −

30. lim ln x = −∞ x→0+

In exercises 31–36, find an M or N corresponding to ε 0.1 for each limit at infinity. 31. lim

x→∞

33.

x2

x2 − 2 =1 +x +1

x2 + 3 = 0.25 x→−∞ 4x 2 − 4 lim

35. lim e

−2x

x→∞

=0

32. lim

x→∞

34.

x2

x −2 =0 +x +1

3x 2 − 2 =3 x→−∞ x 2 + 1 lim

e +x 36. lim x =1 x→∞ e − x 2

x→∞

2 =0 x3

1 = 0, for k > 0 xk 1 − 3 = −3 41. lim x→∞ x2 + 2

39. lim

x→∞

43. lim

x→−3

−2 = −∞ (x + 3)4

4 =∞ x→5 (x − 5)2

45. lim

38. 40.

lim

3 =0 x3

lim

1 = 0, for k > 0 x 2k

x→−∞

x→−∞

42. lim

1 =0 (x − 7)2

44. lim

3 =∞ (x − 7)2

x→∞

x→7

−6 = −∞ x→−4 (x + 4)6

46. lim

In exercises 47–50, identify a specific ε > 0 for which no δ > 0 exists to satisfy the definition to limit. 2x if x < 1, lim f (x) = 2 47. f (x) = x 2 + 3 if x > 1 x→1 48. f (x) =

Formal Definition of the Limit

133

if x < 1, lim f (x) = 2 if x > 1 x→1

x − 1 if x < 2, lim f (x) = 1 x2 if x > 2 x→2

51. A metal washer of (outer) radius r inches weighs 2r 2 ounces. A company manufactures 2-inch washers for different customers who have different error tolerances. If the customer demands a washer of weight 8 ± ε ounces, what is the error tolerance for the radius? That is, find δ such that a radius of r within the interval (2 − δ, 2 + δ) guarantees a weight within (8 − ε, 8 + ε). 52. A fiberglass company ships its glass as spherical marbles. If the volume of each marble must be within ε of π/6, how close does the radius need to be to 1/2? 53. Prove Theorem 3.1 (i). 54. Prove Theorem 3.1 (ii). 55. Prove the Squeeze Theorem, as stated in Theorem 3.5. 56. Given that lim f (x) = L and lim f (x) = L, prove that x→a −

x→a +

lim f (x) = L.

x→a

57. Prove: if lim f (x) = L, then lim [ f (x) − L] = 0. x→a

x→a

x

In exercises 37–46, prove that the limit is correct using the appropriate definition (assume that k is an integer). 37. lim

2x 5 − x2

..

x 2 − 1 if x < 0, lim f (x) = −2 −x − 2 if x > 0 x→0

58. Prove: if lim [ f (x) − L] = 0, then lim f (x) = L. x→a

x→a

59. In this exercise, we explore the definition of lim x 2 = 4 with x→2 √ ε = 0.1. Show that x 2 − 4 < 0.1 if 2 < x < 4.1. This indicates that δ1 = 0.02484 works for x > 2. Show that √ x 2 − 4 > −0.1 if 3.9 < x < 2. This indicates that δ2 = 0.02515 works for x < 2. For the limit definition, is δ = δ1 or δ = δ2 the correct choice? Briefly explain. √ 60. Generalize exercise 59 to find a δ of the form 4 + ε or √ 4 − ε corresponding to any ε > 0.

EXPLORATORY EXERCISES 1. We hope that working through this section has provided you with extra insight into the limit process. However, we have not yet solved any problems we could not already solve in previous sections. We do so now, while investigating an unusual function. Recall that rational numbers can be written as fractions p/q, where p and q are integers. We will assume that p/q has been simplified by dividing out common factors (e.g., 0 if x is irrational 1/2 and not 2/4). Define f (x) = . 1/q if x = qp is rational We will try to show that lim f (x) exists. Without graphics, x→2/3

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we need a good definition to answer this question. We know that f (2/3) = 1/3, but recall that the limit is independent of the actual function value. We need to think about x’s close to 2/3. If such an x is irrational, f (x) = 0. A simple hypothesis would then be lim f (x) = 0. We’ll try this out for x→2/3

ε = 1/6. We would like to guarantee that | f (x)| < 1/6 whenever 0 < |x − 2/3| < δ. Well, how many x’s have a function value greater than 1/6? The only possible function values are 1/5, 1/4, 1/3, 1/2 and 1. The x’s with function value 1/5 are 1/5, 2/5, 3/5, 4/5 and so on. The closest of these x’s to 2/3

1.7

is 3/5. Find the closest x (not counting x = 2/3) to 2/3 with function value 1/4. Repeat for f (x) = 1/3, f (x) = 1/2 and f (x) = 1. Out of all these closest x’s, how close is the absolute closest? Choose δ to be this number, and argue that if 0 < |x − 2/3| < δ, we are guaranteed that | f (x)| < 1/6. Argue that a similar process can find a δ for any ε. 2. State a definition for “ f (x) is continuous at x = a” using Definition 6.1. Use it to prove that the function in exploratory exercise 1 is continuous at every irrational number and discontinuous at every rational number.

LIMITS AND LOSS-OF-SIGNIFICANCE ERRORS “Pay no attention to that man behind the curtain . . . .” (from The Wizard of Oz) Things are not always what they appear to be. We spend much time learning to distinguish reality from mere appearances. Along the way, we develop a healthy level of skepticism. You may have already come to realize that mathematicians are a skeptical lot. This is of necessity, for you simply can’t accept things at face value. People tend to accept a computer’s answer as a fact not subject to debate. However, when we use a computer (or calculator), we must always keep in mind that these devices perform most computations only approximately. Most of the time, this will cause us no difficulty whatsoever. Modern computational devices generally carry out calculations to a very high degree of accuracy. Occasionally, however, the results of round-off errors in a string of calculations are disastrous. In this section, we briefly investigate these errors and learn how to recognize and avoid some of them. We first consider a relatively tame-looking example.

EXAMPLE 7.1

y 9

A Limit with Unusual Graphical and Numerical Behavior 2

(x 3 + 4) − x 6 . x→∞ x3

Evaluate lim 8

7

x 20,000

60,000

100,000

FIGURE 1.57a 2

y=

(x 3 + 4) − x 6 x3

Solution At first glance, the numerator looks like ∞ − ∞, which is indeterminate, while the denominator tends to ∞. Algebraically, the only reasonable step to take is to multiply out the first term in the numerator. Before we do that, let’s draw a graph and compute some function values. (Different computers and different software will produce somewhat different results, but for large values of x, you should see results similar to those shown here.) In Figure 1.57a, the function appears nearly constant, until it begins oscillating around x = 40,000. Notice that the accompanying table of function values is inconsistent with Figure 1.57a. The last two values in the table may have surprised you. Up until that point, the function values seemed to be settling down to 8.0 very nicely. So, what happened here and what is the correct value of the limit? Obviously, something unusual has occurred

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we need a good definition to answer this question. We know that f (2/3) = 1/3, but recall that the limit is independent of the actual function value. We need to think about x’s close to 2/3. If such an x is irrational, f (x) = 0. A simple hypothesis would then be lim f (x) = 0. We’ll try this out for x→2/3

ε = 1/6. We would like to guarantee that | f (x)| < 1/6 whenever 0 < |x − 2/3| < δ. Well, how many x’s have a function value greater than 1/6? The only possible function values are 1/5, 1/4, 1/3, 1/2 and 1. The x’s with function value 1/5 are 1/5, 2/5, 3/5, 4/5 and so on. The closest of these x’s to 2/3

1.7

is 3/5. Find the closest x (not counting x = 2/3) to 2/3 with function value 1/4. Repeat for f (x) = 1/3, f (x) = 1/2 and f (x) = 1. Out of all these closest x’s, how close is the absolute closest? Choose δ to be this number, and argue that if 0 < |x − 2/3| < δ, we are guaranteed that | f (x)| < 1/6. Argue that a similar process can find a δ for any ε. 2. State a definition for “ f (x) is continuous at x = a” using Definition 6.1. Use it to prove that the function in exploratory exercise 1 is continuous at every irrational number and discontinuous at every rational number.

LIMITS AND LOSS-OF-SIGNIFICANCE ERRORS “Pay no attention to that man behind the curtain . . . .” (from The Wizard of Oz) Things are not always what they appear to be. We spend much time learning to distinguish reality from mere appearances. Along the way, we develop a healthy level of skepticism. You may have already come to realize that mathematicians are a skeptical lot. This is of necessity, for you simply can’t accept things at face value. People tend to accept a computer’s answer as a fact not subject to debate. However, when we use a computer (or calculator), we must always keep in mind that these devices perform most computations only approximately. Most of the time, this will cause us no difficulty whatsoever. Modern computational devices generally carry out calculations to a very high degree of accuracy. Occasionally, however, the results of round-off errors in a string of calculations are disastrous. In this section, we briefly investigate these errors and learn how to recognize and avoid some of them. We first consider a relatively tame-looking example.

EXAMPLE 7.1

y 9

A Limit with Unusual Graphical and Numerical Behavior 2

(x 3 + 4) − x 6 . x→∞ x3

Evaluate lim 8

7

x 20,000

60,000

100,000

FIGURE 1.57a 2

y=

(x 3 + 4) − x 6 x3

Solution At first glance, the numerator looks like ∞ − ∞, which is indeterminate, while the denominator tends to ∞. Algebraically, the only reasonable step to take is to multiply out the first term in the numerator. Before we do that, let’s draw a graph and compute some function values. (Different computers and different software will produce somewhat different results, but for large values of x, you should see results similar to those shown here.) In Figure 1.57a, the function appears nearly constant, until it begins oscillating around x = 40,000. Notice that the accompanying table of function values is inconsistent with Figure 1.57a. The last two values in the table may have surprised you. Up until that point, the function values seemed to be settling down to 8.0 very nicely. So, what happened here and what is the correct value of the limit? Obviously, something unusual has occurred

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135

between x = 1 × 104 and x = 1 × 105 . We should look carefully at function values in that interval. A more detailed table is shown below to the right.

y 8.2

Incorrect calculated values

(x 4) − x 6 x3 8.016 8.000016 8.0 8.0 0.0 0.0 3

x

8

7.8 x 20,000

60,000

100,000

FIGURE 1.57b

10 100 1 × 103 1 × 104 1 × 105 1 × 106

2

x 2 × 104 3 × 104 4 × 104 5 × 104

(x 3 4)2 − x 6 x3 8.0 8.14815 7.8125 0

2

y=

(x 3 + 4) − x 6 x3

In Figure 1.57b, we have blown up the graph to enhance the oscillation observed between x = 1 × 104 and x = 1 × 105 . The picture that is emerging is even more confusing. The deeper we look into this limit, the more erratically the function appears to behave. We use the word appears because all of the oscillatory behavior we are seeing is an illusion, created by the finite precision of the computer used to perform the calculations or draw the graph.

Computer Representation of Real Numbers The reason for the unusual behavior seen in example 7.1 boils down to the way in which computers represent real numbers. Without getting into all of the intricacies of computer arithmetic, it suffices to think of computers and calculators as storing real numbers internally in scientific notation. For example, the number 1,234,567 would be stored as 1.234567 × 106 . The number preceding the power of 10 is called the mantissa and the power is called the exponent. Thus, the mantissa here is 1.234567 and the exponent is 6. All computing devices have finite memory and consequently have limitations on the size mantissa and exponent that they can store. (This is called finite precision.) Many calculators carry a 14-digit mantissa and a 3-digit exponent. On a 14-digit computer, this would suggest that the computer would retain only the first 14 digits in the decimal expansion of any given number.

EXAMPLE 7.2

Computer Representation of a Rational Number

1 2 Determine how is stored internally on a 10-digit computer and how is stored internally 3 3 on a 14-digit computer. 1 −1 Solution On a 10-digit computer, is stored internally as 3.333333333 ! " ×10 . On a 3 10 digits

2 −1 14-digit computer, is stored internally as 6.6666666666667 ! " × 10 . 3 14 digits

For most purposes, such finite precision presents no problem. However, we do occasionally come across a disastrous error caused by finite precision. In example 7.3, we subtract two relatively close numbers and examine the resulting catastrophic error.

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EXAMPLE 7.3

A Computer Subtraction of Two “Close” Numbers

Compare the exact value of 18 18 1. 0000000000000 ! " 4 × 10 − 1. 0000000000000 ! " 1 × 10 13 zeros

13 zeros

with the result obtained from a calculator or computer with a 14-digit mantissa. Solution Notice that 18 18 18 1. 0000000000000 ! " 4×10 − 1. 0000000000000 ! " 1×10 = 0. 0000000000000 ! " 3×10 13 zeros

13 zeros

13 zeros

= 30,000.

(7.1)

However, if this calculation is carried out on a computer or calculator with a 14-digit (or smaller) mantissa, both numbers on the left-hand side of (7.1) are stored by the computer as 1 × 1018 and hence, the difference is calculated as 0. Try this calculation for yourself now.

EXAMPLE 7.4

Another Subtraction of Two “Close” Numbers

Compare the exact value of 20 20 1. 0000000000000 ! " 6 × 10 − 1. 0000000000000 ! " 4 × 10 13 zeros

13 zeros

with the result obtained from a calculator or computer with a 14-digit mantissa. Solution Notice that 20 20 20 1.0000000000000 ! " 6 ×10 − 1.0000000000000 ! " 4 ×10 = 0.0000000000000 ! " 2 ×10 13 zeros

13 zeros

13 zeros

= 2,000,000.

However, if this calculation is carried out on a calculator with a 14-digit mantissa, the first number is represented as 1.0000000000001 × 1020 , while the second number is represented by 1.0 × 1020 , due to the finite precision and rounding. The difference between the two values is then computed as 0.0000000000001 × 1020 or 10,000,000, which is, again, a very serious error. In examples 7.3 and 7.4, we witnessed a gross error caused by the subtraction of two numbers whose significant digits are very close to one another. This type of error is called a loss-of-significant-digits error or simply a loss-of-significance error. These are subtle, often disastrous computational errors. Returning now to example 7.1, we will see that it was this type of error that caused the unusual behavior noted.

EXAMPLE 7.5

A Loss-of-Significance Error 2

(x 3 + 4) − x 6 In example 7.1, we considered the function f (x) = . x3 4 Follow the calculation of f (5 × 10 ) one step at a time, as a 14-digit computer would do it.

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137

Solution We have [(5 × 104 )3 + 4]2 − (5 × 104 )6 (5 × 104 )3 (1.25 × 1014 + 4)2 − 1.5625 × 1028 = 1.25 × 1014

f (5 × 104 ) =

=

(125,000,000,000,000 + 4)2 − 1.5625 × 1028 1.25 × 1014

=

(1.25 × 1014 )2 − 1.5625 × 1028 = 0, 1.25 × 1014

REMARK 7.1 If at all possible, avoid subtractions of nearly equal values. Sometimes, this can be accomplished by some algebraic manipulation of the function.

since 125,000,000,000,004 is rounded off to 125,000,000,000,000. Note that the real culprit here was not the rounding of 125,000,000,000,004, but the fact that this was followed by a subtraction of a nearly equal value. Further, note that this is not a problem unique to the numerical computation of limits, but one that occurs in numerical computation, in general.

y

In the case of the function from example 7.5, we can avoid the subtraction and hence, the loss-of-significance error by rewriting the function as follows:

9

2

(x 3 + 4) − x 6 f (x) = x3 6 (x + 8x 3 + 16) − x 6 = x3 3 8x + 16 = , x3

8

7

x 20,000

60,000

100,000

FIGURE 1.58 y=

8x 3 + 16 x3

where we have eliminated the subtraction. Using this new (and equivalent) expression for the function, we can compute a table of function values reliably. Notice, too, that if we redraw the graph in Figure 1.57a using the new expression (see Figure 1.58), we no longer see the oscillation present in Figures 1.57a and 1.57b. From the rewritten expression, we easily obtain 2

8x 16 x3 8.016 8.000016 8.000000016 8.00000000002 8.0 8.0 8.0

(x 3 + 4) − x 6 = 8, x→∞ x3 lim

3

x 10 100 1 × 103 1 × 104 1 × 105 1 × 106 1 × 107

which is consistent with Figure 1.58 and the corrected table of function values. In example 7.6, we examine a loss-of-significance error that occurs for x close to 0.

EXAMPLE 7.6

Loss-of-Significance Involving a Trigonometric Function

1 − cos x 2 . x→0 x4

Evaluate lim

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y

1-66

Solution As usual, we look at a graph (see Figure 1.59) and some function values.

0.5

4

x

2

2

4

1 − cos x x4

2

2

1 − cos x 2 x4

x

1 − cos x 2 x4

0.1 0.01 0.001 0.0001 0.00001

0.499996 0.5 0.5 0.0 0.0

−0.1 −0.01 −0.001 −0.0001 −0.00001

0.499996 0.5 0.5 0.0 0.0

As in example 7.1, note that the function values seem to be approaching 0.5, but then suddenly take a jump down to 0.0. Once again, we are seeing a loss-of-significance error. In this particular case, this occurs because we are subtracting nearly equal values (cos x 2 and 1). We can again avoid the error by eliminating the subtraction. Note that

FIGURE 1.59 y=

x

1 − cos x 2 1 − cos x 2 1 + cos x 2 = · x4 x4 1 + cos x 2 2 2 1 − cos x = 4 x 1 + cos x 2 sin2 x 2 . = 4 x 1 + cos x 2

2

x

sin (x ) x 4 (1 cos x 2 )

±0.1 ±0.01 ±0.001 ±0.0001 ±0.00001

0.499996 0.4999999996 0.5 0.5 0.5

Multiply numerator and denominator by (1 + cos x 2 ). 1 − cos2 (x 2 ) = sin2 (x 2 ).

Since this last (equivalent) expression has no subtraction indicated, we should be able to use it to reliably generate values without the worry of loss-of-significance error. Using this to compute function values, we get the accompanying table. Using the graph and the new table, we conjecture that 1 − cos x 2 1 = . 4 x→0 x 2 lim

We offer one final example where a loss-of-significance error occurs, even though no subtraction is explicitly indicated.

EXAMPLE 7.7

Evaluate lim x[(x 2 + 4)

y

x→−∞

1 108 6 107 2 107

x

1

2

3

FIGURE 1.60 y = x[(x 2 + 4)

1/2

+ x]

A Loss-of-Significance Error Involving a Sum 1/2

+ x].

Solution Initially, you might think that since there is no subtraction (explicitly) indicated, there will be no loss-of-significance error. We first draw a graph (see Figure 1.60) and compute a table of values. #

1/2

x

x (x 2 4)

−100 −1 × 103 −1 × 104 −1 × 105 −1 × 106 −1 × 107 −1 × 108

−1.9998 −1.999998 −2.0 −2.0 −2.0 0.0 0.0

x

$

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139

You should quickly notice the sudden jump in values in the table and the wild oscillation visible in the graph. Although a subtraction is not explicitly indicated, there 1/2 is indeed a subtraction here, since we have x < 0 and (x 2 + 4) > 0. We can again remedy this with some algebraic manipulation, as follows. #

x (x + 4) 2

1/2

# $ 1/2 $ (x 2 + 4) − x + x = x (x + 4) + x # $ (x 2 + 4)1/2 − x $ # 2 (x + 4) − x 2 = x# $ (x 2 + 4)1/2 − x 4x = # $. (x 2 + 4)1/2 − x $

#

1/2

2

Multiply numerator and denominator by the conjugate.

Simplify the numerator.

We use this last expression to generate a graph in the same window as that used for Figure 1.60 and to generate the accompanying table of values. In Figure 1.61, we can see none of the wild oscillation observed in Figure 1.60, and the graph appears to be a horizontal line. y 1 108 6 107 2 107

x

1

2

3

FIGURE 1.61 y=

4x [(x 2

+ 4)1/2 − x]

4x

x

#

−100 −1 × 103 −1 × 104 −1 × 105 −1 × 106 −1 × 107 −1 × 108

−1.9998 −1.999998 −1.99999998 −1.9999999998 −2.0 −2.0 −2.0

(x 2

4)1/2 − x

$

Further, the values displayed in the table no longer show the sudden jump indicative of a loss-of-significance error. We can now confidently conjecture that lim x[(x 2 + 4)

x→−∞

1/2

+ x] = −2.

BEYOND FORMULAS In examples 7.5–7.7, we demonstrated calculations that suffered from catastrophic lossof-significance errors. In each case, we showed how we could rewrite the expression to avoid this error. We have by no means exhibited a general procedure for recognizing and repairing such errors. Rather, we hope that by seeing a few of these subtle errors, and by seeing how to fix even a limited number of them, you will become a more skeptical and intelligent user of technology.

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EXERCISES 1.7 WRITING EXERCISES 1. It is probably clear that caution is important in using technology. Equally important is redundancy. This property is sometimes thought to be a negative (wasteful, unnecessary), but its positive role is one of the lessons of this section. By redundancy, we mean investigating a problem using graphical, numerical and symbolic tools. Why is it important to use multiple methods? Answer this from a practical perspective (think of the problems in this section) and a theoretical perspective (if you have learned multiple techniques, do you understand the mathematics better?). 2. The drawback of caution and redundancy is that they take extra time. In computing limits, when should you stop and take extra time to make sure an answer is correct and when is it safe to go on to the next problem? Should you always look at a graph? compute function values? do symbolic work? an ε−δ proof? Prioritize the techniques in this chapter. f (a + h) − f (a) will be very important in h Chapter 2. For a specific function and specific a, we could compute a table of values of the fraction for smaller and smaller values of h. Why should we be wary of loss-of-significance errors?

3 3 11. lim x 4/3 ( x 2 + 1 − x 2 − 1) x→∞

√ √ 3 3 12. lim x 2/3 ( x + 4 − x − 3) x→∞

In exercises 13 and 14, compare the limits to show that small errors can have disastrous effects. x2 + x − 2 x 2 + x − 2.01 and lim x→1 x→1 x −1 x −1 x −2 x −2 14. lim 2 and lim 2 x→2 x − 4 x→2 x − 4.01 13. lim

15. Compare f (x) = sin π x and g(x) = sin 3.14x for x = 1 (radian), x = 10, x = 100 and x = 1000. 16. For exercise 1, follow the calculation of the function for x = 105 as it would proceed for a machine computing with a 10-digit mantissa. Identify where the round-off error occurs.

3. The limit lim

h→0

4. Notice that we rationalized the numerator in example 7.7. The old rule of rationalizing the denominator is another example of rewriting an expression to try to minimize computational errors. Before computers, square roots were very difficult to compute. To see one reason why you might want the square root in the numerator, suppose that dec√ you can get only one 6 imal place of accuracy, so that 3 ≈ 1.7. Compare 1.7 to √63 and then compare 2(1.7) to √63 . Which of the approximations could you do in your head?

In exercises 1–12, (a) use graphics and numerics to conjecture a value of the limit. (b) Find a computer or calculator graph showing a loss-of-significance error. (c) Rewrite the function to avoid the loss-of-significance error. √ √ 1. lim x 4x 2 + 1 − 2x 2. lim x 4x 2 + 1 + 2x x→∞

√ √ √ 3. lim x x + 4 − x + 2 x→∞

5. lim x x→∞

√

x2 + 4 −

√

x2 + 2

x→−∞

4. lim x 2 x→∞

6. lim x x→∞

1 − cos 2x 12x 2

8. lim

1 − cos x 3 x→0 x6

10. lim

7. lim

x→0

9. lim

x→0

√

√

x4 + 8 − x2

x 3 + 8 − x 3/2

1 − cos x x2

1 − cos x 4 x→0 x8

In exercises 17 and 18, compare the exact answer to one obtained by a computer with a six-digit mantissa. 17. (1.000003 − 1.000001) × 107 18. (1.000006 − 1.000001) × 107 19. If you have access to a CAS, test it on the limits of examples 7.1, 7.6 and 7.7. Based on these results, do you think that your CAS does precise calculations or numerical estimates?

EXPLORATORY EXERCISES 1. In this exercise, we look at one aspect of the mathematical study of chaos. First, iterate the function f (x) = x 2 − 2 starting at x0 = 0.5. That is, compute x1 = f (0.5), then x2 = f (x1 ), then x3 = f (x2 ) and so on. Although the sequence of numbers stays bounded, the numbers never repeat (except by the accident of round-off errors). An impressive property of chaotic functions is the butterfly effect (more properly referred to as sensitive dependence on initial conditions). The butterfly effect applies to the chaotic nature of weather and states that the amount of air stirred by a butterfly flapping its wings in Brazil can create or disperse a tornado in Texas a few days later. Therefore, longrange weather prediction is impossible. To illustrate the butterfly effect, iterate f (x) starting at x0 = 0.5 and x0 = 0.51. How many iterations does it take before the iterations are more than 0.1 apart? Try this again with x0 = 0.5 and x0 = 0.501. Repeat this exercise for the function g(x) = x 2 − 1. Even though the functions are almost identical, g(x) is not chaotic and behaves very differently. This represents an important idea in modern medical research called dynamical diseases: a small change in

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one of the constants in a function (e.g., the rate of an electrical signal within the human heart) can produce a dramatic change in the behavior of the system (e.g., the pumping of blood from the ventricles). 2. Just as we are subject to round-off error in using calculations from a computer, so are we subject to errors in a computergenerated graph. After all, the computer has to compute function values before it can decide where to plot points. On your computer or calculator, graph y = sin x 2 (a disconnected graph—or point plot—is preferable). You should see the oscillations you expect from the sine function, but with the oscillations getting faster as x gets larger. Shift your graphing window to the right several times. At some point, the plot will

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141

become very messy and almost unreadable. Depending on your technology, you may see patterns in the plot. Are these patterns real or an illusion? To explain what is going on, recall that a computer graph is a finite set of pixels, with each pixel representing one x and one y. Suppose the computer is plotting points at x = 0, x = 0.1, x = 0.2 and so on. The y-values would then be sin 02 , sin 0.12 , sin 0.22 and so on. Investigate what will happen between x = 15 and x = 16. Compute all the points (15, sin 152 ), (15.1, sin 15.12 ) and so on. If you were to graph these points, what pattern would emerge? To explain this pattern, argue that there is approximately half a period of the sine curve missing between each point plotted. Also, investigate what happens between x = 31 and x = 32.

Review Exercises WRITING EXERCISES The following list includes terms that are defined and theorems that are stated in this chapter. For each term or theorem, (1) give a precise definition or statement, (2) state in general terms what it means and (3) describe the types of problems with which it is associated. Secant line One-sided limit Removable discontinuity Vertical asymptote Method of bisections Slope of curve

Limit Continuous Horizontal asymptote Squeeze Theorem Length of line segment

Infinite limit Loss-of-significance error Slant asymptote Intermediate Value Theorem

State whether each statement is true or false and briefly explain why. If the statement is false, try to “fix it” by modifying the given statement to make a new statement that is true. 1. In calculus, problems are often solved by first approximating the solution and then improving the approximation. 2. If f (1.1) = 2.1, f (1.01) = 2.01 and so on, then lim f (x) = 2. 3. lim [ f (x) · g(x)] = [lim f (x)][lim g(x)]

x→a

x→a

In exercises 1 and 2, numerically estimate the slope of y f (x) at x a. 1. f (x) = x 2 − 2x, a = 2

x→a

In exercises 3 and 4, numerically estimate the length of the curve using (a) n 4 and (b) n 8 line segments and evenly spaced x-coordinates. 3. f (x) = sin x, 0 ≤ x ≤

π 4

4. f (x) = x 2 − x, 0 ≤ x ≤ 2

x→1

x→a

lim f (x)

4. lim

x→a

2. f (x) = sin 2x, a = 0

TRUE OR FALSE

x→a

8. Small round-off errors typically have only small effects on a calculation. √ 9. lim f (x) = L if and only if lim f (x) = L.

f (x) x→a = g(x) lim g(x) x→a

5. If f (2) = 1 and f (4) = 2, then there exists an x between 2 and 4 such that f (x) = 0. 6. For any polynomial p(x), lim p(x) = ∞. x→∞

p(x) for polynomials p and q with q(a) = 0, then 7. If f (x) = q(x) f has a vertical asymptote at x = a.

In exercises 5–10, use numerical and graphical evidence to conjecture the value of the limit. tan−1 x 2 x→0 x2

5. lim

x +2 |x + 2| 2 x 9. lim 1 + x→∞ x 7. lim

x→−2

6. lim

x→1

x2 − 1 ln x 2

8. lim (1 + 2x)1/x x→0

10. lim x 2/x x→∞

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2-2

Notice that the speed calculation in m/s is the same calculation we would use for the slope between the points (9.85, 100) and (19.79, 200). The connection between slope and speed (and other quantities of interest) is explored in this chapter.

2.1

TANGENT LINES AND VELOCITY

P Point of release

FIGURE 2.1 Path of rock

A traditional slingshot is essentially a rock on the end of a string, which you rotate around in a circular motion and then release. When you release the string, in which direction will the rock travel? An overhead view of this is illustrated in Figure 2.1. Many people mistakenly believe that the rock will follow a curved path, but Newton’s first law of motion tells us that the path in the horizontal plane is straight. In fact, the rock follows a path along the tangent line to the circle at the point of release. Our aim in this section is to extend the notion of tangent line to more general curves. To make our discussion more concrete, suppose that we want to find the tangent line to the curve y = x 2 + 1 at the point (1, 2) (see Figure 2.2). How could we define this? The tangent line hugs the curve near the point of tangency. In other words, like the tangent line to a circle, this tangent line has the same direction as the curve at the point of tangency. So, if you were standing on the curve at the point of tangency, took a small step and tried to stay on the curve, you would step in the direction of the tangent line. Another way to think of this is to observe that, if we zoom in sufficiently far, the graph appears to approximate that of a straight line. In Figure 2.3, we show the graph of y = x 2 + 1 zoomed in on the small rectangular box indicated in Figure 2.2. (Be aware that the “axes” indicated in Figure 2.3 do not intersect at the origin. We provide them only as a guide as to the scale used to produce the figure.) We now choose two points from the curve—for example, (1, 2) and (3, 10)—and compute the slope of the line joining these two points. Such a line is called a secant line and we denote its slope by m sec : m sec =

10 − 2 = 4. 3−1

An equation of the secant line is then determined by y−2 = 4, x −1 y

y 2.2

12

2.1

8

2.0 4 1.9

(1, 2) 4

x

2

2 4

4

1.8 x 0.92 0.96 1.00 1.04 1.08

FIGURE 2.2

FIGURE 2.3

y = x2 + 1

y = x2 + 1

2-3

SECTION 2.1

147

As can be seen in Figure 2.4a, the secant line doesn’t look very much like a tangent line. Taking the second point a little closer to the point of tangency, say (2, 5), gives the slope of the secant line as

y 12 8

m sec = 4 x

2

Tangent Lines and Velocity

y = 4(x − 1) + 2.

so that

4

..

2

4

4

5−2 =3 2−1

and an equation of the secant line as y = 3(x − 1) + 2. As seen in Figure 2.4b, this looks much more like a tangent line, but it’s still not quite there. Choosing our second point much closer to the point of tangency, say (1.05, 2.1025), should give us an even better approximation to the tangent line. In this case, we have

FIGURE 2.4a Secant line joining (1, 2) and (3, 10)

m sec =

2.1025 − 2 = 2.05 1.05 − 1

and an equation of the secant line is y = 2.05(x − 1) + 2. As can be seen in Figure 2.4c, the secant line looks very much like a tangent line, even when zoomed in quite far, as in Figure 2.4d. We continue this process by computing the slope of the secant line joining (1, 2) and the unspecified point (1 + h, f (1 + h)), for some value of h close to 0. The slope of this secant line is m sec =

f (1 + h) − 2 [(1 + h)2 + 1] − 2 = (1 + h) − 1 h

=

(1 + 2h + h 2 ) − 1 2h + h 2 = h h

Multiply out and cancel.

=

h(2 + h) = 2 + h. h

Factor out common h and cancel.

y

y

y 12

12

8

8

4

4

3

2

1 4

x

2

2

4

4

4

x

2

2

4

4

x 0.6

1.0

1.4

FIGURE 2.4b

FIGURE 2.4c

FIGURE 2.4d

Secant line joining (1, 2) and (2, 5)

Secant line joining (1, 2) and (1.05, 2.1025)

Close-up of secant line

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y

2-4

Notice that as h approaches 0, the slope of the secant line approaches 2, which we define to be the slope of the tangent line.

y f (x)

REMARK 1.1 We should make one more observation before moving on to the general case of tangent lines. Unlike the case for a circle, tangent lines may intersect a curve at more than one point, as indicated in Figure 2.5.

x

FIGURE 2.5

The General Case

Tangent line intersecting a curve at more than one point

To find the slope of the tangent line to y = f (x) at x = a, first pick two points on the curve. One point is the point of tangency, (a, f (a)). Call the x-coordinate of the second point x = a + h, for some small number h; the corresponding y-coordinate is then f (a + h). It is natural to think of h as being positive, as shown in Figure 2.6a, although h can also be negative, as shown in Figure 2.6b. y

y

a

ah

x

ah

x a

FIGURE 2.6a

FIGURE 2.6b

Secant line (h > 0)

Secant line (h < 0)

The slope of the secant line through the points (a, f (a)) and (a + h, f (a + h)) is given by y

m sec = Q

P x

FIGURE 2.7 Secant lines approaching the tangent line at the point P

f (a + h) − f (a) f (a + h) − f (a) = . (a + h) − a h

(1.1)

Notice that the expression in (1.1) (called a difference quotient) gives the slope of the secant line for any second point we might choose (i.e., for any h = 0). Recall that in order to obtain an improved approximation to the tangent line, we zoom in closer and closer toward the point of tangency. This makes the two points closer together, which in turn makes h closer to 0. Just how far should we zoom in? The farther, the better; this means that we want h to approach 0. We illustrate this process in Figure 2.7, where we have plotted a number of secant lines for h > 0. Notice that as the point Q approaches the point P (i.e., as h → 0), the secant line approaches the tangent line at P. We define the slope of the tangent line to be the limit of the slopes of the secant lines in (1.1) as h tends to 0, whenever this limit exists.

2-5

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..

Tangent Lines and Velocity

149

DEFINITION 1.1 The slope m tan of the tangent line to y = f (x) at x = a is given by m tan = lim

h→0

f (a + h) − f (a) , h

(1.2)

provided the limit exists. The tangent line is then the line passing through the point (a, f (a)) with slope m tan and so, the point-slope form of the equation of the tangent line is y = m tan (x − a) + f (a).

Equation of tangent line

EXAMPLE 1.1

Finding the Equation of a Tangent Line

Find an equation of the tangent line to y = x 2 + 1 at x = 1. Solution We compute the slope using (1.2): f (1 + h) − f (1) h→0 h [(1 + h)2 + 1] − (1 + 1) lim h→0 h 2 1 + 2h + h + 1 − 2 lim h→0 h 2 2h + h h(2 + h) = lim lim h→0 h→0 h h lim (2 + h) = 2.

m tan = lim y

= 12

=

8

=

4

4

=

x

2

2

4

4

Factor out common h and cancel.

h→0

Notice that the point corresponding to x = 1 is (1, 2) and the line with slope 2 through the point (1, 2) has equation y = 2(x − 1) + 2

FIGURE 2.8 y = x 2 + 1 and the tangent line at x = 1

Multiply out and cancel.

or

y = 2x.

Note how closely this corresponds to the secant lines computed earlier. We show a graph of the function and this tangent line in Figure 2.8.

EXAMPLE 1.2

Tangent Line to the Graph of a Rational Function

Find an equation of the tangent line to y =

2 at x = 2. x

Solution From (1.2), we have m tan

2 −1 f (2 + h) − f (2) 2 + h = lim = lim h→0 h→0 h h 2 − (2 + h) 2−2−h (2 + h) (2 + h) = lim = lim h→0 h→0 h h −h −1 1 = lim = lim =− . h→0 (2 + h)h h→0 2 + h 2

Since f (2 + h) =

2 . 2+h

Add fractions and multiply out.

Cancel h’s.

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The point corresponding to x = 2 is (2, 1), since f (2) = 1. An equation of the tangent line is then 1 y = − (x − 2) + 1. 2 We show a graph of function and this tangent line in Figure 2.9.

y 5 4 3 2

In cases where we cannot (or cannot easily) evaluate the limit for the slope of the tangent line, we can approximate the limit numerically. We illustrate this in example 1.3.

1 x

1 1

1

2

3

4

5

5 x 2

4

5 10

(1, 0)

x −1 x +1

(0.1, −0.8182) (0.01, −0.9802)

y 3

msec

Second Point

0 − (−1) =1 1−0 −0.8182 − (−1) = 1.818 0.1 − 0 −0.9802 − (−1) = 1.98 0.01 − 0

(−0.5, −3) (−0.1, −1.2222) (−0.01, −1.0202)

msec −3 − (−1) = 4.0 −0.5 − 0 −1.2222 − (−1) = 2.22 −0.1 − 0 −1.0202 − (−1) = 2.02 −0.01 − 0

In both columns, as the second point gets closer to (0, −1), the slope of the secant line gets closer to 2. A reasonable estimate of the slope of the curve at the point (0, −1) is then 2.

2 1 1

x −1 is shown in Figure 2.10a. We are interested in the x +1 tangent line at the point (0, −1). We sketch a tangent line in Figure 2.10b, where we have zoomed in to provide better detail. To approximate the slope, we estimate the coordinates of one point on the tangent line other than (0, −1). In Figure 2.10b, it appears that the tangent line passes through the point (1, 1). An estimate of the slope is 1 − (−1) then m tan ≈ = 2. To approximate the slope numerically, we choose several 1−0 points near (0, −1) and compute the slopes of the secant lines. For example, rounding the y-values to four decimal places, we have Second Point

FIGURE 2.10a

2

x −1 at x +1

Solution A graph of y =

10

y=

Graphical and Numerical Approximation of Tangent Lines

x = 0.

y

2

EXAMPLE 1.3

Graphically and numerically approximate the slope of the tangent line to y =

FIGURE 2.9 2 y = and tangent line at (2, 1) x

4

2-6

x 1

1

2 3

FIGURE 2.10b Tangent line

2

Velocity The slopes of tangent lines have many important applications, of which one of the most important is in computing velocity. The term velocity is certainly familiar to you, but can you say precisely what it is? We often describe velocity as a quantity determining the speed and direction of an object, but what exactly is speed? If your car did not have a speedometer, you might determine your speed using the familiar formula distance = rate × time.

(1.3)

Using (1.3), you can find the rate (speed) by simply dividing the distance by the time. However, the rate in (1.3) refers to average speed over a period of time. We are interested in the speed at a specific instant. The following story should indicate the difference. During traffic stops, police officers frequently ask drivers if they know how fast they were going. An overzealous student might answer that during the past, say, 3 years,

2-7

SECTION 2.1

..

Tangent Lines and Velocity

151

2 months, 7 days, 5 hours and 45 minutes, they’ve driven exactly 45,259.7 miles, so that their speed was rate =

distance 45,259.7 miles = ≈ 1.62118 mph. time 27,917.75 hours

Of course, most police officers would not be impressed with this analysis, but, why is it wrong? Certainly there’s nothing wrong with formula (1.3) or the arithmetic. However, it’s reasonable to argue that unless they were in their car during this entire 3-year period, the results are invalid. Suppose that the driver substitutes the following argument instead: “I left home at 6:17 P.M. and by the time you pulled me over at 6:43 P.M., I had driven exactly 17 miles. Therefore, my speed was rate =

17 miles 60 minutes · = 39.2 mph, 26 minutes 1 hour

well under the posted 45-mph speed limit.” While this is a much better estimate of the velocity than the 1.6 mph computed previously, it’s still an average velocity using too long of a time period. More generally, suppose that the function f (t) gives the position at time t of an object moving along a straight line. That is, f (t) gives the displacement (signed distance) from a fixed reference point, so that f (t) < 0 means that the object is located | f (t)| away from the reference point, but in the negative direction. Then, for two times a and b (where a < b), f (b) − f (a) gives the signed distance between positions f (a) and f (b). The average velocity vavg is then given by

vavg =

EXAMPLE 1.4

signed distance f (b) − f (a) = . time b−a

(1.4)

Finding Average Velocity

The position of a car after t minutes driving in a straight line is given by s(t) =

1 2 1 t − t 3, 2 12

0 ≤ t ≤ 4.

Approximate the velocity at time t = 2. Solution Averaging over the 2 minutes from t = 2 to t = 4, we get from (1.4) that 2.666666667 − 1.333333333 s(4) − s(2) ≈ 4−2 2 ≈ 0.666666667 mile/minute ≈ 40 mph. Of course, a 2-minute-long interval is rather long, given that cars can speed up and slow down a great deal in 2 minutes. We get an improved approximation by averaging over just one minute: vavg =

s(3) − s(2) 2.25 − 1.333333333 ≈ 3−2 1 ≈ 0.9166666667 mile/minute ≈ 55 mph.

vavg =

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1.0

s(2 h) − s(2) h 0.9166666667

0.1

0.9991666667

0.01

0.9999916667

0.001

0.999999917

0.0001

1.0

0.00001

1.0

h

2-8

While this latest estimate is certainly better than the first one, we can do better. As we make the time interval shorter and shorter, the average velocity should be getting closer and closer to the velocity at the instant t = 2. It stands to reason that, if we compute the average velocity over the time interval [2, 2 + h] and then let h → 0, the resulting average velocities should be getting closer and closer to the velocity at the instant t = 2. s(2 + h) − s(2) s(2 + h) − s(2) vavg = = . We have (2 + h) − 2 h A sequence of these average velocities is displayed in the accompanying table, for h > 0, with similar results if we allow h to be negative. It appears that the average velocity is approaching 1 mile/minute (60 mph), as h → 0. We refer to this limiting value as the instantaneous velocity. This leads us to make the following definition.

NOTE

DEFINITION 1.2

Notice that if (for example) t is measured in seconds and f (t) is measured in feet, then velocity (average or instantaneous) is measured in feet per second (ft/s). The term velocity is always used to refer to instantaneous velocity.

If f (t) represents the position of an object relative to some fixed location at time t as it moves along a straight line, then the instantaneous velocity at time t = a is given by v(a) = lim

h→0

f (a + h) − f (a) f (a + h) − f (a) = lim , h→0 (a + h) − a h

(1.5)

provided the limit exists.

EXAMPLE 1.5

Finding Average and Instantaneous Velocity

Suppose that the height of a falling object t seconds after being dropped from a height of 64 feet is given by f (t) = 64 − 16t 2 feet. Find the average velocity between times t = 1 and t = 2; the average velocity between times t = 1.5 and t = 2; the average velocity between times t = 1.9 and t = 2 and the instantaneous velocity at time t = 2. Solution The average velocity between times t = 1 and t = 2 is vavg =

f (2) − f (1) 64 − 16(2)2 − [64 − 16(1)2 ] = = −48 ft/s. 2−1 1

The average velocity between times t = 1.5 and t = 2 is vavg =

64 − 16(2)2 − [64 − 16(1.5)2 ] f (2) − f (1.5) = = −56 ft/s. 2 − 1.5 0.5

The average velocity between times t = 1.9 and t = 2 is vavg =

f (2) − f (1.9) 64 − 16(2)2 − [64 − 16(1.9)2 ] = = −62.4 ft/s. 2 − 1.9 0.1

2-9

SECTION 2.1

..

Tangent Lines and Velocity

153

The instantaneous velocity is the limit of such average velocities. From (1.5), we have v(2) = lim

h→0

f (2 + h) − f (2) (2 + h) − 2

[64 − 16(2 + h)2 ] − [64 − 16(2)2 ] h→0 h

= lim

[64 − 16(4 + 4h + h 2 )] − [64 − 16(2)2 ] h→0 h

= lim

−64h − 16h 2 −16h(h + 4) = lim h→0 h→0 h h = lim [−16(h + 4)] = −64 ft/s. = lim

Multiply out and cancel.

Factor out common h and cancel.

h→0

Recall that velocity indicates both speed and direction. In this problem, f (t) measures the height above the ground. So, the negative velocity indicates that the object is moving in the negative (or downward) direction. The speed of the object at the 2-second mark is then 64 ft/s. (Speed is simply the absolute value of velocity.) Observe that the formulas for instantaneous velocity (1.5) and for the slope of a tangent line (1.2) are identical. We want to make this connection as strong as possible, by illustrating example 1.5 graphically. We graph the position function f (t) = 64 − 16t 2 for 0 ≤ t ≤ 2. The average velocity between t = 1 and t = 2 corresponds to the slope of the secant line between the points at t = 1 and t = 2. (See Figure 2.11a.) Similarly, the average velocity between t = 1.5 and t = 2 gives the slope of the corresponding secant line. (See Figure 2.11b.) Finally, the instantaneous velocity at time t = 2 corresponds to the slope of the tangent line at t = 2. (See Figure 2.11c.) y

y

y

80

80

80

60

60

60

40

40

40

20

20

20

t 1

2

t

3

1

2

t

3

1

20

20

20

40

40

40

60

60

60

2

FIGURE 2.11a

FIGURE 2.11b

FIGURE 2.11c

Secant line between t = 1 and t =2

Secant line between t = 1.5 and t = 2

Tangent line at t = 2

3

Velocity is a rate (more precisely, the instantaneous rate of change of position with respect to time). We now generalize this notion of instantaneous rate of change. In general,

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..

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2-10

the average rate of change of a function f (x) between x = a and x = b (a = b) is given by f (b) − f (a) . b−a The instantaneous rate of change of f (x) at x = a is given by lim

h→0

f (a + h) − f (a) , h

provided the limit exists. The units of the instantaneous rate of change are the units of f divided by (or “per”) the units of x.

EXAMPLE 1.6

Interpreting Rates of Change

If the function f (t) gives the population of a city in millions of people t years after January 1, 2000, interpret each of the following quantities, assuming that they f (2) − f (0) equal the given numbers. (a) = 0.34, (b) f (2) − f (1) = 0.31 and 2 f (2 + h) − f (2) (c) lim = 0.3. h→0 h f (b) − f (a) Solution From the preceding, is the average rate of change of the b−a function f between a and b. Expression (a) tells us that the average rate of change of f between a = 0 and b = 2 is 0.34. Stated in more common language, the city’s population grew at an average rate of 0.34 million people per year between 2000 and 2002. Similarly, expression (b) is the average rate of change between a = 1 and b = 2. That is, the city’s population grew at an average rate of 0.31 million people per year in 2001. Finally, expression (c) gives the instantaneous rate of change of the population at time t = 2. As of January 1, 2002, the city’s population was growing at a rate of 0.3 million people per year. Additional applications of the slope of a tangent line are innumerable. These include the rate of a chemical reaction, the inflation rate in economics and learning growth rates in psychology. Rates of change in nearly any discipline you can name can be thought of as slopes of tangent lines. We explore many of these applications as we progress through the text. You hopefully noticed that we tacked the phrase “provided the limit exists” onto the end of the definitions of the slope of a tangent line, the instantaneous velocity and the instantaneous rate of change. This was important, since these defining limits do not always exist, as we see in example 1.7.

y y x

EXAMPLE 1.7

A Graph with No Tangent Line at a Point

Determine whether there is a tangent line to y = |x| at x = 0. Slope 1

Slope 1 x

FIGURE 2.12 y = |x|

Solution We can look at this problem graphically, numerically and symbolically. The graph is shown in Figure 2.12. Our graphical technique is to zoom in on the point of tangency until the graph appears straight. However, no matter how far we zoom in on (0, 0), the graph continues to look like Figure 2.12. (This is one reason why we left off the scale on Figure 2.12.) From this evidence alone, we would conjecture that the tangent line does not exist. Numerically, the slope of the tangent line is the limit of the

2-11

SECTION 2.1

..

Tangent Lines and Velocity

155

slope of a secant line, as the second point approaches the point of tangency. Observe that the secant line through (0, 0) and (1, 1) has slope 1, as does the secant line through (0, 0) and (0.1, 0.1). In fact, if h is any positive number, the slope of the secant line through (0, 0) and (h, |h|) is 1. However, the secant line through (0, 0) and (−1, 1) has slope −1, as does the secant line through (0, 0) and (h, |h|) for any negative number h. We therefore conjecture that the one-sided limits are different, so that the limit (and also the tangent line) does not exist. To prove this conjecture, we take our cue from the numerical work and look at one-sided limits: if h > 0, then |h| = h, so that lim+

h→0

f (0 + h) − f (0) |h| − 0 h = lim+ = lim+ = 1. h→0 h→0 h h h

On the other hand, if h < 0, then |h| = −h (remember that if h < 0, −h > 0), so that lim−

h→0

f (0 + h) − f (0) |h| − 0 −h = lim− = lim− = −1. h→0 h→0 h h h

Since the one-sided limits are different, we conclude that lim

h→0

f (0 + h) − f (0) does not exist h

and hence, the tangent line does not exist.

EXERCISES 2.1 WRITING EXERCISES 1. What does the phrase “off on a tangent” mean? Relate the common meaning of the phrase to the image of a tangent to a circle (use the slingshot example, if that helps). In what way does the zoomed image of the tangent promote the opposite view of the relationship between a curve and its tangent? 2. In general, the instantaneous velocity of an object cannot be computed directly; the limit process is the only way to compute velocity at an instant. Given this, how does a car’s speedometer compute velocity? (Hint: Look this up in a reference book or on the Internet. An important aspect of the car’s ability to do this seemingly difficult task is that it performs analog calculations. For example, the pitch of a fly’s buzz gives us an analog device for computing the speed of a fly’s wings, since pitch is proportional to speed.) 3. Look in the news media (TV, newspaper, Internet) and find references to at least five different rates. We have defined a rate of change as the limit of the difference quotient of a function. For your five examples, state as precisely as possible what the original function is. Is the rate given quantitatively or qualitatively? If it is given quantitatively, is the rate given as a percentage or a number? In calculus, we usually compute rates (quantitatively) as numbers; is this in line with the standard usage? 4. Sketch the graph of a function that is discontinuous at x = 1. Explain why there is no tangent line at x = 1.

In exercises 1–4, sketch in a plausible tangent line at the given point. (Hint: Mentally zoom in on the point and use the zoomed image of the tangent.) 1.

y

x

p

2.

y

x

3.

at x = π

at x = 0

y

x

at x = 0

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Differentiation

y

4.

2-12

y

8. B

C

A x

D

x

at x = 1

1

In exercises 5 and 6, estimate the slope of the tangent line to the curve at x 1. y

5. 1.5

In exercises 9–12, compute the slope of the secant line between the points at (a) x 1 and x 2, (b) x 2 and x 3, (c) x 1.5 and x 2, (d) x 2 and x 2.5, (e) x 1.9 and x 2, (f) x 2 and x 2.1, and (g) use parts (a)–(f) and other calculations as needed to estimate the slope of the tangent line at x 2. √ 9. f (x) = x 3 − x 10. f (x) = x 2 + 1

1.0

11. f (x) = cos x 2

0.5

In exercises 13–16, use a CAS or graphing calculator. 13. On one graph, sketch the secant lines in exercise 9, parts (a)–(d) and the tangent line in part (g).

x 0.5

1.0

12. f (x) = tan (x/4)

1.5

14. On one graph, sketch the secant lines in exercise 10, parts (a)–(d) and the tangent line in part (g). y

6.

15. Animate the secant lines in exercise 9, parts (a), (c) and (e), converging to the tangent line in part (g).

2 x 1

2

16. Animate the secant lines in exercise 9, parts (b), (d) and (f), converging to the tangent line in part (g).

3

In exercises 17–24, find the equation of the tangent line to y f (x) at x a. Graph y f (x) and the tangent line to verify that you have the correct equation.

2 4

In exercises 7 and 8, list the points A, B, C and D in order of increasing slope of the tangent line.

17. f (x) = x 2 − 2, a = 1

18. f (x) = x 2 − 2, a = 0

19. f (x) = x 2 − 3x, a = −2

20. f (x) = x 3 + x, a = 1

2 ,a =1 x +1 √ 23. f (x) = x + 3, a = −2

x ,a =0 x −1 √ 24. f (x) = x 2 + 1, a = 1

21. f (x) =

22. f (x) =

y

7.

In exercises 25–30, use graphical and numerical evidence to determine whether the tangent line to y f (x) exists at x a. If it does, estimate the slope of the tangent; if not, explain why not.

B A

D C

25. f (x) = |x − 1| at a = 1 x

4x at a = 1 x −1 −2x 2 if x < 0 at a = 0 27. f (x) = x3 if x ≥ 0 26. f (x) =

2-13

SECTION 2.1

29. f (x) = 30. f (x) =

−2x if x < 1 at a = 1 x − 3 if x ≥ 1 x2 − 1 x3 + 1 −2x x 2 − 2x

Tangent Lines and Velocity

157

Water level

28. f (x) =

..

if x < 0 at a = 0 if x ≥ 0 if x < 0 at a = 0 if x > 0

24 hours

Time

In exercises 31–34, the function represents the position in feet of an object at time t seconds. Find the average velocity between (a) t 0 and t 2, (b) t 1 and t 2, (c) t 1.9 and t 2, (d) t 1.99 and t 2, and (e) estimate the instantaneous velocity at t 2.

41. Suppose a hot cup of coffee is left in a room for 2 hours. Sketch a reasonable graph of what the temperature would look like as a function of time. Then sketch a graph of what the rate of change of the temperature would look like.

31. f (t) = 16t 2 + 10 √ 33. f (t) = t 2 + 8t

42. Sketch a graph representing the height of a bungee-jumper. Sketch the graph of the person’s velocity (use + for upward velocity and − for downward velocity).

32. f (t) = 3t 3 + t 34. f (t) = 100 sin(t/4)

In exercises 35 and 36, use the position function f (t) meters to find the velocity at time t a seconds. 35. f (t) = −16t 2 + 5, (a) a = 1; (b) a = 2 √ 36. f (t) = t + 16, (a) a = 0; (b) a = 2 37. The table shows the freezing temperature of water in degrees Celsius at various pressures. Estimate the slope of the tangent line at p = 1 and interpret the result. Estimate the slope of the tangent line at p = 3 and interpret the result. p (atm) ◦ C

0 0

1 −7

2 −20

3 −16

4 −11

38. The table shows the range of a soccer kick launched at 30◦ above the horizontal at various initial speeds. Estimate the slope of the tangent line at v = 50 and interpret the result. Distance (yd) Speed (mph)

19 30

28 40

37 50

47 60

58 70

Elevation

39. The graph shows the elevation of a person on a hike up a mountain as a function of time. When did the hiker reach the top? When was the hiker going the fastest on the way up? When was the hiker going the fastest on the way down? What do you think occurred at places where the graph is level?

4 hours

Time

40. The graph shows the amount of water in a city water tank as a function of time. When was the tank the fullest? the emptiest? When was the tank filling up at the fastest rate? When was the tank emptying at the fastest rate? What time of day do you think the level portion represents?

43. Suppose that f (t) represents the balance in dollars of a bank account t years after January 1, 2000. Interpret each of the folf (4) − f (2) lowing. (a) = 21,034, (b) 2[ f (4) − f (3.5)] = 2 f (4 + h) − f (4) = 30,000. 25,036 and (c) lim h→0 h 44. Suppose that f (m) represents the value of a car that has been driven m thousand miles. Interpret each of the following. f (40) − f (38) = −2103, (b) f (40) − f (39) = −2040 (a) 2 f (40 + h) − f (40) and (c) lim = −2000. h→0 h 45. In using a slingshot, it is important to generate a large angular θ(a + h) − θ (a) , velocity. Angular velocity is defined by lim h→0 h where θ(t) is the angle of rotation at time t. If the angle of a slingshot is θ(t) = 0.4t 2 , what is the angular velocity after three rotations? [Hint: Which value of t (seconds) corresponds to three rotations?] 46. Find the angular velocity of the slingshot in exercise 45 after two rotations. Explain why the third rotation is helpful. 47. Sometimes an incorrect method accidentally produces a correct answer. For quadratic functions (but definitely not most other functions), the average velocity between t = r and t = s equals the average of the velocities at t = r and t = s. To show this, assume that f (t) = at 2 + bt + c is the distance function. Show that the average velocity between t = r and t = s equals a(s + r ) + b. Show that the velocity at t = r is 2ar + b and the velocity at t = s is 2as + b. Finally, show that (2ar + b) + (2as + b) a(s + r ) + b = . 2 48. Find a cubic function [try f (t) = t 3 + · · ·] and numbers r and s such that the average velocity between t = r and t = s is different from the average of the velocities at t = r and t = s.

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2-14

f (a + h) − f (a) f (x) − f (a) = lim . (Hint: x→a h x −a

How do reflections normally occur in nature? From water! You would perceive a tree and its reflection in a pool of water. This is the desert mirage!

50. Use the second limit in exercise 49 to recompute the slope in exercises 17 and 19. Which limit do you prefer?

2. You can use a VCR to estimate speed. Most VCRs play at 30 frames per second. So, with a frame-by-frame advance, you can estimate time as the number of frames divided by 30. If you know the distance covered, you can compute the average velocity by dividing distance by time. Try this to estimate how fast you can throw a ball, run 50 yards, hit a tennis ball or whatever speed you find interesting. Some of the possible inaccuracies are explored in exercise 3.

49. Show that lim

h→0

Let h = x − a.)

2 51. A car speeding around a curve in the shape 1 1of y = x (moving from left to right) skids off at the point 2 , 4 . If the car contin ues in a straight path, will it hit a tree located at the point 1, 34 ?

52. For the car in exercise 51, show graphically that there is only 2 one skid point on the curve y = x such that the tangent line passes through the point 1, 34 .

EXPLORATORY EXERCISES 1. Many optical illusions are caused by our brain’s (unconscious) use of the tangent line in determining the positions of objects. Suppose you are in the desert 100 feet from a palm tree. You see a particular spot 10 feet up on the palm tree due to light reflecting from that spot to your eyes. Normally, it is a good approximation to say that the light follows a straight line (top path in the figure).

Two paths of light from tree to person. However, when there is a large temperature difference in the air, light may follow nonlinear paths. If, as in the desert, the air near the ground is much hotter than the air higher up, light will bend as indicated by the bottom path in the figure. Our brains always interpret light coming in straight paths, so you would think the spot on the tree is at y = 10 because of the top path and also at some other y because of the bottom path. If the bottom curve is y = 0.002x 2 − 0.24x + 10, find an equation of the tangent line at x = 100 and show that it crosses the y-axis at y = −10. That is, you would “see” the spot at y = 10 and also at y = −10, a perfect reflection.

Two perceived locations of tree.

3. What is the peak speed for a human being? It has been estimated that Carl Lewis reached a peak speed of 28 mph while winning a gold medal in the 1992 Olympics. Suppose that we have the following data for a sprinter.

Meters 30 40 50 56 58 60

Seconds 3.2 4.2 5.16666 5.76666 5.93333 6.1

Meters 62 64 70 80 90 100

Seconds 6.26666 6.46666 7.06666 8.0 9.0 10.0

We want to estimate peak speed. We could start by computing distance 100 m = = 10 m/s, but this is the average speed time 10 s over the entire race, not the peak speed. Argue that we want to compute average speeds only using adjacent measurements (e.g., 40 and 50 meters, or 50 and 56 meters). Do this for all 11 adjacent pairs and find the largest speed (if you want to convert to mph, divide by 0.447). We will then explore how accurate this estimate might be. Notice that all times are essentially multiples of 1/30, since the data were obtained using the VCR technique in exercise 2. Given this, why is it suspicious that all the distances are whole numbers? To get an idea of how much this might affect your calculations, change some of the distances. For instance, if you change 60 (meters) to 59.8, how much do your average velocity calculations change? One possible way to identify where mistakes have been made is to look at the pattern of average velocities: does it seem reasonable? Would a sprinter speed up and slow down in such a pattern? In places where the pattern seems suspicious, try adjusting the distances and see if you can produce a more realistic pattern. Taking all this into account, try to quantify your error analysis: what is the highest (lowest) the peak speed could be?

2-15

SECTION 2.2

2.2

..

The Derivative

159

THE DERIVATIVE In section 2.1, we investigated two seemingly unrelated concepts: slopes of tangent lines and velocity, both of which are expressed in terms of the same limit. This is an indication of the power of mathematics, that otherwise unrelated notions are described by the same mathematical expression. This particular limit turns out to be so useful that we give it a special name.

DEFINITION 2.1 The derivative of the function f (x) at x = a is defined as f (a + h) − f (a) , (2.1) h provided the limit exists. If the limit exists, we say that f is differentiable at x = a. f (a) = lim

h→0

An alternative form of (2.1) is f (a) = lim

b→a

f (b) − f (a) . b−a

(2.2)

(See exercise 49 in section 2.1.)

EXAMPLE 2.1

Finding the Derivative at a Point

Compute the derivative of f (x) = 3x 3 + 2x − 1 at x = 1. Solution From (2.1), we have f (1 + h) − f (1) h 3(1 + h)3 + 2(1 + h) − 1 − (3 + 2 − 1) = lim h→0 h

f (1) = lim

h→0

3(1 + 3h + 3h 2 + h 3 ) + (2 + 2h) − 1 − 4 h→0 h

Multiply out and cancel.

11h + 9h 2 + 3h 3 h→0 h

Factor out common h and cancel.

= lim = lim

= lim (11 + 9h + 3h 2 ) = 11. h→0

Suppose that in example 2.1 we had also needed to find f (2) and f (3). Must we now repeat the same long limit calculation to find each of f (2) and f (3)? Instead, we compute the derivative without specifying a value for x, leaving us with a function from which we can calculate f (a) for any a, simply by substituting a for x.

EXAMPLE 2.2

Finding the Derivative at an Unspecified Point

Find the derivative of f (x) = 3x 3 + 2x − 1 at an unspecified value of x. Then, evaluate the derivative at x = 1, x = 2 and x = 3.

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2-16

Solution Replacing a with x in the definition of the derivative (2.1), we have f (x + h) − f (x) h→0 h 3(x + h)3 + 2(x + h) − 1 − (3x 3 + 2x − 1) = lim h→0 h

f (x) = lim

3(x 3 + 3x 2 h + 3xh 2 + h 3 ) + (2x + 2h) − 1 − 3x 3 − 2x + 1 h→0 h

= lim

Multiply out and cancel.

9x 2 h + 9xh 2 + 3h 3 + 2h h→0 h = lim (9x 2 + 9xh + 3h 2 + 2)

Factor out common h and cancel.

= lim

h→0

= 9x 2 + 0 + 0 + 2 = 9x 2 + 2. Notice that in this case, we have derived a new function, f (x) = 9x 2 + 2. Simply substituting in for x, we get f (1) = 9 + 2 = 11 (the same as we got in example 2.1!), f (2) = 9(4) + 2 = 38 and f (3) = 9(9) + 2 = 83. Example 2.2 leads us to the following definition.

DEFINITION 2.2 The derivative of f (x) is the function f (x) given by f (x + h) − f (x) , (2.3) h provided the limit exists. The process of computing a derivative is called differentiation. Further, f is differentiable on an interval I if it is differentiable at every point in I . f (x) = lim

h→0

In examples 2.3 and 2.4, observe that the name of the game is to write down the defining limit and then to find some way of evaluating that limit (which initially has the indeterminate form 00 ).

EXAMPLE 2.3

Finding the Derivative of a Simple Rational Function

1 (x = 0), find f (x). x Solution We have

If f (x) =

f (x + h) − f (x) h→0 h

1 1 − x +h x lim h→0 h x − (x + h) x(x + h) lim h→0 h −h lim h→0 hx(x + h) −1 1 lim = − 2, h→0 x(x + h) x

f (x) = lim

=

= = = or f (x) = −x −2 .

Since f (x + h) =

1 . x +h

Add fractions and cancel.

Cancel h’s.

2-17

..

SECTION 2.2

EXAMPLE 2.4 If f (x) =

√

The Derivative

161

The Derivative of the Square Root Function

x (for x ≥ 0), find f (x).

Solution We have f (x + h) − f (x) h √ √ x +h− x = lim h→0 h √ √ √ √ x +h− x x +h+ x = lim √ √ h→0 h x +h+ x

f (x) = lim

h→0

(x + h) − x √ √ x +h+ x h = lim √ √ h→0 h x +h+ x = lim

h→0

h→0

Multiply out and cancel.

h

= lim √

1 x +h+

√

Multiply numerator and denominator by √ √ the conjugate: x + h + x.

Cancel common h’s.

x

1 1 = √ = x −1/2 . 2 2 x Notice that f (x) is defined only for x > 0. y 15 10 5

4

x

2

2

4

The benefits of having a derivative function go well beyond simplifying the computation of a derivative at multiple points. As we’ll see, the derivative function tells us a great deal about the original function. Keep in mind that the value of a derivative at a point is the slope of the tangent line at that point. In Figures 2.13a–2.13c, we have graphed a function along with its tangent lines at three different points. The slope of the tangent line in Figure 2.13a is negative; the slope of the tangent line in Figure 2.13c is positive and the slope of the tangent line in Figure 2.13b is zero. These three tangent lines give us three points on the graph of the derivative function (see Figure 2.13d), by estimating the value of f (x) at the three points. Thus, as x changes, the slope of the tangent line changes and hence f (x) changes.

FIGURE 2.13a m tan < 0 y y

y 4

15

15

10

10

5

5

2

2

x

1

1 2

4

x

2

2

4

4

x

2

2

FIGURE 2.13b

FIGURE 2.13c

m tan = 0

m tan > 0

4

4

FIGURE 2.13d

y = f (x) (three points)

2

162

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2-18

y

Sketching the Graph of f (x ) Given the Graph of f (x )

EXAMPLE 2.5 60

Given the graph of f (x) in Figure 2.14, sketch a plausible graph of f (x).

40 20 4

x

2

2

4

20 40 60

FIGURE 2.14 y = f (x)

y

Solution Rather than worrying about exact values of the slope, we only wish to get the general shape right. As in Figures 2.13a–2.13d, pick a few important points to analyze carefully. You should focus on any discontinuities and any places where the graph of f turns around. The graph levels out at approximately x = −2 and x = 2. At these points, the derivative is 0. As we move from left to right, the graph rises for x < −2, drops for −2 < x < 2 and rises again for x > 2. This means that f (x) > 0 for x < −2, f (x) < 0 for −2 < x < 2 and finally f (x) > 0 for x > 2. We can say even more. As x approaches −2 from the left, observe that the tangent lines get less steep. Therefore, f (x) becomes less positive as x approaches −2 from the left. Moving to the right from x = −2, the graph gets steeper until about x = 0, then gets less steep until it levels out at x = 2. Thus, f (x) gets more negative until x = 0, then less negative until x = 2. Finally, the graph gets steeper as we move to the right from x = 2. Putting this all together, we have the possible graph of f (x) shown in red in Figure 2.15, superimposed on the graph of f (x). The opposite question to that asked in example 2.5 is even more interesting. That is, given the graph of a derivative, what might the graph of the original function look like? We explore this in example 2.6.

60 f ' (x) 40 20 4

x

2

2

4

20 40 f (x)

60

FIGURE 2.15

y = f (x) and y = f (x)

Sketching the Graph of f (x ) Given the Graph of f (x )

EXAMPLE 2.6

Given the graph of f (x) in Figure 2.16, sketch a plausible graph of f (x). Solution Again, do not worry about getting exact values of the function, but rather only the general shape of the graph. Notice from the graph of y = f (x) that f (x) < 0 for x < −2, so that on this interval, the slopes of the tangent lines to y = f (x) are negative and the function is decreasing. On the interval (−2, 1), f (x) > 0, indicating that the tangent lines to the graph of y = f (x) have positive slope and the function is increasing. Further, this says that the graph turns around (i.e., goes from decreasing to increasing) at x = −2. We have drawn a graph exhibiting this behavior in Figure 2.17 y

y 20

20 f (x)

10

10

x

4

2

4

x

4

2

10

10

20

f '(x) 20

FIGURE 2.16 y = f (x)

FIGURE 2.17

4

y = f (x) and a plausible graph of y = f (x)

2-19

SECTION 2.2

..

The Derivative

163

superimposed on the graph of y = f (x). Further, f (x) < 0 on the interval (1, 3), so that the function decreases here. Finally, for x > 3, we have that f (x) > 0, so that the function is increasing here. We show a graph exhibiting all of this behavior in Figure 2.17. We drew the graph of f so that the small “valley” on the right side of the y-axis was not as deep as the one on the left side of the y-axis for a reason. Look carefully at the graph of f (x) and notice that | f (x)| gets much larger on (−2, 1) than on (1, 3). This says that the tangent lines and hence, the graph will be much steeper on the interval (−2, 1) than on (1, 3).

Alternative Derivative Notations

HISTORICAL NOTES Gottfried Leibniz (1646–1716) A German mathematician and philosopher who introduced much of the notation and terminology in calculus and who is credited (together with Sir Isaac Newton) with inventing the calculus. Leibniz was a prodigy who had already received his law degree and published papers on logic and jurisprudence by age 20. A true Renaissance man, Leibniz made important contributions to politics, philosophy, theology, engineering, linguistics, geology, architecture and physics, while earning a reputation as the greatest librarian of his time. Mathematically, he derived many fundamental rules for computing derivatives and helped promote the development of calculus through his extensive communications. The simple and logical notation he invented made calculus accessible to a wide audience and has only been marginally improved upon in the intervening 300 years. He wrote, “In symbols one observes an advantage in discovery which is greatest when they express the exact nature of a thing briefly . . . then indeed the labor of thought is wonderfully diminished.”

We have denoted the derivative function by f (x). There are other commonly used notations, each with advantages and disadvantages. One of the coinventors of the calculus, Gottfried df Leibniz, used the notation (Leibniz notation) for the derivative. If we write y = f (x), dx the following are all alternatives for denoting the derivative: dy df d f (x) = y = = = f (x). dx dx dx d The expression is called a differential operator and tells you to take the derivative of dx whatever expression follows. In section 2.1, we observed that f (x) = |x|, does not have a tangent line at x = 0 (i.e., it is not differentiable at x = 0), although it is continuous everywhere. Thus, there are continuous functions that are not differentiable. You might have already wondered whether the reverse is true. That is, are there differentiable functions that are not continuous? The answer (no) is provided by Theorem 2.1.

THEOREM 2.1 If f (x) is differentiable at x = a, then f (x) is continuous at x = a.

PROOF For f to be continuous at x = a, we need only show that lim f (x) = f (a). We consider x→a f (x) − f (a) lim [ f (x) − f (a)] = lim Multiply and divide by (x − a). (x − a) x→a x→a x −a f (x) − f (a) By Theorem 3.1 (iii) = lim lim (x − a) from section 1.3. x→a x→a x −a = f (a)(0) = 0,

Since f is differentiable at x = a.

where we have used the alternative definition of derivative (2.2) discussed earlier. By Theorem 3.1 in section 1.3, it now follows that 0 = lim [ f (x) − f (a)] = lim f (x) − lim f (a) x→a

x→a

x→a

= lim f (x) − f (a), x→a

which gives us the result. Note that Theorem 2.1 says that if a function is not continuous at a point then it cannot have a derivative at that point. It also turns out that functions are not differentiable at any point where their graph has a “sharp” corner, as is the case for f (x) = |x| at x = 0. (See example 1.7.)

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y

2-20

EXAMPLE 2.7

y f (x)

Show that f (x) =

f(x) 2

4 f(x) 0

Showing That a Function Is Not Differentiable at a Point 4 2x

if x < 2 is not differentiable at x = 2. if x ≥ 2

Solution The graph (see Figure 2.18) indicates a sharp corner at x = 2, so you might expect that the derivative does not exist. To verify this, we investigate the derivative by evaluating one-sided limits. For h > 0, note that (2 + h) > 2 and so, f (2 + h) = 2(2 + h). This gives us

x 2

FIGURE 2.18 lim+

A sharp corner

h→0

y

f (2 + h) − f (2) 2(2 + h) − 4 = lim+ h→0 h h 4 + 2h − 4 = lim+ h→0 h 2h = 2. = lim+ h→0 h

Multiply out and cancel.

Cancel common h’s.

Likewise, if h < 0, (2 + h) < 2 and so, f (2 + h) = 4. Thus, we have lim−

h→0

f (2 + h) − f (2) 4−4 = lim− = 0. h→0 h h

Since the one-sided limits do not agree (0 = 2), f (2) does not exist (i.e., f is not differentiable at x = 2). x

a

FIGURE 2.19a

Figures 2.19a–2.19d show a variety of functions for which f (a) does not exist. In each case, convince yourself that the derivative does not exist.

A jump discontinuity y y

y

a

a

x

a

x

x

FIGURE 2.19b

FIGURE 2.19c

FIGURE 2.19d

A vertical asymptote

A cusp

A vertical tangent line

Numerical Differentiation There are many times in applications when it is not possible or practical to compute derivatives symbolically. This is frequently the case in applications where we have only some data (i.e., a table of values) representing an otherwise unknown function. You will need an understanding of the limit definition to compute reasonable estimates of the derivative.

2-21

SECTION 2.2

EXAMPLE 2.8

..

The Derivative

165

Approximating a Derivative Numerically

√ Numerically estimate the derivative of f (x) = x 2 x 3 + 2 at x = 1.

Solution We are not anxious to struggle through the limit definition for this function. The definition tells us, however, that the derivative at x = 1 is the limit of slopes of secant lines. We compute some of these below: f (1 h) − f (1) h 4.7632 4.3715 4.3342

h 0.1 0.01 0.001

h −0.1 −0.01 −0.001

f (1 h) − f (1) h 3.9396 4.2892 4.3260

Notice that the slopes seem to be converging to approximately 4.33 as h approaches 0. Thus, we make the approximation f (1) ≈ 4.33.

EXAMPLE 2.9

Estimating Velocity Numerically

Suppose that a sprinter reaches the following distances in the given times. Estimate the velocity of the sprinter at the 6-second mark. t(s) f (t) (ft)

5.0 123.7

5.5 141.01

5.8 151.41

5.9 154.90

6.0 158.40

6.1 161.92

6.2 165.42

6.5 175.85

7.0 193.1

Solution The instantaneous velocity is the limit of the average velocity as the time interval shrinks. We first compute the average velocities over the shortest intervals given, from 5.9 to 6.0 and from 6.0 to 6.1.

Time Interval

Average Velocity

Time Interval

Average Velocity

(5.9, 6.0)

35.0 ft/s

(5.5, 6.0)

34.78 ft/s

(6.0, 6.1)

35.2 ft/s

(5.8, 6.0)

34.95 ft/s

(5.9, 6.0)

35.00 ft/s

(6.0, 6.1)

35.20 ft/s

(6.0, 6.2)

35.10 ft/s

(6.0, 6.5)

34.90 ft/s

Since these are the best individual estimates available from the data, we could just split the difference and estimate a velocity of 35.1 ft/s. However, there is useful information in the rest of the data. Based on the accompanying table, we can conjecture that the sprinter was reaching a peak speed at about the 6-second mark. Thus, we might accept the higher estimate of 35.2 ft/s. We should emphasize that there is not a single correct answer to this question, since the data are incomplete (i.e., we know the distance only at fixed times, rather than over a continuum of times).

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BEYOND FORMULAS In sections 2.3–2.8, we derive numerous formulas for computing derivatives. As you learn these formulas, keep in mind the reasons that we are interested in the derivative. Careful studies of the slope of the tangent line to a curve and the velocity of a moving object led us to the same limit, which we named the derivative. In general, the derivative represents the rate of change or the ratio of the change of one quantity to the change in another quantity. The study of change in a quantifiable way led directly to modern science and engineering. If we were limited to studying phenomena with only constant change, how much of the science that you have learned would still exist?

EXERCISES 2.2 WRITING EXERCISES 1. The derivative is important because of its many different uses and interpretations. Describe four aspects of the derivative: graphical (think of tangent lines), symbolic (the derivative function), numerical (approximations) and applications (velocity and others).

In exercises 13–18, match the graphs of the functions on the left with the graphs of their derivatives on the right. 13.

y

2. Mathematicians often use the word “smooth” to describe functions with certain (desirable) properties. Graphically, how are differentiable functions smoother than functions that are continuous but not differentiable, or functions that are not continuous? 3. Briefly describe what the derivative tells you about the original function. In particular, if the derivative is positive at a point, what do you know about the trend of the function at that point? What is different if the derivative is negative at the point?

(a)

y

x x

14.

y

(b)

y

x x

4. Show that the derivative of f (x) = 3x − 5 is f (x) = 3. Explain in terms of slope why this is true. In exercises 1–4, compute f (a) using the limits (2.1) and (2.2). 1. f (x) = 3x + 1, a = 1 3. f (x) =

√ 3x + 1, a = 1

15.

y

(c)

y

2. f (x) = 3x 2 + 1, a = 1 4. f (x) =

x

3 ,a = 2 x +1

x

In exercises 5–12, compute the derivative function f (x) using (2.1) or (2.2). 5. f (x) = 3x 2 + 1

6. f (x) = x 2 − 2x + 1

3 7. f (x) = x +1 √ 9. f (x) = 3x + 1

2 8. f (x) = 2x − 1

11. f (x) = x 3 + 2x − 1

10. f (x) = 2x + 3 12. f (x) = x 4 − 2x 2 + 1

16.

y

(d)

y x

x

2-23

17.

SECTION 2.2

y

y

(e)

..

The Derivative

167

y

22.

x

x

x

18.

y

y

(f)

x x

In exercises 23 and 24, use the given graph of f (x) to sketch a plausible graph of a continuous function f (x). y

23.

In exercises 19–22, use the given graph of f (x) to sketch a graph of f (x). 19.

x

y

y

24.

x

x

20.

25. Graph f (x) = |x| + |x − 2| and identify all x-values at which f (x) is not differentiable.

y

26. Graph f (x) = e−2/(x −x) and identify all x-values at which f (x) is not differentiable. 3

27. Find all real numbers p such that f (0) exists for f (x) = x p . x

28. Prove that if f (x) is differentiable at x = a, then f (a + ch) − f (a) = c f (a). lim h→0 h 29. If

f (x) is differentiable f (x 2 ) − f (a 2 ) . lim x→a x 2 − a2

21.

y

at

x = a = 0,

evaluate

30. Prove that if f (x) is differentiable at x = 0, f (x) ≤ 0 for all x and f (0) = 0, then f (0) = 0.

x

31. The table shows the margin of error in degrees for tennis serves hit at 100 mph with various amounts of topspin (in units of revolutions per second). Estimate the slope of the derivative at x = 60, and interpret it in terms of the benefit of extra spin. (Data adapted from The Physics and Technology of Tennis by Brody, Cross and Lindsey.) Topspin (rps) Margin of error

20 1.8

40 2.4

60 3.1

80 3.9

100 4.6

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32. The table shows the margin of error in degrees for tennis serves hit at 120 mph from various heights. Estimate the slope of the derivative at x = 8.5 and interpret it in terms of hitting a serve from a higher point. (Data adapted from The Physics and Technology of Tennis by Brody, Cross and Lindsey.) Height (ft) Margin of error

7.5 0.3

8.0 0.58

8.5 0.80

9.0 1.04

9.5 1.32

In exercises 33 and 34, use the distances f (t) to estimate the velocity at t 2. 33.

34.

t f (t) t f (t)

1.7 3.1

1.8 3.9

1.7 4.6

1.9 4.8

1.8 5.3

2.0 5.8

1.9 6.1

2.1 6.8

2.0 7.0

2.2 7.7

2.1 7.8

2.3 8.5

2.2 8.6

2.3 9.3

35. The Environmental Protection Agency uses the measurement of ton-MPG to evaluate the power-train efficiency of vehicles. The ton-MPG rating of a vehicle is given by the weight of the vehicle (in tons) multiplied by a rating of the vehicle’s fuel efficiency in miles per gallon. Several years of data for new cars are given in the table. Estimate the rate of change of ton-MPG in (a) 1994 and (b) 2000. Do your estimates imply that cars are becoming more or less efficient? Is the rate of change constant or changing? Year Ton-MPG

1992 44.9

1994 45.7

1996 46.5

1998 47.3

2000 47.7

36. The fuel efficiencies in miles per gallon of cars from 1992 to 2000 are shown in the following table. Estimate the rate of change in MPG in (a) 1994 and (b) 2000. Do your estimates imply that cars are becoming more or less fuel efficient? Comparing your answers to exercise 35, what must be happening to the average weight of cars? If weight had remained constant, what do you expect would have happened to MPG? Year MPG

1992 28.0

1994 28.1

1996 28.3

1998 28.5

2000 28.1

In exercises 37 and 38, use a CAS or graphing calculator. 37. Numerically estimate f (1) for f (x) = x x and verify your answer using a CAS. 38. Numerically estimate f (π) for f (x) = x sin x and verify your answer using a CAS.

In exercises 39 and 40, compute the right-hand derivative f (h) − f (0) D f (0) lim and the left-hand derivative h h→0 f (h) − f (0) D− f (0) lim . h h→0− 2x + 1 if x < 0 39. f (x) = 3x + 1 if x ≥ 0 2 x if x < 0 40. f (x) = 3 x if x ≥ 0 g(x) if x < 0 41. Assume that f (x) = . If f is continuous at k(x) if x ≥ 0 x = 0 and g and k are differentiable at x = 0, prove that D+ f (0) = k (0) and D− f (0) = g (0). Which statement is not true if f has a jump discontinuity at x = 0? 42. Explain why the derivative f (0) exists if and only if the onesided derivatives exist and are equal. 43. If f (x) > 0 for all x, use the tangent line interpretation to argue that f is an increasing function; that is, if a < b, then f (a) < f (b). 44. If f (x) < 0 for all x, use the tangent line interpretation to argue that f is a decreasing function; that is, if a < b, then f (a) > f (b). 45. If f (x) = x 2/3 , show graphically and numerically that f is continuous at x = 0 but f (0) does not exist. 0 if x < 0 , show graphically and numerically 46. If f (x) = 2x if x ≥ 0 that f is continuous at x = 0 but f (0) does not exist. 47. Give an example showing that the following is not true for all functions f : if f (x) ≤ x, then f (x) ≤ 1. 48. Determine whether the following is true for all functions f : if f (0) = 0, f (x) exists for all x and f (x) ≤ x, then f (x) ≤ 1. In exercises 49 and 50, give the units for the derivative function. 49. (a) f (t) represents position, measured in meters, at time t seconds. (b) f (x) represents the demand, in number of items, of a product when the price is x dollars. 50. (a) c(t) represents the amount of a chemical present, in grams, at time t minutes. (b) p(x) represents the mass, in kg, of the first x meters of a pipe. 51. Let f (t) represent the trading value of a stock at time t days. If f (t) < 0, what does that mean about the stock? If you held some shares of this stock, should you sell what you have or buy more?

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169

52. Suppose that there are two stocks with trading values f (t) and g(t), where f (t) > g(t) and 0 < f (t) < g (t). Based on this information, which stock should you buy? Briefly explain.

62. Sketch the graph of a function with the following properties: f (−2) = 4, f (0) = −2, f (2) = 1, f (−2) = −2, f (0) = 2 and f (2) = 1.

53. One model for the spread of a disease assumes that at first the disease spreads very slowly, gradually the infection rate increases to a maximum and then the infection rate decreases back to zero, marking the end of the epidemic. If I (t) represents the number of people infected at time t, sketch a graph of both I (t) and I (t), assuming that those who get infected do not recover.

63. A phone company charges one dollar for the first 20 minutes of a call, then 10 cents per minute for the next 60 minutes and 8 cents per minute for each additional minute (or partial minute). Let f (t) be the price in cents of a t-minute phone call, t > 0. Determine f (t) as completely as possible.

54. One model for urban population growth assumes that at first, the population is growing very rapidly, then the growth rate decreases until the population starts decreasing. If P(t) is the population at time t, sketch a graph of both P(t) and P (t). 55. Use the graph to list the following in increasing order: f (1), f (1.5) − f (1) , f (1). f (2) − f (1), 0.5

64. The table shows the percentage of English Premier League soccer players by birth month, where x = 0 represents November, x = 1 represents December and so on. (The data are adapted from John Wesson’s The Science of Soccer.) If these data come from a differentiable function f (x), estimate f (1). Interpret the derivative in terms of the effect of being a month older but in the same grade of school. Month Percent

0 13

1 11

2 9

3 7

4 7

y 10

EXPLORATORY EXERCISES

8

1. Compute the derivative function for x 2 , x 3 and x 4 . Based on your results, identify the pattern and conjecture a general formula √ for the derivative of x n . Test your conjecture on the functions x = x 1/2 and 1/x = x −1 .

6 4 2 3 2 1

x 1

2

3

Exercises 55 and 56 56. Use the graph to list the following in increasing order: f (0), f (0) − f (−0.5) , f (0). f (0) − f (−1), 0.5 In exercises 57–60, the limit equals f (a) for some function f (x) and some constant a. Determine f (x) and a. (1 + h)2 − (1 + h) h→0 h √ 4+h−2 58. lim h→0 h

57. lim

1 1 − 2 + h 2 59. lim h→0 h (h − 1)2 − 1 h→0 h

60. lim

61. Sketch the graph of a function with the following properties: f (0) = 1, f (1) = 0, f (3) = 6, f (0) = 0, f (1) = −1 and f (3) = 4.

2. In Theorem 2.1, it is stated that a differentiable function is guaranteed to be continuous. The converse is not true: continuous functions are not necessarily differentiable (see example 2.7). This fact is carried to an extreme in Weierstrass’ function, to be explored here. First, graph the function f 4 (x) = cos x + 12 cos 3x + 14 cos 9x + 18 cos 27x + 161 cos 81x in the graphing window 0 ≤ x ≤ 2π and −2 ≤ y ≤ 2. Note that the graph appears to have several sharp corners, where a derivative would not exist. Next, graph the function 1 1 f 6 (x) = f 4 (x) + 32 cos 243x + 64 cos 729x. Note that there are even more places where the graph appears to have sharp corners. Explore graphs of f 10 (x), f 14 (x) and so on, with more terms added. Try to give graphical support to the fact that the Weierstrass function f ∞ (x) is continuous for all x but is not differentiable for any x. More graphical evidence comes from the fractal nature of the Weierstrass function: compare the graphs of f 4 (x) with 0 ≤ x ≤ 2π and −2 ≤ y ≤ 2 and f 6 (x) − cos x − 12 cos 3x with 0 ≤ x ≤ 2π and − 12 ≤ y ≤ 12 . 9 Explain why the graphs are identical. Then explain why this indicates that no matter how much you zoom in on a graph of the Weierstrass function, you will continue to see wiggles and corners. That is, you cannot zoom in to find a tangent line. 3. Suppose there is a function F(x) such that F(1) = 1 and F(0) = f 0 , where 0 < f 0 < 1. If F (1) > 1, show graphically that the equation F(x) = x has a solution q where 0 < q < 1. (Hint: Graph y = x and a plausible F(x) and look for

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intersections.) Sketch a graph where F (1) < 1 and there are no solutions to the equation F(x) = x between 0 and 1 (although x = 1 is a solution). Solutions have a connection with the probability of the extinction of animals or family names. Suppose you and your descendants have children according to the following probabilities: f 0 = 0.2 is the probability of having no children, f 1 = 0.3 is the probability of having exactly one child, and f 2 = 0.5 is the probability of having two children. Define F(x) = 0.2 + 0.3x + 0.5x 2 and show that F (1) > 1. Find the solution of F(x) = x between x = 0 and x = 1; this number is the probability that your “line” will go extinct some time into the future. Find nonzero values of f 0 , f 1 and f 2 such that the corresponding F(x) satisfies F (1) < 1 and hence the probability of your line going extinct is 1.

as h approaches 0. Then compute the limit and compare to the derivative f (1) found in example 1.1. For h = 1, h = 0.5 and h = 0.1, compare the actual values of the symmetric difference f (a + h) − f (a) quotient and the usual difference quotient . h In general, which difference quotient provides a better estimate of the derivative? Next, compare the values of the difference quotients with h = 0.5 and h = −0.5 to the derivative f (1). Explain graphically why one is smaller and one is larger. Compare the average of these two difference quotients to the symmetric difference quotient with h = 0.5. Use this result to explain why the symmetric difference quotient might provide a better estimate of the derivative. Next, compute sev4 if x < 2 eral symmetric difference quotients of f (x) = 2x if x ≥ 2 centered at a = 2. Recall that in example 2.7 we showed that the derivative f (2) does not exist. Given this, discuss one major problem with using the symmetric difference quotient to approximate derivatives. Finally, show that if f (a) exists, f (a + h) − f (a − h) = f (a). then lim h→0 2h

4. The symmetric difference quotient of a function f centered at f (a + h) − f (a − h) . If f (x) = x 2 + 1 x = a has the form 2h and a = 1, illustrate the symmetric difference quotient as a slope of a secant line for h = 1 and h = 0.5. Based on your picture, conjecture the limit of the symmetric difference quotient

2.3

COMPUTATION OF DERIVATIVES: THE POWER RULE You have now computed numerous derivatives using the limit definition. In fact, you may have computed enough that you have started taking some shortcuts. In exploratory exercise 1 in section 2.2, we asked you to compile a list of derivatives of basic functions and to generalize. We continue that process in this section.

The Power Rule We first revisit the limit definition of derivative to compute two very simple derivatives.

y yc c

For any constant c,

a

FIGURE 2.20 A horizontal line

d c = 0. dx

(3.1)

x

Notice that (3.1) says that for any constant c, the horizontal line y = c has a tangent line with zero slope. That is, the tangent line to a horizontal line is the same horizontal line (see Figure 2.20). Let f (x) = c, for all x. From the definition in equation (2.3), we have d f (x + h) − f (x) c = f (x) = lim h→0 dx h c−c = lim = lim 0 = 0. h→0 h→0 h

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intersections.) Sketch a graph where F (1) < 1 and there are no solutions to the equation F(x) = x between 0 and 1 (although x = 1 is a solution). Solutions have a connection with the probability of the extinction of animals or family names. Suppose you and your descendants have children according to the following probabilities: f 0 = 0.2 is the probability of having no children, f 1 = 0.3 is the probability of having exactly one child, and f 2 = 0.5 is the probability of having two children. Define F(x) = 0.2 + 0.3x + 0.5x 2 and show that F (1) > 1. Find the solution of F(x) = x between x = 0 and x = 1; this number is the probability that your “line” will go extinct some time into the future. Find nonzero values of f 0 , f 1 and f 2 such that the corresponding F(x) satisfies F (1) < 1 and hence the probability of your line going extinct is 1.

as h approaches 0. Then compute the limit and compare to the derivative f (1) found in example 1.1. For h = 1, h = 0.5 and h = 0.1, compare the actual values of the symmetric difference f (a + h) − f (a) quotient and the usual difference quotient . h In general, which difference quotient provides a better estimate of the derivative? Next, compare the values of the difference quotients with h = 0.5 and h = −0.5 to the derivative f (1). Explain graphically why one is smaller and one is larger. Compare the average of these two difference quotients to the symmetric difference quotient with h = 0.5. Use this result to explain why the symmetric difference quotient might provide a better estimate of the derivative. Next, compute sev4 if x < 2 eral symmetric difference quotients of f (x) = 2x if x ≥ 2 centered at a = 2. Recall that in example 2.7 we showed that the derivative f (2) does not exist. Given this, discuss one major problem with using the symmetric difference quotient to approximate derivatives. Finally, show that if f (a) exists, f (a + h) − f (a − h) = f (a). then lim h→0 2h

4. The symmetric difference quotient of a function f centered at f (a + h) − f (a − h) . If f (x) = x 2 + 1 x = a has the form 2h and a = 1, illustrate the symmetric difference quotient as a slope of a secant line for h = 1 and h = 0.5. Based on your picture, conjecture the limit of the symmetric difference quotient

2.3

COMPUTATION OF DERIVATIVES: THE POWER RULE You have now computed numerous derivatives using the limit definition. In fact, you may have computed enough that you have started taking some shortcuts. In exploratory exercise 1 in section 2.2, we asked you to compile a list of derivatives of basic functions and to generalize. We continue that process in this section.

The Power Rule We first revisit the limit definition of derivative to compute two very simple derivatives.

y yc c

For any constant c,

a

FIGURE 2.20 A horizontal line

d c = 0. dx

(3.1)

x

Notice that (3.1) says that for any constant c, the horizontal line y = c has a tangent line with zero slope. That is, the tangent line to a horizontal line is the same horizontal line (see Figure 2.20). Let f (x) = c, for all x. From the definition in equation (2.3), we have d f (x + h) − f (x) c = f (x) = lim h→0 dx h c−c = lim = lim 0 = 0. h→0 h→0 h

2-27

..

SECTION 2.3

Computation of Derivatives: The Power Rule

171

Similarly, we have

y yx

d x = 1. dx

a

FIGURE 2.21

x

(3.2)

Notice that (3.2) says that the tangent line to the line y = x is a line of slope one (i.e., y = x; see Figure 2.21). This is unsurprising, since intuitively, it should be clear that the tangent line to any line is that same line. We verify this result as follows. Let f (x) = x. From equation (2.3), we have

Tangent line to y = x

d f (x + h) − f (x) x = f (x) = lim h→0 dx h = lim

h→0

(x + h) − x h

h = lim 1 = 1. h→0 h h→0

= lim f (x)

f (x)

1 x x2 x3 x4

0 1 2x 3x 2 4x 3

The table shown in the margin presents a short list of derivatives calculated previously either as examples or in the exercises using the limit definition. Can you identify any pattern to the derivatives shown in the table? There are two features to note. First, the power of x in the derivative is always one less than the power of x in the original function. Second, the coefficient of x in the derivative is the same as the power of x in the original function. Putting these ideas into symbolic form, we make and then prove a reasonable conjecture.

THEOREM 3.1 (Power Rule) For any integer n > 0,

d n x = nx n−1 . dx

PROOF From the limit definition of derivative given in equation (2.3), if f (x) = x n , then d n f (x + h) − f (x) (x + h)n − x n x = f (x) = lim = lim . h→0 h→0 dx h h

(3.3)

To evaluate the limit, we will need to simplify the expression in the numerator. If n is a positive integer, we can multiply out (x + h)n . Recall that (x + h)2 = x 2 + 2xh + h 2 and (x + h)3 = x 3 + 3x 2 h + 3xh 2 + h 3 . More generally, you may recall from the binomial theorem that (x + h)n = x n + nx n−1 h +

n(n − 1) n−2 2 x h + · · · + nxh n−1 + h n . 2

(3.4)

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Substituting (3.4) into (3.3), we get

f (x) = lim

x n + nx n−1 h +

h→0

n(n − 1) n−2 2 x h + · · · + nxh n−1 + h n − x n 2 h

Cancel x n terms.

n(n − 1) n−2 2 x h + · · · + nxh n−1 + h n 2 = lim h→0 h n(n − 1) Factor out h nx n−1 + x n−2 h 1 + · · · + nxh n−2 + h n−1 2 common h = lim h→0 h and cancel. n(n − 1) n−2 1 = lim nx n−1 + x h + · · · + nxh n−2 + h n−1 = nx n−1 , h→0 2 nx n−1 h +

since every term but the first has a factor of h. The power rule is very easy to apply, as we see in example 3.1.

EXAMPLE 3.1

Using the Power Rule

Find the derivative of f (x) = x 8 and g(t) = t 107 . Solution We have f (x) = Similarly,

g (t) =

d 8 x = 8x 8−1 = 8x 7 . dx

d 107 t = 107t 107−1 = 107t 106 . dt

Recall that in section 2.2, we showed that

d 1 1 = − 2. dx x x

(3.5)

Notice that we can rewrite (3.5) as d −1 x = (−1)x −2 . dx

REMARK 3.1 As we will see, the power rule holds for any power of x. We will not be able to prove this fact for some time now, as the proof of Theorem 3.1 does not generalize, since the expansion in equation (3.4) holds only for positive integer exponents. Even so, we will use the rule freely for any power of x. We state this in Theorem 3.2.

That is, the derivative of x −1 follows the same pattern as the power rule that we just stated and proved for positive integer exponents. Likewise, in section 2.2, we used the limit definition to show that d √ 1 x= √ . dx 2 x We can also rewrite (3.6) as

(3.6)

d 1/2 1 x = x −1/2 . dx 2

Here, notice that the derivative of a rational power of x also follows the same pattern as the power rule that we proved for positive integer exponents.

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..

Computation of Derivatives: The Power Rule

173

THEOREM 3.2 (General Power Rule) For any real number r ,

d r x = r x r −1 . dx

(3.7)

The power rule is simple to use, as we see in example 3.2.

CAUTION Be careful here to avoid a common error: d −19 = −19x −18 . x dx The power rule says to subtract 1 from the exponent (even if the exponent is negative).

EXAMPLE 3.2

Using the General Power Rule

1 √ 3 , x 2 and x π . x 19 Solution From (3.7), we have

d d −19 1 = = −19x −19−1 = −19x −20 . x d x x 19 dx √ 3 If we rewrite x 2 as a fractional power of x, we can use (3.7) to compute the derivative, as follows. d √ d 2/3 2 2 3 x2 = x = x 2/3−1 = x −1/3 . dx dx 3 3 d π x = π x π −1 . Finally, we have dx

Find the derivative of

Notice that there is the additional conceptual problem in example 3.2 (which we resolve in Chapter 4) of deciding what x π means. Since the exponent isn’t rational, what exactly do we mean when we raise a number to the irrational power π ?

General Derivative Rules The power rule gives us a large class of functions whose derivatives we can quickly compute without using the limit definition. The following rules for combining derivatives further expand the number of derivatives we can compute without resorting to the definition. Keep in mind that a derivative is a limit; the differentiation rules in Theorem 3.3 then follow immediately from the corresponding rules for limits (found in Theorem 3.1 in Chapter 1).

THEOREM 3.3 If f (x) and g(x) are differentiable at x and c is any constant, then d [ f (x) + g(x)] = f (x) + g (x), dx d (ii) [ f (x) − g(x)] = f (x) − g (x) and dx d (iii) [c f (x)] = c f (x). dx (i)

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PROOF We prove only part (i). The proofs of parts (ii) and (iii) are left as exercises. Let k(x) = f (x) + g(x). Then, from the limit definition of the derivative (2.3), we get d k(x + h) − k(x) [ f (x) + g(x)] = k (x) = lim h→0 dx h = lim

h→0

[ f (x + h) + g(x + h)] − [ f (x) + g(x)] h

[ f (x + h) − f (x)] + [g(x + h) − g(x)] h→0 h

= lim = lim

h→0

f (x + h) − f (x) g(x + h) − g(x) + lim h→0 h h

By definition of k(x). Grouping the f terms together and the g terms together. By Theorem 3.1 in Chapter 1. Recognizing the derivatives of f and of g.

= f (x) + g (x).

We illustrate Theorem 3.3 by working through the calculation of a derivative step by step, showing all of the details.

EXAMPLE 3.3

Finding the Derivative of a Sum

√ Find the derivative of f (x) = 2x 6 + 3 x. Solution We have

d d √ (2x 6 ) + 3 x dx dx d d = 2 (x 6 ) + 3 (x 1/2 ) dx dx

1 −1/2 5 x = 2(6x ) + 3 2

By Theorem 3.3 (iii).

3 = 12x 5 + √ . 2 x

Simplifying.

f (x) =

EXAMPLE 3.4

By Theorem 3.3 (i).

By the power rule.

Rewriting a Function before Computing the Derivative

√ 4x 2 − 3x + 2 x Find the derivative of f (x) = . x Solution Note that we don’t yet have any rule for computing the derivative of a quotient. So, we must first rewrite f (x) by dividing out the x in the denominator. We have √ 4x 2 3x 2 x f (x) = − + = 4x − 3 + 2x −1/2 . x x x From Theorem 3.3 and the power rule (3.7), we get

d d −1/2 d 1 −3/2 f (x) = 4 (x) − 3 (1) + 2 (x = 4 − x −3/2 . )=4−0+2 − x dx dx dx 2

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..

Computation of Derivatives: The Power Rule

175

y

EXAMPLE 3.5 10

Finding the Equation of the Tangent Line

2 at x = 1. x Solution First, notice that f (x) = 4 − 4x + 2x −1 . From Theorem 3.3 and the power rule, we have Find an equation of the tangent line to f (x) = 4 − 4x +

5 x 1

2

3

5

f (x) = 0 − 4 − 2x −2 = −4 − 2x −2 . At x = 1, the slope of the tangent line is then f (1) = −4 − 2 = −6. The line with slope −6 through the point (1, 2) has equation

10

y − 2 = −6 (x − 1).

FIGURE 2.22 y = f (x) and the tangent line at x = 1

We show a graph of y = f (x) and the tangent line at x = 1 in Figure 2.22.

Higher Order Derivatives One consequence of having the derivative function is that we can compute the derivative of a derivative. It turns out that such higher order derivatives have important applications. Suppose we start with a function f (x) and compute its derivative f (x). We can then compute the derivative of f (x), called the second derivative of f and written f (x). We can then compute the derivative of f (x), called the third derivative of f , written f (x). We can continue to take derivatives indefinitely. Below, we show common notations for the first five derivatives of f [where we assume that y = f (x)]. Order

Prime Notation

Leibniz Notation df dx

1

y = f (x)

2

y = f (x)

d2 f dx2

3

y = f (x)

d3 f dx3

4

y (4) = f (4) (x)

d4 f dx4

5

y (5) = f (5) (x)

d5 f dx5

Computing higher order derivatives is done by simply computing several first derivatives, as we see in example 3.6.

EXAMPLE 3.6

Computing Higher Order Derivatives

If f (x) = 3x 4 − 2x 2 + 1, compute as many derivatives as possible. Solution We have f (x) =

df d = (3x 4 − 2x 2 + 1) = 12x 3 − 4x. dx dx

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d2 f dx2 d3 f f (x) = dx3 d4 f f (4) (x) = dx4 d5 f f (5) (x) = dx5 and so on. It follows that

Then,

f (x) =

d (12x 3 − 4x) = 36x 2 − 4, dx d = (36x 2 − 4) = 72x, dx d = (72x) = 72, dx d = (72) = 0 dx =

f (n) (x) =

dn f = 0, for n ≥ 5. dxn

Acceleration What information does the second derivative of a function give us? Graphically, we get a property called concavity, which we develop in Chapter 3. One important application of the second derivative is acceleration, which we briefly discuss now. You are probably familiar with the term acceleration, which is the instantaneous rate of change of velocity. Consequently, if the velocity of an object at time t is given by v(t), then the acceleration is dv a(t) = v (t) = . dt

EXAMPLE 3.7

Computing the Acceleration of a Skydiver

Suppose that the height of a skydiver t seconds after jumping from an airplane is given by f (t) = 640 − 20t − 16t 2 feet. Find the person’s acceleration at time t. Solution Since acceleration is the derivative of velocity, we first compute velocity: v(t) = f (t) = 0 − 20 − 32t = −20 − 32t ft/s. Computing the derivative of this function gives us a(t) = v (t) = −32. To finish the problem, we need to determine the units of acceleration. Since the distance here is measured in feet and time is measured in seconds, the units of the velocity are feet per second, so that the units of acceleration are feet per second per second, written ft/s/s, or more commonly ft/s2 (feet per second squared). Our answer says that the velocity changes by −32 ft/s every second. In this case, the speed in the downward (negative) direction increases by 32 ft/s every second due to gravity.

BEYOND FORMULAS The power rule gives us a much-needed shortcut for computing many derivatives. Mathematicians always seek the shortest, most efficient computations. By skipping unnecessary lengthy steps and saving brain power, mathematicians free themselves to tackle complex problems with creativity. It’s important to remember,

2-33

SECTION 2.3

..

Computation of Derivatives: The Power Rule

177

however, that shortcuts such as the power rule must always be carefully proved. What shortcuts do you use when solving an equation such as 3x 2 − 4 = 5 and what are the intermediate steps that you are skipping?

EXERCISES 2.3 WRITING EXERCISES 1. Explain to a non-calculus-speaking friend how to (mechanically) use the power rule. Decide whether it is better to give separate explanations for positive and negative exponents; integer and noninteger exponents; other special cases. 2. In the 1700s, mathematical “proofs” were, by modern standards, a bit fuzzy and lacked rigor. In 1734, the Irish metaphysician Bishop Berkeley wrote The Analyst to an “infidel mathematician” (thought to be Edmund Halley of Halley’s comet fame). The accepted proof at the time of the power rule may be described as follows. If x is incremented to x + h, then x n is incremented to (x + h)n . It follows that n 2 − n n−2 (x + h)n − x n n−1 = nx hx + + · · · . Now, let the (x + h) − x 2 increment h vanish, and the derivative is nx n−1 . Bishop Berkeley objected to this argument. “But it should seem that the reasoning is not fair or conclusive. For when it is said, ‘let the increments vanish,’ the former supposition that the increments were something, or that there were increments, is destroyed, and yet a consequence of that supposition is retained. Which . . . is a false way of reasoning. Certainly, when we suppose the increments to vanish, we must suppose . . . everything derived from the supposition of their existence to vanish with them.” Do you think Berkeley’s objection is fair? Is it logically acceptable to assume that something exists to draw one conclusion, and then assume that the same thing does not exist to avoid having to accept other consequences? Mathematically speaking, how does the limit avoid Berkeley’s objection of the increment h both existing and not existing? 3. The historical episode in exercise 2 is just one part of an ongoing conflict between people who blindly use mathematical techniques without proof and those who insist on a full proof before permitting anyone to use the technique. To which side are you sympathetic? Defend your position in an essay. Try to anticipate and rebut the other side’s arguments. 4. Explain the first two terms in the expansion (x + h)n = x n + nhx n−1 + · · ·, where n is a positive integer. Think of multiplying out (x + h)(x + h)(x + h) · · · (x + h); how many terms would include x n ? x n−1 ?

In exercises 1–16, find the derivative of each function. 1. f (x) = x 3 − 2x + 1 √ 3. f (t) = 3t 3 − 2 t 3 5. f (x) = − 8x + 1 x 10 7. h(x) = √ − 2x x 9. f (s) = 2s 3/2 − 3s −1/3 √ 11. f (x) = 2 3 x + 3 √ 13. f (x) = x 3x 2 − x 15. f (x) =

3x 2 − 3x + 1 2x

2. f (x) = x 9 − 3x 5 + 4x 2 − 4x √ 4. f (s) = 5 s − 4s 2 + 3 2 6. f (x) = 4 − x 3 + 2 x 3 8. h(x) = 12x − x 2 − √ x 10. f (t) = 3t π − 2t 1.3 √ 3 12. f (x) = 4x − 3 x 2 14. f (x) = (x + 1)(3x 2 − 4) 16. f (x) =

4x 2 − x + 3 √ x

In exercises 17–24, compute the indicated derivative. 17. f (x) for f (x) = x 4 + 3x 2 − 2 18.

√ d2 f for f (x) = x 6 − x dx2

19.

d2 f 3 for f (x) = 2x 4 − √ dx2 x

20. f (t) for f (t) = 4t 2 − 12 +

4 t2

21. f (4) (x) for f (x) = x 4 + 3x 2 − 2 22. f (5) (x) for f (x) = x 10 − 3x 4 + 2x − 1 x2 − x + 1 √ x √ (4) 2 24. f (t) for f (t) = (t − 1)( t + t)

23. f (x) for f (x) =

In exercises 25–28, use the given position function to find the velocity and acceleration functions. 25. s(t) = −16t 2 + 40t + 10 26. s(t) = 12t 3 − 6t − 1 √ 27. s(t) = t + 2t 2 28. s(t) = 10 −

10 t

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2-34

In exercises 29–32, the given function represents the height of an object. Compute the velocity and acceleration at time t t0 . Is the object going up or down? Is the speed of the object increasing or decreasing?

y

(c)

29. h(t) = −16t 2 + 40t + 5, t0 = 1

x

30. h(t) = −16t 2 + 40t + 5, t0 = 2 31. h(t) = 10t 2 − 24t, t0 = 2 32. h(t) = 10t 2 − 24t, t0 = 1 In exercises 33–36, find an equation of the tangent line to y f (x) at x a. √ 34. f (x) = x 2 − 2x + 1, a = 2 33. f (x) = 4 x − 2x, a = 4 35. f (x) = x 2 − 2, a = 2

y

42. (a) 10

36. f (x) = 3x + 4, a = 2

In exercises 37 and 38, determine the value(s) of x for which the tangent line to y f (x) is horizontal. Graph the function and determine the graphical significance of each point. 37. f (x) = x 3 − 3x + 1

5

38. f (x) = x 4 − 2x 2 + 2

In exercises 39 and 40, determine the value(s) of x for which the slope of the tangent line to y f (x) does not exist. Graph the function and determine the graphical significance of each point. 39. f (x) = x 2/3

41. (a)

1

2

3

1

2

3

5 10

40. f (x) = x 1/3

In exercises 41 and 42, one curve represents a function f (x) and the other two represent f (x) and f (x). Determine which is which.

x

3 2 1

y

(b) 10

y

5

x

3 2 1 5 x

(b)

10

y

(c)

y

10 5

−3

x

−2

−1

−10

x 1

2

3

2-35

SECTION 2.3

In exercises 43 and 44, find a general formula for the nth derivative f (n) (x). 43. f (x) =

√

44. f (x) =

x

2 x

45. Find a second-degree polynomial (of the form ax 2 + bx + c) such that f (0) = −2, f (0) = 2 and f (0) = 3. 46. Find a second-degree polynomial (of the form ax 2 + bx + c) such that f (0) = 0, f (0) = 5 and f (0) = 1. 47. Find the area of the triangle bounded by x = 0, y = 0 and the tangent line to y = x1 at x = 1. Repeat with the triangle bounded by x = 0, y = 0 and the tangent line to y = x1 at x = 2. Show that you get the same area using the tangent line to y = x1 at any x = a > 0. 48. Show that the result of exercise 47 does not hold for y = x12 . That is, the area of the triangle bounded by x = 0, y = 0 and the tangent line to y = x12 at x = a > 0 does depend on the value of a. 49. Assume that a is a real number, f (x) is differentiable for all x ≥ a and g(x) = max f (t) for x ≥ a. Find g (x) in the cases a≤t≤x

(a) f (x) > 0 and (b) f (x) < 0. 50. Assume that a is a real number, f (x) is differentiable for all x ≥ a and g(x) = min f (t) for x ≥ a. Find g (x) in the cases a≤t≤x

(a) f (x) > 0 and (b) f (x) < 0.

..

Computation of Derivatives: The Power Rule

[Hint: to estimate the second derivative, estimate f (1998) and f (1999) and look for a trend.] t

1996

1997

1998

1999

2000

2001

f (t)

7664.8

8004.5

8347.3

8690.7

9016.8

9039.5

56. Let f (t) equal the average weight of a domestic SUV in year t. Several values are given in the table below. Estimate and interpret f (2000) and f (2000). t f (t)

1985 4055

1990 4189

1995 4353

58. Suppose that the daily output of a manufacturing plant is modeled by Q = 1000K 1/2 L 1/3 , where K is the capital investment in thousands of dollars and L is the size of the labor force in worker-hours. Assume that L stays constant and think of output as a function of capital investment, Q(x) = 1000L 1/3 x 1/2 . Find and interpret Q (40). In exercises 59–62, find a function with the given derivative. 59. f (x) = 4x 3 √

60. f (x) = 5x 4 62. f (x) =

61. f (x) =

52. A rod made of an inhomogeneous material extends from x = 0 to x = 4 meters. The mass of the portion of the rod from x = 0 to x = t is given by m(t) = 3t 2 kg. Compute m (t) and explain why it represents the density of the rod.

64. For f (x) = x|x|, show that lim

54. Suppose the function v(d) represents the average speed in m/s of the world record running time for d meters. For example, if the fastest 200-meter time ever is 19.32 s, then v(200) = 200/19.32 ≈ 10.35. Compare the function f (d) = 26.7d −0.177 to the values of v(d), which you will have to research and compute, for distances ranging from d = 400 to d = 2000. Explain what v (d) would represent. 55. Let f (t) equal the gross domestic product (GDP) in billions of dollars for the United States in year t. Several values are given in the table. Estimate and interpret f (2000) and f (2000).

2000 4619

57. If the position of an object is at time t given by f (t), then f (t) represents velocity and f (t) gives acceleration. By Newton’s second law, acceleration is proportional to the net force on the object (causing it to accelerate). Interpret the third derivative f (t) in terms of force. The term jerk is sometimes applied to f (t). Explain why this is an appropriate term.

51. A public official solemnly proclaims, “We have achieved a reduction in the rate at which the national debt is increasing.” If d(t) represents the national debt at time t years, which derivative of d(t) is being reduced? What can you conclude about the size of d(t) itself?

53. For most land animals, the relationship between leg width w and body length b follows an equation of the form w = cb3/2 for some constant c > 0. Show that if b is large enough, w (b) > 1. Conclude that for larger animals, leg width (necessary for support) increases faster than body length. Why does this put a limitation on the size of land animals?

179

x

1 x2

63. Assume that a is a real number and f (a) exists. Then f (a + h) − 2 f (a) + f (a − h) also exists. Find its value. lim h→0 h2 f (h) − 2 f (0) + f (−h) exists h2 but f (0) does not exist. (That is, the converse of exercise 63 is not true.) h→0

EXPLORATORY EXERCISES 1. A plane is cruising at an altitude of 2 miles at a distance of 10 miles from an airport. Choosing the airport to be at the point (0, 0), the plane starts its descent at the point (10, 2) and lands at the airport. Sketch a graph of a reasonable flight path y = f (x), where y represents altitude and x gives the ground distance from the airport. (Think about it as you draw!) Explain what the derivative f (x) represents. (Hint: It’s not velocity.) Explain why it is important and/or necessary to have f (0) = 0, f (10) = 2, f (0) = 0 and f (10) = 0. The simplest polynomial that can meet these requirements is a cubic polynomial f (x) = ax 3 + bx 2 + cx + d (Note: four requirements, four constants). Find values of the constants a, b, c and d to fit the flight path. [Hint: Start by setting f (0) = 0 and then

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2-36

set f (0) = 0. You may want to use your CAS to solve the equations.] Graph the resulting function; does it look right? 2 Suppose that airline regulations prohibit a derivative of 10 or larger. Why might such a regulation exist? Show that the flight path you found is illegal. Argue that in fact all flight paths meeting the four requirements are illegal. Therefore, the descent needs to start farther away than 10 miles. Find a flight path with descent starting at 20 miles away that meets all requirements. 2. We discuss a graphical interpretation of the second derivative in Chapter 3. You can discover the most important aspects of that here. For f (x) = x 4 − 2x 2 − 1, solve the equations f (x) = 0 and f (x) = 0. What do the solutions of the equation f (x) = 0 represent graphically? The solutions of the equation f (x) = 0 are a little harder to interpret. Looking at the graph of f (x) near x = 0, would you say that the graph is curving up or curving down? Compute f (0). Looking at the graph near x = 2 and x = −2, is the graph curving up or down? Compute f (2) and f (−2). Where does the graph change from curving up to curving down and vice versa? Hypothesize a relationship between f (x) and the curving of the graph of y = f (x). Test your hypothesis on a variety of functions. (Try y = x 4 − 4x 3 .) 3. In the enjoyable book Surely You’re Joking Mr. Feynman, physicist Richard Feynman tells the story of a contest he had pitting his brain against the technology of the day (an

2.4

abacus). The contest was to compute the cube root of 1729.03. Feynman came up with 12.002 before the abacus expert gave up. Feynman admits to some luck in the choice of the number 1729.03: he knew that a cubic foot contains 1728 cubic inches. Explain why this told Feynman that the answer is slightly greater than 12. How did he get three digits of accuracy? “I had learned in calculus that for small fractions, the cube root’s excess is one-third of the number’s excess. The excess, 1.03, is only one part in nearly 2000. So all I had to do is find the fraction 1/1728, divide by 3 and multiply by 12.” To see what he did, find an equation of the tangent line to y = x 1/3 at x = 1728 and find the y-coordinate of the tangent line at x = 1729.03. 4. Suppose that you want to find solutions of the equation x 3 − 4x 2 + 2 = 0. Show graphically that there is a solution between x = 0 and x = 1. We will approximate this solution in stages. First, find an equation of the tangent line to y = x 3 − 4x 2 + 2 at x = 1. Then, determine where this tangent line crosses the x-axis. Show graphically that the x-intercept is considerably closer to the solution than is x = 1. Now, repeat the process: for the new x-value, find the equation of the tangent line, determine where it crosses the x-axis and show that this is closer still to the desired solution. This process of using tangent lines to produce continually improved approximations is referred to as Newton’s method. We discuss this in some detail in section 3.1.

THE PRODUCT AND QUOTIENT RULES We have now developed rules for computing the derivatives of a variety of functions, including general formulas for the derivative of a sum or difference of two functions. Given this, you might wonder whether the derivative of a product of two functions is the same as the product of the derivatives. We test this conjecture with a simple example.

Product Rule Consider

d [(x 2 )(x 5 )]. We can compute this derivative by first combining the two terms: dx d d 7 [(x 2 )(x 5 )] = x = 7x 6 . dx dx

Is this derivative the same as the product of the two individual derivatives? Notice that

d 2 d 5 x x = (2x)(5x 4 ) dx dx d = 10x 5 = 7x 6 = [(x 2 )(x 5 )]. (4.1) dx

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2-36

set f (0) = 0. You may want to use your CAS to solve the equations.] Graph the resulting function; does it look right? 2 Suppose that airline regulations prohibit a derivative of 10 or larger. Why might such a regulation exist? Show that the flight path you found is illegal. Argue that in fact all flight paths meeting the four requirements are illegal. Therefore, the descent needs to start farther away than 10 miles. Find a flight path with descent starting at 20 miles away that meets all requirements. 2. We discuss a graphical interpretation of the second derivative in Chapter 3. You can discover the most important aspects of that here. For f (x) = x 4 − 2x 2 − 1, solve the equations f (x) = 0 and f (x) = 0. What do the solutions of the equation f (x) = 0 represent graphically? The solutions of the equation f (x) = 0 are a little harder to interpret. Looking at the graph of f (x) near x = 0, would you say that the graph is curving up or curving down? Compute f (0). Looking at the graph near x = 2 and x = −2, is the graph curving up or down? Compute f (2) and f (−2). Where does the graph change from curving up to curving down and vice versa? Hypothesize a relationship between f (x) and the curving of the graph of y = f (x). Test your hypothesis on a variety of functions. (Try y = x 4 − 4x 3 .) 3. In the enjoyable book Surely You’re Joking Mr. Feynman, physicist Richard Feynman tells the story of a contest he had pitting his brain against the technology of the day (an

2.4

abacus). The contest was to compute the cube root of 1729.03. Feynman came up with 12.002 before the abacus expert gave up. Feynman admits to some luck in the choice of the number 1729.03: he knew that a cubic foot contains 1728 cubic inches. Explain why this told Feynman that the answer is slightly greater than 12. How did he get three digits of accuracy? “I had learned in calculus that for small fractions, the cube root’s excess is one-third of the number’s excess. The excess, 1.03, is only one part in nearly 2000. So all I had to do is find the fraction 1/1728, divide by 3 and multiply by 12.” To see what he did, find an equation of the tangent line to y = x 1/3 at x = 1728 and find the y-coordinate of the tangent line at x = 1729.03. 4. Suppose that you want to find solutions of the equation x 3 − 4x 2 + 2 = 0. Show graphically that there is a solution between x = 0 and x = 1. We will approximate this solution in stages. First, find an equation of the tangent line to y = x 3 − 4x 2 + 2 at x = 1. Then, determine where this tangent line crosses the x-axis. Show graphically that the x-intercept is considerably closer to the solution than is x = 1. Now, repeat the process: for the new x-value, find the equation of the tangent line, determine where it crosses the x-axis and show that this is closer still to the desired solution. This process of using tangent lines to produce continually improved approximations is referred to as Newton’s method. We discuss this in some detail in section 3.1.

THE PRODUCT AND QUOTIENT RULES We have now developed rules for computing the derivatives of a variety of functions, including general formulas for the derivative of a sum or difference of two functions. Given this, you might wonder whether the derivative of a product of two functions is the same as the product of the derivatives. We test this conjecture with a simple example.

Product Rule Consider

d [(x 2 )(x 5 )]. We can compute this derivative by first combining the two terms: dx d d 7 [(x 2 )(x 5 )] = x = 7x 6 . dx dx

Is this derivative the same as the product of the two individual derivatives? Notice that

d 2 d 5 x x = (2x)(5x 4 ) dx dx d = 10x 5 = 7x 6 = [(x 2 )(x 5 )]. (4.1) dx

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SECTION 2.4

..

The Product and Quotient Rules

181

You can now plainly see from (4.1) that the derivative of a product is not generally the product of the corresponding derivatives. In Theorem 4.1, we state a general rule for computing the derivative of a product of two differentiable functions.

THEOREM 4.1 (Product Rule) Suppose that f and g are differentiable at x. Then d [ f (x)g(x)] = f (x)g(x) + f (x)g (x). dx

(4.2)

PROOF Since we are proving a general rule, we have only the limit definition of derivative to use. For p(x) = f (x)g(x), we have d p(x + h) − p(x) [ f (x)g(x)] = p (x) = lim h→0 dx h = lim

h→0

f (x + h)g(x + h) − f (x)g(x) . h

(4.3)

Notice that the elements of the derivatives of f and g are present [limit, f (x + h), f (x) etc.], but we need to get them into the right form. The trick is to add and subtract f (x)g(x + h) in the numerator. From (4.3), we have p (x) = lim

h→0

f (x + h)g(x + h) − f (x)g(x + h) + f (x)g(x + h) − f (x)g(x) h

Subtract and add f (x)g(x + h).

= lim

h→0

f (x + h)g(x + h) − f (x)g(x + h) f (x)g(x + h) − f (x)g(x) + lim h→0 h h

Break into two pieces.

g(x + h) − g(x) f (x + h) − f (x) g(x + h) + lim f (x) h→0 h h f (x + h) − f (x) g(x + h) − g(x) = lim lim g(x + h) + f (x) lim h→0 h→0 h→0 h h

= lim

h→0

= f (x)g(x) + f (x)g (x).

Recognize the derivative of f and the derivative of g.

Here, we identified the limits of the difference quotients as derivatives. There is also a subtle technical detail in the last step: since g is differentiable at x, recall that it must also be continuous at x, so that g(x + h) → g(x) as h → 0. In example 4.1, notice that the product rule saves us from multiplying out a messy product.

EXAMPLE 4.1

Using the Product Rule

√ 2 Find f (x) if f (x) = (2x 4 − 3x + 5) x 2 − x + . x

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2-38

Solution Although we could first multiply out the expression, the product rule will simplify our work:

√ √ d 2 d 2 4 2 4 2 f (x) = (2x − 3x + 5) x − x + + (2x − 3x + 5) x − x+ dx x dx x

√ 2 1 2 = (8x 3 − 3) x 2 − x + + (2x 4 − 3x + 5) 2x − √ − 2 . x x 2 x The product rule usually leaves derivatives in an “unsimplified” form as in example 4.1. Unless you have some particular reason to simplify (or some particularly obvious simplification to make), just leave the results of the product rule alone.

EXAMPLE 4.2

Finding the Equation of the Tangent Line

Find an equation of the tangent line to y = (x 4 − 3x 2 + 2x)(x 3 − 2x + 3) at x = 0. Solution From the product rule, we have y = (4x 3 − 6x + 2)(x 3 − 2x + 3) + (x 4 − 3x 2 + 2x)(3x 2 − 2). Evaluating at x = 0, we have y (0) = (2)(3) + (0)(−2) = 6. The line with slope 6 and passing through the point (0, 0) [why (0, 0)?] has equation y = 6x.

Quotient Rule Given our experience with the product rule, you probably have no expectation that the derivative of a quotient will turn out to be the quotient of the derivatives. Just to be sure, let’s try a simple experiment. Note that

d x5 d 3 = (x ) = 3x 2 , 2 dx x dx

while

d 5

(x ) 5 3 d x5 5x 4 dx 2 . = 1 = x = 3x = d 2 2x 2 dx x2 (x ) dx

Since these are obviously not the same, we know that the derivative of a quotient is generally not the quotient of the corresponding derivatives. Theorem 4.2 provides us with a general rule for computing the derivative of a quotient of two differentiable functions.

THEOREM 4.2 (Quotient Rule) Suppose that f and g are differentiable at x and g(x) = 0. Then d f (x) f (x)g(x) − f (x)g (x) . = d x g(x) [g(x)]2

(4.4)

2-39

SECTION 2.4

..

The Product and Quotient Rules

183

PROOF You should be able to guess the structure of the proof. For Q(x) =

f (x) , we have from the g(x)

limit definition of derivative that d Q(x + h) − Q(x) f (x) = Q (x) = lim h→0 d x g(x) h f (x + h) f (x) − g(x + h) g(x) = lim h→0 h f (x + h)g(x) − f (x)g(x + h) g(x + h)g(x) = lim h→0 h f (x + h)g(x) − f (x)g(x + h) . = lim h→0 hg(x + h)g(x)

Add the fractions.

Simplify.

As in the proof of the product rule, we look for the right term to add and subtract in the numerator, so that we can isolate the limit definitions of f (x) and g (x). In this case, we add and subtract f (x)g(x), to get Q (x) = lim

h→0

f (x + h)g(x) − f (x)g(x + h) hg(x + h)g(x)

f (x + h)g(x) − f (x)g(x) + f (x)g(x) − f (x)g(x + h) Subtract and add f (x)g(x). hg(x + h)g(x) g(x + h) − g(x) f (x + h) − f (x) Group first two and last g(x) − f (x) two terms together h h = lim and factor out common h→0 g(x + h)g(x) = lim

h→0

lim

=

h→0

terms.

f (x + h) − f (x) g(x + h) − g(x) g(x) − f (x) lim h→0 h h lim g(x + h)g(x)

h→0

=

f (x)g(x) − f (x)g (x) , [g(x)]2

where we have again recognized the derivatives of f and g and used the fact that g is differentiable to imply that g is continuous, so that lim g(x + h) = g(x).

h→0

Notice that the numerator in the quotient rule looks very much like the product rule, but with a minus sign between the two terms. For this reason, you need to be very careful with the order.

EXAMPLE 4.3

Using the Quotient Rule

Compute the derivative of f (x) =

x2 − 2 . x2 + 1

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2-40

Solution Using the quotient rule, we have d 2 d (x − 2) (x 2 + 1) − (x 2 − 2) (x 2 + 1) d x d x f (x) = 2 2 (x + 1) =

2x(x 2 + 1) − (x 2 − 2)(2x) (x 2 + 1)2

=

6x . (x 2 + 1)2

In this case, we rewrote the numerator because it simplified significantly. This often occurs with the quotient rule. Now that we have the quotient rule, we can justify the use of the power rule for negative integer exponents. (Recall that we have been using this rule without proof since section 2.3.)

THEOREM 4.3 (Power Rule) For any integer exponent n,

d n x = nx n−1 . dx

PROOF We have already proved this for positive integer exponents. So, suppose that n < 0 and let M = −n > 0. Then, using the quotient rule, we get

d n 1 d −M d 1 x = x = Since x −M = M . x dx dx dx x M d d (1) x M − (1) (x M ) dx dx = By the quotient rule. (x M )2 =

(0)x M − (1)M x M−1 x 2M

By the power rule, since M > 0.

=

−M x M−1 = −M x M−1−2M x 2M

By the usual rules of exponents.

= (−M)x −M−1 = nx n−1 ,

Since n = −M.

d M x = M x M−1 , since M > 0. dx As we see in example 4.4, it is sometimes preferable to rewrite a function, instead of automatically using the product or quotient rule.

where we have used the fact that

EXAMPLE 4.4

A Case Where the Product and Quotient Rules Are Not Needed

√ 2 Compute the derivative of f (x) = x x + 2 . x

2-41

SECTION 2.4

..

The Product and Quotient Rules

185

Solution Although it may be tempting to use the product rule for the first term and the quotient rule for the second term, notice that we can rewrite the function and considerably simplify our work. We can combine the two powers of x in the first term and since the second term is a fraction with a constant numerator, we can more simply write it using a negative exponent. We have √ 2 f (x) = x x + 2 = x 3/2 + 2x −2 . x Using the power rule, we have simply f (x) =

3 1/2 x − 4x −3 . 2

Applications You will see important uses of the product and quotient rules throughout your mathematical and scientific studies. We start you off with a couple of simple applications now.

EXAMPLE 4.5

Investigating the Rate of Change of Revenue

Suppose that a product currently sells for $25, with the price increasing at the rate of $2 per year. At this price, consumers will buy 150 thousand items, but the number sold is decreasing at the rate of 8 thousand per year. At what rate is the total revenue changing? Is the total revenue increasing or decreasing? Solution To answer these questions, we need the basic relationship revenue = quantity × price (e.g., if you sell 10 items at $4 each, you earn $40). Since these quantities are changing in time, we write R(t) = Q(t)P(t), where R(t) is revenue, Q(t) is quantity sold and P(t) is the price, all at time t. We don’t have formulas for any of these functions, but from the product rule, we have R (t) = Q (t)P(t) + Q(t)P (t). We have information about each of these terms: the initial price, P(0), is 25 (dollars); the rate of change of the price is P (0) = 2 (dollars per year); the initial quantity, Q(0), is 150 (thousand items) and the rate of change of quantity is Q (0) = −8 (thousand items per year). Note that the negative sign of Q (0) denotes a decrease in Q. Thus, R (0) = (−8)(25) + (150)(2) = 100 thousand dollars per year. Since the rate of change is positive, the revenue is increasing. This may be a surprise since one of the two factors in the equation is decreasing and the rate of decrease of the quantity is more than the rate of increase in the price.

EXAMPLE 4.6

Using the Derivative to Analyze Sports

A golf ball of mass 0.05 kg struck by a golf club of mass m kg with speed 50 m/s will 83m have an initial speed of u(m) = m/s. Show that u (m) > 0 and interpret this m + 0.05 result in golf terms. Compare u (0.15) and u (0.20).

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y

2-42

Solution From the quotient rule, we have u (m) =

60

40

20

m 0.1

0.2

0.3

FIGURE 2.23 u(m) =

83m m + 0.05

83(m + 0.05) − 83m 4.15 = . 2 (m + 0.05) (m + 0.05)2

Both the numerator and denominator are positive, so u (m) > 0. A positive slope for all tangent lines indicates that the graph of u(m) should rise from left to right (see Figure 2.23). Said a different way, u(m) increases as m increases. In golf terms, this says that (all other things being equal) the greater the mass of the club, the greater the velocity of the ball will be. Finally, we compute u (0.15) = 103.75 and u (0.20) = 66.4. This says that the rate of increase in ball speed is much less for the heavier club than for the lighter one. Since heavier clubs can be harder to control, the relatively small increase in ball speed obtained by making the heavy club even heavier may not compensate for the decrease in control.

EXERCISES 2.4 WRITING EXERCISES 1. The product and quotient rules give you the ability to symbolically calculate the derivative of a wide range of functions. However, many calculators and almost every computer algebra system (CAS) can do this work for you. Discuss why you should learn these basic rules anyway. (Keep example 4.5 in mind.) 2. Gottfried Leibniz is recognized (along with Sir Isaac Newton) as a coinventor of calculus. Many of the fundamental methods and (perhaps more importantly) much of the notation of calculus are due to Leibniz. The product rule was worked out by Leibniz in 1675, in the form d(x y) = (d x)y + x(dy). His “proof,” as given in a letter written in 1699, follows. “If we are to differentiate xy we write: (x + d x)(y + dy) − x y = x dy + y d x + d x d y. But here d x d y is to be rejected as incomparably less than x dy + y d x. Thus, in any particular case the error is less than any finite quantity.” Answer Leibniz’ letter with one describing your own “discovery” of the product rule for d(x yz). Use Leibniz’ notation. 3. In example 4.1, we cautioned you against always multiplying out the terms of the derivative. To see one reason for this warning, suppose that you want to find solutions of the equation f (x) = 0. (In fact, we do this routinely in Chapter 3.) Explain why having a factored form of f (x) is very helpful. Discuss the extent to which the product rule gives you a factored form. 4. Many students prefer the product rule to the quotient rule. Many computer algebra systems actually use the product rule to compute the derivative of f (x)[g(x)]−1 instead of using f (x) (see exercise 18 on the next page). the quotient rule on g(x)

Given the simplifications in problems like example 4.3, explain why the quotient rule can be preferable. In exercises 1–16, find the derivative of each function. 1. f (x) = (x 2 + 3)(x 3 − 3x + 1) 2. f (x) = (x 3 − 2x 2 + 5)(x 4 − 3x 2 + 2)

√ 3 3. f (x) = ( x + 3x) 5x 2 − x

3 4. f (x) = (x 3/2 − 4x) x 4 − 2 + 2 x 3x − 2 6. f (x) = 5. f (x) = 5x + 1 √ 3x − 6 x 8. f (x) = 7. f (x) = 5x 2 − 2 (x + 1)(x − 2) 10. f (x) = 9. f (x) = 2 x − 5x + 1 x 2 + 3x − 2 12. f (x) = 11. f (x) = √ x √ 14. f (x) = 13. f (x) = x( 3 x + 3) 15. f (x) = (x 2 − 1)

x 3 + 3x 2 x2 + 2

x 2 + 2x + 5 x 2 − 5x + 1 6x − 2/x √ x2 + x x 2 − 2x x 2 + 5x 2x x2 + 1 5 x2 + 2 3 x x2 − 1 16. f (x) = (x + 2) 2 x +x

17. Write out the product rule for the function f (x)g(x)h(x). (Hint: Group the first two terms together.) Describe the general product rule: for n functions, what is the derivative of the product f 1 (x) f 2 (x) f 3 (x) · · · f n (x)? How many terms are there? What does each term look like?

2-43

SECTION 2.4

18. Use the quotient rule to show that the derivative of [g(x)]−1 is −g (x)[g(x)]−2 . Then use the product rule to compute the derivative of f (x)[g(x)]−1 . In exercises 19 and 20, find the derivative of each function using the general product rule developed in exercise 17. 19. f (x) = x 2/3 (x 2 − 2)(x 3 − x + 1) 20. f (x) = (x + 4)(x 3 − 2x 2 + 1)(3 − 2/x) In exercises 21–24, assume that f and g are differentiable with f (0) − 1, f (1) − 2, f (0) − 1, f (1) 3, g(0) 3, g(1) 1, g (0) − 1 and g (1) − 2. 21. Find an equation of the tangent line to h(x) = f (x)g(x) at (a) x = 1, (b) x = 0. f (x) at 22. Find an equation of the tangent line to h(x) = g(x) (a) x = 1, (b) x = 0. 23. Find an equation of the tangent line to h(x) = x 2 f (x) at (a) x = 1, (b) x = 0. 24. Find an equation of the tangent line to h(x) = (a) x = 1, (b) x = 0.

x2 at g(x)

25. Suppose that for some toy, the quantity sold Q(t) at time t years decreases at a rate of 4%; explain why this translates to Q (t) = −0.04Q(t). Suppose also that the price increases at a rate of 3%; write out a similar equation for P (t) in terms of P(t). The revenue for the toy is R(t) = Q(t)P(t). Substituting the expressions for Q (t) and P (t) into the product rule R (t) = Q (t)P(t) + Q(t)P (t), show that the revenue decreases at a rate of 1%. Explain why this is “obvious.” 26. As in exercise 25, suppose that the quantity sold decreases at a rate of 4%. By what rate must the price be increased to keep the revenue constant? 27. Suppose the price of an object is $20 and 20,000 units are sold. If the price increases at a rate of $1.25 per year and the quantity sold increases at a rate of 2000 per year, at what rate will revenue increase? 28. Suppose the price of an object is $14 and 12,000 units are sold. The company wants to increase the quantity sold by 1200 units per year, while increasing the revenue by $20,000 per year. At what rate would the price have to be increased to reach these goals? 29. A baseball with mass 0.15 kg and speed 45 m/s is struck by a baseball bat of mass m and speed 40 m/s (in the opposite direction of the ball’s motion). After the collision, the ball has 82.5m − 6.75 initial speed u(m) = m/s. Show that u (m) > 0 m + 0.15 and interpret this in baseball terms. Compare u (1) and u (1.2). 30. In exercise 29, if the baseball has mass M kg at speed 45 m/s and the bat has mass 1.05 kg at speed 40 m/s, the ball’s initial

..

The Product and Quotient Rules

187

86.625 − 45M m/s. Compute u (M) and M + 1.05 interpret its sign (positive or negative) in baseball terms.

speed is u(M) =

31. In example 4.6, it is reasonable to assume that the speed of the golf club at impact decreases as the mass of the club increases. If, for example, the speed of a club of mass m is v = 8.5/m m/s at impact, then the initial speed of the golf ball 14.11 is u(m) = m/s. Show that u (m) < 0 and interpret m + 0.05 this in golf terms. 32. In example 4.6, if the golf club has mass 0.17 kg and strikes the ball with speed v m/s, the ball has initial speed 0.2822v u(v) = m/s. Compute and interpret the derivative 0.217 u (v). 33. Assume that g(x) is continuous at x = 0 and define f (x) = xg(x). Show that f (x) is differentiable at x = 0. Illustrate the result with g(x) = |x|. 34. Determine whether the result of exercise 33 still holds if x = 0 is replaced with x = a = 0. In exercises 35–40, use the symbolic differentiation feature on your CAS or calculator. 35. Repeat example 4.4 with your CAS. If its answer is not in the same form as ours in the text, explain how the CAS computed its answer. 36. Repeat exercise 15 with your CAS. If its answer is not in the same form as ours in the back of the book, explain how the CAS computed its answer. 37. Use your CAS to sketch the derivative of sin x. What function does this look like? Repeat with sin 2x and sin 3x. Generalize to conjecture the derivative of sin kx for any constant k. 38. Repeat exercise 37 with sin x 2 . To identify the derivative, sketch a curve outlining the tops of the curves of the derivative graph and try to identify the amplitude of the derivative. √ 3x 3 + x 2 39. Find the derivative of f (x) = on your CAS. Comx −3 3 for x > 0 and √ for pare its answer to √ 2 3x + 1 2 3x + 1 x < 0. Explain how to get this answer and your CAS’s answer, if it differs.

2x 2 x2 − x − 2 2x − on 40. Find the derivative of f (x) = x −2 x +1 your CAS. Compare its answer to 2. Explain how to get this answer and your CAS’s answer, if it differs. 41. Suppose that F(x) = f (x)g(x) for infinitely differentiable functions f (x) and g(x) (that is, f (x), f (x), etc. exist for all x). Show that F (x) = f (x)g(x) + 2 f (x)g (x) + f (x)g (x). Compute F (x). Compare F (x) to the binomial formula for (a + b)2 and compare F (x) to the formula for (a + b)3 .

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42. Given that (a + b)4 = a 4 + 4a 3 b + 6a 2 b2 + 4ab3 + b4 , write out a formula for F (4) (x). (See exercise 41.) 43. Use the product rule to show that if g(x) = [ f (x)]2 and f (x) is differentiable, then g (x) = 2 f (x) f (x). This is an example of the chain rule, to be discussed in section 2.5. 44. Use the result from exercise 43 and the product rule to show that if g(x) = [ f (x)]3 and f (x) is differentiable, then g (x) = 3[ f (x)]2 f (x). Hypothesize the correct chain rule for the derivative of [ f (x)]n . 45. The relationship among the pressure P, volume V and temperature T of a gas or liquid is given by van der Waals’ equation

n2a P + 2 (V − nb) = n RT for positive constants a, b, n V and R. Solve the equation for P. Treating T as a constant and V as the variable, find the critical point (Tc , Pc , Vc ) such that P (V ) = P (V ) = 0. (Hint: Don’t solve either equation separately, but substitute the result from one equation into the other.) For temperatures above Tc , the substance can exist only in its gaseous form; below Tc , the substance is a gas or liquid (depending on the pressure and volume). For water, take R = 0.08206 l-atm/mo-K, a = 5.464 12 -atm/mo2 and b = 0.03049 l/mo. Find the highest temperature at which n = 1 mo of water may exist as a liquid. (Note: your answer will be in degrees Kelvin; subtract 273.15 to get degrees Celsius.) 46. The amount of an allosteric enzyme is affected by the presence of an activator. If x is the amount of activator and f is the amount of enzyme, then one model of an allosteric acx 2.7 tivation is f (x) = . Find and interpret lim f (x) and x→0 1 + x 2.7 lim f (x). x→∞

47. For the allosteric enzyme of exercise 46, compute and interpret f (x). 48. Enzyme production can also be inhibited. In this situation, the amount of enzyme as a function of the amount of inhibitor 1 . Find and interpret lim f (x), is modeled by f (x) = x→0 1 + x 2.7 lim f (x) and f (x). x→∞

In exercises 49–52, find the derivative of the expression for an unspecified differentiable function f . √ √ x f (x) 3 49. x f (x) 52. x f (x) 50. 51. x2 f (x)

EXPLORATORY EXERCISES 1. Most cars are rated for fuel efficiency by estimating miles per gallon in city driving (c) and miles per gallon in highway driving (h). The Environmental Protection Agency uses the

2-44

1 as its overall rating of gas us0.55/c + 0.45/ h age (this is called the 55/45 combined mpg). You may want to rewrite the function (find a common denominator in the denominator and simplify) to work this exercise. (a) Think of c as the variable and h as a constant, and show dr that > 0. Interpret this result in terms of gas mileage. dc (b) Think of h as the variable and c as a constant, and show dr that > 0. dh (c) Show that if c = h, then r = c. (d) Show that if c < h, then c < r < h. To do this, assume that c is a constant and c < h. Explain why the results of parts (b) and (c) imply that r > c. Next, show that dr < 0.45. Explain why this result along with the result dh of part (b) implies that r < h. Explain why the results of parts (a)–(d) must be true if the EPA’s combined formula is a reasonable way to average the ratings c and h. To get some sense of how the formula works, take c = 20 and graph r as a function of h. Comment on why the EPA might want to use a function whose graph flattens out as this one does. formula r =

2. In many sports, the collision between a ball and a striking implement is central to the game. Suppose the ball has weight w and velocity v before the collision and the striker (bat, tennis racket, golf club, etc.) has weight W and velocity −V before the collision (the negative indicates the striker is moving in the opposite direction from the ball). The velocity of the ball after W V (1 + c) + v(cW − w) the collision will be u = , where W +w the parameter c, called the coefficient of restitution, represents the “bounciness” of the ball in the collision. Treating W as the independent variable (like x) and the other parameters as constants, compute the derivative and verify that du V (1 + c)w + cvw + vw ≥ 0 since all parameters are = dW (W + w)2 nonnegative. Explain why this implies that if the athlete uses a bigger striker (bigger W ) with all other things equal, the speed of the ball increases. Does this match your intuition? What is doubtful about the assumption of all other things being equal? du du du du Similarly compute and interpret , , and . (Hint: dw dv d V dc c is between 0 and 1 with 0 representing a dead ball and 1 the liveliest ball possible.) 3. Suppose that a soccer player strikes the ball with enough energy that a stationary ball would have initial speed 80 mph. Show that the same energy kick on a ball moving directly to the player at 40 mph will launch the ball at approximately 100 mph. (Use the general collision formula in exploratory exercise 2 with c = 0.5 and assume that the ball’s weight is much less than the soccer player’s weight.) In general, what proportion of the ball’s incoming speed is converted by the kick into extra speed in the opposite direction?

2-45

SECTION 2.5

2.5

..

The Chain Rule

189

THE CHAIN RULE √ Suppose that the function P(t) = 100 + 8t models the population of a city after t years. Then the rate of growth of that population after 2 years is given by P (2). At present, we need the limit definition to compute this derivative. However, observe that √ P(t) is the composition of the two functions f (t) = t and g(t) = 100 + 8t, so that P(t) = f (g(t)). Also, notice that both f (t) and g (t) are easily computed using existing derivative rules. We now develop a general rule for the derivative of a composition of two functions. The following simple examples will help us to identify the form of the chain rule. Notice that from the product rule d d [(x 2 + 1)2 ] = [(x 2 + 1)(x 2 + 1)] dx dx = 2x(x 2 + 1) + (x 2 + 1)2x = 2(x 2 + 1)2x. Of course, we can write this as 4x(x 2 + 1), but the unsimplified form helps us to understand the form of the chain rule. Using this result and the product rule, notice that d d [(x 2 + 1)3 ] = [(x 2 + 1)(x 2 + 1)2 ] dx dx = 2x(x 2 + 1)2 + (x 2 + 1)2(x 2 + 1)2x = 3(x 2 + 1)2 2x. We leave it as a straightforward exercise to extend this result to d [(x 2 + 1)4 ] = 4(x 2 + 1)3 2x. dx You should observe that, in each case, we have brought the exponent down, lowered the power by one and then multiplied by 2x, the derivative of x 2 + 1. Notice that we can write (x 2 + 1)4 as the composite function f (g(x)) = (x 2 + 1)4 , where g(x) = x 2 + 1 and f (x) = x 4 . Finally, observe that the derivative of the composite function is d d [ f (g(x))] = [(x 2 + 1)4 ] = 4(x 2 + 1)3 2x = f (g(x))g (x). dx dx This is an example of the chain rule, which has the following general form.

THEOREM 5.1 (Chain Rule) If g is differentiable at x and f is differentiable at g(x), then d [ f (g(x))] = f (g(x)) g (x). dx

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2-46

PROOF At this point, we can prove only the special case where g (x) = 0. Let F(x) = f (g(x)). Then, d F(x + h) − F(x) [ f (g(x))] = F (x) = lim h→0 dx h f (g(x + h)) − f (g(x)) = lim h→0 h = lim

h→0

Since F(x) = f (g(x)).

f (g(x + h)) − f (g(x)) g(x + h) − g(x) h g(x + h) − g(x)

f (g(x + h)) − f (g(x)) g(x + h) − g(x) = lim lim h→0 h→0 g(x + h) − g(x) h =

lim

g(x+h)→g(x)

Multiply numerator and denominator by g(x + h) − g(x). Regroup terms.

f (g(x + h)) − f (g(x)) g(x + h) − g(x) lim h→0 g(x + h) − g(x) h

= f (g(x))g (x), where the next to the last line is valid since as h → 0, g(x + h) → g(x), by the continuity of g. (Recall that since g is differentiable, it is also continuous.) You will be asked in the exercises to fill in some of the gaps in this argument. In particular, you should identify why we need g (x) = 0 in this proof. It is often helpful to think of the chain rule in Leibniz notation. If y = f (u) and u = g(x), then y = f (g(x)) and the chain rule says that dy dy du = . dx du d x

REMARK 5.1 The chain rule should make sense intuitively as follows. dy as the We think of dx (instantaneous) rate of change dy as the of y with respect to x, du (instantaneous) rate of change du as of y with respect to u and dx the (instantaneous) rate of change of u with respect to x. dy = 2 (i.e., y is So, if du changing at twice the rate of u) du = 5 (i.e., u is changing and dx at five times the rate of x), it should make sense that y is changing at 2 × 5 = 10 times dy = 10, the rate of x. That is, dx which is precisely what equation (5.1) says.

EXAMPLE 5.1

(5.1)

Using the Chain Rule

Differentiate y = (x + x − 1)5 . 3

Solution For u = x 3 + x − 1, note that y = u 5 . From (5.1), we have dy dy du d 5 du = = (u ) dx du d x du dx

Since y = u 5 .

d 3 (x + x − 1) dx = 5(x 3 + x − 1)4 (3x 2 + 1).

= 5u 4

It is helpful to think of the chain rule in terms of inside functions and outside functions. For the composition f (g(x)), f is referred to as the outside function and g is referred to as the inside function. The chain rule derivative f (g(x))g (x) can then be viewed as the derivative of the outside function times the derivative of the inside function. In example 5.1, the inside function is x 3 + x − 1 (the expression √ inside the parenof 100 + 8t as comtheses) and the outside function is u 5 . In example 5.2, we think √ posed of the inside function 100 + 8t and the outside function u. Some careful thought about the pieces of a composition of functions will help you use the chain rule effectively.

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SECTION 2.5

EXAMPLE 5.2 Find

..

The Chain Rule

191

Using the Chain Rule on a Radical Function

d √ ( 100 + 8t). dt

Solution Let u = 100 + 8t and note that

√

100 + 8t = u 1/2 . Then, from (5.1),

d √ d 1 du ( 100 + 8t) = (u 1/2 ) = u −1/2 dt dt 2 dt 1 4 d = √ (100 + 8t) = √ . 2 100 + 8t dt 100 + 8t Notice that here, the derivative of the inside is the derivative of the expression under the square root sign. You are now in a position to calculate the derivative of a very large number of functions, by using the chain rule in combination with other differentiation rules.

EXAMPLE 5.3

Derivatives Involving Chain Rules and Other Rules

√ Compute the derivative of f (x) = x 3 4x + 1, g(x) =

8x 8 and h(x) = 3 . (x 3 + 1)2 (x + 1)2

Solution Notice the differences in these three functions. The first function f (x) is a product of two functions, g(x) is a quotient of two functions and h(x) is a constant divided by a function. This tells us to use the product rule for f (x), the quotient rule for g(x) and simply the chain rule for h(x). For the first function, we have √ d √ d 3√ x 4x + 1 = 3x 2 4x + 1 + x 3 4x + 1 dx dx √ 1 d = 3x 2 4x + 1 + x 3 (4x + 1)−1/2 (4x + 1) 2 d x

f (x) =

= 3x

2

√

By the product rule.

By the chain rule.

derivative of the inside

4x + 1 + 2x (4x + 1) 3

−1/2

.

Simplifying.

Next, we have 8(x 3 + 1)2 − 8x d [(x 3 + 1)2 ] 8x dx = (x 3 + 1)2 (x 3 + 1)4 d 3 3 2 3 8(x + 1) − 8x 2(x + 1) (x + 1) d x

d g (x) = dx

=

derivative of the inside

(x 3 + 1)4

=

8(x 3 + 1)2 − 16x(x 3 + 1)3x 2 (x 3 + 1)4

=

8(x 3 + 1) − 48x 3 8 − 40x 3 = . (x 3 + 1)3 (x 3 + 1)3

By the quotient rule.

By the chain rule.

Simplification.

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2-48

For h(x), notice that instead of using the quotient rule, it is simpler to rewrite the function as h(x) = 8(x 3 + 1)−2 . Then h (x) =

d d [8(x 3 + 1)−2 ] = −16(x 3 + 1)−3 (x 3 + 1) = −16(x 3 + 1)−3 (3x 2 ) dx d x derivative of the inside −3

= −48x (x + 1) . 2

3

In example 5.4, we apply the chain rule to a composition of a function with a composition of functions.

EXAMPLE 5.4

A Derivative Involving Multiple Chain Rules

√ Find the derivative of f (x) = ( x 2 + 4 − 3x 2 )3/2 .

Solution We have 1/2 d 3 2 f (x) = x + 4 − 3x 2 x 2 + 4 − 3x 2 2 dx

1/2 3 1 2 d = x 2 + 4 − 3x 2 (x + 4)−1/2 (x 2 + 4) − 6x 2 2 dx

1/2 1 3 2 2 2 −1/2 (x + 4) = x + 4 − 3x (2x) − 6x 2 2 1/2 3 2 x(x 2 + 4)−1/2 − 6x . = x + 4 − 3x 2 2

By the chain rule.

By the chain rule.

Simplification.

We now use the chain rule to compute the derivative of an inverse function in terms of the original function. Recall that we write g(x) = f −1 (x) if g( f (x)) = x for all x in the domain of f and f (g(x)) = x for all x in the domain of g. From this last equation, assuming that f and g are differentiable, it follows that d d [ f (g(x))] = (x). dx dx From the chain rule, we now have f (g(x))g (x) = 1. Solving this for g (x), we get

g (x) =

1 , f (g(x))

assuming we don’t divide by zero. We now state the result as Theorem 5.2.

THEOREM 5.2 If f is differentiable at all x and has an inverse function g(x) = f −1 (x), then 1 g (x) = , f (g(x)) provided f (g(x)) = 0. As we see in example 5.5, in order to use Theorem 5.2, we must be able to compute values of the inverse function.

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SECTION 2.5

EXAMPLE 5.5

..

The Chain Rule

193

The Derivative of an Inverse Function

Given that the function f (x) = x 5 + 3x 3 + 2x + 1 has an inverse function g(x), compute g (7). y 8 6 4 2 −3

−2 −1

x −2

1

2

3

−4 −6 −8

TODAY IN MATHEMATICS Fan Chung (1949– ) A Taiwanese mathematician with a highly successful career in American industry. She says, “As an undergraduate in Taiwan, I was surrounded by good friends and many women mathematicians. . . . A large part of education is learning from your peers, not just the professors.” Collaboration has been a hallmark of her career. “Finding the right problem is often the main part of the work in establishing the connection. Frequently a good problem from someone else will give you a push in the right direction and the next thing you know, you have another good problem.”

FIGURE 2.24 y = x 5 + 3x 3 + 2x + 1

Solution First, notice from Figure 2.24 that f appears to be one-to-one and so, will have an inverse. From Theorem 5.2, we have g (7) =

1 f (g(7))

.

(5.2)

It’s easy to compute f (x) = 5x 4 + 9x 2 + 2, but to use Theorem 5.2 we also need the value of g(7). If we write x = g(7), then x = f −1 (7), so that f (x) = 7. In general, solving the equation f (x) = 7 may be beyond our algebraic abilities. (Try solving x 5 + 3x 3 + 2x + 1 = 7 to see what we mean.) By trial and error, however, it is not hard to see that f (1) = 7, so that g(7) = 1. [Keep in mind that for inverse functions, f (x) = y and g(y) = x are equivalent statements.] Returning to equation (5.2), we now have g (7) =

1 1 = . f (1) 16

Notice that our solution in example 5.5 is dependent on finding an x for which f (x) = 7. This particular example was workable by trial and error, but finding most values other than g(7) would have been quite difficult or impossible to solve exactly.

BEYOND FORMULAS If you think that the method used in example 5.5 is roundabout, then you have the right idea. The chain rule in particular and calculus in general give us methods for determining quantities that are not directly computable. In the case of example 5.5, we use the properties of one function to determine properties of another function. The key to our ability to do this is understanding the theory behind the chain rule. Can you think of situations in your life where your understanding of your friends’ personalities helped you convince them, in a roundabout way, to do something that at first they didn’t want to do?

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EXERCISES 2.5 WRITING EXERCISES 1. If Fred can run 10 mph and Greg can run twice as fast as Fred, how fast can Greg run? The answer is obvious for most people. Formulate this simple problem as a chain rule calculation and conclude that the chain rule (in this context) is obvious.

In exercises 23 and 24, find an equation of the tangent line to y f (x) at x a. √ 23. f (x) = x 2 + 16, a = 3

2. The biggest challenge in computing √ the derivatives of (x 2 + 4)(x 3 − x + 1), (x 2 + 4) x 3 − x + 1 and √ x 2 + 4 x 3 − x + 1 is knowing which rule (product, chain etc.) to use when. Discuss how you know which rule to use when. (Hint: Think of the order in which you would perform operations to compute the value of each function for a specific choice of x.)

24. f (x) =

3. One simple implication of the chain rule is: if g(x) = f (x − a), then g (x) = f (x − a). Explain this derivative graphically: how does g(x) compare to f (x) graphically and why do the slopes of the tangent lines relate as the formula indicates? 4. Another simple implication of the chain rule is: if h(x) = f (2x), then h (x) = 2 f (2x). Explain this derivative graphically: how does h(x) compare to f (x) graphically and why do the slopes of the tangent lines relate as the formula indicates? In exercises 1–4, find the derivative with and without using the chain rule. 1. f (x) = (x 3 − 1)2

2. f (x) = (x 2 + 2x + 1)2

3. f (x) = (x + 1)

4. f (x) = (2x + 1)4

2

3

In exercises 5–22, find the derivative of each function. √ 6. f (x) = (x 3 + x − 1)3 5. f (x) = x 2 + 4 √ √ 7. f (x) = x 5 x 3 + 2 8. f (x) = (x 3 + 2) x 5 x3 x2 + 4 9. f (x) = 2 10. f (x) = 2 (x + 4) (x 3 )2 5

6

11. f (x) = √ x2 + 4 √ 13. f (x) = ( x + 3)4/3 √ 15. f (x) = ( x 3 + 2 + 2x)−2 x 17. f (x) = √ x2 + 1 x 19. f (x) = x2 + 1 21. f (x) = x x 4 + 2x 4 3

22. f (x) =

(x 3 + 4) 12. f (x) = 8 √ 14. f (x) = x(x 4/3 + 3) 16. f (x) = 4x 2 + (8 − x 2 )2 (x 2 − 1)2 x2 + 1 √ 20. f (x) = (x 2 + 1)( x + 1)3

18. f (x) =

x2

6 , a = −2 +4

In exercises 25 and 26, use the position function to find the velocity at time t 2. (Assume units of meters and seconds.) 25. s(t) =

√ t2 + 8

60t 26. s(t) = √ t2 + 1

In exercises 27 and 28, compute f (x), f (x) and f (4) (x), and identify a pattern for the nth derivative f (n) (x). 27. f (x) =

√ 2x + 1

28. f (x) =

In exercises 29–32, use the table of values to estimate the derivative of h(x) f (g(x)) or k(x) g( f (x)). x f (x) g(x) 29. h (1)

−3 −2 6

−2 −1 4

−1 0 2

0 −1 2

30. k (1)

1 −2 4

3x 2 + 2 x 3 + 4/x 4 √ (x 3 − 4) x 2 + 2

2 −3 6

31. k (3)

3 −2 4

4 0 2

5 2 1

32. h (3)

In exercises 33 and 34, use the relevant information to compute the derivative for h(x) f (g(x)). 33. h (1), where f (1) = 3, g(1) = 2, f (1) = 4, f (2) = 3, g (1) = −2 and g (3) = 5 34. h (2), where f (2) = 1, g(2) = 3, f (2) = −1, f (3) = −3, g (1) = 2 and g (2) = 4 In exercises 35–40, f (x) has an inverse g(x). Use Theorem 5.2 to find g (a). 35. f (x) = x 3 + 4x − 1, a = −1 36. f (x) = x 3 + 2x + 1, a = −2 37. f (x) = x 5 + 3x 3 + x, a = 5

8 x +2

2 x +1

38. f (x) = x 5 + 4x − 2, a = −2 √ 39. f (x) = x 3 + 2x + 4, a = 2 √ 40. f (x) = x 5 + 4x 3 + 3x + 1, a = 3

2-51

SECTION 2.5

In exercises 41–44, find a function g(x) such that g (x) f (x). 41. f (x) = (x + 3) (2x) 2

43. f (x) = √

42. f (x) = x (x + 4)

2

2

x

44. f (x) =

x2 + 1

3

..

The Chain Rule

195

y 12

2/3

x (x 2 + 1)2

10 8

45. A function f (x) is an even function if f (−x) = f (x) for all x and is an odd function if f (−x) = − f (x) for all x. Prove that the derivative of an even function is odd and the derivative of an odd function is even.

6 4 2

In exercises 46–49, find the derivative of the expression for an unspecified differentiable function f (x). 46. f (x 2 )

√ 47. f ( x)

48.

4 f (x) + 1

49.

1 1 + [ f (x)]2

50. If the graph of a differentiable function f (x) is symmetric about the line x = a, what can you say about the symmetry of the graph of f (x)? In exercises 51–54, use the given graphs to estimate the derivative. y 10

x

1

2

3

y = c(x) 51. 52. 53. 54.

f (2), where f (x) = b(a(x)) f (0), where f (x) = a(b(x)) f (−1), where f (x) = c(a(x)) f (1), where f (x) = b(c(x))

√ 55. Determine all values of x such that f (x) = 3 x 3 − 3x 2 + 2x is not differentiable. Describe the graphical property that prevents the derivative from existing. 56. Determine all values of x for which f (x) = |2x| + |x − 4| + |x + 4| is not differentiable. Describe the graphical property that prevents the derivative from existing. 57. Which steps in our outline of the proof of the chain rule are not well documented? Where do we use the assumption that g (x) = 0?

8 6 4 2 4

2

EXPLORATORY EXERCISES x 2

4

2

1 T df . Compute the derivative , 2L p dT where we think of T as the independent variable and treat p and L as constants. Interpret this derivative in terms of a guitarist tightening or loosening the string to “tune” it. Compute df the derivative and interpret it in terms of a guitarist playing dL notes by pressing the string against a fret.

at the frequency f =

y = a(x)

y 10 5

4

1. A guitar string of length L, density p and tension T will vibrate

x

2

2 5 10

y = b(x)

4

2. Newton’s second law of motion is F = ma, where m is the mass of the object that undergoes an acceleration a due to an applied force F. This law is accurate at low speeds. At high speeds, we use the corresponding formula from Einstein’s v(t) d , where v(t) theory of relativity, F = m dt 1 − v 2 (t)/c2 is the velocity function and c is the speed of light. Compute v(t) d . What has to be “ignored” to simplify dt 1 − v 2 (t)/c2 this expression to the acceleration a = v (t) in Newton’s second law?

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DERIVATIVES OF TRIGONOMETRIC FUNCTIONS

Displacement u(t) Equilibrium position

FIGURE 2.25 Spring-mass system

Springs are essential components of many mechanical systems, including shock absorbing systems for cars, stereo equipment and other sensitive devices. Imagine a weight hanging from a spring suspended from the ceiling (see Figure 2.25). If we set the weight in motion (e.g., by tapping down on it), the weight will bounce up and down in ever-shortening strokes until it eventually is again at rest (equilibrium). For short periods of time, this motion is nearly periodic. Suppose we measure the vertical displacement of the weight from its natural resting (equilibrium) position (see Figure 2.25). When we pull the weight down, its vertical displacement is negative. The weight then swings up to where the displacement is positive, swings down to a negative displacement and so on. The only functions we’ve experienced that exhibit this kind of behavior are the sine and cosine functions. We calculate the derivatives of these and the other trigonometric functions in this section. We can learn a lot about the derivatives of sin x and cos x from their graphs. From the graph of y = sin x in Figure 2.26a, notice the horizontal tangents at x = −3π/2, −π/2, π/2 and 3π/2. At these x-values, the derivative must equal 0. The tangent lines have positive slope for−2π < x < −3π/2, negative slope for −3π/2 < x < −π/2 and so on. For each interval on which the derivative is positive (or negative), the graph appears to be steepest in the middle of the interval: for example, from x = −π/2, the graph gets steeper until about x = 0 and then gets less steep until leveling out at x = π/2. A sketch of the derivative graph should then look like the graph in Figure 2.26b, which looks like the graph of y = cos x. We show here that this conjecture is, in fact, correct. In the exercises, you are asked to perform a similar graphical analysis to conjecture that the derivative of f (x) = cos x equals −sin x. y

y

1

1

x w

q

q

w

2p

1

x

p

p

2p

1

FIGURE 2.26a

FIGURE 2.26b

y = sin x

The derivative of f (x) = sin x

Before we move to the calculation of the derivatives of the six trigonometric functions, we first consider a few limits involving trigonometric functions. (We refer to these results as lemmas—minor theorems that lead up to some more significant result.) You will see shortly why we must consider these first.

LEMMA 6.1 lim sin θ = 0.

θ →0

2-53

SECTION 2.6

y

..

Derivatives of Trigonometric Functions

197

This result certainly seems reasonable, especially when we consider the graph of y = sin x. In fact, we have been using this for some time now, having stated this (without proof) as part of Theorem 3.4 in section 1.3. We now prove the result. u

1

PROOF

sin u u

x

For 0 < θ <

π , consider Figure 2.27. From the figure, observe that 2 0 ≤ sin θ ≤ θ.

(6.1)

It is a simple matter to see that lim 0 = 0 = lim+ θ.

θ →0+

FIGURE 2.27

θ →0

(6.2)

From the Squeeze Theorem (see section 1.3), it now follows from (6.1) and (6.2) that

Definition of sin θ

lim sin θ = 0,

θ →0+

also. Similarly, you can show that lim sin θ = 0.

θ →0−

This is left as an exercise. Since both one-sided limits are the same, it follows that lim sin θ = 0.

θ →0

y

LEMMA 6.2

Q(1, tan u)

lim cos θ = 1.

P(cos u, sin u)

θ →0

(0, 1)

The proof of this result is straightforward and follows from Lemma 6.1 and the Pythagorean Theorem. We leave this as an exercise. The following result was conjectured to be true (based on a graph and some computations) when we first examined limits in section 1.2. We can now prove the result.

u x

O

R(1, 0)

LEMMA 6.3

FIGURE 2.28a

sin θ = 1. θ →0 θ lim

y

PROOF

1

π . 2 Referring to Figure 2.28a, observe that the area of the circular sector OPR is larger than the area of the triangle OPR, but smaller than the area of the triangle OQR. That is,

Assume 0 < θ < u

x

0 < Area OPR < Area sector OPR < Area OQR.

(6.3)

You can see from Figure 2.28b that FIGURE 2.28b A circular sector

Area sector OPR = π (radius)2 (fraction of circle included) θ θ = π (12 ) = . 2π 2

(6.4)

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1 1 (base) (height) = (1) sin θ 2 2

(6.5)

1 (1) tan θ. 2

(6.6)

1 θ 1 sin θ < < tan θ. 2 2 2

(6.7)

Area OPR =

Also,

Area OQR =

and

Thus, from (6.3), (6.4), (6.5) and (6.6), we have 0< If we divide (6.7) by affected), we get

1 2

sin θ (note that this is positive, so that the inequalities are not 1<

θ tan θ 1 < = . sin θ sin θ cos θ

Taking reciprocals (again, everything here is positive), we find 1>

sin θ > cos θ. θ

(6.8)

The inequality (6.8) also holds if − π2 < θ < 0. (This is left as an exercise.) Finally, note that lim cos θ = 1 = lim 1.

θ →0

θ →0

Thus, it follows from (6.8) and the Squeeze Theorem that sin θ =1 θ →0 θ lim

also.

We need one additional limit result before we tackle the derivatives of the trigonometric functions.

LEMMA 6.4 lim

θ →0

1 − cos θ = 0. θ

y

Before we prove this, we want to make this conjecture seem reasonable. We draw a 1 − cos x graph of y = in Figure 2.29. The tables of function values that follow should x prove equally convincing.

0.8 0.4

4

x

2

2

4

0.8

FIGURE 2.29 1 − cos x y= x

0.1

1 − cos x x 0.04996

−0.1

1 − cos x x −0.04996

0.01

0.00499996

−0.01

−0.00499996

0.001

0.0005

−0.001

−0.0005

0.0001

0.00005

−0.0001

−0.00005

x

x

Now that we have strong evidence for the conjecture, we prove the lemma.

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SECTION 2.6

..

Derivatives of Trigonometric Functions

199

PROOF

1 − cos θ 1 − cos θ 1 + cos θ lim = lim θ→0 θ →0 θ θ 1 + cos θ 1 − cos2 θ θ →0 θ(1 + cos θ)

= lim

Multiply out numerator and denominator.

sin2 θ θ →0 θ(1 + cos θ)

sin θ sin θ = lim θ →0 θ 1 + cos θ

sin θ sin θ = lim lim θ →0 θ θ →0 1 + cos θ

0 = (1) = 0, 1+1 = lim

Multiply numerator and denominator by 1 + cos θ.

Since sin2 θ + cos2 θ = 1.

Split up terms, since both limits exist.

as conjectured. We are finally in a position to compute the derivatives of the sine and cosine functions. The derivatives of the other trigonometric functions will then follow by the quotient rule.

THEOREM 6.1 d sin x = cos x. dx

PROOF From the limit definition of derivative, for f (x) = sin x, we have d sin (x + h) − sin (x) sin x = f (x) = lim h→0 dx h sin x cos h + sin h cos x − sin x = lim h→0 h sin x cos h − sin x sin h cos x + lim = lim h→0 h→0 h h cos h − 1 sin h + (cos x) lim = (sin x) lim h→0 h→0 h h = (sin x)(0) + (cos x)(1) = cos x, from Lemmas 6.3 and 6.4.

THEOREM 6.2 d cos x = − sin x. dx

Trig identity: sin (α + β) = sin α cos β + sin β cos α. Grouping terms with sin x and terms with sin h separately. Factoring sin x from the first term and cos x from the second term.

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The proof of Theorem 6.2 is left as an exercise. For the remaining four trigonometric functions, we can use the quotient rule in conjunction with the derivatives of sin x and cos x.

THEOREM 6.3 d tan x = sec2 x. dx

PROOF We have from the quotient rule that

d d sin x tan x = dx d x cos x d d (sin x)(cos x) − (sin x) (cos x) dx = dx (cos x)2 cos x(cos x) − sin x(−sin x) = (cos x)2 =

cos2 x + sin2 x 1 = = sec2 x, (cos x)2 (cos x)2

where we have used the quotient rule and the preceding results on the derivatives of sin x and cos x (Theorems 6.1 and 6.2). The derivatives of the remaining trigonometric functions are left as exercises. The derivatives of all six trigonometric functions are summarized below.

d sin x = cos x dx

d cos x = −sin x dx

d tan x = sec2 x dx

d cot x = −csc2 x dx

d sec x = sec x tan x dx

d csc x = −csc x cot x dx

Example 6.1 shows where the product rule is necessary.

EXAMPLE 6.1

A Derivative That Requires the Product Rule

Find the derivative of f (x) = x 5 cos x. Solution From the product rule, we have d 5 d d 5 (x cos x) = (x ) cos x + x 5 (cos x) dx dx dx = 5x 4 cos x − x 5 sin x.

2-57

SECTION 2.6

EXAMPLE 6.2

..

Derivatives of Trigonometric Functions

201

Computing Some Routine Derivatives

Compute the derivatives of (a) f (x) = sin2 x and (b) g(x) = 4 tan x − 5 csc x. Solution For (a), we first rewrite the function as f (x) = (sin x)2 and use the chain rule. We have d f (x) = (2 sin x) (sin x) = 2 sin x cos x. d x derivative of the inside

g (x) = 4 sec2 x + 5 csc x cot x.

For (b), we have

You must be very careful to distinguish between similar notations with very different meanings, as you see in example 6.3.

EXAMPLE 6.3

The Derivatives of Some Similar Trigonometric Functions

Compute the derivative of f (x) = cos x 3 , g(x) = cos3 x and h(x) = cos 3x. Solution Note the differences in these three functions. Using the implied parentheses we normally do not bother to include, we have f (x) = cos(x 3 ), g(x) = (cos x)3 and h(x) = cos (3x). For the first function, we have f (x) =

d d 3 cos (x 3 ) = −sin(x 3 ) (x ) = −sin(x 3 )(3x 2 ) = −3x 2 sin x 3 . dx d x derivative of the inside

Next, we have g (x) =

d d (cos x)3 = 3(cos x)2 (cos x) dx d x derivative of the inside

= 3(cos x)2 (−sin x) = −3 sin x cos2 x. Finally, we have h (x) =

d d (cos 3x) = −sin (3x) (3x) = −sin (3x)(3) = −3 sin 3x. dx d x derivative of the inside

By combining our trigonometric rules with the product, quotient and chain rules, we can now differentiate many complicated functions.

EXAMPLE 6.4

A Derivative Involving the Chain Rule and the Quotient Rule

Find the derivative of f (x) = sin Solution We have f (x) = cos

2x . x +1

d 2x 2x x + 1 dx x + 1 derivative of the inside

2(x + 1) − 2x(1) 2x x +1 (x + 1)2

2x 2 = cos . x + 1 (x + 1)2 = cos

By the chain rule.

By the quotient rule.

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Applications The trigonometric functions arise quite naturally in the solution of numerous physical problems of interest. For instance, it can be shown that the vertical displacement of a weight suspended from a spring, in the absence of damping (i.e., when resistance to the motion, such as air resistance, is negligible), is given by u(t) = a cos (ωt) + b sin (ωt), Displacement u(t)

where ω is the frequency, t is time and a and b are constants. (See Figure 2.30 for a depiction of such a spring-mass system.) Equilibrium position

EXAMPLE 6.5 FIGURE 2.30 Spring-mass system

Analysis of a Spring-Mass System

Suppose that u(t) measures the displacement (measured in inches) of a weight suspended from a spring t seconds after it is released and that u(t) = 4 cos t. Find the velocity at any time t and determine the maximum velocity. Solution Since u(t) represents position (displacement), the velocity is given by u (t). We have u (t) = 4(−sin t) = −4 sin t, where u (t) is measured in inches per second. Of course, sin t oscillates between −1 and 1 and hence, the largest that u (t) can be is −4(−1) = 4 inches per second. This occurs when sin t = −1, that is, at t = 3π/2, t = 7π/2 and so on. Notice that at these times u(t) = 0, so that the weight is moving fastest when it is passing through its resting position.

1 farad

EXAMPLE 6.6 1 henry

Analysis of a Simple Electrical Circuit

The diagram in Figure 2.31 shows a simple electrical circuit. If the capacitance is 1 (farad), the inductance is 1 (henry) and the impressed voltage is 2t 2 (volts) at time t, then a model for the total charge Q(t) in the circuit at time t is Q(t) = 2 sin t + 2t 2 − 4 (coulombs).

~ 2t 2 volts

FIGURE 2.31 A simple circuit

The current is defined to be the rate of change of the charge with respect to time and so is given by dQ (amperes). dt Compare the current at times t = 0 and t = 1. I (t) =

Solution In general, the current is given by dQ = 2 cos t + 4t (amperes). dt Notice that at time t = 0, I (0) = 2 (amperes). At time t = 1, I (t) =

I (1) = 2 cos 1 + 4 ≈ 5.08 (amperes). This represents an increase of I (1) − I (0) 3.08 (100)% ≈ (100)% = 154%. I (0) 2

2-59

SECTION 2.6

EXAMPLE 6.7

..

Derivatives of Trigonometric Functions

203

Finding the Equation of the Tangent Line

Find an equation of the tangent line to y = 3 tan x − 2 csc x π at x = . 3 Solution The derivative is

y 10

y = 3 sec2 x − 2(−csc x cot x) = 3 sec2 x + 2 csc x cot x.

5 x 0.4

0.8

1.2

5 10

FIGURE 2.32 y = 3 tan x − 2 csc x and the π tangent line at x = 3

π , we have 3

π 40 1 4 2 y = 3(2)2 + 2 √ ≈ 13.33333. = 12 + = √ 3 3 3 3 3

40 4 π √ The tangent line with slope and point of tangency ,3 3 − √ has equation 3 3 3

√ 40 4 π y= x− +3 3− √ . 3 3 3 At x =

We show a graph of the function and the tangent line in Figure 2.32.

EXERCISES 2.6 WRITING EXERCISES 1. Most people draw sine curves that are very steep and rounded. Given the results of this section, discuss the actual shape of the sine curve. Starting at (0, 0), how steep should the graph be drawn? What is the steepest the graph should be drawn anywhere? In which regions is the graph almost straight and where does it curve a lot? 2. In many common physics and engineering applications, the term sin x makes calculations difficult. A common simplification is to replace sin x with x, accompanied by the justification “sin x approximately equals x for small angles.” Discuss this approximation in terms of the tangent line to y = sin x at x = 0. How small is the “small angle” for which the approximation is good? The tangent line to y = cos x at x = 0 is simply y = 1, but the simplification “cos x approximately equals 1 for small angles” is almost never used. Why would this approximation be less useful than sin x ≈ x? 1. Use a graphical analysis as in the text to argue that the derivative of cos x is −sin x. d 2. In the proof of (sin x) = cos x, where do we use the asdx sumption that x is in radians? If you have access to a CAS, find out what the derivative of sin x ◦ is; explain where the π/180 came from.

In exercises 3–20, find the derivative of each function. 3. f (x) = 4 sin x − x

4. f (x) = x 2 + 2 cos2 x

5. f (x) = tan3 x − csc4 x

6. f (x) = 4 sec x 2 − 3 cot x

7. f (x) = x cos 5x 2 9. f (x) = sin(tan x 2 ) 11. f (x) =

sin x 2 x2

13. f (t) = sin t sec t 1 sin 4x 17. f (x) = 2 sin x cos x √ 19. f (x) = tan x 2 + 1 15. f (x) =

8. f (x) = 4x 2 − 3 tan x 10. f (x) = sin2 x + 2 x2 csc4 x √ 14. f (t) = cos t sec t

12. f (x) =

16. f (x) = x 2 sec2 3x 18. f (x) = 4 sin2 x + 4 cos2 x 20. f (x) = 4x 2 sin x sec 3x

In exercises 21–24, use your CAS or graphing calculator. 21. Repeat exercise 17 with your CAS. If its answer is not in the same form as ours in the back of the book, explain how the CAS computed its answer. 22. Repeat exercise 18 with your CAS. If its answer is not 0, explain how the CAS computed its answer.

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23. Find the derivative of f (x) = 2 sin2 x + cos 2x on your CAS. Compare its answer to 0. Explain how to get this answer and your CAS’s answer, if it differs. tan x on your CAS. Compare its sin x answer to sec x tan x. Explain how to get this answer and your CAS’s answer, if it differs.

24. Find the derivative of f (x) =

In exercises 25–28, find an equation of the tangent line to y f (x) at x a. π 25. f (x) = sin 4x, a = 8 26. f (x) = tan 3x, a = 0 π 27. f (x) = cos x, a = 2 π 28. f (x) = x sin x, a = 2

41. Use the identity cos (x + h) = cos x cos h − sin x sin h to prove Theorem 6.2. 42. Use the quotient rule to derive formulas for the derivatives of cot x, sec x and csc x. 43. Use the basic limits lim

x→0

find the following limits: sin 3x x

(b) lim

cos x − 1 x→0 5x

(d) lim

(a) lim

x→0

t→0

x→0

find the following limits: (a) lim t→0

2t sin t

sin t 4t

sin x 2 x→0 x 2

(c) lim

44. Use the basic limits lim

In exercises 29–32, use the position function to find the velocity at time t t0 . Assume units of feet and seconds.

sin x cos x − 1 = 1 and lim = 0 to x→0 x x

sin x cos x − 1 = 1 and lim = 0 to x→0 x x cos x 2 − 1 x→0 x2

(b) lim

sin 6x tan 2x (d) lim x→0 sin 5x x sin x if x = 0 show that f is continuous and 45. For f (x) = x 1 if x = 0 differentiable for all x. (Hint: Focus on x = 0.) (c) lim

x→0

29. s(t) = t 2 − sin 2t, t0 = 0 30. s(t) = t cos(t 2 + π ), t0 = 0 cos t , t0 = π t 32. s(t) = 4 + 3 sin t, t0 = π 31. s(t) =

33. A spring hanging from the ceiling vibrates up and down. Its vertical position at time t is given by f (t) = 4 sin 3t. Find the velocity of the spring at time t. What is the spring’s maximum speed? What is its location when it reaches its maximum speed?

46. For the function of exercise 45, show that the derivative f (x) is continuous. (In this case, we say that the function f is C 1 .) 47. Show that the function of exercise 45 is C 2 . (That is, f (x) exists and is continuous for all x.)

34. In exercise 33, for what time values is the velocity 0? What is the location of the spring when its velocity is 0? When does the spring change directions?

48. As in exercise 46, show that 3 x sin x1 if x = 0 f (x) = is C 1 . 0 if x = 0

In exercises 35 and 36, refer to example 6.6.

49. Sketch a graph of y = sin x and its tangent line at x = 0. Try to determine how many times they intersect by zooming in on the graph (but don’t spend too much time on this). Show that for f (x) = sin x, f (x) < 1 for 0 < x < 1. Explain why this implies that sin x < x for 0 < x < 1. Use a similar argument to show that sin x > x for −1 < x < 0. Explain why y = sin x intersects y = x at only one point.

35. If the total charge in an electrical circuit at time t is given by Q(t) = 3 sin 2t + t + 4 coulombs, compare the current at times t = 0 and t = 1. 36. If the total charge in an electrical circuit at time t is given by Q(t) = 4 cos 4t − 3t + 1 coulombs, compare the current at times t = 0 and t = 1. 37. For f (x) = sin x, find f

(75)

(x) and f

(150)

(x).

38. For f (x) = cos x, find f (77) (x) and f (120) (x). 39. For Lemma 6.1, show that lim sin θ = 0. θ→0−

40. Use Lemma 6.1 and the identity cos2 θ + sin2 θ = 1 to prove Lemma 6.2.

50. For different positive values of k, determine how many times y = sin kx intersects y = x. In particular, what is the largest value of k for which there is only one intersection? Try to determine the largest value of k for which there are three intersections. 51. On a graphing calculator, graph y = sin x using the following range of x-values: [−50, 50], [−60, 60] and so on. You should find graphs similar to the following.

2-61

SECTION 2.6

y

y

x

..

Derivatives of Trigonometric Functions

205

identify the pattern. (Hint: Write the coefficients— a, b, etc.— in the form 1/n! for some integer n.) We will take a longer look at these approximations, called Taylor polynomials, in Chapter 8. x

y

x

Briefly explain why the graph seems to change so much. In particular, is the calculator showing all of the graph or are there large portions missing?

3. When a ball bounces, we often think of the bounce occurring instantaneously, as in the accompanying figure. The sharp corner in the graph at the point of impact does not take into account that the ball actually compresses and maintains contact with the ground for a brief period of time. As shown in John Wesson’s The Science of Soccer, the amount s that the ball is compressed satisfies the equation s (t) = − cp s(t), where c is m the circumference of the ball, p is the pressure of air in the ball and m is the mass of the ball. Assume that the ball hits the ground at time 0 with vertical speed v m/s. Then s(0) = 0 and s (0) = v. Show that s(t) = vk sin kt satisfies the three condi cp s(t), s(0) = 0 and s (0) = v with k = . tions s (t) = − cp m m Use the properties of the sine function to show that the duration of the bounce is πk seconds and find the maximum compression. For a soccer ball with c = 0.7 m, p = 0.86 × 105 N/m2 , v = 15 m/s, radius R = 0.112 m and m = 0.43 kg, compute the duration of the bounce and the maximum compression. y 62.5 50

EXPLORATORY EXERCISES

1

x 2 sin x if x = 0 has several un0 if x = 0 usual properties. Show that f is continuous and differentiable at x = 0. However, f (x) is discontinuous at x = 0. To see this, 1 1 show that f (x) = −1 for x = ,x = and so on. Then 2π 4π 1 1 and so on. Explain show that f (x) = 1 for x = , x = π 3π why this proves that f (x) cannot be continuous at x = 0.

1. The function f (x) =

2. We have seen that the approximation sin x ≈ x for small x derives from the tangent line to y = sin x at x = 0. We can also sin x = 1. Explain think of it as arising from the result lim x→0 x why the limit implies sin x ≈ x for small x. How can we get a better approximation? Instead of using the tangent line, we can try a quadratic function. To see what to multiply x 2 by, nusin x − x . Then sin x − x ≈ ax 2 merically compute a = lim x→0 x2 or sin x ≈ x + ax 2 for small x. How about a cubic sin x − (x + ax 2 ) . Then approximation? Compute b = lim x→0 x3 sin x − (x + ax 2 ) ≈ bx 3 or sin x ≈ x + ax 2 + bx 3 for small x. cos x − 1 Starting with lim = 0, find a cubic approximation x→0 x of the cosine function. If you have a CAS, take the approximations for sine and cosine out to a seventh-order polynomial and

37.5 25 12.5 0

x 0

1

2

3

4

An idealized bounce y

x

A ball being compressed Putting together the physics for before, during and after the bounce, we obtain the height of the center of mass of

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2-62

a ball of radius R: −4.9t 2 − vt + R R − vk sin kt h(t) = −4.9(t − πk )2 + v(t − πk ) + R

2.7

π k

DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS y

30

20

10

4

if t < 0 if 0 ≤ t ≤ if t > πk .

Determine whether h(t) is continuous for all t and sketch a reasonable graph of this function to replace the figure shown here.

x

2

2

4

FIGURE 2.33a y = 2x y 20

Exponential and logarithmic functions are among the most common functions encountered in applications. We begin with a simple application in business. Suppose that you have an investment that doubles in value every year. If you start with $100, then the value of the investment after 1 year is $100(2) or $200. After 2 years, its value is $100(2)(2) = $400 and after 3 years, its value is $100(23 ) = $800. In general, after t years, the value of your investment is $100(2t ). Since the value doubles every year, you might describe the rate of return on your investment as 100% (usually called the annual percentage yield or APY). To a calculus student, the term rate should suggest the derivative. We first consider f (x) = a x for some constant a > 1 (called the base). Recall from our discussion in Chapter 0 that the graph will look something like the graph of f (x) = 2x , shown in Figure 2.33a. Observe that as you look from left to right, the graph rises. Thus, the slopes of the tangent lines and so, too, the values of the derivative, are always positive. Also, the farther to the right you look, the steeper the graph is and so, the more positive the derivative is. Further, to the left of the origin, the tangent lines are nearly horizontal and hence, the derivative is positive but close to zero. The sketch of y = f (x) shown in Figure 2.33b is consistent with all of the above information. (Use your computer algebra system to generate a series of graphs of y = a x for various values of a > 0, along with the graphs of the corresponding derivatives and you may detect a pattern.) In particular, notice that the sketch of the derivative closely resembles the graph of the function itself.

10

Derivatives of the Exponential Functions 4

2

x 2

4

FIGURE 2.33b The derivative of f (x) = 2x

Let’s start by using the usual limit definition to try to compute the derivative of f (x) = a x , for a > 1. We have f (x + h) − f (x) a x+h − a x = lim h→0 h→0 h h x h x a a −a = lim From the usual rules of exponents. h→0 h ah − 1 . Factor out the common term of a x . = a x lim h→0 h

f (x) = lim

(7.1)

Unfortunately, we have, at present, no means of computing the limit in (7.1). Nonetheless, assuming the limit exists, (7.1) says that d x a = (constant) a x . dx

(7.2)

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2-62

a ball of radius R: −4.9t 2 − vt + R R − vk sin kt h(t) = −4.9(t − πk )2 + v(t − πk ) + R

2.7

π k

DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS y

30

20

10

4

if t < 0 if 0 ≤ t ≤ if t > πk .

Determine whether h(t) is continuous for all t and sketch a reasonable graph of this function to replace the figure shown here.

x

2

2

4

FIGURE 2.33a y = 2x y 20

Exponential and logarithmic functions are among the most common functions encountered in applications. We begin with a simple application in business. Suppose that you have an investment that doubles in value every year. If you start with $100, then the value of the investment after 1 year is $100(2) or $200. After 2 years, its value is $100(2)(2) = $400 and after 3 years, its value is $100(23 ) = $800. In general, after t years, the value of your investment is $100(2t ). Since the value doubles every year, you might describe the rate of return on your investment as 100% (usually called the annual percentage yield or APY). To a calculus student, the term rate should suggest the derivative. We first consider f (x) = a x for some constant a > 1 (called the base). Recall from our discussion in Chapter 0 that the graph will look something like the graph of f (x) = 2x , shown in Figure 2.33a. Observe that as you look from left to right, the graph rises. Thus, the slopes of the tangent lines and so, too, the values of the derivative, are always positive. Also, the farther to the right you look, the steeper the graph is and so, the more positive the derivative is. Further, to the left of the origin, the tangent lines are nearly horizontal and hence, the derivative is positive but close to zero. The sketch of y = f (x) shown in Figure 2.33b is consistent with all of the above information. (Use your computer algebra system to generate a series of graphs of y = a x for various values of a > 0, along with the graphs of the corresponding derivatives and you may detect a pattern.) In particular, notice that the sketch of the derivative closely resembles the graph of the function itself.

10

Derivatives of the Exponential Functions 4

2

x 2

4

FIGURE 2.33b The derivative of f (x) = 2x

Let’s start by using the usual limit definition to try to compute the derivative of f (x) = a x , for a > 1. We have f (x + h) − f (x) a x+h − a x = lim h→0 h→0 h h x h x a a −a = lim From the usual rules of exponents. h→0 h ah − 1 . Factor out the common term of a x . = a x lim h→0 h

f (x) = lim

(7.1)

Unfortunately, we have, at present, no means of computing the limit in (7.1). Nonetheless, assuming the limit exists, (7.1) says that d x a = (constant) a x . dx

(7.2)

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..

Derivatives of Exponential and Logarithmic Functions

207

Further, (7.2) is consistent with what we observed graphically in Figures 2.33a and 2.33b. The question that we now face is: does ah − 1 h→0 h lim

exist for all (or any) values of a > 1? We explore this limit numerically in the following table for a = 2. h 0.01 0.0001 0.000001 0.0000001

2h − 1 h 0.6955550 0.6931712 0.6931474 0.6931470

h −0.01 −0.0001 −0.000001 −0.0000001

2h − 1 h 0.6907505 0.6931232 0.6931469 0.6931472

The numerical evidence suggests that the limit in question exists and that 2h − 1 ≈ 0.693147. h→0 h lim

In the same way, the numerical evidence suggests that 3h − 1 ≈ 1.098612. h→0 h lim

The approximate values of these limits are largely unremarkable until you observe that ln 2 ≈ 0.693147

and

ln 3 ≈ 1.098612.

You’ll find similar results if you consider the limit in (7.1) for other values of a > 1. (Try estimating several of these for yourself.) This suggests that for a > 1,

y 30

d x a = a x ln a. dx 20

10

4

x

2

2

FIGURE 2.34a y = 2−x

4

Next, the exponential function f (x) = a x for 0 < a < 1, for instance, consider x f (x) = 12 . Notice that we may write such a function as f (x) = b−x , where b = a1 > 1. For example, for b = 12 , we have x 1 f (x) = = 2−x . 2 The graph of this type of exponential looks something like the graph of y = 2−x shown in Figure 2.34a. Assuming that the chain rule gives us

d x 2 = 2x ln 2, dx

d 1 x d −x = 2 dx 2 dx d (−x) dx = 2−x (−ln 2) x

1 1 = ln , 2 2

= 2−x ln 2

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Differentiation

y

2-64

where we have used the fact that

1 ln = ln(2−1 ) = −ln 2. 2

x

4

2

4

10

A sketch of the derivative is shown in Figure 2.34b. This suggests the following general result.

20

THEOREM 7.1 FIGURE 2.34b

The derivative of f (x) = 2−x

For any constant a > 0,

d x a = a x ln a. dx

(7.3)

The proof of this result is complete except for the precise evaluation of the limit in (7.1). Unfortunately, we will not be in a position to complete this work until Chapter 4. For the moment, you should be content with the strong numerical, graphical and (nearly complete) algebraic arguments supporting this conjecture. You should notice that (7.3) is consistent with what we observed in Figures 2.33a, 2.33b, 2.34a and 2.34b. In particular, recall that for 0 < a < 1, ln a < 0. This accounts for why the graph of the derivative for 0 < a < 1 resembles the reflection of the graph of the original function through the x-axis.

EXAMPLE 7.1

Finding the Rate of Change of an Investment

If the value of a 100-dollar investment doubles every year, its value after t years is given by v(t) = 100 2t . Find the instantaneous percentage rate of change of the worth. Solution The instantaneous rate of change is the derivative v (t) = 100 2t ln 2. The relative rate of change is then v (t) 100 2t ln 2 = ln 2 ≈ 0.693. = v(t) 100 2t The percentage change is then about 69.3% per year. This is surprising to most people. A percentage rate of 69.3% will double your investment each year if it is compounded “continuously.” We explore the notion of continuous compounding of interest further in exercise 37. The most commonly used base (by far) is the (naturally occurring) irrational number e. You should immediately notice the significance of this choice. Since ln e = 1, the derivative of f (x) = e x is simply d x e = e x ln e = e x . dx We now have the following result.

THEOREM 7.2 d x e = ex . dx

2-65

SECTION 2.7

..

Derivatives of Exponential and Logarithmic Functions

209

PROOF As we have just seen, the proof of this formula follows immediately from Theorem 7.1. You will probably agree that this is the easiest derivative formula to remember. In section 2.6, we looked at a simple model of the oscillations of a weight hanging from a spring. Using the cosine function, we had a model that produced the oscillations of the spring, but did not reproduce the physical reality of the spring eventually coming to rest. Coupling the trigonometric functions with the exponential functions produces a more realistic model.

Displacement u(t) Equilibrium position

EXAMPLE 7.2

Finding the Velocity of a Hanging Weight

If we build damping (i.e., resistance to the motion due to friction, for instance) into our model spring-mass system (see Figure 2.35), the vertical displacement at time t of a weight hanging from a spring can be described by

FIGURE 2.35 Spring-mass system

u(t) = Aeαt cos (ωt) + Beαt sin (ωt),

u 1.2

where A, B, α and ω are constants. For each of

1

(a) u(t) = e−t cos t

0.8

and

(b) v(t) = e−t/6 cos 4t,

0.6

sketch a graph of the motion of the weight and find its velocity at any time t.

0.4 0.2 t 5

0.2

10

FIGURE 2.36a u(t) = e−t cos t

Solution Figure 2.36a displays a graph of u(t) = e−t cos t. Notice that it seems to briefly oscillate and then quickly come to rest at u = 0. You should observe that this is precisely the kind of behavior you expect from your car’s suspension system (the spring-mass system most familiar to you) when you hit a bump in the road. If your car’s suspension system needs repair, you might get the behavior shown in Figure 2.36b, which is a graph of v(t) = e−t/6 cos (4t). The velocity of the weight is given by the derivative. From the product rule, we get

y

d d −t (e ) cos t + e−t (cos t) dt dt d = e−t (−t) cos t − e−t sin t dt

u (t) =

1 0.8 0.6 0.4 0.2

= −e−t (cos t + sin t)

x −0.2 −0.4 −0.6 −0.8 −1

1

3

5

and

FIGURE 2.36b y=e

−t/6

cos (4t)

d d −t/6 (e ) cos (4t) + e−t/6 [cos (4t)] dt dt

d t d = e−t/6 − cos (4t) + e−t/6 [− sin (4t)] (4t) dt 6 dt

v (t) =

1 = − e−t/6 cos (4t) − 4e−t/6 sin (4t). 6 At first glance, the chain rule for exponential functions may look a bit different, in part because the “inside” of the exponential function e f (x) is the exponent f (x). Be careful not to change the exponent when computing the derivative of an exponential function.

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2-66

EXAMPLE 7.3

The Chain Rule with Exponential Functions 2

2

Find the derivative of f (x) = 3e x , g(x) = xe2/x and h(x) = 32x . Solution From the chain rule, we have d 2 2 2 (x ) = 3e x (2x) = 6xe x . dx Using the product rule and the chain rule, we get

2 2/x 2/x d g (x) = (1)e + xe dx x

2 = e2/x + xe2/x − 2 x f (x) = 3e x

2

e2/x x = e2/x (1 − 2/x). = e2/x − 2

h (x) = 32x ln 3 2

Finally, we have

d (2x 2 ) dx

2

= 32x ln 3 (4x) 2

= 4(ln 3) x32x .

Derivative of the Natural Logarithm The natural logarithm function ln x is closely connected with exponential functions. We’ve already seen it arise as a part of the general exponential derivative formula (7.3). Recall from Chapter 0 that the graph of the natural logarithm looks like the one shown in Figure 2.37a. The function is defined only for x > 0 and as you look to the right, the graph always rises. Thus, the slopes of the tangent lines and hence, also the values of the derivative are always positive. Further, as x → ∞, the slopes of the tangent lines become less positive and seem to approach 0. On the other hand, as x approaches 0 from the right, the graph gets steeper and steeper and hence, the derivative gets larger and larger, without bound. The graph shown in Figure 2.37b is consistent with all of these observations. Do you recognize it? y

y 3

5

2

4

1

3 x 2

4

1

6

2 1

2

x

3

1

2

3

4

5

FIGURE 2.37a

FIGURE 2.37b

y = ln x

The derivative of f (x) = ln x

2-67

SECTION 2.7

..

Derivatives of Exponential and Logarithmic Functions

211

Using the definition of derivative, we get the following for f (x) = ln x: f (x) = lim

h→0

f (x + h) − f (x) ln (x + h) − ln x = lim . h→0 h h

Unfortunately, we don’t yet know how to evaluate this limit or even whether it exists. (We’ll develop all of this in Chapter 4.) Our only option at present is to investigate this limit numerically. In the following tables, we compute some values for x = 2.

h 0.1 0.001 0.00001 0.0000001

ln (2 h) − ln 2 h 0.4879016 0.4998750 0.4999988 0.5000002

h −0.1 −0.001 −0.00001 −0.0000001

ln (2 h) − ln 2 h 0.5129329 0.5001250 0.5000013 0.5000000

This leads us to the approximation ln (2 + h) − ln 2 1 ≈ . h→0 h 2

f (2) = lim

Similarly, we compute some values for x = 3.

h 0.1 0.001 0.00001 0.0000001

ln (3 h) − ln 3 h 0.3278982 0.3332778 0.3333328 0.3333333

h −0.1 −0.001 −0.00001 −0.0000001

ln (3 h) − ln 3 h 0.3390155 0.3333889 0.3333339 0.3333333

This leads us to the approximation ln (3 + h) − ln 3 1 ≈ . h→0 h 3

f (3) = lim

You should verify the approximations f (4) ≈ 14 , f (5) ≈ 15 and so on. Look back one more time at Figure 2.37b. Do you recognize the function now? You should convince yourself that the result stated in Theorem 7.3 makes sense.

THEOREM 7.3 For x > 0,

d 1 (ln x) = . dx x

(7.4)

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..

Differentiation

2-68

PROOF The proof of this follows directly from Theorems 5.2 and 7.2. Recall that y = ln x if and only if e y = x. From Theorem 5.2, we have for f (x) = ln x and g(x) = e x f (x) =

1 g (y)

=

1 1 = , y e x

as desired.

EXAMPLE 7.4

Derivatives of Logarithms

Find the derivative of f (x) = x ln x, g(x) = ln x 3 and h(x) = ln(x 2 + 1). Solution Using the product rule, we get

1 f (x) = (1) ln x + x = ln x + 1. x

We could certainly use the chain rule to differentiate g(x). However, using the properties of logarithms, recall that we can rewrite g(x) = ln x 3 = 3 ln x and using (7.4), we get

3 1 d g (x) = 3 (ln x) = 3 = . dx x x Using the chain rule for h(x), we get h (x) =

EXAMPLE 7.5

x2

1 1 2x d 2 (x + 1) = 2 (2x) = 2 . + 1 dx x +1 x +1

Finding the Maximum Concentration of a Chemical

The concentration x of a certain chemical after t seconds of an autocatalytic reaction is 10 . Show that x (t) > 0 and use this information to determine given by x(t) = −20t 9e +1 that the concentration of the chemical never exceeds 10. Solution Before computing the derivative, look carefully at the function x(t). The independent variable is t and the only term involving t is in the denominator. So, we don’t need to use the quotient rule. Instead, first rewrite the function as x(t) = 10(9e−20t + 1)−1 and use the chain rule. We get

x 10

d (9e−20t + 1) dt + 1)−2 (−180e−20t )

x (t) = −10(9e−20t + 1)−2

8 6

= −10(9e−20t

4

= 1800e−20t (9e−20t + 1)−2

2

=

1800e−20t . (9e−20t + 1)2

t 0.2

0.4

0.6

0.8

1.0

FIGURE 2.38 Chemical concentration

Notice that since e−20t > 0 for all t, both the numerator and denominator are positive, so that x (t) > 0. Since all of the tangent lines have positive slope, the graph of y = x(t) rises from left to right, as shown in Figure 2.38.

2-69

SECTION 2.7

..

Derivatives of Exponential and Logarithmic Functions

213

Since the concentration increases for all time, the concentration is always less than the limiting value lim x(t), which is easily computed to be t→∞

lim

10

t→∞ 9e−20t

+1

=

10 = 10. 0+1

Logarithmic Differentiation A clever technique called logarithmic differentiation uses the rules of logarithms to help find derivatives of certain functions for which we don’t presently have derivative formulas. For instance, note that the function f (x) = x x is not a power function because the exponent is not a constant and is not an exponential function because the base is not constant. In example 7.6, we show how to take advantage of the properties of logarithms to find the derivative of such a function.

EXAMPLE 7.6

Logarithmic Differentiation

Find the derivative of f (x) = x x , for x > 0. Solution As already noted, none of our existing derivative rules apply. We begin by taking the natural logarithm of both sides of the equation f (x) = x x . We have ln [ f (x)] = ln (x x ) = x ln x, from the usual properties of logarithms. We now differentiate both sides of this last equation. Using the chain rule on the left side and the product rule on the right side, we get 1 1 f (x) = (1) ln x + x f (x) x or Solving for f (x), we get

f (x) = ln x + 1. f (x) f (x) = (ln x + 1) f (x).

Substituting f (x) = x x gives us f (x) = (ln x + 1)x x .

EXERCISES 2.7 WRITING EXERCISES 1. The graph of f (x) = e x curves upward in the interval from x = −1 to x = 1. Interpreting f (x) = e x as the slopes of tangent lines and noting that the larger x is, the larger e x is, explain why the graph curves upward. For larger values of x, the graph of f (x) = e x appears to shoot straight up with no curve. Using the tangent line, determine whether this is correct or just an optical illusion.

2. The graph of f (x) = ln x appears to get flatter as x gets larger. Interpret the derivative f (x) = x1 as the slopes of tangent lines to determine whether this is correct or just an optical illusion. 3. Graphically compare and contrast the functions x 2 , x 3 , x 4 and e x for x > 0. Sketch the graphs for large x (and very large

214

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..

Differentiation

2-70

y’s) and compare the relative growth rates of the functions. In general, how does the exponential function compare to polynomials? 4. Graphically compare and contrast the functions x 1/2 , x 1/3 , x 1/4 and ln x for x > 1. Sketch the graphs for large x and compare the relative growth √ rates of the functions. In general, how does ln x compare to n x?

relative increase of worth in one year. Find the APY for the following interest rates: (a) 5% (d) 100 ln 2%

2. f (x) = e2x cos 4x

3. f (x) = x + 2 5. f (x) = 2e

4x+1

7. f (x) = (1/3)

x2

9. f (x) = 4−3x+1 e4x x

4. f (x) = x4

3x

6. f (x) = (1/e)

x

−x 2

8. f (x) = 4

10. f (x) = (1/2)1−x

13. f (x) = ln 2x

x e6x √ 14. f (x) = ln 8x

15. f (x) = ln (x 3 + 3x)

16. f (x) = x 3 ln x

17. f (x) = ln (cos x)

18. f (x) = esin 2x

19. f (x) = sin [ln (cos x 3 )] √ ln x 2 21. f (x) = x

20. f (x) = ln (sin x 2 )

11. f (x) =

23. f (x) = ln (sec x + tan x)

(b) 10%

In exercises 39–44, use logarithmic differentiation to find the derivative.

In exercises 1–24, find the derivative of the function.

x

(c) 20%

38. Determine the interest rate needed to obtain an APY of (a) 100%

1. f (x) = x 3 e x

(b) 10% (e) 100%

12. f (x) =

ex 2x √ 3 24. f (x) = e2x x 3 22. f (x) =

In exercises 25–30, find an equation of the tangent line to y f (x) at x 1. 25. f (x) = 3e x

26. f (x) = 2e x−1

27. f (x) = 3x

28. f (x) = 2x

29. f (x) = x 2 ln x

30. f (x) = 2 ln x 3

In exercises 31–34, the value of an investment at time t is given by v(t). Find the instantaneous percentage rate of change. 31. v(t) = 100 3t

32. v(t) = 100 4t

33. v(t) = 100 et

34. v(t) = 100 e−t

35. A bacterial population starts at 200 and triples every day. Find a formula for the population after t days and find the percentage rate of change in population.

2

39. f (x) = x sin x

40. f (x) = x 4−x

41. f (x) = (sin x)x

42. f (x) = (x 2 )4x

43. f (x) = x ln x

44. f (x) = x

√

x

45. The motion of a spring is described by f (t) = e−t cos t. Compute the velocity at time t. Graph the velocity function. When is the velocity zero? What is the position of the spring when the velocity is zero? 46. The motion of a spring is described by f (t) = e−2t sin 3t. Compute the velocity at time t. Graph the velocity function. When is the velocity zero? What is the position of the spring when the velocity is zero? 47. In exercise 45, graphically estimate the value of t > 0 at which the maximum velocity is reached. 48. In exercise 46, graphically estimate the value of t > 0 at which the maximum velocity is reached. In exercises 49–52, involve the hyperbolic sine and hyperbolic e x e− x e x − e− x and cosh x . cosine functions: sinh x 2 2 d d 49. Show that (sinh x) = cosh x and (cosh x) = sinh x. dx dx 50. Find the derivative of the hyperbolic tangent function: sinh x tanh x = . cosh x 51. Show that both sinh x and cosh x have the property that f (x) = f (x). f (x) = sinh (cos x)

52. Find the derivative of (a) (b) f (x) = cosh (x 2 ) − sinh (x 2 ).

and

53. Find the value of a such that the tangent to ln x at x = a is a line through the origin. 54. Find the value of a such that the tangent to e x at x = a is a line through the origin. Compare the slopes of the lines in exercises 53 and 54.

36. A bacterial population starts at 500 and doubles every four days. Find a formula for the population after t days and find the percentage rate of change in population.

In exercises 55–58, use a CAS or graphing calculator.

37. An investment of A dollars receiving 100r percent (per year) interest compounded continuously will be worth f (t) = Aer t dollars after t years. APY can be defined as [ f (1) − A]/A, the

55. Find the derivative of f (x) = eln x on your CAS. Compare its answer to 2x. Explain how to get this answer and your CAS’s answer, if it differs.

2

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SECTION 2.7 2

56. Find the derivative of f (x) = eln (−x ) on your CAS. The correct answer is that it does not exist. Explain how to get this answer and your CAS’s answer, if it differs. √ 57. Find the derivative of f (x) = ln 4e3x on your CAS. Compare 3 its answer to 2 . Explain how to get this answer and your CAS’s answer, if it differs. 4x

e on your CAS. Com58. Find the derivative of f (x) = ln x2 pare its answer to 4 − 2/x. Explain how to get this answer and your CAS’s answer, if it differs. 59. Numerically estimate the limit in (7.1) for a = 3 and compare your answer to ln 3. 60. Numerically estimate the limit in (7.1) for a = your answer to ln 13 .

1 3

and compare

61. The concentration of a certain chemical after t seconds of an 6 autocatalytic reaction is given by x(t) = −8t . Show that 2e + 1 x (t) > 0 and use this information to determine that the concentration of the chemical never exceeds 6. 62. The concentration of a certain chemical after t seconds of an 10 . Show that autocatalytic reaction is given by x(t) = −10t 9e +2 x (t) > 0 and use this information to determine that the concentration of the chemical never exceeds 5. 63. The Pad´e approximation of e x is the function of the form a + bx for which the values of f (0), f (0) and f (0) f (x) = 1 + cx match the corresponding values of e x . Show that these values all equal 1 and find the values of a, b and c that make f (0) = 1, f (0) = 1 and f (0) = 1. Compare the graphs of f (x) and e x . 64. In a World Almanac or Internet resource, look up the population of the United States by decade for as many years as are available. If there are not columns indicating growth by decade, both numerically and by percentage, compute these yourself. (A spreadsheet is helpful here.) The United States has had periods of both linear and exponential growth. Explain why linear growth corresponds to a constant numerical increase; during which decades has the numerical growth been (approximately) constant? Explain why exponential growth corresponds to a constant percentage growth. During which decades has the percentage growth been (approximately) constant? 65. In statistics, the function f (x) = e−x /2 is used to analyze random quantities that have a bell-shaped distribution. Solutions of the equation f (x) = 0 give statisticians a measure of the variability of the random variable. Find all solutions. 2

66. Repeat exercise 65 for the function f (x) = e−x /8 . Comparing the graphs of the two functions, explain why you would say that this distribution is more spread out than that of exercise 65. 2

67. Repeat exercise 65 for the general 2 2 f (x) = e−(x−m) /2c , where m and c are constants.

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215

68. In exercise 67, find the solution to the equation f (x) = 0. This value is known as the mode (or average) of the distribution.

EXPLORATORY EXERCISES Ax n for positive constants A, n θn + xn and θ are used to model a variety of chemical and biological processes. In this exercise, we explore a technique for determin ing the values of the constants. First, show that

f (x) > 0 for f (x)/A x > 0 and lim f (x) = A. Second, if u = ln x→∞ 1 − f (x)/A and v = ln x, show that u is a linear function of v. To see why these facts are important, consider the following data collected in a study of the binding of oxygen to hemoglobin. Here, x is the percentage of oxygen in the air and y is the percentage of hemoglobin saturated with oxygen.

1. The Hill functions f (x) =

x y

1 2

2 13

3 32

4 52

5 67

6 77

7 84

8 88

9 91

Plot these data points. As you already showed, Hill functions are increasing and level off at a horizontal asymptote. Explain why it would be reasonable to try to find a Hill function to match these data. Use the limiting value of f (x) to explain why in this data set, A = 100. Next, for each x-y pair, compute u and v as defined above. Plot the u-v points and show that they are (almost exactly) linear. Find the slope and use this to determine the values of n and θ. 2. In Chapter 0, we defined e as the limit e = lim (1 + 1/n)n . n→∞

In this section, e is given as the value of the base a such that d x a = a x . There are a variety of other interesting properties dx of this important number. We discover one here. Graph the functions x 2 and 2x . For x > 0, how many times do they intersect? Graph the functions x 3 and 3x . For x > 0, how many times do they intersect? Try the pair x 2.5 and 2.5x and the pair x 4 and 4x . What would be a reasonable conjecture for the number of intersections (x > 0) of the functions x a and a x ? Explain why x = a is always a solution. Under what circumstances is there another solution less than a? greater than a? By trial and error, verify that e is the value of a at which the “other” solution changes. e−1/x 3. For n = 1 and n = 2, investigate lim n numerically and x→0 x e−1/x graphically. Conjecture the value of lim n for any posix→0 x tive integer n and use your conjecture for the remainder of the 0 if x ≤ 0 , show that f is differexercise. For f (x) = e−1/x if x > 0 entiable at each x and that f (x) is continuous for all x. Then show that f (0) exists and compare the work needed to show that f (x) is continuous at x = 0 and to show that f (0) exists.

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IMPLICIT DIFFERENTIATION AND INVERSE TRIGONOMETRIC FUNCTIONS y

Compare the following two equations describing familiar curves:

2

y = x 2 + 3 (parabola) and

x 2 + y 2 = 4 (circle).

x

2

2

2

FIGURE 2.39 The tangent line √ at the point (1, − 3)

The first equation defines y as a function of x explicitly, since for each x, the equation gives an explicit formula y = f (x) for finding the corresponding value of y. On the other hand, the second equation does not define a function, since the circle in Figure 2.39 doesn’t pass the However, √ vertical line test. √ you can solve for y and find at least two functions y = 4 − x 2 and y = − 4 − x 2 that are defined implicitly by the equation x 2 + y 2 = 4. 2 2 Suppose √ that we want to find the slope of the tangent line to the circle x + y = 4 at the point 1, − 3 (see Figure 2.39). We can think of the circle as the graph of two semicircles, √ √ √ in the point (1, − 3), we use y = 4 − x 2 and y = − 4 − x 2 . Since we are interested √ the equation describing the bottom semicircle, y = − 4 − x 2 to compute the derivative 1 x y (x) = − √ (−2x) = √ . 2 4 − x2 4 − x2 √ 1 So, the slope of the tangent line at the point 1, − 3 is then y (1) = √ . 3 This calculation was not especially challenging, although we will soon see an easier way to do it. However, it’s not always possible to explicitly solve for a function defined implicitly by a given equation. For example, van der Waals’ equation relating the pressure P, volume V and temperature T of a gas has the form

an 2 P + 2 (V − nb) = nRT, (8.1) V where a, n, b and R are constants. Notice the difficulty in solving this for V as a function of P. If we want the derivative dV , we will need a method for computing the derivative dP directly from the implicit representation given in (8.1). Consider each of the following calculations: d 3 (x + 4)2 = 2(x 3 + 4)(3x 2 ), dx d (sin x − 3x)2 = 2(sin x − 3x)(cos x − 3) dx and

d (tan x + 2)2 = 2(tan x + 2) sec2 x. dx

Notice that each of these calculations has the form d [y(x)]2 = 2[y(x)]y (x), dx for some choice of the function y(x). This last equation is simply an expression of the chain rule. We can use this notion to find the derivatives of functions defined implicitly by equations.

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217

We first return to the simple case of the circle x 2 + y 2 = 4. Assuming this equation defines one or more differentiable functions of x: y = y(x), the equation is x 2 + [y(x)]2 = 4.

(8.2)

Differentiating both sides of equation (8.2) with respect to x, we obtain d 2 d x + [y(x)]2 = (4). dx dx From the chain rule,

d [y(x)]2 = 2y(x)y (x), as above and so, we have dx 2x + 2y(x)y (x) = 0.

(8.3)

Subtracting 2x from both sides of (8.3) gives us 2y(x)y (x) = −2x and dividing by 2y(x) (assuming this is not zero), we have solved for y (x): y (x) =

−2x −x = . 2y(x) y(x)

of both x and y. To get the slope at the point Notice √that here y (x) is expressed in terms √ (1, − 3), substitute x = 1 and y = − 3. We have −x −1 1 y (1) = = √ =√ . y(x) x=1 − 3 3

Notice that this is the same as we had found earlier by first solving for y explicitly and then differentiating. This process of differentiating both sides of an equation with respect to x and then solving for y (x) is called implicit differentiation. When faced with an equation implicitly defining one or more differentiable functions y = y(x), differentiate both sides with respect to x, being careful to recognize that differentiating any function of y will require the chain rule: d g(y) = g (y)y (x). dx Then, gather any terms with a factor of y (x) on one side of the equation, with the remaining terms on the other side of the equation and solve for y (x). We illustrate this process in the examples that follow.

EXAMPLE 8.1

Finding a Slope Implicitly

Find y (x) for x + y − 2y = 3. Then, find the slope of the tangent line at the point (2, 1). 2

3

Solution Since we can’t (easily) solve for y in terms of x explicitly, we compute the derivative implicitly. Differentiating both sides with respect to x, we get d 2 d (x + y 3 − 2y) = (3) dx dx and so,

2x + 3y 2 y (x) − 2y (x) = 0.

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To solve for y (x), simply write all terms involving y (x) on one side of the equation and all other terms on the other side. We have 3y 2 y (x) − 2y (x) = −2x

Subtracting 2x from both sides.

and hence, after factoring, we have (3y 2 − 2)y (x) = −2x. Solving for y (x), we get

y

y (x) =

3 2

2

x 2

−2x . 3y 2 − 2

Dividing by (3y 2 − 2).

Substituting x = 2 and y = 1, we find that the slope of the tangent line at the point (2, 1) is

(2, 1)

1 4

Factoring y (x) from both terms on the left side.

y (2) =

4

1 2

−4 = −4. 3−2

The equation of the tangent line is then

3

y − 1 = −4(x − 2).

FIGURE 2.40

We have plotted a graph of the equation and the tangent line in Figure 2.40 using the implicit plot mode of our computer algebra system.

Tangent line at (2, 1)

EXAMPLE 8.2

Finding a Tangent Line by Implicit Differentiation

Find y (x) for x 2 y 2 − 2x = 4 − 4y. Then, find an equation of the tangent line at the point (2, −2). Solution Differentiating both sides with respect to x, we get d 2 2 d (x y − 2x) = (4 − 4y). dx dx Since the first term is the product of x 2 and y 2 , we must use the product rule. We get 2x y 2 + x 2 (2y)y (x) − 2 = 0 − 4y (x). Grouping the terms with y (x) on one side, we get (2x 2 y + 4)y (x) = 2 − 2x y 2 ,

y 2

so that x

3

2

4

2 4

y (x) =

Substituting x = 2 and y = −2, we get the slope of the tangent line, y (2) =

2 − 16 7 = . −16 + 4 6

Finally, an equation of the tangent line is given by y+2=

FIGURE 2.41 Tangent line at (2, −2)

2 − 2x y 2 . 2x 2 y + 4

7 (x − 2). 6

We have plotted the curve and the tangent line at (2, −2) in Figure 2.41 using the implicit plot mode of our computer algebra system.

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219

You can use implicit differentiation to find a needed derivative from virtually any equation you can write down. We illustrate this next for an application.

EXAMPLE 8.3

Rate of Change of Volume with Respect to Pressure

Suppose that van der Waals’ equation for a specific gas is

5 P + 2 (V − 0.03) = 9.7. V

(8.4)

Thinking of the volume V as a function of pressure P, use implicit differentiation to find dV the derivative at the point (5, 1). dP Solution Differentiating both sides of (8.4) with respect to P, we have d d [(P + 5V −2 )(V − 0.03)] = (9.7). dP dP From the product rule and the chain rule, we get

dV −3 dV 1 − 10V (V − 0.03) + (P + 5V −2 ) = 0. dP dP

V 6

Grouping the terms containing

4

dV , we get dP

[−10V −3 (V − 0.03) + P + 5V −2 ] 2

dV 0.03 − V . = dP −10V −3 (V − 0.03) + P + 5V −2

so that P 2

4

6

FIGURE 2.42 Graph of van der Waals’ equation and the tangent line at the point (5, 1)

dV = 0.03 − V, dP

We now have V (5) =

−0.97 97 0.03 − 1 = =− . −10(1)(0.97) + 5 + 5(1) 0.3 30

(The units are in terms of volume per unit pressure.) We show a graph of van der Waals’ equation, along with the tangent line to the graph at the point (5, 1) in Figure 2.42. Of course, since we can find one derivative implicitly, we can also find second and higher order derivatives implicitly. In example 8.4, notice that you can choose which equation to use for the second derivative. A smart choice can save you time and effort.

EXAMPLE 8.4

Finding a Second Derivative Implicitly

Find y (x) implicitly for y 2 + 2e−x y = 6. Then find the value of y at the point (0, 2). Solution As always, start by differentiating both sides of the equation with respect to x. We have d 2 d (y + 2e−x y ) = (6). dx dx By the chain rule, we have 2yy (x) + 2e−x y [−y − x y (x)] = 0.

(8.5)

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2-76

Notice that we don’t need to solve this for y (x). Dividing out the common factor of 2 and differentiating again, we get y (x)y (x) + yy (x) − e−x y [−y − x y (x)][y + x y (x)] − e−x y [y (x) + y (x) + x y (x)] = 0.

y 2.4 2.2 2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 3 2 1

Grouping all the terms involving y (x) on one side of the equation gives us yy (x) − xe−x y y (x) = −[y (x)]2 − e−x y [y + x y (x)]2 + 2e−x y y (x). Factoring out the y (x) on the left, we get (y − xe−x y )y (x) = −[y (x)]2 − e−x y [y + x y (x)]2 + 2e−x y y (x), so that x 1

2

3

FIGURE 2.43 y 2 + 2e−x y = 6

y (x) =

−[y (x)]2 − e−x y [y + x y (x)]2 + 2e−x y y (x) . y − xe−x y

(8.6)

Notice that (8.6) gives us a (rather messy) formula for y (x) in terms of x, y and y (x). If we need to have y (x) in terms of x and y only, we can solve (8.5) for y (x) and substitute into (8.6). However, we don’t need to do this to find y (0). Instead, first substitute x = 0 and y = 2 into (8.5) to get 4y (0) + 2(−2) = 0, from which we conclude that y (0) = 1. Then substitute x = 0, y = 2 and y (0) = 1 into (8.6) to get y (0) =

−1 − (2)2 + 2 3 =− . 2 2

See Figure 2.43 for a graph of y 2 + 2e−x y = 6 near the point (0, 2).

TODAY IN MATHEMATICS Dusa McDuff (1945– ) A British mathematician who has won prestigious awards for her research in multidimensional geometry. McDuff was inspired by Russian mathematician Israel Gelfand. “Gelfand amazed me by talking of mathematics as if it were poetry. . . . I had always thought of mathematics as being much more straightforward: a formula is a formula, and an algebra is an algebra, but Gelfand found hedgehogs lurking in the rows of his spectral sequences!’’ McDuff has made important contributions to undergraduate teaching and Women in Science and Engineering, lectures around the world and has coauthored several research monographs.

Recall that, up to this point, we have proved the power rule d r x = r x r −1 dx only for integer exponents (see Theorems 3.1 and 4.3), although we have been freely using this result for any real exponent, r . Now that we have developed implicit differentiation, however, we have the tools we need to prove the power rule for the case of any rational exponent.

THEOREM 8.1 For any rational exponent, r,

d r x = r x r −1 . dx

PROOF Suppose that r is any rational number. Then r =

p , for some integers p and q. Let q

y = x r = x p/q .

(8.7)

Then, raising both sides of equation (8.7) to the qth power, we get yq = x p .

(8.8)

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Differentiating both sides of equation (8.8) with respect to x, we get d q d p (y ) = (x ). dx dx From the chain rule, we have Solving for

qy q−1

dy = px p−1 . dx

dy , we have dx dy px p−1 px p−1 = = dx qy q−1 q(x p/q )q−1 px p−1 p = x p−1− p+ p/q q x p− p/q q p = x p/q−1 = r x r −1 , q =

Since y = x p/q .

Using the usual rules of exponents.

Since

p = r. q

as desired.

Derivatives of the Inverse Trigonometric Functions The inverse trigonometric functions are useful in any number of applications and are essential for solving equations. We now develop derivative rules for these functions. Recall from our discussion in Chapter 0 that you must pay very close attention to the domains and ranges for these functions. In particular, the inverse sine (or function is defined by ! arcsine) π π" restricting the domain of the sine function to the interval − , . Specifically, we had 2 2 π π y = sin−1 x if and only if sin y = x and − ≤ y ≤ . 2 2 Differentiating the equation sin y = x implicitly, we have d d sin y = x dx dx dy cos y = 1. dx

and so, Solving this for

dy , we find (for cos y = 0) that dx dy 1 = . dx cos y

This is not entirely satisfactory, though, since this gives us the derivative in terms of y. Notice that for − π2 ≤ y ≤ π2 , cos y ≥ 0 and hence, cos y = 1 − sin2 y = 1 − x 2 . dy 1 1 = =√ , dx cos y 1 − x2

This leaves us with for −1 < x < 1. That is,

d 1 sin−1 x = √ , dx 1 − x2

for −1 < x < 1.

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2-78

Alternatively, we can derive this formula using Theorem 5.2 in section 2.5. Similarly, we can show that d −1 , cos−1 x = √ dx 1 − x2 To find

for −1 < x < 1.

d tan−1 x, recall that we have dx y = tan−1 x if and only if tan y = x

and

−

π π

Using implicit differentiation, we then have d d tan y = x dx dx (sec2 y)

and so, We solve this for

dy = 1. dx

dy , to obtain dx dy 1 = dx sec2 y 1 = 1 + tan2 y 1 . = 1 + x2

That is,

d 1 . tan−1 x = dx 1 + x2

The derivatives of the remaining inverse trigonometric functions are left as exercises. The derivatives of all six inverse trigonometric functions are summarized here.

d sin−1 x dx d cos−1 x dx d tan−1 x dx d cot−1 x dx d sec−1 x dx

= √ = √ = = =

d csc−1 x = dx

1 1 − x2 −1

,

for −1 < x < 1

, for −1 < x < 1 1 − x2 1 1 + x2 −1 1 + x2 1 , for |x| > 1 √ |x| x 2 − 1 −1 , for |x| > 1 √ |x| x 2 − 1

2-79

SECTION 2.8

EXAMPLE 8.5

..

Implicit Differentiation and Inverse Trigonometric Functions

223

Finding the Derivative of an Inverse Trigonometric Function

Compute the derivative of (a) cos−1 (3x 2 ), (b) (sec−1 x)2 and (c) tan−1 (x 3 ). Solution From the chain rule, we have (a)

d d −1 cos−1 (3x 2 ) = (3x 2 ) 2 2 dx 1 − (3x ) d x = √

(b)

and (c)

1 − 9x 4

.

d d (sec−1 x)2 = 2(sec−1 x) (sec−1 x) dx dx 1 = 2(sec−1 x) √ |x| x 2 − 1 d d 3 1 [tan−1 (x 3 )] = (x ) dx 1 + (x 3 )2 d x =

EXAMPLE 8.6

−6x

3x 2 . 1 + x6

Modeling the Rate of Change of a Ballplayer’s Gaze

One of the guiding principles of most sports is to “keep your eye on the ball.” In baseball, a batter stands 2 feet from home plate as a pitch is thrown with a velocity of 130 ft/s (about 90 mph). At what rate does the batter’s angle of gaze need to change to follow the ball as it crosses home plate? Solution First, look at the triangle shown in Figure 2.44. We denote the distance from the ball to home plate by d and the angle of gaze by θ . Since the distance is changing with time, we write d = d(t). The velocity of 130 ft/s means that d (t) = −130. [Why Overhead view

d

θ 2

FIGURE 2.44 A ballplayer’s gaze

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2-80

would d (t) be negative?] From Figure 2.44, notice that d(t) θ(t) = tan−1 . 2 The rate of change of the angle is then θ (t) =

1+

=

d (t) 2 2 d(t)

1

2

2d (t) radians/second. 4 + [d(t)]2

When d(t) = 0 (i.e., when the ball is crossing home plate), the rate of change is then 2(−130) = −65 radians/second. 4 One problem with this is that most humans can accurately track objects only at the rate of about 3 radians/second. Keeping your eye on the ball in this case is thus physically impossible. (See Watts and Bahill, Keep Your Eye on the Ball.) θ (t) =

BEYOND FORMULAS Implicit differentiation allows us to find the derivative of functions even when we don’t have a formula for the function. This remarkable result means that if we have any equation for the relationship between two quantities, we can find the rate of change of one with respect to the other. Here is a case where mathematics requires creative thinking beyond formula memorization. In what other situations have you seen the need for creativity in mathematics?

EXERCISES 2.8 WRITING EXERCISES 1. For implicit differentiation, we assume that y is a function of x: we write y(x) to remind ourselves of this. However, for the 2 circle x√ + y 2 = 1, it is not true that y is a function of x. Since y = ± 1 − x 2 , there are actually (at least) two functions of x defined implicitly. Explain why this is not really a contradiction; that is, explain exactly what we are assuming when we do implicit differentiation. 2. To perform implicit differentiation on an equation such as x 2 y 2 + 3 = x, we start by differentiating all terms. We get 2x y 2 + x 2 (2y)y = 1. Many students learn the rules this way: take “regular” derivatives of all terms, and tack on a y every time you take a y-derivative. Explain why this works, and rephrase the rule in a more accurate and understandable form. 3. In implicit differentiation, the derivative is typically a function of both x and y; for example, on the circle x 2 + y 2 = r 2 , we have y = −x/y. If we take the derivative −x/y and plug in

any x and y, will it always be the slope of a tangent line? That is, are there any requirements on which x’s and y’s we can plug in? 4. In each example in this section, after we differentiated the given equation, we were able to rewrite the resulting equation in the form f (x, y)y (x) = g(x, y) for some functions f (x, y) and g(x, y). Explain why this can always be done; that is, why doesn’t the chain rule ever produce a term like [y (x)]2 1 or ? y (x) In exercises 1–4, compute the slope of the tangent line at the given point both explicitly (first solve for y as a function of x) and implicitly. 1. x 2 + 4y 2 = 8 at (2, 1) √ √ 2. x 3 y − 4 x = x 2 y at (2, 2)

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SECTION 2.8

3. y − 3x 2 y = cos x at (0, 1) 4. y 2 + 2x y + 4 = 0 at (−2, 2) In exercises 5–16, find the derivative y (x) implicitly. 5. x 2 y 2 + 3y = 4x √ 7. x y − 4y 2 = 12 9.

6. 3x y 3 − 4x = 10y 2 8. sin x y = x 2 − 3

x +3 = 4x + y 2 y

10. 3x + y 3 − 4y = 10x 2

2

11. e x y − e y = x √ 13. x + y − 4x 2 = y

12. xe y − 3y sin x = 1 14. cos y − y 2 = 8

15. e4y − ln y = 2x

16. e x y − 3y = x 2 + 1

2

In exercises 17–22, find an equation of the tangent line at the given point. If you have a CAS that will graph implicit curves, sketch the curve and the tangent line. 17. x 2 − 4y 3 = 0 at (2, 1)

18. x 2 y 2 = 4x at (1, 2)

19. x y = 4y at (2, 1)

20. x y = −3x y at (−1, −3) √ 22. x 4 = 8(x 2 − y 2 ) at (2, − 2)

2 2

21. x 4 = 4(x 2 − y 2 ) at (1,

3 2

√ 3 ) 2

In exercises 23 and 24, find the locations of all horizontal and vertical tangents. 23. x 2 + y 3 − 3y = 4

24. x y 2 − 2y = 2

In exercises 25–28, find the second derivative y (x). 25. x 2 y 2 + 3x − 4y = 5 27. y = x − 6x + 4 cos y 2

3

26. x 2/3 + y 2/3 = 4 28. e

xy

+ 2y − 3x = sin y

In exercises 29–38, find the derivative of the given function. √ 29. f (x) = tan−1 x 30. f (x) = sin−1 (x 3 + 1) 31. f (x) = tan−1 (cos x) 33. f (x) = 4 sec (x 4 ) −1

35. f (x) = etan 37. f (x) =

x

tan−1 x x2 + 1

32. f (x) = 4 sec−1 (x 4 ) √ 34. f (x) = 2 + tan−1 x 36. f (x) =

x2 cot−1 x

38. f (x) = sin−1 (sin x)

39. In example 8.1, it is easy to find a y-value for x = 2, but other y-values are not so easy to find. Try solving for y if x = 1.9. Use the tangent line found in example 8.1 to estimate a y-value. Repeat for x = 2.1. 40. Use the tangent line found in example 8.2 to estimate a y-value corresponding to x = 1.9; x = 2.1.

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41. For elliptic curves, there are nice ways of finding points with rational coordinates (see Ezra Brown’s article “Three Fermat Trails to Elliptic Curves” in the May 2000 College Mathematics Journal for more information). If you have access to an implicit plotter, graph the elliptic curve defined by y 2 = x 3 − 6x + 9. Show that the points (−3, 0) and (0, 3) are on the curve. Find the line through these two points and show that the line intersects the curve in another point with rational (in this case, integer) coordinates. 42. For the elliptic curve y 2 = x 3 − 6x + 4, show that the point (−1, 3) is on the curve. Find the tangent line to the curve at this point and show that it intersects the curve at another point with rational coordinates. e x − e−x e x + e−x and coth x = x both x −x e +e e − e−x have inverses on the appropriate domains and ranges. Show 1 d 1 d tanh−1 x = coth−1 x = and . Identhat dx 1 − x2 dx 1 − x2 tify the set of x-values for which each formula is valid. Is it fair to say that the derivatives are equal?

43. The functions tanh x =

44. Use a CAS to plot the set of points for which (cos x)2 + (sin y)2 = 1. Determine whether the segments plotted are straight or not. 45. Find and simplify the derivative of sin−1 x + cos−1 x. Use the result to write out an equation relating sin−1 x and cos−1 x.

x 46. Find and simplify the derivative of sin−1 √ . Use the x 2 + 1

x −1 and result to write out an equation relating sin √ x2 + 1 −1 tan x. 47. Use implicit differentiation to find y (x) for x 2 y − 2y = 4. Based on this equation, why would you expect to find vertical √ tangents at x = ± 2 and horizontal tangents at y = 0? Show that there are no points for these values. To see what’s going on, solve the original equation for √ y and sketch the graph. Describe what’s happening at x = ± 2 and y = 0. 48. Show that any curve of the form x y = c for some constant c intersects any curve of the form x 2 − y 2 = k for some constant k at right angles (that is, the tangent lines to the curves at the intersection points are perpendicular). In this case, we say that the families of curves are orthogonal. In exercises 49–52, show that the families of curves are orthogonal (see exercise 48). c 49. y = and y 2 = x 2 + k x 50. x 2 + y 2 = cx and x 2 + y 2 = ky 51. y = cx 3 and x 2 + 3y 2 = k 52. y = cx 4 and x 2 + 4y 2 = k

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53. Based on exercises 51 and 52, make a conjecture for a family of functions that is orthogonal to y = cx n . Show that your conjecture is correct. Are there any values of n that must be excluded? 54. Suppose that a circle of radius r and center (0, c) is inscribed in the parabola y = x 2 . At the point of tangency, the slopes must be the same. Find the slope of the circle implicitly and show that at the point of tangency, y = c − 12 . Then use the equations of the circle and parabola to show that c = r 2 + 14 . y

x

55. In example 8.6, it was shown that by the time the baseball reached home plate, the rate of rotation of the player’s gaze (θ ) was too fast for humans to track. Given a maximum rotational rate of θ = −3 radians per second, find d such that θ = −3. That is, find how close to the plate a player can track the ball. In a major league setting, the player must start swinging by the time the pitch is halfway (30 ) to home plate. How does this correspond to the distance at which the player loses track of the ball? 56. Suppose the pitching speed d in example 8.6 is different. Then θ will be different and the value of d for which θ = −3 will change. Find d as a function of d for d ranging from 30 ft/s (slowpitch softball) to 140 ft/s (major league fastball), and sketch the graph. 57. Suppose a painting hangs on a wall. The frame extends from 6 feet to 8 feet above the floor. A person stands x feet from the wall and views the painting, with a viewing angle A formed by the ray from the person’s eye (5 feet above the floor) to the top of the frame and the ray from the person’s eye to the bottom of the frame. Find the value of x that maximizes the viewing angle A.

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58. What changes in exercise 57 if the person’s eyes are 6 feet above the floor?

EXPLORATORY EXERCISES 1. Suppose a slingshot (see section 2.1) rotates counterclockwise along the circle x 2 + y 2 = 9 and the rock is released at the point (2.9, 0.77). If the rock travels 300 feet, where does it land? [Hint: Find the tangent line at (2.9, 0.77), and find the point (x, y) on that line such that the distance is (x − 2.9)2 + (y − 0.77)2 = 300.] 2. A landowner’s property line runs along the path y = 6 − x. The landowner wants to run an irrigation ditch from a reservoir bounded by the ellipse 4x 2 + 9y 2 = 36. The landowner wants to build the shortest ditch possible from the reservoir to the closest point on the property line. We explore how to find the best path. Sketch the line and ellipse, and draw in a tangent line to the ellipse that is parallel to the property line. Argue that the ditch should start at the point of tangency and run perpendicular to the two lines. We start by identifying the point on the right side of the ellipse with tangent line parallel to y = 6 − x. Find the slope of the tangent line to the ellipse at (x, y) and set it equal to −1. Solve for x and substitute into the equation of the ellipse. Solve for y and you have the point on the ellipse at which to start the ditch. Find an equation of the (normal) line through this point perpendicular to y = 6 − x and find the intersection of the normal line and y = 6 − x. This point is where the ditch ends. 3. In this exercise, you will design a movie theater with all seats having an equal view of the screen. Suppose the screen extends vertically from 10 feet to 30 feet above the floor. The first row of seats is 15 feet from the screen. Your task is to determine a function h(x) such that if seats x feet from the screen are raised h(x) feet above floor level, then the angle from the bottom of the screen to the viewer to the top of the screen will be the same as for a viewer sitting in the first row. You will be able to accomplish this only for a limited range of x-values. Beyond the maximum such x, find the height that maximizes the viewing angle. [Hint: Write the angle as a difference of inverse tan a − tan b tangents and use the formula tan (a − b) = .] 1 + tan a tan b

THE MEAN VALUE THEOREM In this section, we present the Mean Value Theorem, which is so significant that we will be deriving new ideas from it for many chapters to come. Before considering the main result, we look at a special case, called Rolle’s Theorem. The idea behind Rolle’s Theorem is really quite simple. For any function f that is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) and where f (a) = f (b), there must be at least one point between x = a and x = b where the

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53. Based on exercises 51 and 52, make a conjecture for a family of functions that is orthogonal to y = cx n . Show that your conjecture is correct. Are there any values of n that must be excluded? 54. Suppose that a circle of radius r and center (0, c) is inscribed in the parabola y = x 2 . At the point of tangency, the slopes must be the same. Find the slope of the circle implicitly and show that at the point of tangency, y = c − 12 . Then use the equations of the circle and parabola to show that c = r 2 + 14 . y

x

55. In example 8.6, it was shown that by the time the baseball reached home plate, the rate of rotation of the player’s gaze (θ ) was too fast for humans to track. Given a maximum rotational rate of θ = −3 radians per second, find d such that θ = −3. That is, find how close to the plate a player can track the ball. In a major league setting, the player must start swinging by the time the pitch is halfway (30 ) to home plate. How does this correspond to the distance at which the player loses track of the ball? 56. Suppose the pitching speed d in example 8.6 is different. Then θ will be different and the value of d for which θ = −3 will change. Find d as a function of d for d ranging from 30 ft/s (slowpitch softball) to 140 ft/s (major league fastball), and sketch the graph. 57. Suppose a painting hangs on a wall. The frame extends from 6 feet to 8 feet above the floor. A person stands x feet from the wall and views the painting, with a viewing angle A formed by the ray from the person’s eye (5 feet above the floor) to the top of the frame and the ray from the person’s eye to the bottom of the frame. Find the value of x that maximizes the viewing angle A.

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58. What changes in exercise 57 if the person’s eyes are 6 feet above the floor?

EXPLORATORY EXERCISES 1. Suppose a slingshot (see section 2.1) rotates counterclockwise along the circle x 2 + y 2 = 9 and the rock is released at the point (2.9, 0.77). If the rock travels 300 feet, where does it land? [Hint: Find the tangent line at (2.9, 0.77), and find the point (x, y) on that line such that the distance is (x − 2.9)2 + (y − 0.77)2 = 300.] 2. A landowner’s property line runs along the path y = 6 − x. The landowner wants to run an irrigation ditch from a reservoir bounded by the ellipse 4x 2 + 9y 2 = 36. The landowner wants to build the shortest ditch possible from the reservoir to the closest point on the property line. We explore how to find the best path. Sketch the line and ellipse, and draw in a tangent line to the ellipse that is parallel to the property line. Argue that the ditch should start at the point of tangency and run perpendicular to the two lines. We start by identifying the point on the right side of the ellipse with tangent line parallel to y = 6 − x. Find the slope of the tangent line to the ellipse at (x, y) and set it equal to −1. Solve for x and substitute into the equation of the ellipse. Solve for y and you have the point on the ellipse at which to start the ditch. Find an equation of the (normal) line through this point perpendicular to y = 6 − x and find the intersection of the normal line and y = 6 − x. This point is where the ditch ends. 3. In this exercise, you will design a movie theater with all seats having an equal view of the screen. Suppose the screen extends vertically from 10 feet to 30 feet above the floor. The first row of seats is 15 feet from the screen. Your task is to determine a function h(x) such that if seats x feet from the screen are raised h(x) feet above floor level, then the angle from the bottom of the screen to the viewer to the top of the screen will be the same as for a viewer sitting in the first row. You will be able to accomplish this only for a limited range of x-values. Beyond the maximum such x, find the height that maximizes the viewing angle. [Hint: Write the angle as a difference of inverse tan a − tan b tangents and use the formula tan (a − b) = .] 1 + tan a tan b

THE MEAN VALUE THEOREM In this section, we present the Mean Value Theorem, which is so significant that we will be deriving new ideas from it for many chapters to come. Before considering the main result, we look at a special case, called Rolle’s Theorem. The idea behind Rolle’s Theorem is really quite simple. For any function f that is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) and where f (a) = f (b), there must be at least one point between x = a and x = b where the

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tangent line to y = f (x) is horizontal. In Figures 2.45a to 2.45c, we draw a number of graphs satisfying the above criteria. Notice that each one has at least one point where there is a horizontal tangent line. Draw your own graphs, to convince yourself that, under these circumstances, it’s not possible to connect the two points (a, f (a)) and (b, f (b)) without having at least one horizontal tangent line. Note that since f (x) = 0 at a horizontal tangent, this says that there is at least one point c in (a, b), for which f (c) = 0 (see Figures 2.45a to 2.45c). y

y

y

c2 x a

c

a

x

b

a

c

x

c1

b

b

FIGURE 2.45a

FIGURE 2.45b

FIGURE 2.45c

Graph initially rising

Graph initially falling

Graph with two horizontal tangents

THEOREM 9.1 (Rolle’s Theorem)

HISTORICAL NOTES Michel Rolle (1652–1719) A French mathematician who proved Rolle’s Theorem for polynomials. Rolle came from a poor background, being largely self-taught and struggling through a variety of jobs including assistant attorney, scribe and elementary school teacher. He was a vigorous member of the French Academy of Sciences, arguing against such luminaries as Descartes that if a < b then −b < −a (so, for instance, −2 < −1). Oddly, Rolle was known as an opponent of the newly developed calculus, calling it a “collection of ingenious fallacies.”

Suppose that f is continuous on the interval [a, b], differentiable on the interval (a, b) and f (a) = f (b). Then there is a number c ∈ (a, b) such that f (c) = 0.

SKETCH OF PROOF We present the main ideas of the proof from a graphical perspective. First, note that if f (x) is constant on [a, b], then f (x) = 0 for all x’s between a and b. On the other hand, if f (x) is not constant on [a, b], then, as you look from left to right, the graph must at some point start to either rise or fall (see Figures 2.46a and 2.46b). For the case where the graph starts to rise, notice that in order to return to the level at which it started, it will need to turn around at some point and start to fall. (Think about it this way: if you start to climb up a mountain—so that your altitude rises—if you are to get back down to where you started, you will need to turn around at some point—where your altitude starts to fall.) So, there is at least one point where the graph turns around, changing from rising to falling (see Figure 2.46a). Likewise, in the case where the graph first starts to fall, the graph y

y f (c) 0

(a, f (a))

(a, f (a))

(b, f(b))

(b, f (b))

f (c) 0 x

c

x c

FIGURE 2.46a

FIGURE 2.46b

Graph rises and turns around to fall back to where it started.

Graph falls and then turns around to rise back to where it started.

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must turn around from falling to rising (see Figure 2.46b). We name this point x = c. Since we know that f (c) exists, we have that either f (c) > 0, f (c) < 0 or f (c) = 0. We want to argue that f (c) = 0, as Figures 2.46a and 2.46b suggest. To establish this, it is easier to show that it is not true that f (c) > 0 or f (c) < 0. If it were true that f (c) > 0, then from the alternative definition of the derivative given in equation (2.2) in section 2.2, we have f (c) = lim

x→c

f (x) − f (c) > 0. x −c

This says that for every x sufficiently close to c, f (x) − f (c) > 0. x −c

(9.1)

In particular, for the case where the graph first rises, if x − c > 0 (i.e., x > c), this says that f (x) − f (c) > 0 or f (x) > f (c), which can’t happen for every x > c (with x sufficiently close to c) if the graph has turned around at c and started to fall. From this, we conclude that it can’t be true that f (c) > 0. Similarly, we can show that it is not true that f (c) < 0. Therefore, f (c) = 0, as desired. The proof for the case where the graph first falls is nearly identical and is left to the reader. We now give an illustration of the conclusion of Rolle’s Theorem.

EXAMPLE 9.1

An Illustration of Rolle’s Theorem

Find a value of c satisfying the conclusion of Rolle’s Theorem for f (x) = x 3 − 3x 2 + 2x + 2 on the interval [0, 1]. Solution First, we verify that the hypotheses of the theorem are satisfied: f is differentiable and continuous for all x [since f (x) is a polynomial and all polynomials are continuous and differentiable everywhere]. Also, f (0) = f (1) = 2. We have f (x) = 3x 2 − 6x + 2. We now look for values of c such that f (c) = 3c2 − 6c + 2 = 0. √ By the quadratic formula, we get c = 1 + 13 3 ≈ 1.5774 [not in the interval (0, 1)] and √ c = 1 − 13 3 ≈ 0.42265 ∈ (0, 1).

REMARK 9.1 We want to emphasize that example 9.1 is merely an illustration of Rolle’s Theorem. Finding the number(s) c satisfying the conclusion of Rolle’s Theorem is not the point of our discussion. Rather, Rolle’s Theorem is of interest to us primarily because we use it to prove one of the fundamental results of elementary calculus, the Mean Value Theorem.

Although Rolle’s Theorem is a simple result, we can use it to derive numerous properties of functions. For example, we are often interested in finding the zeros of a function f (that is, solutions of the equation f (x) = 0). In practice, it is often difficult to determine how many zeros a given function has. Rolle’s Theorem can be of help here.

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THEOREM 9.2 If f is continuous on the interval [a, b], differentiable on the interval (a, b) and f (x) = 0 has two solutions in [a, b], then f (x) = 0 has at least one solution in (a, b).

PROOF This is just a special case of Rolle’s Theorem. Identify the two zeros of f (x) as x = s and x = t, where s < t. Since f (s) = f (t), Rolle’s Theorem guarantees that there is a number c such that s < c < t (and hence a < c < b) where f (c) = 0. We can easily generalize the result of Theorem 9.2, as in the following theorem.

THEOREM 9.3 For any integer n > 0, if f is continuous on the interval [a, b] and differentiable on the interval (a, b) and f (x) = 0 has n solutions in [a, b], then f (x) = 0 has at least (n − 1) solutions in (a, b).

PROOF From Theorem 9.2, between every pair of solutions of f (x) = 0 is at least one solution of f (x) = 0. In this case, there are (n − 1) consecutive pairs of solutions of f (x) = 0 and so, the result follows. We can use Theorems 9.2 and 9.3 to investigate the number of zeros a given function has. (Here, we consider only real zeros of a function and not complex zeros.) y

EXAMPLE 9.2 10

Prove that x 3 + 4x + 1 = 0 has exactly one solution.

5

2

x

1

Determining the Number of Zeros of a Function

1 5 10

FIGURE 2.47 y = x 3 + 4x + 1

2

Solution The graph shown in Figure 2.47 makes the result seem reasonable: we can see one solution, but how can we be sure there are no others outside of the displayed window? Notice that if f (x) = x 3 + 4x + 1, then f (x) = 3x 2 + 4 > 0 for all x. By Theorem 9.2, if f (x) = 0 had two solutions, then f (x) = 0 would have at least one solution. However, since f (x) = 0 for all x, it can’t be true that f (x) = 0 has two (or more) solutions. Therefore, f (x) = 0 has exactly one solution. We now generalize Rolle’s Theorem to one of the most significant results of elementary calculus.

THEOREM 9.4 (Mean Value Theorem) Suppose that f is continuous on the interval [a, b] and differentiable on the interval (a, b). Then there exists a number c ∈ (a, b) such that f (c) =

f (b) − f (a) . b−a

(9.2)

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PROOF

NOTE Note that in the special case where f (a) = f (b), (9.2) simplifies to the conclusion of Rolle’s Theorem, that f (c) = 0.

Note that the hypotheses are identical to those of Rolle’s Theorem, except that there is no f (b) − f (a) assumption about the values of f at the endpoints. The expression is the slope b−a of the secant line connecting the endpoints, (a, f (a)) and (b, f (b)). The theorem states that there is a line tangent to the curve at some point x = c in (a, b) that has the same slope as (and hence, is parallel to) the secant line (see Figures 2.48 and 2.49). If you tilt your head so that the line segment looks horizontal, Figure 2.49 will look like a figure for Rolle’s Theorem (Figures 2.46a and 2.46b). The proof is to “tilt” the function and then apply Rolle’s Theorem. y

y m

y f (x)

f (b) f (a) ba

x

x a

y f (x)

m f (c)

a

b

c

b

FIGURE 2.48

FIGURE 2.49

Secant line

Mean Value Theorem

The equation of the secant line through the endpoints is y − f (a) = m(x − a), where

m=

f (b) − f (a) . b−a

Define the “tilted” function g to be the difference between f and the function whose graph is the secant line: g(x) = f (x) − [m(x − a) + f (a)].

(9.3)

Note that g is continuous on [a, b] and differentiable on (a, b), since f is. Further, g(a) = f (a) − [0 + f (a)] = 0 and

g(b) = f (b) − [m(b − a) + f (a)] = f (b) − [ f (b) − f (a) + f (a)] = 0.

Since m =

f (b) − f (a) . b−a

Since g(a) = g(b), we have by Rolle’s Theorem that there exists a number c in the interval (a, b) such that g (c) = 0. Differentiating (9.3), we get 0 = g (c) = f (c) − m. Finally, solving (9.4) for f (c) gives us f (c) = m = as desired.

f (b) − f (a) , b−a

(9.4)

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Before we demonstrate some of the power of the Mean Value Theorem, we first briefly illustrate its conclusion.

EXAMPLE 9.3

An Illustration of the Mean Value Theorem

Find a value of c satisfying the conclusion of the Mean Value Theorem for f (x) = x 3 − x 2 − x + 1 on the interval [0, 2]. Solution Notice that f is continuous on [0, 2] and differentiable on (0, 2). The Mean Value Theorem then says that there is a number c in (0, 2) for which f (c) =

y 3

f (2) − f (0) 3−1 = = 1. 2−0 2−0

To find this number c, we set

2

f (c) = 3c2 − 2c − 1 = 1

1

3c2 − 2c − 2 = 0. √ 1± 7 From the quadratic formula, we get c = . In this case, only one of these, 3 √ 1+ 7 c= , is in the interval (0, 2). In Figure 2.50, we show the graphs of y = f (x), 3 the secant line joining the endpoints of the portion of the curve on the interval [0, 2] and √ 1+ 7 the tangent line at x = . 3 or

x 1 1 2

FIGURE 2.50 Mean Value Theorem

2

The illustration in example 9.3, where we found the number c whose existence is guaranteed by the Mean Value Theorem, is not the point of the theorem. In fact, these c’s usually remain unknown. The significance of the Mean Value Theorem is that it relates a difference of function values to the difference of the corresponding x-values, as in equation (9.5) below. Note that if we take the conclusion of the Mean Value Theorem (9.2) and multiply both sides by the quantity (b − a), we get f (b) − f (a) = f (c)(b − a).

(9.5)

As it turns out, many of the most important results in the calculus (including one known as the Fundamental Theorem of Calculus) follow from the Mean Value Theorem. For now, we derive a result essential to our work in Chapter 4. The question concerns how many functions share the same derivative. Recall that for any constant c, d (c) = 0. dx A question that you probably haven’t thought to ask is: Are there any other functions whose derivative is zero? The answer is no, as we see in Theorem 9.5.

THEOREM 9.5 Suppose that f (x) = 0 for all x in some open interval I . Then, f (x) is constant on I .

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PROOF Pick any two numbers, say a and b, in I , with a < b. Since f is differentiable in I and (a, b) ⊂ I , f is continuous on [a, b] and differentiable on (a, b). By the Mean Value Theorem, we know that f (b) − f (a) = f (c), b−a

(9.6)

for some number c ∈ (a, b) ⊂ I . Since, f (x) = 0 for all x ∈ I , f (c) = 0 and it follows from (9.6) that f (b) − f (a) = 0

or

f (b) = f (a).

Since a and b were arbitrary points in I , this says that f is constant on I , as desired. A question closely related to Theorem 9.5 is the following. We know, for example, that d 2 (x + 2) = 2x. dx But are there any other functions with the same derivative? You should quickly come up with several. For instance, x 2 + 3 and x 2 − 4 also have the derivative 2x. In fact, d 2 (x + c) = 2x, dx for any constant c. Are there any other functions, though, with the derivative 2x? Corollary 9.1 says that there are no such functions.

COROLLARY 9.1 Suppose that g (x) = f (x) for all x in some open interval I . Then, for some constant c, g(x) = f (x) + c, for all x ∈ I.

PROOF Define h(x) = g(x) − f (x). Then h (x) = g (x) − f (x) = 0 for all x in I . From Theorem 9.5, h(x) = c, for some constant c. The result then follows immediately from the definition of h(x). We see in Chapter 4 that Corollary 9.1 has significant implications when we try to reverse the process of differentiation (called antidifferentiation). We take a look ahead to this in example 9.4.

EXAMPLE 9.4

Finding Every Function with a Given Derivative

Find all functions that have a derivative equal to 3x 2 + 1. Solution We first write down (from our experience with derivatives) one function with the correct derivative: x 3 + x. Then, Corollary 9.1 tells us that any other function with the same derivative differs by at most a constant. So, every function whose derivative equals 3x 2 + 1 has the form x 3 + x + c, for some constant c.

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As our final example, we demonstrate how the Mean Value Theorem can be used to establish a useful inequality.

EXAMPLE 9.5

Proving an Inequality for sin x |sin a| ≤ |a| for all a.

Prove that

Solution First, note that f (x) = sin x is continuous and differentiable on any interval and that for any a, |sin a| = |sin a − sin 0|, since sin 0 = 0. From the Mean Value Theorem, we have that (for a = 0) sin a − sin 0 = f (c) = cos c, (9.7) a−0 for some number c between a and 0. Notice that if we multiply both sides of (9.7) by a and take absolute values, we get |sin a| = |sin a − sin 0| = |cos c| |a − 0| = |cos c| |a|.

(9.8)

But, |cos c| ≤ 1, for all real numbers c and so, from (9.8), we have |sin a| = |cos c| |a| ≤ (1) |a| = |a|, as desired.

BEYOND FORMULAS The Mean Value Theorem is subtle, but its implications are far-reaching. Although the illustration in Figure 2.49 makes the result seem obvious, the consequences of the Mean Value Theorem, such as example 9.4, are powerful and not at all obvious. For example, most of the rest of the calculus developed in this book depends on the Mean Value Theorem either directly or indirectly. A thorough understanding of the theory of calculus can lead you to important conclusions, particularly when the problems are beyond what your intuition alone can handle. What other theorems have you learned that continue to provide insight beyond their original context?

EXERCISES 2.9 WRITING EXERCISES 1. Notice that for both Rolle’s Theorem and the Mean Value Theorem, we have assumed that the function is continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Recall that if a function is differentiable at x = a, then it is also continuous at x = a. Therefore, if we had assumed that the function was differentiable on the closed interval, we would not have had to mention continuity. We do not do so because the assumption of being differentiable at the endpoints is not

necessary. The “ethics” of the statement of a theorem is to include only assumptions that are absolutely necessary. Discuss the virtues of this tradition. Is this common practice in our social dealings, such as financial obligations or personal gossip? 2. One of the results in this section is that if f (x) = g (x), then g(x) = f (x) + c for some constant c. Explain this result graphically.

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3. Explain the result of Corollary 9.1 in terms of position and velocity functions. That is, if two objects have the same velocity functions, what can you say about the relative positions of the two objects?

26. Two runners start a race at time 0. At some time t = a, one runner has pulled ahead, but the other runner has taken the lead by time t = b. Prove that at some time t = c > 0, the runners were going exactly the same speed.

4. As we mentioned, you can derive Rolle’s Theorem from the Mean Value Theorem simply by setting f (a) = f (b). Given this, it may seem odd that Rolle’s Theorem rates its own name and portion of the book. To explain why we do this, discuss ways in which Rolle’s Theorem is easier to understand than the Mean Value Theorem.

27. If f and g are differentiable functions on the interval [a, b] with f (a) = g(a) and f (b) = g(b), prove that at some point in the interval [a, b], f and g have parallel tangent lines.

In exercises 1–6, check the hypotheses of Rolle’s Theorem and the Mean Value Theorem and find a value of c that makes the appropriate conclusion true. Illustrate the conclusion with a graph.

28. Prove that the result of exercise 27 still holds if the assumptions f (a) = g(a) and f (b) = g(b) are relaxed to requiring f (b) − f (a) = g(b) − g(a). In exercises 29–34, find all functions g such that g (x) f (x). 29. f (x) = x 2 31. f (x) = 1/x 2

30. f (x) = 9x 4 √ 32. f (x) = x 34. f (x) = cos x

1. f (x) = x 2 + 1, [−2, 2]

2. f (x) = x 2 + 1, [0, 2]

33. f (x) = sin x

3. f (x) = x 3 + x 2 , [0, 1]

4. f (x) = x 3 + x 2 , [−1, 1]

5. f (x) = sin x, [0, π/2]

6. f (x) = sin x, [−π, 0]

In exercises 35–38, explain why it is not valid to use the Mean Value Theorem. When the hypotheses are not true, the theorem does not tell you anything about the truth of the conclusion. In three of the four cases, show that there is no value of c that makes the conclusion of the theorem true. In the fourth case, find the value of c. 1 1 35. f (x) = , [−1, 1] 36. f (x) = 2 , [−1, 2] x x 37. f (x) = tan x, [0, π] 38. f (x) = x 1/3 , [−1, 1]

7. If f (x) > 0 for all x, prove that f (x) is an increasing function: that is, if a < b, then f (a) < f (b). 8. If f (x) < 0 for all x, prove that f (x) is a decreasing function: that is, if a < b, then f (a) > f (b). In exercises 9–16, determine whether the function is increasing, decreasing or neither. 9. f (x) = x 3 + 5x + 1 10. f (x) = x 5 + 3x 3 − 1 11. f (x) = −x 3 − 3x + 1

12. f (x) = x 4 + 2x 2 + 1

13. f (x) = e x

14. f (x) = e−x

15. f (x) = ln x

16. f (x) = ln x 2

17. Prove that x 3 + 5x + 1 = 0 has exactly one solution. 18. Prove that x 3 + 4x − 3 = 0 has exactly one solution. 19. Prove that x 4 + 3x 2 − 2 = 0 has exactly two solutions. 20. Prove that x 4 + 6x 2 − 1 = 0 has exactly two solutions. 21. Prove that x 3 + ax + b = 0 has exactly one solution for a > 0. 22. Prove that x 4 + ax 2 − b = 0 (a > 0, b > 0) has exactly two solutions.

39. Assume that f is a differentiable function such that f (0) = f (0) = 0 and f (0) > 0. Argue that there exists a positive constant a > 0 such that f (x) > 0 for all x in the interval (0, a). Can anything be concluded about f (x) for negative x’s? 40. Show that for any real numbers u and v, | cos u − cos v| ≤ |u − v|. (Hint: Use the Mean Value Theorem.) 41. Prove that | sin a| < |a| for all a = 0 and use the result to show that the only solution to the equation sin x = x is x = 0. What happens if you try to find all intersections with a graphing calculator? 42. Use the Mean Value Theorem to show that |tan−1 a| < |a| for all a = 0 and use this inequality to find all solutions of the equation tan−1 x = x. 43. Prove that |x| < |sin−1 x| for 0 < |x| < 1.

23. Prove that x 5 + ax 3 + bx + c = 0 has exactly one solution for a > 0, b > 0.

44. Prove that |x| ≤ |tan x| for |x| <

24. Prove that a third-degree (cubic) polynomial has at most three zeros. (You may use the quadratic formula.)

45. For f (x) =

25. Suppose that s(t) gives the position of an object at time t. If s is differentiable on the interval [a, b], prove that at some time t = c, the instantaneous velocity at t = c equals the average velocity between times t = a and t = b.

π . 2

2x if x ≤ 0 show that f is continuous on 2x − 4 if x > 0 the interval (0, 2), differentiable on the interval (0, 2) and has f (0) = f (2). Show that there does not exist a value of c such that f (c) = 0. Which hypothesis of Rolle’s Theorem is not satisfied?

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46. Assume that f is a differentiable function such that f (0) = f (0) = 0. Show by example that it is not necessarily true that f (x) = 0 for all x. Find the flaw in the following bogus “proof.” Using the Mean Value Theorem with a = x f (x) − f (0) . Since f (0) = 0 and b = 0, we have f (c) = x −0 f (x) so that f (x) = 0. and f (c) = 0, we have 0 = x

EXPLORATORY EXERCISES 1. In section 2.1, we gave an example of computing the velocity of a moving car. The point of the story was that computing instantaneous velocity requires us to compute a limit. However, we left an interesting question unanswered. If you have an average velocity of 60 mph over 1 hour and the speed limit is 65 mph, you are unable to prove that you never exceeded the speed limit. What is the longest time interval over which you can average 60 mph and still guarantee no speeding? We can use the Mean Value Theorem to answer the question after clearing up a couple of preliminary questions. First, argue that we need to know the maximum acceleration of a car and the maximum positive acceleration may differ from the maximum negative acceleration. Based on your experience, what is the fastest your car could accelerate (speed up)? What is the fastest your car could decelerate (slow down)? Back up your estimates with some real data (e.g., my car goes from 0 to 60 in 15 seconds). Call the larger number A (use units of mph per second). Next, argue that if acceleration (the derivative of velocity) is constant, then the velocity function is linear. Therefore, if the velocity varies from 55 mph to 65 mph at constant acceleration, the average

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velocity will be 60 mph. Now, apply the Mean Value Theorem to the velocity function v(t) on a time interval [0, T ], where the velocity changes from 55 mph to 65 mph at constant accel65 − 55 v(T ) − v(0) eration A: v (c) = becomes A = and so T −0 T −0 T = 10/A. For how long is the guarantee good? 2. Suppose that a pollutant is dumped into a lake at the rate of p (t) = t 2 − t + 4 tons per month. The amount of pollutant dumped into the lake in the first two months is A = p(2) − p(0). Using c = 1 (the midpoint of the interval), estimate A by applying the Mean Value Theorem to p(t) on the interval [0, 2]. To get a better estimate, apply the Mean Value Theorem to the intervals [0, 1/2], [1/2, 1], [1, 3/2] and [3/2, 2]. [Hint: A = p(1/2) − p(0) + p(1) − p(1/2) + p(3/2) − p(1) + p(2) − p(3/2).] If you have access to a CAS, get better estimates by dividing the interval [0, 2] into more and more pieces and try to conjecture the limit of the estimates. You will be well on your way to understanding Chapter 4 on integration. 3. A result known as the Cauchy Mean Value Theorem states that if f and g are differentiable on the interval (a, b) and continuous on [a, b], then there exists a number c with a < c < b and [ f (b) − f (a)]g (c) = [g(b) − g(a)] f (c). Find all flaws in the following invalid attempt to prove the result, and then find a correct proof. Invalid attempt: The hypotheses of the Mean Value Theorem are satisfied by both functions, so there exists a number c with a < c < b f (b) − f (a) g(b) − g(a) and f (c) = and g (c) = . Then b−a b−a g(b) − g(a) f (b) − f (a) = and thus b−a = f (c) g (c) [ f (b) − f (a)]g (c) = [g(b) − g(a)] f (c).

Review Exercises TRUE OR FALSE

WRITING EXERCISES The following list includes terms that are defined and theorems that are stated in this chapter. For each term or theorem, (1) give a precise definition or statement, (2) state in general terms what it means and (3) describe the types of problems with which it is associated. Tangent line Derivative Product rule Implicit differentiation

Velocity Power rule Quotient rule Mean Value Theorem

Average velocity Acceleration Chain rule Rolle’s Theorem

State the derivative of each function: sin x, cos x, tan x, cot x, sec x, csc x, sin−1 x, cos−1 x, tan−1 x, −1 cot x, sec−1 x, csc−1 x, e x , b x , ln x, logb x

State whether each statement is true or false and briefly explain why. If the statement is false, try to “fix it” by modifying the given statement to make a new statement that is true. 1. If a function is continuous at x = a, then it has a tangent line at x = a. 2. The average velocity between t = a and t = b is the average of the velocities at t = a and t = b. 3. The derivative of a function gives its slope. 4. Given the graph of f (x), you can construct the graph of f (x).

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Jupiter and an asteroid is a third system. The clusters of asteroids at the L4 and L5 Lagrange points of the Sun-Jupiter system are called Trojan asteroids. For a given system, the locations of the five Lagrange points can be determined by solving equations. As you will see in the section 3.1 exercises, the equation for the location of SOHO is a difficult fifth-order polynomial equation. For a fifth-order equation, we usually are forced to gather graphical and numerical evidence to approximate solutions. The graphing and analysis of complicated functions and the solution of equations involving these functions are the emphases of this chapter.

3.1

LINEAR APPROXIMATIONS AND NEWTON’S METHOD There are two distinctly different tasks for which you use scientific calculators. First, they perform basic arithmetic operations much faster than any of us could. We all know how to multiply 1024 by 1673, but it is time-consuming to carry out this calculation with pencil and paper. For such problems, calculators are a tremendous convenience. More significantly, we also use calculators to compute values of transcendental functions such as sine, cosine, tangent, exponentials and logarithms. In this case, the calculator is much more than a mere convenience. How would you calculate sin(1.2345678) without a calculator? Don’t worry if you don’t know how to do this. The problem is that the sine function is not algebraic. That is, there is no formula for sin x involving only the arithmetic operations. So, how does your calculator “know” that sin(1.2345678) ≈ 0.9440056953? In short, it doesn’t know this at all. Rather, the calculator has a built-in program that generates approximate values of the sine and other transcendental functions. In this section, we develop a simple approximation method. Although somewhat crude, it points the way toward more sophisticated approximation techniques to follow later in the text.

Linear Approximations Suppose we wanted to find an approximation for f (x1 ), where f (x1 ) is unknown, but where f (x0 ) is known for some x0 “close” to x1 . For instance, the value of cos(1) is unknown, but we do know that cos(π/3) = 12 (exactly) and π/3 ≈ 1.047 is “close” to 1. While we could use 12 as an approximation to cos(1), we can do better. Referring to Figure 3.1, notice that if x1 is “close” to x0 and we follow the tangent line at x = x0 to the point corresponding to x = x1 , then the y-coordinate of that point (y1 ) should be “close” to the y-coordinate of the point on the curve y = f (x) [i.e., f (x1 )]. Since the slope of the tangent line to y = f (x) at x = x0 is f (x0 ), the equation of the tangent line to y = f (x) at x = x0 is found from y − f (x0 ) . x − x0

(1.1)

y = f (x0 ) + f (x0 )(x − x0 ).

(1.2)

m tan = f (x0 ) = Solving equation (1.1) for y gives us

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y y f (x) f (x 1) y f (x0 ) f (x0)(x x 0) y1 f (x 0)

x x0

x1

FIGURE 3.1 Linear approximation of f (x1 )

Notice that (1.2) is the equation of the tangent line to the graph of y = f (x) at x = x0 . We give the linear function defined by this equation a name, as follows.

DEFINITION 1.1 The linear (or tangent line) approximation of f (x) at x = x0 is the function L(x) = f (x0 ) + f (x0 )(x − x0 ).

Observe that the y-coordinate y1 of the point on the tangent line corresponding to x = x1 is simply found by substituting x = x1 in equation (1.2). This gives us y1 = f (x0 ) + f (x0 )(x1 − x0 ).

(1.3)

We define the increments x and y by x = x1 − x0 and

y = f (x1 ) − f (x0 ).

Using this notation, equation (1.3) gives us the approximation f (x1 ) ≈ y1 = f (x0 ) + f (x0 )x.

(1.4)

We illustrate this in Figure 3.2. We sometimes rewrite (1.4) by subtracting f (x0 ) from both sides, to yield y = f (x1 ) − f (x0 ) ≈ f (x0 ) x = dy,

(1.5)

where dy = f (x0 )x is called the differential of y. When using this notation, we also define d x, the differential of x, by d x = x, so that by (1.5), dy = f (x0 ) d x.

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y y f (x) f (x 1)

y1

y f (x0 ) f (x0)(x x 0)

y dy

f (x 0) x x x0

x1

FIGURE 3.2 Increments and differentials

We can use linear approximations to produce approximate values of transcendental functions, as in example 1.1.

EXAMPLE 1.1

Finding a Linear Approximation

Find the linear approximation to f (x) = cos x at x0 = π/3 and use it to approximate cos(1). y

p u

FIGURE 3.3 y = cos x and its linear approximation at x0 = π/3

Solution From Definition 1.1, the linear approximation is defined as L(x) = f (x0 ) + f (x0 )(x − x0 ). Here, x0 = π/3, f (x) = cos x and f (x) = − sin x. So, we have √ π π π 1 π 3 L(x) = cos − sin x− = − x− . 3 3 3 2 2 3 x

In Figure 3.3, we show a graph of y = cos x and the linear approximation to cos x for x0 = π/3. Notice that the linear approximation (i.e., the tangent line at x0 = π/3) stays close to the graph of y = cos x only for x close to π/3. In fact, for x < 0 or x > π , the linear approximation is obviously quite poor. It is typical of linear approximations (tangent lines) to stay close to the curve only nearby the point of tangency. Observe that we chose x0 = π3 since π3 is the value closest to 1 at which we know the value of the cosine exactly. Finally, an estimate of cos(1) is √ 1 3 π L(1) = − 1− ≈ 0.5409. 2 2 3 Your calculator gives you cos(1) ≈ 0.5403 and so, we have found a fairly good approximation to the desired value. In example 1.2, we derive a useful approximation to sin x, valid for x close to 0. This approximation is often used in applications in physics and engineering to simplify equations involving sin x.

EXAMPLE 1.2

Linear Approximation of sin x

Find the linear approximation of f (x) = sin x, for x close to 0.

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245

Solution Here, f (x) = cos x, so that from Definition 1.1, we have

y

sin x ≈ L(x) = f (0) + f (0) (x − 0) = sin 0 + cos 0 (x) = x.

1

x

1

..

1 1

FIGURE 3.4 y = sin x and y = x

This says that for x close to 0, sin x ≈ x. We illustrate this in Figure 3.4, where we show graphs of both y = sin x and y = x. Observe from Figure 3.4 that the graph of y = x stays close to the graph of y = sin x only in the vicinity of x = 0. Thus, the approximation sin x ≈ x is valid only for x close to 0. Also note that the farther x gets from 0, the worse the approximation becomes. This becomes even more apparent in example 1.3, where we also illustrate the use of the increments x and y.

EXAMPLE 1.3

Linear Approximation to Some Cube Roots

√ √ √ 8.02, 3 8.07, 3 8.15 and 3 25.2. √ Solution Here we are approximating values of the function f (x) = 3 x = x 1/3 . So, f (x) = 13 x −2/3 . The closest number to any of 8.02, 8.07 or 8.15 whose cube root we know exactly is 8. So, we write

Use a linear approximation to approximate

√ 3

f (8.02) = f (8) + [ f (8.02) − f (8)]

Add and substract f (8).

= f (8) + y.

(1.6)

From (1.5), we have y ≈ dy = f (8)x 1 −2/3 1 = 8 . (8.02 − 8) = 3 600

Since x = 8.02 − 8.

(1.7)

Using (1.6) and (1.7), we get f (8.02) ≈ f (8) + dy = 2 + while your calculator accurately returns

√ 3

1 ≈ 2.0016667, 600

8.02 ≈ 2.0016653. Similarly, we get

1 f (8.07) ≈ f (8) + 8−2/3 (8.07 − 8) ≈ 2.0058333 3 1 f (8.15) ≈ f (8) + 8−2/3 (8.15 − 8) ≈ 2.0125, 3 √ √ while your calculator√returns 3 8.07 ≈ 2.005816 and 3 8.15 ≈ 2.012423, respectively. To approximate 3 25.2, observe that 8 is not the closest number to 25.2 whose cube root we know exactly. Since 25.2 is much closer to 27 than to 8, we write and

f (25.2) = f (27) + y ≈ f (27) + dy = 3 + dy. In this case, 1 1 dy = f (27)x = 27−2/3 (25.2 − 27) = 3 3

and we have

f (25.2) ≈ 3 + dy = 3 −

1 1 (−1.8) = − 9 15

1 ≈ 2.9333333, 15

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y

3-6

compared to the value of 2.931794, produced by your calculator. It is important to recognize here that the farther the value of x gets from the point of tangency, the worse the approximation tends to be. You can see this clearly in Figure 3.5.

2

x 8

FIGURE 3.5 √ y = 3 x and the linear approximation at x0 = 8

x f (x)

6 84

10 60

Our first three examples were intended to familiarize you with the technique and to give you a feel for how good (or bad) linear approximations tend to be. In example 1.4, there is no exact answer to compare with the approximation. Our use of the linear approximation here is referred to as linear interpolation.

EXAMPLE 1.4

Using a Linear Approximation to Perform Linear Interpolation

The price of an item affects consumer demand for that item. Suppose that based on market research, a company estimates that f (x) thousand small cameras can be sold at the price of $x, as given in the accompanying table. Estimate the number of cameras that can be sold at $7.

14 32

Solution The closest x-value to x = 7 in the table is x = 6. [In other words, this is the closest value of x at which we know the value of f (x).] The linear approximation of f (x) at x = 6 would look like L(x) = f (6) + f (6)(x − 6). From the table, we know that f (6) = 84, but we do not know f (6). Further, we can’t compute f (x), since we don’t have a formula for f (x). The best we can do with the given data is to approximate the derivative by

Number of cameras sold

y

80

f (6) ≈

60 40

f (10) − f (6) 60 − 84 = = −6. 10 − 6 4

The linear approximation is then L(x) ≈ 84 − 6(x − 6).

20 0

x 0

5 7 10 15 Price of cameras

FIGURE 3.6 Linear interpolation

Using this, we estimate that the number of cameras sold at x = 7 would be L(7) ≈ 84 − 6 = 78 thousand. That is, we would expect to sell approximately 78 thousand cameras at a price of $7. We show a graphical interpretation of this in Figure 3.6, where the straight line is the linear approximation (in this case, the secant line joining the first two data points).

Newton’s Method We now return to the question of finding zeros of a function. In section 1.4, we introduced the method of bisections as one procedure for finding zeros of a continuous function. Here, we explore a method that is usually much more efficient than bisections. We are again looking for values of x such that f (x) = 0. These values are called roots of the equation f (x) = 0 or zeros of the function f . It’s easy to find the zeros of f (x) = ax 2 + bx + c, but how would you find zeros of f (x) = tan x − x?

3-7

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247

y

y 5

y f (x) x

3

3

x 5

x0

x1

x2

FIGURE 3.7

FIGURE 3.8

y = tan x − x

Newton’s method

This function is not algebraic and there are no formulas available for finding the zeros. Even so, we can clearly see zeros in Figure 3.7. (In fact, there are infinitely many of them.) The question is, how are we to find them? In general, to find approximate solutions to f (x) = 0, we first make an initial guess, denoted x0 , of the location of a solution. Since the tangent line to y = f (x) at x = x0 tends to hug the curve, we follow the tangent line to where it intersects the x-axis (see Figure 3.8). This appears to provide an improved approximation to the zero. The equation of the tangent line to y = f (x) at x = x0 is given by the linear approximation at x0 [see equation (1.2)], y = f (x0 ) + f (x0 )(x − x0 ).

(1.8)

We denote the x-intercept of the tangent line by x1 [found by setting y = 0 in (1.8)]. We then have

HISTORICAL NOTES Sir Isaac Newton (1642–1727) An English mathematician and scientist known as the co-inventor of calculus. In a 2-year period from 1665 to 1667, Newton made major discoveries in several areas of calculus, as well as optics and the law of gravitation. Newton’s mathematical results were not published in a timely fashion. Instead, techniques such as Newton’s method were quietly introduced as useful tools in his scientific papers. Newton’s Mathematical Principles of Natural Philosophy is widely regarded as one of the greatest achievements of the human mind.

0 = f (x0 ) + f (x0 )(x1 − x0 ) and, solving this for x1 , we get x1 = x0 −

f (x0 ) . f (x0 )

If we repeat this process, using x1 as our new guess, we should produce a further improved approximation, x2 = x1 −

f (x1 ) f (x1 )

and so on (see Figure 3.8). In this way, we generate a sequence of successive approximations determined by

xn+1 = xn −

f (xn ) , for n = 0, 1, 2, 3, . . . . f (xn )

(1.9)

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This procedure is called the Newton-Raphson method, or simply Newton’s method. If Figure 3.8 is any indication, xn should get closer and closer to a zero as n increases. Newton’s method is generally a very fast, accurate method for approximating the zeros of a function, as we illustrate with example 1.5.

EXAMPLE 1.5

Using Newton’s Method to Approximate a Zero

Find a zero of f (x) = x 5 − x + 1. y

3

2

x

1

1

Solution From Figure 3.9, it appears as if the only zero of f is located between x = −2 and x = −1. We further observe that f (−1) = 1 > 0 and f (−2) = −29 < 0. Since f is continuous, the Intermediate Value Theorem (Theorem 4.4 in section 1.4) says that f must have a zero on the interval (−2, −1). Because the zero appears to be closer to x = −1, we make the initial guess x0 = −1. Finally, f (x) = 5x 4 − 1 and so, Newton’s method gives us xn+1 = xn −

3

= xn −

FIGURE 3.9 y = x −x +1 5

f (xn ) f (xn ) xn5 − xn + 1 , 5xn4 − 1

n = 0, 1, 2, . . . .

Using the initial guess x0 = −1, we get x1 = −1 − = −1 −

(−1)5 − (−1) + 1 5(−1)4 − 1 1 5 =− . 4 4

5 Likewise, from x1 = − , we get the improved approximation 4 5 5 5 − − +1 − 5 4 4 x2 = − − 4 5 4 5 − −1 4 ≈ −1.178459394 and so on. We find that

x3 ≈ −1.167537389, x4 ≈ −1.167304083

and

x5 ≈ −1.167303978 ≈ x6 .

Since x5 = x6 , we will make no further progress by calculating additional steps. As a final check on the accuracy of our approximation, we compute f (x6 ) ≈ 1 × 10−13 . Since this is very close to zero, we say that x6 ≈ −1.167303978 is an approximate zero of f . You can bring Newton’s method to bear on a variety of approximation problems. As we illustrate in example 1.6, you may first need to rephrase the problem as a rootfinding problem.

3-9

SECTION 3.1

EXAMPLE 1.6

Linear Approximations and Newton’s Method

249

Using Newton’s Method to Approximate a Cube Root

Use Newton’s method to approximate y

.. √ 3

7.

Solution Recall that we can use a linear approximation to do this. On the other hand, Newton’s method is used to solve equations of the form√f (x) = 0. We can rewrite the current problem in this form, as follows. Suppose x = 3 7. Then, x 3 = 7, which can be rewritten as

30

f (x) = x 3 − 7 = 0.

x 2

Here, f (x) = 3x 2 and we obtain an initial guess from a graph of y = f (x) (see Figure 3.10). Notice that there is a zero near x = 2 and so we take x0 = 2. Newton’s method then yields

30

FIGURE 3.10

x1 = 2 −

y = x3 − 7

23 − 7 23 = ≈ 1.916666667. 2 3(2 ) 12

Continuing this process, we have x2 ≈ 1.912938458 and Further,

x3 ≈ 1.912931183 ≈ x4 . f (x4 ) ≈ 1 × 10−13

and so, x4 is an approximate zero of f . This also says that √ 3 7 ≈ 1.912931183, √ which compares very favorably with the value of 3 7 produced by your calculator.

REMARK 1.1 Although it seemed to be very efficient in examples 1.5 and 1.6, Newton’s method does not always work. We urge you to make sure that the values coming from the method are getting progressively closer and closer together (zeroing in, we hope, on the desired solution). Don’t stop until you’ve reached the limits of accuracy of your computing device. Also, be sure to compute the value of the function at the suspected approximate zero. If the function value is not close to zero, do not accept the value as an approximate zero.

As we illustrate in example 1.7, Newton’s method needs a good initial guess to find an accurate approximation.

y 8

EXAMPLE 1.7 x 2

3

8

FIGURE 3.11 y = x 3 − 3x 2 + x − 1

The Effect of a Bad Guess on Newton’s Method

Use Newton’s method to find an approximate zero of f (x) = x 3 − 3x 2 + x − 1. Solution From the graph in Figure 3.11, there appears to be a zero on the interval (2, 3). Using the (not particularly good) initial guess x0 = 1, we get x1 = 0, x2 = 1, x3 = 0 and so on. Try this for yourself. Newton’s method is sensitive to the initial guess and x0 = 1 is just a bad initial guess. If we had instead started with the improved initial guess x0 = 2, Newton’s method would have quickly converged to the approximate zero 2.769292354. (Again, try this for yourself.)

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As we see in example 1.7, making a good initial guess is essential with Newton’s method. However, this alone will not guarantee rapid convergence. By rapid convergence, we mean that it takes only a few iterations to obtain an accurate approximation.

n

xn

1 2 3 4

−9.5 −65.9 −2302 −2,654,301

5

−3.5 × 1012

EXAMPLE 1.8

6

−6.2 × 1024

Use Newton’s method with (a) x0 = −2, (b) x0 = −1 and (c) x0 = 0 to try to locate the (x − 1)2 zero of f (x) = 2 . x +1 Solution Of course, there’s no mystery here: f has only one zero, located at x = 1. However, watch what happens when we use Newton’s method with the specified guesses. (a) If we take x0 = −2, and apply Newton’s method, we obtain the values in the table found in the margin. Obviously, the successive iterations are blowing up with this initial guess. To see why, look at Figure 3.12, which shows the graphs of both y = f (x) and the tangent line at x = −2. If you follow the tangent line to where it intersects the x-axis, you will be going away from the zero (far away). Since all of the tangent lines for x ≤ −2 have positive slope [compute f (x) to see why this is true], each subsequent step takes you farther from the zero. (b) If we use the improved initial guess x0 = −1, we cannot even compute x1 . In this case, f (x0 ) = 0 and so, Newton’s method fails. Graphically, this means that the tangent line to y = f (x) at x = −1 is horizontal (see Figure 3.13), so that the tangent line never intersects the x-axis. (c) With the even better initial guess x0 = 0, we obtain the successive approximations in the following table.

Newton’s method iterations for x0 = −2 y

1

x

2

FIGURE 3.12 (x − 1)2 and the tangent line y= 2 x +1 at x = −2

Unusually Slow Convergence for Newton’s Method

y

1

1

x

FIGURE 3.13 y=

(x − 1) and the tangent line x2 + 1 at x = −1

n

xn

n

xn

1 2 3 4 5 6

0.5 0.70833 0.83653 0.912179 0.95425 0.976614

7 8 9 10 11 12

0.9881719 0.9940512 0.9970168 0.9985062 0.9992525 0.9996261

Newton’s method iterations for x0 = 0

2

Finally, we happened upon an initial guess for which Newton’s method converges to the root x = 1. What is unusual here is that the successive approximations shown in the table are converging to 1 much more slowly than in previous examples. By comparison, note that in example 1.5, the iterations stop changing at x5 . Here, x5 is not particularly close to the desired zero of f (x). In fact, in this example, x12 is not as close to the zero as x5 is in example 1.5. We look further into this type of behavior in the exercises. Despite the minor problems experienced in examples 1.7 and 1.8, you should view Newton’s method as a generally reliable and efficient method of locating zeros approximately. Just use a bit of caution and common sense. If the successive approximations are converging to some value that does not appear consistent with the graph, then you need to scrutinize your results more carefully and perhaps try some other initial guesses.

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251

BEYOND FORMULAS Approximations are at the heart of calculus. To find the slope of a tangent line, for example, we start by approximating the tangent line with secant lines. The fact that there are numerous simple derivative formulas to help us compute exact slopes is an unexpected bonus. In this section, the tangent line is thought of as an approximation of a curve and is used to approximate solutions of equations for which algebra fails. Although this doesn’t provide us the luxury of simply computing an exact answer, we can make the approximation as accurate as we like and so, for most practical purposes, we can “solve” the equation. Think about a situation where you need the time of day. How often do you need the exact time?

EXERCISES 3.1 WRITING EXERCISES 1. Briefly explain in terms of tangent lines why the approximation in example 1.3 gets worse as x gets farther from 8. 2. We constructed a variety of linear approximations in this section. Approximations can be “good” approximations or “bad” approximations. Explain why it can be said that y = x is a good approximation to y = sin x near x = 0 but y = 1 is not a good approximation to y = cos x near x = 0. (Hint: Look at the graphs of y = sin x and y = x on the same axes, then do the same with y = cos x and y = 1.) 3. In example 1.6, we mentioned that you might think of using a linear approximation instead of Newton’s method.√Discuss the relationship between a linear approximation to 3 7 and a √ 3 Newton’s method approximation to 7. (Hint: Compare the first step of Newton’s method to a linear approximation.) 4. Explain why Newton’s method fails computationally if f (x0 ) = 0. In terms of tangent lines intersecting the x-axis, explain why having f (x0 ) = 0 is a problem.

In exercises 1–6, find the linear approximation to f (x) at x x0 . Graph the function and its linear approximation. √ 2. f (x) = (x + 1)1/3 , x0 = 0 1. f (x) = x, x0 = 1 √ 4. f (x) = 2/x, x0 = 1 3. f (x) = 2x + 9, x0 = 0 5. f (x) = sin 3x, x0 = 0

6. f (x) = sin x, x0 = π

7. (a) Find the linear approximation at x = 0 to each of f (x) = (x + 1)2 , g(x) = 1 + sin(2x) and h(x) = e2x . Compare your results. (b) Graph each function in part (a) together with its linear approximation derived in part (a). Which function has the closest fit with its linear approximation?

8. (a) Find the linear approximation at x = 0 to each of f (x) = e x − e−x sin x, g(x) = tan−1 x and h(x) = sinh x = . 2 Compare your results. (b) Graph each function in part (a) together with its linear approximation derived in part (a). Which function has the closest fit with its linear approximation? In exercises 9 and 10, use linear approximations to estimate the quantity. √ √ √ 9. (a) 4 16.04 (b) 4 16.08 (c) 4 16.16 10. (a) sin (0.1) (b) sin (1.0) (c) sin 94 11. For exercise 9, compute the errors (the absolute value of the difference between the exact values and the linear approximations). 12. √ Thinking of exercises 9a–9c as numbers of the form 4 16 + x, denote the errors as e(x) (where x = 0.04, x = 0.08 and x = 0.16). Based on these three computations, determine a constant c such that e(x) ≈ c(x)2 . In exercises 13–16, use linear interpolation to estimate the desired quantity. 13. A company estimates that f (x) thousand software games can be sold at the price of $x as given in the table. x f (x)

20 18

30 14

40 12

Estimate the number of games that can be sold at (a) $24 and (b) $36.

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14. A vending company estimates that f (x) cans of soft drink can be sold in a day if the temperature is x ◦ F as given in the table. x f (x)

60 84

80 120

100 168

Estimate the number of cans that can be sold at (a) 72◦ and (b) 94◦ . 15. An animation director enters the position f (t) of a character’s head after t frames of the movie as given in the table.

3-12

In exercises 21–24, use Newton’s method with the given x0 to (a) compute x1 and x2 by hand and (b) use a computer or calculator to find the root to at least five decimal places of accuracy. 21. x 3 + 3x 2 − 1 = 0, x0 = 1 22. x 3 + 4x 2 − x − 1 = 0, x0 = −1 23. x 4 − 3x 2 + 1 = 0, x0 = 1 24. x 4 − 3x 2 + 1 = 0, x0 = −1 In exercises 25–32, use Newton’s method to find an approximate root (accurate to six decimal places). Sketch the graph and explain how you determined your initial guess. 25. x 3 + 4x 2 − 3x + 1 = 0

26. x 4 − 4x 3 + x 2 − 1 = 0

27. x + 3x + x − 1 = 0 29. sin x = x 2 − 1

28. cos x − x = 0 30. cos x 2 = x √ 32. e−x = x

5

t f (t)

200 128

220 142

240 136

31. e x = −x

If the computer software uses interpolation to determine the intermediate positions, determine the position of the head at frame numbers (a) 208 and (b) 232. 16. A sensor measures the position f (t) of a particle t microseconds after a collision as given in the table. t f (t)

5 8

10 14

33. Show that Newton’s method applied to x 2 − c = 0 (where c > 0 is some constant) produces the √ iterative scheme xn+1 = 12 (xn + c/xn ) for approximating c. This scheme has been known for over 2000 years. To understand why it works, √ suppose that your initial guess√(x0 ) for c is a little too small. How would c/x0 compare to c? Explain why the√average of x0 and c/x0 would give a better approximation to c. 34. Show that Newton’s method applied to x n − c = 0 (where n and c are positive constants) produces the iterative √ scheme xn+1 = n1 [(n − 1)xn + cxn1−n ] for approximating n c.

15 18

Estimate the position of the particle at times (a) t = 8 and (b) t = 12. 17. Given the graph of y = f (x), draw in the tangent lines used in Newton’s method to determine x1 and x2 after starting at x0 = 2. Which of the zeros will Newton’s method converge to? y

2

2

3

x 2

18. Repeat exercise 17 with x0 = −2 and x0 = 0.4. 19. What would happen to Newton’s method in exercise 17 if you had a starting value of x0 = 0? 20. Consider the use of Newton’s method in exercise 17 with x0 = 0.2 and x0 = 10. Obviously, x0 = 0.2 is much closer to a zero of the function, but which initial guess would work better in Newton’s method? Explain.

In exercises 35–40, use Newton’s method [state the function f (x) you use] to estimate the given number. (Hint: See exercises 33 and 34.) √ √ √ √ 35. 11 36. 23 37. 3 11 38. 3 23 √ √ 40. 46 24 39. 44 24 In exercises 41–46, Newton’s method fails. Explain why the method fails and, if possible, find a root by correcting the problem. 41. 4x 3 − 7x 2 + 1 = 0, x0 = 0 42. 4x 3 − 7x 2 + 1 = 0, x0 = 1 43. x 2 + 1 = 0, x0 = 0 44. x 2 + 1 = 0, x0 = 1 4x 2 − 8x + 1 = 0, x0 = −1 45. 4x 2 − 3x − 7 1/3 x +1 46. = 0, x0 = 0.5 x −2 47. Use Newton’s method with (a) x0 = 1.2 and (b) x0 = 2.2 to find a zero of f (x) = x 3 − 5x 2 + 8x − 4. Discuss the difference in the rates of convergence in each case. 48. Use Newton’s method with (a) x0 = 0.2 and (b) x0 = 3.0 to find a zero of f (x) = x sin x. Discuss the difference in the rates of convergence in each case. 49. Use Newton’s method with (a) x0 = −1.1 and (b) x0 = 2.1 to find a zero of f (x) = x 3 − 3x − 2. Discuss the difference in the rates of convergence in each case.

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253

50. Factor the polynomials in exercises 47 and 49. Find a relationship between the factored polynomial and the rate at which Newton’s method converges to a zero. Explain how the function in exercise 48, which does not factor, fits into this relationship. (Note: The relationship will be explored further in exploratory exercise 2.)

61. The spruce budworm is an enemy of the balsam fir tree. In one model of the interaction between these organisms, possible long-term populations of the budworm are solutions of the equation r (1 − x/k) = x/(1 + x 2 ), for positive constants r and k (see Murray’s Mathematical Biology). Find all positive solutions of the equation with r = 0.5 and k = 7.

In exercises 51–54, find the linear approximation at x 0 to show that the following commonly used approximations are valid for “small” x. Compare the approximate and exact values for x 0.01, x 0.1 and x 1. √ 51. tan x ≈ x 52. 1 + x ≈ 1 + 12 x √ 54. e x ≈ 1 + x 53. 4 + x ≈ 2 + 14 x

62. Repeat exercise 61 with r = 0.5 and k = 7.5. For a small change in the environmental constant k (from 7 to 7.5), how did the solution change from exercise 61 to exercise 62? The largest solution corresponds to an “infestation” of the spruce budworm.

55. Use a computer algebra system (CAS) to determine the range of x’s in exercise 51 for which the approximation is accurate to within 0.01. That is, find x such that | tan x − x| < 0.01. 56. Use a CAS to determine the range of x’s in exercise 54 for which the approximation is accurate to within 0.01. That is, find x such that |e x − (1 + x)| < 0.01. 57. A water wave of length L meters in water of depth d meters has velocity v satisfying the equation v2 =

4.9L e2π d/L − e−2π d/L . π e2π d/L + e−2π d/L

Treating L as a constant and thinking of v 2 as a function f (d), use a linear approximation to show that f (d) ≈ 9.8d for small values of d. That √is, for small depths, the velocity of the wave is approximately 9.8d and is independent of the wavelength L. 58. Planck’s law states that the energy density of blackbody radiation of wavelength x is given by f (x) =

8π hcx −5 . ehc/(kT x) − 1

Use the linear approximation in exercise 54 to show that f (x) ≈ 8πkT /x 4 , which is known as the Rayleigh-Jeans law. 59. Suppose that a species reproduces as follows: with probability p0 , an organism has no offspring; with probability p1 , an organism has one offspring; with probability p2 , an organism has two offspring and so on. The probability that the species goes extinct is given by the smallest nonnegative solution of the equation p0 + p1 x + p2 x 2 + · · · = x (see Sigmund’s Games of Life). Find the positive solutions of the equations 0.1 + 0.2x + 0.3x 2 + 0.4x 3 = x and 0.4 + 0.3x + 0.2x 2 + 0.1x 3 = x. Explain in terms of species going extinct why the first equation has a smaller solution than the second. 60. In Einstein’s theory of relativity, the length of an object depends on its velocity. If L 0 is the length of the object at rest, v is the object’s velocity and c is the speed of light, the Lorentz contraction formula for the length of the object is L = L 0 1 − v 2 /c2 . Treating L as a function of v, find the linear approximation of L at v = 0.

63. Newton’s theory of gravitation states that the weight of a person at elevation x feet above sea level is W (x) = P R 2 /(R + x)2 , where P is the person’s weight at sea level and R is the radius of the earth (approximately 20,900,000 feet). Find the linear approximation of W (x) at x = 0. Use the linear approximation to estimate the elevation required to reduce the weight of a 120-pound person by 1%. 64. One important aspect of Einstein’s theory of relativity is that mass is not constant. For a person with mass m 0 at rest, the mass will equal m = m 0 / 1 − v 2 /c2 at velocity v (where c is the speed of light). Thinking of m as a function of v, find the linear approximation of m(v) at v = 0. Use the linear approximation to show that mass is essentially constant for small velocities. 65. In the figure, a beam is clamped at its left end and simply supported at its right end. A force P (called an axial load) is applied at the right end.

P

If enough force is applied, the beam will buckle. Obviously, it is important for engineers to be able to compute this buckling load. For certain types of beams, the buckling√load √ is the smallest positive solution of the equation tan √x = √x. The √ shape of √ the buckled √ beam is given by y = L − L x − L cos L x + sin L x, where L is the buckling load. Find L and the shape of the buckled beam. 66. The spectral radiancy S of an ideal radiator at constant temperature can be thought of as a function S( f ) of the radiant frequency f . The function S( f ) attains its maximum when 3e−c f + c f − 3 = 0 for the constant c = 10−13 . Use Newton’s method to approximate the solution. 67. In the diagram (on the following page), a hockey player is D feet from the net on the central axis of the rink. The goalie blocks off a segment of width w and stands d feet from the net. The shooting angle to the left of the goalie is given by 3(1 − d/D) − w/2 −1 φ = tan . Use a linear approximation D−d −1 . of tan x at x = 0 to show that if d = 0, then φ ≈ 3−w/2 D Based on this, describe how φ changes if there is an increase in (a) w or (b) D.

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3-14

D

w

Exercise 67

the iterates√converge to 0. (Note: For complex numbers, |a + bi| = a 2 + b2 .) Color the pixel at the point (a, b) black if the iterates converge to 0 and white if not. You can change the ranges of a and b and the step size to “zoom in” on interesting regions. The accompanying pictures show the basin of attraction (in black) for x = 1. In the first figure, we display the region with −1.5 ≤ x ≤ 3.5. In the second figure, we have zoomed in to the portion for 0.26 ≤ x ≤ 0.56. The third shows an even tighter zoom: 0.5 ≤ x ≤ 0.555.

68. The shooter in exercise 67 is assumed to be in the center of the ice. Suppose that the line from the shooter to the center of the goal makes an angle of θ with center line. For the goalie to completely block the goal, he must stand d feet away from the net where d = D(1 − w/6 cos θ ). Show that for small angles, d ≈ D(1 − w/6). In exercises 69 and 70, we explore the convergence of Newton’s method for f (x) x 3 − 3x 2 2x. 69. The zeros of f are x = 0, x = 1 and x = 2. Determine which of the three zeros Newton’s method iterates converge to for (a) x0 = 0.1, (b) x0 = 1.1 and (c) x0 = 2.1. 70. Determine which of the three zeros Newton’s method iterates converge to for (a) x0 = 0.54, (b) x0 = 0.55 and (c) x0 = 0.56.

EXPLORATORY EXERCISES 1. In this exercise, you will extend the work of exercises 69 and 70. First, a definition: The basin of attraction of a zero is the set of starting values x0 for which Newton’s method iterates converge to the zero. As exercises 69 and 70 indicate, the basin boundaries are more complicated than you might expect. For example, you have seen that the interval [0.54, 0.56] contains points in all three basins of attraction. Show that the same is true of the interval [0.552, 0.553]. The picture gets even more interesting when you use complex numbers. These √ are numbers of the form a + bi where i = −1. The remainder of the exercise requires a CAS or calculator that is programmable and performs calculations with complex numbers. First, try Newton’s method with starting point x0 = 1 + i. The formula is exactly the same! Use your computer to show x 3 − 3x02 + 2x0 that x1 = x0 − 0 2 = 1 + 12 i. Then verify that 3x0 − 6x0 + 2 1 x2 = 1 + 17 i and x3 = 1 + 182 i. It certainly appears that the iterates are converging to the zero x = 1. Now, for some programming: set up a double loop with the parameter a running from 0 to 2 in steps of 0.02 and b running from −1 to 1 in steps of 0.02. Within the double loop, set x0 = a + bi and compute 10 Newton’s method iterates. If x10 is close to 0, say |x10 − 0| < 0.1, then we can conjecture that

2. Another important question involving Newton’s method is how fast it converges to a given zero. Intuitively, we can distinguish between the rate of convergence for f (x) = x 2 − 1 (with x0 = 1.1) and that for g(x) = x 2 − 2x + 1 (with x0 = 1.1). But how can we measure this? One method is to take successive approximations xn−1 and xn and compute the difference n = xn − xn−1 . To discover the importance of this quantity, run Newton’s method with x0 = 1.5 and then compute the ratios 3 /2 , 4 /3 , 5 /4 and so on, for each of the following functions: F1 (x) F2 (x) F3 (x) F4 (x)

= = = =

(x (x (x (x

− 1)(x + 2)3 = x 4 + 5x 3 + 6x 2 − 4x − 8, − 1)2 (x + 2)2 = x 4 + 2x 3 − 3x 2 − 4x + 4, − 1)3 (x + 2) = x 4 − x 3 − 3x 2 + 5x − 2 and − 1)4 = x 4 − 4x 3 + 6x 2 − 4x + 1.

n+1 . n If the limit exists and is nonzero, we say that Newton’s method converges linearly. How does r relate to your intuitive sense of how fast the method converges? For f (x) = (x − 1)4 , In each case, conjecture a value for the limit r = lim

n→∞

3-15

SECTION 3.2

we say that the zero x = 1 has multiplicity 4. For f (x) = (x − 1)3 (x + 2), x = 1 has multiplicity 3 and so on. How does r relate to the multiplicity of the zero? Based on this analysis, why did Newton’s method converge faster for f (x) = x 2 − 1 than for g(x) = x 2 − 2x + 1? Finally, use Newton’s method to compute the rate r and hypothesize the multiplicity of the zero x = 0 for f (x) = x sin x and g(x) = x sin x 2 . 3. This exercise looks at a special case of the three-body problem, in which there is a large object A of mass m A , a much smaller object B of mass m B m A and an object C of negligible mass. (Here, a b means that a is much smaller than b.) Assume that object B orbits in a circular path around the common center of mass. There are five circular orbits for object C that maintain constant relative positions of the three objects. These are called Lagrange points L 1 , L 2 , L 3 , L 4 and L 5 , as shown in the figure. L4

B

A L3

L1

L2

L5

3.2

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Indeterminate Forms and L’Hˆ opital’s Rule

255

To derive equations for the Lagrange points, set up a coordinate system with object A at the origin and object B at the point (1, 0). Then L 1 is at the point (x1 , 0), where x1 is the solution of (1 + k)x 5 − (3k + 2)x 4 + (3k + 1)x 3 − x 2 + 2x − 1 = 0; L 2 is at the point (x2 , 0), where x2 is the solution of (1 + k)x 5 − (3k + 2)x 4 + (3k + 1)x 3 − (2k + 1)x 2 + 2x − 1 = 0 and L 3 is at the point (−x3 , 0), where x3 is the solution of (1 + k)x 5 + (3k + 2)x 4 + (3k + 1)x 3 − x 2 − 2x − 1 = 0, mB . Use Newton’s method to find approximate mA solutions of the following.

where k =

(a) Find L 1 for the Earth-Sun system with k = 0.000002. This point has an uninterrupted view of the sun and is the location of the solar observatory SOHO. (b) Find L 2 for the Earth-Sun system with k = 0.000002. This is the future location of NASA’s Microwave Anistropy Probe. (c) Find L 3 for the Earth-Sun system with k = 0.000002. This point is invisible from the Earth and is the location of Planet X in many science fiction stories. (d) Find L 1 for the Moon-Earth system with k = 0.01229. This point has been suggested as a good location for a space station to help colonize the moon. (e) The points L 4 and L 5 form equilateral triangles with objects A and B. Explain why means that polar coordinates this for L 4 are (r, θ) = 1, π6 . Find (x, y)-coordinates for L 4 and L 5 . In the Jupiter-Sun system, these are locations of numerous Trojan asteroids.

ˆ INDETERMINATE FORMS AND L’H OPITAL’S RULE In this section, we reconsider the problem of computing limits. You have frequently seen limits of the form lim

x→a

f (x) , g(x)

where lim f (x) = lim g(x) = 0 or where lim f (x) = lim g(x) = ∞ (or −∞). Recall that x→a x→a x→a x→a from either of these forms 00 or ∞ , called indeterminate forms , we cannot determine ∞ the value of the limit, or even whether the limit exists. For instance, note that lim

x2 − 1 (x − 1)(x + 1) x +1 2 = lim = lim = = 2, x→1 x→1 x −1 x −1 1 1

lim

x −1 x −1 1 1 = lim = lim = 2 x→1 (x − 1)(x + 1) x→1 x + 1 x −1 2

x→1

x→1

3-15

SECTION 3.2

we say that the zero x = 1 has multiplicity 4. For f (x) = (x − 1)3 (x + 2), x = 1 has multiplicity 3 and so on. How does r relate to the multiplicity of the zero? Based on this analysis, why did Newton’s method converge faster for f (x) = x 2 − 1 than for g(x) = x 2 − 2x + 1? Finally, use Newton’s method to compute the rate r and hypothesize the multiplicity of the zero x = 0 for f (x) = x sin x and g(x) = x sin x 2 . 3. This exercise looks at a special case of the three-body problem, in which there is a large object A of mass m A , a much smaller object B of mass m B m A and an object C of negligible mass. (Here, a b means that a is much smaller than b.) Assume that object B orbits in a circular path around the common center of mass. There are five circular orbits for object C that maintain constant relative positions of the three objects. These are called Lagrange points L 1 , L 2 , L 3 , L 4 and L 5 , as shown in the figure. L4

B

A L3

L1

L2

L5

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..

Indeterminate Forms and L’Hˆ opital’s Rule

255

To derive equations for the Lagrange points, set up a coordinate system with object A at the origin and object B at the point (1, 0). Then L 1 is at the point (x1 , 0), where x1 is the solution of (1 + k)x 5 − (3k + 2)x 4 + (3k + 1)x 3 − x 2 + 2x − 1 = 0; L 2 is at the point (x2 , 0), where x2 is the solution of (1 + k)x 5 − (3k + 2)x 4 + (3k + 1)x 3 − (2k + 1)x 2 + 2x − 1 = 0 and L 3 is at the point (−x3 , 0), where x3 is the solution of (1 + k)x 5 + (3k + 2)x 4 + (3k + 1)x 3 − x 2 − 2x − 1 = 0, mB . Use Newton’s method to find approximate mA solutions of the following.

where k =

(a) Find L 1 for the Earth-Sun system with k = 0.000002. This point has an uninterrupted view of the sun and is the location of the solar observatory SOHO. (b) Find L 2 for the Earth-Sun system with k = 0.000002. This is the future location of NASA’s Microwave Anistropy Probe. (c) Find L 3 for the Earth-Sun system with k = 0.000002. This point is invisible from the Earth and is the location of Planet X in many science fiction stories. (d) Find L 1 for the Moon-Earth system with k = 0.01229. This point has been suggested as a good location for a space station to help colonize the moon. (e) The points L 4 and L 5 form equilateral triangles with objects A and B. Explain why means that polar coordinates this for L 4 are (r, θ) = 1, π6 . Find (x, y)-coordinates for L 4 and L 5 . In the Jupiter-Sun system, these are locations of numerous Trojan asteroids.

ˆ INDETERMINATE FORMS AND L’H OPITAL’S RULE In this section, we reconsider the problem of computing limits. You have frequently seen limits of the form lim

x→a

f (x) , g(x)

where lim f (x) = lim g(x) = 0 or where lim f (x) = lim g(x) = ∞ (or −∞). Recall that x→a x→a x→a x→a from either of these forms 00 or ∞ , called indeterminate forms , we cannot determine ∞ the value of the limit, or even whether the limit exists. For instance, note that lim

x2 − 1 (x − 1)(x + 1) x +1 2 = lim = lim = = 2, x→1 x→1 x −1 x −1 1 1

lim

x −1 x −1 1 1 = lim = lim = 2 x→1 (x − 1)(x + 1) x→1 x + 1 x −1 2

x→1

x→1

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and

lim

x→1

3-16

x2

x −1 x −1 1 = lim = lim , which does not exist, 2 x→1 x→1 − 2x + 1 x −1 (x − 1)

even though all three limits initially look like 00 . The lesson here is that the expression 00 is mathematically meaningless. It indicates only that both the numerator and denominator tend to zero and that we’ll need to dig deeper to find the value of the limit. Similarly, each of the following limits has the indeterminate form ∞ : ∞ 1 1 1 + 3 x2 + 1 0 x3 x x = lim lim = lim = = 0, x→∞ x 3 + 5 x→∞ x→∞ 5 1 1 1+ 3 (x 3 + 5) 3 x x

(x 2 + 1)

CAUTION will frequently write 00 or We ∞ next to an expression, for ∞ instance, x −1 0 lim 2 . x→1 x − 1 0 We use this shorthand to indicate that the limit has the indicated indeterminate form. This notation does not mean that the value of the limit is 00 . You should take care to avoid writing lim f (x) = 00 or ∞ , ∞ x→a

as these are meaningless expressions.

1 5 (x + 5) x+ 2 x3 + 5 x2 x = lim lim = lim =∞ x→∞ x 2 + 1 x→∞ x→∞ 1 1 1 + (x 2 + 1) x2 x2

3

and

1 (2x + 3x − 5) 2+ 2x 2 + 3x − 5 x2 = lim lim 2 = lim x→∞ x + 4x − 11 x→∞ x→∞ 1 1+ (x 2 + 4x − 11) x2 2

5 3 − 2 2 x x = = 2. 4 11 1 − 2 x x

So, as with limits of the form 00 , if a limit has the form ∞ , we must dig deeper to determine the ∞ value of the limit or whether the limit even exists. Unfortunately, limits with indeterminate forms are frequently more difficult than those just given. For instance, back at section sin x 2.6, where we struggled with the limit lim , ultimately resolving it only with an x→0 x intricate geometric argument. This limit has the indeterminate form 00 , but there is no way to manipulate the numerator or denominator to simplify the expression. In the case of f (x) lim , where lim f (x) = lim g(x) = 0, we can use linear approximations to suggest a x→c g(x) x→c x→c solution, as follows. If both functions are differentiable at x = c, then they are also continuous at x = c, so that f (c) = lim f (x) = 0 and g(c) = lim g(x) = 0. We now have the linear approximations x→c

x→c

f (x) ≈ f (c) + f (c)(x − c) = f (c)(x − c) g(x) ≈ g(c) + g (c)(x − c) = g (c)(x − c),

and

since f (c) = 0 and g(c) = 0. As we have seen, the approximation should improve as x approaches c, so we would expect that if the limits exist, lim

x→c

f (c) f (x) f (c)(x − c) f (c) = lim = lim = , x→c g (c)(x − c) x→c g (c) g(x) g (c)

assuming that g (c) = 0. Note that if f (x) and g (x) are continuous at x = c and g (c) = 0, f (c) f (x) then = lim . This suggests the following result. x→c g (x) g (c)

3-17

SECTION 3.2

..

Indeterminate Forms and L’Hˆ opital’s Rule

257

THEOREM 2.1 (L’Hˆ opital’s Rule)

HISTORICAL NOTES Guillaume de l’Hˆ opital (1661–1704) A French mathematician who first published the result now known as l’Hˆ opital’s Rule. Born into nobility, l’Hˆ opital was taught calculus by the brilliant mathematician Johann Bernoulli. A competent mathematician, l’Hˆ opital is best known as the author of the first calculus textbook. L’Hˆ opital was a friend and patron of many of the top mathematicians of the seventeenth century.

Suppose that f and g are differentiable on the interval (a, b), except possibly at some fixed point c ∈ (a, b) and that g (x) = 0 on (a, b), except possibly at c. f (x) 0 ∞ Suppose further that lim has the indeterminate form or and that x→c g(x) 0 ∞ f (x) lim = L (or ±∞). Then, x→c g (x) f (x) f (x) lim = lim . x→c g(x) x→c g (x)

PROOF Here, we prove only the 00 case where f, f , g and g are all continuous on all of (a, b) and g (c) = 0, while leaving the more intricate general 00 case for Appendix A. First, recall the alternative form of the definition of derivative (found in section 2.2): f (c) = lim

x→c

f (x) − f (c) . x −c

Working backward, we have by continuity that f (x) f (c) lim = = x→c g (x) g (c)

f (x) − f (c) f (x) − f (c) f (x) − f (c) x −c x −c = lim = lim . g(x) − g(c) x→c g(x) − g(c) x→c g(x) − g(c) lim x→c x −c x −c

lim

x→c

Further, since f and g are continuous at x = c, we have that f (c) = lim f (x) = 0 x→c

and

g(c) = lim g(x) = 0. x→c

It now follows that lim

x→c

f (x) f (x) − f (c) f (x) = lim = lim , x→c g(x) − g(c) x→c g(x) g (x)

which is what we wanted. We leave the proof for the

∞ ∞

case to more advanced texts.

REMARK 2.1 f (x) is replaced with any of the g(x) f (x) f (x) f (x) f (x) limits lim+ , lim , lim or lim . (In each case, we must make x→−∞ g(x) x→c g(x) x→c− g(x) x→∞ g(x) appropriate adjustments to the hypotheses.)

The conclusion of Theorem 2.1 also holds if lim

x→c

EXAMPLE 2.1 Evaluate lim

x→0

The Indeterminate Form

0 0

1 − cos x . sin x

Solution This has the indeterminate form 00 , and both (1 − cos x) and sin x are continuous and differentiable everywhere. Further, ddx sin x = cos x = 0 in some

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Applications of Differentiation

interval containing x = 0. (Can you determine one such interval?) From the graph of 1 − cos x seen in Figure 3.14, it appears that f (x) → 0, as x → 0. We can f (x) = sin x confirm this with l’Hˆopital’s Rule, as follows: d (1 − cos x) 0 1 − cos x sin x dx lim = lim = = 0. = lim d x→0 x→0 x→0 cos x sin x 1 (sin x) dx

y 3 2 1 x

2

1

3-18

2

1 2

L’Hˆopital’s Rule is equally easy to apply with limits of the form

3

EXAMPLE 2.2

FIGURE 3.14 y=

1 − cos x sin x

The Indeterminate Form

∞ . ∞

∞ ∞

ex . x→∞ x Solution This has the form ∞ and from the graph in Figure 3.15, it appears that the ∞ function grows larger and larger, without bound, as x → ∞. Applying l’Hˆopital’s Rule confirms our suspicions, as d x (e ) ex ex dx lim = lim = ∞. = lim x→∞ x x→∞ d x→∞ 1 (x) dx

Evaluate lim

y 30

20

10

For some limits, you may need to apply l’Hˆopital’s Rule repeatedly. Just be careful to verify the hypotheses at each step. x 1

2

3

4

5

EXAMPLE 2.3 FIGURE 3.15 y=

ex x

4

6

0.6

0.4

0.2

x 8

FIGURE 3.16 y=

x2 ex

x2 . x→∞ e x Solution First, note that this limit has the form ∞ . From the graph in Figure 3.16, it ∞ seems that the function tends to 0 as x → ∞ (and does so very rapidly, at that). Applying l’Hˆopital’s Rule twice, we get d 2 (x ) x2 2x ∞ dx lim x = lim = lim x x→∞ e x→∞ d x→∞ e ∞ (e x ) dx d (2x) 2 dx = lim = lim x = 0, x→∞ d x→∞ e (e x ) dx as expected.

Evaluate lim

y

2

A Limit Requiring Two Applications of L’Hˆ opital’s Rule

10

REMARK 2.2 A very common error is to apply l’Hˆopital’s Rule indiscriminately, without first checking that the limit has the indeterminate form 00 or ∞ . Students also sometimes ∞ incorrectly compute the derivative of the quotient, rather than the quotient of the derivatives. Be very careful here.

3-19

SECTION 3.2

EXAMPLE 2.4 y

..

Indeterminate Forms and L’Hˆ opital’s Rule

259

An Erroneous Use of L’Hˆ opital’s Rule

Find the mistake in the string of equalities x2 2x 2 2 = lim x = lim x = = 2. x→0 e x − 1 x→0 e x→0 e 1 lim

x 3

3

FIGURE 3.17 y=

Solution From the graph in Figure 3.17, we can see that the limit is approximately 0, x2 so 2 appears to be incorrect. The first limit, lim x , has the form 00 and the functions x→0 e − 1 f (x) = x 2 and g(x) = e x − 1 satisfy the hypotheses of l’Hˆopital’s Rule. Therefore, the x2 2x 2x 0 first equality, lim x = lim x , holds. However, notice that lim x = = 0 and x→0 e − 1 x→0 e x→0 e 1 l’Hˆopital’s Rule does not apply here. The correct evaluation is then

x2 ex − 1

lim

x→0 e x

x2 2x 0 = lim x = = 0. − 1 x→0 e 1

Sometimes an application of l’Hˆopital’s Rule must be followed by some simplification, as we see in example 2.5.

EXAMPLE 2.5 y

Evaluate lim+ x→0

Simplification of the Indeterminate Form

∞ ∞

ln x . csc x

0.4 0.2 x 0.4

0.8

1.2

0.2 0.4

FIGURE 3.18 y=

ln x csc x

Solution First, notice that this limit has the form ∞ . From the graph in Figure 3.18, it ∞ appears that the function tends to 0 as x → 0+ . Applying l’Hˆopital’s Rule, we have d 1 (ln x) ∞ ln x dx x lim+ = lim+ . = lim+ d x→0 csc x x→0 x→0 −csc x cot x ∞ (csc x) dx This last limit still has the indeterminate form ∞ , but rather than apply l’Hˆopital’s Rule ∞ again, observe that we can rewrite the expression. (Do this wherever possible when a limit expression gets too complicated.) We have 1 sin x ln x x lim = lim+ = lim+ − tan x = (−1)(0) = 0, x→0+ csc x x→0 −csc x cot x x→0 x as expected, where we have used the fact (established in section 2.6) that sin x lim = 1. x→0 x (You can also establish this by using l’Hˆopital’s Rule.) Notice that if we had simply blasted away with further applications of l’Hˆopital’s Rule to lim+ would never have resolved the limit. (Why not?)

x→0

1 x

−csc x cot x

, we

Other Indeterminate Forms There are five additional indeterminate forms to consider: ∞ − ∞, 0 · ∞, 00 , 1∞ and ∞0 . Look closely at each of these to see why they are indeterminate. When evaluating a limit of this type, the objective is to somehow reduce it to one of the indeterminate forms 00 or ∞ , at which point we can apply l’Hˆopital’s Rule. ∞

260

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Applications of Differentiation

3-20

y

EXAMPLE 2.6

x

3 2

2

3

Evaluate lim

x→0

The Indeterminate Form ∞ − ∞

1 1 − 4 . x2 x

Solution First, notice that the limit has the form (∞ − ∞). From the graph of the function in Figure 3.19, it appears that the function values tend to −∞ as x → 0. If we add the fractions, we get 2 1 x −1 1 lim = lim = −∞, − x→0 x 2 x→0 x4 x4

5

10

FIGURE 3.19

as conjectured, where we have resolved the limit without using l’Hˆopital’s Rule, which does not apply here. (Why not?)

1 1 y= 2 − 4 x x

EXAMPLE 2.7

The Indeterminate Form ∞ − ∞

1 1 Evaluate lim − . x→0 ln (x + 1) x

y 1.0 0.8 0.6 0.4 0.2 x

1

1

2

FIGURE 3.20 y=

1 1 − ln (x + 1) x

3

Solution In this case, the limit has the form (∞ − ∞). From the graph in Figure 3.20, it appears that the limit is somewhere around 0.5. If we add the fractions, we get a form to which we can apply l’Hˆopital’s Rule. We have 1 1 0 x − ln (x + 1) lim − = lim x→0 ln (x + 1) x→0 ln (x + 1)x x 0 d [x − ln (x + 1)] dx = lim By l’Hˆopital’s Rule. d x→0 [ln (x + 1)x] dx 1 1− 0 x +1 = lim . x→0 1 0 x + ln (x + 1)(1) x +1 Rather than apply l’Hˆopital’s Rule to this last expression, we first simplify the expression, by multiplying top and bottom by (x + 1). We now have 1 1− 1 1 x +1 x +1 lim − = lim x→0 ln (x + 1) x→0 1 x x +1 x + ln (x + 1)(1) x +1 (x + 1) − 1 0 = lim x→0 x + (x + 1) ln (x + 1) 0 d (x) dx = lim By l’Hˆopital’s Rule. x→0 d [x + (x + 1) ln (x + 1)] dx 1 1 = , = lim 1 x→0 2 1 + (1) ln (x + 1) + (x + 1) (x + 1) which is consistent with Figure 3.20.

3-21

..

SECTION 3.2

EXAMPLE 2.8

Evaluate lim

x→∞

y 0.4 0.3 0.2 0.1 x 20

40

60

80

100

FIGURE 3.21 y=

1 ln x x

Indeterminate Forms and L’Hˆ opital’s Rule

261

The Indeterminate Form 0 · ∞

1 ln x . x

Solution This limit has the indeterminate form (0 · ∞). From the graph in Figure 3.21, it appears that the function is decreasing very slowly toward 0 as x → ∞. It’s easy to rewrite this in the form ∞ , from which we can use l’Hˆopital’s Rule. Note that ∞ 1 ∞ ln x lim ln x = lim x→∞ x x→∞ x ∞ d ln x dx = lim By l’Hˆopital’s Rule. x→∞ d x dx 1 0 x = lim = = 0. x→∞ 1 1 Note: If lim [ f (x)]g(x) has one of the indeterminate forms 00 , ∞0 or 1∞ , then, letting x→c

y = [ f (x)]g(x) , we have for f (x) > 0 that ln y = ln[ f (x)]g(x) = g(x) ln [ f (x)], so that lim ln y = lim {g(x) ln [ f (x)]} will have the indeterminate form 0 · ∞, which we x→c x→c can deal with as in example 2.8.

EXAMPLE 2.9

The Indeterminate Form 1∞

1

Evaluate lim+ x x−1 . x→1

Solution First, note that this limit has the indeterminate form (1∞ ). From the graph in 1 Figure 3.22, it appears that the limit is somewhere around 3. We define y = x x−1 , so that

y 20

1

ln y = ln x x−1 =

15

1 ln x. x −1

We now consider the limit

10

1 ln x (∞ · 0) x→1 x→1 x − 1 0 ln x = lim+ x→1 x − 1 0 d (ln x) x −1 dx = lim+ = 1. By l’Hˆopital’s Rule. = lim+ d x→1 x→1 1 (x − 1) dx Be careful; we have found that lim+ ln y = 1, but this is not the original limit. We want lim+ ln y = lim+

5 x 0.5

1.0

1.5

FIGURE 3.22 1

y = x x−1

2.0

x→1

lim y = lim+ eln y = e1 ,

x→1+

which is consistent with Figure 3.22.

x→1

262

CHAPTER 3

..

Applications of Differentiation

y

3-22

Often, the computation of a limit in this form requires several applications of l’Hˆopital’s Rule. Just be careful (in particular, verify the hypotheses at every step) and do not lose sight of the original problem.

1.0 0.9

EXAMPLE 2.10

0.8

The Indeterminate Form 00

Evaluate lim+ (sin x)x . x→0

0.7 x 0.2

0.4

0.6

0.8

Solution This limit has the indeterminate form (00 ). In Figure 3.23, it appears that the limit is somewhere around 1. We let y = (sin x)x , so that ln y = ln (sin x)x = x ln (sin x).

1.0

FIGURE 3.23 y = (sin x)x

Now consider the limit lim ln y = lim+ ln (sin x)x = lim+ [x ln (sin x)] (0 · ∞) x→0 x→0 ln (sin x) ∞ = lim+ 1 x→0 ∞ x d [ln (sin x)] dx = lim+ By l’Hˆopital’s Rule. d −1 x→0 (x ) dx ∞ (sin x)−1 cos x . = lim+ −2 x→0 −x ∞

x→0+

TODAY IN MATHEMATICS Vaughan Jones (1952– ) A New Zealand mathematician whose work has connected apparently disjoint areas of mathematics. He was awarded the Fields Medal in 1990 for mathematics that was described by peers as ‘astonishing’. One of his major accomplishments is a discovery in knot theory that has given biologists insight into the replication of DNA. A strong supporter of science and mathematics education in New Zealand, Jones’ “style of working is informal, and one which encourages the free and open interchange of ideas . . . His openness and generosity in this regard have been in the best tradition and spirit of mathematics.” His ideas have “served as a rich source of ideas for the work of others.”

As we have seen earlier, we should rewrite the expression before proceeding. Here, we multiply top and bottom by x 2 sin x to get (sin x)−1 cos x x 2 sin x lim ln y = lim+ x→0+ x→0 −x −2 x 2 sin x 2 −x cos x 0 = lim+ x→0 sin x 0 d (−x 2 cos x) dx = lim+ By l’Hˆopital’s Rule. d x→0 (sin x) dx −2x cos x + x 2 sin x 0 = lim+ = = 0. x→0 cos x 1 Again, we have not yet found the original limit. However, lim y = lim+ eln y = e0 = 1,

x→0+

x→0

which is consistent with Figure 3.23.

EXAMPLE 2.11

The Indeterminate Form ∞0

Evaluate lim (x + 1)2/x . x→∞

Solution This limit has the indeterminate form (∞0 ). From the graph in Figure 3.24, it appears that the function tends to a limit around 1 as x → ∞. We let y = (x + 1)2/x

3-23

..

SECTION 3.2

y

and consider

lim ln y = lim ln (x + 1)

4

x→∞

3

2/x

x→∞

2 ln (x + 1) x→∞ x

= lim

2

x 40

60

80

= lim

x→∞

∞ ∞

100

FIGURE 3.24 y = (x + 1)2/x

We now have that

263

2 ln (x + 1) x

d [2 ln (x + 1)] 2(x + 1)−1 dx = lim = lim x→∞ x→∞ d 1 x dx 2 = lim = 0. x→∞ x + 1

1 20

Indeterminate Forms and L’Hˆ opital’s Rule

(0 · ∞)

By l’Hˆopital’s Rule.

lim y = lim eln y = e0 = 1,

x→∞

x→∞

as expected.

BEYOND FORMULAS On several occasions, we have had the benefit of rewriting a function or an equation into a more convenient form. L’Hˆopital’s Rule gives us a way to rewrite limits that are indeterminate. As we have seen, sometimes just the act of rewriting a function as a single fraction lets us determine the limit. What are other examples where an equation is made easier by rewriting, perhaps with a trigonometric identity or exponential rule?

EXERCISES 3.2 WRITING EXERCISES 1. L’Hˆopital’s Rule states that, in certain situations, the ratios of function values approach the same limits as the ratios of corresponding derivatives (rates of change). Graphically, this may f (x) be hard to understand. To get a handle on this, consider g(x) where both f (x) = ax + b and g(x) = cx + d are linear funcf (x) tions. Explain why the value of lim should depend on x→∞ g(x) the relative sizes of the slopes of the lines; that is, it should be f (x) equal to lim . x→∞ g (x) 2. Think of a limit of 0 as actually meaning “getting very small” and a limit of ∞ as meaning “getting very large.” Discuss whether the following limit forms are indeterminate or not and explain your answer: ∞ − ∞, 10 , 0 · ∞, ∞ · ∞, ∞0 , 0∞ and 00 . 3. A friend is struggling with l’Hˆopital’s Rule. When asked to work a problem, your friend says, “First, I plug in for x and

get 0 over 0. Then I use the quotient rule to take the derivative. Then I plug x back in.” Explain to your friend what the mistake is and how to correct it. 4. Suppose that two runners begin a race from the starting line, with one runner initially going twice as fast as the other. If f (t) and g(t) represent the positions of the runners at time t ≥ 0, explain why we can assume that f (0) = g(0) = 0 and f (t) lim = 2. Explain in terms of the runners’ positions why + t→0 g (t) f (t) = 2. l’Hˆopital’s Rule holds: that is, lim t→0 g(t) In exercises 1–38, find the indicated limits. x +2 x2 − 4 1. lim 2 2. lim 2 x→−2 x − 4 x→2 x − 3x + 2 3x 2 + 2 x +1 3. lim 2 4. lim 2 x→∞ x − 4 x→−∞ x + 4x + 3

264

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Applications of Differentiation

e2x − 1 x→0 x tan−1 x lim x→0 sin x sin 2x lim x→π sin x sin x − x lim x→0 x3 √ x −1 lim x→1 x − 1 x3 lim x x→∞ e x cos x − sin x lim x→0 x sin2 x sin π x lim x→1 x − 1 ln x lim x→∞ x 2 lim xe−x

sin x e3x − 1 sin x lim x→0 sin−1 x cos−1 x lim 2 x→−1 x − 1 tan x − x lim x→0 x3 ln x lim x→1 x − 1 ex lim 4 x→∞ x 1 2 lim cot x − 2 x→0 x e x−1 − 1 lim x→1 x 2 − 1 ln x lim √ x→∞ x lim x sin (1/x)

5. lim

6. lim

7.

8.

9. 11. 13. 15. 17. 19. 21. 23.

x→∞

ln (ln x) 25. lim x→1 ln x

10. 12. 14. 16. 18. 20. 22. 24.

ln x cot x 31. lim ( x 2 + 1 − x)

29. lim

x→0+

1 x 33. lim 1 + x→∞ x 1 x 35. lim √ − x→0+ x +1 x x 37. lim (1/x)

x→0+

32. lim (ln x − x) x→∞

√ 2

x + 1 x −4

34. lim

x→∞ x − 2

√ 5−x −2 36. lim √ x→1 10 − x − 3 38. lim (cos x)1/x x→0+

In exercises 39 and 40, find the error(s) in the incorrect calculations. cos x −sin x −cos x 1 39. lim 2 = lim = lim =− x→0 x x→0 x→0 2x 2 2 1 ex − 1 ex ex = lim = = lim 40. lim x→0 x→0 2x x→0 2 x2 2 41. Find all errors in the string x2 x2 2x 2 = lim = lim = lim x→0 ln x 2 x→0 2 ln x x→0 2/x x→0 −2/x 2 = lim (−x 2 ) = 0. lim

x→0

Then, determine the correct value of the limit. 42. Find all errors in the string lim

44. 45.

46.

47.

48.

x→0

x→0+

x→0

x→0

x→∞

sin (sin x) sin x sin x 28. lim √ x→0+ x √ x 30. lim x→0+ ln x

sin x cos x −sin x = lim = 0. = lim x→0 2x x→0 x2 2

Then, determine the correct value of the limit.

sin 3x 3x , cancel sin to get lim , then cancel x→0 2x sin 2x 3 x’s to get 2 . This answer is correct. Is either of the steps used valid? Use linear approximations to argue that the first step is likely to give a correct answer. sin nx Evaluate lim for nonzero constants n and m. x→0 sin mx sin x 2 sin x (a) Compute lim and compare your result to lim . x→0 x 2 x→0 x 2 1 − cos x and compare your result to (b) Compute lim x→0 x4 1 − cos x . lim x→0 x2 sin x 3 Use your results from exercise 45 to evaluate lim and x→0 x 3 1 − cos x 3 lim without doing any calculations. x→0 x6 sin kx 2 has the indeterminate form 00 and then Show that lim x→0 x2 evaluate the limit (where k is some real number). What is the range of values that a limit of the indeterminate form 00 can have? cot kx 2 has the indeterminate form ∞ and then Show that lim ∞ x→0 csc x 2 evaluate the limit (where k is some real number). What is the range of values that a limit of the indeterminate form ∞ can ∞ have?

43. Starting with lim

x→0

26. lim

27. lim x ln x

x→∞

3-24

In exercises 49 and 50, determine which function “dominates,” where we say that the function f (x) dominates the function g(x) as x → ∞ if lim f (x) lim g(x) ∞ and either x→∞

lim

x→∞

x→∞

f (x) g(x) ∞ or lim 0. x→∞ f (x) g(x)

49. e x or x n (n = any positive integer) 50. ln x or x p (for any number p > 0) ecx − 1 for any constant c. 51. Evaluate lim x→0 x tan cx − cx 52. Evaluate lim for any constant c. x→0 x3 f (x) f (x 2 ) = L , what can be said about lim ? Explain 53. If lim x→0 g(x) x→0 g(x 2 ) f (x) = L for a = 0, 1 does not tell why knowing that lim x→a g(x) 2 f (x ) you anything about lim . x→a g(x 2 ) f (x 2 ) 54. Give an example of functions f and g for which lim x→0 g(x 2 ) f (x) exists, but lim does not exist. x→0 g(x) 55. In section 1.2, we briefly discussed the position of a baseball thrown with the unusual knuckleball pitch. The left/right position (in feet) of a ball thrown with spin rate ω and a particular grip at time t seconds is f (ω) = (2.5/ω)t − (2.5/4ω2 ) sin 4ωt. Treating t as a constant and ω as the variable (change to x if you

3-25

SECTION 3.3

like), show that lim f (ω) = 0 for any value of t. (Hint: Find ω→0

a common denominator and use l’Hˆopital’s Rule.) Conclude that this pitch does not move left or right at all. 56. In this exercise, we look at a knuckleball thrown with a different grip than that of exercise 55. The left or right position (in feet) of a ball thrown with spin rate ω and this new grip at time t seconds is f (ω) = (2.5/4ω2 ) − (2.5/4ω2 ) sin (4ωt + π/2). Treating t as a constant and ω as the variable (change to x if you like), find lim f (ω). Your answer should depend on t. By ω→0

graphing this function of t, you can see the path of the pitch (use a domain of 0 ≤ t ≤ 0.68). Describe this pitch. 57. Find functions f such that lim f (x) has the indeterminate x→∞

form ∞ , but where the limit (a) does not exist; (b) equals 0; ∞ (c) equals 3 and (d) equals −4. 58. Find functions f such that lim f (x) has the indeterminate x→∞

form ∞ − ∞, but where the limit (a) does not exist; (b) equals 0 and (c) equals 2. 59. In the figure shown here, a section of the unit circle is determined by angle θ . Region 1 is the triangle ABC. Region 2 is bounded by the line segments AB and BC and the arc of the circle. As the angle θ decreases, the difference between the two regions decreases, also. You might expect that the areas of the regions become nearly equal, in which case the ratio of the areas approaches 1. To see what really happens, show that the area of region 1 divided by the area of region 2 equals sin θ − 12 sin 2θ (1 − cos θ ) sin θ = and find the limit of this θ − cos θ sin θ θ − 12 sin 2θ expression as θ → 0. Surprise! y 1

A

θ

C B

Exercise 59

3.3

1

x

..

Maximum and Minimum Values

265

60. The size of an animal’s pupils expand and contract depending 160x −0.4 + 90 on the amount of light available. Let f (x) = be 8x −0.4 + 10 the size in mm of the pupils at light intensity x. Find lim f (x) x→0+

and lim f (x), and argue that these represent the largest and x→∞

smallest possible sizes of the pupils, respectively.

EXPLORATORY EXERCISES 1. In this exercise, you take a quick look at what we call Taylor sin x = 1. Briefly series in Chapter 8. Start with the limit lim x→0 x explain why this means that for x close to 0, sin x ≈ x. Graph y = sin x and y = x to see why this is true. If you look far enough away from x = 0, the graph of y = sin x eventually curves noticeably. Let’s see if we can find polynomials of higher order to match this curving. Show that sin x − x lim = 0. This means that sin x − x ≈ 0 or (again) x→0 x2 sin x − x 1 = − . This says that if x is sin x ≈ x. Show that lim x→0 x3 6 1 1 close to 0, then sin x − x ≈ − x 3 or sin x ≈ x − x 3 . Graph 6 6 these two functions to see how well they match up. To consin x − (x − x 3 /6) sin x − f (x) tinue, compute lim and lim 4 x→0 x→0 x x5 for the appropriate approximation f (x). At this point, look at the pattern of terms you have (Hint: 6 = 3! and 120 = 5!). Using this pattern, approximate sin x with an 11th-degree polynomial and graph the two functions. 2. A zero of a function f (x) is a solution of the equation f (x) = 0. Clearly, not all zeros are created equal. For example, x = 1 is a zero of f (x) = x − 1, but in some ways it seems that x = 1 should count as two zeros of f (x) = (x − 1)2 . To quantify this, we say that x = 1 is a zero of multiplicity 2 of f (x) = (x − 1)2 . The precise definition is: x = c is a zero of f (x) exists and multiplicity n of f (x) if f (c) = 0 and lim x→c (x − c)n is nonzero. Thus, x = 0 is a zero of multiplicity 2 of x sin x x sin x sin x = 1. Find the multiplicity = lim since lim x→0 x→0 x x2 of each zero of the following functions: x 2 sin x, x sin x 2 , x 4 sin x 3 , (x − 1) ln x, ln (x − 1)2 , e x − 1 and cos x − 1.

MAXIMUM AND MINIMUM VALUES To remain competitive in today’s global economy, businesses need to minimize waste and maximize the return on their investment. In the extremely competitive personal computer industry, companies must continually evaluate how low they can afford to set their prices and still earn a profit. With this backdrop, it should be apparent that it is increasingly important to

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SECTION 3.3

like), show that lim f (ω) = 0 for any value of t. (Hint: Find ω→0

a common denominator and use l’Hˆopital’s Rule.) Conclude that this pitch does not move left or right at all. 56. In this exercise, we look at a knuckleball thrown with a different grip than that of exercise 55. The left or right position (in feet) of a ball thrown with spin rate ω and this new grip at time t seconds is f (ω) = (2.5/4ω2 ) − (2.5/4ω2 ) sin (4ωt + π/2). Treating t as a constant and ω as the variable (change to x if you like), find lim f (ω). Your answer should depend on t. By ω→0

graphing this function of t, you can see the path of the pitch (use a domain of 0 ≤ t ≤ 0.68). Describe this pitch. 57. Find functions f such that lim f (x) has the indeterminate x→∞

form ∞ , but where the limit (a) does not exist; (b) equals 0; ∞ (c) equals 3 and (d) equals −4. 58. Find functions f such that lim f (x) has the indeterminate x→∞

form ∞ − ∞, but where the limit (a) does not exist; (b) equals 0 and (c) equals 2. 59. In the figure shown here, a section of the unit circle is determined by angle θ . Region 1 is the triangle ABC. Region 2 is bounded by the line segments AB and BC and the arc of the circle. As the angle θ decreases, the difference between the two regions decreases, also. You might expect that the areas of the regions become nearly equal, in which case the ratio of the areas approaches 1. To see what really happens, show that the area of region 1 divided by the area of region 2 equals sin θ − 12 sin 2θ (1 − cos θ ) sin θ = and find the limit of this θ − cos θ sin θ θ − 12 sin 2θ expression as θ → 0. Surprise! y 1

A

θ

C B

Exercise 59

3.3

1

x

..

Maximum and Minimum Values

265

60. The size of an animal’s pupils expand and contract depending 160x −0.4 + 90 on the amount of light available. Let f (x) = be 8x −0.4 + 10 the size in mm of the pupils at light intensity x. Find lim f (x) x→0+

and lim f (x), and argue that these represent the largest and x→∞

smallest possible sizes of the pupils, respectively.

EXPLORATORY EXERCISES 1. In this exercise, you take a quick look at what we call Taylor sin x = 1. Briefly series in Chapter 8. Start with the limit lim x→0 x explain why this means that for x close to 0, sin x ≈ x. Graph y = sin x and y = x to see why this is true. If you look far enough away from x = 0, the graph of y = sin x eventually curves noticeably. Let’s see if we can find polynomials of higher order to match this curving. Show that sin x − x lim = 0. This means that sin x − x ≈ 0 or (again) x→0 x2 sin x − x 1 = − . This says that if x is sin x ≈ x. Show that lim x→0 x3 6 1 1 close to 0, then sin x − x ≈ − x 3 or sin x ≈ x − x 3 . Graph 6 6 these two functions to see how well they match up. To consin x − (x − x 3 /6) sin x − f (x) tinue, compute lim and lim 4 x→0 x→0 x x5 for the appropriate approximation f (x). At this point, look at the pattern of terms you have (Hint: 6 = 3! and 120 = 5!). Using this pattern, approximate sin x with an 11th-degree polynomial and graph the two functions. 2. A zero of a function f (x) is a solution of the equation f (x) = 0. Clearly, not all zeros are created equal. For example, x = 1 is a zero of f (x) = x − 1, but in some ways it seems that x = 1 should count as two zeros of f (x) = (x − 1)2 . To quantify this, we say that x = 1 is a zero of multiplicity 2 of f (x) = (x − 1)2 . The precise definition is: x = c is a zero of f (x) exists and multiplicity n of f (x) if f (c) = 0 and lim x→c (x − c)n is nonzero. Thus, x = 0 is a zero of multiplicity 2 of x sin x x sin x sin x = 1. Find the multiplicity = lim since lim x→0 x→0 x x2 of each zero of the following functions: x 2 sin x, x sin x 2 , x 4 sin x 3 , (x − 1) ln x, ln (x − 1)2 , e x − 1 and cos x − 1.

MAXIMUM AND MINIMUM VALUES To remain competitive in today’s global economy, businesses need to minimize waste and maximize the return on their investment. In the extremely competitive personal computer industry, companies must continually evaluate how low they can afford to set their prices and still earn a profit. With this backdrop, it should be apparent that it is increasingly important to

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use mathematical methods to maximize and minimize various quantities of interest. In this section, we investigate the notion of maximum and minimum from a purely mathematical standpoint. In section 3.7, we examine how to apply these notions to problems of an applied nature. We begin by giving careful mathematical definitions of some familiar terms.

DEFINITION 3.1 For a function f defined on a set S of real numbers and a number c ∈ S, (i) f (c) is the absolute maximum of f on S if f (c) ≥ f (x) for all x ∈ S and (ii) f (c) is the absolute minimum of f on S if f (c) ≤ f (x) for all x ∈ S. An absolute maximum or an absolute minimum is referred to as an absolute extremum. If a function has more than one extremum, we refer to these as extrema (the plural form of extremum).

The first question you might ask is whether every function has an absolute maximum and an absolute minimum. The answer is no, as we can see from Figures 3.25a and 3.25b. y y

Has no absolute maximum

Absolute maximum f (c)

c

x c

Absolute minimum

f (c)

FIGURE 3.25a

EXAMPLE 3.1

x

Has no absolute minimum

FIGURE 3.25b

Absolute Maximum and Minimum Values

(a) Locate any absolute extrema of f (x) = x 2 − 9 on the interval (−∞, ∞). (b) Locate any absolute extrema of f (x) = x 2 − 9 on the interval (−3, 3). (c) Locate any absolute extrema of f (x) = x 2 − 9 on the interval [−3, 3]. Solution (a) In Figure 3.26, notice that f has an absolute minimum value of f (0) = −9, but has no absolute maximum value. (b) In Figure 3.27a, we see that f has an absolute minimum value of f (0) = −9, but still has no absolute maximum value. Your initial reaction might be to say that f has an absolute maximum of 0, but f (x) = 0 for any x ∈ (−3, 3), since this is an open interval and hence, does not include the endpoints −3 and 3.

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SECTION 3.3

Maximum and Minimum Values

y

No absolute maximum

y

..

y No absolute maximum 3

3

267

x

Absolute maximum f (3) f (3) 0 3

3

x

x

3

3 f (0) 9 (Absolute minimum) f(0) 9 (Absolute minimum)

FIGURE 3.26

f(0) 9 (Absolute minimum)

FIGURE 3.27a

FIGURE 3.27b

y = x 2 − 9 on (−3, 3)

y = x 2 − 9 on [−3, 3]

(c) In this case, the endpoints 3 and −3 are in the interval [−3, 3]. Here, f assumes its absolute maximum at two points: f (3) = f (−3) = 0 (see Figure 3.27b).

y = x 2 − 9 on (−∞, ∞)

We have seen that a function may or may not have absolute extrema, depending on the interval on which we’re looking. In example 3.1, the function failed to have an absolute maximum, except on the closed, bounded interval [−3, 3]. This provides some clues, and example 3.2 provides another piece of the puzzle.

EXAMPLE 3.2 y

A Function with No Absolute Maximum or Minimum

Locate any absolute extrema of f (x) = 1/x, on the interval [−3, 3]. Solution From the graph in Figure 3.28, f clearly fails to have either an absolute maximum or an absolute minimum on [−3, 3]. The following table of values for f (x) for x close to 0 suggests the same conclusion.

2 3

x

x 1 0.1 0.01 0.001 0.0001 0.00001 0.000001

3

FIGURE 3.28 y = 1/x

1/x 1 10 100 1000 10,000 100,000 1,000,000

x −1 −0.1 −0.01 −0.001 −0.0001 −0.00001 −0.000001

1/x −1 −10 −100 −1000 −10,000 −100,000 −1,000,000

The most obvious difference between the functions in examples 3.1 and 3.2 is that f (x) = 1/x is discontinuous at a point in the interval [−3, 3]. We offer the following theorem without proof.

THEOREM 3.1 (Extreme Value Theorem) A continuous function f defined on a closed, bounded interval [a, b] attains both an absolute maximum and an absolute minimum on that interval.

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While you do not need to have a continuous function or a closed interval to have an absolute extremum, Theorem 3.1 says that continuous functions are guaranteed to have an absolute maximum and an absolute minimum on a closed, bounded interval. In example 3.3, we revisit the function from example 3.2, but look on a different interval. y

EXAMPLE 3.3

Finding Absolute Extrema of a Continuous Function

Find the absolute extrema of f (x) = 1/x on the interval [1, 3].

1

x 1

3

Solution Notice that on the interval [1, 3], f is continuous. Consequently, the Extreme Value Theorem guarantees that f has both an absolute maximum and an absolute minimum on [1, 3]. Judging from the graph in Figure 3.29, it appears that f (x) reaches its maximum value of 1 at x = 1 and its minimum value of 1/3 at x = 3.

FIGURE 3.29 y = 1/x on [1, 3]

Our objective is to determine how to locate the absolute extrema of a given function. Before we do this, we need to consider an additional type of extremum.

DEFINITION 3.2 (i) f (c) is a local maximum of f if f (c) ≥ f (x) for all x in some open interval containing c. (ii) f (c) is a local minimum of f if f (c) ≤ f (x) for all x in some open interval containing c. In either case, we call f (c) a local extremum of f . Local maxima and minima (the plural forms of maximum and minimum, respectively) are sometimes referred to as relative maxima and minima, respectively. Notice from Figure 3.30 that each local extremum seems to occur either at a point where the tangent line is horizontal [i.e., where f (x) = 0], at a point where the tangent line is vertical [where f (x) is undefined] or at a corner [again, where f (x) is undefined]. We can see this behavior quite clearly in examples 3.4 and 3.5. y Local maximum [ f (d) is undefined] Local maximum [ f (b) 0]

a

c b

x d

Local minimum [ f (a) 0] Local minimum [ f (c) is undefined]

FIGURE 3.30 Local extrema

3-29

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..

Maximum and Minimum Values

269

y

EXAMPLE 3.4

A Function with a Zero Derivative at a Local Maximum

Locate any local extrema for f (x) = 9 − x 2 and describe the behavior of the derivative at the local extremum.

5 x

2

2

Solution We can see from Figure 3.31 that there is a local maximum at x = 0. Further, note that f (x) = −2x and so, f (0) = 0. Note that this says that the tangent line to y = f (x) at x = 0 is horizontal, as indicated in Figure 3.31.

10

EXAMPLE 3.5 FIGURE 3.31 y = 9 − x 2 and the tangent line at x = 0 y

A Function with an Undefined Derivative at a Local Minimum

Locate any local extrema for f (x) = |x| and describe the behavior of the derivative at the local extremum. Solution We can see from Figure 3.32 that there is a local minimum at x = 0. As we have noted in section 2.1, the graph has a corner at x = 0 and hence, f (0) is undefined. [See example 1.7 in Chapter 2.]

3

x 2

FIGURE 3.32

The graphs shown in Figures 3.30–3.32 are not unusual. Here is a small challenge: spend a little time now drawing graphs of functions with local extrema. It should not take long to convince yourself that local extrema occur only at points where the derivative is either zero or undefined. Because of this, we give these points a special name.

y = |x|

DEFINITION 3.3 A number c in the domain of a function f is called a critical number of f if f (c) = 0 or f (c) is undefined.

HISTORICAL NOTES Pierre de Fermat (1601–1665) A French mathematician who discovered many important results, including the theorem named for him. Fermat was a lawyer and member of the Toulouse supreme court, with mathematics as a hobby. The “Prince of Amateurs” left an unusual legacy by writing in the margin of a book that he had discovered a wonderful proof of a clever result, but that the margin of the book was too small to hold the proof. Fermat’s Last Theorem confounded many of the world’s best mathematicians for more than 300 years before being proved by Andrew Wiles in 1995.

It turns out that our earlier observation regarding the location of extrema is correct. That is, local extrema occur only at points where the derivative is zero or undefined. We state this formally in Theorem 3.2.

THEOREM 3.2 (Fermat’s Theorem) Suppose that f (c) is a local extremum (local maximum or local minimum). Then c must be a critical number of f .

PROOF Suppose that f is differentiable at x = c. (If not, c is a critical number of f and we are done.) Suppose further that f (c) = 0. Then, either f (c) > 0 or f (c) < 0. If f (c) > 0, we have by the definition of derivative that f (c + h) − f (c) f (c) = lim > 0. h→0 h So, for all h sufficiently small, f (c + h) − f (c) > 0. (3.1) h

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For h > 0, (3.1) says that

f (c + h) − f (c) > 0 f (c + h) > f (c).

and so, Thus, f (c) is not a local maximum. For h < 0, (3.1) says that

TODAY IN MATHEMATICS Andrew Wiles (1953– ) A British mathematician who in 1995 published a proof of Fermat’s Last Theorem, the most famous unsolved problem of the 20th century. Fermat’s Last Theorem states that there is no integer solution x , y and z of the equation x n + y n = zn for integers n > 2. Wiles had wanted to prove the theorem since reading about it as a 10-year-old. After more than ten years as a successful research mathematician, Wiles isolated himself from colleagues for seven years as he developed the mathematics needed for his proof. “I realised that talking to people casually about Fermat was impossible because it generated too much interest. You cannot focus yourself for years unless you have this kind of undivided concentration which too many spectators would destroy.” The last step of his proof came, after a year of intense work on this one step, as “this incredible revelation” that was “so indescribably beautiful, it was so simple and elegant.”

f (c + h) − f (c) < 0 and so,

f (c + h) < f (c).

Thus, f (c) is not a local minimum. Since we had assumed that f (c) was a local extremum, this is a contradiction. This rules out the possibility that f (c) > 0. Similarly, if f (c) < 0, we obtain the same contradiction. This is left as an exercise. The only remaining possibility is to have f (c) = 0 and this proves the theorem.

We can use Fermat’s Theorem and calculator- or computer-generated graphs to find local extrema, as in examples 3.6 and 3.7.

EXAMPLE 3.6

Finding Local Extrema of a Polynomial

Find the critical numbers and local extrema of f (x) = 2x 3 − 3x 2 − 12x + 5.

y

20

1

x 2

20

FIGURE 3.33 y = 2x 3 − 3x 2 − 12x + 5

Solution Here,

f (x) = 6x 2 − 6x − 12 = 6(x 2 − x − 2) = 6(x − 2)(x + 1).

Thus, f has two critical numbers, x = −1 and x = 2. Notice from the graph in Figure 3.33 that these correspond to the locations of a local maximum and a local minimum, respectively.

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Maximum and Minimum Values

271

y

EXAMPLE 3.7

An Extremum at a Point Where the Derivative Is Undefined

Find the critical numbers and local extrema of f (x) = (3x + 1)2/3 .

4

Solution Here, we have f (x) = x

2

2 2 (3x + 1)−1/3 (3) = . 3 (3x + 1)1/3

Of course, f (x) = 0 for all x, but f (x) is undefined at x = − 13 . Be sure to note that − 13 is in the domain of f . Thus, x = − 13 is the only critical number of f . From the graph in Figure 3.34, we see that this corresponds to the location of a local minimum (also the absolute minimum). If you use your graphing utility to try to produce a graph of y = f (x), you may get only half of the graph displayed in Figure 3.34. The reason is that the algorithms used by most calculators and many computers will return a complex number (or an error) when asked to compute certain fractional powers of negative numbers. While this annoying shortcoming presents only occasional difficulties, we mention this here only so that you are aware that technology has limitations.

2

FIGURE 3.34 y = (3x + 1)2/3

REMARK 3.1 Fermat’s Theorem says that local extrema can occur only at critical numbers. This does not say that there is a local extremum at every critical number. In fact, this is false, as we illustrate in examples 3.8 and 3.9.

EXAMPLE 3.8

A Horizontal Tangent at a Point That Is Not a Local Extremum

Find the critical numbers and local extrema of f (x) = x 3 .

y

Solution It should be clear from Figure 3.35 that f has no local extrema. However, f (x) = 3x 2 = 0 for x = 0 (the only critical number of f ). In this case, f has a horizontal tangent line at x = 0, but does not have a local extremum there.

2 x

2

EXAMPLE 3.9

2 2

A Vertical Tangent at a Point That Is Not a Local Extremum

Find the critical numbers and local extrema of f (x) = x 1/3 . Solution As in example 3.8, f has no local extrema (see Figure 3.36). Here, f (x) = 13 x −2/3 and so, f has a critical number at x = 0 (in this case the derivative is undefined at x = 0). However, f does not have a local extremum at x = 0.

FIGURE 3.35 y = x3 y

You should always check that a given value is in the domain of the function before declaring it a critical number, as in example 3.10.

2 x

2

2 2

FIGURE 3.36 y = x 1/3

EXAMPLE 3.10

Finding Critical Numbers of a Rational Function

2x 2 . x +2 Solution You should note that the domain of f consists of all real numbers other than x = −2. Here, we have

Find all the critical numbers of f (x) =

4x(x + 2) − 2x 2 (1) (x + 2)2 2x(x + 4) = . (x + 2)2

f (x) =

From the quotient rule.

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Notice that f (x) = 0 for x = 0, −4 and f (x) is undefined for x = −2. However, −2 is not in the domain of f and consequently, the only critical numbers are x = 0 and x = −4.

REMARK 3.2

We have observed that local extrema occur only at critical numbers and that continuous functions must have an absolute maximum and an absolute minimum on a closed, bounded interval. However, we haven’t yet really been able to say how to find these extrema. Theorem 3.3 is particularly useful.

When we use the terms maximum, minimum or extremum without specifying absolute or local, we will always be referring to absolute extrema.

THEOREM 3.3 Suppose that f is continuous on the closed interval [a, b]. Then, the absolute extrema of f must occur at an endpoint (a or b) or at a critical number.

PROOF First, recall that by the Extreme Value Theorem, f will attain its maximum and minimum values on [a, b], since f is continuous. Let f (c) be an absolute extremum. If c is not an endpoint (i.e., c = a and c = b), then c must be in the open interval (a, b). Thus, f (c) must be a local extremum, also. Finally, by Fermat’s Theorem, c must be a critical number, since local extrema occur only at critical numbers.

REMARK 3.3 Theorem 3.3 gives us a simple procedure for finding the absolute extrema of a continuous function on a closed, bounded interval: 1. Find all critical numbers in the interval and compute function values at these points. 2. Compute function values at the endpoints. 3. The largest function value is the absolute maximum and the smallest function value is the absolute minimum. We illustrate Theorem 3.3 for the case of a polynomial function in example 3.11.

EXAMPLE 3.11

Finding Absolute Extrema on a Closed Interval

Find the absolute extrema of f (x) = 2x 3 − 3x 2 − 12x + 5 on the interval [−2, 4].

y

Solution From the graph in Figure 3.37, the maximum appears to be at the endpoint x = 4, while the minimum appears to be at a local minimum near x = 2. In example 3.6, we found that the critical numbers of f are x = −1 and x = 2. Further, both of these are in the interval [−2, 4]. So, we compare the values at the endpoints:

40

20

f (−2) = 1 x

2

2

4

20

FIGURE 3.37 y = 2x 3 − 3x 2 − 12x + 5

and

f (4) = 37,

and the values at the critical numbers: f (−1) = 12 and f (2) = −15. Since f is continuous on [−2, 4], Theorem 3.3 says that the absolute extrema must be among these four values. Thus, f (4) = 37 is the absolute maximum and f (2) = −15 is the absolute minimum. Note that these values are consistent with what we see in the graph in Figure 3.37.

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273

Of course, most real problems of interest are unlikely to result in derivatives with integer zeros. Consider the following somewhat less user-friendly example. y

EXAMPLE 3.12

Finding Extrema for a Function with Fractional Exponents

Find the absolute extrema of f (x) = 4x 5/4 − 8x 1/4 on the interval [0, 4].

10

x 2

4

Solution First, we draw a graph of the function to get an idea of where the extrema are located (see Figure 3.38). From the graph, it appears that the maximum occurs at the endpoint x = 4 and the minimum near x = 12 . Next, observe that 5x − 2 . x 3/4

Thus, the critical numbers are x = 25 since f 25 = 0 and x = 0 (since f (0) is undefined and 0 is in the domain of f ). We now need only compare f (0) = 0, f (4) ≈ 11.3137 and f 25 ≈ −5.0897.

5

f (x) = 5x 1/4 − 2x −3/4 =

FIGURE 3.38 y = 4x 5/4 − 8x 1/4

So, the absolute maximum is f (4) ≈ 11.3137 and the absolute minimum is f 25 ≈ −5.0897, which is consistent with what we expected from Figure 3.38.

y 8

In practice, the critical numbers are not always as easy to find as they were in examples 3.11 and 3.12. In example 3.13, it is not even known how many critical numbers there are. We can, however, estimate the number and locations of these from a careful analysis of computer-generated graphs. x

2

3

EXAMPLE 3.13

Finding Absolute Extrema Approximately

Find the absolute extrema of f (x) = x 3 − 5x + 3 sin x 2 on the interval [ −2, 2.5]. 4

Solution We first draw a graph to get an idea of where the extrema will be located (see Figure 3.39). From the graph, we can see that the maximum seems to occur near x = −1, while the minimum seems to occur near x = 2. Next, we compute

FIGURE 3.39 y = f (x) = x 3 − 5x + 3 sin x 2

f (x) = 3x 2 − 5 + 6x cos x 2 .

y

20

x

2

3 10

a ≈ −1.26410884789,

FIGURE 3.40

y = f (x) = 3x − 5 + 6x cos x 2

Unlike examples 3.11 and 3.12, there is no algebra we can use to find the zeros of f . Our only alternative is to find the zeros approximately. You can do this by using Newton’s method to solve f (x) = 0. (You can also use any other rootfinding method built into your calculator or computer.) First, we’ll need adequate initial guesses. We obtain these from the graph of y = f (x) found in Figure 3.40. From the graph, it appears that there are four zeros of f (x) on the interval in question, located near x = −1.3, 0.7, 1.2 and 2.0. Further, referring back to Figure 3.39, these four zeros correspond with the four local extrema seen in the graph of y = f (x). We now apply Newton’s method to solve f (x) = 0, using the preceding four values as our initial guesses. This leads us to four approximate critical numbers of f on the interval [−2, 2.5]. We have

2

c ≈ 1.2266828947 and

b ≈ 0.674471354085, d ≈ 2.01830371473.

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We now need only compare the values of f at the endpoints and the approximate critical numbers: f (a) ≈ 7.3, f (d) ≈ −4.3,

f (b) ≈ −1.7, f (−2) ≈ −0.3

f (c) ≈ −1.3 f (2.5) ≈ 3.0.

and

Thus, the absolute maximum is approximately f (−1.26410884789) ≈ 7.3 and the absolute minimum is approximately f (2.01830371473) ≈ −4.3. It is important (especially in light of how much of our work here was approximate and graphical) to verify that the approximate extrema correspond with what we expect from the graph of y = f (x). Since these correspond closely, we have great confidence in their accuracy. We have now seen how to locate the absolute extrema of a continuous function on a closed interval. In section 3.4, we see how to find local extrema.

BEYOND FORMULAS The Extreme Value Theorem is an important but subtle result. Think of it this way. If the hypotheses of the theorem are met, you will never waste your time looking for the maximum of a function that does not have a maximum. That is, the problem is always solvable. The technique described in Remark 3.3 always works. If you are asked to find a novel with a certain plot, does it help to know that there actually is such a novel?

EXERCISES 3.3 WRITING EXERCISES 1. Using one or more graphs, explain why the Extreme Value Theorem is true. Is the conclusion true if we drop the hypothesis that f is a continuous function? Is the conclusion true if we drop the hypothesis that the interval is closed? 2. Using one or more graphs, argue that Fermat’s Theorem is true. Discuss how Fermat’s Theorem is used. Restate the theorem in your own words to make its use clearer. 3. Suppose that f (t) represents your elevation after t hours on a mountain hike. If you stop to rest, explain why f (t) = 0. Discuss the circumstances under which you would be at a local maximum, local minimum or neither.

converse always true? Sometimes true? Apply this logic to both the Extreme Value Theorem and Fermat’s Theorem: state the converse and decide whether it is sometimes true or always true.

In exercises 1–4, use the graph to locate the absolute extrema (if they exist) of the function on the given interval. 1. f (x) =

1 on (a) (−∞, ∞), (b) [−1, 1] and (c) (0, 1) x2 − 1 y 10

4. Mathematically, an if/then statement is usually strictly onedirectional. When we say “If A, then B” it is generally not the case that “If B, then A” is also true: when both are true, we say “A if and only if B,” which is abbreviated to “A iff B.” Unfortunately, common English usage is not always this precise. This occasionally causes a problem interpreting a mathematical theorem. To get this straight, consider the statement, “If you wrote a best-selling book, then you made a lot of money.” Is this true? How does this differ from its converse, “If you made a lot of money, then you wrote a best-selling book.” Is the

5

4

x

2

2 5 10

4

3-35

SECTION 3.3

2. f (x) =

x2 on (a) (−∞, ∞), (b) [−1, 1] and (c) (0, 1) (x − 1)2 y 10 8

..

Maximum and Minimum Values

In exercises 11–30, find all critical numbers by hand. If available, use graphing technology to determine whether the critical number represents a local maximum, local minimum or neither. 11. f (x) = x 4 − 3x 3 + 2

12. f (x) = x 4 + 6x 2 − 2

13. f (x) = x 3/4 − 4x 1/4

14. f (x) = (x 2/5 − 3x 1/5 )2 √ 16. f (x) = 3 sin x + cos x

15. f (x) = sin x cos x, [0, 2π] 6

x2 − 2 x +2 x 19. f (x) = 2 x +1

4 2 4

x

2

2

4

3. f (x) = sin x on (a) (−∞, ∞), (b) 0, π4 and (c) π4 , 3π4 y 1 0.5

10

x

5

5

10

0.5

x2 − x + 4 x −1 3x 20. f (x) = 2 x −1

17. f (x) =

18. f (x) =

21. f (x) = 12 (e x + e−x )

22. f (x) = xe−2x

23. f (x) = x 4/3 + 4x 1/3 + 4x −2/3 √ 25. f (x) = 2x x + 1

24. f (x) = x 7/3 − 28x 1/3 √ 26. f (x) = x/ x 2 + 1

27. f (x) = |x 2 − 1| √ 28. f (x) = 3 x 3 − 3x 2 + 2x x 2 + 2x − 1 if x < 0 29. f (x) = x 2 − 4x + 3 if x ≥ 0 sin x if −π < x < π 30. f (x) = − tan x if |x| ≥ π In exercises 31–38, find the absolute extrema of the given function on each indicated interval.

1

4. f (x) = x 3 − 3x + 1 on (a) (−∞, ∞), (b) [−2, 2] and (c) (0, 2) y

35. f (x) = e−x on (a) [0, 2] and (b) [−3, 2] 2

x 2

4

5 10

36. f (x) = x 2 e−4x on (a) [−2, 0] and (b) [0, 4] 3x 2 37. f (x) = on (a) [−2, 2] and (b) [2, 8] x −3 38. f (x) = tan−1 (x 2 ) on (a) [0, 1] and (b) [−3, 4] In exercises 39–44, numerically estimate the absolute extrema of the given function on the indicated intervals.

In exercises 5–10, find all critical numbers by hand. Use your knowledge of the type of graph (i.e., parabola or cubic) to determine whether the critical number represents a local maximum, local minimum or neither. 5. f (x) = x 2 + 5x − 1

6. f (x) = −x 2 + 4x + 2

7. f (x) = x 3 − 3x + 1

8. f (x) = −x 3 + 6x 2 + 2

9. f (x) = x 3 − 3x 2 + 6x

32. f (x) = x 4 − 8x 2 + 2 on (a) [−3, 1] and (b) [−1, 3]

34. f (x) = sin x + cos x on (a) [0, 2π] and (b) [π/2, π ]

5

2

31. f (x) = x 3 − 3x + 1 on (a) [0, 2] and (b) [−3, 2]

33. f (x) = x 2/3 on (a) [−4, −2] and (b) [−1, 3]

10

4

275

10. f (x) = x 3 − 3x 2 + 3x

39. f (x) = x 4 − 3x 2 + 2x + 1 on (a) [−1, 1] and (b) [−3, 2] 40. f (x) = x 6 − 3x 4 − 2x + 1 on (a) [−1, 1] and (b) [−2, 2] 41. f (x) = x 2 − 3x cos x on (a) [−2, 1] and (b) [−5, 0] 42. f (x) = xecos 2x on (a) [−2, 2] and (b) [2, 5]

43. f (x) = x sin x + 3 on (a) −π , π2 and (b) [0, 2π ] 2 44. f (x) = x 2 + e x on (a) [0, 1] and (b) [−2, 2]

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45. Repeat exercises 31–38, except instead of finding extrema on the closed interval, find the extrema on the open interval, if they exist. 46. Briefly outline a procedure for finding extrema on an open interval (a, b), a procedure for the half-open interval (a, b] and a procedure for the half-open interval [a, b). 47. Sketch a graph of a function f such that the absolute maximum of f (x) on the interval [−2, 2] equals 3 and the absolute minimum does not exist. 48. Sketch a graph of a continuous function f such that the absolute maximum of f (x) on the interval (−2, 2) does not exist and the absolute minimum equals 2. 49. Sketch a graph of a continuous function f such that the absolute maximum of f (x) on the interval (−2, 2) equals 4 and the absolute minimum equals 2.

3-36

59. If two soccer teams each score goals at a rate of r goals per minute, the probability that n goals will be scored in t minutes is n 1 P = (rn!t) e−r t . Take r = 25 . Show that for a 90-minute game, P is maximized with n = 3. Briefly explain why this makes sense. 60. In the situation of exercise 59, find t to maximize the probability that exactly 1 goal has been scored. Briefly explain why your answer makes sense. 61. If f is differentiable on the interval [a, b] and f (a) < 0 < f (b), prove that there is a c with a < c < b for which f (c) = 0. (Hint: Use the Extreme Value Theorem and Fermat’s Theorem.) 62. Sketch a graph showing that y = f (x) = x 2 + 1 and y = g(x) = ln x do not intersect. Find x to minimize f (x) − g(x). At this value of x, show that the tangent lines to y = f (x) and y = g(x) are parallel. Explain graphically why it makes sense that the tangent lines are parallel. x2 for x > 0 and determine x2 + 1 where the graph is steepest. (That is, find where the slope is a maximum.)

50. Sketch a graph of a function f such that the absolute maximum of f (x) on the interval [−2, 2] does not exist and the absolute minimum does not exist.

63. Sketch a graph of f (x) =

51. Give an example showing that the following statement is false (not always true): between any two local minima of f (x) there is a local maximum.

64. Sketch a graph of f (x) = e−x and determine where the graph is steepest. (Note: This is an important problem in probability theory.)

52. If you have won three out of four matches against someone, does that mean that the probability that you will win the next one is 34 ? In general, if you have a probability p of winning each match, the probability of winning m out of n matches n! p m (1 − p)n−m . Find p to maximize f . is f ( p) = (n − m)! m! This value of p is called the maximum likelihood estimator of the probability. Briefly explain why your answer makes sense. 53. In this exercise, we will explore the family of functions f (x) = x 3 + cx + 1, where c is constant. How many and what types of local extrema are there? (Your answer will depend on the value of c.) Assuming that this family is indicative of all cubic functions, list all types of cubic functions. 54. Prove that any fourth-order polynomial must have at least one local extremum and can have a maximum of three local extrema. Based on this information, sketch several possible graphs of fourth-order polynomials. 55. Show that f (x) = x 3 + bx 2 + cx + d has both a local maximum and a local minimum if c < 0.

2

65. A section of roller coaster is in the shape of y = x 5 − 4x 3 − x + 10, where x is between −2 and 2. Find all local extrema and explain what portions of the roller coaster they represent. Find the location of the steepest part of the roller coaster. 66. Suppose a large computer file is sent over the Internet. If the probability that it reaches its destination without any errors is x, then the probability that an error is made is 1 − x. The field of information theory studies such situations. An important quantity is entropy (a measure of unpredictability), defined by H = −x ln x − (1 − x) ln(1 − x), for 0 < x < 1. Find the value of x that maximizes this quantity. Explain why this probability would maximize the measure of unpredictability of errors. (Hint: If x = 0 or x = 1, are errors unpredictable?) 67. Researchers in a number of fields (including population biology, economics and the study of animal tumors) make use of −t the Gompertz growth curve, W (t) = ae−be . As t → ∞, show that W (t) → a and W (t) → 0. Find the maximum growth rate.

57. For the family of functions f (x) = x 4 + cx 2 + 1, find all local extrema (your answer will depend on the value of the constant c).

68. The rate R of an enzymatic reaction as a function of the sub[S]Rm strate concentration [S] is given by R = , where Rm K m + [S] and K m are constants. K m is called the Michaelis constant and Rm is referred to as the maximum reaction rate. Show that Rm is not a proper maximum in that the reaction rate can never be equal to Rm .

58. For the family of functions f (x) = x 4 + cx 3 + 1, find all local extrema. (Your answer will depend on the value of the constant c.)

69. Suppose a painting hangs on a wall as in the figure. The frame extends from 6 feet to 8 feet above the floor. A person whose eyes are 5 feet above the ground stands x feet from the wall

56. In exercise 55, show that the sum of the critical numbers is − 2b3 .

3-37

SECTION 3.4

and views the painting, with a viewing angle A formed by the ray from the person’s eye to the top of the frame and the ray from the person’s eye to the bottom of the bottom of the frame. Find the value of x that maximizes the viewing angle A. 8'

6' A

5'

..

Increasing and Decreasing Functions

277

and h = 2x so V = 2π y 2 x. Call the rod measurement z. By the Pythagorean Theorem, x 2 + (2y)2 = z 2 . Kepler’s mystery was how to compute V given only z. The key observation made by Kepler was that Austrian wine casks were made with the same height-to-diameter ratio (for us, x/y). Let t = x/y 2 and show that z 2 /y 2 = t 2 + 4. Use this to replace y in the 2 2 volume formula. Then replace x with z + 4y . Show that 2π z 3 t . In this formula, t is a constant, so the vintV = (4 + t 2 )3/2 ner could measure z and quickly estimate the volume. We haven’t told you yet what t equals. Kepler assumed that the vintners would have made a smart choice for this ratio. Find the value of t that maximizes the volume for a given z. This is, in fact, the ratio used in the construction of Austrian wine casks!

x

70. What changes in exercise 69 if the person’s eyes are 6 feet above the ground? z

EXPLORATORY EXERCISES 1. Explore the graphs of e−x , xe−x , x 2 e−x and x 3 e−x . Find all local extrema and use l’Hˆopital’s Rule to determine the behavior as x → ∞. You can think of the graph of x n e−x as showing the results of a tug-of-war: x n → ∞ as x → ∞ but e−x → 0 as x → ∞. Describe the graph of x n e−x in terms of this tug-of-war. 2. Johannes Kepler (1571–1630) is best known as an astronomer, especially for his three laws of planetary motion. However, his discoveries were primarily due to his brilliance as a mathematician. While serving in Austrian Emperor Matthew I’s court, Kepler observed the ability of Austrian vintners to quickly and mysteriously compute the capacities of a variety of wine casks. Each cask (barrel) had a hole in the middle of its side (see Figure a). The vintner would insert a rod in the hole until it hit the far corner and then announce the volume. Kepler first analyzed the problem for a cylindrical barrel (see Figure b). The volume of a cylinder is V = πr 2 h. In Figure b, r = y

3.4

FIGURE a

z

2y 2x

FIGURE b 3. Suppose that a hockey player is shooting at a 6-foot-wide net from a distance of d feet away from the goal line and 4 feet to the side of the center line. (a) Find the distance d that maximizes the shooting angle. (b) Repeat part (a) with the shooter 2 feet to the side of the center line. Explain why the answer is so different. (c) Repeat part (a) with the goalie blocking all but the far 2 feet of the goal.

INCREASING AND DECREASING FUNCTIONS In section 3.3, we determined that local extrema occur only at critical numbers. However, not all critical numbers correspond to local extrema. In this section, we see how to determine which critical numbers correspond to local extrema. At the same time, we’ll learn more about the connection between the derivative and graphing.

3-37

SECTION 3.4

and views the painting, with a viewing angle A formed by the ray from the person’s eye to the top of the frame and the ray from the person’s eye to the bottom of the bottom of the frame. Find the value of x that maximizes the viewing angle A. 8'

6' A

5'

..

Increasing and Decreasing Functions

277

and h = 2x so V = 2π y 2 x. Call the rod measurement z. By the Pythagorean Theorem, x 2 + (2y)2 = z 2 . Kepler’s mystery was how to compute V given only z. The key observation made by Kepler was that Austrian wine casks were made with the same height-to-diameter ratio (for us, x/y). Let t = x/y 2 and show that z 2 /y 2 = t 2 + 4. Use this to replace y in the 2 2 volume formula. Then replace x with z + 4y . Show that 2π z 3 t . In this formula, t is a constant, so the vintV = (4 + t 2 )3/2 ner could measure z and quickly estimate the volume. We haven’t told you yet what t equals. Kepler assumed that the vintners would have made a smart choice for this ratio. Find the value of t that maximizes the volume for a given z. This is, in fact, the ratio used in the construction of Austrian wine casks!

x

70. What changes in exercise 69 if the person’s eyes are 6 feet above the ground? z

EXPLORATORY EXERCISES 1. Explore the graphs of e−x , xe−x , x 2 e−x and x 3 e−x . Find all local extrema and use l’Hˆopital’s Rule to determine the behavior as x → ∞. You can think of the graph of x n e−x as showing the results of a tug-of-war: x n → ∞ as x → ∞ but e−x → 0 as x → ∞. Describe the graph of x n e−x in terms of this tug-of-war. 2. Johannes Kepler (1571–1630) is best known as an astronomer, especially for his three laws of planetary motion. However, his discoveries were primarily due to his brilliance as a mathematician. While serving in Austrian Emperor Matthew I’s court, Kepler observed the ability of Austrian vintners to quickly and mysteriously compute the capacities of a variety of wine casks. Each cask (barrel) had a hole in the middle of its side (see Figure a). The vintner would insert a rod in the hole until it hit the far corner and then announce the volume. Kepler first analyzed the problem for a cylindrical barrel (see Figure b). The volume of a cylinder is V = πr 2 h. In Figure b, r = y

3.4

FIGURE a

z

2y 2x

FIGURE b 3. Suppose that a hockey player is shooting at a 6-foot-wide net from a distance of d feet away from the goal line and 4 feet to the side of the center line. (a) Find the distance d that maximizes the shooting angle. (b) Repeat part (a) with the shooter 2 feet to the side of the center line. Explain why the answer is so different. (c) Repeat part (a) with the goalie blocking all but the far 2 feet of the goal.

INCREASING AND DECREASING FUNCTIONS In section 3.3, we determined that local extrema occur only at critical numbers. However, not all critical numbers correspond to local extrema. In this section, we see how to determine which critical numbers correspond to local extrema. At the same time, we’ll learn more about the connection between the derivative and graphing.

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Salary

3-38

We are all familiar with the terms increasing and decreasing. If your employer informs you that your salary will be increasing steadily over the term of your employment, you have in mind that as time goes on, your salary will rise something like Figure 3.41. If you take out a loan to purchase a car, once you start paying back the loan, your indebtedness will decrease over time. If you plotted your debt against time, the graph might look something like Figure 3.42. We now carefully define these notions. Notice that Definition 4.1 is merely a formal statement of something you already understand. Time

DEFINITION 4.1

FIGURE 3.41

A function f is (strictly) increasing on an interval I if for every x1 , x2 ∈ I with x1 < x2 , f (x1 ) < f (x2 ) [i.e., f (x) gets larger as x gets larger]. A function f is (strictly) decreasing on the interval I if for every x1 , x2 ∈ I with x1 < x2 , f (x1 ) > f (x2 ) [i.e., f (x) gets smaller as x gets larger].

Increasing salary

Debt

Time

FIGURE 3.42 Decreasing debt

Why do we bother with such an obvious definition? Of course, anyone can look at a graph of a function and immediately see where that function is increasing and decreasing. The real challenge is to determine where a function is increasing and decreasing, given only a mathematical formula for the function. For example, can you determine where f (x) = x 2 sin x is increasing and decreasing, without looking at a graph? Look carefully at Figure 3.43 to see if you can notice what happens at every point at which the function is increasing or decreasing. f increasing (tangent lines have positive slope)

y

y f (x)

x

f decreasing (tangent lines have negative slope)

FIGURE 3.43 Increasing and decreasing

Observe that on intervals where the tangent lines have positive slope, f is increasing, while on intervals where the tangent lines have negative slope, f is decreasing. Of course, the slope of the tangent line at a point is given by the value of the derivative at that point. So, whether a function is increasing or decreasing on an interval seems to be connected to the sign of its derivative on that interval. This conjecture, although it’s based on only a single picture, sounds like a theorem and it is.

3-39

SECTION 3.4

..

Increasing and Decreasing Functions

279

THEOREM 4.1 Suppose that f is differentiable on an interval I . (i) If f (x) > 0 for all x ∈ I , then f is increasing on I . (ii) If f (x) < 0 for all x ∈ I , then f is decreasing on I .

PROOF (i) Pick any two points x1 , x2 ∈ I , with x1 < x2 . Applying the Mean Value Theorem (Theorem 9.4 in section 2.9) to f on the interval (x1 , x2 ), we get f (x2 ) − f (x1 ) = f (c), x2 − x1

(4.1)

for some number c ∈ (x1 , x2 ). (Why can we apply the Mean Value Theorem here?) By hypothesis, f (c) > 0 and since x1 < x2 (so that x2 − x1 > 0), we have from (4.1) that 0 < f (x2 ) − f (x1 ) f (x1 ) < f (x2 ).

or

(4.2)

Since (4.2) holds for all x1 < x2 , f is increasing on I . The proof of (ii) is nearly identical and is left as an exercise.

What You See May Not Be What You Get One aim here and in sections 3.5 and 3.6 is to learn how to draw representative graphs of functions (i.e., graphs that display all of the significant features of a function: where it is increasing or decreasing, any extrema, asymptotes and two features we’ll introduce in section 3.5, concavity and inflection points). When we draw a graph, we are drawing in a particular viewing window (i.e., a particular range of x- and y-values). In the case of computer- or calculator-generated graphs, the window is often chosen by the machine. So, how do we know when significant features are hidden outside of a given window? Further, how do we determine the precise locations of features that we can see in a given window? As we’ll discover, the only way we can resolve these questions is with some calculus.

EXAMPLE 4.1

y

Draw a graph of f (x) = 2x 3 + 9x 2 − 24x − 10 showing all local extrema.

10

x

10

Drawing a Graph

10 10

FIGURE 3.44 y = 2x 3 + 9x 2 − 24x − 10

Solution Many graphing calculators use the default window defined by −10 ≤ x ≤ 10 and −10 ≤ y ≤ 10. Using this window, the graph of y = f (x) looks like that displayed in Figure 3.44, although the three segments shown are not particularly revealing. Instead of blindly manipulating the window in the hope that a reasonable graph will magically appear, we stop briefly to determine where the function is increasing and decreasing. First, observe that f (x) = 6x 2 + 18x − 24 = 6(x 2 + 3x − 4) = 6(x − 1)(x + 4).

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0

0

..

Applications of Differentiation

Note that the critical numbers (1 and −4) are the only possible locations for local extrema. We can see where the two factors and consequently the derivative are positive and negative from the number lines displayed in the margin. From this, note that

6(x 1)

1

(x 4)

4

0

0

4

6(x 1)(x 4)

f (x) > 0 on (−∞, −4) ∪ (1, ∞)

f increasing.

f (x) < 0 on (−4, 1).

f decreasing.

1

and

y

100

x

8

4 50

FIGURE 3.45a y = 2x 3 + 9x 2 − 24x − 10 y f(x)

f '(x)

For convenience, we have placed arrows indicating where the function is increasing and decreasing beneath the last number line. In Figure 3.45a, we redraw the graph in the window defined by −8 ≤ x ≤ 4 and −50 ≤ y ≤ 125. Here, we have selected the y-range so that the critical points (−4, 102) and (1, −23) are displayed. Since f is increasing on all of (−∞, −4), we know that the function is still increasing to the left of the portion displayed in Figure 3.45a. Likewise, since f is increasing on all of (1, ∞), we know that the function continues to increase to the right of the displayed portion. In Figure 3.45b, we have plotted both y = f (x) (shown in blue) and y = f (x) (shown in red). Notice the connection between the two graphs. When f (x) > 0, f is increasing; when f (x) < 0, f is decreasing and also notice what happens to f (x) at the local extrema of f . (We’ll say more about this shortly.)

100

x

8

3-40

4 50

You may be tempted to think that you can draw graphs by machine and with a little fiddling with the graphing window, get a reasonable looking graph. Unfortunately, this frequently isn’t enough. For instance, while it’s clear that the graph in Figure 3.44 is incomplete, the initial graph in example 4.2 has a familiar shape and may look reasonable, but it is incorrect. The calculus tells you what features you should expect to see in a graph. Without it, you’re simply taking a shot in the dark.

FIGURE 3.45b

y = f (x) and y = f (x)

EXAMPLE 4.2

Uncovering Hidden Behavior in a Graph

Graph f (x) = 3x 4 + 40x 3 − 0.06x 2 − 1.2x showing all local extrema.

0

0.1

0

0

(x 10)

10

0 10

0

0.1

f (x) = 12x 3 + 120x 2 − 0.12x − 1.2 = 12(x 2 − 0.01)(x + 10)

(x 0.1)

0.1

12(x 0.1)

Solution We first show the default graph drawn by our computer algebra system (see Figure 3.46a). We show a common default graphing calculator graph in Figure 3.46b. You can certainly make Figure 3.46b look more like Figure 3.46a by adjusting the window some. But with some calculus, you can discover features of the graph that are hidden in both graphs. First, notice that

0 0.1

= 12(x − 0.1)(x + 0.1)(x + 10).

f'(x)

We show number lines for the three factors in the margin. Observe that f (x)

> 0 on (−10, −0.1) ∪ (0.1, ∞) < 0 on (−∞, −10) ∪ (−0.1, 0.1).

f increasing. f decreasing.

3-41

SECTION 3.4

..

Increasing and Decreasing Functions

281

y

6000

y 10

3000

x

4

4 3000

y

10,000

x

15

5

10,000

FIGURE 3.47a The global behavior of f (x) = 3x 4 + 40x 3 − 0.06x 2 − 1.2x

x

10

10 10

FIGURE 3.46a

FIGURE 3.46b

Default CAS graph of y = 3x 4 + 40x 3 − 0.06x 2 − 1.2x

Default calculator graph of y = 3x 4 + 40x 3 − 0.06x 2 − 1.2x

This says that neither of the machine-generated graphs seen in Figures 3.46a or 3.46b is adequate, as the behavior on (−∞, −10) ∪ (−0.1, 0.1) cannot be seen in either graph. As it turns out, no single graph captures all of the behavior of this function. However, by increasing the range of x-values to the interval [−15, 5], we get the graph seen in Figure 3.47a. This shows the big picture, what we refer to as the global behavior of the function. Here, you can see the local minimum at x = −10, which was missing in our earlier graphs, but the behavior for values of x close to zero is not clear. To see this, we need a separate graph, restricted to a smaller range of x-values, as seen in Figure 3.47b. Notice that here, we can see the behavior of the function for x close to zero quite clearly. In particular, the local maximum at x = −0.1 and the local minimum at x = 0.1 are clearly visible. We often say that a graph such as Figure 3.47b shows the local behavior of the function. In Figures 3.48a and 3.48b, we show graphs indicating the global and local behavior of f (x) (in blue) and f (x) (in red) on the same set of axes. Pay particular attention to the behavior of f (x) in the vicinity of local extrema of f (x). y

y

y

f '(x)

0.4

f (x)

0.4

10,000

x

0.3

0.3

f '(x) x

0.3

0.3

x

15

5

10,000 0.4

FIGURE 3.48a

y = f (x) and y = f (x) (global behavior)

FIGURE 3.47b Local behavior of f (x) = 3x 4 + 40x 3 − 0.06x 2 − 1.2x

f (x) 1.2

FIGURE 3.48b

y = f (x) and y = f (x) (local behavior)

You may have already noticed a connection between local extrema and the intervals on which a function is increasing and decreasing. We state this in Theorem 4.2.

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3-42

y

THEOREM 4.2 (First Derivative Test) Local maximum f (x) 0 f increasing

Suppose that f is continuous on the interval [a, b] and c ∈ (a, b) is a critical number. f (x) 0 f decreasing

(i) If f (x) > 0 for all x ∈ (a, c) and f (x) < 0 for all x ∈ (c, b) (i.e., f changes from increasing to decreasing at c), then f (c) is a local maximum. (ii) If f (x) < 0 for all x ∈ (a, c) and f (x) > 0 for all x ∈ (c, b) (i.e., f changes from decreasing to increasing at c), then f (c) is a local minimum. (iii) If f (x) has the same sign on (a, c) and (c, b), then f (c) is not a local extremum.

x

c

It’s easiest to think of this result graphically. If f is increasing to the left of a critical number and decreasing to the right, then there must be a local maximum at the critical number (see Figure 3.49a). Likewise, if f is decreasing to the left of a critical number and increasing to the right, then there must be a local minimum at the critical number (see Figure 3.49b). This suggests a proof of the theorem; the job of writing out all of the details is left as an exercise.

FIGURE 3.49a Local maximum y

c

x

EXAMPLE 4.3

Finding Local Extrema Using the First Derivative Test

Find the local extrema of the function from example 4.1, f (x) = 2x 3 + 9x 2 − 24x − 10. f (x) 0 f decreasing

f (x) 0 f increasing

Solution We had found in example 4.1 that

f (x) Local minimum

FIGURE 3.49b

> 0, on (−∞, −4) ∪ (1, ∞)

f increasing.

< 0, on (−4, 1).

f decreasing.

It now follows from the First Derivative Test that f has a local maximum located at x = −4 and a local minimum located at x = 1.

Local minimum

Theorem 4.2 works equally well for a function with critical points where the derivative is undefined.

EXAMPLE 4.4

Finding Local Extrema of a Function with Fractional Exponents

Find the local extrema of f (x) = x 5/3 − 3x 2/3 . f (x) =

Solution We have

5 2/3 2 x −3 x −1/3 3 3

5x − 6 , 3x 1/3 so that the critical numbers are 65 [ f 65 = 0] and 0 [ f (0) is undefined]. Again drawing number lines for the factors, we determine where f is increasing and decreasing. Here, we have placed an above the 0 on the number line for f (x) to indicate that f (x) is not defined at x = 0. From this, we can see at a glance where f is positive and negative: =

0

(5x 6)

6/5

0

3x1/3

0

0

0 6/5

f (x)

> 0, on (−∞, 0) ∪ f (x) < 0, on 0, 65 .

6 5

,∞

f increasing. f decreasing.

3-43

SECTION 3.4

..

Increasing and Decreasing Functions

Consequently, f has a local maximum at x = 0 and a local minimum at x = 65 . These local extrema are both clearly visible in the graph in Figure 3.50.

y

1 x 2

EXAMPLE 4.5

Finding Local Extrema Approximately

Find the local extrema of f (x) = x 4 + 4x 3 − 5x 2 − 31x + 29 and draw a graph. Solution A graph of y = f (x) using the most common graphing calculator default window appears in Figure 3.51. Without further analysis, we do not know whether this graph shows all of the significant behavior of the function. [Note that some fourth-degree polynomials (e.g., f (x) = x 4 ) have graphs that look very much like the one in Figure 3.51.] First, we compute

FIGURE 3.50 y = x 5/3 − 3x 2/3

f (x) = 4x 3 + 12x 2 − 10x − 31.

y 10

x

10

10

However, this derivative does not easily factor. A graph of y = f (x) (see Figure 3.52) reveals three zeros, one near each of x = −3, −1.5 and 1.5. Since a cubic polynomial has at most three zeros, there are no others. Using Newton’s method or some other rootfinding method [applied to f (x)], we can find approximations to the three zeros of f . We get a ≈ −2.96008, b ≈ −1.63816 and c ≈ 1.59824. From Figure 3.52, we can see that

10

and FIGURE 3.51 f (x) = x 4 + 4x 3 − 5x 2 − 31x + 29

f (x) > 0 on (a, b) ∪ (c, ∞)

f increasing.

f (x) < 0 on (−∞, a) ∪ (b, c).

f decreasing.

You can quickly read off the local extrema: a local minimum at a ≈ −2.96008, a local maximum at b ≈ −1.63816 and a local minimum at c ≈ 1.59824. Since only the local minimum at x = c is visible in the graph in Figure 3.51, this graph is clearly not representative of the behavior of the function. By narrowing the range of displayed x-values and widening the range of displayed y-values, we obtain the far more useful graph seen in Figure 3.53. You should convince yourself, using the preceding analysis, that the local minimum at x = c ≈ 1.59824 is also the absolute minimum.

y

y

100

100

50 a

b

c

4

x 4

50

FIGURE 3.52

f (x) = 4x 3 + 12x 2 − 10x − 31

c 4

a

b

x 4

50

FIGURE 3.53 f (x) = x 4 + 4x 3 − 5x 2 − 31x + 29

283

284

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..

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3-44

EXERCISES 3.4 WRITING EXERCISES 1. Suppose that f (0) = 2 and f is an increasing function. To sketch the graph of y = f (x), you could start by plotting the point (0, 2). Filling in the graph to the left, would you move your pencil up or down? How does this fit with the definition of increasing? 2. Suppose you travel east on an east-west interstate highway. You reach your destination, stay a while and then return home. Explain the First Derivative Test in terms of your velocities (positive and negative) on this trip. 3. Suppose that you have a differentiable function f with two critical numbers. Your computer has shown you a graph that looks like the one in the figure. y

In exercises 11–20, find (by hand) all critical numbers and use the First Derivative Test to classify each as the location of a local maximum, local minimum or neither. 11. y = x 4 + 4x 3 − 2

12. y = x 5 − 5x 2 + 1

13. y = xe−2x

14. y = x 2 e−x 16. y = sin−1 1 −

15. y = tan−1 (x 2 ) x 1 + x3 √ 19. y = x 3 + 3x 2

17. y =

23. y =

4

4

x 1 + x4

20. y = x 4/3 + 4x 1/3

x2 −1

x2

x −1

22. y =

x2

x2 − 4x + 3

24. y =

x 1 − x4

26. y =

x2 + 2 (x + 1)2

25. y = √

x

In exercises 21–26, find (by hand) all asymptotes and extrema, and sketch a graph. 21. y =

10

18. y =

1 x2

x x2 + 1

x2

In exercises 27–34, find the x-coordinates of all extrema and sketch graphs showing global and local behavior of the function.

10

27. y = x 3 − 13x 2 − 10x + 1 Discuss the possibility that this is a representative graph: that is, is it possible that there are any important points not shown in this window? 4. Suppose that the function in exercise 3 has three critical numbers. Explain why the graph is not a representative graph. Explain how you would change the graphing window to show the rest of the graph.

28. y = x 3 + 15x 2 − 70x + 2 29. y = x 4 − 15x 3 − 2x 2 + 40x − 2 30. y = x 4 − 16x 3 − 0.1x 2 + 0.5x − 1 31. y = x 5 − 200x 3 + 605x − 2 32. y = x 4 − 0.5x 3 − 0.02x 2 + 0.02x + 1 33. y = (x 2 + x + 0.45)e−2x

In exercises 1–10, find (by hand) the intervals where the function is increasing and decreasing. Use this information to sketch a graph. 1. y = x 3 − 3x + 2

2. y = x 3 + 2x 2 + 1

3. y = x 4 − 8x 2 + 1

4. y = x 3 − 3x 2 − 9x + 1

5. y = (x + 1)2/3

6. y = (x − 1)1/3

7. y = sin x + cos x

8. y = sin2 x

9. y = e

x 2 −1

10. y = ln(x − 1) 2

34. y = x 5 ln 8x 2 In exercises 35–38, sketch a graph of a function with the given properties. 35. f (0) = 1, f (2) = 5, f (x) < 0 for x < 0 and x > 2, f (x) > 0 for 0 < x < 2. 36. f (−1) = 1, f (2) = 5, f (x) < 0 for x < −1 and x > 2, f (x) > 0 for −1 < x < 2, f (−1) = 0, f (2) does not exist. 37. f (3) = 0, f (x) < 0 for x < 0 and x > 3, f (x) > 0 for 0 < x < 3, f (3) = 0, f (0) and f (0) do not exist.

3-45

SECTION 3.4

38. f (1) = 0, lim f (x) = 2, f (x) < 0 for x < 1, f (x) > 0 for

x→∞

x > 1, f (1) = 0. In exercises 39–42, estimate critical numbers and sketch graphs showing both global and local behavior. 39. y =

x − 30 x4 − 1

40. y =

x2 − 8 x4 − 1

41. y =

x + 60 x2 + 1

42. y =

x − 60 x2 − 1

43. For f (x) =

x + 2x 2 sin 1/x if x = 0 0

if x = 0 show that f (0) > 0,

but that f is not increasing in any interval around 0. Explain why this does not contradict Theorem 4.1. 44. For f (x) = x 3 , show that f is increasing in any interval around 0, but f (0) = 0. Explain why this does not contradict Theorem 4.1. 45. Prove Theorem 4.2 (the First Derivative Test). 46. Give a graphical argument that if f (a) = g(a) and f (x) > g (x) for all x > a, then f (x) > g(x) for all x > a. Use the Mean Value Theorem to prove it. In exercises 47–50, use the result of exercise 46 to verify the inequality. √ 1 47. 2 x > 3 − for x > 1 48. x > sin x for x > 0 x x 50. x − 1 > ln x for x > 1 49. e > x + 1 for x > 0 51. Give an example showing that the following statement is false. If f (0) = 4 and f (x) is a decreasing function, then the equation f (x) = 0 has exactly one solution. 52. Assume that f is an increasing function with inverse function f −1 . Show that f −1 is also an increasing function.

..

Increasing and Decreasing Functions

285

function with local extrema so close together that they’re difficult to see. For instance, suppose you want local extrema at x = −0.1 and x = 0.1. Explain why you could start with f (x) = (x − 0.1)(x + 0.1) = x 2 − 0.01. Look for a function whose derivative is as given. Graph your function to see if the extrema are “hidden.” Next, construct a polynomial of degree 4 with two extrema very near x = 1 and another near x = 0. 60. Suppose that f and g are differentiable functions and x = c is a critical number of both functions. Either prove (if it is true) or disprove (with a counterexample) that the composition f ◦ g also has a critical number at x = c. 61. Show that f (x) = x 3 + bx 2 + cx + d is an increasing function if b2 ≤ 3c. 62. Find a condition on the coefficients b and c similar to exercise 61 that guarantees that f (x) = x 5 + bx 3 + cx + d is an increasing function. 63. The table shows the coefficient of friction µ of ice as a function of temperature. The lower µ is, the more “slippery” the ice is. Estimate µ (C) at (a) C = −10 and (b) C = −6. If skating warms the ice, does it get easier or harder to skate? Briefly explain. ◦

C −12 −10 −8 −6 −4 −2 µ 0.0048 0.0045 0.0043 0.0045 0.0048 0.0055

64. For a college football field with the dimensions shown, the angle θ for kicking a field goal from a (horizontal) distance of x feet from the goal post is given by θ(x) = tan−1 (29.25/x) − tan−1 (10.75/x). Show that t f (t) = 2 is increasing for a > t and use this fact to show a + t2 that θ(x) is a decreasing function for x ≥ 30. Announcers often say that for a short field goal (50 ≤ x ≤ 60), a team can improve the angle by backing up 5 yards with a penalty. Is this true?

53. State the domain for sin−1 x and determine where it is increasing and decreasing. 54. State the domain for sin−1 π2 tan−1 x and determine where it is increasing and decreasing. 55. If f and g are both increasing functions, is it true that f (g(x)) is also increasing? Either prove that it is true or give an example that proves it false. 56. If f and g are both increasing functions with f (5) = 0, find the maximum and minimum of the following values: g(1), g(4), g( f (1)), g( f (4)). 57. Suppose that the√total sales of a product after t months is given by s(t) = t + 4 thousand dollars. Compute and interpret s (t). 58. In exercise 57, show that s (t) > 0 for all t > 0. Explain in business terms why it is impossible to have s (t) < 0. 59. In this exercise, you will play the role of professor and construct a tricky graphing exercise. The first goal is to find a

18.5'

40' x

EXPLORATORY EXERCISES 1. In this exercise, we look at the ability of fireflies to synchronize their flashes. (To see a remarkable demonstration of this ability, see David Attenborough’s video series Trials of Life.) Let the function f represent an individual firefly’s

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3-46

other hand, if T (t) is large enough, then T (t) < 0 and the T-cell population will decrease. For what value of T (t) is T (t) = 0? Even though this population would remain stable, explain why this would be bad news for the infected human.

rhythm, so that the firefly flashes whenever f (t) equals an integer. Let e(t) represent the rhythm of a neighboring firefly, where again e(t) = n, for some integer n, whenever the neighbor flashes. One model of the interaction between fireflies is f (t) = ω + A sin [e(t) − f (t)] for constants ω and A. If the fireflies are synchronized [e(t) = f (t)], then f (t) = ω, so the fireflies flash every 1/ω time units. Assume that the difference between e(t) and f (t) is less than π. Show that if f (t) < e(t), then f (t) > ω. Explain why this means that the individual firefly is speeding up its flash to match its neighbor. Similarly, discuss what happens if f (t) > e(t).

3. In a sport like soccer or hockey where ties are possible, the probability that the stronger team wins depends in an interesting way on the number of goals scored. Suppose that at any point, the probability that team A scores the next goal is p, where 0 < p < 1. If 2 goals are scored, a 1-1 tie could result from team A scoring first (probability p) and then team B tieing the score (probability 1 − p), or vice versa. The probability of a tie in a 2-goal game is then 2 p(1 − p). Similarly, the probability of a 2-2 tie in a 4-goal game is 4·3 2 p (1 − p)2 , the probability of a 3-3 tie in a 6-goal game 2·1

2. The HIV virus attacks specialized T-cells that trigger the human immune system response to a foreign substance. If T (t) is the population of uninfected T-cells at time t (days) and V (t) is the population of infectious HIV in the bloodstream, a model that has been used to study AIDS is given by the following differential equation that describes the rate at which the population of T cells changes. 1 T (t)V (t) T (t) = 10 1 + − 0.02T (t) + 0.01 1 + V (t) 100 + V (t)

is 63 ·· 52 ·· 41 p 3 (1 − p)3 and so on. As the number of goals increases, does the probability of a tie increase or decrease? < 4 for x > 0 and To find out, first show that (2x+2)(2x+1) (x+1)2 x(1 − x) ≤ 14 for 0 ≤ x ≤ 1. Use these inequalities to show that the probability of a tie decreases as the (even) number of goals increases. In a 1-goal game, the probability that team A wins is p. In a 2-goal game, the probability that team A wins is p 2 . In a 3-goal game, the probability that team A wins is p 3 + 3 p 2 (1 − p). In a 4-goal game, the probability that team A wins is p 4 + 4 p 3 (1 − p). In a 5-goal game, the probability that team A wins is p 5 + 5 p 4 (1 − p) + 52 ·· 41 p 3 (1 − p)2 . Explore the extent to which the probability that team A wins increases as the number of goals increases.

− 0.000024 T (t)V (t). If there is no HIV present [that is, V (t) = 0] and T (t) = 1000, show that T (t) = 0. Explain why this means that the T-cell count will remain constant at 1000 (cells per cubic mm). Now, suppose that V (t) = 100. Show that if T (t) is small enough, then T (t) > 0 and the T-cell population will increase. On the

3.5

CONCAVITY AND THE SECOND DERIVATIVE TEST In section 3.4, we saw how to determine where a function is increasing and decreasing and how this relates to drawing a graph of the function. First, recognize that simply knowing where a function increases and decreases is not sufficient to draw a good graph. In Figures 3.54a and 3.54b, we show two very different shapes of increasing functions joining the same two points. y

y

x a

b

x a

b

FIGURE 3.54a

FIGURE 3.54b

Increasing function

Increasing function

286

CHAPTER 3

..

Applications of Differentiation

3-46

other hand, if T (t) is large enough, then T (t) < 0 and the T-cell population will decrease. For what value of T (t) is T (t) = 0? Even though this population would remain stable, explain why this would be bad news for the infected human.

rhythm, so that the firefly flashes whenever f (t) equals an integer. Let e(t) represent the rhythm of a neighboring firefly, where again e(t) = n, for some integer n, whenever the neighbor flashes. One model of the interaction between fireflies is f (t) = ω + A sin [e(t) − f (t)] for constants ω and A. If the fireflies are synchronized [e(t) = f (t)], then f (t) = ω, so the fireflies flash every 1/ω time units. Assume that the difference between e(t) and f (t) is less than π. Show that if f (t) < e(t), then f (t) > ω. Explain why this means that the individual firefly is speeding up its flash to match its neighbor. Similarly, discuss what happens if f (t) > e(t).

3. In a sport like soccer or hockey where ties are possible, the probability that the stronger team wins depends in an interesting way on the number of goals scored. Suppose that at any point, the probability that team A scores the next goal is p, where 0 < p < 1. If 2 goals are scored, a 1-1 tie could result from team A scoring first (probability p) and then team B tieing the score (probability 1 − p), or vice versa. The probability of a tie in a 2-goal game is then 2 p(1 − p). Similarly, the probability of a 2-2 tie in a 4-goal game is 4·3 2 p (1 − p)2 , the probability of a 3-3 tie in a 6-goal game 2·1

2. The HIV virus attacks specialized T-cells that trigger the human immune system response to a foreign substance. If T (t) is the population of uninfected T-cells at time t (days) and V (t) is the population of infectious HIV in the bloodstream, a model that has been used to study AIDS is given by the following differential equation that describes the rate at which the population of T cells changes. 1 T (t)V (t) T (t) = 10 1 + − 0.02T (t) + 0.01 1 + V (t) 100 + V (t)

is 63 ·· 52 ·· 41 p 3 (1 − p)3 and so on. As the number of goals increases, does the probability of a tie increase or decrease? < 4 for x > 0 and To find out, first show that (2x+2)(2x+1) (x+1)2 x(1 − x) ≤ 14 for 0 ≤ x ≤ 1. Use these inequalities to show that the probability of a tie decreases as the (even) number of goals increases. In a 1-goal game, the probability that team A wins is p. In a 2-goal game, the probability that team A wins is p 2 . In a 3-goal game, the probability that team A wins is p 3 + 3 p 2 (1 − p). In a 4-goal game, the probability that team A wins is p 4 + 4 p 3 (1 − p). In a 5-goal game, the probability that team A wins is p 5 + 5 p 4 (1 − p) + 52 ·· 41 p 3 (1 − p)2 . Explore the extent to which the probability that team A wins increases as the number of goals increases.

− 0.000024 T (t)V (t). If there is no HIV present [that is, V (t) = 0] and T (t) = 1000, show that T (t) = 0. Explain why this means that the T-cell count will remain constant at 1000 (cells per cubic mm). Now, suppose that V (t) = 100. Show that if T (t) is small enough, then T (t) > 0 and the T-cell population will increase. On the

3.5

CONCAVITY AND THE SECOND DERIVATIVE TEST In section 3.4, we saw how to determine where a function is increasing and decreasing and how this relates to drawing a graph of the function. First, recognize that simply knowing where a function increases and decreases is not sufficient to draw a good graph. In Figures 3.54a and 3.54b, we show two very different shapes of increasing functions joining the same two points. y

y

x a

b

x a

b

FIGURE 3.54a

FIGURE 3.54b

Increasing function

Increasing function

3-47

SECTION 3.5

..

Concavity and the Second Derivative Test

287

Given that a curve joins two particular points and is increasing, we need further information to determine which of the two shapes shown (if either) we should draw. Realize that this is an important distinction to make. For example, suppose that Figure 3.54a or 3.54b depicts the balance in your bank account. Both indicate a balance that is growing. However, the rate of growth in Figure 3.54a, is increasing, while the rate of growth depicted in Figure 3.54b is decreasing. Which would you want to have describe your bank balance? Why? Figures 3.55a and 3.55b are the same as Figures 3.54a and 3.54b, respectively, but with a few tangent lines drawn in. y

y

x a

x

b

a

b

FIGURE 3.55a

FIGURE 3.55b

Concave up, increasing

Concave down, increasing

Although all of the tangent lines have positive slope [since f (x) > 0], the slopes of the tangent lines in Figure 3.55a are increasing, while those in Figure 3.55b are decreasing. We call the graph in Figure 3.55a concave up and the graph in Figure 3.55b concave down. The situation is similar for decreasing functions. In Figures 3.56a and 3.56b, we show two different shapes of decreasing functions. The one shown in Figure 3.56a is concave up (slopes of tangent lines increasing) and the one shown in Figure 3.56b is concave down (slopes of tangent lines decreasing). We summarize this in Definition 5.1. y

y

x a

b

x a

b

FIGURE 3.56a

FIGURE 3.56b

Concave up, decreasing

Concave down, decreasing

DEFINITION 5.1 For a function f that is differentiable on an interval I , the graph of f is (i) concave up on I if f is increasing on I or (ii) concave down on I if f is decreasing on I .

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3-48

Note that you can tell when f is increasing or decreasing from the derivative of f (i.e., f ). Theorem 5.1 connects concavity with what we already know about increasing and decreasing functions. The proof is a straightforward application of Theorem 4.1 to Definition 5.1.

THEOREM 5.1 Suppose that f exists on an interval I . (i) If f (x) > 0 on I , then the graph of f is concave up on I . (ii) If f (x) < 0 on I , then the graph of f is concave down on I .

EXAMPLE 5.1

Determining Concavity

Determine where the graph of f (x) = 2x 3 + 9x 2 − 24x − 10 is concave up and concave down, and draw a graph showing all significant features of the function.

y 100

Solution Here, we have Inflection point x

4

4 50

and from our work in example 4.3, we have > 0 on (−∞, −4) ∪ (1, ∞) f (x) < 0 on (−4, 1).

Further, we have f (x) = 12x + 18

FIGURE 3.57 y = 2x + 9x − 24x − 10 3

2

NOTES If (c, f (c)) is an inflection point, then either f (c) = 0 or f (c) is undefined. So, finding all points where f (x) is zero or is undefined gives you all possible candidates for inflection points. But beware: not all points where f (x) is zero or undefined correspond to inflection points.

0

0

3

0

0

3

0 0

0 3

< 0, for x <

− 32 .

f decreasing.

Concave up. Concave down.

Using all of this information, we areable to draw the graph shown in Figure 3.57. Notice that at the point − 32 , f − 32 , the graph changes from concave down to concave up. Such points are called inflection points, which we define more precisely in Definition 5.2.

DEFINITION 5.2 Suppose that f is continuous on the interval (a, b) and that the graph changes concavity at a point c ∈ (a, b) (i.e., the graph is concave down on one side of c and concave up on the other). Then, the point (c, f (c)) is called an inflection point of f .

EXAMPLE 5.2

( x 3 )

Determine where the graph of f (x) = x 4 − 6x 2 + 1 is concave up and concave down, find any inflection points and draw a graph showing all significant features.

( x 3 )

3

> 0, for x > − 32

f increasing.

4x

0

f (x) = 6x 2 + 18x − 24

f (x)

Determining Concavity and Locating Inflection Points

Solution Here, we have f (x) = 4x 3 − 12x = 4x(x 2 − 3) √ √ = 4x(x − 3)(x + 3).

3-49

SECTION 3.5

0

0

(x 1)

1

0

0

1

0

3

0

f (x)

0

f (x) = 12x 2 − 12 = 12(x − 1)(x + 1).

f'(x)

3

0

0

289

1

Next, we have

Concavity and the Second Derivative Test

We have drawn number lines for the factors of f (x) in the margin. From this, we can see that √ √ > 0, on (− 3, 0) ∪ ( 3, ∞) f increasing. √ √ f (x) < 0, on (−∞, − 3) ∪ (0, 3). f decreasing.

12(x 1)

1

..

1

We have drawn number lines for the two factors in the margin. From this, we can see that

0

f''(x)

1

f (x)

> 0, on (−∞, −1) ∪ (1, ∞)

Concave up.

< 0, on (−1, 1).

Concave down.

y 10 5 x

3

3

For convenience, we have indicated the concavity below the bottom number line, with small concave up and concave down segments. Finally, observe that since the graph changes concavity at x = −1 and x = 1, there are inflection points located at (−1, −4) and (1, −4). Using all of this information, we are able to draw the graph shown in Figure 3.58. For your convenience, we have reproduced the number lines for f (x) and f (x) together above the figure. As we see in example 5.3, having f (x) = 0 does not imply the existence of an inflection point.

5 10

EXAMPLE 5.3 FIGURE 3.58

Determine the concavity of f (x) = x 4 and locate any inflection points.

y = x 4 − 6x 2 + 1

Solution There’s nothing tricky about this function. We have f (x) = 4x 3 and f (x) = 12x 2 . Since f (x) > 0 for x > 0 and f (x) < 0 for x < 0, we know that f is increasing for x > 0 and decreasing for x < 0. Further, f (x) > 0 for all x = 0, while f (0) = 0. So, the graph is concave up for x = 0. Further, even though f (0) = 0, there is no inflection point at x = 0. We show a graph of the function in Figure 3.59.

y

4

We now explore a connection between second derivatives and extrema. Suppose that f (c) = 0 and that the graph of f is concave down in some open interval containing c. Then, near x = c, the graph looks like that in Figure 3.60a and hence, f (c) is a local maximum. Likewise, if f (c) = 0 and the graph of f is concave up in some open interval containing c, then near x = c, the graph looks like that in Figure 3.60b and hence, f (c) is a local minimum.

2

2

A Graph with No Inflection Points

x

1

1

2

y

y

FIGURE 3.59

f (c) 0

y = x4

f (c) 0

f (c) 0 c

f (c) 0 x

c

FIGURE 3.60a

FIGURE 3.60b

Local maximum

Local minimum

x

290

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3-50

We state this more precisely in Theorem 5.2.

THEOREM 5.2 (Second Derivative Test) Suppose that f is continuous on the interval (a, b) and f (c) = 0, for some number c ∈ (a, b). (i) If f (c) < 0, then f (c) is a local maximum. (ii) If f (c) > 0, then f (c) is a local minimum.

We leave a formal proof of this theorem as an exercise. When applying the theorem, simply think about Figures 3.60a and 3.60b.

EXAMPLE 5.4

Using the Second Derivative Test to Find Extrema

Use the Second Derivative Test to find the local extrema of f (x) = x 4 − 8x 2 + 10.

y

Solution Here, 20

f (x) = 4x 3 − 16x = 4x(x 2 − 4) = 4x(x − 2)(x + 2). Thus, the critical numbers are x = 0, 2 and −2. We also have f (x) = 12x 2 − 16

4

x

2

2

4

and so,

f (0) = −16 < 0, f (−2) = 32 > 0

10

and

FIGURE 3.61

f (2) = 32 > 0.

So, by the Second Derivative Test, f (0) is a local maximum and f (−2) and f (2) are local minima. We show a graph of y = f (x) in Figure 3.61.

y = x 4 − 8x 2 + 10

REMARK 5.1 If f (c) = 0 or f (c) is undefined, the Second Derivative Test yields no conclusion. That is, f (c) may be a local maximum, a local minimum or neither. In this event, we must rely solely on first derivative information (i.e., the First Derivative Test) to determine whether f (c) is a local extremum. We illustrate this with example 5.5.

y 30

4

x

2

2

30

FIGURE 3.62a y = x3

4

EXAMPLE 5.5

Functions for Which the Second Derivative Test Is Inconclusive

Use the Second Derivative Test to try to classify any local extrema for (a) f (x) = x 3 , (b) g(x) = x 4 and (c) h(x) = −x 4 . Solution (a) Note that f (x) = 3x 2 and f (x) = 6x. So, the only critical number is x = 0 and f (0) = 0, also. We leave it as an exercise to show that the point (0, 0) is not a local extremum (see Figure 3.62a).

3-51

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..

Concavity and the Second Derivative Test

291

y

y

2

x

1

1

2

4 2 2 4 2

x

1

1

2

FIGURE 3.62b

FIGURE 3.62c

y = x4

y = −x 4

(b) We have g (x) = 4x 3 and g (x) = 12x 2 . Again, the only critical number is x = 0 and g (0) = 0. In this case, though, g (x) < 0 for x < 0 and g (x) > 0 for x > 0. So, by the First Derivative Test, (0, 0) is a local minimum (see Figure 3.62b). (c) Finally, we have h (x) = −4x 3 and h (x) = −12x 2 . Once again, the only critical number is x = 0, h (0) = 0 and we leave it as an exercise to show that (0, 0) is a local maximum for h (see Figure 3.62c). We can use first and second derivative information to help produce a meaningful graph of a function, as in example 5.6.

EXAMPLE 5.6

Drawing a Graph of a Rational Function

25 , showing all significant features. x Solution The domain of f consists of all real numbers other than x = 0. Then,

Draw a graph of f (x) = x +

25 x 2 − 25 = 2 x x2 (x − 5)(x + 5) = . x2

f (x) = 1 −

Add the fractions.

So, the only two critical numbers are x = −5 and x = 5. (Why is x = 0 not a critical number?) Looking at the three factors in f (x), we get the number lines shown in the margin. Thus, f (x)

0

(x 5)

> 0, on (−∞, −5) ∪ (5, ∞)

f increasing.

< 0, on (−5, 0) ∪ (0, 5).

f decreasing.

5

0

(x 5)

5

0

0

5

0

0 5

f (x) =

50 x3

> 0, on (0, ∞)

Concave up.

< 0, on (−∞, 0).

Concave down.

x2

0

Further,

f (x)

Be careful here. There is no inflection point on the graph, even though the graph is concave up on one side of x = 0 and concave down on the other. (Why not?) We can now use either the First Derivative Test or the Second Derivative Test to determine the

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3-52

f (5) =

local extrema. Since

f (−5) = −

and

y 20

50 >0 125 50 < 0, 125

there is a local minimum at x = 5 and a local maximum at x = −5 by the Second Derivative Test. Finally, before we can draw a representative graph, we need to know what happens to the graph near x = 0, since 0 is not in the domain of f. We have

10

x

15 10 5

5

10

lim f (x) = lim+

15

x→0+

x→0

10

20

FIGURE 3.63 y=x+

lim f (x) = lim−

and

x→0−

x→0

25 x+ x

25 x+ x

=∞

= −∞,

so that there is a vertical asymptote at x = 0. Putting together all of this information, we get the graph shown in Figure 3.63.

25 x

In example 5.6, we computed lim+ f (x) and lim− f (x) to uncover the behavior of the x→0

x→0

function near x = 0, since x = 0 was not in the domain of f. In example 5.7, we’ll see that since x = −2 is not in the domain of f (although it is in the domain of f ), we must compute lim + f (x) and lim − f (x) to uncover the behavior of the tangent lines near x = −2. x→−2

x→−2

EXAMPLE 5.7

A Function with a Vertical Tangent Line at an Inflection Point

Draw a graph of f (x) = (x + 2)1/5 + 4, showing all significant features. Solution First, notice that the domain of f is the entire real line. We also have f (x) =

So, f is increasing everywhere, except at x = −2 [the only critical number, where f (−2) is undefined]. This also says that f has no local extrema. Further, > 0, on (−∞, −2) Concave up. 4 −9/5 f (x) = − (x + 2) 25 < 0, on (−2, ∞). Concave down.

y 6

So, there is an inflection point at x = −2. In this case, f (x) is undefined at x = −2. Since −2 is in the domain of f , but not in the domain of f , we consider

4

1 lim − f (x) = lim − (x + 2)−4/5 = ∞ x→−2 x→−2 5

2

4 3 2 1

1 (x + 2)−4/5 > 0, for x = −2. 5

x 1

FIGURE 3.64 y = (x + 2)1/5 + 4

2

and

1 lim f (x) = lim + (x + 2)−4/5 = ∞. x→−2 5

x→−2+

This says that the graph has a vertical tangent line at x = −2. Putting all of this information together, we get the graph shown in Figure 3.64.

3-53

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293

EXERCISES 3.5 WRITING EXERCISES 1. It is often said that a graph is concave up if it “holds water.” This is certainly true for parabolas like y = x 2 , but is it true for graphs like y = 1/x 2 ? It is always helpful to put a difficult concept into everyday language, but the danger is in oversimplification. Do you think that “holds water” is helpful or can it be confusing? Give your own description of concave up, using everyday language. (Hint: One popular image involves smiles and frowns.)

In exercises 15–26, determine all significant features by hand and sketch a graph.

19. f (x) = sin x + cos x

20. f (x) = e−x sin x

2. Find a reference book with the population of the United States since 1800. From 1800 to 1900, the numerical increase by decade increased. Argue that this means that the population curve is concave up. From 1960 to 1990, the numerical increase by decade has been approximately constant. Argue that this means that the population curve is near a point where the curve is neither concave up nor concave down. Why does this not necessarily mean that we are at an inflection point? Argue that we should hope, in order to avoid overpopulation, that it is indeed an inflection point.

21. f (x) = x 3/4 − 4x 1/4

22. f (x) = x 2/3 − 4x 1/3

23. f (x) = x|x|

24. f (x) = x 2 |x| √ x 26. f (x) = √ 1+ x

3. The goal of investing in the stock market is to buy low and sell high. But, how can you tell whether a price has peaked or not? Once a stock price goes down, you can see that it was at a peak, but then it’s too late to do anything about it! Concavity can help. Suppose a stock price is increasing and the price curve is concave up. Why would you suspect that it will continue to rise? Is this a good time to buy? Now, suppose the price is increasing but the curve is concave down. Why should you be preparing to sell? Finally, suppose the price is decreasing. If the curve is concave up, should you buy or sell? What if the curve is concave down? 4. Suppose that f (t) is the amount of money in your bank account at time t. Explain in terms of spending and saving what would cause f (t) to be decreasing and concave down; increasing and concave up; decreasing and concave up.

In exercises 1–8, determine the intervals where the graph of the given function is concave up and concave down. 1. f (x) = x 3 − 3x 2 + 4x − 1

2. f (x) = x 4 − 6x 2 + 2x + 3

3. f (x) = x + 1/x

4. f (x) = x + 3(1 − x)1/3

5. f (x) = sin x − cos x

6. f (x) = tan−1 (x 2 )

7. f (x) = x 4/3 + 4x 1/3

8. f (x) = xe−4x

15. f (x) = (x 2 + 1)2/3 17. f (x) =

10. f (x) = x 4 + 4x 2 + 1 2 12. f (x) = e−x 2 x −1 14. f (x) = x

x x2 − 9

18. f (x) =

25. f (x) = x 1/5 (x + 1)

x x +2

In exercises 27–36, determine all significant features (approximately if necessary) and sketch a graph. 27. f (x) = x 4 − 26x 3 + x 28. f (x) = 2x 4 − 11x 3 + 17x 2 √ 29. f (x) = 3 2x 2 − 1 √ 30. f (x) = x 3 + 1 31. f (x) = x 4 − 16x 3 + 42x 2 − 39.6x + 14 32. f (x) = x 4 + 32x 3 − 0.02x 2 − 0.8x √ 33. f (x) = x x 2 − 4 1 35. f (x) = tan−1 x2 − 1

34. f (x) = √

2x x2 + 4

36. f (x) = e−2x cos x

In exercises 37–40, estimate the intervals where the function is concave up and concave down. (Hint: Estimate where the slope is increasing and decreasing.) y

37. 20

10

In exercises 9–14, find all critical numbers and use the Second Derivative Test to determine all local extrema. 9. f (x) = x 4 + 4x 3 − 1 11. f (x) = xe−x x 2 − 5x + 4 13. f (x) = x

16. f (x) = x ln x

2

−3 −2

x 2

3

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45. Show that any cubic f (x) = ax 3 + bx 2 + cx + d has one inflection point. Find conditions on the coefficients a−e that guarantee that the quartic f (x) = ax 4 + bx 3 + cx 2 + d x + e has two inflection points.

y

38. 1

−4

46. If f and g are functions with two derivatives for all x, f (0) = g(0) = f (0) = g (0) = 0, f (0) > 0 and g (0) < 0, state as completely as possible what can be said about whether f (x) > g(x) or f (x) < g(x).

x

−2

2

4

In exercises 47 and 48, estimate the intervals of increase and decrease, the locations of local extrema, intervals of concavity and locations of inflection points.

−1

y

39.

3-54

y

47.

10

8

5

6 4

x

−2

2

4

2

−5 −3 −2

− 10

x

−1

1

2

3

−2

y

40.

y

48.

10 4

5

−3

x

−1

1

2

3

x

−1

−5

1

2

3

−4

−10

In exercises 41–44, sketch a graph with the given properties. 41. f (0) = 0, f (x) > 0 for x < −1 and −1 < x < 1, f (x) < 0 for x > 1, f (x) > 0 for x < −1, 0 < x < 1 and x > 1, f (x) < 0 for −1 < x < 0 42. f (0) = 2, f (x) > 0 for all x, f (0) = 1, f (x) > 0 for x < 0, f (x) < 0 for x > 0 43. f (0) = 0, f (−1) = −1, f (1) = 1, f (x) > 0 for x < −1 and 0 < x < 1, f (x) < 0 for −1 < x < 0 and x > 1, f (x) < 0 for x < 0 and x > 0 44. f (1) = 0, f (x) < 0 for x < 1, f (x) < 0 for x < 1 and x > 1

f (x) > 0 for x > 1,

49. Repeat exercises 47 and 48 if the given graph is of f (x) instead of f (x). 50. Repeat exercises 47 and 48 if the given graph is of f (x) instead of f (x). 51. Suppose that w(t) is the depth of water in a city’s water reservoir at time t. Which would be better news at time t = 0, w (0) = 0.05 or w (0) = −0.05, or would you need to know the value of w (0) to determine which is better? 52. Suppose that T (t) is a sick person’s temperature at time t. Which would be better news at time t, T (0) = 2 or T (0) = −2, or would you need to know the value of T (0) and T (0) to determine which is better?

3-55

SECTION 3.5

53. Suppose that a company that spends $x thousand on advertising sells $s(x) of merchandise, where s(x) = −3x 3 + 270x 2 − 3600x + 18,000. Find the value of x that maximizes the rate of change of sales. (Hint: Read the question carefully!) Find the inflection point and explain why in advertising terms this is the “point of diminishing returns.” 54. The number of units Q that a worker has produced in a day is related to the number of hours t since the work day began. Suppose that Q(t) = −t 3 + 6t 2 + 12t. Explain why Q (t) is a measure of the efficiency of the worker at time t. Find the time at which the worker’s efficiency is a maximum. Explain why it is reasonable to call the inflection point the “point of diminishing returns.” 55. Suppose that it costs a company C(x) = 0.01x + 40x + 3600 dollars to manufacture x units of a product. For this cost funcC(x) tion, the average cost function is C(x) = . Find the x value of x that minimizes the average cost. The cost function can be related to the efficiency of the production process. Explain why a cost function that is concave down indicates better efficiency than a cost function that is concave up. 2

56. The antiarrhythmic drug lidocaine slowly decays after entering the bloodstream. The plasma concentration t minutes after administering the drug can be modeled by c(t) = 92.8 −0.129e−t/6.55 + 0.218e−t/65.7 − 0.089e−t/13.3 .

..

Concavity and the Second Derivative Test

295

points and asymptotes (for x ≥ 0) for each function and comparing to the graph. Each of these functions is called a Hill function. 58. Two functions with similar properties to the Hill functions of 1 exercise 57 are g1 (x) = tan−1 x and g2 (x) = . As 1 + 99e−x/2 in exercise 57, test these functions for use as a model of the hemoglobin data. 59. Give an example of a function showing that the following statement is false. If the graph of y = f (x) is concave down for all x, the equation f (x) = 0 has at least one solution. 60. Determine whether the following statement is true or false. If f (0) = 1, f (x) exists for all x and the graph of y = f (x) is concave down for all x, the equation f (x) = 0 has at least one solution. 61. A basic principle of physics is that light follows the path of minimum time. Assuming that the speed of light in the earth’s atmosphere decreases as altitude decreases, argue that the path that light follows is concave down. Explain why this means that the setting sun appears higher in the sky than it really is.

Use a CAS to estimate the time of maximum concentration and inflection point (t > 0). Suppose that f (t) represents the concentration of another drug. If the graph of f (t) has a similar shape and the same maximum point as the graph of c(t) but the inflection point occurs at a larger value of t, would this drug be more or less effective than lidocaine? Briefly explain. 57. The plot shows the relationship between the specific partial pressure of oxygen (pO2 , measured in mm Hg) and the saturation level of hemoglobin (y = 1 would mean that no more oxygen can bind). 1.0

0.98

EXPLORATORY EXERCISES

0.9 0.8

Saturation (%)

62. Prove Theorem 5.2 (the Second Derivative Test). (Hint: Think about what the definition of f (c) says when f (c) > 0 or f (c) < 0.)

0.75

0.7 0.6 0.50

0.5 0.4 0.3 0.2 0.1

27

40

103

0.0 0

10

20

30

40

50

60

70

80

90

100

110

120

pO2 (mm Hg)

x x3 or f 2 (x) = 3 is 27 + x 27 + x 3 a better model for these data by finding extrema, inflection Determine whether f 1 (x) =

1. The linear approximation that we defined in section 3.1 is the line having the same location and the same slope as the function being approximated. Since two points determine a line, two requirements (point, slope) are all that a linear function can satisfy. However, a quadratic function can satisfy three requirements, since three points determine a parabola (and there are three constants in a general quadratic function ax 2 + bx + c). Suppose we want to define a quadratic approximation to f (x) at x = a. Building on the linear approximation, the general form is g(x) = f (a) + f (a)(x − a) + c(x − a)2 for some constant c to be determined. In this way, show that g(a) = f (a) and g (a) = f (a). That is, g(x) has the right position and slope at x = a. The third requirement is that g(x) have the right

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3-56

concavity at x = a, so that g (a) = f (a). Find the constant c that makes this true. Then, find such a quadratic approximation for each of the functions sin x, cos x and e x at x = 0. In each case, graph the original function, linear approximation and quadratic approximation, and describe how close the approximations are to the original functions. 2. In this exercise, we explore a basic problem in genetics. Suppose that a species reproduces according to the following probabilities: p0 is the probability of having no children, p1 is the probability of having one offspring, p2 is the probability of having two offspring, . . . , pn is the probability of having n offspring and n is the largest number of offspring possible. Explain why for each i, we have 0 ≤ pi ≤ 1 and p0 + p1 + p2 + · · · + pn = 1. We define the function

3.6

F(x) = p0 + p1 x + p2 x 2 + · · · + pn x n . The smallest nonnegative solution of the equation F(x) = x for 0 ≤ x ≤ 1 represents the probability that the species becomes extinct. Show graphically that if p0 > 0 and F (1) > 1, then there is a solution of F(x) = x with 0 < x < 1. Thus, there is a positive probability of survival. However, if p0 > 0 and F (1) < 1, show that there are no solutions of F(x) = x with 0 < x < 1. (Hint: First show that F is increasing and concave up.) x +c x2 − 1 as possible. In particular, find the values of c for which there are two critical points (or one critical point, or no critical points) and identify any extrema. Similarly, determine how the existence or not of inflection points depends on the value of c.

3. Give as complete a description of the graph of f (x) =

OVERVIEW OF CURVE SKETCHING Graphing calculators and computer algebra systems are powerful tools in the study or application of mathematics. However, they do not actually draw graphs. What they do is plot points (albeit lots of them) and then connect the points as smoothly as possible. While this is very helpful, it often leaves something to be desired. The problem boils down to knowing the window in which you should draw a given graph and how many points you plot in that window. The only way to know how to choose this is to use the calculus to determine the properties of the graph that you are interested in seeing. We have already made this point a number of times. We begin this section by summarizing the various tests that you should perform on a function when trying to draw a graph of y = f (x). r Domain: You should always determine the domain of f first. r Vertical Asymptotes: For any isolated point not in the domain of f, check the r r

r r r

limiting value of the function as x approaches that point, to see if there is a vertical asymptote or a jump or removable discontinuity at that point. First Derivative Information: Determine where f is increasing and decreasing, and find any local extrema. Vertical Tangent Lines: At any isolated point not in the domain of f , but in the domain of f , check the limiting values of f (x), to determine whether there is a vertical tangent line at that point. Second Derivative Information: Determine where the graph is concave up and concave down, and locate any inflection points. Horizontal Asymptotes: Check the limit of f (x) as x → ∞ and as x → −∞. Intercepts: Locate x- and y-intercepts, if any. If this can’t be done exactly, then do so approximately (e.g., using Newton’s method). We start with a very straightforward example.

EXAMPLE 6.1

Drawing a Graph of a Polynomial

Draw a graph of f (x) = x 4 + 6x 3 + 12x 2 + 8x + 1, showing all significant features.

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3-56

concavity at x = a, so that g (a) = f (a). Find the constant c that makes this true. Then, find such a quadratic approximation for each of the functions sin x, cos x and e x at x = 0. In each case, graph the original function, linear approximation and quadratic approximation, and describe how close the approximations are to the original functions. 2. In this exercise, we explore a basic problem in genetics. Suppose that a species reproduces according to the following probabilities: p0 is the probability of having no children, p1 is the probability of having one offspring, p2 is the probability of having two offspring, . . . , pn is the probability of having n offspring and n is the largest number of offspring possible. Explain why for each i, we have 0 ≤ pi ≤ 1 and p0 + p1 + p2 + · · · + pn = 1. We define the function

3.6

F(x) = p0 + p1 x + p2 x 2 + · · · + pn x n . The smallest nonnegative solution of the equation F(x) = x for 0 ≤ x ≤ 1 represents the probability that the species becomes extinct. Show graphically that if p0 > 0 and F (1) > 1, then there is a solution of F(x) = x with 0 < x < 1. Thus, there is a positive probability of survival. However, if p0 > 0 and F (1) < 1, show that there are no solutions of F(x) = x with 0 < x < 1. (Hint: First show that F is increasing and concave up.) x +c x2 − 1 as possible. In particular, find the values of c for which there are two critical points (or one critical point, or no critical points) and identify any extrema. Similarly, determine how the existence or not of inflection points depends on the value of c.

3. Give as complete a description of the graph of f (x) =

OVERVIEW OF CURVE SKETCHING Graphing calculators and computer algebra systems are powerful tools in the study or application of mathematics. However, they do not actually draw graphs. What they do is plot points (albeit lots of them) and then connect the points as smoothly as possible. While this is very helpful, it often leaves something to be desired. The problem boils down to knowing the window in which you should draw a given graph and how many points you plot in that window. The only way to know how to choose this is to use the calculus to determine the properties of the graph that you are interested in seeing. We have already made this point a number of times. We begin this section by summarizing the various tests that you should perform on a function when trying to draw a graph of y = f (x). r Domain: You should always determine the domain of f first. r Vertical Asymptotes: For any isolated point not in the domain of f, check the r r

r r r

limiting value of the function as x approaches that point, to see if there is a vertical asymptote or a jump or removable discontinuity at that point. First Derivative Information: Determine where f is increasing and decreasing, and find any local extrema. Vertical Tangent Lines: At any isolated point not in the domain of f , but in the domain of f , check the limiting values of f (x), to determine whether there is a vertical tangent line at that point. Second Derivative Information: Determine where the graph is concave up and concave down, and locate any inflection points. Horizontal Asymptotes: Check the limit of f (x) as x → ∞ and as x → −∞. Intercepts: Locate x- and y-intercepts, if any. If this can’t be done exactly, then do so approximately (e.g., using Newton’s method). We start with a very straightforward example.

EXAMPLE 6.1

Drawing a Graph of a Polynomial

Draw a graph of f (x) = x 4 + 6x 3 + 12x 2 + 8x + 1, showing all significant features.

3-57

..

SECTION 3.6

Overview of Curve Sketching

297

y y

1600 10

1200 800

− 4 −3 −2 −1

x

−10

400 x 1

2

3

10 −10

4

FIGURE 3.65a

FIGURE 3.65b

y = x 4 + 6x 3 + 12x 2 + 8x + 1 (one view)

y = x 4 + 6x 3 + 12x 2 + 8x + 1 (standard calculator view)

Solution Computer algebra systems and graphing calculators usually do one of two things to determine the window in which they will display a graph. One method is to compute a set number of function values over a given standard range of x-values. The y-range is then chosen so that all of the calculated points can be displayed. This might result in a graph that looks like the one in Figure 3.65a. Another method is to draw a graph in a fixed, default window. For instance, most graphing calculators use the default window defined by −10 ≤ x ≤ 10

and

−10 ≤ y ≤ 10.

Using this window, we get the graph shown in Figure 3.65b. Of course, these two graphs are very different. Without the calculus, it’s difficult to tell which, if either, of these is truly representative of the behavior of f . Some analysis will clear up the situation. First, note that the domain of f is the entire real line. Further, since f (x) is a polynomial, it doesn’t have any vertical or horizontal asymptotes. Next, note that

0

f (x) = 4x 3 + 18x 2 + 24x + 8 = 2(2x + 1)(x + 2)2 .

2(2x 1)

Q

0

(x 2)2

2

0

0

2

f (x)

Q

This also tells us that there is a local minimum at x = − 12 and that there are no local maxima. Next, we have

0

12(x 2)

2

0

(x 1)

1

0 2

f (x)

1

0

2

0

0 2

0

f (x)

Q

0 1

Drawing number lines for the individual factors of f (x), we have that > 0, on − 12 , −∞ f increasing. f (x) < 0, on (−∞, −2) ∪ −2, − 12 . f decreasing.

f (x)

f (x) = 12x 2 + 36x + 24 = 12(x + 2)(x + 1). Drawing number lines for the factors of f (x), we have > 0, on (−∞, −2) ∪ (−1, ∞) f (x) < 0, on (−2, −1).

Concave up. Concave down.

From this, we see that there are inflection points at x = −2 and at x = −1. Finally, to find the x-intercepts, we need to solve f (x) = 0 approximately. Doing this (we leave the details as an exercise: use Newton’s method or your calculator’s solver), we find that there are two x-intercepts: x = −1 (exactly) and x ≈ −0.160713. Notice that the significant x-values that we have identified are x = −2, x = −1 and x = − 12 . Computing the corresponding y-values from y = f (x), we get the points (−2, 1), (−1, 0) and − 12 , − 11 . We summarize the first and second derivative 16

298

..

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Applications of Differentiation

y

3-58

information in the number lines in the margin. In Figure 3.66, we include all of these important points by setting the x-range to be −4 ≤ x ≤ 1 and the y-range to be −2 ≤ y ≤ 8.

8 6

In example 6.2, we examine a function that has local extrema, inflection points and both vertical and horizontal asymptotes.

4 2 −4

−3

−2

x

−1

1 −2

Drawing a Graph of a Rational Function

x2 − 3 , showing all significant features. x3 Solution The default graph drawn by our computer algebra system appears in Figure 3.67a. Notice that this doesn’t seem to be a particularly useful graph, since very little is visible (or at least distinguishable from the axes). The graph drawn using the most common graphing calculator default window is seen in Figure 3.67b. This is arguably an improvement over Figure 3.67a, but does this graph convey all that it could about the function (e.g., about local extrema, inflection points, etc.)? We can answer this question only after we do some calculus. We follow the outline given at the beginning of the section. First, observe that the domain of f includes all real numbers x = 0. Since x = 0 is an isolated point not in the domain of f, we scrutinize the limiting behavior of f as x approaches 0. We have

Draw a graph of f (x) =

FIGURE 3.66 y = x 4 + 6x 3 + 12x 2 + 8x + 1

y 3e + 24 1e + 24 −4

EXAMPLE 6.2

x 4

−1e + 24 −3e + 24

−

2 3 = −∞ lim+ f (x) = lim+ x − 3 x→0 x→0 x

FIGURE 3.67a

+

x2 − 3 x3

y=

−

2 3 = ∞. lim− f (x) = lim− x − 3 x→0 x→0 x

and

From (6.1) and (6.2), we see that the graph has a vertical asymptote at x = 0. Next, we look for whatever information the first derivative will yield. We have

10

x

−10

10

x2 − 3 y= x3 +

0

−

(3 − x)

3 +

(3 + x)

−3 +

0

x4

0 −

0 3 −3

+

× 0

+

0 3

−

2x(x 3 ) − (x 2 − 3)(3x 2 ) (x 3 )2

x 2 [2x 2 − 3(x 2 − 3)] x6 9 − x2 = x4 (3 − x)(3 + x) = . x4

FIGURE 3.67b

+

f (x) = =

−10

0

(6.2)

−

y

−

(6.1)

f'(x)

Quotient rule.

Factor out x 2 .

Combine terms.

Factor difference of two squares.

Looking at the individual factors in f (x), we have the number lines shown in the margin. Thus, f increasing. > 0, on (−3, 0) ∪ (0, 3) f (x) (6.3) < 0, on (−∞, −3) ∪ (3, ∞). f decreasing. Note that this says that f has a local minimum at x = −3 and a local maximum at x = 3.

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299

Next, we look at f (x) =

−2x(x 4 ) − (9 − x 2 )(4x 3 ) (x 4 )2

−2x 3 [x 2 + (9 − x 2 )(2)] x8 −2(18 − x 2 ) = x5 √ √ 2(x − 18)(x + 18) = . x5 =

0

2 ( x 18 )

18

0

0

x5

0

0 18

f (x)

0

f (x)

18

0

Factor out −2x 3 .

Combine terms.

Factor difference of two squares.

Looking at the individual factors in f (x), we obtain the number lines shown in the margin. Thus, we have

( x 18 )

18

Quotient rule.

√ √ > 0, on (− 18, 0) ∪ ( 18, ∞) √ √ < 0, on (−∞, − 18) ∪ (0, 18).

Concave up.

(6.4)

Concave down.

√ This says that there are inflection points at x = ± 18. (Why is there no inflection point at x = 0?) To determine the limiting behavior as x → ±∞, we consider x2 − 3 x→∞ x3 1 3 − 3 = 0. = lim x→∞ x x

lim f (x) = lim

x→∞

0

3

0

0 18

0

f (x)

3

0

f (x)

18

0

0.4

x 5 0.4

FIGURE 3.68 y=

lim f (x) = 0.

x→−∞

x2 − 3 x3

10

(6.6)

So, the line y = 0 is a horizontal asymptote both as x → ∞ and as x → −∞. Finally, the x-intercepts are where 0 = f (x) =

y

5

Likewise, we have

(6.5)

x2 − 3 , x3

√ that is, at x = ± 3. Notice that there are no y-intercepts, since x = 0 is not in the domain of the function. We now have all of the information that we need to draw a representative graph. With some experimentation, you can set the x- and y-ranges so that most of the significant features of the graph (i.e., vertical and horizontal asymptotes, local extrema, inflection points, etc.) are displayed, as in Figure 3.68, which is consistent with all of the information that we accumulated about the function in (6.1)–(6.6). Although the existence of the inflection points is clearly indicated by the change in concavity, their precise location is as yet a bit fuzzy in this graph. Notice, however, that both vertical and horizontal asymptotes and the local extrema are clearly indicated, something that cannot be said about either of Figures 3.67a or 3.67b. In example 6.3, there are multiple vertical asymptotes, only one extremum and no inflection points.

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3-60

y

EXAMPLE 6.3

A Graph with Two Vertical Asymptotes

200

Draw a graph of f (x) =

150 100 50 x

4

4

x2 showing all significant features. x2 − 4

Solution The default graph produced by our computer algebra system is seen in Figure 3.69a, while the default graph drawn by most graphing calculators looks like the graph seen in Figure 3.69b. Notice that the domain of f includes all x except x = ±2 (since the denominator is zero at x = ±2). Figure 3.69b suggests that there are vertical asymptotes at x = ±2, but let’s establish this carefully. We have

50

+

x2 x2 lim+ 2 = lim+ = ∞. (x − 2) (x + 2) x→2 x − 4 x→2

FIGURE 3.69a y=

+

2

x x2 − 4

Similarly, we get

y

lim−

x→2

5 10

5

x2 = −∞, x2 − 4

and

lim −

x→−2

x 5

5

lim +

x→−2

x2 = −∞ x2 − 4

x2 = ∞. x2 − 4

(6.8) (6.9)

Thus, there are vertical asymptotes at x = ±2. Next, we have

10

f (x) = FIGURE 3.69b y=

(6.7)

+

2x(x 2 − 4) − x 2 (2x) (x 2 − 4)

2

=

−8x (x 2 − 4)2

.

Since the denominator is positive for x = ±2, it is a simple matter to see that > 0, on (−∞, −2) ∪ (−2, 0) f increasing. f (x) < 0, on (0, 2) ∪ (2, ∞). f decreasing.

x2 x2 − 4

(6.10)

In particular, notice that the only critical number is x = 0 (since x = −2, 2 are not in the domain of f ). Thus, the only local extremum is the local maximum located at x = 0. Next, we have

0

−8(x 2 − 4)2 + (8x)2(x 2 − 4)1 (2x) (x 2 − 4)4

Quotient rule.

=

8(x 2 − 4)[−(x 2 − 4) + 4x 2 ] (x 2 − 4)4

Factor out 8(x 2 − 4).

=

8(3x 2 + 4) (x 2 − 4)3

Combine terms.

=

8(3x 2 + 4) . (x − 2)3 (x + 2)3

f (x) = (x 2)3

2

0

(x 2)3

2

2

2

2

f (x)

2

0

0

2

2

f (x)

f (x)

Factor difference of two squares.

Since the numerator is positive for all x, we need only consider the terms in the denominator, as seen in the margin. We then have > 0, on (−∞, −2) ∪ (2, ∞) Concave up. f (x) (6.11) < 0, on (−2, 2). Concave down. However, since x = 2, −2 are not in the domain of f , there are no inflection points. It is

3-61

SECTION 3.6

y

Overview of Curve Sketching

301

an easy exercise to verify that

6

lim

x2 =1 x2 − 4

(6.12)

lim

x2 = 1. x2 − 4

(6.13)

x→∞

4 2 x

6 4

..

4

2

and

x→−∞

6

From (6.12) and (6.13), we have that y = 1 is a horizontal asymptote, both as x → ∞ and as x → −∞. Finally, we observe that the only x-intercept is at x = 0. We summarize the information in (6.7)–(6.13) in the graph seen in Figure 3.70.

4 6

FIGURE 3.70

In example 6.4, we need to use computer-generated graphs, as well as a rootfinding method to determine the behavior of the function.

x2 y= 2 x −4

EXAMPLE 6.4

y 10

Graphing Where the Domain and Extrema Must Be Approximated

1 showing all significant features. x 3 + 3x 2 + 3x + 3 Solution The default graph drawn by most graphing calculators and computer algebra systems looks something like the one shown in Figure 3.71. As we have already seen, we can only determine all the significant features by doing some calculus. Since f is a rational function, it is defined for all x, except for where the denominator is zero, that is, where

Draw a graph of f (x) = x

10

10 10

FIGURE 3.71 y=

1 x 3 + 3x 2 + 3x + 3

x 3 + 3x 2 + 3x + 3 = 0. If you don’t see how to factor the expression to find the zeros exactly, you must rely on approximate methods. First, to get an idea of where the zero(s) might be, draw a graph of the cubic (see Figure 3.72). The graph does not need to be elaborate, merely detailed enough to get an idea of where and how many zeros there are. In the present case, we see that there is only one zero, around x = −2. We can verify that this is the only zero, since

y

20 10 4

2

d 3 (x + 3x 2 + 3x + 3) = 3x 2 + 6x + 3 = 3(x + 1)2 ≥ 0. dx

x 2 10 20

FIGURE 3.72

Since the derivative is never negative, observe that the function cannot decrease to cross the x-axis a second time. You can get the approximate zero x = a ≈ −2.25992 using Newton’s method or your calculator’s solver. We can use the graph in Figure 3.72 to help us compute the limits

y = x 3 + 3x 2 + 3x + 3

+

lim f (x) = lim+

x→a +

x→a

1 =∞ x 3 + 3x 2 + 3x + 3

(6.14)

+

+

and

lim f (x) = lim−

x→a −

x→a

1 = −∞. x + 3x + 3x + 3 3

2

(6.15)

−

From (6.14) and (6.15), f has a vertical asymptote at x = a. Turning to the derivative

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3-62

information, we have f (x) = −(x 3 + 3x 2 + 3x + 3)−2 (3x 2 + 6x + 3) (x + 1)2 = −3 (x 3 + 3x 2 + 3x + 3)2 2 x +1 = −3 x 3 + 3x 2 + 3x + 3 < 0, for x = a or −1

(6.16)

and f (−1) = 0. Thus, f is decreasing for x < a and x > a. Also, notice that the only critical number is x = −1, but since f is decreasing everywhere except at x = a, there are no local extrema. Turning to the second derivative, we get x +1 1(x 3 + 3x 2 + 3x + 3) − (x + 1)(3x 2 + 6x + 3) f (x) = −6 x 3 + 3x 2 + 3x + 3 (x 3 + 3x 2 + 3x + 3)2 = =

(x 3

−6(x + 1) (−2x 3 − 6x 2 − 6x) + 3x 2 + 3x + 3)3

12x(x + 1)(x 2 + 3x + 3) . (x 3 + 3x 2 + 3x + 3)3

Since (x 2 + 3x + 3) > 0 for all x (why is that?), we need not consider this factor. Considering the remaining factors, we have the number lines shown here.

0

12x

0

0

(x 1)

1

0

(x 3 3x 2 3x 3)

a 2.2599…

0 1

a

0

f (x)

0

Thus, we have that

f (x) y 2 1 3

1 1

x 1

> 0, on (a, −1) ∪ (0, ∞) < 0, on (−∞, a) ∪ (−1, 0)

2

3

FIGURE 3.73 1 x 3 + 3x 2 + 3x + 3

Concave up.

(6.17)

Concave down.

It now follows that there are inflection points at x = 0 and at x = −1. Notice that in Figure 3.71, the concavity information is not very clear and the inflection points are difficult to discern. We note the obvious fact that the function is never zero and hence, there are no x-intercepts. Finally, we consider the limits lim

1 =0 x 3 + 3x 2 + 3x + 3

(6.18)

lim

1 = 0. x 3 + 3x 2 + 3x + 3

(6.19)

x→∞

2

y=

and

x→−∞

Using all of the information in (6.14)–(6.19), we draw the graph seen in Figure 3.73. Here, we can clearly see the vertical and horizontal asymptotes, the inflection points and the fact that the function is decreasing across its entire domain.

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303

In example 6.5, we consider the graph of a transcendental function with a vertical asymptote.

EXAMPLE 6.5

Graphing Where Some Features Are Difficult to See

Draw a graph of f (x) = e1/x showing all significant features. Solution The default graph produced by our computer algebra system is not particularly helpful (see Figure 3.74a). The default graph produced by most graphing calculators (see Figure 3.74b) is certainly better, but we can’t be sure this is adequate without further analysis. First, notice that the domain of f is (−∞, 0) ∪ (0, ∞). Thus, we consider

y 5e08

lim e1/x = ∞,

2.5e08

x→0+

(6.20)

since 1/x → ∞ as x → 0+ . Also, since 1/x → −∞ as x → 0− (and et → 0 as t → −∞), we have 4

x

2

2

4

lim e1/x = 0.

x→0−

FIGURE 3.74a

(6.21)

From (6.20) and (6.21), there is a vertical asymptote at x = 0, but an unusual one, in that f (x) → ∞ on one side of 0 and f (x) → 0 on the other side. Next, 1 1/x d f (x) = e dx x −1 < 0, for all x = 0, (6.22) = e1/x x2

y = e1/x y 10

x

10

10 10

FIGURE 3.74b y = e1/x

since e1/x > 0, for all x = 0. From (6.22), we have that f is decreasing for all x = 0. We also have −1 −1 2 1/x f (x) = e1/x + e 2 2 x x x3 1 1 + 2x 2 1/x 1/x =e + 3 =e x4 x x4 < 0, on −∞, − 12 Concave down. (6.23) 1 > 0, on − 2 , 0 ∪ (0, ∞). Concave up. Since x = 0 is not in the domain of f , the only inflection point is at x = − 12 . Next, note that lim e1/x = 1,

x→∞

(6.24)

since 1/x → 0 as x → ∞ and et → 1 as t → 0. Likewise, lim e1/x = 1.

x→−∞

From (6.24) and (6.25), y = 1 is a horizontal asymptote, both as x → ∞ and as x → −∞. Finally, since e1/x > 0,

(6.25)

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y 5 4 3 2 1 4

x

2

2

4

3-64

for all x = 0, there are no x-intercepts. It is worthwhile noting that all of these features of the graph were discernible from Figure 3.74b, except for the inflection point at x = − 12 . Note that in almost any graph you draw, it is difficult to see all of the features of the function. This happens because the inflection point − 12 , e−2 or (−0.5, 0.135335 . . .) is so close to the x-axis. Since the horizontal asymptote is the line y = 1, it is difficult to see both of these features on the same graph (without drawing the graph on a very large piece of paper). We settle for the graph seen in Figure 3.75, which shows 1 all of the features except the inflection point and the concavity on the interval − 2 , 0 . To clearly see the behavior near the inflection point, we draw a graph that is zoomed-in on the area of the inflection point (see Figure 3.76). Here, while we have resolved the problem of the concavity near x = 0 and the inflection point, we have lost the details of the “big picture.”

FIGURE 3.75 y = e1/x

In our final example, we consider the graph of a function that is the sum of a trigonometric function and a polynomial.

y 0.3

EXAMPLE 6.6 0.2

Graphing the Sum of a Polynomial and a Trigonometric Function

Draw a graph of f (x) = cos x − x, showing all significant features. y

0.1 4 0.8 0.6 0.4 0.2

y

x 0.2

2

10

FIGURE 3.76 y = e1/x

4

5

x

2

2

4 10

2

5

x 5

5

10

10

4

FIGURE 3.77a

FIGURE 3.77b

y = cos x − x

y = cos x − x

Solution The default graph provided by our computer algebra system can be seen in Figure 3.77a. The graph produced by most graphing calculators looks like that in Figure 3.77b. As always, we will use the calculus to determine the behavior of the function more precisely. First, notice that the domain of f is the entire real line. Consequently, there are no vertical asymptotes. Next, we have f (x) = − sin x − 1 ≤ 0,

for all x.

(6.26)

Further, f (x) = 0 if and only if sin x = −1. So, there are critical numbers (here, these are all locations of horizontal tangent lines), but since f (x) does not change sign, there are no local extrema. Even so, it is still of interest to find the locations of the horizontal tangent lines. Recall that sin x = −1

for x =

3π 2

3-65

SECTION 3.6

x=

and more generally, for

..

Overview of Curve Sketching

305

3π + 2nπ, 2

for any integer n. Next, we see that f (x) = − cos x and on the interval [0, 2π ], we have π 3π ∪ , 2π > 0, on 0, 2 2 cos x π 3π < 0, on , . 2 2

So,

π 3π ∪ , 2π < 0, on 0, 2 2 f (x) = − cos x π 3π > 0, on , . 2 2

Concave down.

(6.27) Concave up.

Outside of [0, 2π ], f (x) simply repeats this pattern. In particular, this says that the graph has infinitely many inflection points, located at odd multiples of π/2. To determine the behavior as x → ±∞, we examine the limits lim (cos x − x) = −∞

(6.28)

lim (cos x − x) = ∞,

(6.29)

x→∞

and

y

since −1 ≤ cos x ≤ 1, for all x and since lim x = ∞. x→∞ Finally, to determine the x-intercept(s), we need to solve

15 10 5 15 10 5 5

x→−∞

f (x) = cos x − x = 0.

x 5

10

10 15

FIGURE 3.78 y = cos x − x

15

This can’t be solved exactly, however. Since f (x) ≤ 0 for all x and Figures 3.77a and 3.77b show a zero around x = 1, there is only one zero and we must approximate this. (Use Newton’s method or your calculator’s solver.) We get x ≈ 0.739085 as an approximation to the only x-intercept. Assembling all of the information in (6.26)–(6.29), we can draw the graph seen in Figure 3.78. Notice that Figure 3.77b shows the behavior just as clearly as Figure 3.78, but for a smaller range of x- and y-values. Which of these is more “representative” is open to discussion.

BEYOND FORMULAS The main characteristic of the examples in sections 3.4–3.6 is the interplay between graphing and equation solving. To analyze the graph of a function, you will go back and forth several times between solving equations (for critical numbers and inflection points and so on) and identifying graphical features of interest. Even if you have access to graphing technology, the equation solving may lead you to uncover hidden features of the graph. What types of graphical features can sometimes be hidden?

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3-66

EXERCISES 3.6 WRITING EXERCISES 1. We have talked about sketching representative graphs, but it is often impossible to draw a graph correctly to scale that shows all of the properties we might be interested in. For example, try to generate a computer or calculator graph that shows all three local extrema of x 4 − 25x 3 − 2x 2 + 80x − 3. When two extrema have y-coordinates of approximately −60 and 50, it takes a very large graph to also show a point with y = −40,000! If an accurate graph cannot show all the points of interest, perhaps a freehand sketch like the one shown below is needed. y

4 x 7. f (x) = x ln x √ 9. f (x) = x 2 + 1 5. f (x) = x +

x2 − 1 x 8. f (x) = x ln x 2 √ 10. f (x) = 2x − 1 6. f (x) =

4x x2 − x + 1 √ 13. f (x) = 3 x 3 − 3x 2 + 2x

4x 2 x2 − x + 1 √ 14. f (x) = x 3 − 3x 2 + 2x

15. f (x) = x 5 − 5x

16. f (x) = x 3 −

17. f (x) = e−2/x

18. f (x) = e1/x

11. f (x) =

12. f (x) =

2

3 x 400

In exercises 19–32, determine all significant features (approximately if necessary) and sketch a graph. 19. f (x) = (x 3 − 3x 2 + 2x)2/3 x

20. f (x) = x 6 − 10x 5 − 7x 4 + 80x 3 + 12x 2 − 192x x2 + 1 3x 2 − 1 5x 23. f (x) = 3 x −x +1 25. f (x) = x 2 x 2 − 9

2x 2 x3 + 1 4x 24. f (x) = 2 x +x +1 3 26. f (x) = 2x 2 − 1

27. f (x) = e−2x sin x

28. f (x) = sin x −

21. f (x) =

There is no scale shown on the graph because we have distorted different portions of the graph in an attempt to show all of the interesting points. Discuss the relative merits of an “honest” graph with a consistent scale but not showing all the points of interest versus a caricature graph that distorts the scale but does show all the points of interest. 2. While studying for a test, a friend of yours says that a graph is not allowed to intersect an asymptote. While it is often the case that graphs don’t intersect asymptotes, there is definitely not any rule against it. Explain why graphs can intersect a horizontal asymptote any number of times (Hint: Look at the graph of e−x sin x), but can’t pass through a vertical asymptote. 3. Explain why polynomials never have vertical or horizontal asymptotes. 4. Explain how the graph of f (x) = cos x − x in example 6.6 relates to the graphs of y = cos x and y = −x. Based on this discussion, explain how to sketch the graph of y = x + sin x.

In exercises 1–18, graph the function and completely discuss the graph as in example 6.2. 1. f (x) = x 3 − 3x 2 + 3x

2. f (x) = x 4 − 3x 2 + 2x

3. f (x) = x 5 − 2x 3 + 1

4. f (x) = sin x − cos x

22. f (x) =

29. f (x) = x 4 − 16x 3 + 42x 2 − 39.6x + 14

1 sin 2x 2

30. f (x) = x 4 + 32x 3 − 0.02x 2 − 0.8x √ 1 25 − 50 x 2 + 0.25 −1 32. f (x) = tan 31. f (x) = x x2 − 1 In exercises 33–38, the “family of functions” contains a parameter c. The value of c affects the properties of the functions. Determine what differences, if any, there are for c being zero, positive or negative. Then determine what the graph would look like for very large positive c’s and for very large negative c’s. 33. f (x) = x 4 + cx 2 x2 35. f (x) = 2 x + c2 37. f (x) = sin(cx)

34. f (x) = x 4 + cx 2 + x 36. f (x) = e−x /c √ 38. f (x) = x 2 c2 − x 2 2

39. In a variety of applications, researchers model a phenomenon whose graph starts at the origin, rises to a single maximum and then drops off to a horizontal asymptote of y = 0. For example, the probability density function of events such as the time from conception to birth of an animal and the amount of time surviving after contracting a fatal disease might have these properties. Show that the family of functions xe−bx has these properties for all positive constants b. What effect does

3-67

SECTION 3.6

b have on the location of the maximum? In the case of the time since conception, what would b represent? In the case of survival time, what would b represent? 40. The “FM” in FM radio stands for frequency modulation, a method of transmitting information encoded in a radio wave by modulating (or varying) the frequency. A basic example of such a modulated wave is f (x) = cos (10x + 2 cos x). Use computer-generated graphs of f (x), f (x) and f (x) to try to locate all local extrema of f (x). p(x) , where p(x) q(x) and q(x) are polynomials. Is it true that all rational functions have vertical asymptotes? Is it true that all rational functions have horizontal asymptotes?

41. A rational function is a function of the form

42. It can be useful to identify asymptotes other than vertical and horizontal. For example, the parabola y = x 2 is an asymptote of f (x) if lim [ f (x) − x 2 ] = 0 and/or lim [ f (x) − x 2 ] = 0. x→∞

x→−∞ 4

x − x2 + 1 . Graph x2 − 1 y = f (x) and zoom out until the graph looks like a parabola. (Note: The effect of zooming out is to emphasize large values of x.)

Show that x is an asymptote of f (x) = 2

A function f has a slant asymptote y mx b (m 0) if lim [ f (x) − (mx b)] 0 and/or lim [ f (x) − (mx b)] 0. x→∞

x→− ∞

In exercises 43–48, find the slant asymptote. (Use long division to rewrite the function.) Then, graph the function and its asymptote on the same axes. 3x 2 − 1 x x 3 − 2x 2 + 1 45. f (x) = x2 4 x 47. f (x) = 3 x +1 43. f (x) =

3x 2 − 1 x −1 x3 − 1 46. f (x) = 2 x −1 x4 − 1 48. f (x) = 3 x +x 44. f (x) =

In exercises 49–52, find a function whose graph has the given asymptotes. 49. x = 1, x = 2 and y = 3

50. x = −1, x = 1 and y = 0

51. x = −1, x = 1, y = −2 and y = 2 52. x = 1, y = 2 and x = 3 53. Find all extrema and inflection points, and sketch the graphs e x + e−x e x − e−x and y = cosh x = . of y = sinh x = 2 2 54. On the same axes as the graphs of exercise 53, sketch in the graphs of y = 12 e x and y = 12 e−x . Explain why these graphs serve as an envelope for the graphs in exercise 53. (Hint: As x → ±∞, what happens to e x and e−x ?)

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307

EXPLORATORY EXERCISES 1. For each function, find a polynomial p(x) such that lim [ f (x) − p(x)] = 0. x→∞

(a)

x4 x +1

(b)

x5 − 1 x +1

(c)

x6 − 2 x +1

Show by zooming out that f (x) and p(x) look similar for large x. The first term of a polynomial is the term with the highest power (e.g., x 3 is the first term of x 3 − 3x + 1). Can you zoom out enough to make the graph of f (x) look like the first term of its polynomial asymptote? State a very quick rule enabling you to look at a rational function and determine the first term of its polynomial asymptote (if one exists). 2. One of the natural enemies of the balsam fir tree is the spruce budworm, which attacks the leaves of the fir tree in devastating outbreaks. Define N (t) to be the number of worms on a particular tree at time t. A mathematical model of the population dynamics of the worm must include a term to indicate the worm’s death rate due to its predators (e.g., birds). The form B[N (t)]2 of this term is often taken to be 2 for positive conA + [N (t)]2 2 2x 2 x2 x , , stants A and B. Graph the functions 2 2 4 + x 1 + x 9 + x2 3x 2 and for x > 0. Based on these graphs, discuss why 1 + x2 B[N (t)]2 is a plausible model for the death rate by predaA2 + [N (t)]2 tion. What role do the constants A and B play? The possible stable population levels for the spruce budworms are determined by intersections of the graphs of y = r (1 − x/k) and x y= . Here, x = N /A, r is proportional to the birthrate 1 + x2 of the budworms and k is determined by the amount of food available to the budworms. Note that y = r (1 − x/k) is a line with y-intercept r and x-intercept k. How many solutions are x there to the equation r (1 − x/k) = ? (Hint: The answer 1 + x2 depends on the values of r and k.) One current theory is that outbreaks are caused in situations where there are three solutions and the population of budworms jumps from a small population to a large population. 3. Suppose that f (x) is a function with two derivatives and that f (a) = f (a) = 0 but f (a) = 0 for some number a. Show that f (x) has a local extremum at x = a. Next, suppose that f (x) is a function with three derivatives and that f (a) = f (a) = f (a) = 0 but f (a) = 0 for some number a. Show that f (x) does not have a local extremum at x = a. Generalize your work to the case where f (k) (a) = 0 for k = 0, 1, . . . , n − 1, but f (n) (a) = 0, keeping in mind that there are different conclusions depending on whether n is odd or even. Use this result to determine whether f (x) = x sin x 2 or g(x) = x 2 sin(x 2 ) has a local extremum at x = 0.

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OPTIMIZATION Everywhere in business and industry today, we see people struggling to minimize waste and maximize productivity. We are now in a position to bring the power of the calculus to bear on problems involving finding a maximum or a minimum. This section contains a number of illustrations of such problems. Pay close attention to how we solve the problems. We start by giving a few general guidelines. (Notice that we said guidelines and not rules.) Do not memorize them, but rather keep these in mind as you work through the section. r If there’s a picture to draw, draw it! Don’t try to visualize how things look in your

head. Put a picture down on paper and label it.

r Determine what the variables are and how they are related. r Decide what quantity needs to be maximized or minimized. r Write an expression for the quantity to be maximized or minimized in terms of only

one variable. To do this, you may need to solve for any other variables in terms of this one variable. r Determine the minimum and maximum allowable values (if any) of the variable you’re using. r Solve the problem. (Be sure to answer the question that is asked.) We begin with a simple example where the goal is to accomplish what businesses face every day: getting the most from limited resources.

EXAMPLE 7.1

Constructing a Rectangular Garden of Maximum Area

You have 40 (linear) feet of fencing with which to enclose a rectangular space for a garden. Find the largest area that can be enclosed with this much fencing and the dimensions of the corresponding garden.

OR

Solution First, note that there are lots of possibilities. We could enclose a plot that is very long but narrow, or one that is very wide but not very long (see Figure 3.79). How are we to decide which configuration is optimal? We first draw a picture and label it appropriately (see Figure 3.80). The variables for a rectangular plot are length and width, which we name x and y, respectively. We want to find the dimensions of the largest possible area, that is, maximize FIGURE 3.79 Possible plots

A = x y. Notice immediately that this function has two variables and so, cannot be dealt with via the means we have available. However, there is another piece of information that we can use here. If we want the maximum area, then all of the fencing must be used. This says that the perimeter of the resulting fence must be 40 and hence,

y

40 = perimeter = 2x + 2y. x

FIGURE 3.80 Rectangular plot

(7.1)

Notice that we can use (7.1) to solve for one variable (either one) in terms of the other. We have 2y = 40 − 2x

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y = 20 − x.

and hence, Substituting for y, we get that

A = x y = x(20 − x). So, our job is to find the maximum value of the function A(x) = x(20 − x). y

Hold it! You may want to multiply out this expression. Unless you have a clear picture of how it will affect your later work, leave the function alone! Before we maximize A(x), we need to determine the interval in which x must lie. Since x is a distance, we must have 0 ≤ x. Further, since the perimeter is 40 , we must have x ≤ 20. (Why don’t we have x ≤ 40?) So, we want to find the maximum value of A(x) on the closed interval [0, 20]. This is now a simple problem. As a check on what a reasonable answer should be, we draw a graph of y = A(x) (see Figure 3.81) on the interval [0, 20]. The maximum value appears to occur around x = 10. Now, let’s analyze the problem carefully. We have

100 80 60 40 20 x 5

10

15

FIGURE 3.81 y = x(20 − x)

20

A (x) = 1(20 − x) + x(−1) = 20 − 2x = 2(10 − x). So, the only critical number is x = 10 and this is in the interval under consideration. Recall that the maximum and minimum values of a continuous function on a closed and bounded interval must occur at either the endpoints or a critical number. This says that we need only compare the function values A(0) = 0,

A(20) = 0

and

A(10) = 100.

Thus, the maximum area that can be enclosed with 40 of fencing is 100 ft2 . We also want the dimensions of the plot. (This result is only of theoretical value if we don’t know how to construct the rectangle with the maximum area.) We have that x = 10 and y = 20 − x = 10. That is, the rectangle of perimeter 40 with maximum area is a square 10 on a side. A more general problem that you can now solve is to show that (given a fixed perimeter) the rectangle of maximum area is a square. This is virtually identical to example 7.1 and is left as an exercise. It’s worth noting here that this more general problem is one that cannot be solved by simply using a calculator to draw a graph. You’ll need to use some calculus here. Manufacturing companies routinely make countless decisions that affect the efficiency of their production processes. One decision that is surprisingly important is how to economically package products for shipping. Example 7.2 provides a simple illustration of this problem.

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18

EXAMPLE 7.2

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Constructing a Box of Maximum Volume

A square sheet of cardboard 18 on a side is made into an open box (i.e., there’s no top), by cutting squares of equal size out of each corner (see Figure 3.82a) and folding up the sides along the dotted lines (see Figure 3.82b). Find the dimensions of the box with the maximum volume.

18 2x

x

Solution Recall that the volume of a rectangular parallelepiped (a box) is given by V = l × w × h.

18 2x x

From Figure 3.82b, we can see that the height is h = x, while the length and width are l = w = 18 − 2x. Thus, we can write the volume in terms of the one variable x as

FIGURE 3.82a A sheet of cardboard

V = V (x) = (18 − 2x)2 (x) = 4x(9 − x)2 .

x 18 2x 18 2x

Once again, don’t multiply this out, just out of habit. Notice that since x is a distance, we have x ≥ 0. Further, we have x ≤ 9, since cutting squares of side 9 out of each corner will cut up the entire sheet of cardboard. Thus, we are faced with finding the absolute maximum of the continuous function V (x) = 4x(9 − x)2

FIGURE 3.82b

on the closed interval 0 ≤ x ≤ 9. This should be a simple matter. The graph of y = V (x) on the interval [0, 9] is seen in Figure 3.83. From the graph, the maximum volume seems to be somewhat over 400 and seems to occur around x = 3. Now, we solve the problem precisely. We have

Rectangular box y 500 400

V (x) = 4(9 − x)2 + 4x(2)(9 − x)(−1)

300

= 4(9 − x)[(9 − x) − 2x]

200

= 4(9 − x)(9 − 3x).

100 x 2

4

6

FIGURE 3.83 y = 4x(9 − x)2

8

Product rule and chain rule. Factor out 4(9 − x).

So, V has two critical numbers, 3 and 9, and these are both in the interval under consideration. We now need only compare the value of the function at the endpoints and the critical numbers. We have V (0) = 0,

V (9) = 0

and

V (3) = 432.

Obviously, the maximum possible volume is 432 cubic inches. We can achieve this volume if we cut squares of side 3 out of each corner. You should note that this corresponds with what we expected from the graph of y = V (x) in Figure 3.83. Finally, observe that the dimensions of this optimal box are 12 long by 12 wide by 3 deep. When a new building is built, it must be connected to existing telephone and power cables, water and sewer lines and paved roads. If the cables, water or sewer lines or road are curved, then it may not be obvious how to make the shortest (i.e., least expensive) connection possible. In examples 7.3 and 7.4, we consider the general problem of finding the shortest distance from a point to a curve.

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y

EXAMPLE 7.3 9

Finding the Closest Point on a Parabola

(3, 9)

Find the point on the parabola y = 9 − x 2 closest to the point (3, 9) (see Figure 3.84).

(x, y)

Solution Using the usual distance formula, we find that the distance between the point (3, 9) and any point (x, y) is d = (x − 3)2 + (y − 9)2 .

y 9 x2

x

4

4

FIGURE 3.84

If the point (x, y) is on the parabola, note that its coordinates satisfy the equation y = 9 − x 2 and so, we can write the distance in terms of the single variable x as follows d(x) =

y = 9 − x2

= =

(x − 3)2 + [(9 − x 2 ) − 9]2 (x − 3)2 + (−x 2 )2 (x − 3)2 + x 4 .

Although we can certainly solve the problem in its present form, we can simplify our work by observing that d(x) is minimized if and only if the quantity under the square root is minimized. (We leave it as an exercise to show why this is true.) So, instead of minimizing d(x) directly, we minimize the square of d(x):

y 80 60

f (x) = [d(x)]2 = (x − 3)2 + x 4

40 20 x 1

2

FIGURE 3.85 y = (x − 3)2 + x 4

instead. Notice from Figure 3.84 that any point on the parabola to the left of the y-axis is farther away from (3, 9) than is the point (0, 9). Likewise, any point on the parabola below the x-axis is farther from (3, 9) than is the point (3, 0). So, it suffices to look for the closest point with

3

0 ≤ x ≤ 3. See Figure 3.85 for a graph of y = f (x) over the interval of interest. Observe that the minimum value of f (the square of the distance) seems to be around 5 and seems to occur near x = 1. We have f (x) = 2(x − 3)1 + 4x 3 = 4x 3 + 2x − 6. Notice that f (x) factors. [One way to see this is to recognize that x = 1 is a zero of f (x), which makes (x − 1) a factor.] We have f (x) = 2(x − 1)(2x 2 + 2x + 3). So, x = 1 is a critical number. In fact, it’s the only critical number, since (2x 2 + 2x + 3) has no zeros. (Why not?) We now need only compare the value of f at the endpoints and the critical number. We have f (0) = 9,

f (3) = 81

and

f (1) = 5.

Thus, the minimum value of f√(x) is 5. This says that the minimum distance from the point (3, 9) to the parabola is 5 and the closest point on the parabola is (1, 8). Again, notice that this corresponds with what we expected from the graph of y = f (x).

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Example 7.4 is very similar to example 7.3, except that we need to use approximate methods to find the critical number.

(5, 11) 9 (x, y) y 9 x2

EXAMPLE 7.4

Finding Minimum Distance Approximately

Find the point on the parabola y = 9 − x 2 closest to the point (5, 11) (see Figure 3.86). x

4

4

FIGURE 3.86

Solution As in example 7.3, we want to minimize the distance from a fixed point [in this case, the point (5, 11)] to a point (x, y) on the parabola. Using the distance formula, the distance from any point (x, y) on the parabola to the point (5, 11) is d=

y = 9 − x2

= y

=

(x − 5)2 + (y − 11)2 (x − 5)2 + [(9 − x 2 ) − 11]2 (x − 5)2 + (x 2 + 2)2 .

600

Again, it is equivalent (and simpler) to minimize the quantity under the square root: 400

f (x) = [d(x)]2 = (x − 5)2 + (x 2 + 2)2 .

200

As in example 7.3, we can see from Figure 3.86 that any point on the parabola to the left of the y-axis is farther from (5, 11) than is (0, 9). Likewise, any point on the parabola to the right of x = 5 is farther from (5, 11) than is (5, −16). Thus, we minimize f (x) for 0 ≤ x ≤ 5. In Figure 3.87 we see a graph of y = f (x) on the interval of interest. The minimum value of f seems to occur around x = 1. We can make this more precise as follows:

x 1

2

3

4

5

FIGURE 3.87 y = f (x) = [d(x)]2

f (x) = 2(x − 5) + 2(x 2 + 2)(2x) = 4x 3 + 10x − 10.

y

Unlike in example 7.3, the expression for f (x) has no obvious factorization. Our only choice then is to find zeros of f (x) approximately. First, we draw a graph of y = f (x) on the interval of interest (see Figure 3.88). The only zero appears to be slightly less than 1. Using x0 = 1 as an initial guess in Newton’s method (applied to f (x) = 0) or using your calculator’s solver, you should get the approximate root xc ≈ 0.79728. We now compare function values:

300

200

100

x 1

2

3

4

FIGURE 3.88 y = f (x)

f (0) = 29,

f (5) = 729

and

f (xc ) ≈ 24.6.

5

Thus, the minimum distance from (5, 11) to the parabola is approximately √ 24.6 ≈ 4.96 and the closest point on the parabola is located at approximately (0.79728, 8.364).

Notice that in both Figures 3.84 and 3.86, the shortest path appears to be perpendicular to the tangent line to the curve at the point where the path intersects the curve. We leave it as an exercise to prove that this is always the case. This observation is an important geometric principle that applies to many problems of this type.

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REMARK 7.1 At this point you might be tempted to forgo the comparison of function values at the endpoints and at the critical numbers. After all, in all of the examples we have seen so far, the desired maximizer or minimizer (i.e., the point at which the maximum or minimum occurred) was the only critical number in the interval under consideration. You might just suspect that if there is only one critical number, it will correspond to the maximizer or minimizer for which you are searching. Unfortunately, this is not always the case. In 1945, two prominent aeronautical engineers derived a function to model the range of an aircraft. Their intention was to use this function to discover how to maximize the range. They found a critical number of this function (corresponding to distributing virtually all of the plane’s weight in the wings) and reasoned that it gave the maximum range. The result was the famous “Flying Wing” aircraft. Some years later, it was argued that the critical number they found corresponded to a local minimum of the range function. In the engineers’ defense, they did not have easy, accurate computational power at their fingertips, as we do today. Remarkably, this design strongly resembles the modern B-2 Stealth bomber. This story came out as controversy brewed over the production of the B-2 (see Science, 244, pp. 650–651, May 12, 1989; also see the Monthly of the Mathematical Association of America, October, 1993, pp. 737–738). The moral should be crystal clear: check the function values at the critical numbers and at the endpoints. Do not simply assume (even by virtue of having only one critical number) that a given critical number corresponds to the extremum you are seeking.

Next, we consider an optimization problem that cannot be restricted to a closed interval. We will use the fact that for a continuous function, a single local extremum must be an absolute extremum. (Think about why this is true.)

EXAMPLE 7.5

A soda can is to hold 12 fluid ounces. Find the dimensions that will minimize the amount of material used in its construction, assuming that the thickness of the material is uniform (i.e., the thickness of the aluminum is the same everywhere in the can).

r

h

FIGURE 3.89 Soda can

Designing a Soda Can That Uses a Minimum Amount of Material

Solution First, we draw and label a picture of a typical soda can (see Figure 3.89). Here we have drawn a right circular cylinder of height h and radius r . Assuming uniform thickness of the aluminum, notice that we minimize the amount of material by minimizing the surface area of the can. We have area = area of top + area of bottom + curved surface area = 2πr 2 + 2πr h.

(7.2)

We can eliminate one of the variables by using the fact that the volume (using 1 fluid ounce ≈ 1.80469 in.3 ) must be 12 fluid ounces ≈ 12 fl oz × 1.80469

in.3 = 21.65628 in.3 . fl oz

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Further, the volume of a right circular cylinder is vol = πr 2 h and so, y

h=

vol 21.65628 ≈ . πr 2 πr 2

(7.3)

Thus, from (7.2) and (7.3), the surface area is approximately 21.65628 A(r ) = 2πr 2 + 2πr πr 2 21.65628 = 2π r 2 + . πr

150 100 50 x 1

2

3

4

5

FIGURE 3.90 y = A(r )

6

So, our job is to minimize A(r ), but here, there is no closed and bounded interval of allowable values. In fact, all we can say is that r > 0. We can have r as large or small as you can imagine, simply by taking h to be correspondingly small or large, respectively. That is, we must find the absolute minimum of A(r ) on the open and unbounded interval (0, ∞). To get an idea of what a plausible answer might be, we graph y = A(r ) (see Figure 3.90). There appears to be a local minimum (slightly less than 50) located between r = 1 and r = 2. Next, we compute d 21.65628 A (r ) = 2π r 2 + dr πr 21.65628 = 2π 2r − πr 2 2πr 3 − 21.65628 = 2π . πr 2 Notice that the only critical numbers are those for which the numerator of the fraction is zero: 0 = 2πr 3 − 21.65628. r3 =

This occurs if and only if

21.65628 2π

and hence, the only critical number is r = rc =

3

21.65628 ≈ 1.510548. 2π

Further, notice that for 0 < r < rc , A (r ) < 0 and for rc < r, A (r ) > 0. That is, A(r ) is decreasing on the interval (0, rc ) and increasing on the interval (rc , ∞). Thus, A(r ) has not only a local minimum, but also an absolute minimum at r = rc . Notice, too, that this corresponds with what we expected from the graph of y = A(r ) in Figure 3.90. This says that the can that uses a minimum of material has radius rc ≈ 1.510548 and height h=

21.65628 ≈ 3.0211. πrc2

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Note that the optimal can from example 7.5 is “square,” in the sense that the height (h) equals the diameter (2r ). Also, we should observe that example 7.5 is not completely realistic. A standard 12-ounce soda can has a radius of about 1.156 . You should review example 7.5 to find any unrealistic assumptions we made. We study the problem of designing a soda can further in the exercises. In our final example, we consider a problem where most of the work must be done numerically and graphically.

EXAMPLE 7.6

Minimizing the Cost of Highway Construction

The state wants to build a new stretch of highway to link an existing bridge with a turnpike interchange, located 8 miles to the east and 8 miles to the south of the bridge. There is a 5-mile-wide stretch of marshland adjacent to the bridge that must be crossed (see Figure 3.91). Given that the highway costs $10 million per mile to build over the marsh and only $7 million to build over dry land, how far to the east of the bridge should the highway be when it crosses out of the marsh?

Bridge

5

Marsh x

8x 3 Interchange

FIGURE 3.91 A new highway

Solution You might guess that the highway should cut directly across the marsh, so as to minimize the amount built over marshland. We will use the calculus to decide this question. We let x represent the distance in question (see Figure 3.91). Then, the interchange lies (8 − x) miles to the east of the point where the highway leaves the marsh. Thus, the total cost (in millions of dollars) is

y

cost = 10(distance across marsh) + 7(distance across dry land ).

120

Using the Pythagorean Theorem on the two right triangles seen in Figure 3.91, we get the cost function C(x) = 10 x 2 + 25 + 7 (8 − x)2 + 9.

110

100 x 2

4

6

FIGURE 3.92 y = C(x)

8

Observe from Figure 3.91 that we must have 0 ≤ x ≤ 8. So, we have the routine problem of minimizing a continuous function C(x) over the closed and bounded interval [0, 8]. Or is it really that routine? First, we draw a graph of y = C(x) on the interval in question to get an idea of a plausible answer (see Figure 3.92). From the graph, the minimum appears to be slightly less than 100 and occurs around x = 4.

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We now compare d 2 10 x + 25 + 7 (8 − x)2 + 9 dx 7 = 5(x 2 + 25)−1/2 (2x) + [(8 − x)2 + 9]−1/2 (2)(8 − x)1 (−1) 2 10x 7(8 − x) = √ − . x 2 + 25 (8 − x)2 + 9

10

C (x) = 5 x 2

4

6

8

5 10

FIGURE 3.93 y = C (x)

First, note that the only critical numbers are where C (x) = 0. (Why?) The only way to find these is to approximate them. From the graph of y = C (x) seen in Figure 3.93, the only zero of C (x) on the interval [0, 8] appears to be between x = 3 and x = 4. We approximate this zero numerically (e.g., with bisections or your calculator’s solver), to obtain the approximate critical number xc ≈ 3.560052. Now, we need only compare the value of C(x) at the endpoints and at this one critical number: C(0) ≈ $109.8 million, C(8) ≈ $115.3 million and

C(xc ) ≈ $98.9 million.

So, by using a little calculus, we can save the taxpayers more than $10 million over cutting directly across the marsh and more than $16 million over cutting diagonally across the marsh (not a bad reward for a few minutes of work). The examples that we’ve presented in this section together with the exercises should give you the basis for solving a wide range of applied optimization problems. When solving these problems, be careful to draw good pictures, as well as graphs of the functions involved. Make sure that the answer you obtain computationally is consistent with what you expect from the graphs. If not, further analysis is required to see what you have missed. Also, make sure that the solution makes physical sense, when appropriate. All of these multiple checks on your work will reduce the likelihood of error.

EXERCISES 3.7 WRITING EXERCISES 1. Suppose some friends complain to you that they can’t work any of the problems in this section. When you ask to see their work, they say that they couldn’t even get started. In the text, we have emphasized sketching a picture and defining variables. Part of the benefit of this is to help you get started writing something (anything) down. Do you think this advice helps? What do you think is the most difficult aspect of these problems? Give your friends the best advice you can.

2. We have neglected one important aspect of optimization problems, an aspect that might be called “common sense.” For example, suppose you are finding the optimal dimensions for a fence and the √mathematical solution is to build a square fence of length 10 5 feet on each side. At the meeting with the carpenter who is going √to build the fence, what length fence do you order? Why is 10 5 probably not √ the best way to express the length? We can approximate 10 5 ≈ 22.36. What would you

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tell the carpenter? Suppose the carpenter accepts measurements down to the inch. Assuming that the building constraint was that the perimeter of the fence could not exceed a certain figure, why should you truncate to 22 4 instead of rounding up to 22 5 ? √ 3. In example 7.3, we stated that d(x) = f (x) is minimized √ by exactly the same x-value(s) as f (x). Use the fact that x is an increasing function to explain why this is true. 4. Suppose that f (x) is a continuous function with a single critical number and f (x) has a local minimum at that critical number. Explain why f (x) also has an absolute minimum at the critical number. 1. Give an example showing that f (x) and sin ( f (x)) need not be minimized by the same x-values. 2. True or false: e f (x) is minimized by exactly the same x-value(s) as f (x). 3. A three-sided fence is to be built next to a straight section of river, which forms the fourth side of a rectangular region. The enclosed area is to equal 1800 ft2 . Find the minimum perimeter and the dimensions of the corresponding enclosure. 4. A three-sided fence is to be built next to a straight section of river, which forms the fourth side of a rectangular region. There is 96 feet of fencing available. Find the maximum enclosed area and the dimensions of the corresponding enclosure. 5. A two-pen corral is to be built. The outline of the corral forms two identical adjoining rectangles. If there is 120 ft of fencing available, what dimensions of the corral will maximize the enclosed area? 6. A showroom for a department store is to be rectangular with walls on three sides, 6-ft door openings on the two facing sides and a 10-ft door opening on the remaining wall. The showroom is to have 800 ft2 of floor space. What dimensions will minimize the length of wall used? 7. Show that the rectangle of maximum area for a given perimeter P is always a square. 8. Show that the rectangle of minimum perimeter for a given area A is always a square. 9. Find the point on the curve y = x 2 closest to the point (0, 1). 10. Find the point on the curve y = x 2 closest to the point (3, 4). 11. Find the point on the curve y = cos x closest to the point (0, 0). 12. Find the point on the curve y = cos x closest to the point (1, 1). 13. In exercises 9 and 10, find the slope of the line through the given point and the closest point on the given curve. Show that in each case, this line is perpendicular to the tangent line to the curve at the given point. 14. Sketch the graph of some function y = f (x) and mark a point not on the curve. Explain why the result of exercise 13

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is true. (Hint: Pick a point for which the joining line is not perpendicular and explain why you can get closer.) 15. A box with no top is to be built by taking a 6 -by-10 sheet of cardboard and cutting x-in. squares out of each corner and folding up the sides. Find the value of x that maximizes the volume of the box. 16. A box with no top is to be built by taking a 12 -by-16 sheet of cardboard and cutting x-in. squares out of each corner and folding up the sides. Find the value of x that maximizes the volume of the box. 17. A water line runs east-west. A town wants to connect two new housing developments to the line by running lines from a single point on the existing line to the two developments. One development is 3 miles south of the existing line; the other development is 4 miles south of the existing line and 5 miles east of the first development. Find the place on the existing line to make the connection to minimize the total length of new line. 18. A company needs to run an oil pipeline from an oil rig 25 miles out to sea to a storage tank that is 5 miles inland. The shoreline runs east-west and the tank is 8 miles east of the rig. Assume it costs $50 thousand per mile to construct the pipeline under water and $20 thousand per mile to construct the pipeline on land. The pipeline will be built in a straight line from the rig to a selected point on the shoreline, then in a straight line to the storage tank. What point on the shoreline should be selected to minimize the total cost of the pipeline? 19. A city wants to build a new section of highway to link an existing bridge with an existing highway interchange, which lies 8 miles to the east and 10 miles to the south of the bridge. The first 4 miles south of the bridge is marshland. Assume that the highway costs $5 million per mile over marsh and $2 million per mile over dry land. The highway will be built in a straight line from the bridge to the edge of the marsh, then in a straight line to the existing interchange. At what point should the highway emerge from the marsh in order to minimize the total cost of the new highway? How much is saved over building the new highway in a straight line from the bridge to the interchange? (Hint: Use similar triangles to find the point on the boundary corresponding to a straight path and evaluate your cost function at that point.) 20. After construction has begun on the highway in exercise 19, the cost per mile over marshland is reestimated at $6 million. Find the point on the marsh/dry land boundary that would minimize the total cost of the highway with the new cost function. If the construction is too far along to change paths, how much extra cost is there in using the path from exercise 19? 21. After construction has begun on the highway in exercise 19, the cost per mile over dry land is reestimated at $3 million. Find the point on the marsh/dry land boundary that would minimize the total cost of the highway with the new cost function. If the construction is too far along to change paths, how much extra cost is there in using the path from exercise 19?

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3-78

22. In an endurance contest, contestants 2 miles at sea need to reach a location 2 miles inland and 3 miles east (the shoreline runs east-west). Assume a contestant can swim 4 mph and run 10 mph. To what point on the shoreline should the person swim to minimize the total time? Compare the amount of time spent in the water and the amount of time spent on land. 23. Suppose that light travels from point A to point B as shown in the figure. (Recall that light always follows the path that minimizes time.) Assume that the velocity of light above the boundary line is v1 and the velocity of light below the boundary is v2 . Show that the total time to get from point A to point B is T (x) =

√ 1 + x2 1 + (2 − x)2 + . v1 v2

Write out the equation T (x) = 0, replace the square roots using the sines of the angles in the figure and derive Snell’s sin θ1 v1 Law = . sin θ2 v2 A

25. A soda can is to hold 12 fluid ounces. Suppose that the bottom and top are twice as thick as the sides. Find the dimensions of the can that minimize the amount of material used. (Hint: Instead of minimizing surface area, minimize the cost, which is proportional to the product of the thickness and the area.) 26. Following example 7.5, we mentioned that real soda cans have a radius of about 1.156 . Show that this radius minimizes the cost if the top and bottom are 2.23 times as thick as the sides. 27. The human cough is intended to increase the flow of air to the lungs, by dislodging any particles blocking the windpipe and changing the radius of the pipe. Suppose a windpipe under no pressure has radius r0 . The velocity of air through the windpipe at radius r is approximately V (r ) = cr 2 (r0 − r ) for some constant c. Find the radius that maximizes the velocity of air through the windpipe. Does this mean the windpipe expands or contracts? 28. To supply blood to all parts of the body, the human artery system must branch repeatedly. Suppose an artery of radius r branches off from an artery of radius R (R > r ) at an angle θ . The energy lost due to friction is approximately E(θ) =

1

u1

2x x

csc θ 1 − cot θ + . r4 R4

Find the value of θ that minimizes the energy loss. 29. In an electronic device, individual circuits may serve many purposes. In some cases, the flow of electricity must be controlled by reducing the power instead of amplifying it. In the circuit shown here, a voltage V volts and resistance R ohms are given. We want to determine the size of the remaining resistor (x ohms). The power absorbed by the circuit is

u2 1

B

p(x) =

Exercise 23 24. Suppose that light reflects off a mirror to get from point A to point B as indicated in the figure. Assuming a constant velocity of light, we can minimize time by minimizing the distance traveled. Find the point on the mirror that minimizes the distance traveled. Show that the angles in the figure are equal (the angle of incidence equals the angle of reflection).

V 2x . (R + x)2

Find the value of x that maximizes the power absorbed. R V

x

A

B 2 u1

u2 4x

x

Exercise 24

1

30. In an AC circuit with voltage V (t) = v sin 2π ft, a voltmeter actually shows the average (root-mean-square) voltage of √ v/ 2. If the frequency is f = 60 (Hz) and the meter registers 115 volts, find the maximum voltage reached. [Hint: This is “obvious” if you determine v and think about the graph of V (t).] 31. A Norman window has the outline of a semicircle on top of a rectangle, as shown in the figure. Suppose there is 8 + π feet of wood trim available. Discuss why a window designer might want to maximize the area of the window. Find the dimensions

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SECTION 3.7

of the rectangle (and, hence, the semicircle) that will maximize the area of the window.

..

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319

42. Suppose that Q(t) represents the output of a worker, as in exercise 41. If T is the length of a workday, then the graph of Q(t) should be increasing for 0 ≤ t ≤ T . Suppose that the graph of Q(t) has a single inflection point for 0 ≤ t ≤ T , called the point of diminishing returns. Show that the worker’s efficiency is maximized at the point of diminishing returns. 43. Suppose that group tickets to a concert are priced at $40 per ticket if 20 tickets are ordered, but cost $1 per ticket less for each extra ticket ordered, up to a maximum of 50 tickets. (For example, if 22 tickets are ordered, the price is $38 per ticket.) Find the number of tickets that maximizes the total cost of the tickets.

32. Suppose a wire 2 ft long is to be cut into two pieces, each of which will be formed into a square. Find the size of each piece to maximize the total area of the two squares. 33. An advertisement consists of a rectangular printed region plus 1-in. margins on the sides and 2-in. margins at top and bottom. If the area of the printed region is to be 92 in.2 , find the dimensions of the printed region and overall advertisement that minimize the total area. 34. An advertisement consists of a rectangular printed region plus 1-in. margins on the sides and 1.5-in. margins at top and bottom. If the total area of the advertisement is to be 120 in.2 , what dimensions should the advertisement be to maximize the area of the printed region? 35. A hallway of width a = 5 ft meets a hallway of width b = 4 ft at a right angle. Find the length of the longest ladder that could be carried around the corner. (Hint: Express the length of the ladder as a function of the angle θ in the figure.) u

b

44. In exercise 43, if management wanted the solution to be 50 (that is, ordering the maximum number of tickets produces the maximum cost), how much should the price be discounted for extra tickets ordered? 45. In sports where balls are thrown or hit, the ball often finishes at a different height than it starts. Examples include a downhill golf shot and a basketball shot. In the diagram, a ball is released at an angle θ and finishes at an angle β above the horizontal (for downhill trajectories, β would be negative). Neglecting air resistance and spin, the horizontal range is given by R=

2v 2 cos2 θ (tan θ − tan β) g

if the initial velocity is v and g is the gravitational constant. In the following cases, find θ to maximize R (treat v and g as constants): (a) β = 10◦ , (b) β = 0◦ and (c) β = −10◦ . Verify that θ = 45◦ + β ◦ /2 maximizes the range.

u

b

a

36. In exercise 35, show that the maximum ladder length for general a and b equals (a 2/3 + b2/3 )3/2 . 37. In exercise 35, suppose that a = 5 and the ladder is 8 ft long. Find the minimum value of b such that the ladder can turn the corner. 38. Solve exercise 37 for a general a and ladder length L. 39. A company’s revenue for selling x (thousand) items is given 35x − x 2 . Find the value of x that maximizes the by R(x) = 2 x + 35 revenue and find the maximum revenue. 40. The function in exercise 39 has a special form. For any positive cx − x 2 constant c, find x to maximize R(x) = 2 . x +c 41. In t hours, a worker makes Q(t) = −t 3 + 12t 2 + 60t items. Graph Q (t) and explain why it can be interpreted as the efficiency of the worker. Find the time at which the worker’s efficiency is maximum.

46. For your favorite sport in which it is important to throw or hit a ball a long way, explain the result of exercise 45 in the language of your sport. 47. A ball is thrown from s = b to s = a (where a < b) with initial speed v0 . Assuming that air resistance is proportional to speed, the time it takes the ball to reach s = a is b−a 1 , T = − ln 1 − c c v0 where c is a constant of proportionality. A baseball player is 300 ft from home plate and throws a ball directly toward home plate with an initial speed of 125 ft/s. Suppose that c = 0.1. How long does it take the ball to reach home plate? Another player standing x feet from home plate has the option of catching the ball and then, after a delay of 0.1 s, relaying the ball

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toward home plate with an initial speed of 125 ft/s. Find x to minimize the total time for the ball to reach home plate. Is the straight throw or the relay faster? What, if anything, changes if the delay is 0.2 s instead of 0.1 s?

3-80

imperfections in the dimensions of the cask would have little effect on the volume. Explain why V (y) = 0 means that small variations in y would convert to small errors in the volume V .

48. For the situation in exercise 47, for what length delay is it equally fast to have a relay and not have a relay? Do you think that you could catch and throw a ball in such a short time? Why do you think it is considered important to have a relay option in baseball?

z 2x

FIGURE a

49. Repeat exercises 47 and 48 if the second player throws the ball with initial speed 100 ft/s. 50. For a delay of 0.1 s in exercise 47, find the value of the initial speed of the second player’s throw for which it is equally fast to have a relay and not have a relay. x2 y2 + 2 = 1 defines an ellipse with 2 a b −a ≤ x ≤ a and −b ≤ y ≤ b. The area enclosed by the ellipse equals πab. Find the maximum area of a rectangle inscribed in the ellipse (that is, a rectangle with sides parallel to the x-axis and y-axis and vertices on the ellipse). Show that the ratio of the maximum inscribed area to the area of the ellipse to the area of the circumscribed rectangle is 1 : π2 : 2.

z

53. Find the maximum area of an isosceles triangle of given perimeter p. [Hint: Use Heron’s formula for the area of a √ triangle of sides a, b and c : A = s(s − a)(s − b)(s − c), where s = 12 (a + b + c).]

EXPLORATORY EXERCISES 1. In exploratory exercise 2 in section 3.3, you did a preliminary investigation of Kepler’s wine cask √ problem. You showed that a height-to-diameter ratio (x/y) of 2 for a cylindrical barrel will maximize the volume (see Figure a). However, real wine casks are bowed out (like beer kegs). Kepler continued his investigation of wine cask construction by approximating a cask with the straight-sided barrel in Figure b. It can be shown (we told you Kepler was good!) that the√volume of this barrel is V = 23 π[y 2 + (w − y)2 + y(w − y)] z 2 − w 2 . Treating w and z as constants, show that V (y) = 0 if y = w/2. Recall that such a critical point can correspond to a maximum or minimum of V (y), but it also could correspond to something else (e.g., an inflection point). To discover which one we have here, redraw Figure b to scale (show the correct relationship between 2y and w). In physical terms (think about increasing and decreasing y), argue that this critical point is neither a maximum nor minimum. Interestingly enough, such a nonextreme critical point would have a definite advantage to the Austrian vintners. Recall that their goal was to convert the measurement z into an estimate of the volume. The vintners would hope that small

w

2y

FIGURE b

51. The equation

52. Show that the maximum volume enclosed by a right circular cylinder inscribed in a sphere equals √13 times the volume of the sphere.

2y

2. The following problem is fictitious, but involves the kind of ambiguity that can make technical jobs challenging. The Band Candy Company decides to liquidate one of its candies. The company has 600,000 bags in inventory that it wants to sell. The candy had cost 35 cents per bag to manufacture and originally sold for 90 cents per bag. A marketing study indicates that if the candy is priced at p cents per bag, approximately Q( p) = − p 2 + 40 p + 250 thousand bags will be sold. Your task as consultant is to recommend the best selling price for the candy. As such, you should do the following: (a) find p to maximize Q( p); (b) find p to maximize 10 p Q( p), which is the actual revenue brought in by selling the candy. Then form your opinion, based on your evaluation of the relative importance of getting rid of as much candy as possible and making the most money possible. 3. A wonderful article by Timothy J. Pennings in the May 2003 issue of The College Mathematics Journal asks the question, “Do Dogs Know Calculus?” A slightly simplified version of exercises 17–22 can be solved in a very general form. Suppose that a ball is thrown from point A on the edge of the water and lands at point B, which is x meters into the water and z meters downshore from point A. (See the diagram.) At what point C should a dog enter the water to minimize the time to reach the ball? Assume that the dog’s running speed is r m/s and the dog’s swimming speed is s m/s. Find y as a function of x to minimize the time to reach the ball. Show that the answer is independent of z! Explain why your solution is invalid if r ≤ s and explain what the dog should do in this case. Dr. Pennings’ dog Elvis was clocked at r = 6.4 m/s and s = 0.9 m/s. Show that Elvis should follow the rule y = 0.144x. In fact, Elvis’ actual entry points are very close to these values for a variety of throws! B x A

C z

y

3-81

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3.8

..

Related Rates

321

RELATED RATES In this section, we present a group of problems known as related rates problems. The common thread in each problem is an equation relating two or more quantities that are all changing with time. In each case, we will use the chain rule to find derivatives of all terms in the equation (much as we did in section 2.8 with implicit differentiation). The differentiated equation allows us to determine how different derivatives (rates) are related.

EXAMPLE 8.1

A Related Rates Problem

An oil tanker has an accident and oil pours out at the rate of 150 gallons per minute. 1 Suppose that the oil spreads onto the water in a circle at a thickness of 10 (see Figure 3.94). Given that 1 ft3 equals 7.5 gallons, determine the rate at which the radius of the spill is increasing when the radius reaches 500 feet. FIGURE 3.94

Solution Since the area of a circle of radius r is πr 2 , the volume of oil is given by

Oil spill

V = (depth)(area) = since the depth is

1 10

=

1 πr 2 , 120

1 120

ft. Both volume and radius are functions of time, so π V (t) = [r (t)]2 . 120 Differentiating both sides of the equation with respect to t, we get π V (t) = 2r (t)r (t). 120 We are given a radius of 500 feet. The volume increases at a rate of 150 gallons per minute, or 150 = 20 ft3 /min. Substituting in V (t) = 20 and r = 500, we have 7.5 π 2(500)r (t). 120 Finally, solving for r (t), we find that the radius is increasing at the rate of 2.4 ≈ 0.76394 feet per minute. π 20 =

Although the details change from problem to problem, the general pattern of solution is the same for all related rates problems. Looking back, you should be able to identify each of the following steps in example 8.1.

1. 2. 3. 4. 5.

Make a simple sketch, if appropriate. Set up an equation relating all of the relevant quantities. Differentiate (implicitly) both sides of the equation with respect to time (t). Substitute in values for all known quantities and derivatives. Solve for the remaining rate.

EXAMPLE 8.2

A Sliding Ladder

A 10-foot ladder leans against the side of a building. If the top of the ladder begins to slide down the wall at the rate of 2 ft/sec, how fast is the bottom of the ladder sliding away from the wall when the top of the ladder is 8 feet off the ground?

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y

10 x

FIGURE 3.95

3-82

Solution First, we make a sketch of the problem, as seen in Figure 3.95. We have denoted the height of the top of the ladder as y and the distance from the wall to the bottom of the ladder as x. Since the ladder is sliding down the wall at the rate of 2 ft/sec, dy we must have that = −2. (Note the minus sign here.) Observe that both x and y are dt functions of time, t. We can relate the variables by observing that, since the ladder is 10 feet long, the Pythagorean Theorem gives us

Sliding ladder

[x(t)]2 + [y(t)]2 = 100. Differentiating both sides of this equation with respect to time gives us d d (100) = [x(t)]2 + [y(t)]2 dt dt = 2x(t)x (t) + 2y(t)y (t).

0=

We can solve for x (t), to obtain x (t) = −

y(t) y (t). x(t)

To make use of this, we need values for x(t), y(t) and y (t) at the point in question. Since we know that the height above ground of the top of the ladder is 8 feet, we have y = 8 and from the Pythagorean Theorem, we get 100 = x 2 + 82 , so that x = 6. We now have that at the point in question, x (t) = −

y(t) 8 8 y (t) = − (−2) = . x(t) 6 3

So, the bottom of the ladder is sliding away from the building at the rate of

EXAMPLE 8.3 y

50

x 40

FIGURE 3.96 Cars approaching an intersection

8 3

ft/sec.

Another Related Rates Problem

A car is traveling at 50 mph due south at a point 12 mile north of an intersection. A police car is traveling at 40 mph due west at a point 14 mile east of the same intersection. At that instant, the radar in the police car measures the rate at which the distance between the two cars is changing. What does the radar gun register? Solution First, we sketch a picture and denote the vertical distance of the first car from the center of the intersection y and the horizontal distance of the police car x dx (see Figure 3.96). Notice that this says that = −40, since the police car is moving dt dy in the direction of the negative x-axis and = −50, since the other car is moving in dt the direction of the negative y-axis. From the Pythagorean Theorem, the distance between the two cars is d = x 2 + y 2 . Since all quantities are changing with time, our equation is d(t) = [x(t)]2 + [y(t)]2 = {[x(t)]2 + [y(t)]2 }1/2 .

3-83

SECTION 3.8

..

Related Rates

323

Differentiating both sides with respect to t, we have by the chain rule that 1 {[x(t)]2 + [y(t)]2 }−1/2 2[x(t)x (t) + y(t)y (t)] 2 x(t)x (t) + y(t)y (t) = . [x(t)]2 + [y(t)]2

d (t) =

Substituting in x(t) = 14 , x (t) = −40, y(t) = d (t) =

1 (−40) 4

1 4

1 2

+ 12 (−50) +

1 16

and y (t) = −50, we have −140 = √ ≈ −62.6, 5

so that the radar gun registers 62.6 mph. Note that this is a poor estimate of the car’s actual speed. For this reason, police nearly always take radar measurements from a stationary position. In some problems, the variables are not related by a geometric formula, in which case you will not need to follow the first two steps of our outline. In example 8.4, the third step is complicated by the lack of a given value for one of the rates of change.

EXAMPLE 8.4

Estimating a Rate of Change in Economics

A small company estimates that when it spends x thousand dollars for advertising in a year, its annual sales will be described by s = 60 − 40e−0.05x thousand dollars. The four most recent annual advertising totals are given in the following table. Year Dollars

1 14,500

2 16,000

3 18,000

4 20,000

Estimate the current (year 4) value of x (t) and the current rate of change of sales. Solution From the table, we see that the recent trend is for advertising to increase by $2000 per year. A good estimate is then x (4) ≈ 2. Starting with the sales equation s(t) = 60 − 40e−0.05x(t) , we use the chain rule to obtain s (t) = −40e−0.05x(t) [−0.05x (t)] = 2x (t)e−0.05x(t) . Using our estimate that x (4) ≈ 2 and since x(4) = 20, we get s (4) ≈ 2(2)e−1 ≈ 1.472. Thus, sales are increasing at the rate of approximately $1472 per year. Notice that example 8.5 is similar to example 8.6 of section 2.8.

EXAMPLE 8.5

Tracking a Fast Jet

A spectator at an air show is trying to follow the flight of a jet. The jet follows a straight path in front of the observer at 540 mph. At its closest approach, the jet passes 600 feet in front of the person. Find the maximum rate of change of the angle between the spectator’s line of sight and a line perpendicular to the flight path, as the jet flies by.

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y

3-84

Solution Place the spectator at the origin (0, 0) and the jet’s path left to right on the line y = 600, and call the angle between the positive y-axis and the line of sight θ (see Figure 3.97). If we measure distance in feet and time in seconds, we first need to convert the jet’s speed to feet per second. We have 1 h ft 540 mi 5280 mi = 792 fts . = 540 mi h h 3600 s

Path of plane 600

θ

Observer

x

FIGURE 3.97 Path of jet

From triangle trigonometry (see Figure 3.97), an equation relating the angle θ with x x and y is tan θ = . Be careful with this; since we are measuring θ from the vertical, this y equation may not be what you expect. Since all quantities are changing with time, we have x(t) tan θ (t) = . y(t) Differentiating both sides with respect to time, we have [sec2 θ (t)] θ (t) =

x (t)y(t) − x(t)y (t) . [y(t)]2

With the jet moving left to right along the line y = 600, we have x (t) = 792, y(t) = 600 and y (t) = 0. Substituting these quantities, we have [sec2 θ(t)] θ (t) =

792(600) = 1.32. 6002

Solving for the rate of change θ (t), we get θ (t) =

1.32 = 1.32 cos2 θ(t). sec2 θ(t)

Observe that the rate of change is a maximum when cos2 θ(t) is a maximum. Since the maximum of the cosine function is 1, the maximum value of cos2 θ (t) is 1, occurring when θ = 0. We conclude that the maximum rate of angle change is 1.32 radians/second. This occurs when θ = 0, that is, when the jet reaches its closest point to the observer. (Think about this; it should match your intuition!) Since humans can track objects at up to about 3 radians/second, this means that we can visually follow even a fast jet at a very small distance.

EXERCISES 3.8 WRITING EXERCISES 1. As you read examples 8.1–8.3, to what extent do you find the pictures helpful? In particular, would it be clear what x and y represent in example 8.3 without a sketch? Also, in example 8.3 explain why the derivatives x (t), y (t) and d (t) are all negative. Does the sketch help in this explanation?

1 ft3 equals 7.5 gallons, determine the rate at which the radius of the spill is increasing when the radius reaches (a) 100 ft and (b) 200 ft. Explain why the rate decreases as the radius increases.

2. In example 8.4, the increase in advertising dollars from year 1 to year 2 was $1500. Explain why this amount is not especially relevant to the approximation of s (4).

2. Oil spills out of a tanker at the rate of 90 gallons per minute. The oil spreads in a circle with a thickness of 18 . Determine the rate at which the radius of the spill is increasing when the radius reaches 100 feet.

1. Oil spills out of a tanker at the rate of 120 gallons per minute. The oil spreads in a circle with a thickness of 14 . Given that

3. Oil spills out of a tanker at the rate of g gallons per minute. The oil spreads in a circle with a thickness of 14 . Given that the radius of the spill is increasing at a rate of

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..

Related Rates

325

0.6 ft/min when the radius equals 100 feet, determine the value of g.

15. Rework example 8.3 if the police car is not moving. Does this make the radar gun’s measurement more accurate?

4. In exercises 1–3 and example 8.1, if the thickness of the oil is doubled, how does the rate of increase of the radius change?

16. Show that the radar gun of example 8.3 gives the correct speed if the police car is located at the origin.

5. Assume that the infected area of an injury is circular. If the radius of the infected area is 3 mm and growing at a rate of 1 mm/hr, at what rate is the infected area increasing? 6. For the injury of exercise 5, find the rate of increase of the infected area when the radius reaches 6 mm. Explain in commonsense terms why this rate is larger than that of exercise 5. 7. Suppose that a raindrop evaporates in such a way that it maintains a spherical shape. Given that the volume of a sphere of radius r is V = 43 πr 3 and its surface area is A = 4πr 2 , if the radius changes in time, show that V = Ar . If the rate of evaporation (V ) is proportional to the surface area, show that the radius changes at a constant rate. 8. Suppose a forest fire spreads in a circle with radius changing at a rate of 5 feet per minute. When the radius reaches 200 feet, at what rate is the area of the burning region increasing? 9. A 10-foot ladder leans against the side of a building as in example 8.2. If the bottom of the ladder is pulled away from the wall at the rate of 3 ft/s and the ladder remains in contact with the wall, find the rate at which the top of the ladder is dropping when the bottom is 6 feet from the wall. 10. In exercise 9, find the rate at which the angle between the ladder and the horizontal is changing when the bottom of the ladder is 6 feet from the wall. 11. Two buildings of height 20 feet and 40 feet, respectively, are 60 feet apart. Suppose that the intensity of light at a point between the buildings is proportional to the angle θ in the figure. If a person is moving from right to left at 4 ft/s, at what rate is θ changing when the person is exactly halfway between the two buildings?

20'

40'

θ

17. Show that the radar gun of example 8.3 gives the correct speed if the police car is at x = 12 moving at a speed of √ ( 2 − 1) 50 mph. 18. Find a position and speed for which the radar gun of example 8.3 has a slower reading than the actual speed. 19. Suppose that the average yearly cost per item for producing x items of a business product is C(x) = 10 + 100 . If the curx rent production is x = 10 and production is increasing at a rate of 2 items per year, find the rate of change of the average cost. 20. Suppose that the average yearly cost per item for producing x items of a business product is C(x) = 12 + 94 . The three most x recent yearly production figures are given in the table. Year Prod. (x)

0 8.2

1 8.8

2 9.4

Estimate the value of x (2) and the current (year 2) rate of change of the average cost. 21. For a small company spending $x thousand per year in advertising, suppose that annual sales in thousands of dollars equal s = 60 − 40e−0.05x . The three most recent yearly advertising figures are given in the table. Year Adver.

0 16,000

1 18,000

2 20,000

Estimate the value of x (2) and the current (year 2) rate of change of sales. 22. For a small company spending $x thousand per year in advertising, suppose that annual sales in thousands of dollars equal s = 80 − 20e−0.04x . If the current advertising budget is x = 40 and the budget is increasing at a rate of $1500 per year, find the rate of change of sales.

60'

12. Find the location in exercise 11 where the angle θ is maximum. 13. A plane is located x = 40 miles (horizontally) away from an airport at an altitude of h miles. Radar at the airport detects that the distance s(t) between the plane and airport is changing at the rate of s (t) = −240 mph. If the plane flies toward the airport at the constant altitude h = 4, what is the speed |x (t)| of the airplane? 14. Repeat exercise 13 with a height of 6 miles. Based on your answers, how important is it to know the actual height of the airplane?

23. A baseball player stands 2 feet from home plate and watches a pitch fly by. In the diagram, x is the distance from the ball to home plate and θ is the angle indicating the direction of the player’s gaze. Find the rate θ at which his eyes must move to watch a fastball with x (t) = −130 ft/s as it crosses home plate at x = 0. x

Plate u

2 Player

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..

Applications of Differentiation

24. In the situation of exercise 23, humans can maintain focus only when θ ≤ 3 (see Watts and Bahill’s book Keep Your Eye on the Ball). Find the fastest pitch that you could actually watch cross home plate. 25. A camera tracks the launch of a vertically ascending spacecraft. The camera is located at ground level 2 miles from the launchpad. If the spacecraft is 3 miles up and traveling at 0.2 mile per second, at what rate is the camera angle (measured from the horizontal) changing? 26. Repeat exercise 25 for the spacecraft at 1 mile up (assume the same velocity). Which rate is higher? Explain in commonsense terms why it is larger. 27. Suppose a 6-ft-tall person is 12 ft away from an 18-ft-tall lamppost (see the figure). If the person is moving away from the lamppost at a rate of 2 ft/s, at what rate is the length of the x +s s shadow changing? Hint: Show that = . 18 6

3-86

constant rate of 5 m3 /s, at what rate is the height of the pile increasing when the height is 2 meters? 33. The frequency at which a guitar string vibrates (which determines the pitch of the note we hear) is related to the tension T to which the string is tightened, the density ρ of the string and the effective length L of the string by the equation 1 T . By running his finger along a string, a guitarist 2L ρ can change L by changing the distance between the bridge f =

T = 220 ft/s so ρ that the units of f are Hertz (cycles per second). If the guitarist’s hand slides so that L (t) = −4, find f (t). At this rate, how long will it take to raise the pitch one octave (that is, double f )? and his finger. Suppose that L =

1 2

ft and

18 ft

6 ft

x

s

Exercise 27 28. Rework exercise 27 if the person is 6 ft away from the lamppost and is walking toward the lamppost at a rate of 3 ft/s. 29. Boyle’s law for a gas at constant temperature is PV = c, where P is pressure, V is volume and c is a constant. Assume that both P and V are functions of time. Show that P (t)/V (t) = −c/V 2 . 30. In exercise 29, solve for P as a function of V . Treating V as an independent variable, compute P (V ). Compare P (V ) and P (t)/V (t) from exercise 29. 31. A dock is 6 feet above water. Suppose you stand on the edge of the dock and pull a rope attached to a boat at the constant rate of 2 ft/s. Assume that the boat remains at water level. At what speed is the boat approaching the dock when it is 20 feet from the dock? 10 feet from the dock? Isn’t it surprising that the boat’s speed is not constant? 32. Sand is poured into a conical pile with the height of the pile equalling the diameter of the pile. If the sand is poured at a

34. Suppose that you are blowing up a balloon by adding air at the rate of 1 ft3 /s. If the balloon maintains a spherical shape, the volume and radius are related by V = 43 πr 3 . Compare the rate at which the radius is changing when r = 0.01 ft versus when r = 0.1 ft. Discuss how this matches the experience of a person blowing up a balloon. 35. Water is being pumped into a spherical tank of radius 60 feet at the constant rate of 10 ft3 /s. Find the rate at which the radius of the top level of water in the tank changes when the tank is half full. 36. For the water tank in exercise 35, find the height at which the height of the water in the tank changes at the same rate as the radius. 37. Sand is dumped such that the shape of the sandpile remains a cone with height equal to twice the radius. If the sand is dumped at the constant rate of 20 ft3 /s, find the rate at which the radius is increasing when the height reaches 6 feet. 38. Repeat exercise 37 for a sandpile for which the edge of the sandpile forms an angle of 45◦ with the horizontal. 39. To start skating, you must angle your foot and push off the ice. Alain Hach´e’s The Physics of Hockey derives the relationship between the skate angle θ, the sideways stride distance s, the stroke period 2s T and the forward speed v of the skater, with θ = tan−1 vT . For T = 1 second, s = 60 cm and an acceleration of 1 m/s2 , find the rate of change of the angle θ when the

3-87

SECTION 3.9

skater reaches (a) 1 m/s and (b) 2 m/s. Interpret the sign and size of θ in terms of skating technique. v

s

θ

push

..

Rates of Change in Economics and the Sciences

327

are analyzed here. In the diagram below, a camera follows an object directly from left to right. If the camera is at the origin, the object moves with speed 1 m/s and the line of motion is at y = c, find an expression for θ as a function of the position of the object. In the diagram to the right, the camera looks down into a parabolic mirror and indirectly views the object. If the mirror has polar coordinates (in this case, the angle θ 1 − sin θ is measured from the horizontal) equation r = and 2 cos2 θ x = r cos θ, find an expression for θ as a function of the position of the object. Compare values of θ at x = 0 and other x-values. If a large value of θ causes the image to blur, which camera system is better? Does the distance y = c affect your preference? (x, y)

EXPLORATORY EXERCISES

(x, y)

1. A thin rectangular advertising sign rotates clockwise at the rate of 6 revolutions per minute. The sign is 4 feet wide and an observer stands 40 feet away. Find the rate of change of the viewing angle θ as a function of the sign angle A. At which position of the sign is the rate of change maximum?

θ

θ

3. A particle moves down a ramp subject only to the force of gravity. Let y0 be the maximum height of the particle. Then conservation of energy gives A θ

2. Vision has proved to be the biggest challenge for building functional robots. Robot vision can either be designed to mimic human vision or follow a different design. Two possibilities

3.9

1 2 mv + mgy = mgy0 2 (a) From the definition v(t) = [x (t)]2 + [y (t)]2 , conclude that |y (t)| ≤ |v(t)|. (b) Show that |v (t)| ≤ g. (c) What shape must the ramp have to get equality in part (b)? Briefly explain in physical terms why g is the maximum value of |v (t)|.

RATES OF CHANGE IN ECONOMICS AND THE SCIENCES It has often been said that mathematics is the language of nature. Today, the concepts of calculus are being applied in virtually every field of human endeavor. The applications in this section represent but a small sampling of some elementary uses of the derivative. These are not all of the uses of the derivative nor are they necessarily the most important uses, but rather, represent some interesting applications in a variety of fields. Recall that the derivative of a function gives the instantaneous rate of change of that function. So, when you see the word rate, you should be thinking derivative. You can hardly pick up a newspaper without finding reference to some rates (e.g., inflation rate, interest rate, unemployment rate, etc.). These can be thought of as derivatives. There are also many quantities with which you are familiar, but that you might not recognize as rates of change. Our first example, which comes from economics, is of this type.

3-87

SECTION 3.9

skater reaches (a) 1 m/s and (b) 2 m/s. Interpret the sign and size of θ in terms of skating technique. v

s

θ

push

..

Rates of Change in Economics and the Sciences

327

are analyzed here. In the diagram below, a camera follows an object directly from left to right. If the camera is at the origin, the object moves with speed 1 m/s and the line of motion is at y = c, find an expression for θ as a function of the position of the object. In the diagram to the right, the camera looks down into a parabolic mirror and indirectly views the object. If the mirror has polar coordinates (in this case, the angle θ 1 − sin θ is measured from the horizontal) equation r = and 2 cos2 θ x = r cos θ, find an expression for θ as a function of the position of the object. Compare values of θ at x = 0 and other x-values. If a large value of θ causes the image to blur, which camera system is better? Does the distance y = c affect your preference? (x, y)

EXPLORATORY EXERCISES

(x, y)

1. A thin rectangular advertising sign rotates clockwise at the rate of 6 revolutions per minute. The sign is 4 feet wide and an observer stands 40 feet away. Find the rate of change of the viewing angle θ as a function of the sign angle A. At which position of the sign is the rate of change maximum?

θ

θ

3. A particle moves down a ramp subject only to the force of gravity. Let y0 be the maximum height of the particle. Then conservation of energy gives A θ

2. Vision has proved to be the biggest challenge for building functional robots. Robot vision can either be designed to mimic human vision or follow a different design. Two possibilities

3.9

1 2 mv + mgy = mgy0 2 (a) From the definition v(t) = [x (t)]2 + [y (t)]2 , conclude that |y (t)| ≤ |v(t)|. (b) Show that |v (t)| ≤ g. (c) What shape must the ramp have to get equality in part (b)? Briefly explain in physical terms why g is the maximum value of |v (t)|.

RATES OF CHANGE IN ECONOMICS AND THE SCIENCES It has often been said that mathematics is the language of nature. Today, the concepts of calculus are being applied in virtually every field of human endeavor. The applications in this section represent but a small sampling of some elementary uses of the derivative. These are not all of the uses of the derivative nor are they necessarily the most important uses, but rather, represent some interesting applications in a variety of fields. Recall that the derivative of a function gives the instantaneous rate of change of that function. So, when you see the word rate, you should be thinking derivative. You can hardly pick up a newspaper without finding reference to some rates (e.g., inflation rate, interest rate, unemployment rate, etc.). These can be thought of as derivatives. There are also many quantities with which you are familiar, but that you might not recognize as rates of change. Our first example, which comes from economics, is of this type.

328

CHAPTER 3

..

Applications of Differentiation

3-88

In economics, the term marginal is used to indicate a rate. Thus, marginal cost is the derivative of the cost function, marginal profit is the derivative of the profit function and so on. We introduce marginal cost in some detail here, with further applications given in the exercises. Suppose that you are manufacturing an item, where your start-up costs are $4000 and productions costs are $2 per item. The total cost of producing x items would then be 4000 + 2x. Of course, the assumption that the cost per item is constant is unrealistic. Efficient mass-production techniques could reduce the cost per item, but machine maintenance, labor, plant expansion and other factors could drive costs up as production (x) increases. In example 9.1, a quadratic cost function is used to take into account some of these extra factors. In practice, you would find a cost function by making some observations of the cost of producing a number of different quantities and then fitting the data to the graph of a known function. (This is one way in which the calculus is brought to bear on real-world problems.) When the cost per item is not constant, an important question for managers to answer is how much it will cost to increase production. This is the idea behind marginal cost.

EXAMPLE 9.1

Analyzing the Marginal Cost of Producing a Commercial Product

Suppose that C(x) = 0.02x 2 + 2x + 4000 is the total cost (in dollars) for a company to produce x units of a certain product. Compute the marginal cost at x = 100 and compare this to the actual cost of producing the 100th unit. Solution The marginal cost function is the derivative of the cost function: C (x) = 0.04x + 2 and so, the marginal cost at x = 100 is C (100) = 4 + 2 = 6 dollars per unit. On the other hand, the actual cost of producing item number 100 would be C(100) − C(99). (Why?) We have C(100) − C(99) = 200 + 200 + 4000 − (196.02 + 198 + 4000) = 4400 − 4394.02 = 5.98 dollars. Note that this is very close to the marginal cost of $6. Also notice that the marginal cost is easier to compute. Another quantity that businesses use to analyze production is average cost. You can easily remember the formula for average cost by thinking of an example. If it costs a total of $120 to produce 12 items, then the average cost would be $10 $ 120 per item. In 12 general, the total cost is given by C(x) and the number of items by x, so average cost is defined by C(x) =

C(x) . x

A business manager would want to know the level of production that minimizes average cost.

3-89

SECTION 3.9

EXAMPLE 9.2

..

Rates of Change in Economics and the Sciences

329

Minimizing the Average Cost of Producing a Commercial Product C(x) = 0.02x 2 + 2x + 4000

Suppose that

is t

CHAPTER 0

..

Preliminaries

0-2

y 35 30 25 20 15 10 5 0

x 1

2

3

4

5

6

7

8

FIGURE 0.1 The Fibonacci sequence

these is the attempt to find patterns to help us better describe the world. The other theme is the interplay between graphs and functions. By connecting the powerful equation-solving techniques of algebra with the visual images provided by graphs, you will significantly improve your ability to make use of your mathematical skills in solving real-world problems.

0.1

POLYNOMIALS AND RATIONAL FUNCTIONS

The Real Number System and Inequalities Although mathematics is far more than just a study of numbers, our journey into calculus begins with the real number system. While this may seem to be a fairly mundane starting place, we want to give you the opportunity to brush up on those properties that are of particular interest for calculus. The most familiar set of numbers is the set of integers, consisting of the whole numbers and their additive inverses: 0, ±1, ±2, ±3, . . . . A rational number is any number of the p 27 form q , where p and q are integers and q = 0. For example, 23 , − 73 and 125 are all rational numbers. Notice that every integer n is also a rational number, since we can write it as the n quotient of two integers: n = . 1 p The irrational numbers are all those real numbers that cannot be written in the form q , where p and q are integers. Recall that rational numbers have decimal expansions that either ¯ 1 = 0.125 and 1 = 0.166666¯ are terminate or repeat. For instance, 12 = 0.5, 13 = 0.33333, 8 6 all rational numbers. By contrast, irrational numbers have decimal expansions that do not repeat or terminate. For instance, three familiar irrational numbers and their decimal expansions are √ 2 = 1.41421 35623 . . . , π = 3.14159 26535 . . . and

e = 2.71828 18284 . . . .

We picture the real numbers arranged along the number line displayed in Figure 0.2 (the real line). The set of real numbers is denoted by the symbol R.

0-3

SECTION 0.1

..

2 5 4 3 2 1

Polynomials and Rational Functions

3 0

1

3

p

2

3

4

5

e

FIGURE 0.2 The real line

For real numbers a and b, where a < b, we define the closed interval [a, b] to be the set of numbers between a and b, including a and b (the endpoints), that is, a

FIGURE 0.3 A closed interval

a

[a, b] = {x ∈ R | a ≤ x ≤ b},

b

as illustrated in Figure 0.3, where the solid circles indicate that a and b are included in [a, b]. Similarly, the open interval (a, b) is the set of numbers between a and b, but not including the endpoints a and b, that is,

b

FIGURE 0.4 An open interval

(a, b) = {x ∈ R | a < x < b}, as illustrated in Figure 0.4, where the open circles indicate that a and b are not included in (a, b). You should already be very familiar with the following properties of real numbers.

THEOREM 1.1 If a and b are real numbers and a < b, then (i) (ii) (iii) (iv)

For any real number c, a + c < b + c. For real numbers c and d, if c < d, then a + c < b + d. For any real number c > 0, a · c < b · c. For any real number c < 0, a · c > b · c.

REMARK 1.1 We need the properties given in Theorem 1.1 to solve inequalities. Notice that (i) says that you can add the same quantity to both sides of an inequality. Part (iii) says that you can multiply both sides of an inequality by a positive number. Finally, (iv) says that if you multiply both sides of an inequality by a negative number, the inequality is reversed.

We illustrate the use of Theorem 1.1 by solving a simple inequality.

EXAMPLE 1.1

Solving a Linear Inequality

Solve the linear inequality 2x + 5 < 13. Solution We can use the properties in Theorem 1.1 to isolate the x. First, subtract 5 from both sides to obtain (2x + 5) − 5 < 13 − 5

4

..

CHAPTER 0

Preliminaries

0-4

2x < 8.

or

Finally, divide both sides by 2 (since 2 > 0, the inequality is not reversed) to obtain x < 4. We often write the solution of an inequality in interval notation. In this case, we get the interval (−∞, 4). You can deal with more complicated inequalities in the same way.

EXAMPLE 1.2

Solving a Two-Sided Inequality

Solve the two-sided inequality 6 < 1 − 3x ≤ 10. Solution First, recognize that this problem requires that we find values of x such that 6 < 1 − 3x

and

1 − 3x ≤ 10.

Here, we can use the properties in Theorem 1.1 to isolate the x by working on both inequalities simultaneously. First, subtract 1 from each term, to get 6 − 1 < (1 − 3x) − 1 ≤ 10 − 1 5 < −3x ≤ 9.

or

Now, divide by −3, but be careful. Since −3 < 0, the inequalities are reversed. We have 5 −3x 9 > ≥ −3 −3 −3 5 > x ≥ −3. 3 5 −3 ≤ x < − , 3 −

or We usually write this as

or in interval notation as [−3, − 53 ). y

You will often need to solve inequalities involving fractions. We present a typical example in the following.

8 4

4

EXAMPLE 1.3 x

2

1

2

4 8

FIGURE 0.5 y=

x −1 x +2

4

Solving an Inequality Involving a Fraction

x −1 ≥ 0. x +2 Solution In Figure 0.5, we show a graph of the function, which appears to indicate that the solution includes all x < −2 and x ≥ 1. Carefully read the inequality and observe that there are only three ways to satisfy this: either both numerator and denominator are positive, both are negative or the numerator is zero. To visualize this, we draw number lines for each of the individual terms, indicating where each is positive, negative or zero and use these to draw a third number line indicating the value of the quotient, as shown in the margin. In the third number line, we have placed an “ ”

Solve the inequality

0-5

SECTION 0.1

0

0

x2

2

0

2

5

For inequalities involving a polynomial of degree 2 or higher, factoring the polynomial and determining where the individual factors are positive and negative, as in example 1.4, will lead to a solution.

x1 x2

1

Polynomials and Rational Functions

above the −2 to indicate that the quotient is undefined at x = −2. From this last number line, you can see that the quotient is nonnegative whenever x < −2 or x ≥ 1. We write the solution in interval notation as (−∞, −2) ∪ [1, ∞).

x1

1

..

y

EXAMPLE 1.4

20

Solving a Quadratic Inequality

Solve the quadratic inequality

x

6 4 2

2

4

FIGURE 0.6

0

(1.2)

x3

3

(x + 3)(x − 2) > 0.

This can happen in only two ways: when both factors are positive or when both factors are negative. As in example 1.3, we draw number lines for both of the individual factors, indicating where each is positive, negative or zero and use these to draw a number line representing the product. We show these in the margin. Notice that the third number line indicates that the product is positive whenever x < −3 or x > 2. We write this in interval notation as (−∞, −3) ∪ (2, ∞).

y = x2 + x − 6 0

Solution In Figure 0.6, we show a graph of the polynomial on the left side of the inequality. Since this polynomial factors, (1.1) is equivalent to

6

10

(1.1)

x 2 + x − 6 > 0.

10

No doubt, you will recall the following standard definition. x2

2

0

3

0

(x 3)(x 2)

2

DEFINITION 1.1

The absolute value of a real number x is |x| =

x, −x,

if x ≥ 0. if x < 0

Make certain that you read Definition 1.1 correctly. If x is negative, then −x is positive. This says that |x| ≥ 0 for all real numbers x. For instance, using the definition, |− 4 | = −(−4) = 4, notice that for any real numbers a and b, |a · b| = |a| · |b|.

NOTES

However,

For any two real numbers a and b, |a − b| gives the distance between a and b. (See Figure 0.7.)

|a + b| = |a| + |b|,

in general. (To verify this, simply take a = 5 and b = −2 and compute both quantities.) However, it is always true that |a + b| ≤ |a| + |b|.

a b a

b

FIGURE 0.7 The distance between a and b

This is referred to as the triangle inequality. The interpretation of |a − b| as the distance between a and b (see the note in the margin) is particularly useful for solving inequalities involving absolute values. Wherever possible, we suggest that you use this interpretation to read what the inequality means, rather than merely following a procedure to produce a solution.

6

..

CHAPTER 0

Preliminaries

EXAMPLE 1.5

0-6

Solving an Inequality Containing an Absolute Value

Solve the inequality |x − 2| < 5. 5

5

2 5 3

257

2

FIGURE 0.8 |x − 2| < 5

(1.3)

Solution Before you start trying to solve this, take a few moments to read what it says. Since |x − 2| gives the distance from x to 2, (1.3) says that the distance from x to 2 must be less than 5. So, find all numbers x whose distance from 2 is less than 5. We indicate the set of all numbers within a distance 5 of 2 in Figure 0.8. You can now read the solution directly from the figure: −3 < x < 7 or in interval notation: (−3, 7). Many inequalities involving absolute values can be solved simply by reading the inequality correctly, as in example 1.6.

EXAMPLE 1.6

Solving an Inequality with a Sum Inside an Absolute Value

Solve the inequality |x + 4| ≤ 7. 7

7

4 7 11 4

(1.4)

Solution To use our distance interpretation, we must first rewrite (1.4) as

4 7 3

|x − (−4)| ≤ 7. This now says that the distance from x to −4 is less than or equal to 7. We illustrate the solution in Figure 0.9, from which it follows that −11 ≤ x ≤ 3 or [−11, 3].

FIGURE 0.9 | x + 4 |≤ 7

Recall that for any real number r > 0, |x| < r is equivalent to the following inequality not involving absolute values: −r < x < r. In example 1.7, we use this to revisit the inequality from example 1.5.

EXAMPLE 1.7

An Alternative Method for Solving Inequalities

Solve the inequality |x − 2| < 5. y

Solution This is equivalent to the two-sided inequality (x2, y2)

y2

−5 < x − 2 < 5. Adding 2 to each term, we get the solution

Distance

y2 y1

−3 < x < 7, or in interval notation (−3, 7), as before.

y1

(x1, y1)

x2 x1

x1

x2

FIGURE 0.10 Distance

x

Recall that the distance between two points (x1 , y1 ) and (x2 , y2 ) is a simple consequence of the Pythagorean Theorem and is given by d{(x1 , y1 ), (x2 , y2 )} = (x2 − x1 )2 + (y2 − y1 )2 . We illustrate this in Figure 0.10.

0-7

SECTION 0.1

EXAMPLE 1.8

..

Polynomials and Rational Functions

7

Using the Distance Formula

Find the distance between the points (1, 2) and (3, 4). Solution The distance between (1, 2) and (3, 4) is d{(1, 2), (3, 4)} =

(3 − 1)2 + (4 − 2)2 =

√

4+4=

√

8.

Equations of Lines Year 1960 1970 1980 1990

x

The federal government conducts a nationwide census every 10 years to determine the population. Population data for the last several decades are shown in the accompanying table. One difficulty with analyzing these data is that the numbers are so large. This problem is remedied by transforming the data. We can simplify the year data by defining x to be the number of years since 1960. Then, 1960 corresponds to x = 0, 1970 corresponds to x = 10 and so on. The population data can be simplified by rounding the numbers to the nearest million. The transformed data are shown in the accompanying table and a scatter plot of these data points is shown in Figure 0.11. Most people would say that the points in Figure 0.11 appear to form a straight line. (Use a ruler and see if you agree.) To determine whether the points are, in fact, on the same line (such points are called colinear), we might consider the population growth in each of the indicated decades. From 1960 to 1970, the growth was 24 million. (That is, to move from the first point to the second, you increase x by 10 and increase y by 24.) Likewise, from 1970 to 1980, the growth was 24 million. However, from 1980 to 1990, the growth was only 22 million. Since the rate of growth is not constant, the data points do not fall on a line. Notice that to stay on the same line, y would have had to increase by 24 again. The preceding argument involves the familiar concept of slope.

U.S. Population 179,323,175 203,302,031 226,542,203 248,709,873

y

0

179

10

203

20

227

30

249

Transformed data y 250 200

DEFINITION 1.2 For x1 = x2 , the slope of the straight line through the points (x1 , y1 ) and (x2 , y2 ) is the number y2 − y1 m= . (1.5) x2 − x1

150 100 50 x 10

20

FIGURE 0.11 Population data

30

When x1 = x2 , the line through (x1 , y1 ) and (x2 , y2 ) is vertical and the slope is undefined. y , We often describe slope as “the change in y divided by the change in x,” written x Rise or more simply as (see Figure 0.12a on the following page). Run The slope of a straight line is the same no matter which two points on the line you select. Referring to Figure 0.12b (where the line has positive slope), notice that for any four points A, B, D and E on the line, the two right triangles ABC and DEF are similar. Recall that for similar triangles, the ratios of corresponding sides must be the same. In this case, this says that y y = x x

8

CHAPTER 0

..

Preliminaries

0-8

y

y

B (x2, y2)

y2

y A

y y2 y1 Rise y1

x

E

C

y

(x1, y1) D

x x2 x1

x1

F

x

Run x

x2

x

FIGURE 0.12a

FIGURE 0.12b

Slope

Similar triangles and slope

and so, the slope is the same no matter which two points on the line are selected. Furthermore, a line is the only curve with constant slope. Notice that a line is horizontal if and only if its slope is zero.

EXAMPLE 1.9

Finding the Slope of a Line

Find the slope of the line through the points (4, 3) and (2, 5). Solution From (1.5), we get m=

EXAMPLE 1.10

y2 − y1 5−3 2 = = = −1. x2 − x1 2−4 −2

Using Slope to Determine if Points Are Colinear

Use slope to determine whether the points (1, 2), (3, 10) and (4, 14) are colinear. Solution First, notice that the slope of the line joining (1, 2) and (3, 10) is m1 =

y2 − y1 10 − 2 8 = = = 4. x2 − x1 3−1 2

Similarly, the slope through the line joining (3, 10) and (4, 14) is m2 =

y2 − y1 14 − 10 = = 4. x2 − x1 4−3

Since the slopes are the same, the points must be colinear. Recall that if you know the slope and a point through which the line must pass, you have enough information to graph the line. The easiest way to graph a line is to plot two points and then draw the line through them. In this case, you need only to find a second point.

0-9

SECTION 0.1

..

Polynomials and Rational Functions

9

y

EXAMPLE 1.11

Graphing a Line

If a line passes through the point (2, 1) with slope 23 , find a second point on the line and then graph the line. y2 − y1 Solution Since slope is given by m = , we take m = 23 , y1 = 1 and x1 = 2, x2 − x1 to obtain

4 3 2 1

2 y2 − 1 = . 3 x2 − 2

x 1

2

3

4

5

You are free to choose the x-coordinate of the second point. For instance, to find the point at x2 = 5, substitute this in and solve. From

1

2 y2 − 1 y2 − 1 = = , 3 5−2 3

FIGURE 0.13a Graph of straight line

we get 2 = y2 − 1 or y2 = 3. A second point is then (5, 3). The graph of the line is shown in Figure 0.13a. An alternative method for finding a second point is to use the slope

y

m=

4

y 2

3 2 1 x 3 1

2

3

4

x 5

1

FIGURE 0.13b

2 y = . 3 x

The slope of 23 says that if we move three units to the right, we must move two units up to stay on the line, as illustrated in Figure 0.13b. In example 1.11, the choice of x = 5 was entirely arbitrary; you can choose any x-value you want to find a second point. Further, since x can be any real number, you can leave x as a variable and write out an equation satisfied by any point (x, y) on the line. In the general case of the line through the point (x0 , y0 ) with slope m, we have from (1.5) that y − y0 m= . (1.6) x − x0 Multiplying both sides of (1.6) by (x − x0 ), we get

Using slope to find a second point

y − y0 = m(x − x0 ) or

POINT-SLOPE FORM OF A LINE y = m(x − x0 ) + y0 .

y 7

Equation (1.7) is called the point-slope form of the line.

6 5

EXAMPLE 1.12

4

Finding the Equation of a Line Given Two Points

Find an equation of the line through the points (3, 1) and (4, −1), and graph the line.

3 2

−1 − 1 −2 = = −2. Using (1.7) with slope 4−3 1 m = −2, x-coordinate x0 = 3 and y-coordinate y0 = 1, we get the equation of the line:

Solution From (1.5), the slope is m =

1 x 1

(1.7)

1

2

3

4

y = −2(x − 3) + 1. FIGURE 0.14 y = −2(x − 3) + 1

(1.8)

To graph the line, plot the points (3, 1) and (4, −1), and you can easily draw the line seen in Figure 0.14.

10

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CHAPTER 0

Preliminaries

0-10

In example 1.12, you may be tempted to simplify the expression for y given in (1.8). As it turns out, the point-slope form of the equation is often the most convenient to work with. So, we will typically not ask you to rewrite this expression in other forms. At times, a form of the equation called the slope-intercept form is more convenient. This has the form y = mx + b, where m is the slope and b is the y-intercept (i.e., the place where the graph crosses the y-axis). In example 1.12, you simply multiply out (1.8) to get y = −2x + 6 + 1 or y = −2x + 7. As you can see from Figure 0.14, the graph crosses the y-axis at y = 7. Theorem 1.2 presents a familiar result on parallel and perpendicular lines.

THEOREM 1.2 Two (nonvertical) lines are parallel if they have the same slope. Further, any two vertical lines are parallel. Two (nonvertical) lines of slope m 1 and m 2 are perpendicular whenever the product of their slopes is −1 (i.e., m 1 · m 2 = −1). Also, any vertical line and any horizontal line are perpendicular. y

Since we can read the slope from the equation of a line, it’s a simple matter to determine when two lines are parallel or perpendicular. We illustrate this in examples 1.13 and 1.14.

20 10

4

x

2

2

EXAMPLE 1.13

4

Find an equation of the line parallel to y = 3x − 2 and through the point (−1, 3).

10

Solution It’s easy to read the slope of the line from the equation: m = 3. The equation of the parallel line is then

20

y = 3[x − (−1)] + 3

FIGURE 0.15

or simply y = 3(x + 1) + 3. We show a graph of both lines in Figure 0.15.

Parallel lines y

EXAMPLE 1.14

Finding the Equation of a Perpendicular Line

Find an equation of the line perpendicular to y = −2x + 4 and intersecting the line at the point (1, 2).

4 2 x

2

Finding the Equation of a Parallel Line

2

4

2 4

FIGURE 0.16 Perpendicular lines

Solution The slope of y = −2x + 4 is −2. The slope of the perpendicular line is then −1/(−2) = 12 . Since the line must pass through the point (1, 2), the equation of the perpendicular line is 1 (x − 1) + 2. 2 We show a graph of the two lines in Figure 0.16. y=

We now return to this subsection’s introductory example and use the equation of a line to estimate the population in the year 2000.

0-11

SECTION 0.1

EXAMPLE 1.15

..

Polynomials and Rational Functions

11

Using a Line to Estimate Population

Given the population data for the census years 1960, 1970, 1980 and 1990, estimate the population for the year 2000. Solution We began this subsection by showing that the points in the corresponding table are not colinear. Nonetheless, they are nearly colinear. So, why not use the straight line connecting the last two points (20, 227) and (30, 249) (corresponding to the populations in the years 1980 and 1990) to estimate the population in 2000? (This is a simple example of a more general procedure called extrapolation.) The slope of the line joining the two data points is

y 300

200

m= 100

249 − 227 22 11 = = . 30 − 20 10 5

The equation of the line is then y=

x 10

20

30

40

50

11 (x − 30) + 249. 5

See Figure 0.17 for a graph of the line. If we follow this line to the point corresponding to x = 40 (the year 2000), we have the estimated population

FIGURE 0.17 Population

11 (40 − 30) + 249 = 271. 5 That is, the estimated population is 271 million people. The actual census figure for 2000 was 281 million, which indicates that the U.S. population has grown at a rate that is faster than linear.

Functions A

B f

x

For any two subsets A and B of the real line, we make the following familiar definition.

y

DEFINITION 1.3

REMARK 1.2 Functions can be defined by simple formulas, such as f (x) = 3x + 2, but in general, any correspondence meeting the requirement of matching exactly one y to each x defines a function.

A function f is a rule that assigns exactly one element y in a set B to each element x in a set A. In this case, we write y = f (x). We call the set A the domain of f. The set of all values f (x) in B is called the range of f . That is, the range of f is { f (x) | x ∈ A}. Unless explicitly stated otherwise, the domain of a function f is the largest set of real numbers for which the function is defined. We refer to x as the independent variable and to y as the dependent variable.

By the graph of a function f, we mean the graph of the equation y = f (x). That is, the graph consists of all points (x, y), where x is in the domain of f and where y = f (x). Notice that not every curve is the graph of a function, since for a function, only one y-value corresponds to a given value of x. You can graphically determine whether a curve is the graph of a function by using the vertical line test: if any vertical line intersects the graph in more than one point, the curve is not the graph of a function.

12

CHAPTER 0

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Preliminaries

0-12

EXAMPLE 1.16

Using the Vertical Line Test

Determine which of the curves in Figures 0.18a and 0.18b correspond to functions. y

y 1 x

1

1

x 0.5

1

2

1

FIGURE 0.18a y

Solution Notice that the circle in Figure 0.18a is not the graph of a function, since a vertical line at x = 0.5 intersects the circle twice (see Figure 0.19a). The graph in Figure 0.18b is the graph of a function, even though it swings up and down repeatedly. Although horizontal lines intersect the graph repeatedly, vertical lines, such as the one at x = 1.2, intersect only once (see Figure 0.19b). 0.5

1

FIGURE 0.18b

x 1

You are already familiar with a number of different types of functions, and we will only briefly review these here and in sections 0.4 and 0.5. The functions that you are probably most familiar with are polynomials. These are the simplest functions to work with because they are defined entirely in terms of arithmetic.

FIGURE 0.19a Curve fails vertical line test

DEFINITION 1.4

y

A polynomial is any function that can be written in the form f (x) = an x n + an−1 x n−1 + · · · + a1 x + a0 ,

1

x 0.5

1

2

1

FIGURE 0.19b

where a0 , a1 , a2 , . . . , an are real numbers (the coefficients of the polynomial) with an = 0 and n ≥ 0 is an integer (the degree of the polynomial). Note that the domain of every polynomial function is the entire real line. Further, recognize that the graph of the linear (degree 1) polynomial f (x) = ax + b is a straight line.

Curve passes vertical line test

EXAMPLE 1.17

Sample Polynomials

The following are all examples of polynomials: f (x) = 2 (polynomial of degree 0 or constant), f (x) = 3x + 2 (polynomial of degree 1 or linear polynomial), f (x) = 5x 2 − 2x + 1 (polynomial of degree 2 or quadratic polynomial), f (x) = x 3 − 2x + 1 (polynomial of degree 3 or cubic polynomial), f (x) = −6x 4 + 12x 2 − 3x + 13 (polynomial of degree 4 or quartic polynomial),

0-13

SECTION 0.1

..

Polynomials and Rational Functions

13

and f (x) = 2x 5 + 6x 4 − 8x 2 + x − 3 (polynomial of degree 5 or quintic polynomial). We show graphs of these six functions in Figures 0.20a–0.20f. y

y 120

15 y

10

4 1

x

2

2

4 40

5 10 15

x

3 2

80

5

3

x

6 4 2

2

4

6

FIGURE 0.20a

FIGURE 0.20b

FIGURE 0.20c

f (x) = 2

f (x) = 3x + 2

f (x) = 5x 2 − 2x + 1

y

y

10

20

5

10

1

x 1

2

2

3

y 20 10

x

1

1

5

10

10

20

2 3

2

x

1

1 10

FIGURE 0.20d

FIGURE 0.20e

FIGURE 0.20f

f (x) = x 3 − 2x + 1

f (x) = −6x 4 + 12x 2 − 3x + 13

f (x) = 2x 5 + 6x 4 − 8x 2 + x − 3

DEFINITION 1.5 Any function that can be written in the form f (x) =

p(x) , q(x)

where p and q are polynomials, is called a rational function.

Notice that since p(x) and q(x) are polynomials, they are both defined for all x, and p(x) so, the rational function f (x) = is defined for all x for which q(x) = 0. q(x)

14

CHAPTER 0

..

Preliminaries

0-14

y

EXAMPLE 1.18 10

A Sample Rational Function

Find the domain of the function

5

3

f (x) =

x 2 + 7x − 11 . x2 − 4

x

1

1

3

Solution Here, f (x) is a rational function. We show a graph in Figure 0.21. Its domain consists of those values of x for which the denominator is nonzero. Notice that

5

x 2 − 4 = (x − 2)(x + 2)

10

and so, the denominator is zero if and only if x = ±2. This says that the domain of f is

FIGURE 0.21

{x ∈ R | x = ±2} = (−∞, −2) ∪ (−2, 2) ∪ (2, ∞).

x 2 + 7x − 11 f (x) = x2 − 4

√ The square root function is defined in the usual way. When we√ write y = x, we mean that y is that number for which y 2 =√x and y ≥ 0. In particular, 4 = 2. Be careful not to write erroneous statements such as 4 = ±2. In particular, be careful to write √ x 2 = |x|. √ Since x 2 is asking for the nonnegative number whose square is x 2 , we are looking for |x| and not x. We can say √ x 2 = x, only for x ≥ 0. √ Similarly, for any integer n ≥ 2, y = n x whenever y n = x, where for n even, x ≥ 0 and y ≥ 0.

EXAMPLE 1.19

Finding the Domain of a Function Involving a Square Root or a Cube Root

Find the domains of f (x) =

√

x 2 − 4 and g(x) =

√ 3

x 2 − 4.

Solution Since even roots are defined only for nonnegative values, f (x) is defined only for x 2 − 4 ≥ 0. Notice that this is equivalent to having x 2 ≥ 4, which occurs when x ≥ 2 or x ≤ −2. The domain of f is then (−∞, −2] ∪ [2, ∞). On the other hand, odd roots are defined for both positive and negative values. Consequently, the domain of g(x) is the entire real line, (−∞, ∞). We often find it useful to label intercepts and other significant points on a graph. Finding these points typically involves solving equations. A solution of the equation f (x) = 0 is called a zero of the function f or a root of the equation f (x) = 0. Notice that a zero of the function f corresponds to an x-intercept of the graph of y = f (x).

y 10 8

2

EXAMPLE 1.20

Finding Zeros by Factoring

6

Find all x- and y-intercepts of f (x) = x 2 − 4x + 3.

4

Solution To find the y-intercept, set x = 0 to obtain

2

y = 0 − 0 + 3 = 3. x 1

2

3

4

FIGURE 0.22 y = x 2 − 4x + 3

6

To find the x-intercepts, solve the equation f (x) = 0. In this case, we can factor to get f (x) = x 2 − 4x + 3 = (x − 1)(x − 3) = 0. You can now read off the zeros: x = 1 and x = 3, as indicated in Figure 0.22.

0-15

SECTION 0.1

..

Polynomials and Rational Functions

15

Unfortunately, factoring is not always so easy. Of course, for the quadratic equation ax 2 + bx + c = 0 (for a = 0), the solution(s) are given by the familiar quadratic formula: x=

EXAMPLE 1.21

−b ±

√

b2 − 4ac . 2a

Finding Zeros Using the Quadratic Formula

Find the zeros of f (x) = x 2 − 5x − 12. Solution You probably won’t have much luck trying to factor this. However, from the quadratic formula, we have √ √ −(−5) ± (−5)2 − 4 · 1 · (−12) 5 ± 25 + 48 5 ± 73 x= = = . 2·1 2 2 √

So, the two solutions are given by x = 52 + 273 ≈ 6.772 and x = (No wonder you couldn’t factor the polynomial!)

5 2

−

√

73 2

≈ −1.772.

Finding zeros of polynomials of degree higher than 2 and other functions is usually trickier and is sometimes impossible. At the least, you can always find an approximation of any zero(s) by using a graph to zoom in closer to the point(s) where the graph crosses the x-axis, as we’ll illustrate shortly. A more basic question, though, is to determine how many zeros a given function has. In general, there is no way to answer this question without the use of calculus. For the case of polynomials, however, Theorem 1.3 (a consequence of the Fundamental Theorem of Algebra) provides a clue.

THEOREM 1.3 A polynomial of degree n has at most n distinct zeros.

REMARK 1.3 Polynomials may also have complex zeros. For instance, f (x) = x 2 + 1 has only the complex zeros x = ±i, where i is the imaginary number defined √ by i = −1.

Notice that Theorem 1.3 does not say how many zeros a given polynomial has, but rather, that the maximum number of distinct (i.e., different) zeros is the same as the degree. A polynomial of degree n may have anywhere from 0 to n distinct real zeros. However, polynomials of odd degree must have at least one real zero. For instance, for the case of a cubic polynomial, we have one of the three possibilities illustrated in Figures 0.23a, 0.23b and 0.23c on the following page. In these three figures, we show the graphs of cubic polynomials with 1, 2 and 3 distinct, real zeros, respectively. These are the graphs of the functions f (x) = x 3 − 2x 2 + 3 = (x + 1)(x 2 − 3x + 3), g(x) = x 3 − x 2 − x + 1 = (x + 1)(x − 1)2 and

h(x) = x 3 − 3x 2 − x + 3 = (x + 1)(x − 1)(x − 3),

respectively. Note that you can see from the factored form where the zeros are (and how many there are).

16

CHAPTER 0

..

Preliminaries

0-16

y

y

x1

y

x2

x

x1

x2

x3

x

x

x1

FIGURE 0.23a

FIGURE 0.23b

FIGURE 0.23c

One zero

Two zeros

Three zeros

y 4

Theorem 1.4 provides an important connection between factors and zeros of polynomials.

2

THEOREM 1.4 (Factor Theorem) 2

x

1

1

2

For any polynomial f, f (a) = 0 if and only if (x − a) is a factor of f (x).

3

2

EXAMPLE 1.22

Find the zeros of f (x) = x 3 − x 2 − 2x + 2.

FIGURE 0.24a y = x 3 − x 2 − 2x + 2

0.2

1.41

1.39

x

0.2

FIGURE 0.24b Zoomed in on zero near x = −1.4 0.02 x 1.40

1.42

0.02

Solution By calculating f (1), you can see that one zero of this function is x = 1, but how many other zeros are there? A graph of the function (see Figure 0.24a) shows that there are two other zeros of f , one near x = −1.5 and one near x = 1.5. You can find these zeros more precisely by using your graphing calculator or computer algebra system to zoom in on the locations of these zeros (as shown in Figures 0.24b and 0.24c). From these zoomed graphs it is clear that the two remaining zeros of f are near x = 1.41 and x = −1.41. You can make these estimates more precise by zooming in even more closely. Most graphing calculators and computer algebra systems can also find approximate zeros, using a built-in “solve” program. In Chapter 3, we present a versatile method (called Newton’s method) for obtaining accurate approximations to zeros. The only way to find the exact solutions is to factor the expression (using either long division or synthetic division). Here, we have √ √ f (x) = x 3 − x 2 − 2x + 2 = (x − 1)(x 2 − 2) = (x − 1)(x − 2)(x + 2), √ √ from which you can see that the zeros are x = 1, x = 2 and x = − 2. Recall that to find the points of intersection of two curves defined by y = f (x) and y = g(x), we set f (x) = g(x) to find the x-coordinates of any points of intersection.

EXAMPLE 1.23

0.04

FIGURE 0.24c Zoomed in on zero near x = 1.4

Finding the Zeros of a Cubic Polynomial

Finding the Intersections of a Line and a Parabola

Find the points of intersection of the parabola y = x 2 − x − 5 and the line y = x + 3. Solution A sketch of the two curves (see Figure 0.25) shows that there are two intersections, one near x = −2 and the other near x = 4. To determine these precisely,

0-17

SECTION 0.1

Polynomials and Rational Functions

17

we set the two functions equal and solve for x:

y

x 2 − x − 5 = x + 3.

20

Subtracting (x + 3) from both sides leaves us with 0 = x 2 − 2x − 8 = (x − 4)(x + 2).

10

4 2

..

This says that the solutions are exactly x = −2 and x = 4. We compute the corresponding y-values from the equation of the line y = x + 3 (or the equation of the parabola). The points of intersection are then (−2, 1) and (4, 7). Notice that these are consistent with the intersections seen in Figure 0.25.

x 4

6

10

Unfortunately, you won’t always be able to solve equations exactly, as we did in examples 1.20–1.23. We explore some options for dealing with more difficult equations in section 0.2.

FIGURE 0.25 y = x + 3 and y = x 2 − x − 5

EXERCISES 0.1 WRITING EXERCISES 1. If the slope of the line passing through points A and B equals the slope of the line passing through points B and C, explain why the points A, B and C are colinear.

In exercises 11–16, find a second point on the line with slope m and point P, graph the line and find an equation of the line.

2. If a graph fails the vertical line test, it is not the graph of a function. Explain this result in terms of the definition of a function.

11. m = 2, P = (1, 3)

3. You should not automatically write the equation of a line in slope-intercept form. Compare the following forms of the same line: y = 2.4(x − 1.8) + 0.4 and y = 2.4x − 3.92. Given x = 1.8, which equation would you rather use to compute y? How about if you are given x = 0? For x = 8, is there any advantage to one equation over the other? Can you quickly read off the slope from either equation? Explain why neither form of the equation is “better.” 4. To understand Definition 1.1, you must believe that | x |= −x for negative x’s. Using x = −3 as an example, explain in words why multiplying x by −1 produces the same result as taking the absolute value of x.

12. m = −2, P = (1, 4) 13. m = 0, P = (−1, 1) 14. m = 12 , P = (2, 1) 15. m = 1.2, P = (2.3, 1.1) 16. m = − 14 , P = (−2, 1) In exercises 17–22, determine if the lines are parallel, perpendicular, or neither. 17. y = 3(x − 1) + 2 and y = 3(x + 4) − 1 18. y = 2(x − 3) + 1 and y = 4(x − 3) + 1 19. y = −2(x + 1) − 1 and y = 12 (x − 2) + 3

In exercises 1–4, determine if the points are colinear.

20. y = 2x − 1 and y = −2x + 2

1. (2, 1), (0, 2), (4, 0)

2. (3, 1), (4, 4), (5, 8)

21. y = 3x + 1 and y = − 13 x + 2

3. (4, 1), (3, 2), (1, 3)

4. (1, 2), (2, 5), (4, 8)

22. x + 2y = 1 and 2x + 4y = 3

In exercises 5–10, find the slope of the line through the given points. 5. (1, 2), (3, 6)

6. (1, 2), (3, 3)

In exercises 23–26, find an equation of a line through the given point and (a) parallel to and (b) perpendicular to the given line.

7. (3, −6), (1, −1)

8. (1, −2), (−1, −3)

23. y = 2(x + 1) − 2 at (2, 1)

24. y = 3(x − 2) + 1 at (0, 3)

25. y = 2x + 1 at (3, 1)

26. y = 1 at (0, −1)

9. (0.3, −1.4), (−1.1, −0.4)

10. (1.2, 2.1), (3.1, 2.4)

18

..

CHAPTER 0

Preliminaries

In exercises 27–30, find an equation of the line through the given points and compute the y-coordinate of the point on the line corresponding to x 4.

0-18

In exercises 31–34, use the vertical line test to determine whether the curve is the graph of a function. y

31.

y

27.

10 5

5

4 3

x

3 2

2

2

3

5

1

10 x 1

2

3

4

5

y

28.

y

32. 6

10

5

5

4

4

3

2

x 2

2

4

5

1 x 1

2

3

4

5

10

6

y

29. 4

y

33.

3

6

2

4

1

2 x 0.5

1.0

1.5

x

3 2 1

2.0

1

2

y

30.

1

2.0

0.5

1.0

2

1

y

34.

3.0

x

x 1

0.5

1

1.5

2

3

0-19

SECTION 0.1

In exercises 35–40, identify the given function as polynomial, rational, both or neither.

Speed

35. f (x) = x 3 − 4x + 1 x 2 + 2x − 1 x +1 √ 39. f (x) = x 2 + 1 37. f (x) =

..

4 x2 − 1

38. f (x) =

x 3 + 4x − 1 x4 − 1

40. f (x) = 2x − x 2/3 − 6

46. f (x) =

4x x 2 + 2x − 6

In exercises 47–50, find the indicated function values. 47. f (x) = x 2 − x − 1;

f (0), f (2), f (−3), f (1/2)

x +1 ; f (0), f (2), f (−2), f (1/2) x −1 √ 49. f (x) = x + 1; f (0), f (3), f (−1), f (1/2) 48. f (x) =

50. f (x) =

3 ; x

19

36. f (x) = 3 − 2x + x 4

In exercises 41–46, find the domain of the function. √ √ 41. f (x) = x + 2 42. f (x) = 2x + 1 √ √ 44. f (x) = x 2 − 4 43. f (x) = 3 x − 1 45. f (x) =

Polynomials and Rational Functions

Time

FIGURE A Bicycle speed

60. Figure B shows the population of a small country as a function of time. During the time period shown, the country experienced two influxes of immigrants, a war and a plague. Identify these important events. Population

f (1), f (10), f (100), f (1/3)

In exercises 51–54, a brief description is given of a physical situation. For the indicated variable, state a reasonable domain.

Time

51. A parking deck is to be built; x = width of deck (in feet).

FIGURE B

52. A parking deck is to be built on a 200 -by-200 lot; x = width of deck (in feet). 53. A new candy bar is to be sold; x = number of candy bars sold in the first month. 54. A new candy bar is to be sold; x = cost of candy bar (in cents). In exercises 55–58, discuss whether you think y would be a function of x. 55. y = grade you get on an exam, x = number of hours you study

Population In exercises 61–66, find all intercepts of the given graph. 61. y = x 2 − 2x − 8

62. y = x 2 + 4x + 4

63. y = x 3 − 8

64. y = x 3 − 3x 2 + 3x − 1

65. y =

x2 − 4 x +1

66. y =

2x − 1 x2 − 4

56. y = probability of getting lung cancer, x = number of cigarettes smoked per day

In exercises 67–74, factor and/or use the quadratic formula to find all zeros of the given function.

57. y = a person’s weight, x = number of minutes exercising per day

67. f (x) = x 2 − 4x + 3

68. f (x) = x 2 + x − 12

69. f (x) = x 2 − 4x + 2

70. f (x) = 2x 2 + 4x − 1

71. f (x) = x 3 − 3x 2 + 2x

72. f (x) = x 3 − 2x 2 − x + 2

73. f (x) = x 6 + x 3 − 2

74. f (x) = x 3 + x 2 − 4x − 4

58. y = speed at which an object falls, x = weight of object 59. Figure A shows the speed of a bicyclist as a function of time. For the portions of this graph that are flat, what is happening to the bicyclist’s speed? What is happening to the bicyclist’s speed when the graph goes up? down? Identify the portions of the graph that correspond to the bicyclist going uphill; downhill.

75. The boiling point of water (in degrees Fahrenheit) at elevation h (in thousands of feet above sea level) is given by B(h) = −1.8h + 212. Find h such that water boils at 98.6◦ . Why would this altitude be dangerous to humans?

20

..

CHAPTER 0

Preliminaries

0-20

76. The spin rate of a golf ball hit with a 9 iron has been measured at 9100 rpm for a 120-compression ball and at 10,000 rpm for a 60-compression ball. Most golfers use 90-compression balls. If the spin rate is a linear function of compression, find the spin rate for a 90-compression ball. Professional golfers often use 100-compression balls. Estimate the spin rate of a 100-compression ball. 77. The chirping rate of a cricket depends on the temperature. A species of tree cricket chirps 160 times per minute at 79◦ F and 100 times per minute at 64◦ F. Find a linear function relating temperature to chirping rate. 78. When describing how to measure temperature by counting cricket chirps, most guides suggest that you count the number of chirps in a 15-second time period. Use exercise 77 to explain why this is a convenient period of time. 79. A person has played a computer game many times. The statistics show that she has won 415 times and lost 120 times, and the winning percentage is listed as 78%. How many times in a

0.2

40 20 x

2

2

2. Solve the equation | x − 2 | + | x − 3 | = 1. (Hint: It’s an unusual solution, in that it’s more than just a couple of numbers.) Then, solve the equation √ √ x + 3 − 4 x − 1 + x + 8 − 6 x − 1 = 1. (Hint: If you make the correct substitution, you can use your solution to the previous equation.)

4

EXAMPLE 2.1

FIGURE 0.26a y

8 4 x 1

FIGURE 0.26b y = 3x 2 − 1

Generating a Calculator Graph

Use your calculator or computer to sketch a graph of f (x) = 3x 2 − 1.

y = 3x 2 − 1

1

1. Suppose you have a machine that will proportionally enlarge a photograph. For example, it could enlarge a 4 × 6 photograph to 8 × 12 by doubling the width and height. You could make an 8 × 10 picture by cropping 1 inch off each side. Explain how you would enlarge a 3 12 × 5 picture to an 8 × 10. A friend returns from Scotland with a 3 12 × 5 picture showing the Loch Ness monster in the outer 14 on the right. If you use your procedure to make an 8 × 10 enlargement, does Nessie make the cut?

The relationships between functions and their graphs are central topics in calculus. Graphing calculators and user-friendly computer software allow you to explore these relationships for a much wider variety of functions than you could with pencil and paper alone. This section presents a general framework for using technology to explore the graphs of functions. Recall that the graphs of linear functions are straight lines and the graphs of quadratic polynomials are parabolas. One of the goals of this section is for you to become more familiar with the graphs of other functions. The best way to become familiar is through experience, by working example after example.

60

2

EXPLORATORY EXERCISES

GRAPHING CALCULATORS AND COMPUTER ALGEBRA SYSTEMS

y

4

row must she win to raise the reported winning percentage to 80%?

2

Solution You should get an initial graph that looks something like that in Figure 0.26a. This is simply a parabola opening upward. A graph is often used to search for important points, such as x-intercepts, y-intercepts or peaks and troughs. In this case, we could see these points better if we zoom in, that is, display a smaller range of x- and y-values than the technology has initially chosen for us. The graph in Figure 0.26b shows x-values from x = −2 to x = 2 and y-values from y = −2 to y = 10. You can see more clearly in Figure 0.26b that the parabola bottoms out roughly at the point (0, −1) and crosses the x-axis at approximately x = −0.5 and x = 0.5. You can make this more precise by doing some algebra. Recall that an x-intercept is a point where y = 0 or f (x) = 0. Solving 3x 2 − 1 = 0 gives 3x 2 = 1 or x 2 = 13 , so that x = ± 13 ≈ ±0.57735.

20

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0-20

76. The spin rate of a golf ball hit with a 9 iron has been measured at 9100 rpm for a 120-compression ball and at 10,000 rpm for a 60-compression ball. Most golfers use 90-compression balls. If the spin rate is a linear function of compression, find the spin rate for a 90-compression ball. Professional golfers often use 100-compression balls. Estimate the spin rate of a 100-compression ball. 77. The chirping rate of a cricket depends on the temperature. A species of tree cricket chirps 160 times per minute at 79◦ F and 100 times per minute at 64◦ F. Find a linear function relating temperature to chirping rate. 78. When describing how to measure temperature by counting cricket chirps, most guides suggest that you count the number of chirps in a 15-second time period. Use exercise 77 to explain why this is a convenient period of time. 79. A person has played a computer game many times. The statistics show that she has won 415 times and lost 120 times, and the winning percentage is listed as 78%. How many times in a

0.2

40 20 x

2

2

2. Solve the equation | x − 2 | + | x − 3 | = 1. (Hint: It’s an unusual solution, in that it’s more than just a couple of numbers.) Then, solve the equation √ √ x + 3 − 4 x − 1 + x + 8 − 6 x − 1 = 1. (Hint: If you make the correct substitution, you can use your solution to the previous equation.)

4

EXAMPLE 2.1

FIGURE 0.26a y

8 4 x 1

FIGURE 0.26b y = 3x 2 − 1

Generating a Calculator Graph

Use your calculator or computer to sketch a graph of f (x) = 3x 2 − 1.

y = 3x 2 − 1

1

1. Suppose you have a machine that will proportionally enlarge a photograph. For example, it could enlarge a 4 × 6 photograph to 8 × 12 by doubling the width and height. You could make an 8 × 10 picture by cropping 1 inch off each side. Explain how you would enlarge a 3 12 × 5 picture to an 8 × 10. A friend returns from Scotland with a 3 12 × 5 picture showing the Loch Ness monster in the outer 14 on the right. If you use your procedure to make an 8 × 10 enlargement, does Nessie make the cut?

The relationships between functions and their graphs are central topics in calculus. Graphing calculators and user-friendly computer software allow you to explore these relationships for a much wider variety of functions than you could with pencil and paper alone. This section presents a general framework for using technology to explore the graphs of functions. Recall that the graphs of linear functions are straight lines and the graphs of quadratic polynomials are parabolas. One of the goals of this section is for you to become more familiar with the graphs of other functions. The best way to become familiar is through experience, by working example after example.

60

2

EXPLORATORY EXERCISES

GRAPHING CALCULATORS AND COMPUTER ALGEBRA SYSTEMS

y

4

row must she win to raise the reported winning percentage to 80%?

2

Solution You should get an initial graph that looks something like that in Figure 0.26a. This is simply a parabola opening upward. A graph is often used to search for important points, such as x-intercepts, y-intercepts or peaks and troughs. In this case, we could see these points better if we zoom in, that is, display a smaller range of x- and y-values than the technology has initially chosen for us. The graph in Figure 0.26b shows x-values from x = −2 to x = 2 and y-values from y = −2 to y = 10. You can see more clearly in Figure 0.26b that the parabola bottoms out roughly at the point (0, −1) and crosses the x-axis at approximately x = −0.5 and x = 0.5. You can make this more precise by doing some algebra. Recall that an x-intercept is a point where y = 0 or f (x) = 0. Solving 3x 2 − 1 = 0 gives 3x 2 = 1 or x 2 = 13 , so that x = ± 13 ≈ ±0.57735.

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21

Notice that in example 2.1, the graph suggested approximate values for the two x-intercepts, but we needed the algebra to find the values exactly. We then used those values to obtain a view of the graph that highlighted the features that we wanted. Before investigating other graphs, we should say a few words about what a computeror calculator-generated graph really is. Although we call them graphs, what the computer actually does is light up some tiny screen elements called pixels. If the pixels are small enough, the image appears to be a continuous curve or graph. By graphing window, we mean the rectangle defined by the range of x- and y-values displayed. The graphing window can dramatically affect the look of a graph. For example, suppose the x’s run from x = −2 to x = 2. If the computer or calculator screen is wide enough for 400 columns of pixels from left to right, then points will be displayed for x = −2, x = −1.99, x = −1.98, . . . . If there is an interesting feature of this function located between x = −1.99 and x = −1.98, you will not see it unless you zoom in some. In this case, zooming in would reduce the difference between adjacent x’s. Similarly, suppose that the y’s run from y = 0 to y = 3 and that there are 600 rows of pixels from top to bottom. Then, there will be pixels corresponding to y = 0, y = 0.005, y = 0.01, . . . . Now, suppose that f (−2) = 0.0049 and f (−1.99) = 0.0051. Before points are plotted, function values are rounded to the nearest y-value, in this case 0.005. You won’t be able to see any difference in the y-values of these points. If the actual difference is important, you will have to zoom in some to see it.

REMARK 2.1 Most calculators and computer drawing packages use one of the following two schemes for defining the graphing window for a given function. r Fixed graphing window: Most calculators follow this method. Graphs are plotted

in a preselected range of x- and y-values, unless you specify otherwise. For example, the Texas Instruments graphing calculators’ default graphing window plots points in the rectangle defined by −10 ≤ x ≤ 10 and −10 ≤ y ≤ 10. r Automatic graphing window: Most computer drawing packages and some calculators use this method. Graphs are plotted for a preselected range of x-values and the computer calculates the range of y-values so that all of the calculated points will fit in the window. Get to know how your calculator or computer software operates, and use it routinely as you progress through this course. You should always be able to reproduce the computer-generated graphs used in this text by adjusting your graphing window appropriately.

REMARK 2.2 To be precise, f (M) is a local maximum of f if there exist numbers a and b with a < M < b such that f (M) ≥ f (x) for all x such that a < x < b.

Graphs are drawn to provide visual displays of the significant features of a function. What qualifies as significant will vary from problem to problem, but often the x- and y-intercepts and points known as extrema are of interest. The function value f (M) is called a local maximum of the function f if f (M) ≥ f (x) for all x’s “nearby” x = M. Similarly, the function value f (m) is a local minimum of the function f if f (m) ≤ f (x) for all x’s “nearby” x = m. A local extremum is a function value that is either a local maximum or local minimum. Whenever possible, you should produce graphs that show all intercepts and extrema.

EXAMPLE 2.2

Sketching a Graph

Sketch a graph of f (x) = x 3 + 4x 2 − 5x − 1 showing all intercepts and extrema.

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0-22

Solution Depending on your calculator or computer software, you may initially get a graph that looks like one of those in Figure 0.27a or 0.27b. y

y 10

200

100

4 y 30 20 10 4

2

x 2

4

FIGURE 0.28 y = x 3 + 4x 2 − 5x − 1 y

y a1x a0

x

FIGURE 0.29a Line, a1 < 0 y

x

2

x

10

2

4

5

10

10

FIGURE 0.27a

FIGURE 0.27b

y = x 3 + 4x 2 − 5x − 1

y = x 3 + 4x 2 − 5x − 1

Neither graph is completely satisfactory, although both should give you the idea of a graph that (reading left to right) rises to a local maximum near x = −3, drops to a local minimum near x = 1, and then rises again. To get a better graph, notice the scales on the x- and y-axes. The graphing window for Figure 0.27a is the rectangle defined by −5 ≤ x ≤ 5 and −6 ≤ y ≤ 203. The graphing window for Figure 0.27b is defined by the rectangle −10 ≤ x ≤ 10 and −10 ≤ y ≤ 10. From either graph, it appears that we need to show y-values larger than 10, but not nearly as large as 203, to see the local maximum. Since all of the significant features appear to lie between x = −6 and x = 6, one choice for a better window is −5 ≤ x ≤ 5 and −6 ≤ y ≤ 30, as seen in Figure 0.28. There, you can clearly see the three x-intercepts, the local maximum and the local minimum. The graph in example 2.2 was produced by a process of trial and error with thoughtful corrections. You are unlikely to get a perfect picture on the first try. However, you can enlarge the graphing window (i.e., zoom out) if you need to see more, or shrink the graphing window (i.e., zoom in) if the details are hard to see. You should get comfortable enough with your technology that this revision process is routine (and even fun!). In the exercises, you will be asked to graph a variety of functions and discuss the shapes of the graphs of polynomials of different degrees. Having some knowledge of the general shapes will help you decide whether you have found an acceptable graph. To get you started, we now summarize the different shapes of linear, quadratic and cubic polynomials. Of course, the graphs of linear functions [ f (x) = a1 x + a0 ] are simply straight lines of slope a1 . Two possibilities are shown in Figures 0.29a and 0.29b. The graphs of quadratic polynomials [ f (x) = a2 x 2 + a1 x + a0 ] are parabolas. The parabola opens upward if a2 > 0 and opens downward if a2 < 0. We show typical parabolas in Figures 0.30a and 0.30b. y

y y a2x2 a1x a0

y a2x 2 a1x a0

y a1x a0

x x

x

FIGURE 0.29b Line, a1 > 0

FIGURE 0.30a

FIGURE 0.30b

Parabola, a2 > 0

Parabola, a2 < 0

0-23

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Graphing Calculators and Computer Algebra Systems

23

The graphs of cubic functions [ f (x) = a3 x 3 + a2 x 2 + a1 x + a0 ] are somewhat S-shaped. Reading from left to right, the function begins negative and ends positive if a3 > 0, and begins positive and ends negative if a3 < 0, as indicated in Figures 0.31a and 0.31b, respectively. y y y a3x3 a2x 2 a1x a0

y a3x3 a2x 2 a1x a0

x

x

y y a3x3 a2x 2 a1x a0

Inflection point

FIGURE 0.31a

FIGURE 0.31b

Cubic: one max, min, a3 > 0

Cubic: one max, min, a3 < 0

Some cubics have one local maximum and one local minimum, as do those in Figures 0.31a and 0.31b. Many curves (including all cubics) have what’s called an inflection point, where the curve changes its shape (from being bent upward, to being bent downward, or vice versa), as indicated in Figures 0.32a and 0.32b. You can already use your knowledge of the general shapes of certain functions to see how to adjust the graphing window, as in example 2.3.

x

FIGURE 0.32a Cubic: no max or min, a3 > 0

EXAMPLE 2.3

y

Sketching the Graph of a Cubic Polynomial

Sketch a graph of the cubic polynomial f (x) = x 3 − 20x 2 − x + 20. y

a3x3 a2x 2

a1x a0

Solution Your initial graph probably looks like Figure 0.33a or 0.33b.

Inflection point y

y 10

x

x

4

FIGURE 0.32b Cubic: no max or min, a3 < 0

4 200 400 600

10

x

5

5

10

10

FIGURE 0.33a

FIGURE 0.33b

f (x) = x 3 − 20x 2 − x + 20

f (x) = x 3 − 20x 2 − x + 20

However, you should recognize that neither of these graphs looks like a cubic; they look more like parabolas. To see the S-shape behavior in the graph, we need to consider a larger range of x-values. To determine how much larger, we need some of the concepts

24

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y 20 10

x

10 400 800

0-24

of calculus. For the moment, we use trial and error, until the graph resembles the shape of a cubic. You should recognize the characteristic shape of a cubic in Figure 0.33c. Although we now see more of the big picture (often referred to as the global behavior of the function), we have lost some of the details (such as the x-intercepts), which we could clearly see in Figures 0.33a and 0.33b (often referred to as the local behavior of the function).

1200

Rational functions have some properties not found in polynomials, as we see in examples 2.4, 2.5 and 2.6.

FIGURE 0.33c f (x) = x 3 − 20x 2 − x + 20

EXAMPLE 2.4

Sketching the Graph of a Rational Function

x −1 and describe the behavior of the graph near x = 2. x −2 Solution Your initial graph should look something like Figure 0.34a or 0.34b. From either graph, it should be clear that something unusual is happening near x = 2. Zooming in closer to x = 2 should yield a graph like that in Figure 0.35.

Sketch a graph of f (x) =

y

y

4

1e08

10

5e07

5 x

20

5e07

10

1e08 x 2

10

10

x −1 x −2

y 10 5 x 4

x

5

5

5

10

10

FIGURE 0.34a

FIGURE 0.34b y=

x −1 x −2

In Figure 0.35, it appears that as x increases up to 2, the function values get more and more negative, while as x decreases down to 2, the function values get more and more positive. This is also observed in the following table of function values.

FIGURE 0.35

4

4

x −1 y= x −2

20

y=

y

x

f(x)

x

f(x)

1.8

−4

2.2

6

1.9

−9

2.1

11

1.99

−99

2.01

101

1.999

−999

2.001

1001

1.9999

−9999

2.0001

10,001

8

5 10

FIGURE 0.36 Vertical asymptote

Note that at x = 2, f (x) is undefined. However, as x approaches 2 from the left, the graph veers down sharply. In this case, we say that f (x) tends to −∞. Likewise, as x approaches 2 from the right, the graph rises sharply. Here, we say that f (x) tends to ∞ and there is a vertical asymptote at x = 2. (We’ll define this more carefully in Chapter 1.) It is common to draw a vertical dashed line at x = 2 to indicate this (see Figure 0.36). Since f (2) is undefined, there is no point plotted at x = 2.

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Graphing Calculators and Computer Algebra Systems

25

Many rational functions have vertical asymptotes. Notice that there is no point plotted on the vertical asymptote since the function is undefined at such an x-value (due to the division by zero when that value of x is substituted in). Given a rational function, you can locate possible vertical asymptotes by finding where the denominator is zero. It turns out that if the numerator is not zero at that point, there is a vertical asymptote at that point.

EXAMPLE 2.5

A Graph with Several Vertical Asymptotes

x −1 . − 5x + 6 Solution Note that the denominator factors as

Find all vertical asymptotes for f (x) =

x2

x 2 − 5x + 6 = (x − 2)(x − 3), so that the only possible locations for vertical asymptotes are x = 2 and x = 3. Since neither x-value makes the numerator (x − 1) equal to zero, there are vertical asymptotes at both x = 2 and x = 3. A computer-generated graph gives little indication of how the function behaves near the asymptotes. (See Figure 0.37a and note the scale on the y-axis.) y y 2e08 5

1e08 x 1

x

4

1e08

1

4

5

5

2e08 10

3e08

FIGURE 0.37a x −1 y= 2 x − 5x + 6

FIGURE 0.37b y=

x −1 x 2 − 5x + 6

y 0.2

x

10

10

We can improve the graph by zooming in in both the x- and y-directions. Figure 0.37b shows a graph of the same function using the graphing window defined by the rectangle −1 ≤ x ≤ 5 and −13 ≤ y ≤ 7. This graph clearly shows the vertical asymptotes at x = 2 and x = 3.

20

As we see in example 2.6, not all rational functions have vertical asymptotes. 0.2

EXAMPLE 2.6 0.4

FIGURE 0.38 y=

x −1 x2 + 4

A Rational Function with No Vertical Asymptotes

Find all vertical asymptotes of

x −1 . x2 + 4

Solution Notice that x 2 + 4 = 0 has no (real) solutions, since x 2 + 4 > 0 for all real numbers, x. So, there are no vertical asymptotes. The graph in Figure 0.38 is consistent with this observation.

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0-26

Graphs are useful for finding approximate solutions of difficult equations, as we see in examples 2.7 and 2.8. y

EXAMPLE 2.7 12

Find approximate solutions of the equation x 2 =

8

4

4

Finding Zeros Approximately

x

2

2

4

FIGURE 0.39a y = x2 −

√

x +3

y 10 8

√

x + 3. √ Solution You could rewrite√this equation as x 2 − x + 3 = 0 and then look for zeros in the graph of f (x) = x 2 − x + 3, seen in Figure 0.39a. Note that two zeros are clearly indicated: one near −1, the√other near 1.5. However, since you know very little of the nature of the function x 2 − x + 3, you cannot say whether or not there are any other zeros, ones that don’t show up in the window seen in Figure 0.39a. On the other hand, if you graph the two functions on either side of the equation on the same set of axes, as in Figure 0.39b, you can clearly see two points where the graphs intersect (corresponding to the two zeros seen in Figure 0.39a). Further, since you know the general shapes of both of the graphs, you can infer from Figure 0.39b that there are no other intersections (i.e., there are no other zeros of f ). This is important information that you cannot obtain from Figure 0.39a. Now that you know how many solutions there are, you need to estimate their values. One method is to zoom in on the zeros graphically. We leave it as an exercise to verify that the zeros are approximately x = 1.4 and x = −1.2. If your calculator or computer algebra system has a solve command, you can use it to quickly obtain an accurate approximation. In this case, we get x ≈ 1.452626878 and x ≈ −1.164035140.

6

When using the solve command on your calculator or computer algebra system, be sure to check that the solutions make sense. If the results don’t match what you’ve seen in your preliminary sketches and zooms, beware! Even high-tech equation solvers make mistakes occasionally.

4 2 4

x

2

2

4

EXAMPLE 2.8

FIGURE 0.39b √

y = x and y =

x +3

2

Find all points of intersection of the graphs of y = 2 cos x and y = 2 − x.

y 4

1

x 1

3

5

2

Finding Intersections by Calculator: An Oversight

Solution Notice that the intersections correspond to solutions of the equation 2 cos x = 2 − x. Using the solve command on the Tl-92 graphing calculator, we found intersections at x ≈ 3.69815 and x = 0. So, what’s the problem? A sketch of the graphs of y = 2 − x and y = 2 cos x (we discuss this function in the next section) clearly shows three intersections (see Figure 0.40). The middle solution, x ≈ 1.10914, was somehow passed over by the calculator’s solve routine. The lesson here is to use graphical evidence to support your solutions, especially when using software and/or functions with which you are less than completely familiar.

FIGURE 0.40 y = 2 cos x and y = 2 − x

You need to look skeptically at the answers provided by your calculator’s solver program. While such solvers provide a quick means of approximating solutions of equations, these programs will sometimes return incorrect answers, as we illustrate with example 2.9. So, how do you know if your solver is giving you an accurate answer or one that’s incorrect? The only answer to this is that you must carefully test your calculator’s solution, by separately calculating both sides of the equation (by hand) at the calculated solution.

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EXAMPLE 2.9

..

Graphing Calculators and Computer Algebra Systems

27

Solving an Equation by Calculator: An Erroneous Answer

1 1 = . x x Solution Certainly, you don’t need a calculator to solve this equation, but consider what happens when you use one. Most calculators report a solution that is very close to zero, while others report that the solution is x = 0. Not only are these answers incorrect, but the given equation has no solution, as follows. First, notice that the 1 equation makes sense only when x = 0. Subtracting from both sides of the equation x leaves us with x = 0, which can’t possibly be a solution, since it does not satisfy the original equation. Notice further that, if your calculator returns the approximate solution x = 1 × 10−7 and you use your calculator to compute the values on both sides of the equation, the calculator will compute 1 x + = 1 × 10−7 + 1 × 107 , x 1 which it approximates as 1 × 107 = , since calculators carry only a finite number of x digits. In other words, although

Use your calculator’s solver program to solve the equation x +

1 × 10−7 + 1 × 107 = 1 × 107 , your calculator treats these numbers as the same and so incorrectly reports that the equation is satisfied. The moral of this story is to be an intelligent user of technology and don’t blindly accept everything a calculator tells you. We want to emphasize again that graphing should be the first step in the equation-solving process. A good graph will show you how many solutions to expect, as well as give their approximate locations. Whenever possible, you should factor or use the quadratic formula to get exact solutions. When this is impossible, approximate the solutions by zooming in on them graphically or by using your calculator’s solve command. Always compare your results to a graph to see if there’s anything you’ve missed.

EXERCISES 0.2 WRITING EXERCISES 1. Explain why there is a significant difference among Figures 0.33a, 0.33b and 0.33c. 2. In Figure 0.36, the graph approaches the lower portion of the vertical asymptote from the left, whereas the graph approaches the upper portion of the vertical asymptote from the right. Use the table of function values found in example 2.4 to explain how to determine whether a graph approaches a vertical asymptote by dropping down or rising up. 3. In the text, we discussed the difference between graphing with a fixed window versus an automatic window. Discuss the advantages and disadvantages of each. (Hint: Consider the case of a first graph of a function you know nothing about and the case of hoping to see the important details of a graph for which you know the general shape.)

x3 + 1 with each of the following x graphing windows: (a) −10 ≤ x ≤ 10, (b) − 1000 ≤ x ≤ 1000. Explain why the graph in (b) doesn’t show the details that the graph in (a) does.

4. Examine the graph of y =

In exercises 1–30, sketch a graph of the function showing all extrema, intercepts and asymptotes. 1. f (x) = x 2 − 1

2. f (x) = 3 − x 2

3. f (x) = x 2 + 2x + 8

4. f (x) = x 2 − 20x + 11

5. f (x) = x 3 + 1

6. f (x) = 10 − x 3

7. f (x) = x 3 + 2x − 1

8. f (x) = x 3 − 3x + 1

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9. f (x) = x 4 − 1

0-28

10. f (x) = 2 − x 4

11. f (x) = x 4 + 2x − 1

12. f (x) = x 4 − 6x 2 + 3

13. f (x) = x 5 + 2

14. f (x) = 12 − x 5

15. f (x) = x 5 − 8x 3 + 20x − 1 3 17. f (x) = x −1

16. f (x) = x 5 + 5x 4 + 2x 3 + 1 4 18. f (x) = x +2

3x 19. f (x) = x −1

4x 20. f (x) = x +2

3x 2 21. f (x) = x −1

4x 2 22. f (x) = x +2

23. f (x) =

2 x2 − 4

24. f (x) =

6 x2 − 9

25. f (x) =

3 x2 + 4

26. f (x) =

6 x2 + 9

27. f (x) =

x2

29. f (x) = √

x +2 +x −6 3x

x2 + 4

28. f (x) =

x2

30. f (x) = √

55. cos x = x 2 − 1

56. sin x = x 2 + 1

60. f (x) = x 4 − 2x + 1

2x

61. f (x) = x 4 − 7x 3 − 15x 2 − 10x − 1410

x2 + 1

62. f (x) = x 6 − 4x 4 + 2x 3 − 8x − 2

x +4 x2 − 9

33. f (x) =

4x x 2 + 3x − 10

34. f (x) =

x +2 x 2 − 2x − 15

35. f (x) =

4x x2 + 4

36. f (x) = √

3x

63. Graph y = x 2 in the graphing window −10 ≤ x ≤ 10, −10 ≤ y ≤ 10, without drawing the x- and y-axes. Adjust the graphing window for y = 2(x − 1)2 + 3 so that (without the axes showing) the graph looks identical to that of y = x 2. 64. Graph y = x 2 in the graphing window −10 ≤ x ≤ 10, −10 ≤ y ≤ 10. Separately graph y = x 4 with the same graphing window. Compare and contrast the graphs. Then graph the two functions on the same axes and carefully examine the differences in the intervals −1 < x < 1 and x > 1.

x2 − 9

3x x4 − 1

In exercises 39–42, a standard graphing window will not reveal all of the important details of the graph. Adjust the graphing window to find the missing details. 39. f (x) = 13 x 3 −

54. (x + 1)2/3 = 2 − x

59. f (x) = x 4 − 3x 3 − x + 1

x −1 + 4x + 3

32. f (x) =

38. f (x) =

53. (x 2 − 1)2/3 = 2x + 1

58. f (x) = x 3 − 4x 2 + 2

3x x2 − 4

x2 + 1 + 3x 2 + 2x

52. x 3 + 1 = −3x 2 − 3x

57. f (x) = x 3 − 3x + 1

31. f (x) =

x3

51. x 3 − 3x 2 = 1 − 3x

In exercises 57–62, use a graphing calculator or computer graphing utility to estimate all zeros.

In exercises 31–38, find all vertical asymptotes.

37. f (x) =

In exercises 49–56, determine the number of (real) solutions. Solve for the intersection points exactly if possible and estimate the points if necessary. √ √ 49. x − 1 = x 2 − 1 50. x 2 + 4 = x 2 + 2

1 x 400

40. f (x) = x 4 − 11x 3 + 5x − 2 √ 41. f (x) = x 144 − x 2

65. In this exercise, you will find an equation describing all points equidistant from the x-axis and the point (0, 2). First, see if you can sketch a picture of what this curve ought to look like. For a point (x, y) that is on the curve, explain why y 2 = x 2 + (y − 2)2 . Square both sides of this equation and solve for y. Identify the curve. 66. Find an equation describing all points equidistant from the x-axis and (1, 4) (see exercise 65).

42. f (x) = 15 x 5 − 78 x 4 + 13 x 3 + 72 x 2 − 6x In exercises 43–48, adjust the graphing window to identify all vertical asympotes. 4x x2 − 1

43. f (x) =

3 x −1

44. f (x) =

46. f (x) =

2x x +4

x2 − 1 47. f (x) = √ x4 + x

45. f (x) =

2

3x x2 − 1

48. f (x) = √

2x x2 + x

EXPLORATORY EXERCISES 1. Suppose that a graphing calculator is set up with pixels corresponding to x = 0, 0.1, 0.2, 0.3, . . . , 2.0 and y = 0, 0.1, 0.2, 0.3, . . . , 4.0. For the function f (x) = x 2 , compute the indicated function values and round off to give pixel coordinates [e.g., the point (1.19, 1.4161) has pixel coordinates (1.2, 1.4)].

0-29

SECTION 0.3

(a) f (0.4), (b) f (0.39), (c) f (1.17), (d) f (1.20), (e) f (1.8), (f) f (1.81). Repeat (c)–(d) if the graphing window is zoomed in so that x = 1.00, 1.01, . . . , 1.20 and y = 1.30, 1.31, . . . , 1.50. Repeat (e)–(f ) if the graphing window is zoomed in so that x = 1.800, 1.801, . . . , 1.820 and y = 3.200, 3.205, . . . , 3.300. 2. Graph y = x 2 − 1, y = x 2 + x − 1, y = x 2 + 2x − 1, y = x 2 − x − 1, y = x 2 − 2x − 1 and other functions of the form y = x 2 + cx − 1. Describe the effect(s) a change in c has on the graph.

0.3

Inverse Functions

29

3. Figures 0.31 and 0.32 provide a catalog of the possible types of graphs of cubic polynomials. In this exercise, you will compile a catalog of graphs of fourth-order polynomials (i.e., y = ax 4 + bx 3 + cx 2 + d x + e). Start by using your calculator or computer to sketch graphs with different values of a, b, c, d and e. Try y = x 4 , y = 2x 4 , y = −2x 4 , y = x 4 + x 3 , y = x 4 + 2x 3 , y = x 4 − 2x 3 , y = x 4 + x 2 , y = x 4 − x 2 , y = x 4 − 2x 2 , y = x 4 + x, y = x 4 − x and so on. Try to determine what effect each constant has.

INVERSE FUNCTIONS f

x

y

Domain { f}

Range { f } g

FIGURE 0.41 g(x) = f −1 (x) y 8 y x3

6 4 2 2

..

x 1

The notion of an inverse relationship is basic to many areas of science. The number of common inverse problems is immense. As only one example, take the case of the electrocardiogram (EKG). In an EKG, technicians connect a series of electrodes to a patient’s chest and use measurements of electrical activity on the surface of the body to infer something about the electrical activity on the surface of the heart. This is referred to as an inverse problem, since physicians are attempting to determine what inputs (i.e., the electrical activity on the surface of the heart) cause an observed output (the measured electrical activity on the surface of the chest). The mathematical notion of inverse is much the same as that just described. Given an output (in this case, a value in the range of a given function), we wish to find the input (the value in the domain) that produced that output. That is, given a y ∈ Range{ f }, find the x ∈ Domain{ f } for which y = f (x). (See the illustration of the inverse function g shown in Figure 0.41.) For instance, suppose that f (x) = x 3 and y = 8. Can you find an x such that x 3 = 8? That is, can you find the x-value corresponding the √ to y = 8? (See Figure 0.42.) Of course, √ solution of this particular equation is x = 3 8 = 2. In general, if x 3 = y, then x = 3 y. In light of this, we say that the cube root function is the inverse of f (x) = x 3 .

2

2 4

EXAMPLE 3.1

If f (x) = x 3 and g(x) = x 1/3 , show that

FIGURE 0.42 Finding the x-value corresponding to y = 8

Two Functions That Reverse the Action of Each Other

f (g(x)) = x

and

g( f (x)) = x,

for all x. Solution For all real numbers x, we have f (g(x)) = f (x 1/3 ) = (x 1/3 )3 = x and

g( f (x)) = g(x 3 ) = (x 3 )1/3 = x.

Notice in example 3.1 that the action of f undoes the action of g and vice versa. We take this as the definition of an inverse function. (Again, think of Figure 0.41.)

0-29

SECTION 0.3

(a) f (0.4), (b) f (0.39), (c) f (1.17), (d) f (1.20), (e) f (1.8), (f) f (1.81). Repeat (c)–(d) if the graphing window is zoomed in so that x = 1.00, 1.01, . . . , 1.20 and y = 1.30, 1.31, . . . , 1.50. Repeat (e)–(f ) if the graphing window is zoomed in so that x = 1.800, 1.801, . . . , 1.820 and y = 3.200, 3.205, . . . , 3.300. 2. Graph y = x 2 − 1, y = x 2 + x − 1, y = x 2 + 2x − 1, y = x 2 − x − 1, y = x 2 − 2x − 1 and other functions of the form y = x 2 + cx − 1. Describe the effect(s) a change in c has on the graph.

0.3

Inverse Functions

29

3. Figures 0.31 and 0.32 provide a catalog of the possible types of graphs of cubic polynomials. In this exercise, you will compile a catalog of graphs of fourth-order polynomials (i.e., y = ax 4 + bx 3 + cx 2 + d x + e). Start by using your calculator or computer to sketch graphs with different values of a, b, c, d and e. Try y = x 4 , y = 2x 4 , y = −2x 4 , y = x 4 + x 3 , y = x 4 + 2x 3 , y = x 4 − 2x 3 , y = x 4 + x 2 , y = x 4 − x 2 , y = x 4 − 2x 2 , y = x 4 + x, y = x 4 − x and so on. Try to determine what effect each constant has.

INVERSE FUNCTIONS f

x

y

Domain { f}

Range { f } g

FIGURE 0.41 g(x) = f −1 (x) y 8 y x3

6 4 2 2

..

x 1

The notion of an inverse relationship is basic to many areas of science. The number of common inverse problems is immense. As only one example, take the case of the electrocardiogram (EKG). In an EKG, technicians connect a series of electrodes to a patient’s chest and use measurements of electrical activity on the surface of the body to infer something about the electrical activity on the surface of the heart. This is referred to as an inverse problem, since physicians are attempting to determine what inputs (i.e., the electrical activity on the surface of the heart) cause an observed output (the measured electrical activity on the surface of the chest). The mathematical notion of inverse is much the same as that just described. Given an output (in this case, a value in the range of a given function), we wish to find the input (the value in the domain) that produced that output. That is, given a y ∈ Range{ f }, find the x ∈ Domain{ f } for which y = f (x). (See the illustration of the inverse function g shown in Figure 0.41.) For instance, suppose that f (x) = x 3 and y = 8. Can you find an x such that x 3 = 8? That is, can you find the x-value corresponding the √ to y = 8? (See Figure 0.42.) Of course, √ solution of this particular equation is x = 3 8 = 2. In general, if x 3 = y, then x = 3 y. In light of this, we say that the cube root function is the inverse of f (x) = x 3 .

2

2 4

EXAMPLE 3.1

If f (x) = x 3 and g(x) = x 1/3 , show that

FIGURE 0.42 Finding the x-value corresponding to y = 8

Two Functions That Reverse the Action of Each Other

f (g(x)) = x

and

g( f (x)) = x,

for all x. Solution For all real numbers x, we have f (g(x)) = f (x 1/3 ) = (x 1/3 )3 = x and

g( f (x)) = g(x 3 ) = (x 3 )1/3 = x.

Notice in example 3.1 that the action of f undoes the action of g and vice versa. We take this as the definition of an inverse function. (Again, think of Figure 0.41.)

30

..

CHAPTER 0

Preliminaries

REMARK 3.1

0-30

DEFINITION 3.1

Pay close attention to the notation. Notice that f −1 (x) 1 . We write does not mean f (x) the reciprocal of f (x) as

Assume that f and g have domains A and B, respectively, and that f (g(x)) is defined for all x ∈ B and g( f (x)) is defined for all x ∈ A. If

1 = [ f (x)]−1 . f (x)

f (g(x)) = x,

for all x ∈ B

g( f (x)) = x,

for all x ∈ A,

and

we say that g is the inverse of f, written g = f −1 . Equivalently, f is the inverse of g, f = g −1 .

Observe that many familiar functions have no inverse. y

EXAMPLE 3.2 20

Show that f (x) = x 2 has no inverse on the interval (−∞, ∞). Solution Notice that f (4) = 16 and f (−4) = 16. That is, there are two x-values that produce the same y-value. So, if we were to try to define an inverse of f, how would we define f −1 (16)? Look at the graph of y = x 2 (see Figure 0.43) to see what the problem is. For each y > 0, there are two x-values for which y = x 2 . Because of this, the function does not have an inverse.

12 8 4

4

A Function with No Inverse

x

2

2

4

FIGURE 0.43 y = x2

√ For f (x) = x 2 , it is tempting to jump √ to2the conclusion that g(x) = x is the inverse of f (x). Notice that although f (g(x)) = ( x) = x√for all x ≥ 0 (i.e., for all x in the domain of g(x)), it is not generally true that g( f (x)) = x 2 = x. In fact, this last equality holds only for x √ ≥ 0. However, for f (x) = x 2 restricted to the domain x ≥ 0, we do have that −1 f (x) = x.

DEFINITION 3.2 A function f is called one-to-one when for every y ∈ Range{ f }, there is exactly one x ∈ Domain{ f } for which y = f (x).

y

REMARK 3.2

y f (x)

x a

Observe that an equivalent definition of one-to-one is the following. A function f (x) is one-to-one if and only if the equality f (a) = f (b) implies a = b. This version of the definition is often useful for proofs involving one-to-one functions.

b

FIGURE 0.44a f (a) = f (b), for a = b So, f does not pass the horizontal line test and is not one-to-one.

It is helpful to think of the concept of one-to-one in graphical terms. Notice that a function f is one-to-one if and only if every horizontal line intersects the graph in at most one point. This is referred to as the horizontal line test. We illustrate this in Figures 0.44a and 0.44b. The following result should now make sense.

0-31

SECTION 0.3

..

Inverse Functions

31

y

THEOREM 3.1

y f (x)

A function f has an inverse if and only if it is one-to-one.

This theorem simply says that every one-to-one function has an inverse and every function that has an inverse is one-to-one. However, it says nothing about how to find an inverse. For very simple functions, we can find inverses by solving equations. x

a

EXAMPLE 3.3

Find the inverse of f (x) = x 3 − 5.

FIGURE 0.44b

Solution Note that it is not entirely clear from the graph (see Figure 0.45) whether f passes the horizontal line test. To find the inverse function, write y = f (x) and solve for x (i.e., solve for the input x that produced the observed output y). We have

Every horizontal line intersects the curve in at most one point. So, f passes the horizontal line test and is one-to-one.

y = x 3 − 5.

y

Adding 5 to both sides and taking the cube root gives us (y + 5)1/3 = (x 3 )1/3 = x.

40

So, x = f −1 (y) = (y + 5)1/3 . Reversing the variables x and y gives us

20

4

Finding an Inverse Function

f −1 (x) = (x + 5)1/3 .

x

2

2

4

20

EXAMPLE 3.4

A Function That Is Not One-to-One

Show that f (x) = 10 − x 4 does not have an inverse.

40

Solution You can see from a graph (see Figure 0.46) that f is not one-to-one; for instance, f (1) = f (−1) = 9. Consequently, f does not have an inverse.

FIGURE 0.45 y = x3 − 5

Most often, we cannot find a formula for an inverse function and must be satisfied with simply knowing that the inverse function exists. Example 3.5 is typical of this situation.

y 20 4

2

2 20 40 60 80 100

FIGURE 0.46 y = 10 − x 4

4

x

EXAMPLE 3.5

Finding Values of an Inverse Function

Given that f (x) = x 5 + 8x 3 + x + 1 has an inverse, find f −1 (1) and f −1 (11). Solution First, notice that from the graph shown in Figure 0.47 (on the following page), the function looks like it might be one-to-one, but how can we be certain of this? (Remember that graphs can be deceptive!) Until we develop some calculus, we will be unable to verify this. Ideally, we would show that f has an inverse by finding a formula for f −1 , as in example 3.3. However, in this case, we must solve the equation y = x 5 + 8x 3 + x + 1 for x. Think about this for a moment: you should realize that we can’t solve for x in terms of y here. We need to assume that the inverse exists, as indicated in the instructions.

32

CHAPTER 0

..

Preliminaries

0-32

Turning to the problem of finding f −1 (1) and f −1 (11), you might wonder if this is possible, since we were unable to find a formula for f −1 (x). While it’s certainly true that we have no such formula, you might observe that f (0) = 1, so that f −1 (1) = 0. By trial and error, you might also discover that f (1) = 11 and so, f −1 (11) = 1.

y 3 2 1 3 2

1 1

x 1

2

3

2

In example 3.5, we examined a function that has an inverse, although we could not find that inverse algebraically. Even when we can’t find an inverse function explicitly, we can say something graphically. Notice that if (a, b) is a point on the graph of y = f (x) and f has an inverse, f −1 , then since

3

b = f (a),

FIGURE 0.47

That is, (b, a) is a point on the graph of y = f −1 (x). This tells us a great deal about the inverse function. In particular, we can immediately obtain any number of points on the graph of y = f −1 (x), simply by inspection. Further, notice that the point (b, a) is the reflection of the point (a, b) through the line y = x (see Figure 0.48). It now follows that given the graph of any one-to-one function, you can draw the graph of its inverse simply by reflecting the entire graph through the line y = x. In example 3.6, we illustrate the symmetry of a function and its inverse.

y yx a

f −1 (b) = f −1 ( f (a)) = a.

we have that

y = x 5 + 8x 3 + x + 1

(b, a)

(a, b)

b

EXAMPLE 3.6 x b

a

FIGURE 0.48 Reflection through y = x

The Graph of a Function and Its Inverse

Draw a graph of f (x) = x 3 and its inverse. Solution From example 3.1, the inverse of f (x) = x 3 is f −1 (x) = x 1/3 . Notice the symmetry of their graphs shown in Figure 0.49.

y y

yx

yx 1

y x 1/3

y f (x)

yx

3

x

1

1

y f 1(x)

1

FIGURE 0.49 y = x 3 and y = x 1/3

x

FIGURE 0.50

Graphs of f and f −1

Observe that we can use this symmetry principle to draw the graph of an inverse function, even when we don’t have a formula for that function (see Figure 0.50).

0-33

SECTION 0.3

..

Inverse Functions

33

y 3

y f (x)

EXAMPLE 3.7

yx

Draw a graph of f (x) = x 5 + 8x 3 + x + 1 and its inverse.

2 y f 1(x)

1 3 2

1 1

Drawing the Graph of an Unknown Inverse Function

x 1

2

3

2 3

Solution In example 3.5, we were unable to find a formula for the inverse function. Despite this, we can draw a graph of f −1 with ease. We simply take the graph of y = f (x) seen in Figure 0.47 and reflect it across the line y = x, as shown in Figure 0.51. (When we introduce parametric equations in section 9.1, we will see a clever way to draw this graph with a graphing calculator.) In example 3.8, we apply our theoretical knowledge of inverse functions in a medical setting.

FIGURE 0.51

y = f (x) and y = f −1 (x)

EXAMPLE 3.8

Determining the Proper Dosage of a Drug

Suppose that the injection of a certain drug raises the level of a key hormone in the body. Physicians want to determine the dosage that produces a healthy hormone level. Dosages of 1, 2, 3 and 4 mg produce hormone levels of 12, 20, 40 and 76, respectively. If the desired hormone level is 30, what is the proper dosage?

y

y

TODAY IN MATHEMATICS Kim Rossmo (1955– ) A Canadian criminologist who developed the Criminal Geographic Targeting algorithm that indicates the most probable area of residence for serial murderers, rapists and other criminals. Rossmo served 21 years with the Vancouver Police Department. His mentors were Professors Paul and Patricia Brantingham of Simon Fraser University. The Brantinghams developed Crime Pattern Theory, which predicts crime locations from where criminals live, work and play. Rossmo inverted their model and used the crime sites to determine where the criminal most likely lives. The premiere episode of the television drama Numb3rs was based on Rossmo’s work.

80

80

60

60

40

40

20

20 x 1

2

3

4

x 1

2

3

FIGURE 0.52a

FIGURE 0.52b

Hormone data

Approximate curve

4

Solution A plot of the points (1, 12), (2, 20), (3, 40) and (4, 76) summarizes the data (see Figure 0.52a). The problem is an inverse problem: given y = 30, what is x? It is tempting to argue the following: since 30 is halfway between 20 and 40, the x-value should be halfway between 2 and 3: x = 2.5. This method of solution is called linear interpolation, since the point (x, y) = (2.5, 30) lies on the line through the points (2, 20) and (3, 40). However, this estimate does not take into account all of the information we have. The points in Figure 0.52a suggest a graph that is curving up. If this is the case, x = 2.6 or x = 2.7 may be a better estimate of the required dosage. In Figure 0.52b, we have sketched a smooth curve through the data points and indicated a graphical solution of the problem. More advanced techniques (e.g., polynomial interpolation) have been developed by mathematicians to make the estimate of such quantities as accurate as possible.

34

CHAPTER 0

..

Preliminaries

0-34

EXERCISES 0.3 WRITING EXERCISES 1. Explain in words (and a picture) why the following is true: if f (x) is increasing for all x, then f has an inverse. 2. Suppose the graph of a function passes the horizontal line test. Explain why you know that the function has an inverse (defined on the range of the function). 3. Radar works by bouncing a high-frequency electromagnetic pulse off of a moving object, then measuring the disturbance in the pulse as it is bounced back. Explain why this is an inverse problem by identifying the input and output.

17. f (x) = 18. f (x) =

√ √

(b) f −1 (1)

y 4 2

4

x

2

2

4

2

4

2

4

2 4

y

20. 4 2

1 − 2x 1 and g(x) = (x = 0, x = −2) x +2 x

4

5. f (x) = x 3 − 2

6. f (x) = x 3 + 4

7. f (x) = x 5 − 1

8. f (x) = x 5 + 4 10. f (x) = x 4 − 2x − 1 √ 12. f (x) = x 2 + 1

2 4

y

21. 4

In exercises 13–18, assume that the function has an inverse. Without solving for the inverse, find the indicated function values. 13. f (x) = x 3 + 4x − 1,

(a) f −1 (−1),

(b) f −1 (4)

14. f (x) = x 3 + 2x + 1,

(a) f −1 (1),

(b) f −1 (13)

15. f (x) = x 5 + 3x 3 + x,

(a) f −1 (−5),

(b) f −1 (5)

16. f (x) = x 5 + 4x − 2,

(a) f −1 (38),

(b) f −1 (3)

x

2

In exercises 5–12, determine whether the function is one-to-one. If it is, find the inverse and graph both the function and its inverse.

9. f (x) = x 4 + 2 √ 11. f (x) = x 3 + 1

x 5 + 4x 3 + 3x + 1, (a) f −1 (3),

19.

In exercises 1–4, show that f (g(x)) x and g( f (x)) x for all x:

4. f (x) =

(b) f −1 (2)

In exercises 19–22, use the given graph to graph the inverse function.

4. Each human disease has a set of symptoms associated with it. Physicians attempt to solve an inverse problem: given the symptoms, they try to identify the disease causing the symptoms. Explain why this is not a well-defined inverse problem (i.e., logically it is not always possible to correctly identify diseases from symptoms alone).

1. f (x) = x 5 and g(x) = x 1/5 1/3 2. f (x) = 4x 3 and g(x) = 14 x x −1 3 3. f (x) = 2x + 1 or g(x) = 3 2

(a) f −1 (4),

x 3 + 2x + 4,

2

4

x

2 2 4

0-35

SECTION 0.3

4

35

41. Graph f (x) = (x − 2)2 and find an interval on which it is oneto-one. Find the inverse of the function restricted to that interval. Graph both functions.

2

4

Inverse Functions

40. Graph f (x) = x 2 + 2 for x ≤ 0 and verify that it is one-to-one. Find its inverse. Graph both functions.

y

22.

..

x

2

2

4

2 4

In exercises 23–26, use linear interpolation (see example 3.8) to estimate f − 1 (b). Use the apparent curving of the graph to conjecture whether the estimate is too high or too low. 23. (1, 12), (2, 20), (3, 26), (4, 30), b = 23 24. (1, 12), (2, 10), (3, 6), (4, 0), b = 8

42. Graph f (x) = (x + 1)4 and find an interval on which it is oneto-one. Find the inverse of the function restricted to that interval. Graph both functions. √ 43. Graph f (x) = x 2 − 2x and find an interval on which it is one-to-one. Find the inverse of the function restricted to that interval. Graph both functions. x 44. Graph f (x) = 2 and find an interval on which it is one-tox −4 one. Find the inverse of the function restricted to that interval. Graph both functions. 45. Graph f (x) = sin x and find an interval on which it is one-toone. Find the inverse of the function restricted to that interval. Graph both functions.

26. (1, 12), (2, 20), (3, 36), (4, 50), b = 32

46. Graph f (x) = cos x and find an interval on which it is one-toone. Find the inverse of the function restricted to that interval. Graph both functions.

In exercises 27–36, use a graph to determine whether the function is one-to-one. If it is, graph the inverse function.

In exercises 47–52, discuss whether the function described has an inverse.

27. f (x) = x 3 − 5

47. The income of a company varies with time.

28. f (x) = x − 3

48. The height of a person varies with time.

29. f (x) = x + 2x − 1

49. For a dropped ball, its height varies with time.

30. f (x) = x − 2x − 1

50. For a ball thrown upward, its height varies with time.

31. f (x) = x − 3x − 1

51. The shadow made by an object depends on its threedimensional shape.

25. (1, 12), (2, 6), (3, 2), (4, 0), b = 5

2 3 3 5

3

32. f (x) = x + 4x − 2 5

33. f (x) =

3

1 x +1

4 x2 + 1 x 35. f (x) = x +4 x 36. f (x) = √ 2 x +4 34. f (x) =

52. The number of calories burned depends on how fast a person runs. 53. Suppose that your boss informs you that you have been awarded a 10% raise. The next week, your boss announces that due to circumstances beyond her control, all employees will have their salaries cut by 10%. Are you as well off now as you were two weeks ago? Show that increasing by 10% and decreasing by 10% are not inverse processes. Find the inverse for adding 10%. (Hint: To add 10% to a quantity you can multiply the quantity by 1.10.)

Exercises 37–46 involve inverse functions on restricted domains. √ 37. Show that f (x) = x 2 (x ≥ 0) and g(x) = x (x ≥ 0) are inverse functions. Graph both functions. √ 38. Show that f (x) = x 2 − 1 (x ≥ 0) and g(x) = x + 1 (x ≥ −1) are inverse functions. Graph both functions.

1. Find all values of k such that f (x) = x 3 + kx + 1 is one-toone.

39. Graph f (x) = x 2 for x ≤ 0 and verify that it is one-to-one. Find its inverse. Graph both functions.

2. Find all values of k such that f (x) = x 3 + 2x 2 + kx − 1 is one-to-one.

EXPLORATORY EXERCISES

36

CHAPTER 0

0.4

..

Preliminaries

0-36

TRIGONOMETRIC AND INVERSE TRIGONOMETRIC FUNCTIONS Many phenomena encountered in your daily life involve waves. For instance, music is transmitted from radio stations in the form of electromagnetic waves. Your radio receiver decodes these electromagnetic waves and causes a thin membrane inside the speakers to vibrate, which, in turn, creates pressure waves in the air. When these waves reach your ears, you hear the music from your radio (see Figure 0.53). Each of these waves is periodic, meaning that the basic shape of the wave is repeated over and over again. The mathematical description of such phenomena involves periodic functions, the most familiar of which are the trigonometric functions. First, we remind you of a basic definition.

FIGURE 0.53 Radio and sound waves

NOTES DEFINITION 4.1

When we discuss the period of a function, we most often focus on the fundamental period.

A function f is periodic of period T if f (x + T ) = f (x)

y

for all x such that x and x + T are in the domain of f. The smallest such number T > 0 is called the fundamental period.

(cos u, sin u)

u

1 u

sin u

cos u

FIGURE 0.54 Definition of sin θ and cos θ : cos θ = x and sin θ = y

x

There are several equivalent ways of defining the sine and cosine functions. We want to emphasize a simple definition from which you can easily reproduce many of the basic properties of these functions. Referring to Figure 0.54, begin by drawing the unit circle x 2 + y 2 = 1. Let θ be the angle measured (counterclockwise) from the positive x-axis to the line segment connecting the origin to the point (x, y) on the circle. Here, we measure θ in radians, where the radian measure of the angle θ is the length of the arc indicated in the figure. Again referring to Figure 0.54, we define sin θ to be the y-coordinate of the point on the circle and cos θ to be the x-coordinate of the point. From this definition, it follows that sin θ and cos θ are defined for all values of θ , so that each has domain −∞ < θ < ∞, while the range for each of these functions is the interval [−1, 1].

0-37

SECTION 0.4

..

Trigonometric and Inverse Trigonometric Functions

37

REMARK 4.1 Unless otherwise noted, we always measure angles in radians.

Note that since the circumference of a circle (C = 2πr ) of radius 1 is 2π , we have that 360◦ corresponds to 2π radians. Similarly, 180◦ corresponds to π radians, 90◦ corresponds to π/2 radians, and so on. In the accompanying table, we list some common angles as measured in degrees, together with the corresponding radian measures.

Angle in degrees

0◦

30◦

45◦

60◦

90◦

135◦

180◦

270◦

360◦

Angle in radians

0

π 6

π 4

π 3

π 2

3π 4

π

3π 2

2π

THEOREM 4.1 The functions f (θ ) = sin θ and g(θ ) = cos θ are periodic, of period 2π .

PROOF Referring to Figure 0.54, since a complete circle is 2π radians, adding 2π to any angle takes you all the way around the circle and back to the same point (x, y). This says that sin(θ + 2π) = sin θ cos(θ + 2π ) = cos θ,

and

for all values of θ. Furthermore, 2π is the smallest angle for which this is true. You are likely already familiar with the graphs of f (x) = sin x and g(x) = cos x shown in Figures 0.55a and 0.55b, respectively. y

y

1

1

x r

w

q

q

1

w

r

2p

x

p

p 1

FIGURE 0.55a

FIGURE 0.55b

y = sin x

y = cos x

2p

38

CHAPTER 0

x

sin x

cos x

0

0

1

π 6

π 3

1 2 √ 2 2 √ 3 2

π 2

1

π 4

Preliminaries

0-38

Notice that you could slide the graph of y = sin x slightly to the left or right and get an exact copy of the graph of y = cos x. Specifically, we have the relationship

√

3 2 √ 2 2

π sin x + = cos x. 2

1 2

The accompanying table lists some common values of sine and cosine. Notice that many of these can be read directly from Figure 0.54.

0

3π 4

√ 3 2 √ 2 2

5π 6

1 2

√ 2 2 √ − 23

π

0

−1

3π 2

−1

0

2π

0

1

2π 3

..

− 12

EXAMPLE 4.1

−

Solving Equations Involving Sines and Cosines

Find all solutions of the equations (a) 2 sin x − 1 = 0 and (b) cos2 x − 3 cos x + 2 = 0. Solution For (a), notice that 2 sin x − 1 = 0 if 2 sin x = 1 or sin x = 12 . From the unit circle, we find that sin x = 12 if x = π6 or x = 5π . Since sin x has period 2π, additional 6 π 5π π solutions are 6 + 2π, 6 + 2π, 6 + 4π and so on. A convenient way of indicating that any integer multiple of 2π can be added to either solution is to write x = π6 + 2nπ or x = 5π + 2nπ, for any integer n. Part (b) may look rather difficult at first. However, 6 notice that it looks like a quadratic equation using cos x instead of x. With this clue, you can factor the left-hand side to get

REMARK 4.2 Instead of writing (sin θ)2 or (cos θ )2 , we usually use the notation sin2 θ and cos2 θ , respectively.

0 = cos2 x − 3 cos x + 2 = (cos x − 1)(cos x − 2), from which it follows that either cos x = 1 or cos x = 2. Since −1 ≤ cos x ≤ 1 for all x, the equation cos x = 2 has no solution. However, we get cos x = 1 if x = 0, 2π or any integer multiple of 2π . We can summarize all the solutions by writing x = 2nπ , for any integer n. We now give definitions of the remaining four trigonometric functions.

DEFINITION 4.2 sin x . cos x cos x The cotangent function is defined by cot x = . sin x 1 The secant function is defined by sec x = . cos x

The tangent function is defined by tan x =

REMARK 4.3 Most calculators have keys for the functions sin x, cos x and tan x, but not for the other three trigonometric functions. This reflects the central role that sin x, cos x and tan x play in applications. To calculate function values for the other three trigonometric functions, you can simply use the identities cot x = and

1 , tan x

sec x =

csc x =

1 . sin x

1 cos x

The cosecant function is defined by csc x =

1 . sin x

We show graphs of these functions in Figures 0.56a, 0.56b, 0.56c and 0.56d. Notice in each graph the locations of the vertical asymptotes. For the “co” functions cot x and csc x, the division by sin x causes vertical asymptotes at 0, ±π, ±2π and so on (where sin x = 0). For tan x and sec x, the division by cos x produces vertical asymptotes at ±π/2, ±3π/2, ±5π/2 and so on (where cos x = 0). Once you have determined the vertical asymptotes, the graphs are relatively easy to draw. Notice that tan x and cot x are periodic, of period π , while sec x and csc x are periodic, of period 2π.

0-39

SECTION 0.4

..

Trigonometric and Inverse Trigonometric Functions

39

y y

x 2p w

q

p

q

p

w

x 2p w

2p

p

q

q

FIGURE 0.56a

FIGURE 0.56b

y = tan x

y = cot x

1

q

p

q 1

2p w

2p

y

y

p

w

p

q

x w

2p w

2p

w

1

x

1

p

q

2p

p

FIGURE 0.56c

FIGURE 0.56d

y = sec x

y = csc x

It is important to learn the effect of slight modifications of these functions. We present a few ideas here and in the exercises.

EXAMPLE 4.2

Altering Amplitude and Period

Graph y = 2 sin x and y = sin 2x, and describe how each differs from the graph of y = sin x (see Figure 0.57a). y

y

2

y

2

1

2

1 w

q

1 w

q x

w

q

x w

q

1

1

2

2

2p

x

p

p 1 2

FIGURE 0.57a

FIGURE 0.57b

FIGURE 0.57c

y = sin x

y = 2 sin x

y = sin (2x)

2p

40

CHAPTER 0

..

Preliminaries

0-40

Solution The graph of y = 2 sin x is given in Figure 0.57b. Notice that this graph is similar to the graph of y = sin x, except that the y-values oscillate between −2 and 2 instead of −1 and 1. Next, the graph of y = sin 2x is given in Figure 0.57c. In this case, the graph is similar to the graph of y = sin x except that the period is π instead of 2π (so that the oscillations occur twice as fast). The results in example 4.2 can be generalized. For A > 0, the graph of y = A sin x oscillates between y = −A and y = A. In this case, we call A the amplitude of the sine curve. Notice that for any positive constant c, the period of y = sin cx is 2π/c. Similarly, for the function A cos cx, the amplitude is A and the period is 2π/c. The sine and cosine functions can be used to model sound waves. A pure tone (think of a single flute note) is a pressure wave described by the sinusoidal function A sin ct. (Here, we are using the variable t, since the air pressure is a function of time.) The amplitude A determines how loud the tone is perceived to be and the period determines the pitch of the note. In this setting, it is convenient to talk about the frequency f = c/2π. The higher the frequency is, the higher the pitch of the note will be. (Frequency is measured in hertz, where 1 hertz equals 1 cycle per second.) Note that the frequency is simply the reciprocal of the period.

EXAMPLE 4.3

Finding Amplitude, Period and Frequency

Find the amplitude, period and frequency of (a) f (x) = 4 cos 3x and (b) g(x) = 2 sin(x/3). Solution (a) For f (x), the amplitude is 4, the period is 2π/3 and the frequency is 3/(2π) (see Figure 0.58a). (b) For g(x), the amplitude is 2, the period is 2π/(1/3) = 6π and the frequency is 1/(6π ) (see Figure 0.58b). y

y 2

4

x 2p

o

i

i

o

2p

3p 2p p

x p

2p 3p

2

4

FIGURE 0.58a

FIGURE 0.58b

y = 4 cos 3x

y = 2 sin (x/3)

There are numerous formulas or identities that are helpful in manipulating the trigonometric functions. You should observe that, from the definition of sin θ and cos θ (see Figure 0.54), the Pythagorean Theorem gives us the familiar identity sin2 θ + cos2 θ = 1, since the hypotenuse of the indicated triangle is 1. This is true for any angle θ . In addition, sin(−θ ) = − sin θ

and

cos(−θ) = cos θ

0-41

SECTION 0.4

..

Trigonometric and Inverse Trigonometric Functions

41

We list several important identities in Theorem 4.2.

THEOREM 4.2 For any real numbers α and β, the following identities hold: sin (α + β) = sin α cos β + sin β cos α cos (α + β) = cos α cos β − sin α sin β sin2 α = 12 (1 − cos 2α)

(4.1) (4.2) (4.3)

cos2 α = 12 (1 + cos 2α).

(4.4)

From the basic identities summarized in Theorem 4.2, numerous other useful identities can be derived. We derive two of these in example 4.4.

EXAMPLE 4.4

Deriving New Trigonometric Identities

Derive the identities sin 2θ = 2 sin θ cos θ and cos 2θ = cos2 θ − sin2 θ . Solution These can be obtained from formulas (4.1) and (4.2), respectively, by substituting α = θ and β = θ . Alternatively, the identity for cos 2θ can be obtained by subtracting equation (4.3) from equation (4.4). y

The Inverse Trigonometric Functions

1 x q

q 1

FIGURE 0.59

y = sin x on − π2 , π2

We now expand the set of functions available to you by defining inverses to the trigonometric functions. To get started, look at a graph of y = sin x (see Figure 0.57a). Notice that we cannot define an inverse function, since sin x is not one-to-one. Although the sine function does not have an inverse function, we can define one by modifying the domain of the sine. We do this by choosing a portion of the sine curve that passes the horizontal line test.

If we restrict the domain to the interval − π2 , π2 , then y = sin x is one-to-one there (see Figure 0.59) and, hence, has an inverse. We thus define the inverse sine function by y = sin−1 x

if and only if

sin y = x and − π2 ≤ y ≤

π . 2

(4.5)

Think of this definition as follows: if y = sin−1 x, then y is the angle between − π2 and π2 for which sin y = x. Note that we could have selected any interval on which sin x is one-toone, but − π2 , π2 is the most convenient. To verify that these are inverse functions, observe that sin (sin−1 x) = x,

REMARK 4.4 and Mathematicians often use the notation arcsin x in place of sin−1 x. People read sin−1 x interchangeably as “inverse sine of x” or “arcsine of x.”

sin−1 (sin x) = x,

for all x ∈ [−1, 1] π π for all x ∈ − , . 2 2

(4.6)

−1 Read equation (4.6) very carefully. It does not say

that sin (sin x) = x for all x, but rather, only for those in the restricted domain, − π2 , π2 . For instance, sin−1 (sin π ) = π , since

sin−1 (sin π ) = sin−1 (0) = 0.

42

..

CHAPTER 0

Preliminaries

0-42

EXAMPLE 4.5 Evaluate (a) sin−1

Evaluating the Inverse Sine Function

√ 3 2

and (b) sin−1 − 12 .

Solution For (a), we look for the angle θ in the interval − π2 , π2 for which √ √

sin θ = 23 . Note that since sin π3 = 23 and π3 ∈ − π2 , π2 , we have that √

sin−1 23 = π3 . For (b), note that sin − π6 = − 12 and − π6 ∈ − π2 , π2 . Thus, sin−1 − 12 = − π6 . Judging by example 4.5, you might think that (4.5) is a roundabout way of defining a function. If so, you’ve got the idea exactly. In fact, we want to emphasize that what we know about the inverse sine function is principally through reference to the sine function. Recall from our discussion in section 0.3 that we can draw of y = sin−1 x π aπ graph

simply by reflecting the graph of y = sin x on the interval − 2 , 2 (from Figure 0.59) through the line y = x (see Figure 0.60).

Turning to y = cos x, observe that restricting the domain to the interval − π2 , π2 , as we did for the inverse sine function, will not work here. (Why not?) The simplest way to make cos x one-to-one is to restrict its domain to the interval [0, π ] (see Figure 0.61). Consequently, we define the inverse cosine function by

y q

x

1

1

q

FIGURE 0.60 y = sin−1 x

y = cos−1 x

y

if and only if

cos y = x and 0 ≤ y ≤ π.

Note that here, we have

1

cos (cos−1 x) = x,

for all x ∈ [−1, 1]

cos−1 (cos x) = x,

for all x ∈ [0, π ].

x q

p

and

1

As with the definition of arcsine, it is helpful to think of cos−1 x as that angle θ in [0, π ] for which cos θ = x. As with sin−1 x, it is common to use cos−1 x and arccos x interchangeably.

FIGURE 0.61 y = cos x on [0, π]

EXAMPLE 4.6

√ Evaluate (a) cos−1 (0) and (b) cos−1 − 22 .

y p

q

1

Evaluating the Inverse Cosine Function

x 1

FIGURE 0.62 y = cos−1 x

Solution For (a), you will need to find that angle θ in [0, π ] for which cos θ = 0. It’s not hard to see that cos−1 (0) = π2 . (If you calculate this on your calculator and get 90, your calculator is in degrees mode. In this event, you should immediately √ change it to radians mode.) For (b), look for the angle θ ∈ [0, π ] for which cos θ = − 22 . Notice √ = − 22 and 3π that cos 3π ∈ [0, π ]. Consequently, 4 4 √ cos−1 − 22 = 3π . 4 Once again, we obtain the graph of this inverse function by reflecting the graph of y = cos x on the interval [0, π ] (seen in Figure 0.61) through the line y = x (see Figure 0.62).

0-43

SECTION 0.4

..

Trigonometric and Inverse Trigonometric Functions

43

We can define inverses for each of the four remaining trigonometric functions in similar ways. For y = tan x, we restrict the domain to the interval − π2 , π2 . Think about why the endpoints of this interval are not included (see Figure 0.63). Having done this, you should readily see that we define the inverse tangent function by y = tan−1 x

tan y = x and − π2 < y <

if and only if

π . 2

The graph of y = tan−1 x is then as seen in Figure 0.64, found by reflecting the graph in Figure 0.63 through the line y = x. y 6 4 y

2 q

x q

q 2 4

y 10

4

2

4

6

q

FIGURE 0.63 y = tan x on − π2 , π2

p

1 1

x

2

6

5

FIGURE 0.64 y = tan−1 x

x

q

EXAMPLE 4.7

5

Evaluating an Inverse Tangent

Evaluate tan−1 (1).

Solution You must look for the angle θ on the interval − π2 , π2 for which tan θ = 1. This is easy enough. Since tan π4 = 1 and π4 ∈ − π2 , π2 , we have that tan−1 (1) = π4 .

10

FIGURE 0.65 y = sec x on [0, π ]

We now turn to defining an inverse for sec x. First, we must issue a disclaimer. There are several reasonable ways in which to suitably restrict the domain and different authors restrict We have (somewhat arbitrarily) chosen to restrict the domain to be π itdifferently.

0, 2 ∪ π2 , π . Why not use all of [0, π ]? You need only think about the definition of sec x to see why we needed to exclude the value x = π2 . See Figure 0.65 for a graph of sec x on this domain. Note the vertical asymptote at x = π2 . Consequently, we define the inverse secant function by

y p

q

10

6

5

1 1

y = sec−1 x

x 5

FIGURE 0.66 y = sec−1 x

if and only if

sec y = x and y ∈ 0, π2 ∪ π2 , π .

A graph of sec−1 x is shown in Figure 0.66.

10

EXAMPLE 4.8

√

Evaluating an Inverse Secant

Evaluate sec−1 (− 2).

44

CHAPTER 0

..

Preliminaries

REMARK 4.5 We can likewise define inverses to cot x and csc x. As these functions are used only infrequently, we will omit them here and examine them in the exercises.

Function Domain sin−1 x [−1, 1] cos−1 x −1

tan

x

Range π π

−2, 2

[0, π] (−∞, ∞) − π2 , π2 [−1, 1]

0-44

Solution You must look for the angle θ with θ ∈ 0, π2 ∪ π2 , π , for which √ √ √ sec θ = − 2. Notice that if sec θ = − 2, then cos θ = − √12 = − 22 . Recall from √

example 4.6 that cos 3π = − 22 . Further, the angle 3π is in the interval π2 , π and so, 4 4 √ sec−1 (− 2) = 3π . 4 Calculators do not usually have built-in functions for sec x or sec−1 x. In this case, you must convert the desired secant value to a cosine value and use the inverse cosine function, as we did in example 4.8. We summarize the domains and ranges of the three main inverse trigonometric functions in the margin. In many applications, we need to calculate the length of one side of a right triangle using the length of another side and an acute angle (i.e., an angle between 0 and π2 radians). We can do this rather easily, as in example 4.9.

EXAMPLE 4.9

Finding the Height of a Tower

A person 100 feet from the base of a tower measures an angle of 60◦ from the ground to the top of the tower (see Figure 0.67). (a) Find the height of the tower. (b) What angle is measured if the person is 200 feet from the base? h

sin θ 100 ft

l θ cos θ

Solution For (a), we first convert 60◦ to radians: π π 60◦ = 60 = radians. 180 3 We are given that the base of the triangle in Figure 0.67 is 100 feet. We must now compute the height h of the tower. Using the similar triangles indicated in Figure 0.67, we have sin θ h = , cos θ 100

FIGURE 0.67 Height of a tower

so that the height of the tower is √ sin θ π = 100 tan θ = 100 tan = 100 3 ≈ 173 feet. cos θ 3 For part (b), the similar triangles in Figure 0.67 give us √ √ h 3 100 3 tan θ = = = . 200 200 2 π Since 0 < θ < , we have 2

√ 3 −1 θ = tan ≈ 0.7137 radians (about 41 degrees). 2 h = 100

In example 4.10, we simplify expressions involving both trigonometric and inverse trigonometric functions.

EXAMPLE 4.10

Simplifying Expressions Involving Inverse Trigonometric Functions

Simplify (a) sin (cos−1 x) and (b) tan (cos−1 x). Solution Do not look for some arcane formula to help you out. Think first: cos−1 x is the angle (call it θ ) for which x = cos θ . First, consider the case where x > 0. Looking

0-45

SECTION 0.4

..

Trigonometric and Inverse Trigonometric Functions

45

at Figure 0.68, we have drawn a right triangle, with hypotenuse 1 and adjacent angle θ . From the definition of the sine and cosine, then, we have that the base of the triangle is cos θ = x and the altitude is sin θ, which by the Pythagorean Theorem is sin (cos−1 x) = sin θ = 1 − x 2 . sin u 1 x2 1

u cos1x

Wait! We have not yet finished part (a). Figure 0.68 shows 0 < θ < π2 , but by definition, θ = cos−1 x could range from 0 to π . Does our answer change if π2 < θ < π? To see that it doesn’t change, note that if 0 ≤ θ ≤ π, then sin θ ≥ 0. From the Pythagorean identity sin2 θ + cos2 θ = 1, we get sin θ = ± 1 − cos2 θ = ± 1 − x 2 . Since sin θ ≥ 0, we must have

cos u x

FIGURE 0.68 θ = cos−1 x

sin θ =

1 − x 2,

for all values of x. For part (b), you can read from Figure 0.68 that

√ 1 − x2 sin θ tan (cos x) = tan θ = = . cos θ x Note that this last identity is valid, regardless of whether x = cos θ is positive or negative. −1

EXERCISES 0.4 WRITING EXERCISES 1. Many students are comfortable using degrees to measure angles and don’t understand why they must learn radian measures. As discussed in the text, radians directly measure distance along the unit circle. Distance is an important aspect of many applications. In addition, we will see later that many calculus formulas are simpler in radians form than in degrees. Aside from familiarity, discuss any and all advantages of degrees over radians. On balance, which is better? 2. A student graphs f (x) = cos x on a graphing calculator and gets what appears to be a straight line at height y = 1 instead of the usual cosine curve. Upon investigation, you discover that the calculator has graphing window −10 ≤ x ≤ 10, −10 ≤ y ≤ 10 and is in degrees mode. Explain what went wrong and how to correct it.

5. In example 4.3, f (x) = 4 cos 3x has period 2π/3 and g(x) = 2 sin (x/3) has period 6π. Explain why the sum h(x) = 4 cos 3x + 2 sin (x/3) has period 6π. 6. Give a different range for sec−1 x than that given in the text. For which x’s would the value of sec−1 x change? Using the calculator discussion in exercise 4, give one reason why we might have chosen the range that we did. In exercises 1 and 2, convert the given radians measure to degrees. 1. (a) 2. (a)

π 4 3π 5

π 3 (b) π7

(b)

(c)

π 6

(c) 2

(d)

4π 3

(d) 3

3. Inverse functions are necessary for solving equations. The restricted range we had to use to define inverses of the trigonometric functions also restricts their usefulness in equation solving. Explain how to use sin−1 x to find all solutions of the equation sin u = x.

In exercises 3 and 4, convert the given degrees measure to radians. 3. (a) 180◦ (b) 270◦ (c) 120◦ (d) 30◦

4. Discuss how to compute sec−1 x, csc−1 x and cot−1 x on a calculator that has built-in functions only for sin−1 x, cos−1 x and tan−1 x.

In exercises 5–14, find all solutions of the given equation.

4. (a) 40◦

(b) 80◦

5. 2 cos x − 1 = 0 √ 7. 2 cos x − 1 = 0

(c) 450◦

(d) 390◦

6. 2 sin x + 1 = 0 √ 3=0

8. 2 sin x −

46

CHAPTER 0

..

Preliminaries

0-46

10. sin2 x − 2 sin x − 3 = 0

51. f (x) = sin 2x − cos 5x

11. sin2 x + cos x − 1 = 0

12. sin 2x − cos x = 0

52. f (x) = cos 3x − sin 7x

13. cos x + cos x = 0

14. sin x − sin x = 0

9. sin2 x − 4 sin x + 3 = 0

2

2

In exercises 15–24, sketch a graph of the function.

In exercises 53–56, use the range for θ to determine the indicated function value.

15. f (x) = sin 2x

16. f (x) = cos 3x

53. sin θ = 13 , 0 ≤ θ ≤

π ; 2

find cos θ.

17. f (x) = tan 2x

18. f (x) = sec 3x 20. f (x) = 4 cos (x + π)

π ; 2

find sin θ.

19. f (x) = 3 cos (x − π/2)

54. cos θ = 45 , 0 ≤ θ ≤

21. f (x) = sin 2x − 2 cos 2x

22. f (x) = cos 3x − sin 3x

23. f (x) = sin x sin 12x

24. f (x) = sin x cos 12x

55. sin θ = 12 ,

π 2

≤ θ ≤ π;

find cos θ.

56. sin θ = 12 ,

π 2

≤ θ ≤ π;

find tan θ.

25. f (x) = 3 sin 2x

26. f (x) = 2 cos 3x

In exercises 57–64, use a triangle to simplify each expression. Where applicable, state the range of x’s for which the simplification holds.

27. f (x) = 5 cos 3x

28. f (x) = 3 sin 5x

57. cos (sin−1 x)

58. cos (tan−1 x)

29. f (x) = 3 cos (2x − π/2)

30. f (x) = 4 sin (3x + π)

31. f (x) = −4 sin x

32. f (x) = −2 cos 3x

59. tan (sec−1 x)

60. cot (cos−1 x)

61. sin cos−1 12

62. cos sin−1 12

63. tan cos−1 35

64. csc sin−1 23

In exercises 25–32, identify the amplitude, period and frequency.

In exercises 33–36, prove that the given trigonometric identity is true. 33. sin (α − β) = sin α cos β − sin β cos α 34. cos (α − β) = cos α cos β + sin α sin β 35. (a) cos (2θ ) = 2 cos2 θ − 1

(b) cos (2θ) = 1 − 2 sin2 θ

36. (a) sec2 θ = tan2 θ + 1

(b) csc2 θ = cot2 θ + 1

In exercises 37–46, evaluate the inverse function by sketching a unit circle and locating the correct angle on the circle. 37. cos−1 0

38. tan−1 0

39. sin−1 (−1)

40. cos−1 (1)

41. sec−1 1

42. tan−1 (−1)

43. sec−1 2

44. csc−1 2 √ 46. tan−1 3

45. cot−1 1 47. Prove that, for some constant β,

4 cos x − 3 sin x = 5 cos (x + β). Then, estimate the value of β. 48. Prove that, for some constant β, √ 2 sin x + cos x = 5 sin (x + β). Then, estimate the value of β. In exercises 49–52, determine whether the function is periodic. If it is periodic, find the smallest (fundamental) period. 49. f (x) = cos 2x + 3 sin π x √ 50. f (x) = sin x − cos 2x

In exercises 65–68, use a graphing calculator or computer to determine the number of solutions of each equation, and numerically estimate the solutions (x is in radians). 65. 2 cos x = 2 − x

66. 3 sin x = x

67. cos x = x − 2

68. sin x = x 2

2

69. A person sitting 2 miles from a rocket launch site measures 20◦ up to the current location of the rocket. How high up is the rocket? 70. A person who is 6 feet tall stands 4 feet from the base of a light pole and casts a 2-foot-long shadow. How tall is the light pole? 71. A surveyor stands 80 feet from the base of a building and measures an angle of 50◦ to the top of the steeple on top of the building. The surveyor figures that the center of the steeple lies 20 feet inside the front of the structure. Find the distance from the ground to the top of the steeple. 72. Suppose that the surveyor of exercise 71 estimates that the center of the steeple lies between 20 and 21 inside the front of the structure. Determine how much the extra foot would change the calculation of the height of the building. 73. A picture hanging in an art gallery has a frame 20 inches high, and the bottom of the frame is 6 feet above the floor. A person whose eyes are 6 feet above the floor stands x feet from the wall. Let A be the angle formed by the ray from the person’s eye to the bottom of the frame and the ray from the person’s

0-47

SECTION 0.4

eye to the top of the frame. Write A as a function of x and graph y = A(x).

20"

..

Trigonometric and Inverse Trigonometric Functions

47

The player can catch the ball by running to keep the angle ψ constant (this makes it appear that the ball is moving in a straight line). Assuming that all triangles shown are right tan α triangles, show that tan ψ = and then solve for ψ. tan β

A

Ball

6'

c

x

74. In golf, the goal is to hit a ball into a hole of diameter 4.5 inches. Suppose a golfer stands x feet from the hole trying to putt the ball into the hole. A first approximation of the margin of error in a putt is to measure the angle A formed by the ray from the ball to the right edge of the hole and the ray from the ball to the left edge of the hole. Find A as a function of x. 75. In an AC circuit, the voltage is given by v(t) = v p sin 2π ft, where v p is the peak voltage and f is the frequency in Hz. A voltmeter actually measures an√average (called the root-meansquare) voltage, equal to v p / 2. If the voltage has amplitude 170 and period π/30, find the frequency and meter voltage. 76. An old-style LP record player rotates records at 33 13 rpm (revolutions per minute). What is the period (in minutes) of the rotation? What is the period for a 45-rpm record? 77. Suppose that the ticket sales of an airline (in of thousands dollars) is given by s(t) = 110 + 2t + 15 sin 16 πt , where t is measured in months. What real-world phenomenon might cause the fluctuation in ticket sales modeled by the sine term? Based on your answer, what month corresponds to t = 0? Disregarding seasonal fluctuations, by what amount is the airline’s sales increasing annually? 78. Piano tuners sometimes start by striking a tuning fork and then the corresponding piano key. If the tuning fork and piano note each have frequency 8, then the resulting sound is sin 8t + sin 8t. Graph this. If the piano is slightly out-of-tune at frequency 8.1, the resulting sound is sin 8t + sin 8.1t. Graph this and explain how the piano tuner can hear the small difference in frequency.

Home plate

a b Outfielder

EXPLORATORY EXERCISES 1. In his book and video series The Ring of Truth, physicist Philip Morrison performed an experiment to estimate the circumference of the earth. In Nebraska, he measured the angle to a bright star in the sky, then drove 370 miles due south into Kansas and measured the new angle to the star. Some geometry shows that the difference in angles, about 5.02◦ , equals the angle from the center of the earth to the two locations in Nebraska and Kansas. If the earth is perfectly spherical (it’s not) and the circumference of the portion of the circle measured out by 5.02◦ is 370 miles, estimate the circumference of the earth. This experiment was based on a similar experiment by the ancient Greek scientist Eratosthenes. The ancient Greeks and the Spaniards of Columbus’ day knew that the earth was round, they just disagreed about the circumference. Columbus argued for a figure about half of the actual value, since a ship couldn’t survive on the water long enough to navigate the true distance. 2. An oil tank with circular cross sections lies on its side. A stick is inserted in a hole at the top and used to measure the depth d of oil in the tank. Based on this measurement, the goal is to compute the percentage of oil left in the tank.

79. Give precise definitions of csc−1 x and cot−1 x. 80. In baseball, outfielders are able to easily track down and catch fly balls that have very long and high trajectories. Physicists have argued for years about how this is done. A recent explanation involves the following geometry.

d

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To simplify calculations, suppose the circle is a unit circle with center at (0, 0). Sketch radii extending from the origin to the top of the oil. The area of oil at the bottom equals the area of the portion of the circle bounded by the radii minus the area of the triangle formed above the oil in the figure.

1

1 u d

Start with the triangle, which has area one-half base times height. Explain why the height is 1 − d. Find a right triangle in the figure (there are two of them) with hypotenuse 1 (the radius of the circle) and one vertical side of length 1 − d. The horizontal side has length equal to one-half the base of the larger triangle. Show that this equals 1 − (1 − d)2 . The area of the portion of the circle equals π θ/2π = θ/2, where

0.5

θ is the angle at the top of the triangle. Find this angle as a function of d. (Hint: Go back to the right triangle used above with upper angle θ/2.) Then find the area filled with oil and divide by π to get the portion of the tank filled with oil. 3. Computer graphics can be misleading. This exercise works best using a “disconnected” graph (individual dots, not connected). Graph y = sin x 2 using a graphing window for which each pixel represents a step of 0.1 in the x- or y-direction. You should get the impression of a sine wave that oscillates more and more rapidly as you move to the left and right. Next, change the graphing window so that the middle of the original screen (probably x = 0) is at the far left of the new screen. You will likely see what appears to be a random jumble of dots. Continue to change the graphing window by increasing the x-values. Describe the patterns or lack of patterns that you see. You should find one pattern that looks like two rows of dots across the top and bottom of the screen; another pattern looks like the original sine wave. For each pattern that you find, pick adjacent points with x-coordinates a and b. Then change the graphing window so that a ≤ x ≤ b and find the portion of the graph that is missing. Remember that, whether the points are connected or not, computer graphs always leave out part of the graph; it is part of your job to know whether or not the missing part is important.

EXPONENTIAL AND LOGARITHMIC FUNCTIONS Some bacteria reproduce very quickly, as you may have discovered if you have ever had an infected cut or strep throat. Under the right circumstances, the number of bacteria in certain cultures will double in as little as an hour. In this section, we discuss some functions that can be used to model such rapid growth. Suppose that initially there are 100 bacteria at a given site and the population doubles every hour. Call the population function P(t), where t represents time (in hours) and start the clock running at time t = 0. Since the initial population is 100, we have P(0) = 100. After 1 hour, the population will double to 200, so that P(1) = 200. After another hour, the population will have doubled again to 400, making P(2) = 400 and so on. To compute the bacterial population after 10 hours, you could calculate the population at 4 hours, 5 hours and so on, or you could use the following shortcut. To find P(1), double the initial population, so that P(1) = 2 · 100. To find P(2), double the population at time t = 1, so that P(2) = 2 · 2 · 100 = 22 · 100. Similarly, P(3) = 23 · 100. This pattern leads us to P(10) = 210 · 100 = 102,400. Observe that the population can be modeled by the function P(t) = 2t · 100. We call P(t) an exponential function because the variable t is in the exponent. There is a subtle question here: what is the domain of this function? We have so far used only integer

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To simplify calculations, suppose the circle is a unit circle with center at (0, 0). Sketch radii extending from the origin to the top of the oil. The area of oil at the bottom equals the area of the portion of the circle bounded by the radii minus the area of the triangle formed above the oil in the figure.

1

1 u d

Start with the triangle, which has area one-half base times height. Explain why the height is 1 − d. Find a right triangle in the figure (there are two of them) with hypotenuse 1 (the radius of the circle) and one vertical side of length 1 − d. The horizontal side has length equal to one-half the base of the larger triangle. Show that this equals 1 − (1 − d)2 . The area of the portion of the circle equals π θ/2π = θ/2, where

0.5

θ is the angle at the top of the triangle. Find this angle as a function of d. (Hint: Go back to the right triangle used above with upper angle θ/2.) Then find the area filled with oil and divide by π to get the portion of the tank filled with oil. 3. Computer graphics can be misleading. This exercise works best using a “disconnected” graph (individual dots, not connected). Graph y = sin x 2 using a graphing window for which each pixel represents a step of 0.1 in the x- or y-direction. You should get the impression of a sine wave that oscillates more and more rapidly as you move to the left and right. Next, change the graphing window so that the middle of the original screen (probably x = 0) is at the far left of the new screen. You will likely see what appears to be a random jumble of dots. Continue to change the graphing window by increasing the x-values. Describe the patterns or lack of patterns that you see. You should find one pattern that looks like two rows of dots across the top and bottom of the screen; another pattern looks like the original sine wave. For each pattern that you find, pick adjacent points with x-coordinates a and b. Then change the graphing window so that a ≤ x ≤ b and find the portion of the graph that is missing. Remember that, whether the points are connected or not, computer graphs always leave out part of the graph; it is part of your job to know whether or not the missing part is important.

EXPONENTIAL AND LOGARITHMIC FUNCTIONS Some bacteria reproduce very quickly, as you may have discovered if you have ever had an infected cut or strep throat. Under the right circumstances, the number of bacteria in certain cultures will double in as little as an hour. In this section, we discuss some functions that can be used to model such rapid growth. Suppose that initially there are 100 bacteria at a given site and the population doubles every hour. Call the population function P(t), where t represents time (in hours) and start the clock running at time t = 0. Since the initial population is 100, we have P(0) = 100. After 1 hour, the population will double to 200, so that P(1) = 200. After another hour, the population will have doubled again to 400, making P(2) = 400 and so on. To compute the bacterial population after 10 hours, you could calculate the population at 4 hours, 5 hours and so on, or you could use the following shortcut. To find P(1), double the initial population, so that P(1) = 2 · 100. To find P(2), double the population at time t = 1, so that P(2) = 2 · 2 · 100 = 22 · 100. Similarly, P(3) = 23 · 100. This pattern leads us to P(10) = 210 · 100 = 102,400. Observe that the population can be modeled by the function P(t) = 2t · 100. We call P(t) an exponential function because the variable t is in the exponent. There is a subtle question here: what is the domain of this function? We have so far used only integer

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values of t, but for what other values of t does P(t) make √sense? Certainly, rational powers make sense, as in P(1/2) = 21/2 · 100, where 21/2 = 2. This says that the number of bacteria in the culture after a half hour is approximately √ P(1/2) = 21/2 · 100 = 2 · 100 ≈ 141. It’s a simple matter to interpret fractional powers as roots. For instance, √ x 1/2 = x, √ x 1/3 = 3 x, √ x 1/4 = 4 x, √ √ 3 x 2/3 = x 2 = ( 3 x)2 , √ 10 x 3.1 = x 31/10 = x 31 and so on. But, what about irrational powers? They are harder to define, but they work exactly the way you would want them to. For instance, since π is between 3.14 and 3.15, 2π is between 23.14 and 23.15 . In this way, we define 2x for x irrational to fill in the gaps in the graph of y = 2x for x rational. That is, if x is irrational and a < x < b, for rational numbers a and b, then 2a < 2x < 2b . This is the logic behind the definition of irrational powers. If for some reason you wanted to find the bacterial population after π hours, you can use your calculator or computer to obtain the approximate population: P(π ) = 2π · 100 ≈ 882. For your convenience, we now summarize the usual rules of exponents.

RULES OF EXPONENTS r For any integers m and n,

x m/n =

√ n

√ x m = ( n x)m .

r For any real number p,

x−p = r For any real numbers p and q,

1 . xp

(x p )q = x p·q . r For any real numbers p and q,

x p · x q = x p+q .

Throughout your calculus course, you will need to be able to quickly convert back and forth between exponential form and fractional or root form.

EXAMPLE 5.1

Converting Expressions to Exponential Form

√ 5 3x 2 Convert each to exponential form: (a) 3 x 5 , (b) √ , (c) √ and (d) (2x · 23+x )2 . 3 x 2 x

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Solution For (a), simply leave the 3 alone and convert the power: √ 3 x 5 = 3x 5/2 . For (b), use a negative exponent to write x in the numerator: 5 = 5x −1/3 . √ 3 x For (c), first separate the constants from the variables and then simplify: 3x 2 3 3 3 x2 = x 2−1/2 = x 3/2 . √ = 1/2 2x 2 2 2 x For (d), first work inside the parentheses and then square: (2x · 23+x )2 = (2x+3+x )2 = (22x+3 )2 = 24x+6 . The function in part (d) of example 5.1 is called an exponential function with a base of 2.

DEFINITION 5.1 For any constant b > 0, the function f (x) = b x is called an exponential function. Here, b is called the base and x is the exponent. Be careful to distinguish between algebraic functions such as f (x) = x 3 and g(x) = x 2/3 and exponential functions. For exponential functions such as h(x) = 2x , the variable is in the exponent (hence the name), instead of in the base. Also, notice that the domain of an exponential function is the entire real line, (−∞, ∞), while the range is the open interval (0, ∞), since b x > 0 for all x. While any positive real number can be used as a base for an exponential function, three bases are the most commonly used in practice. Base 2 arises naturally when analyzing processes that double at regular intervals (such as the bacteria at the beginning of this section). Our standard counting system is base 10, so this base is commonly used. However, far and away the most useful base is the irrational number e. Like π , the number e has a surprising tendency to occur in important calculations. We define e by e = lim

n→∞

1+

1 n

n .

(5.1)

Note that equation (5.1) has at least two serious shortcomings. First, we have not yet said what the notation lim means. (In fact, we won’t define this until Chapter 1.) Second, it’s n→∞ unclear why anyone would ever define a number in such a strange way. We will not be in a position to answer the second question until Chapter 4 (but the answer is worth the wait). It suffices for the moment to say that equation (5.1) means that e can be approximated by calculating values of (1 + 1/n)n for large values of n and that the larger the value of n, the closer the approximation will be to the actual value of e. In particular, if you look at the sequence of numbers (1 + 1/2)2 , (1 + 1/3)3 , (1 + 1/4)4 and so on, they will get progressively closer and closer to (i.e., home in on) the irrational number e.

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To get an idea of the value of e, compute several of these numbers: 1 10 1+ = 2.5937 . . . , 10 1000 1 1+ = 2.7169 . . . , 1000 10,000 1 1+ = 2.7181 . . . 10,000 and so on. You should compute enough of these values to convince yourself that the first few digits of the decimal representation of e (e ≈ 2.718281828459 . . .) are correct.

EXAMPLE 5.2

Computing Values of Exponentials

Approximate e4 , e−1/5 and e0 . Solution From a calculator, we find that e4 = e · e · e · e ≈ 54.598. From the usual rules of exponents, e−1/5 =

1 e1/5

1 =√ ≈ 0.81873. 5 e

(On a calculator, it is convenient to replace −1/5 with −0.2.) Finally, e0 = 1. The graphs of the exponential functions summarize many of their important properties.

EXAMPLE 5.3

Sketching Graphs of Exponentials

Sketch the graphs of the exponential functions y = 2x , y = e x , y = e2x , y = e x/2 , y = (1/2)x and y = e−x . Solution Using a calculator or computer, you should get graphs similar to those that follow. y

4

y

30

30

20

20

10

10 x

2

2

4

4

x

2

2

FIGURE 0.69a

FIGURE 0.69b

y = 2x

y = ex

4

Notice that each of the graphs in Figures 0.69a, 0.69b, 0.70a and 0.70b starts very near the x-axis (reading left to right), passes through the point (0, 1) and then rises steeply. This is true for all exponentials with base greater than 1 and with a positive coefficient in the exponent. Note that the larger the base (e > 2) or the larger the coefficient in the

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0-52

y

4

y

30

30

20

20

10

10 x

2

2

4

4

x

2

2

FIGURE 0.70a

FIGURE 0.70b

y = e2x

y = e x/2

y

4

4

y

30

30

20

20

10

10 x

2

2

4

4

FIGURE 0.71a

x

2

2

4

FIGURE 0.71b y = e−x

y = (1/2)x

exponent (2 > 1 > 1/2), the more quickly the graph rises to the right (and drops to the left). Note that the graphs in Figures 0.71a and 0.71b are the mirror images in the y-axis of Figures 0.69a and 0.69b, respectively. The graphs rise as you move to the left and drop toward the x-axis as you move to the right. It’s worth noting that by the rules of exponents, (1/2)x = 2−x and (1/e)x = e−x . In Figures 0.69–0.71, each exponential function is one-to-one and, hence, has an inverse function. We define the logarithmic functions to be inverses of the exponential functions.

DEFINITION 5.2 For any positive number b = 1, the logarithm function with base b, written logb x, is defined by y = logb x

if and only if

x = by .

That is, the logarithm logb x gives the exponent to which you must raise the base b to get the given number x. For example, log10 10 = 1 (since 101 = 10), log10 100 = 2 (since 102 = 100), log10 1000 = 3 (since 103 = 1000) and so on. The value of log10 45 is less clear than the preceding three values, but the idea is the same: you need to find the number y such that 10 y = 45. The answer is between

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1 and 2, but to be more precise, you will need to employ trial and error. You should get log10 45 ≈ 1.6532. Observe from Definition 5.2 that for any base b > 0 (b = 1), if y = logb x, then x = b y > 0. That is, the domain of f (x) = logb x is the interval (0, ∞). Likewise, the range of f is the entire real line, (−∞, ∞). As with exponential functions, the most useful bases turn out to be 2, 10, and e. We usually abbreviate log10 x by log x. Similarly, loge x is usually abbreviated ln x (for natural logarithm).

EXAMPLE 5.4

Evaluating Logarithms

Without using your calculator, determine log(1/10), log(0.001), ln e and ln e3 . Solution Since 1/10 = 10−1 , log(1/10) = −1. Similarly, since 0.001 = 10−3 , we have that log(0.001) = −3. Since ln e = loge e1 , ln e = 1. Similarly, ln e3 = 3. We want to emphasize the inverse relationship defined by Definition 5.2. That is, b x and logb x are inverse functions for any b > 0 (b = 1). In particular, for the base e, we have eln x = x

for any x > 0

and

ln (e x ) = x

for any x.

(5.2)

We demonstrate this as follows. Let y = ln x = loge x. By Definition 5.2, we have that x = e y = eln x . We can use this relationship between natural logarithms and exponentials to solve equations involving logarithms and exponentials, as in examples 5.5 and 5.6.

EXAMPLE 5.5

Solving a Logarithmic Equation

Solve the equation ln(x + 5) = 3 for x. Solution Taking the exponential of both sides of the equation and writing things backward (for convenience), we have e3 = eln(x+5) = x + 5, from (5.2). Subtracting 5 from both sides gives us e3 − 5 = x.

EXAMPLE 5.6

Solving an Exponential Equation

Solve the equation e x+4 = 7 for x. Solution Taking the natural logarithm of both sides and writing things backward (for simplicity), we have from (5.2) that ln 7 = ln (e x+4 ) = x + 4. Subtracting 4 from both sides yields ln 7 − 4 = x.

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y

0-54

As always, graphs provide excellent visual summaries of the important properties of a function.

2 1 x 1

1

2

3

4

5

EXAMPLE 5.7

Sketching Graphs of Logarithms

Sketch graphs of y = log x and y = ln x, and briefly discuss the properties of each.

2

Solution From a calculator or computer, you should obtain graphs resembling those in Figures 0.72a and 0.72b. Notice that both graphs appear to have a vertical asymptote at x = 0 (why would that be?), cross the x-axis at x = 1 and very gradually increase as x increases. Neither graph has any points to the left of the y-axis, since log x and ln x are defined only for x > 0. The two graphs are very similar, although not identical.

3

FIGURE 0.72a y = log x

The properties just described graphically are summarized in Theorem 5.1.

y 2

THEOREM 5.1

1 x 1

1

2

3

4

5

For any positive base b = 1, (i) logb x is defined only for x > 0, (ii) logb 1 = 0 and (iii) if b > 1, then logb x < 0 for 0 < x < 1 and logb x > 0 for x > 1.

2 3

FIGURE 0.72b y = ln x

PROOF (i) Note that since b > 0, b y > 0 for any y. So, if logb x = y, then x = b y > 0. (ii) Since b0 = 1 for any number b = 0, logb 1 = 0 (i.e., the exponent to which you raise the base b to get the number 1 is 0). (iii) We leave this as an exercise. All logarithms share a set of defining properties, as stated in Theorem 5.2.

THEOREM 5.2 For any positive base b = 1 and positive numbers x and y, we have (i) logb (x y) = logb x + logb y, (ii) logb (x/y) = logb x − logb y and (iii) logb (x y ) = y logb x.

As with most algebraic rules, each one of these properties can dramatically simplify calculations when it applies.

EXAMPLE 5.8

Simplifying Logarithmic Expressions

Write each as a single logarithm: (a) log2 27x − log2 3x and (b) ln 8 − 3 ln (1/2). Solution First, note that there is more than one order in which to work each problem. For part (a), we have 27 = 33 and so, 27x = (33 )x = 33x . This gives us log2 27x − log2 3x = log2 33x − log2 3x = 3x log2 3 − x log2 3 = 2x log2 3 = log2 32x .

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55

For part (b), note that 8 = 23 and 1/2 = 2−1 . Then, ln 8 − 3 ln (1/2) = 3 ln 2 − 3(− ln 2) = 3 ln 2 + 3 ln 2 = 6 ln 2 = ln 26 = ln 64. In some circumstances, it is beneficial to use the rules of logarithms to expand a given expression, as in example 5.9.

EXAMPLE 5.9

Expanding a Logarithmic Expression

Use the rules of logarithms to expand the expression ln

x 3 y4 . z5

Solution From Theorem 5.2, we have that 3 4 x y ln = ln (x 3 y 4 ) − ln (z 5 ) = ln (x 3 ) + ln (y 4 ) − ln (z 5 ) z5 = 3 ln x + 4 ln y − 5 ln z. Using the rules of exponents and logarithms, we can rewrite any exponential as an exponential with base e, as follows. For any base a > 0, we have x

a x = eln (a ) = e x ln a .

(5.3)

This follows from Theorem 5.2 (iii) and the fact that eln y = y, for all y > 0.

EXAMPLE 5.10

Rewriting Exponentials as Exponentials with Base e

Rewrite the exponentials 2x , 5x and (2/5)x as exponentials with base e. Solution From (5.3), we have x

2x = eln (2 ) = e x ln 2 , x

and

5x = eln (5 ) = e x ln 5 x 2 x = eln [(2/5) ] = e x ln (2/5) . 5

Just as we can rewrite an exponential with any positive base in terms of an exponential with base e, we can rewrite any logarithm in terms of natural logarithms, as follows. For any positive base b (b = 1), we will show that logb x =

ln x . ln b

(5.4)

Let y = logb x. Then by Definition 5.2, we have that x = b y . Taking the natural logarithm of both sides of this equation, we get by Theorem 5.2 (iii) that ln x = ln(b y ) = y ln b. Dividing both sides by ln b (since b = 1, ln b = 0) gives us y= establishing (5.4).

ln x , ln b

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0-56

Equation (5.4) is useful for computing logarithms with bases other than e or 10. This is important since, more than likely, your calculator has keys only for ln x and log x. We illustrate this idea in example 5.11.

EXAMPLE 5.11

Approximating the Value of Logarithms

Approximate the value of log7 12. Solution From (5.4), we have log7 12 =

ln 12 ≈ 1.2769894. ln 7

Hyperbolic Functions There are two special combinations of exponential functions, called the hyperbolic sine and hyperbolic cosine functions, that have important applications. For instance, the Gateway Arch in Saint Louis was built in the shape of a hyperbolic cosine graph. (See the photograph in the margin.) The hyperbolic sine function [denoted by sinh (x)] and the hyperbolic cosine function [denoted by cosh (x)] are defined by sinh x =

Saint Louis Gateway Arch

e x − e−x 2

and

cosh x =

e x + e−x . 2

Graphs of these functions are shown in Figures 0.73a and 0.73b. The hyperbolic functions (including the hyperbolic tangent, tanh x, defined in the expected way) are often convenient to use when solving equations. For now, we verify several basic properties that the hyperbolic functions satisfy in parallel with their trigonometric counterparts. y

4

y

10

10

5

5 x

2

2

4

4

x

2

2

5

5

10

10

FIGURE 0.73a

FIGURE 0.73b

y = sinh x

y = cosh x

EXAMPLE 5.12

4

Computing Values of Hyperbolic Functions

Compute f (0), f (1) and f (−1), and determine how f (x) and f (−x) compare for each function: (a) f (x) = sinh x and (b) f (x) = cosh x.

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57

e0 − e−0 1−1 = = 0. Note that this 2 2 1 −1 e −e means that sinh 0 = sin 0 = 0. Also, we have sinh 1 = ≈ 1.18, while 2 e−1 − e1 sinh (−1) = ≈ −1.18. Notice that sinh (−1) = − sinh 1. In fact, for any x, 2 e−x − e x −(e x − e−x ) sinh (−x) = = = − sinh x. 2 2

Solution For part (a), we have sinh 0 =

[The same rule holds for the sine function: sin (−x) = −sin x.] For part (b), we have e0 + e−0 1+1 cosh 0 = = = 1. Note that this means that cosh 0 = cos 0 = 1. Also, 2 2 1 −1 e +e e−1 + e1 we have cosh 1 = ≈ 1.54, while cosh (−1) = ≈ 1.54. Notice that 2 2 cosh (−1) = cosh 1. In fact, for any x, cosh (−x) =

e−x + e x e x + e−x = = cosh x. 2 2

[The same rule holds for the cosine function: cos (−x) = cos x.]

Fitting a Curve to Data You are familiar with the idea that two points determine a straight line. As we see in example 5.13, two points will also determine an exponential function.

EXAMPLE 5.13

Matching Data to an Exponential Curve

Find the exponential function of the form f (x) = aebx that passes through the points (0, 5) and (3, 9). Solution We must solve for a and b, using the properties of logarithms and exponentials. First, for the graph to pass through the point (0, 5), this means that 5 = f (0) = aeb · 0 = a, so that a = 5. Next, for the graph to pass through the point (3, 9), we must have 9 = f (3) = ae3b = 5e3b . To solve for b, we divide both sides of the equation by 5 and take the natural logarithm of both sides, which yields 9 ln = ln e3b = 3b, 5 Year

U.S. Population

1790

3,929,214

1800

5,308,483

1810

7,239,881

1820

9,638,453

1830

12,866,020

1840

17,069,453

1850

23,191,876

1860

31,443,321

from (5.2). Finally, dividing by 3 gives us the value for b: 1 9 b = ln . 3 5 1

Thus, f (x) = 5e 3 ln (9/5) x. Consider the population of the United States from 1790 to 1860, found in the accompanying table. A plot of these data points can be seen in Figure 0.74 (where the vertical scale represents the population in millions). This shows that the population was increasing, with larger and larger increases each decade. If you sketch an imaginary curve through these points, you will probably get the impression of a parabola or perhaps the right half of a

U.S. population (in millions)

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Preliminaries

cubic or exponential. And that’s the question: are these data best modeled by a quadratic function, a cubic function, an exponential function or what? We can use the properties of logarithms from Theorem 5.2 to help determine whether a given set of data is modeled better by a polynomial or an exponential function, as follows. Suppose that the data actually come from an exponential, say, y = aebx (i.e., the data points lie on the graph of this exponential). Then,

35 30 25 20 15 10

ln y = ln (aebx ) = ln a + ln ebx = ln a + bx.

5 1 2 3 4 5 6 7 8 Number of decades since 1780

FIGURE 0.74

If you draw a new graph, where the horizontal axis shows values of x and the vertical axis corresponds to values of ln y, then the graph will be the line ln y = bx + c (where the constant c = ln a). On the other hand, suppose the data actually came from a polynomial. If y = bx n (for any n), then observe that

U.S. Population 1790–1860

Natural logarithm of U.S. population (millions)

0-58

ln y = ln (bx n ) = ln b + ln x n = ln b + n ln x.

3

In this case, a graph with horizontal and vertical axes corresponding to x and ln y, respectively, will look like the graph of a logarithm, ln y = n ln x + c. Such semi-log graphs (i.e., graphs of ln y versus x) let us distinguish the graph of an exponential from that of a polynomial: graphs of exponentials become straight lines, while graphs of polynomials (of degree ≥ 1) become logarithmic curves. Scientists and engineers frequently use semi-log graphs to help them gain an understanding of physical phenomena represented by some collection of data.

2

EXAMPLE 5.14

1

Determine whether the population of the United States from 1790 to 1860 was increasing exponentially or as a polynomial.

4

1 2 3 4 5 6 7 8 Number of decades since 1780

FIGURE 0.75 Semi-log plot of U.S. population

Using a Semi-Log Graph to Identify a Type of Function

Solution As already indicated, the trick is to draw a semi-log graph. That is, instead of plotting (1, 3.9) as the first data point, plot (1, ln 3.9) and so on. A semi-log plot of this data set is seen in Figure 0.75. Although the points are not exactly colinear (how would you prove this?), the plot is very close to a straight line with ln y-intercept of 1 and slope 0.3. You should conclude that the population is well modeled by an exponential function. The exponential model would be y = P(t) = aebt , where t represents the number of decades since 1780. Here, b is the slope and ln a is the ln y-intercept of the line in the semi-log graph. That is, b ≈ 0.3 and ln a ≈ 1 (why?), so that a ≈ e. The population is then modeled by P(t) = e · e0.3t million.

EXERCISES 0.5 WRITING EXERCISES 1. Starting from a single cell, a human being is formed by 50 generations of cell division. Explain why after n divisions there are 2n cells. Guess how many cells will be present after 50 divisions, then compute 250 . Briefly discuss how rapidly exponential functions increase.

2. Explain why the graphs of f (x) = 2−x and g(x) = the same.

1 x 2

are

3. Compare f (x) = x 2 and g(x) = 2x for x = 12 , x = 1, x = 2, x = 3 and x = 4. In general, which function is bigger for large values of x? For small values of x?

0-59

SECTION 0.5

4. Compare f (x) = 2x and g(x) = 3x for x = −2, x = − 12 , x = 12 and x = 2. In general, which function is bigger for negative values of x? For positive values of x? In exercises 1–6, convert each exponential expression into fractional or root form. −3

−2

2. 4

1. 2

2/5

2/3

4. 6

5. 5

3. 3

−2/3

6. 4

In exercises 17–20, use a calculator or computer to estimate each value. 18. 4e−2/3

12 e

14 20. √ e

In exercises 21–30, sketch a graph of the given function. 21. f (x) = e2x

22. f (x) = e3x

23. f (x) = 2e x/4

24. f (x) = e−x

25. f (x) = 3e−2x

26. f (x) = 10e−x/3

27. f (x) = ln 2x

28. f (x) = ln x 2

29. f (x) = e2 ln x

30. f (x) = e−x/4 sin x

2

In exercises 31–40, solve the given equation for x. 31. e2x = 2

32. e4x = 3

33. e x (x 2 − 1) = 0

34. xe−2x + 2e−2x = 0

35. ln 2x = 4

36. 2 ln 3x = 1

37. 4 ln x = −8

38. x 2 ln x − 9 ln x = 0

39. e2 ln x = 4

40. ln (e2x ) = 6

In exercises 41 and 42, use the definition of logarithm to determine the value. 41. (a) log3 9 42. (a) log4

1 16

1 27

(b) log4 64

(c) log3

(b) log4 2

(c) log9 3

59

In exercises 43 and 44, use equation (5.4) to approximate the value. 43. (a) log3 7 44. (a) log4

1 10

1 24

(b) log4 60

(c) log3

(b) log4 3

(c) log9 8

In exercises 45–50, rewrite the expression as a single logarithm. 46. 2 ln 4 − ln 3

ln 4 − ln 2

48. 3 ln 2 − ln 12

49. ln 34 + 4 ln 2

50. ln 9 − 2 ln 3

47.

In exercises 13–16, find the integer value of the given expression without using a calculator. √ 8 2 3/2 2/3 13. 4 14. 8 15. 1/2 16. 2 (1/3)2

19.

Exponential and Logarithmic Functions

45. ln 3 − ln 4

1/2

In exercises 7–12, convert each expression into exponential form. √ 1 2 3 7. 2 8. x 2 9. 3 x x 4 1 3 10. 2 11. √ 12. √ x 2 x 2 x3

17. 2e−1/2

..

1 2

In exercises 51–54, find a function of the form f (x) aebx with the given function values. 51. f (0) = 2, f (2) = 6

52. f (0) = 3, f (3) = 4

53. f (0) = 4, f (2) = 2

54. f (0) = 5, f (1) = 2

55. A fast-food restaurant gives every customer a game ticket. With each ticket, the customer has a 1-in-10 chance of winning a free meal. If you go 10 times, estimate your chances of at winning 10 least one free meal. The exact probability is 1 − 109 . Compute this number and compare it to your guess. 56. In exercise 55, if you had 20 tickets with a 1-in-20 chance of winning, would you expect your probability of winning at least once to increase or decrease? Compute the probability 20 1 − 19 to find out. 20 57. In general, if you have n chances of winning with a 1-in-n chance on each n try, the probability of winning at least once is 1 − 1 − n1 . As n gets larger, what number does this probability approach? (Hint: There is a very good reason that this question is in this section!) 58. If y = a · x m , show that ln y = ln a + m ln x. If v = ln y, u = ln x and b = ln a, show that v = mu + b. Explain why the graph of v as a function of u would be a straight line. This graph is called the log-log plot of y and x. 59. For the given data, compute v = ln y and u = ln x, and plot points (u, v). Find constants m and b such that v = mu + b and use the results of exercise 58 to find a constant a such that y = a · xm. x y

2.2 14.52

2.4 17.28

2.6 20.28

2.8 23.52

3.0 27.0

3.2 30.72

3.6 13.66

3.8 14.81

60. Repeat exercise 59 for the given data. x y

2.8 9.37

3.0 10.39

3.2 11.45

3.4 12.54

61. Construct a log-log plot (see exercise 58) of the U.S. population data in example 5.14. Compared to the semi-log plot of the data in Figure 0.75, does the log-log plot look linear? Based on this, are the population data modeled better by an exponential function or a polynomial (power) function?

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62. Construct a semi-log plot of the data in exercise 59. Compared to the log-log plot already constructed, does this plot look linear? Based on this, are these data better modeled by an exponential or power function? 63. The concentration [H+ ] of free hydrogen ions in a chemical solution determines the solution’s pH, as defined by pH = − log [H+ ]. Find [H+ ] if the pH equals (a) 7, (b) 8 and (c) 9. For each increase in pH of 1, by what factor does [H+ ] change? 64. Gastric juice is considered an acid, with a pH of about 2.5. Blood is considered alkaline, with a pH of about 7.5. Compare the concentrations of hydrogen ions in the two substances (see exercise 63). 65. The Richter magnitude M of an earthquake is defined in terms of the energy E in joules released by the earthquake, with log10 E = 4.4 + 1.5M. Find the energy for earthquakes with magnitudes (a) 4, (b) 5 and (c) 6. For each increase in M of 1, by what factor does E change? 66. It puzzles some people who have not grown up around earthquakes that a magnitude 6 quake is considered much more severe than a magnitude 3 quake. Compare the amount of energy released in the two quakes. (See exercise 65.) 67. The decibel level of a noise is defined in terms of the intensity I of the noise, with dB = 10 log (I /I0 ). Here, I0 = 10−12 W/m2 is the intensity of a barely audible sound. Compute the intensity levels of sounds with (a) dB = 80, (b) dB = 90 and (c) dB = 100. For each increase of 10 decibels, by what factor does I change? 68. At a basketball game, a courtside decibel meter shows crowd noises ranging from 60 dB to 110 dB. Compare the intensity level of the 110-dB crowd noise with that of the 60-dB noise. (See exercise 67.) 69. Use a graphing calculator to graph y = xe−x , y = xe−2x , y = xe−3x and so on. Estimate the location of the maximum for each. In general, state a rule for the location of the maximum of y = xe−kx . 70. In golf, the task is to hit a golf ball into a small hole. If the ground near the hole is not flat, the golfer must judge how much the ball’s path will curve. Suppose the golfer is at the point (−13, 0), the hole is at the point (0, 0) and the path of the ball is, for −13 ≤ x ≤ 0, y = −1.672x + 72 ln (1 + 0.02x). Show that the ball goes in the hole and estimate the point on the y-axis at which the golfer aimed. Exercises 71–76 refer to the hyperbolic functions. 71. Show that the range of the hyperbolic cosine is cosh x ≥ 1 and the range of the hyperbolic sine is the entire real line. 72. Show that cosh2 x − sinh2 x = 1 for all x. 73. The Saint Louis Gateway Arch is both 630 feet wide and 630 feet tall. (Most people think that it looks taller than

0-60

it is wide.) One model xfor the outline of the arch is y = 757.7 − 127.7 cosh 127.7 for y ≥ 0. Use a graphing calculator to approximate the x- and y-intercepts and determine if the model has the correct horizontal and vertical measurements. 74. To model the outline of the Gateway Arch with a parabola, you can start with y = −c(x + 315)(x − 315) for some constant c. Explain why this gives the correct x-intercepts. Determine the constant c that gives a y-intercept of 630. Graph this parabola and the hyperbolic cosine in exercise 73 on the same axes. Are the graphs nearly identical or very different? 75. Find all solutions of sinh (x 2 − 1) = 0. 76. Find all solutions of cosh (3x + 2) = 0. 77. On a standard piano, the A below middle C produces a sound wave with frequency 220 Hz (cycles per second). The frequency of the A one octave higher is 440 Hz. In general, doubling the frequency produces the same note an octave higher. Find an exponential formula for the frequency f as a function of the number of octaves x above the A below middle C. 78. There are 12 notes in an octave on a standard piano. Middle C is 3 notes above A (see exercise 77). If the notes are tuned equally, this means that middle C is a quarter-octave above A. Use x = 14 in your formula from exercise 77 to estimate the frequency of middle C.

EXPLORATORY EXERCISES 1. Graph y = x 2 and y = 2x and approximate the two positive solutions of the equation x 2 = 2x . Graph y = x 3 and y = 3x , and approximate the two positive solutions of the equation x 3 = 3x . Explain why x = a will always be a solution of x a = a x , a > 0. What is different about the role of x = 2 as a solution of x 2 = 2x compared to the role of x = 3 as a solution of x 3 = 3x ? To determine the a-value at which the change occurs, graphically solve x a = a x for a = 2.1, 2.2, . . . , 2.9, and note that a = 2.7 and a = 2.8 behave differently. Continue to narrow down the interval of change by testing a = 2.71, 2.72, . . . , 2.79. Then guess the exact value of a. 2. Graph y = ln x and describe the behavior near x = 0. Then graph y = x ln x and describe the behavior near x = 0. Repeat this for y = x 2 ln x, y = x 1/2 ln x and y = x a ln x for a variety of positive constants a. Because the function “blows up” at x = 0, we say that y = ln x has a singularity at x = 0. The order of the singularity at x = 0 of a function f (x) is the smallest value of a such that y = x a f (x) doesn’t have a singularity at x = 0. Determine the order of the singularity at x = 0 for (a) f (x) = x1 , (b) f (x) = 12 and (c) f (x) = 13 . The x x higher the order of the singularity, the “worse” the singularity is. Based on your work, how bad is the singularity of y = ln x at x = 0?

0-61

SECTION 0.6

0.6

..

Transformations of Functions

61

TRANSFORMATIONS OF FUNCTIONS You are now familiar with a long list of functions: polynomials, rational functions, trigonometric functions, exponentials and logarithms. One important goal of this course is to more fully understand the properties of these functions. To a large extent, you will build your understanding by examining a few key properties of functions. We expand on our list of functions by combining them. We begin in a straight-forward fashion with Definition 6.1.

DEFINITION 6.1 Suppose that f (x) and g(x) are functions with domains D1 and D2 , respectively. The functions f + g, f − g and f · g are defined by ( f + g)(x) = f (x) + g(x), ( f − g)(x) = f (x) − g(x) ( f · g)(x) = f (x) · g(x),

and

f for all x in D1 ∩ D2 (i.e., x ∈ D1 , and x ∈ D2 ). The function is defined by g f f (x) (x) = , g g(x) for all x in D1 ∩ D2 such that g(x) = 0.

In example 6.1, we examine various combinations of several simple functions.

EXAMPLE 6.1

Combinations of Functions

If f (x) = x − 3 and g(x) =

√

x − 1, determine the functions f + g, 3 f − g and

f , g

stating the domains of each. Solution First, note that the domain of f is the entire real line and the domain of g is the set of all x ≥ 1. Now, √ ( f + g)(x) = x − 3 + x − 1 √ √ and (3 f − g)(x) = 3(x − 3) − x − 1 = 3x − 9 − x − 1. Notice that the domain of both ( f + g) and (3 f − g) is {x|x ≥ 1}. For f f (x) x −3 , (x) = =√ g g(x) x −1 the domain is {x|x > 1}, where we have added the restriction x = 1 to avoid dividing by 0.

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0-62

Definition 6.1 and example 6.1 show us how to do arithmetic with functions. An operation on functions that does not directly correspond to arithmetic is the composition of two functions.

DEFINITION 6.2

f

f (g(x))

The composition of functions f and g, written f ◦ g, is defined by ( f ◦ g)(x) = f (g(x)), for all x such that x is in the domain of g and g(x) is in the domain of f .

g(x)

x

g

The composition of two functions is a two-step process, as indicated in the margin schematic. Be careful to notice what this definition is saying. In particular, for f (g(x)) to be defined, you first need g(x) to be defined, so x must be in the domain of g. Next, f must be defined at the point g(x), so that the number g(x) will need to be in the domain of f.

( f ◦ g)(x) = f (g(x))

EXAMPLE 6.2

Finding the Composition of Two Functions

For f (x) = x 2 + 1 and g(x) = identify the domain of each.

√

x − 2, find the compositions f ◦ g and g ◦ f and

Solution First, we have

√ ( f ◦ g)(x) = f (g(x)) = f ( x − 2) √ = ( x − 2)2 + 1 = x − 2 + 1 = x − 1.

It’s tempting to write that the domain of f ◦ g is the entire real line, but look more carefully. Note that for x to be in the domain of g, we must have x ≥ 2. The domain of f is the whole real line, so this places no further restrictions on the domain of f ◦ g. Even though the final expression x − 1 is defined for all x, the domain of ( f ◦ g) is {x|x ≥ 2}. For the second composition, (g ◦ f )(x) = g( f (x)) = g(x 2 + 1) = (x 2 + 1) − 2 = x 2 − 1. The resulting square root requires x 2 − 1 ≥ 0 or |x| ≥ 1. Since the “inside” function f is defined for all x, the domain of g ◦ f is {x ∈ R|x| ≥ 1}, which we write in interval notation as (−∞, −1] ∪ [1, ∞). As you progress through the calculus, you will often find yourself needing to recognize that a given function is a composition of simpler functions. For now, it is an important skill to practice.

EXAMPLE 6.3

Identifying Compositions of Functions

Identify functions f and g√such that the given function can be written as ( f ◦ g)(x) for √ each of (a) x 2 + 1, (b) ( x + 1)2 , (c) sin x 2 and (d) cos2 x. Note that more than one answer is possible for each function. 2 Solution (a) Notice that x√ + 1 is inside the square root. So, one choice is to have 2 g(x) = x + 1 and f (x) = x.

0-63

SECTION 0.6

Transformations of Functions

63

√ √ (b) Here, x + 1 is inside the square. So, one choice is g(x) = x + 1 and f (x) = x 2 . (c) The function can be rewritten as sin (x 2 ), with x 2 clearly inside the sine function. Then, g(x) = x 2 and f (x) = sin x is one choice. (d) The function as written is shorthand for (cos x)2 . So, one choice is g(x) = cos x and f (x) = x 2 .

y

In general, it is quite difficult to take the graphs of f (x) and g(x) and produce the graph of f (g(x)). If one of the functions f and g is linear, however, there is a simple graphical procedure for graphing the composition. Such linear transformations are explored in the remainder of this section. The first case is to take the graph of f (x) and produce the graph of f (x) + c for some constant c. You should be able to deduce the general result from example 6.4.

10 8 6 4 2 4

..

EXAMPLE 6.4 x

2

2

4

Graph y = x 2 and y = x 2 + 3; compare and contrast the graphs. Solution You can probably sketch these by hand. You should get graphs like those in Figures 0.76a and 0.76b. Both figures show parabolas opening upward. The main obvious difference is that x 2 has a y-intercept of 0 and x 2 + 3 has a y-intercept of 3. In fact, for any given value of x, the point on the graph of y = x 2 + 3 will be plotted exactly 3 units higher than the corresponding point on the graph of y = x 2 . This is shown in Figure 0.77a. In Figure 0.77b, the two graphs are shown on the same set of axes. To many people, it does not look like the top graph is the same as the bottom graph moved up 3 units.

FIGURE 0.76a y=x

Vertical Translation of a Graph

2

y 10 8

y

y

6 25 4 2 4

2

x 2

25

Move graph up 3 units

20

20

15

15

10

10

5

5

4

FIGURE 0.76b y = x2 + 3

4

2

x 2

4

4

2

x 2

FIGURE 0.77a

FIGURE 0.77b

Translate graph up

y = x 2 and y = x 2 + 3

4

This is an unfortunate optical illusion. Humans usually mentally judge distance between curves as the shortest distance between the curves. For these parabolas, the shortest distance is vertical at x = 0 but becomes increasingly horizontal as you move away from the y-axis. The distance of 3 between the parabolas is measured vertically. In general, the graph of y = f (x) + c is the same as the graph of f (x) shifted up (if c > 0) or down (if c < 0) by |c| units. We usually refer to f (x) + c as a vertical translation (up or down, by |c| units).

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0-64

In example 6.5, we explore what happens if a constant is added to x.

EXAMPLE 6.5

A Horizontal Translation

Compare and contrast the graphs of y = x 2 and y = (x − 1)2 . Solution The graphs are shown in Figures 0.78a and 0.78b, respectively. y

y

Move graph to the right one unit

4

6

2

x 2

FIGURE 0.79 Translation to the right

4

10

8

8

6

6

4

4

x

2

2

4

4

2

x 2

FIGURE 0.78a

FIGURE 0.78b

y = x2

y = (x − 1)2

4

4

10

2

10 8

y

4

Notice that the graph of y = (x − 1)2 appears to be the same as the graph of y = x 2 , except that it is shifted 1 unit to the right. This should make sense for the following reason. Pick a value of x, say, x = 13. The value of (x − 1)2 at x = 13 is 122 , the same as the value of x 2 at x = 12, 1 unit to the left. Observe that this same pattern holds for any x you choose. A simultaneous plot of the two functions (see Figure 0.79) shows this. In general, for c > 0, the graph of y = f (x − c) is the same as the graph of y = f (x) shifted c units to the right. Likewise (again, for c > 0), you get the graph of f (x + c) by moving the graph of y = f (x) to the left c units. We usually refer to f (x − c) and f (x + c) as horizontal translations (to the right and left, respectively, by c units). To avoid confusion on which way to translate the graph of y = f (x), focus on what makes the argument (the quantity inside the parentheses) zero. For f (x), this is x = 0, but for f (x − c) you must have x = c to get f (0) [i.e., the same y-value as f (x) when x = 0]. This says that the point on the graph of y = f (x) at x = 0 corresponds to the point on the graph of y = f (x − c) at x = c.

EXAMPLE 6.6

Comparing Vertical and Horizontal Translations

Given the graph of y = f (x) shown in Figure 0.80a, sketch the graphs of y = f (x) − 2 and y = f (x − 2). Solution To graph y = f (x) − 2, simply translate the original graph down 2 units, as shown in Figure 0.80b. To graph y = f (x − 2), simply translate the original graph to

0-65

SECTION 0.6

..

Transformations of Functions

65

the right 2 units (so that the x-intercept at x = 0 in the original graph corresponds to an x-intercept at x = 2 in the translated graph), as seen in Figure 0.80c. y

3

y

y

15

15

15

10

10

10

5

5

5

x

1 5

2

x

3 2 1 5

3

10

10

15

15

1

2

3

1

x 1

2

3

4

5

15

FIGURE 0.80a

FIGURE 0.80b

FIGURE 0.80c

y = f (x)

y = f (x) − 2

y = f (x − 2)

Example 6.7 explores the effect of multiplying or dividing x or y by a constant.

EXAMPLE 6.7

Comparing Some Related Graphs

Compare and contrast the graphs of y = x 2 − 1, y = 4(x 2 − 1) and y = (4x)2 − 1. Solution The first two graphs are shown in Figures 0.81a and 0.81b, respectively. y

y

y

10

40

10

8

32

8

6

24

4

16

2

8

y 4(x 2 1)

6 4

3 2 1 2

x 1

2

3

3 2 1 8

2 x 1

2

3 2 1

y x2 1 x 1

2

3

3 4

FIGURE 0.81a

FIGURE 0.81b

FIGURE 0.81c

y = x2 − 1

y = 4(x 2 − 1)

y = x 2 − 1 and y = 4(x 2 − 1)

These graphs look identical until you compare the scales on the y-axes. The scale in Figure 0.81b is four times as large, reflecting the multiplication of the original function by 4. The effect looks different when the functions are plotted on the same scale, as in Figure 0.81c. Here, the parabola y = 4(x 2 − 1) looks thinner and has a different y-intercept. Note that the x-intercepts remain the same. (Why would that be?) The graphs of y = x 2 − 1 and y = (4x)2 − 1 are shown in Figures 0.82a and 0.82b, respectively.

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0-66

y

y

y

10

10

10

8

8

8

6

6

6

4

4

4

2

2 x

3 2 1 2

1

2

3

0.75

0.25 2

y (4x)2 1

y x2 1 x 0.25

0.75

3 2 1 2

x 1

2

3

FIGURE 0.82a

FIGURE 0.82b

FIGURE 0.82c

y = x2 − 1

y = (4x)2 − 1

y = x 2 − 1 and y = (4x)2 − 1

Can you spot the difference here? In this case, the x-scale has now changed, by the same factor of 4 as in the function. To see this, note that substituting x = 1/4 into (4x)2 − 1 produces (1)2 − 1, exactly the same as substituting x = 1 into the original function. When plotted on the same set of axes (as in Figure 0.82c), the parabola y = (4x)2 − 1 looks thinner. Here, the x-intercepts are different, but the y-intercepts are the same.

y 20

10

4

x

2

2

4

FIGURE 0.83a y = x2 y

EXAMPLE 6.8

40

2

A Translation and a Stretching

Describe how to get the graph of y = 2x 2 − 3 from the graph of y = x 2 . Solution You can get from x 2 to 2x 2 − 3 by multiplying by 2 and then subtracting 3. In terms of the graph, this has the effect of multiplying the y-scale by 2 and then shifting the graph down by 3 units (see the graphs in Figures 0.83a and 0.83b).

20

4

We can generalize the observations made in example 6.7. Before reading our explanation, try to state a general rule for yourself. How are the graphs of the functions c f (x) and f (cx) related to the graph of y = f (x)? Based on example 6.7, notice that to obtain a graph of y = c f (x) for some constant c > 0, you can take the graph of y = f (x) and multiply the scale on the y-axis by c. To obtain a graph of y = f (cx) for some constant c > 0, you can take the graph of y = f (x) and multiply the scale on the x-axis by 1/c. These basic rules can be combined to understand more complicated graphs.

x 2

FIGURE 0.83b y = 2x 2 − 3

4

EXAMPLE 6.9

A Translation in Both x - and y -Directions

Describe how to get the graph of y = x 2 + 4x + 3 from the graph of y = x 2 . Solution We can again relate this (and the graph of every quadratic) to the graph of y = x 2 . We must first complete the square. Recall that in this process, you take the coefficient of x (4), divide by 2 (4/2 = 2) and square the result (22 = 4). Add and subtract this number and then, rewrite the x-terms as a perfect square. We have y = x 2 + 4x + 3 = (x 2 + 4x + 4) − 4 + 3 = (x + 2)2 − 1.

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SECTION 0.6

..

Transformations of Functions

67

To graph this function, take the parabola y = x 2 (see Figure 0.84a) and translate the graph 2 units to the left and 1 unit down (see Figure 0.84b). y

y

4

20

20

10

10

x

2

2

4

6

4

2

FIGURE 0.84a

FIGURE 0.84b

y = x2

y = (x + 2)2 − 1

x 2

The following table summarizes our discoveries in this section. Transformations of f (x) Transformation

Form

Effect on Graph

Vertical translation

f (x) + c

|c| units up (c > 0) or down (c < 0)

Horizontal translation

f (x + c)

|c| units left (c > 0) or right (c < 0)

Vertical scale

c f (x) (c > 0)

multiply vertical scale by c

Horizontal scale

f (cx) (c > 0)

divide horizontal scale by c

You will explore additional transformations in the exercises.

EXERCISES 0.6 WRITING EXERCISES 1. The restricted domain of example 6.2 may be puzzling. Consider the following analogy. Suppose you have an airplane flight from New York to Los Angeles with a stop for refueling in Minneapolis. If bad weather has closed the airport in Minneapolis, explain why your flight will be canceled (or at least rerouted) even if the weather is great in New York and Los Angeles. 2. Explain why the graphs of y = 4(x 2 − 1) and y = (4x)2 − 1 in Figures 0.81c and 0.82c appear “thinner” than the graph of y = x 2 − 1.

3. As illustrated in example 6.9, completing the square can be used to rewrite any quadratic function in the form a(x − d)2 + e. Using the transformation rules in this section, explain why this means that all parabolas (with a > 0) will look essentially the same. 4. Explain why the graph of y = f (x + 4) is obtained by moving the graph of y = f (x) four units to the left, instead of to the right.

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Preliminaries

0-68

In exercises 1–6, find the compositions f ◦ g and g ◦ f , and identify their respective domains. √ 1. f (x) = x + 1, g(x) = x − 3 √ 2. f (x) = x − 2, g(x) = x + 1 3. f (x) = e , g(x) = ln x √ 4. f (x) = 1 − x, g(x) = ln x

y 10 5

x

4

5. f (x) = x 2 + 1, g(x) = sin x 1 6. f (x) = 2 , g(x) = x 2 − 2 x −1

x

2

2

4

5 10

In exercises 7–16, identify functions f(x) and g(x) such that the given function equals ( f ◦g)(x). √ √ 1 7. x 4 + 1 8. 3 x + 3 9. 2 x +1 1 11. (4x + 1)2 + 3 12. 4 (x + 1)2 + 3 10. 2 + 1 x 2 13. sin3 x 14. sin x 3 15. e x +1 16. e4x−2

In exercises 39–44, complete the square and explain how to transform the graph of y x 2 into the graph of the given function. 39. f (x) = x 2 + 2x + 1

40. f (x) = x 2 − 4x + 4

41. f (x) = x 2 + 2x + 4

42. f (x) = x 2 − 4x + 2

43. f (x) = 2x + 4x + 4

44. f (x) = 3x 2 − 6x + 2

2

In exercises 17–22, identify functions f (x), g(x) and h(x) such that the given function equals [ f ◦(g◦h)] (x). √ 3 17. √ 18. e4x + 1 sin x + 2 √ 19. cos3 (4x − 2) 20. ln x 2 + 1

2 2 21. 4e x − 5 22. tan−1 (3x + 1) In exercises 23–30, use the graph of y f (x) given in the figure to graph the indicated function. 23. f (x) − 3

24. f (x + 2)

25. f (x − 3)

26. f (x) + 2

27. f (2x)

28. 3 f (x)

29. 4 f (x) − 1

30. 3 f (x + 2)

46. f (x) = −3(x 2 − 1) 47. f (x) = −3(x 2 − 1) + 2 48. f (x) = −2(x 2 − 1) − 1 In exercises 49–52, graph the given function and compare to the graph of y (x − 1)2 − 1 x 2 − 2x. 49. f (x) = (−x)2 − 2(−x) 50. f (x) = (−2x)2 − 2(−2x) 52. f (x) = (−3x)2 − 2(−3x) − 3

10

53. Based on exercises 45–48, state a rule for transforming the graph of y = f (x) into the graph of y = c f (x) for c < 0.

8

54. Based on exercises 49–52, state a rule for transforming the graph of y = f (x) into the graph of y = f (cx) for c < 0.

6 4

55. Sketch the graph of y = |x|3 . Explain why the graph of y = |x|3 is identical to that of y = x 3 to the right of the y-axis. For y = |x|3 , describe how the graph to the left of the y-axis compares to the graph to the right of the y-axis. In general, describe how to draw the graph of y = f (|x|) given the graph of y = f (x).

2 2

45. f (x) = −2(x 2 − 1)

51. f (x) = (−x)2 − 2(−x) + 1

y

4

In exercises 45–48, graph the given function and compare to the graph of y x 2 – 1.

x 2

4

2

In exercises 31–38, use the graph of y f (x) given in the figure to graph the indicated function. 31. f (x − 4)

32. f (x + 3)

33. f (2x)

34. f (2x − 4)

35. f (3x + 3)

36. 3 f (x)

37. 2 f (x) − 4

38. 3 f (x) + 3

56. For y = x 3 , describe how the graph to the left of the y-axis compares to the graph to the right of the y-axis. Show that for f (x) = x 3 , we have f (−x) = − f (x). In general, if you have the graph of y = f (x) to the right of the y-axis and f (−x) = − f (x) for all x, describe how to graph y = f (x) to the left of the y-axis.

0-69

CHAPTER 0

57. Iterations of functions are important in a variety of applications. To iterate f (x), start with an initial value x0 and compute x1 = f (x0 ), x2 = f (x1 ), x3 = f (x2 ) and so on. For example, with f (x) = cos x and x0 = 1, the iterates are x1 = cos 1 ≈ 0.54, x2 = cos x1 ≈ cos 0.54 ≈ 0.86, x3 ≈ cos 0.86 ≈ 0.65 and so on. Keep computing iterates and show that they get closer and closer to 0.739085. Then pick your own x0 (any number you like) and show that the iterates with this new x0 also converge to 0.739085. 58. Referring to exercise 57, show that the iterates of a function can be written as x1 = f (x0 ), x2 = f ( f (x0 )), x3 = f ( f ( f (x0 ))) and so on. Graph y = cos (cos x), y = cos (cos (cos x)) and y = cos (cos (cos (cos x))). The graphs should look more and more like a horizontal line. Use the result of exercise 57 to identify the limiting line. 59. Compute several iterates of f (x) = sin x (see exercise 57) with a variety of starting values. What happens to the iterates in the long run? 60. Repeat exercise 59 for f (x) = x 2 . 61. In cases where the iterates of a function (see exercise 57) repeat a single number, that number is called a fixed point. Explain why any fixed point must be a solution of the equation f (x) = x. Find all fixed points of f (x) = cos x by solving the equation cos x = x. Compare your results to that of exercise 57. 62. Find all fixed points of f (x) = sin x (see exercise 61). Compare your results to those of exercise 59.

EXPLORATORY EXERCISES 1. You have explored how completing the square can transform any quadratic function into the form y = a(x − d)2 + e. We

..

Review Exercises

69

concluded that all parabolas with a > 0 look alike. To see that the same statement is not true of cubic polynomials, graph y = x 3 and y = x 3 − 3x. In this exercise, you will use completing the cube to determine how many different cubic graphs there are. To see what “completing the cube” would look like, first show that (x + a)3 = x 3 + 3ax 2 + 3a 2 x + a 3 . Use this result to transform the graph of y = x 3 into the graphs of (a) y = x 3 − 3x 2 + 3x − 1 and (b) y = x 3 − 3x 2 + 3x + 2. Show that you can’t get a simple transformation to y = x 3 −3x 2 +4x −2. However, show that y = x 3 −3x 2 + 4x −2 can be obtained from y = x 3 + x by basic transformations. Show that the following statement is true: any cubic (y = ax 3 + bx 2 + cx + d) can be obtained with basic transformations from y = ax 3 + kx for some constant k. 2. In many applications, it is important to take a section of a graph (e.g., some data) and extend it for predictions or other analysis. For example, suppose you have an electronic signal equal to f (x) = 2x for 0 ≤ x ≤ 2. To predict the value of the signal at x = −1, you would want to know whether the signal was periodic. If the signal is periodic, explain why f (−1) = 2 would be a good prediction. In some applications, you would assume that the function is even. That is, f (x) = f (−x) for all x. In this case, you want f (x) = 2(−x) = −2x for −2 ≤ x ≤ 0. −2x if − 2 ≤ x ≤ 0 . Graph the even extension f (x) = 2x if 0 ≤ x ≤ 2 2 Find the even extension for (a) f (x) = x + 2x + 1, 0 ≤ x ≤ 2 and (b) f (x) = e−x , 0 ≤ x ≤ 2. 3. Similar to the even extension discussed in exploratory exercise 2, applications sometimes require a function to be odd; that is, f (−x) = − f (x). For f (x) = x 2 , 0 ≤ x ≤ 2, the odd extension requires that for −2≤ x ≤ 0, f (x) = − f (−x) = −x 2 if − 2 ≤ x ≤ 0 −(−x)2 = −x 2 so that f (x) = . Graph x2 if 0 ≤ x ≤ 2 y = f (x) and discuss how to graphically rotate the right half of the graph to get the left half of the graph. Find the odd extension for (a) f (x) = x 2 + 2x, 0 ≤ x ≤ 2 and (b) f (x) = e−x − 1, 0 ≤ x ≤ 2.

Review Exercises WRITING EXERCISES The following list includes terms that are defined and theorems that are stated in this chapter. For each term or theorem, (1) give a precise definition or statement, (2) state in general terms what it means and (3) describe the types of problems with which it is associated.

Slope of a line Domain Graphing window Inverse function Sine function e Composition

Parallel lines Intercepts Local maximum One-to-one function Cosine function Exponential function

Perpendicular lines Zeros of a function Vertical asymptote Periodic function Arcsine function Logarithm

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1.1

..

Limits and Continuity

1-2

A BRIEF PREVIEW OF CALCULUS: TANGENT LINES AND THE LENGTH OF A CURVE In this section, we approach the boundary between precalculus mathematics and the calculus by investigating several important problems requiring the use of calculus. Recall that the slope of a straight line is the change in y divided by the change in x. This fraction is the same regardless of which two points you use to compute the slope. For example, the points (0, 1), (1, 4), and (3, 10) all lie on the line y = 3x + 1. The slope of 3 can be obtained from any two of the points. For instance, 4−1 10 − 1 =3 or m= = 3. 1−0 3−0 In the calculus, we generalize this problem to find the slope of a curve at a point. For instance, suppose we wanted to find the slope of the curve y = x 2 + 1 at the point (1, 2). You might think of picking a second point on the parabola, say (2, 5). The slope of the line through these two points (called a secant line; see Figure 1.2a) is easy enough to compute. We have m=

m sec =

5−2 = 3. 2−1

However, using the points (0, 1) and (1, 2), we get a different slope (see Figure 1.2b): m sec =

2−1 = 1. 1−0

y

y

6

6

4

4

2

2

0.5

x 0.5

1

1.5

2

2

y

2.5

0.5

x 0.5

1

1.5

2

2.5

2

FIGURE 1.2a

FIGURE 1.2b

Secant line, slope = 3

Secant line, slope = 1

2.10 2.05 2.00 1.95 1.90 x 0.96 0.98 1.00 1.02 1.04

FIGURE 1.3 y = x2 + 1

For curves other than straight lines, the slopes of secant lines joining different points are generally not the same, as seen in Figures 1.2a and 1.2b. If you get different slopes using different pairs of points, then what exactly does it mean for a curve to have a slope at a point? The answer can be visualized by graphically zooming in on the specified point. Take the graph of y = x 2 + 1 and zoom in tight on the point (1, 2). You should get a graph something like the one in Figure 1.3. The graph looks very much like a straight line. In fact, the more you zoom in, the straighter the curve appears to be and the less it matters which two points are used to compute a slope. So, here’s the strategy: pick several points on the parabola, each closer to the point (1, 2) than the previous one. Compute the slopes of the lines through (1, 2) and each of the points. The closer the second point gets to (1, 2), the closer the computed slope is to the answer you seek.

1-3

SECTION 1.1

..

A Brief Preview of Calculus

75

For example, the point (1.5, 3.25) is on the parabola fairly close to (1, 2). The slope of the line joining these points is m sec =

3.25 − 2 = 2.5. 1.5 − 1

The point (1.1, 2.21) is even closer to (1, 2). The slope of the secant line joining these two points is m sec =

2.21 − 2 = 2.1. 1.1 − 1

Continuing in this way, observe that the point (1.01, 2.0201) is closer yet to the point (1, 2). The slope of the secant lines through these points is m sec =

2.0201 − 2 = 2.01. 1.01 − 1

The slopes of the secant lines (2.5, 2.1, 2.01) are getting closer and closer to the slope of the parabola at the point (1, 2). Based on these calculations, it seems reasonable to say that the slope of the curve is approximately 2. Example 1.1 takes our introductory example just a bit further.

EXAMPLE 1.1

Estimating the Slope of a Curve

Estimate the slope of y = x 2 + 1 at x = 1. Solution We focus on the point whose coordinates are x = 1 and y = 12 + 1 = 2. To estimate the slope, choose a sequence of points near (1, 2) and compute the slopes of the secant lines joining those points with (1, 2). (We showed sample secant lines in Figures 1.2a and 1.2b.) Choosing points with x > 1 (x-values of 2, 1.1 and 1.01) and points with x < 1 (x-values of 0, 0.9 and 0.99), we compute the corresponding y-values using y = x 2 + 1 and get the slopes shown in the following table. Second Point (2, 5) (1.1, 2.21) (1.01, 2.0201)

msec 5−2 =3 2−1 2.21 − 2 = 2.1 1.1 − 1 2.0201 − 2 = 2.01 1.01 − 1

Second Point (0, 1) (0.9, 1.81) (0.99, 1.9801)

msec 1−2 =1 0−1 1.81 − 2 = 1.9 0.9 − 1 1.9801 − 2 = 1.99 0.99 − 1

Observe that in both columns, as the second point gets closer to (1, 2), the slope of the secant line gets closer to 2. A reasonable estimate of the slope of the curve at the point (1, 2) is then 2. In Chapter 2, we develop a powerful technique for computing such slopes exactly (and easily). Note what distinguishes the calculus problem from the corresponding algebra problem. The calculus problem involves a process we call a limit. While we presently can

76

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Limits and Continuity

1-4

only estimate the slope of a curve using a sequence of approximations, the limit allows us to compute the slope exactly.

EXAMPLE 1.2

Estimating the Slope of a Curve

Estimate the slope of y = sin x at x = 0. Solution This turns out to be a very important problem, one that we will return to later. For now, choose a sequence of points near (0, 0) and compute the slopes of the secant lines joining those points with (0, 0). The following table shows one set of choices.

y

q

q x

Second Point

msec

Second Point

msec

(1, sin 1) (0.1, sin 0.1) (0.01, sin 0.01)

0.84147 0.99833 0.99998

(−1, sin (−1)) (−0.1, sin (−0.1)) (−0.01, sin (−0.01))

0.84147 0.99833 0.99998

Note that as the second point gets closer and closer to (0, 0), the slope of the secant line (m sec ) appears to get closer and closer to 1. A good estimate of the slope of the curve at the point (0, 0) would then appear to be 1. Although we presently have no way of computing the slope exactly, this is consistent with the graph of y = sin x in Figure 1.4. Note that near (0, 0), the graph resembles that of y = x, a straight line of slope 1.

FIGURE 1.4 y = sin x

A second problem requiring the power of calculus is that of computing distance along a curved path. While this problem is of less significance than our first example (both historically and in the development of the calculus), it provides a good indication of the need for mathematics beyond simple algebra. You should pay special attention to the similarities between the development of this problem and our earlier work with slope. Recall that the (straight-line) distance between two points (x1 , y1 ) and (x2 , y2 ) is d{(x1 , y1 ), (x2 , y2 )} =

(x2 − x1 )2 + (y2 − y1 )2 .

For instance, the distance between the points (0, 1) and (3, 4) is y

d{(0, 1), (3, 4)} =

(3, 4)

4 3 2 1

(0, 1) x 1

2

3

FIGURE 1.5a y = (x − 1)2

4

√ (3 − 0)2 + (4 − 1)2 = 3 2 ≈ 4.24264.

However, this is not the only way we might want to compute the distance between these two points. For example, suppose that you needed to drive a car from (0, 1) to (3, 4) along a road that follows the curve y = (x − 1)2 (see Figure 1.5a). In this case, you don’t care about the straight-line distance connecting the two points, but only about how far you must drive along the curve (the length of the curve or arc length). √ Notice that the distance along the curve must be greater than 3 2 (the straight-line distance). Taking a cue from the slope problem, we can formulate a strategy for obtaining a sequence of increasingly accurate approximations. Instead of using just one line segment √ to get the approximation of 3 2, we could use two line segments, as in Figure 1.5b. Notice that the sum of the lengths of the two line segments appears to be a much better approximation to the actual length of the curve than the straight-line distance used previously. This

1-5

SECTION 1.1

y

A Brief Preview of Calculus

77

y (3, 4)

4

(3, 4)

4

3

3

2

2

1

..

(0, 1)

1

(0, 1)

(2, 1)

x 1 1.5 2

3

4

x 1

2

3

FIGURE 1.5b

FIGURE 1.5c

Two line segments

Three line segments

4

distance is d2 = d{(0, 1), (1.5, 0.25)} + d{(1.5, 0.25), (3, 4)} = (1.5 − 0)2 + (0.25 − 1)2 + (3 − 1.5)2 + (4 − 0.25)2 ≈ 5.71592. You’re probably way ahead of us by now. If approximating the length of the curve with two line segments gives an improved approximation, why not use three or four or more? Using the three line segments indicated in Figure 1.5c, we get the further improved approximation

No. of Segments 1 2 3 4 5 6 7

d3 = d{(0, 1), (1, 0)} + d{(1, 0), (2, 1)} + d{(2, 1), (3, 4)} = (1 − 0)2 + (0 − 1)2 + (2 − 1)2 + (1 − 0)2 + (3 − 2)2 + (4 − 1)2 √ √ = 2 2 + 10 ≈ 5.99070.

Distance 4.24264 5.71592 5.99070 6.03562 6.06906 6.08713 6.09711

Note that the more line segments we use, the better the approximation appears to be. This process will become much less tedious with the development of the definite integral in Chapter 4. For now we list a number of these successively better approximations (produced using points on the curve with evenly spaced x-coordinates) in the table found in the margin. The table suggests that the length of the curve is approximately 6.1 (quite far from the straight-line distance of 4.2). If we continued this process using more and more line segments, the sum of their lengths would approach the actual length of the curve (about 6.126). As in the problem of computing the slope of a curve, the exact arc length is obtained as a limit.

y 1

y = sin x

EXAMPLE 1.3

Estimating the Arc Length of a Curve

Estimate the arc length of the curve y = sin x for 0 ≤ x ≤ π (see Figure 1.6a).

q

p

x

FIGURE 1.6a Approximating the curve with two line segments

Solution The endpoints of the curve on this interval are (0, 0) and (π, 0). The distance between these points is d1 = π . The point on the graph of y = sin x corresponding to the midpoint of the interval [0, π ] is (π /2, 1). The distance from (0, 0) to (π/2, 1) plus the distance from (π/2, 1) to (π , 0) (illustrated in Figure 1.6a) is π 2 π 2 d2 = +1+ + 1 ≈ 3.7242. 2 2

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..

Limits and Continuity

√ √ Using the five points (0, 0), (π/4, 1/ 2), (π/2, 1), (3π/4, 1/ 2), and (π, 0) (i.e., four line segments, as indicated in Figure 1.6b), the sum of the lengths of these line segments is π 2 1 2 π 2 1 d4 = 2 + +2 + 1− √ ≈ 3.7901. 4 2 4 2

y y = sin x

d

q

w

1-6

p

Using nine points (i.e., eight line segments), you need a good calculator and some patience to compute the distance of 3.8125. A table showing further approximations is given in the margin. At this stage, it would be reasonable to estimate the length of the sine curve on the interval [0, π ] as slightly more than 3.8.

x

FIGURE 1.6b Approximating the curve with four line segments

BEYOND FORMULAS

Number of Line Segments 8 16 32 64

In the process of estimating both the slope of a curve and the length of a curve, we make some reasonably obvious (straight-line) approximations and then systematically improve on those approximations. In each case, the shorter the line segments are, the closer the approximations are to the desired value. The essence of this is the concept of limit, which separates precalculus mathematics from the calculus. At first glance, this limit idea might seem of little practical importance, since in our examples we never compute the exact solution. In the chapters to come, we will find remarkably simple shortcuts to exact answers. Can you think of ways to find the exact slope in example 1.1?

Sum of Lengths 3.8125 3.8183 3.8197 3.8201

EXERCISES 1.1 WRITING EXERCISES 1. Explain why each approximation of arc length in example 1.3 is less than the actual arc length. 2. To estimate the slope of f (x) = x 2 + 1 at x = 1, you would compute the slopes of various secant lines. Note that y = x 2 + 1 curves up. Explain why the secant line connecting (1, 2) and (1.1, 2.21) will have slope greater than the slope of the tangent line. Discuss how the slope of the secant line between (1, 2) and (0.9, 1.81) compares to the slope of the tangent line.

9. f (x) = e x , a = 0 11. f (x) = ln x, a = 1

1. f (x) = x 2 + 1, a = 1

2. f (x) = x 2 + 1, a = 2

3. f (x) = cos x, a = 0

4. f (x) = cos x, a = π/2

5. f (x) = x 3 + 2, a = 1

6. f (x) = x 3 + 2, a = 2

7. f (x) =

√

x + 1, a = 0

8. f (x) =

√

x + 1, a = 3

12. f (x) = ln x, a = 2

In exercises 13–20, estimate the length of the curve y f (x) on the given interval using (a) n 4 and (b) n 8 line segments. (c) If you can program a calculator or computer, use larger n’s and conjecture the actual length of the curve. 13. f (x) = x 2 + 1, 0 ≤ x ≤ 2 14. f (x) = x 3 + 2, 0 ≤ x ≤ 1

In exercises 1–12, estimate the slope (as in example 1.1) of y f (x) at x a.

10. f (x) = e x , a = 1

15. f (x) = cos x, 0 ≤ x ≤ π/2 16. f (x) = sin x, 0 ≤ x ≤ π/2 √ 17. f (x) = x + 1, 0 ≤ x ≤ 3 18. f (x) = 1/x, 1 ≤ x ≤ 2 19. f (x) = x 2 + 1, −2 ≤ x ≤ 2 20. f (x) = x 3 + 2, −1 ≤ x ≤ 1

1-7

SECTION 1.2

The Concept of Limit

79

26. Use the technique of exercises 21–23 to estimate the area below y = x 3 and above the x-axis between x = 0 and x = 2.

21. An important problem in calculus is finding the area of a region. Sketch the parabola y = 1 − x 2 and shade in the region above the x-axis between x = −1 and x = 1. Then sketch in the following rectangles: (1) height f (− 34 ) and width 12 extending from x = −1 to x = − 12 . (2) height f (− 14 ) and width 1 extending from x = − 12 to x = 0. (3) height f ( 14 ) and width 2 1 extending from x = 0 to x = 12 . (4) height f ( 34 ) and width 12 2 extending from x = 12 to x = 1. Compute the sum of the areas of the rectangles. Based on your sketch, does this give you a good approximation of the area under the parabola?

EXPLORATORY EXERCISE 1. Several central concepts of calculus have been introduced in this section. An important aspect of our future development of calculus is to derive simple techniques for computing quantities such as slope and arc length. In this exercise, you will learn how to directly compute the slope of a curve at a point. Suppose you want the slope of y = x 2 at x = 1. You could start by computing slopes of secant lines connecting the point (1, 1) with nearby points. Suppose the nearby point has x-coordinate 1 + h, where h is a small (positive or negative) number. Explain why the corresponding y-coordinate is (1 + h)2 . Show (1 + h)2 − 1 = 2 + h. As that the slope of the secant line is 1+h−1 h gets closer and closer to 0, this slope better approximates the slope of the tangent line. Letting h approach 0, show that the slope of the tangent line equals 2. In a similar way, show that the slope of y = x 2 at x = 2 is 4 and find the slope of y = x 2 at x = 3. Based on your answers, conjecture a formula for the slope of y = x 2 at x = a, for any unspecified value of a.

22. To improve the approximation of exercise 21, divide the interval [−1, 1] into 8 pieces and construct a rectangle of the appropriate height on each subinterval. Compared to the approximation in exercise 21, explain why you would expect this to be a better approximation of the actual area under the parabola. 23. Use a computer or calculator to compute an approximation of the area in exercise 21 using (a) 16 rectangles, (b) 32 rectangles, (c) 64 rectangles. Use these calculations to conjecture the exact value of the area under the parabola. 24. Use the technique of exercises 21–23 to estimate the area below y = sin x and above the x-axis between x = 0 and x = π. 25. Use the technique of exercises 21–23 to estimate the area below y = x 3 and above the x-axis between x = 0 and x = 1.

1.2

..

THE CONCEPT OF LIMIT In this section, we develop the notion of limit using some common language and illustrate the idea with some simple examples. The notion turns out to be a rather subtle one, easy to think of intuitively, but a bit harder to pin down in precise terms. We present the precise definition of limit in section 1.6. There, we carefully define limits in considerable detail. The more informal notion of limit that we introduce and work with here and in sections 1.3, 1.4 and 1.5 is adequate for most purposes. As a start, consider the functions

y f(x) 4 f(x)

f (x) =

2

x

2

x

2

FIGURE 1.7a y=

x2 − 4 x −2

x

x2 − 4 x −2

and

g(x) =

x2 − 5 . x −2

Notice that both functions are undefined at x = 2. So, what does this mean, beyond saying that you cannot substitute 2 for x? We often find important clues about the behavior of a function from a graph (see Figures 1.7a and 1.7b). Notice that the graphs of these two functions look quite different in the vicinity of x = 2. Although we can’t say anything about the value of these functions at x = 2 (since this is outside the domain of both functions), we can examine their behavior in the vicinity of

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SECTION 1.2

The Concept of Limit

79

26. Use the technique of exercises 21–23 to estimate the area below y = x 3 and above the x-axis between x = 0 and x = 2.

21. An important problem in calculus is finding the area of a region. Sketch the parabola y = 1 − x 2 and shade in the region above the x-axis between x = −1 and x = 1. Then sketch in the following rectangles: (1) height f (− 34 ) and width 12 extending from x = −1 to x = − 12 . (2) height f (− 14 ) and width 1 extending from x = − 12 to x = 0. (3) height f ( 14 ) and width 2 1 extending from x = 0 to x = 12 . (4) height f ( 34 ) and width 12 2 extending from x = 12 to x = 1. Compute the sum of the areas of the rectangles. Based on your sketch, does this give you a good approximation of the area under the parabola?

EXPLORATORY EXERCISE 1. Several central concepts of calculus have been introduced in this section. An important aspect of our future development of calculus is to derive simple techniques for computing quantities such as slope and arc length. In this exercise, you will learn how to directly compute the slope of a curve at a point. Suppose you want the slope of y = x 2 at x = 1. You could start by computing slopes of secant lines connecting the point (1, 1) with nearby points. Suppose the nearby point has x-coordinate 1 + h, where h is a small (positive or negative) number. Explain why the corresponding y-coordinate is (1 + h)2 . Show (1 + h)2 − 1 = 2 + h. As that the slope of the secant line is 1+h−1 h gets closer and closer to 0, this slope better approximates the slope of the tangent line. Letting h approach 0, show that the slope of the tangent line equals 2. In a similar way, show that the slope of y = x 2 at x = 2 is 4 and find the slope of y = x 2 at x = 3. Based on your answers, conjecture a formula for the slope of y = x 2 at x = a, for any unspecified value of a.

22. To improve the approximation of exercise 21, divide the interval [−1, 1] into 8 pieces and construct a rectangle of the appropriate height on each subinterval. Compared to the approximation in exercise 21, explain why you would expect this to be a better approximation of the actual area under the parabola. 23. Use a computer or calculator to compute an approximation of the area in exercise 21 using (a) 16 rectangles, (b) 32 rectangles, (c) 64 rectangles. Use these calculations to conjecture the exact value of the area under the parabola. 24. Use the technique of exercises 21–23 to estimate the area below y = sin x and above the x-axis between x = 0 and x = π. 25. Use the technique of exercises 21–23 to estimate the area below y = x 3 and above the x-axis between x = 0 and x = 1.

1.2

..

THE CONCEPT OF LIMIT In this section, we develop the notion of limit using some common language and illustrate the idea with some simple examples. The notion turns out to be a rather subtle one, easy to think of intuitively, but a bit harder to pin down in precise terms. We present the precise definition of limit in section 1.6. There, we carefully define limits in considerable detail. The more informal notion of limit that we introduce and work with here and in sections 1.3, 1.4 and 1.5 is adequate for most purposes. As a start, consider the functions

y f(x) 4 f(x)

f (x) =

2

x

2

x

2

FIGURE 1.7a y=

x2 − 4 x −2

x

x2 − 4 x −2

and

g(x) =

x2 − 5 . x −2

Notice that both functions are undefined at x = 2. So, what does this mean, beyond saying that you cannot substitute 2 for x? We often find important clues about the behavior of a function from a graph (see Figures 1.7a and 1.7b). Notice that the graphs of these two functions look quite different in the vicinity of x = 2. Although we can’t say anything about the value of these functions at x = 2 (since this is outside the domain of both functions), we can examine their behavior in the vicinity of

80

..

CHAPTER 1

Limits and Continuity

1-8

x2 − 4 , we compute x −2 some values of the function for x close to 2, as in the following tables.

y

this point. We consider these functions one at a time. First, for f (x) =

10 5 10

x 5 5 10

FIGURE 1.7b y=

x2 − 5 x −2

10

x

f (x)

1.9 1.99 1.999 1.9999

3.9 3.99 3.999 3.9999

x2 − 4 x− 2

x

f (x)

2.1 2.01 2.001 2.0001

4.1 4.01 4.001 4.0001

x2 − 4 x− 2

Notice that as you move down the first column of the table, the x-values get closer to 2, but are all less than 2. We use the notation x → 2− to indicate that x approaches 2 from the left side. Notice that the table and the graph both suggest that as x gets closer and closer to 2 (with x < 2), f (x) is getting closer and closer to 4. In view of this, we say that the limit of f(x) as x approaches 2 from the left is 4, written lim f (x) = 4.

x→2−

Likewise, we need to consider what happens to the function values for x close to 2 but larger than 2. Here, we use the notation x → 2+ to indicate that x approaches 2 from the right side. We compute some of these values in the second table. Again, the table and graph both suggest that as x gets closer and closer to 2 (with x > 2), f (x) is getting closer and closer to 4. In view of this, we say that the limit of f(x) as x approaches 2 from the right is 4, written lim f (x) = 4.

x→2+

We call lim− f (x) and lim+ f (x) one-sided limits. Since the two one-sided limits of x→2

x→2

f (x) are the same, we summarize our results by saying that the limit of f(x) as x approaches 2 is 4, written lim f (x) = 4.

x→2

The notion of limit as we have described it here is intended to communicate the behavior of a function near some point of interest, but not actually at that point. We finally observe that we can also determine this limit algebraically, as follows. Notice that since the expression x2 − 4 in the numerator of f (x) = factors, we can write x −2 x2 − 4 x→2 x − 2 (x − 2)(x + 2) = lim x→2 x −2 = lim (x + 2) = 4,

lim f (x) = lim

x→2

x→2

Cancel the factors of (x − 2). As x approaches 2, (x + 2) approaches 4.

where we can cancel the factors of (x − 2) since in the limit as x → 2, x is close to 2, but x = 2, so that x − 2 = 0.

1-9

SECTION 1.2

x 1.9 1.99 1.999 1.9999

..

The Concept of Limit

81

x2 − 5 , as x → 2. Based on the x −2 graph in Figure 1.7b and the table of approximate function values shown in the margin, observe that as x gets closer and closer to 2 (with x < 2), g(x) increases without bound. Since there is no number that g(x) is approaching, we say that the limit of g(x) as x approaches 2 from the left does not exist, written Similarly, we consider one-sided limits for g(x) =

x2 − 5 g(x) x− 2 13.9 103.99 1003.999 10,003.9999

lim g(x) does not exist.

x→2−

Similarly, the graph and the table of function values for x > 2 (shown in the margin) suggest that g(x) decreases without bound as x approaches 2 from the right. Since there is no number that g(x) is approaching, we say that

x2 − 5 x− 2

x

g(x)

2.1 2.01 2.001 2.0001

−5.9 −95.99 −995.999 −9995.9999

lim g(x) does not exist.

x→2+

Finally, since there is no common value for the one-sided limits of g(x) (in fact, neither limit exists), we say that the limit of g(x) as x approaches 2 does not exist, written lim g(x) does not exist.

x→2

Before moving on, we should summarize what we have said about limits.

A limit exists if and only if both corresponding one-sided limits exist and are equal. That is, lim f (x) = L , for some number L, if and only if lim− f (x) = lim+ f (x) = L .

x→a

x→a

x→a

In other words, we say that lim f (x) = L if we can make f (x) as close as we might like to x→a L, by making x sufficiently close to a (on either side of a), but not equal to a. Note that we can think about limits from a purely graphical viewpoint, as in example 2.1. y

EXAMPLE 2.1 2

Use the graph in Figure 1.8 to determine lim− f (x), lim+ f (x), lim f (x) and lim f (x). x→1

1 2

1

Determining Limits Graphically

x 1 1 2

2

x→1

x→1

x→−1

Solution For lim− f (x), we consider the y-values as x gets closer to 1, with x < 1. x→1

That is, we follow the graph toward x = 1 from the left (x < 1). Observe that the graph dead-ends into the open circle at the point (1, 2). Therefore, we say that lim− f (x) = 2. x→1

For lim+ f (x), we follow the graph toward x = 1 from the right (x > 1). In this case, x→1

FIGURE 1.8 y = f (x)

the graph dead-ends into the solid circle located at the point (1, −1). For this reason, we say that lim+ f (x) = −1. Because lim− f (x) = lim+ f (x), we say that lim f (x) does x→1

x→1

x→1

x→1

not exist. Finally, we have that lim f (x) = 1, since the graph approaches a y-value of x→−1

1 as x approaches −1 both from the left and from the right.

82

..

CHAPTER 1

Limits and Continuity

1-10

y

EXAMPLE 2.2 3

3

x

f(x)

x

Q

x 2

f(x)

Further, note that

3

lim −

FIGURE 1.9 lim

x→−3

x→−3

1 3x + 9 =− x2 − 9 2

−3.1 −3.01 −3.001 −3.0001

−2.9 −2.99 −2.999 −2.9999

lim

x→−3

1 3x + 9 =− . x2 − 9 2

In example 2.2, the limit exists because both one-sided limits exist and are equal. In example 2.3, neither one-sided limit exists.

y

EXAMPLE 2.3

30

Cancel factors of (x + 3).

Finally, since the function approaches the same value as x → −3 both from the right and from the left (i.e., the one-sided limits are equal), we write

3x 9 x2 − 9 −0.508475 −0.500835 −0.500083 −0.500008

x

3x + 9 3(x + 3) = lim − 2 x→−3 (x + 3)(x − 3) x −9 3 1 = lim − =− , x→−3 x − 3 2

since (x − 3) → −6 as x → −3. Again, the cancellation of the factors of (x + 3) is valid since in the limit as x → −3, x is close to −3, but x = −3, so that x + 3 = 0. Likewise, 3x + 9 1 lim =− . x→−3+ x 2 − 9 2

3x 9 x2 − 9 −0.491803 −0.499168 −0.499917 −0.499992

x

A Limit Where Two Factors Cancel

3x + 9 Evaluate lim 2 . x→−3 x − 9 Solution We examine a graph (see Figure 1.9) and compute some function values for x near −3. Based on this numerical and graphical evidence, it’s reasonable to conjecture that 3x + 9 3x + 9 1 lim + 2 = lim − 2 =− . x→−3 x − 9 x→−3 x − 9 2

f(x)

A Limit That Does Not Exist

3x + 9 exists. x2 − 9 Solution We first draw a graph (see Figure 1.10) and compute some function values for x close to 3. 3x + 9 Based on this numerical and graphical evidence, it appears that, as x → 3+ , 2 x −9 is increasing without bound. Thus,

Determine whether lim

x→3

x 3

x

f(x) 30

FIGURE 1.10 y=

3x + 9 x2 − 9

x

3x + 9 does not exist. x2 − 9 Similarly, from the graph and the table of values for x < 3, we can say that lim

x→3+

3x + 9 does not exist. x2 − 9 Since neither one-sided limit exists, we say lim

x→3−

x 3.1 3.01 3.001 3.0001

3x 9 x2 − 9 30 300 3000 30,000

3x + 9 does not exist. x2 − 9 Here, we considered both one-sided limits for the sake of completeness. Of course, you should keep in mind that if either one-sided limit fails to exist, then the limit does not exist. lim

x→3

1-11

SECTION 1.2

3x 9 x2 − 9 −30 −300 −3000 −30,000

x 2.9 2.99 2.999 2.9999

EXAMPLE 2.4

sin x . x Solution Unlike some of the limits considered previously, there is no algebra that will simplify this expression. However, we can still draw a graph (see Figure 1.11) and compute some function values.

0.1 0.01 0.001 0.0001 0.00001

f (x)

x x 2

4

lim

sin x x 0.998334 0.999983 0.99999983 0.9999999983 0.999999999983

x −0.1 −0.01 −0.001 −0.0001 −0.00001

sin x =1 x from which we conjecture that x→0+

sin x =1 x

sin x x 0.998334 0.999983 0.99999983 0.9999999983 0.999999999983

The graph and the tables of values lead us to the conjectures: lim

FIGURE 1.11 x→0

Approximating the Value of a Limit

x→0

x

2

83

Evaluate lim

1

4

The Concept of Limit

Many limits cannot be resolved using algebraic methods. In these cases, we can approximate the limit using graphical and numerical evidence, as we see in example 2.4.

y

x

..

and

lim

x→0−

sin x = 1, x

sin x = 1. x In Chapter 2, we examine these limits with greater care (and prove that these conjectures are correct). lim

x→0

REMARK 2.1 Computer or calculator computation of limits is unreliable. We use graphs and tables of values only as (strong) evidence pointing to what a plausible answer might be. To be certain, we need to obtain careful verification of our conjectures. We see how to do this in sections 1.3–1.7.

EXAMPLE 2.5 y

Evaluate lim

x→0

1

x

4

4

A Case Where One-Sided Limits Disagree

x . |x|

Solution The computer-generated graph shown in Figure 1.12a is incomplete. Since x is undefined at x = 0, there is no point at x = 0. The graph in Figure 1.12b |x| correctly shows open circles at the intersections of the two halves of the graph with the y-axis. We also have

1

lim

x→0+

FIGURE 1.12a y=

x |x|

x x = lim+ x→0 x |x| = lim+ 1 x→0

=1

Since |x| = x, when x > 0.

84

CHAPTER 1

..

Limits and Continuity

1-12

y

and

lim

x→0−

1

x

x→0

= −1.

x 2

f(x)

It now follows that

lim

x→0

1

FIGURE 1.12b lim

x→0

Since |x| = −x, when x < 0.

= lim− −1

f(x) x 2

x x = lim− x→0 −x |x|

x does not exist. |x|

x does not exist, |x|

since the one-sided limits are not the same. You should also keep in mind that this observation is entirely consistent with what we see in the graph.

EXAMPLE 2.6

A Limit Describing the Movement of a Baseball Pitch

The knuckleball is one of the most exotic pitches in baseball. Batters describe the ball as unpredictably moving left, right, up and down. For a typical knuckleball speed of 60 mph, the left/right position of the ball (in feet) as it crosses the plate is given by f (ω) =

1.7 5 − sin(2.72ω) ω 8ω2

(derived from experimental data in Watts and Bahill’s book Keeping Your Eye on the Ball), where ω is the rotational speed of the ball in radians per second and where f (ω) = 0 corresponds to the middle of home plate. Folk wisdom among baseball pitchers has it that the less spin on the ball, the better the pitch. To investigate this theory, we consider the limit of f (ω) as ω → 0+ . As always, we look at a graph (see Figure 1.13) and generate a table of function values. The graphical and numerical evidence suggests that lim+ f (ω) = 0. ω→0

y 1.5

1.0

0.5

v 2

4

6

8

10

FIGURE 1.13 5 1.7 − y= sin(2.72ω) ω 8ω2

ω

f (ω)

10 1 0.1 0.01 0.001 0.0001

0.1645 1.4442 0.2088 0.021 0.0021 0.0002

The limit indicates that a knuckleball with absolutely no spin doesn’t move at all (and therefore would be easy to hit). According to Watts and Bahill, a very slow rotation rate of about 1 to 3 radians per second produces the best pitch (i.e., the most movement). Take another look at Figure 1.13 to convince yourself that this makes sense.

1-13

SECTION 1.2

..

The Concept of Limit

85

EXERCISES 1.2 WRITING EXERCISES 1. Suppose your professor says, “You can think of the limit of f (x) as x approaches a as what f (a) should be.” Critique this statement. What does it mean? Does it provide important insight? Is there anything misleading about it? Replace the phrase in italics with your own best description of what the limit is. 2. Your friend’s professor says, “The limit is a prediction of what f (a) will be.” Compare and contrast this statement to the one in exercise 1. Does the inclusion of the word prediction make the limit idea seem more useful and important?

2. For the function graphed below, identify each limit or state that it does not exist. (a) lim f (x)

(b) lim f (x)

(c) lim f (x)

(d) lim f (x)

(e) lim f (x)

(f) lim f (x)

(g) lim f (x)

(h) lim f (x)

(i) lim f (x)

(j) lim f (x)

x→0−

x→0+ x→2−

x→0

x→1−

x→−2 x→1+

x→1

x→−1

x→3

3. We have observed that lim f (x) does not depend on the actual x→a

y

value of f (a), or even onwhether f (a) exists. In principle, x 2 if x = 2 are as “normal” as functions such as f (x) = 13 if x = 2 functions such as g(x) = x 2 . With this in mind, explain why it is important that the limit concept is independent of how (or whether) f (a) is defined. 4. The most common limit encountered in everyday life is the speed limit. Describe how this type of limit is very different from the limits discussed in this section.

4 2 6

2

x 2

4

6

2 4

1. For the function graphed below, identify each limit or state that it does not exist. (a) lim f (x)

(b) lim f (x)

(c) lim f (x)

(d) lim f (x)

(e) lim f (x)

(f) lim f (x)

(g) lim f (x)

(h) lim f (x)

x→0−

x→0+ x→1−

x→0

x→2−

x→−1 x→2+

x→2

(i) lim f (x)

(j) lim f (x)

x→−2

x→3

y

3. Sketch the graph of f (x) = (a) lim f (x)

(b) lim f (x)

(c) lim f (x)

(d) lim f (x)

x→2− x→2

2 2 4

x→2+ x→1

3 x −1 4. Sketch the graph of f (x) = 0 √ x +1−2 identify each limit. (a) lim f (x)

(b) lim f (x)

(d) lim f (x)

(e) lim f (x)

x→0−

x→−1

2 6

6

5. Sketch the graph of f (x) =

lim f (x)

x→−1−

(c) lim f (x) x→−1

x <0 x = 0 and x >0

x→0

x→3

x2 + 1 3x + 1

if if

x < −1 and idenx ≥ −1

tify each limit. (a)

if if if

(c) lim f (x)

x→0+

4

x <2 and identify x ≥2

if if

each limit.

4

x

2x x2

(b)

lim f (x)

x→−1+

(d) lim f (x) x→1

86

CHAPTER 1

..

Limits and Continuity

2x + 1 if if 6. Sketch the graph of f (x) = 3 2x + 1 if identify each limit. (a)

lim f (x)

x→−1−

(d) lim f (x) x→1

(b)

1-14

x < −1 −1 ≤ x < 1 and x >1

lim f (x)

(c) lim f (x)

x→−1+

x→−1

x→0

8. Evaluate f (−1.5), f (−1.1), f (−1.01) and f (−1.001), and x +1 conjecture a value for lim f (x) for f (x) = 2 . Evaluate x→−1− x −1 f (−0.5), f (−0.9), f (−0.99) and f (−0.999), and conjecture x +1 . Does lim f (x) a value for lim f (x) for f (x) = 2 x→−1 x→−1+ x −1 exist? In exercises 9–14, use numerical and graphical evidence to conjecture values for each limit. x2 + x 10. lim 2 x→−1 x − x − 2 sin x 12. lim x→π x − π 14. lim e−1/x

2

x→0

In exercises 15–26, use numerical and graphical evidence to conjecture whether the limit at x a exists. If not, describe what is happening at x a graphically. x2 − 1 x→1 x 2 − 2x + 1 √ 5−x −2 17. lim √ x→1 10 − x − 3 1 19. lim sin x→0 x 15. lim

x −2 x→2 |x − 2|

16. lim

x→−1

x2 − 1 x +1

x 2 + 4x 18. lim √ x→0 x3 + x2 1 20. lim x sin x→0 x |x + 1| x→−1 x 2 − 1

21. lim

22. lim

23. lim ln x

24. lim x ln (x 2 )

x→0

tan x x→0 x

25. lim

x→0

tan−1 x x→0 x

26. lim

x2 + 1 x +1 , lim and similar limits to investix→1 x − 1 x→2 x 2 − 4 gate the following. Suppose that f (x) and g(x) are polynomials with g(a) = 0 and f (a) = 0. What can you conjecture about f (x) ? lim x→a g(x)

27. Compute lim

x→−1

(e) lim f (x)

7. Evaluate f (1.5), f (1.1), f (1.01) and f (1.001), and conjecx −1 . Evaluate ture a value for lim f (x) for f (x) = √ x→1+ x −1 f (0.5), f (0.9), f (0.99) and f (0.999), and conjecture a value x −1 for lim f (x) for f (x) = √ . Does lim f (x) exist? x→1 x→1− x −1

x2 − 1 9. lim x→1 x − 1 x2 + x 11. lim x→0 sin x tan x 13. lim x→0 sin x

x +1 sin x , lim and similar limits to invesx 2 + 1 x→π x tigate the following. Suppose that f (x) and g(x) are functions with f (a) = 0 and g(a) = 0. What can you conjecture about f (x) lim ? x→a g(x)

28. Compute lim

In exercises 29–32, sketch a graph of a function with the given properties. 29. f (−1) = 2, f (0) = −1, f (1) = 3 and lim f (x) does not exist. x→1

30. f (x) = 1 for −2 ≤ x ≤ 1, lim f (x) = 3 and lim f (x) = 1. x→1+

x→−2

31. f (0) = 1, lim f (x) = 2 and lim f (x) = 3. x→0−

x→0+

32. lim f (x) = −2, f (0) = 1, f (2) = 3 and lim f (x) does not x→0

x→2

exist. 33. As we see in Chapter 2, the slope of the tangent √ line to the √ 1+h−1 . curve y = x at x = 1 is given by m = lim h→0 h √ Estimate the slope m. Graph y = x and the line with slope m through the point (1, 1). 34. As we see √ in Chapter 2, the velocity of an object that has traveled x √ miles in x hours at the x = 1 hour mark is given x −1 . Estimate this limit. by v = lim x→1 x − 1 π 35. Consider the following arguments concerning lim sin . x→0+ x π increases without bound; First, as x > 0 approaches 0, x since sin t oscillates for increasing t, the limit does not exist. Second: taking x = 1, 0.1, 0.01 and so on, we compute sin π = sin 10π = sin 100π = · · · = 0; therefore the limit equals 0. Which argument sounds better to you? Explain. Explore the limit and determine which answer is correct. 36. Consider the following argument concerning lim e−1/x . As x x→0 1 −1 approaches 0, increases without bound and decreases x x t without bound. Since e approaches 0 as t decreases without bound, the limit equals 0. Discuss all the errors made in this argument. 37. Numerically estimate lim (1 + x)1/x and lim (1 + x)1/x . x→0+

x→0−

Note that the function values for x > 0 increase as x decreases, while for x < 0 the function values decrease as x increases. Explain why this indicates that, if lim (1 + x)1/x x→0

exists, it is between function values for positive and negative x’s. Approximate this limit correct to eight digits. 38. Explain what is wrong with the following logic (you may use exercise 37 to convince yourself that the answer is wrong, but discuss the logic without referring to exercise 37): as x → 0, it is clear that (1 + x) → 1. Since 1 raised to any power is 1, lim (1 + x)1/x = lim (1)1/x = 1. x→0

x→0

1-15

SECTION 1.3

39. Numerically estimate lim x sec x . Try to numerically estimate x→0+

lim x sec x . If your computer has difficulty evaluating the func-

x→0−

tion for negative x’s, explain why. 40. Explain what is wrong with the following logic (note from exercise 39 that the answer is accidentally correct): since 0 to any power is 0, lim x sec x = lim 0sec x = 0. x→0

x→0

41. Give an example of a function f such that lim f (x) exists but x→0

f (0) does not exist. Give an example of a function g such that g(0) exists but lim g(x) does not exist. x→0

42. Give an example of a function f such that lim f (x) exists and x→0

f (0) exists, but lim f (x) = f (0). x→0

43. In the text, we described lim f (x) = L as meaning “as x gets x→a

closer and closer to a, f (x) is getting closer and closer to L.” As x gets closer and closer to 0, it is true that x 2 gets closer and closer to −0.01, but it is certainly not true that lim x 2 = −0.01. Try to modify the description of limit to x→0

make it clear that lim x 2 = −0.01. We explore a very precise x→0

definition of limit in section 1.6. 44. In Figure 1.13, the final position of the knuckleball at time t = 0.68 is shown as a function of the rotation rate ω. The batter must decide at time t = 0.4 whether to swing at the pitch. At t = 0.4, the left/right position of the ball is given 1 5 by h(ω) = − sin (1.6ω). Graph h(ω) and compare to ω 8ω2 Figure 1.13. Conjecture the limit of h(ω) as ω → 0. For ω = 0, is there any difference in ball position between what the batter sees at t = 0.4 and what he tries to hit at t = 0.68? 45. A parking lot charges $2 for each hour or portion of an hour, with a maximum charge of $12 for all day. If f (t) equals the total parking bill for t hours, sketch a graph of y = f (t) for 0 ≤ t ≤ 24. Determine the limits lim f (t) and lim f (t), if t→3.5

t→4

they exist. 46. For the parking lot in exercise 45, determine all values of a with 0 ≤ a ≤ 24 such that lim f (t) does not exist. Briefly t→a

discuss the effect this has on your parking strategy (e.g., are there times where you would be in a hurry to move your car or times where it doesn’t matter whether you move your car?).

1.3

..

Computation of Limits

87

EXPLORATORY EXERCISES 1. In a situation similar to that of example 2.6, the left/right position of a knuckleball pitch in baseball can be modeled by 5 (1 − cos 4ωt), where t is time measured in seconds P= 8ω2 (0 ≤ t ≤ 0.68) and ω is the rotation rate of the ball measured in radians per second. In example 2.6, we chose a specific t-value and evaluated the limit as ω → 0. While this gives us some information about which rotation rates produce hardto-hit pitches, a clearer picture emerges if we look at P over its entire domain. Set ω = 10 and graph the resulting func1 tion (1 − cos 40t) for 0 ≤ t ≤ 0.68. Imagine looking at a 160 pitcher from above and try to visualize a baseball starting at the pitcher’s hand at t = 0 and finally reaching the batter, at t = 0.68. Repeat this with ω = 5, ω = 1, ω = 0.1 and whatever values of ω you think would be interesting. Which values of ω produce hard-to-hit pitches? 2. In this exercise, the results you get will depend on the accuracy of your computer or calculator. Work this exercise and compare your results with your classmates’ results. We will incos x − 1 vestigate lim . Start with the calculations presented x→0 x2 in the table (your results may vary): x

f(x)

0.1 0.01 0.001

−0.499583. . . −0.49999583. . . −0.4999999583. . .

Describe as precisely as possible the pattern shown here. What would you predict for f (0.0001)? f (0.00001)? Does your computer or calculator give you this answer? If you continue trying powers of 0.1 (0.000001, 0.0000001 etc.) you should eventually be given a displayed result of −0.5. Do you think this is exactly correct or has the answer just been rounded off? Why is rounding off inescapable? It turns out that −0.5 is the exact value for the limit, so the round-off here is somewhat helpful. However, if you keep evaluating the function at smaller and smaller values of x, you will eventually see a reported function value of 0. This round-off error is not so benign; we discuss this error in section 1.7. For now, evaluate cos x at the current value of x and try to explain where the 0 came from.

COMPUTATION OF LIMITS Now that you have an idea of what a limit is, we need to develop some means of calculating limits of simple functions. In this section, we present some basic rules for dealing with common limit problems. We begin with two simple limits.

1-15

SECTION 1.3

39. Numerically estimate lim x sec x . Try to numerically estimate x→0+

lim x sec x . If your computer has difficulty evaluating the func-

x→0−

tion for negative x’s, explain why. 40. Explain what is wrong with the following logic (note from exercise 39 that the answer is accidentally correct): since 0 to any power is 0, lim x sec x = lim 0sec x = 0. x→0

x→0

41. Give an example of a function f such that lim f (x) exists but x→0

f (0) does not exist. Give an example of a function g such that g(0) exists but lim g(x) does not exist. x→0

42. Give an example of a function f such that lim f (x) exists and x→0

f (0) exists, but lim f (x) = f (0). x→0

43. In the text, we described lim f (x) = L as meaning “as x gets x→a

closer and closer to a, f (x) is getting closer and closer to L.” As x gets closer and closer to 0, it is true that x 2 gets closer and closer to −0.01, but it is certainly not true that lim x 2 = −0.01. Try to modify the description of limit to x→0

make it clear that lim x 2 = −0.01. We explore a very precise x→0

definition of limit in section 1.6. 44. In Figure 1.13, the final position of the knuckleball at time t = 0.68 is shown as a function of the rotation rate ω. The batter must decide at time t = 0.4 whether to swing at the pitch. At t = 0.4, the left/right position of the ball is given 1 5 by h(ω) = − sin (1.6ω). Graph h(ω) and compare to ω 8ω2 Figure 1.13. Conjecture the limit of h(ω) as ω → 0. For ω = 0, is there any difference in ball position between what the batter sees at t = 0.4 and what he tries to hit at t = 0.68? 45. A parking lot charges $2 for each hour or portion of an hour, with a maximum charge of $12 for all day. If f (t) equals the total parking bill for t hours, sketch a graph of y = f (t) for 0 ≤ t ≤ 24. Determine the limits lim f (t) and lim f (t), if t→3.5

t→4

they exist. 46. For the parking lot in exercise 45, determine all values of a with 0 ≤ a ≤ 24 such that lim f (t) does not exist. Briefly t→a

discuss the effect this has on your parking strategy (e.g., are there times where you would be in a hurry to move your car or times where it doesn’t matter whether you move your car?).

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..

Computation of Limits

87

EXPLORATORY EXERCISES 1. In a situation similar to that of example 2.6, the left/right position of a knuckleball pitch in baseball can be modeled by 5 (1 − cos 4ωt), where t is time measured in seconds P= 8ω2 (0 ≤ t ≤ 0.68) and ω is the rotation rate of the ball measured in radians per second. In example 2.6, we chose a specific t-value and evaluated the limit as ω → 0. While this gives us some information about which rotation rates produce hardto-hit pitches, a clearer picture emerges if we look at P over its entire domain. Set ω = 10 and graph the resulting func1 tion (1 − cos 40t) for 0 ≤ t ≤ 0.68. Imagine looking at a 160 pitcher from above and try to visualize a baseball starting at the pitcher’s hand at t = 0 and finally reaching the batter, at t = 0.68. Repeat this with ω = 5, ω = 1, ω = 0.1 and whatever values of ω you think would be interesting. Which values of ω produce hard-to-hit pitches? 2. In this exercise, the results you get will depend on the accuracy of your computer or calculator. Work this exercise and compare your results with your classmates’ results. We will incos x − 1 vestigate lim . Start with the calculations presented x→0 x2 in the table (your results may vary): x

f(x)

0.1 0.01 0.001

−0.499583. . . −0.49999583. . . −0.4999999583. . .

Describe as precisely as possible the pattern shown here. What would you predict for f (0.0001)? f (0.00001)? Does your computer or calculator give you this answer? If you continue trying powers of 0.1 (0.000001, 0.0000001 etc.) you should eventually be given a displayed result of −0.5. Do you think this is exactly correct or has the answer just been rounded off? Why is rounding off inescapable? It turns out that −0.5 is the exact value for the limit, so the round-off here is somewhat helpful. However, if you keep evaluating the function at smaller and smaller values of x, you will eventually see a reported function value of 0. This round-off error is not so benign; we discuss this error in section 1.7. For now, evaluate cos x at the current value of x and try to explain where the 0 came from.

COMPUTATION OF LIMITS Now that you have an idea of what a limit is, we need to develop some means of calculating limits of simple functions. In this section, we present some basic rules for dealing with common limit problems. We begin with two simple limits.

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1-16

y

For any constant c and any real number a,

yc

c

lim c = c.

(3.1)

x→a

x

x

x

a

FIGURE 1.14

In other words, the limit of a constant is that constant. This certainly comes as no surprise, since the function f (x) = c does not depend on x and so, stays the same as x → a (see Figure 1.14). Another simple limit is the following.

lim c = c

x→a

For any real number a, lim x = a.

(3.2)

x→a

y f(x)

Again, this is not a surprise, since as x → a, x will approach a (see Figure 1.15). Be sure that you are comfortable enough with the limit notation to recognize how obvious the limits in (3.1) and (3.2) are. As simple as they are, we use them repeatedly in finding more complex limits. We also need the basic rules contained in Theorem 3.1.

yx

a f(x)

THEOREM 3.1 x

x a

FIGURE 1.15 lim x = a

x→a

x

Suppose that lim f (x) and lim g(x) both exist and let c be any constant. The x→a x→a following then apply: (i) lim [c · f (x)] = c · lim f (x), x→a

x→a

(ii) lim [ f (x) ± g(x)] = lim f (x) ± lim g(x), x→a x→a x→a

(iii) lim [ f (x) · g(x)] = lim f (x) lim g(x) and x→a

x→a

x→a

f (x) x→a = if lim g(x) = 0 . x→a g(x) lim g(x) lim f (x)

(iv) lim

x→a

x→a

The proof of Theorem 3.1 is found in Appendix A and requires the formal definition of limit discussed in section 1.6. You should think of these rules as sensible results that you would certainly expect to be true, given your intuitive understanding of what a limit is. Read them in plain English. For instance, part (ii) says that the limit of a sum (or a difference) equals the sum (or difference) of the limits, provided the limits exist. Think of this as follows. If as x approaches a, f (x) approaches L and g(x) approaches M, then f (x) + g(x) should approach L + M. Observe that by applying part (iii) of Theorem 3.1 with g(x) = f (x), we get that, whenever lim f (x) exists, x→a

lim [ f (x)]2 = lim [ f (x) · f (x)] x→a

2 = lim f (x) lim f (x) = lim f (x) .

x→a

x→a

x→a

x→a

Likewise, for any positive integer n, we can apply part (iii) of Theorem 3.1 repeatedly, to yield

n lim [ f (x)]n = lim f (x) (3.3) x→a

(see exercises 67 and 68).

x→a

1-17

SECTION 1.3

..

Computation of Limits

89

Notice that taking f (x) = x in (3.3) gives us that for any integer n > 0 and any real number a, lim x n = a n .

(3.4)

x→a

That is, to compute the limit of any positive power of x, you simply substitute in the value of x being approached.

EXAMPLE 3.1

Finding the Limit of a Polynomial

Apply the rules of limits to evaluate lim (3x 2 − 5x + 4). x→2

Solution We have lim (3x 2 − 5x + 4) = lim (3x 2 ) − lim (5x) + lim 4

x→2

x→2

x→2

= 3 lim x 2 − 5 lim x + 4

By Theorem 3.1 (i).

= 3 · (2)2 − 5 · 2 + 4 = 6.

By (3.4).

x→2

EXAMPLE 3.2

By Theorem 3.1 (ii).

x→2

x→2

Finding the Limit of a Rational Function

Apply the rules of limits to evaluate lim

x→3

x 3 − 5x + 4 . x2 − 2

Solution We get lim (x 3 − 5x + 4) x 3 − 5x + 4 x→3 lim = x→3 x2 − 2 lim (x 2 − 2)

By Theorem 3.1 (iv).

x→3

= =

lim x 3 − 5 lim x + lim 4

x→3

x→3 x→3 2 lim x − lim 2 x→3 x→3

33 − 5 · 3 + 4 16 = . 32 − 2 7

By Theorem 3.1 (i) and (ii).

By (3.4).

You may have noticed that in examples 3.1 and 3.2, we simply ended up substituting the value for x, after taking many intermediate steps. In example 3.3, it’s not quite so simple.

EXAMPLE 3.3

Finding a Limit by Factoring

x2 − 1 . x→1 1 − x Solution Notice right away that

Evaluate lim

lim (x 2 − 1) x2 − 1 x→1 lim = , x→1 1 − x lim (1 − x) x→1

since the limit in the denominator is zero. (Recall that the limit of a quotient is the quotient of the limits only when both limits exist and the limit in the denominator is not

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1-18

zero.) We can resolve this problem by observing that x2 − 1 (x − 1)(x + 1) Factoring the numerator and = lim factoring −1 from denominator. x→1 1 − x x→1 −(x − 1) (x + 1) Simplifying and = lim = −2, substituting x = 1. x→1 −1 where the cancellation of the factors of (x − 1) is valid because in the limit as x → 1, x is close to 1, but x = 1, so that x − 1 = 0. lim

In Theorem 3.2, we show that the limit of a polynomial is simply the value of the polynomial at that point; that is, to find the limit of a polynomial, we simply substitute in the value that x is approaching.

THEOREM 3.2 For any polynomial p(x) and any real number a, lim p(x) = p(a).

x→a

PROOF Suppose that p(x) is a polynomial of degree n ≥ 0, p(x) = cn x n + cn−1 x n−1 + · · · + c1 x + c0 . Then, from Theorem 3.1 and (3.4), lim p(x) = lim (cn x n + cn−1 x n−1 + · · · + c1 x + c0 )

x→a

x→a

= cn lim x n + cn−1 lim x n−1 + · · · + c1 lim x + lim c0 x→a

x→a

x→a

x→a

= cn a n + cn−1 a n−1 + · · · + c1 a + c0 = p(a). Evaluating the limit of a polynomial is now easy. Many other limits are evaluated just as easily.

THEOREM 3.3 Suppose that lim f (x) = L and n is any positive integer. Then, x→a lim n f (x) = n lim f (x) = n L , x→a

x→a

where for n even, we assume that L > 0. The proof of Theorem 3.3 is given in Appendix A. Notice that this result says that we may (under the conditions outlined in the hypotheses) bring limits “inside” nth roots. We can then use our existing rules for computing the limit inside.

EXAMPLE 3.4 Evaluate lim

x→2

√ 5

3x 2

Evaluating the Limit of an nth Root of a Polynomial − 2x.

Solution By Theorems 3.2 and 3.3, we have √ 5 5 lim 3x 2 − 2x = 5 lim (3x 2 − 2x) = 8. x→2

x→2

1-19

REMARK 3.1 In general, in any case where the limits of both the numerator and the denominator are 0, you should try to algebraically simplify the expression, to get a cancellation, as we do in examples 3.3 and 3.5.

SECTION 1.3

EXAMPLE 3.5 √

Evaluate lim

x→0

..

Computation of Limits

91

Finding a Limit by Rationalizing

x +2− x

√

2

.

√ √ Solution First, notice that both the numerator ( x + 2 − 2) and the denominator (x) approach 0 as x approaches 0. Unlike example 3.3, we can’t factor the numerator. However, we can rationalize the numerator, as follows: √ √ √ √ √ √ x +2− 2 ( x + 2 − 2)( x + 2 + 2) x +2−2 = = √ √ √ √ x x( x + 2 + 2) x( x + 2 + 2) x 1 = √ √ =√ √ , x( x + 2 + 2) x +2+ 2 where the last equality holds if x = 0 (which is the case in the limit as x → 0). So, we have √ √ x +2− 2 1 1 1 lim = lim √ √ =√ √ = √ . x→0 x→0 x x +2+ 2 2+ 2 2 2 So that we are not restricted to discussing only the algebraic functions (i.e., those that can be constructed by using addition, subtraction, multiplication, division, exponentiation and by taking nth roots), we state the following result now, without proof.

THEOREM 3.4 For any real number a, we have (i) lim sin x = sin a,

(v) lim sin−1 x = sin−1 a, for −1 < a < 1,

(ii) lim cos x = cos a,

(vi) lim cos−1 x = cos−1 a, for −1 < a < 1,

(iii) lim e x = ea and

(vii) lim tan−1 x = tan−1 a, for −∞

x→a x→a x→a

x→a x→a x→a

(iv) lim ln x = ln a, for a > 0. (viii) if p is a polynomial and lim f (x) = L , x→a

then lim f ( p(x)) = L .

x→ p(a)

x→a

Notice that Theorem 3.4 says that limits of the sine, cosine, exponential, natural logarithm, inverse sine, inverse cosine and inverse tangent functions are found simply by substitution. A more thorough discussion of functions with this property (called continuity) is found in section 1.4.

EXAMPLE 3.6 Evaluate lim sin x→0

−1

Evaluating a Limit of an Inverse Trigonometric Function

x +1 . 2

Solution By Theorem 3.4, we have x +1 1 π lim sin−1 = sin−1 = . x→0 2 2 6 So much for limits that we can compute using elementary rules. Many limits can be found only by using more careful analysis, often by an indirect approach. For instance, consider the following problem.

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y

EXAMPLE 3.7

A Limit of a Product That Is Not the Product of the Limits

Evaluate lim (x cot x). x→0

Solution Your first reaction might be to say that this is a limit of a product and so, must be the product of the limits: lim (x cot x) = lim x lim cot x This is incorrect!

x p

q

q

x→0

p

x→0

x→0

= 0 · ? = 0,

(3.5) where we’ve written a “?” since you probably don’t know what to do with lim cot x. x→0

Since the first limit is 0, do we really need to worry about the second limit? The problem here is that we are attempting to apply the result of Theorem 3.1 in a case where the hypotheses are not satisfied. Specifically, Theorem 3.1 says that the limit of a product is the product of the respective limits when all of the limits exist. The graph in Figure 1.16 suggests that lim cot x does not exist. You should compute some function

FIGURE 1.16 y = cot x

x→0

values, as well, to convince yourself that this is in fact the case. So, equation (3.5) does not hold and we’re back to square one. Since none of our rules seem to apply here, the most reasonable step is to draw a graph (see Figure 1.17) and compute some function values. Based on these, we conjecture that

y 1

lim (x cot x) = 1,

0.99

x→0

0.98

0.97 x

0.3

0.3

FIGURE 1.17 y = x cot x

x ±0.1 ±0.01 ±0.001 ±0.0001 ±0.00001

x cot x 0.9967 0.999967 0.99999967 0.9999999967 0.999999999967

which is definitely not 0, as you might have initially suspected. You can also think about this limit as follows: cos x x lim (x cot x) = lim x = lim cos x x→0 x→0 x→0 sin x sin x x = lim lim cos x x→0 sin x x→0 lim cos x

1 = = 1, sin x 1 lim x→0 x since lim cos x = 1 and where we have used the conjecture we made in example 2.4 x→0 sin x = 1. (We verify this last conjecture in section 2.6, using the Squeeze that lim x→0 x Theorem, which follows.) =

x→0

At this point, we introduce a tool that will help us determine a number of important limits.

THEOREM 3.5 (Squeeze Theorem) Suppose that f (x) ≤ g(x) ≤ h(x) for all x in some interval (c, d), except possibly at the point a ∈ (c, d) and that lim f (x) = lim h(x) = L ,

x→a

x→a

for some number L. Then, it follows that lim g(x) = L , also.

x→a

1-21

SECTION 1.3

y

FIGURE 1.18 The Squeeze Theorem

REMARK 3.2 The Squeeze Theorem also applies to one-sided limits.

Computation of Limits

x

EXAMPLE 3.8

Using the Squeeze Theorem to Verify the Value of a Limit

1 Determine the value of lim x cos . x→0 x 2

Solution Your first reaction might be to observe that this is a limit of a product and so, might be the product of the limits: 1 1 ? 2 2 lim x cos lim cos = lim x . This is incorrect! (3.6) x→0 x→0 x→0 x x However, the graph of y = cos x1 found in Figure 1.19 suggests that cos x1 oscillates back and forth between −1 and 1. Further, the closer x gets to 0, the more rapid the oscillations become. You should compute some function values, as well, to convince yourself that lim cos x1 does not exist. Equation (3.6) then does not hold and x→0

we’re back to square one. Since none of our rules seem to apply here, the most reasonable step is to draw a graph and compute some function values in an effort to see what is going on. The graph of y = x 2 cos x1 appears in Figure 1.20 and a table of function values is shown in the margin. y

y

1

0.03

x

0.2

0.2

1

y = cos

x 2 cos (1/x) −0.008 8.6 × 10−5 5.6 × 10−7 −9.5 × 10−9 −9.99 × 10−11

x

0.3

0.3

0.03

FIGURE1.19

x ±0.1 ±0.01 ±0.001 ±0.0001 ±0.00001

93

The proof of Theorem 3.5 is given in Appendix A, since it depends on the precise definition of limit found in section 1.6. However, if you refer to Figure 1.18, you should clearly see that if g(x) lies between f (x) and h(x), except possibly at a itself and both f (x) and h(x) have the same limit as x → a, then g(x) gets squeezed between f (x) and h(x) and therefore should also have a limit of L. The challenge in using the Squeeze Theorem is in finding appropriate functions f and h that bound a given function g from below and above, respectively, and that have the same limit as x → a.

y h(x) y g(x) y f(x)

a

..

1 x

FIGURE 1.20 y = x 2 cos

1 x

The graph and the table of function values suggest the conjecture 1 2 = 0, lim x cos x→0 x which we prove using the Squeeze Theorem. First, we need to find functions f and h such that 1 2 f (x) ≤ x cos ≤ h(x), x

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for all x = 0 and where lim f (x) = lim h(x) = 0. Recall that

y 0.03

x→0

y x2

x

0.3

0.3

0.03

1-22

y x2

x→0

1 −1 ≤ cos ≤ 1, x

for all x = 0. If we multiply (3.7) through by x 2 (notice that since x 2 ≥ 0, this multiplication preserves the inequalities), we get 1 −x 2 ≤ x 2 cos ≤ x 2, x for all x = 0. We illustrate this inequality in Figure 1.21. Further, lim (−x 2 ) = 0 = lim x 2 .

FIGURE 1.21

y = x 2 cos x1 , y = x 2 and y = −x 2

(3.7)

x→0

x→0

So, from the Squeeze Theorem, it now follows that 1 lim x 2 cos = 0, x→0 x also, as we had conjectured.

BEYOND FORMULAS

TODAY IN MATHEMATICS Michael Freedman (1951– ) An American mathematician who first solved one of the most famous problems in mathematics, the four-dimensional Poincar´e conjecture. A winner of the Fields Medal, the mathematical equivalent of the Nobel Prize, Freedman says, “Much of the power of mathematics comes from combining insights from seemingly different branches of the discipline. Mathematics is not so much a collection of different subjects as a way of thinking. As such, it may be applied to any branch of knowledge.” Freedman finds mathematics to be an open field for research, saying that, “It isn’t necessary to be an old hand in an area to make a contribution.”

To resolve the limit in example 3.8, we could not apply the rules for limits contained in Theorem 3.1. So, we resorted to an indirect method of finding the limit. This tour de force of graphics plus calculation followed by analysis is sometimes referred to as the Rule of Three. (The Rule of Three presents a general strategy for attacking new problems. The basic idea is to look at problems graphically, numerically and analytically.) In the case of example 3.8, the first two elements of this “rule” (the graphics in Figure 1.20 and the accompanying table of function values) suggest a plausible conjecture, while the third element provides us with a careful mathematical verification of the conjecture. In what ways does this sound like the scientific method? Functions are often defined by different expressions on different intervals. Such piecewise-defined functions are important and we illustrate such a function in example 3.9.

EXAMPLE 3.9

A Limit for a Piecewise-Defined Function

Evaluate lim f (x), where f is defined by x→0 2 x + 2 cos x + 1, for x < 0 f (x) = . for x ≥ 0 e x − 4, Solution Since f is defined by different expressions for x < 0 and for x ≥ 0, we must consider one-sided limits. We have lim f (x) = lim− (x 2 + 2 cos x + 1) = 2 cos 0 + 1 = 3,

x→0−

x→0

by Theorem 3.4. Also, we have lim f (x) = lim+ (e x − 4) = e0 − 4 = 1 − 4 = −3.

x→0+

x→0

Since the one-sided limits are different, we have that lim f (x) does not exist. x→0

1-23

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Computation of Limits

95

We end this section with an example of the use of limits in computing velocity. In section 2.1, we see that for an object moving in a straight line, whose position at time t is given by the function f (t), the instantaneous velocity of that object at time t = 1 (i.e., the velocity at the instant t = 1, as opposed to the average velocity over some period of time) is given by the limit f (1 + h) − f (1) lim . h→0 h

EXAMPLE 3.10

Evaluating a Limit Describing Velocity

Suppose that the position function for an object at time t (seconds) is given by f (t) = t 2 + 2 (feet), find the instantaneous velocity of the object at time t = 1. Solution Given what we have just learned about limits, this is now an easy problem to solve. We have f (1 + h) − f (1) [(1 + h)2 + 2] − 3 lim = lim . h→0 h→0 h h While we can’t simply substitute h = 0 (why not?), we can write [(1 + h)2 + 2] − 3 (1 + 2h + h 2 ) − 1 = lim h→0 h→0 h h lim

Expanding the squared term.

2h + h 2 h(2 + h) = lim h→0 h→0 h h 2+h = lim = 2. h→0 1

= lim

Canceling factors of h.

So, the instantaneous velocity of this object at time t = 1 is 2 feet per second.

EXERCISES 1.3 WRITING EXERCISES 1. Given your knowledge of the graphs of polynomials, explain why equations (3.1) and (3.2) and Theorem 3.2 are obvious. Name five non-polynomial functions for which limits can be evaluated by substitution. 2. Suppose that you can draw the graph of y = f (x) without lifting your pencil from your paper. Explain why lim f (x) = f (a), for every value of a. x→a

3. In one or two sentences, explain the Squeeze Theorem. Use a real-world analogy (e.g., having the functions represent the locations of three people as they walk) to indicate why it is true. 4. Given the graph in Figure 1.20 and the calculations that follow, it may be unclear why we insist on using the Squeeze Theorem before concluding that lim [x 2 cos (1/x)] is indeed 0. Review x→0

section 1.2 to explain why we are being so fussy.

In exercises 1–34, evaluate the indicated limit, if it exists. Assume sin x that lim 1. x→0 x √ 3 2. lim 2x + 1 1. lim (x 2 − 3x + 1) x→0

x→2

−1

2

3. lim cos (x ) x→0

x2 − x − 6 x→3 x −3 2 x −x −2 7. lim x→2 x2 − 4 sin x 9. lim x→0 tan x xe−2x+1 11. lim 2 x→0 x + x 5. lim

x −5 x2 + 4 x2 + x − 2 6. lim 2 x→1 x − 3x + 2 x3 − 1 8. lim 2 x→1 x + 2x − 3 tan x 10. lim x→0 x 4. lim

x→2

12. lim x 2 csc2 x x→0+

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Limits and Continuity

√

x +4−2 2x 14. lim √ x→0 x 3− x +9 √ x −1 x −2 lim √ 16. lim x→1 x→4 x − 4 x −1 2 2 1 2 lim − 2 18. lim − x→1 x→0 x −1 x −1 x |x| 2x 1−e −1/x 2 lim 20. lim sin(e ) x→0 1 − e x x→0 sin |x| sin2 (x 2 ) lim 22. lim x→0 x→0 x x4 2x if x < 2 lim f (x), where f (x) = x 2 if x ≥ 2 x→2 2 x + 1 if x < −1 lim f (x), where f (x) = 3x + 1 if x ≥ −1 x→−1 2 x + 2 if x < −1 lim f (x), where f (x) = 3x + 1 if x ≥ −1 x→0 2x if x < 2 lim f (x), where f (x) = x 2 if x ≥ 2 x→1 2x + 1 if x < −1 if −1 < x < 1 lim f (x), where f (x) = 3 x→−1 2x + 1 if x > 1 2x + 1 if x < −1 if −1 < x < 1 lim f (x), where f (x) = 3 x→1 2x + 1 if x > 1 (2 + h)2 − 4 (1 + h)3 − 1 lim 30. lim h→0 h→0 h √ h h2 x2 + x + 4 − 2 lim √ 32. lim √ h→0 x→0 x2 + x h2 + h + 3 − h + 3 1 1 +t tan 2x lim 2 34. lim t→−2 2 + t x→0 5x Use numerical and graphical evidence to conjecture the value of lim x 2 sin (1/x). Use the Squeeze Theorem to x→0 prove that you are correct: identify the functions f and h, show graphically that f (x) ≤ x 2 sin (1/x) ≤ h(x) and justify lim f (x) = lim h(x).

13. lim

x→0

15. 17. 19. 21. 23. 24. 25. 26. 27.

28. 29. 31. 33. 35.

x→0

x→0

36. Why can’t you use the Squeeze Theorem as in exercise 35 to prove that lim x 2 sec (1/x) = 0? Explore this limit graphically. x→0 √ 37. Use the Squeeze Theorem to prove that lim [ x cos2 (1/x)] = 0. x→0+

Identify √the functions f and h, show graphically that f (x) ≤ x cos2 (1/x) ≤ h(x) for all x > 0, and justify lim f (x) = 0 and lim h(x) = 0. x→0+

1 1 − cos x 43. Given that lim = , quickly evaluate 2 + x→0 x 2 √ 1 − cos x . lim x→0+ x sin x 1 − cos2 x 44. Given that lim = 1, quickly evaluate lim . x→0 x x→0 x2 g(x) if x < a for polynomials g(x) and 45. Suppose f (x) = h(x) if x > a h(x). Explain why lim f (x) = g(a) and determine lim f (x). x→a −

x→0

In exercises 39–42, either find the limit or explain why it does not exist. 39. lim 16 − x 2 40. lim 16 − x 2 + − x→4 x→4 41. lim x 2 + 3x + 2 42. lim x 2 + 3x + 2 x→−2+

x→a +

46. Explain how to determine lim f (x) if g and h are polynomials x→a g(x) if x < a if x = a . and f (x) = c h(x) if x > a 47. Evaluate each limit and justify each step by citing the appropriate theorem or equation. x −2 (a) lim (x 2 − 3x + 1) (b) lim 2 x→2 x→0 x + 1 48. Evaluate each limit and justify each step by citing the appropriate theorem or equation. xe x (a) lim [(x + 1) sin x] (b) lim x→−1 x→1 tan x In exercises 49–52, use the given position function f (t) to find the velocity at time t a. 49. f (t) = t 2 + 2, a = 2

50. f (t) = t 2 + 2, a = 0

51. f (t) = t , a = 0

52. f (t) = t 3 , a = 1

3

√ 53. In Chapter 2, the slope of the tangent √ line to the curve y = x 1+h−1 . Compute the at x = 1 is defined by m = lim h→0 h √ slope m. Graph y = x and the line with slope m through the point (1, 1). 54. In Chapter 2, √ an alternative form for the limit in exercise 53 is x −1 given by lim . Compute this limit. x→1 x − 1 In exercises 55–62, use numerical evidence to conjecture the value of the limit if it exists. Check your answer with your Computer Algebra System (CAS). If you disagree, which one of you is correct? 55. lim (1 + x)1/x x→0+

58. lim x ln x x→0+

x→0+

38. Suppose that f (x) is bounded: that is, there exists a constant M such that | f (x)| ≤ M for all x. Use the Squeeze Theorem to prove that lim x 2 f (x) = 0.

x→−2−

1-24

−1

61. lim tan x→0

1 x

57. lim x −x

56. lim e1/x x→0+ 1 59. lim sin x→0 x 1 62. lim ln x→0 x

2

x→0+

60. lim e1/x x→0

In exercises 63–66, use lim f (x) 2, lim g(x) − 3 and x→a

x→a

lim h(x) 0 to determine the limit, if possible.

x→a

63. lim [2 f (x) − 3g(x)] x→a f (x) + g(x) 65. lim x→a h(x)

64. lim [3 f (x)g(x)] x→a 3 f (x) + 2g(x) 66. lim x→a h(x)

1-25

SECTION 1.4

67. Assume that lim f (x) = L. Use Theorem 3.1 to prove that x→a 3

lim [ f (x)] = L . Also, show that lim [ f (x)] = L . 3

4

x→a

4

x→a

68. How did you work exercise 67? You probably used Theorem 3.1 to work from lim [ f (x)]2 = L 2 to lim [ f (x)]3 = L 3 , and x→a

x→a

then used lim [ f (x)]3 = L 3 to get lim [ f (x)]4 = L 4 . Going one x→a

x→a

step at a time, we should be able to reach lim [ f (x)]n = L n x→a

for any positive integer n. This is the idea of mathematical induction. Formally, we need to show the result is true for a specific value of n = n 0 [we show n 0 = 2 in the text], then assume the result is true for a general n = k ≥ n 0 . If we show that we can get from the result being true for n = k to the result being true for n = k + 1, we have proved that the result is true for any positive integer n. In one sentence, explain why this is true. Use this technique to prove that lim [ f (x)]n = L n

..

Continuity and Its Consequences

97

76. Suppose a state’s income tax code states that tax liability is 12% on the first $20,000 of taxable earnings and 16% on the remainder. Find constants a and b for the tax function a + 0.12x if x ≤ 20,000 T (x) = b + 0.16(x − 20,000) if x > 20,000 such that lim T (x) = 0 and lim T (x) exists. Why is it x→0+

x→20,000

important for these limits to exist? 77. The greatest integer function is denoted by f (x) = [x] and equals the greatest integer that is less than or equal to x. Thus, [2.3] = 2, [−1.2] = −2 and [3] = 3. In spite of this last fact, show that lim [x] does not exist. x→3

78. Investigate the existence of (a) lim [x], (b) (c) lim [2x], and (d) lim (x − [x]). x→1.5

x→1

lim [x],

x→1.5

x→1

x→a

for any positive integer n. 69. Find all the errors in the following incorrect string of equalities: 1 x 1 = lim 2 = lim x lim 2 = 0 · ? = 0. x→0 x x→0 x→0 x x 70. Find all the errors in the following incorrect string of equalities: lim

x→0

0 sin 2x = = 1. x 0 71. Give an example of functions f and g such that lim [ f (x) + g(x)] exists but lim f (x) and lim g(x) do not exist. lim

x→0

x→0

x→0

x→0

72. Give an example of functions f and g such that lim [ f (x) · g(x)] exists but at least one of lim f (x) and lim g(x) x→0

x→0

x→0

does not exist. 73. If lim f (x) exists and lim g(x) does not exist, is it always true x→a

x→a

that lim [ f (x) + g(x)] does not exist? Explain. x→a

74. Is the following true or false? If lim f (x) does not exist, then x→0

lim

x→0

Compute lim T (x); why is this good? Compute

1.4

sin x = 1. Using graphical x sin 2x , and numerical evidence, conjecture the value of lim x→0 x sin 3x sin π x sin x/2 , lim and lim . In general, conjeclim x→0 x→0 x→0 x x x sin cx for any constant c. Given that ture the value of lim x→0 x sin cx = 1 for any constant c = 0, prove that your conlim x→0 cx jecture is correct. x→0

75. Suppose a state’s income tax code states the tax liability on x dollars of taxable income is given by 0.14x if 0 ≤ x < 10,000 . T (x) = 1500 + 0.21x if 10,000 ≤ x x→0+

1. The value x = 0 is called a zero of multiplicity n (n ≥ 1) f (x) for the function f if lim n exists and is nonzero but x→0 x f (x) = 0. Show that x = 0 is a zero of multiplicity 2 lim x→0 x n−1 2 for x , x = 0 is a zero of multiplicity 3 for x 3 and x = 0 is a zero of multiplicity 4 for x 4 . For polynomials, what does multiplicity describe? The reason the definition is not as straightforward as we might like is so that it can apply to nonpolynomial functions, as well. Find the multiplicity of x = 0 for f (x) = sin x; f (x) = x sin x; f (x) = sin x 2 . If you know that x = 0 is a zero of multiplicity m for f (x) and multiplicity n for g(x), what can you say about the multiplicity of x = 0 for f (x) + g(x)? f (x) · g(x)? f (g(x))? 2. We have conjectured that lim

1 does not exist. Explain. f (x)

why is this bad?

EXPLORATORY EXERCISES

lim

x→10,000

T (x);

CONTINUITY AND ITS CONSEQUENCES When you describe something as continuous, just what do you have in mind? For example, if told that a machine has been in continuous operation for the past 60 hours, most of us would interpret this to mean that the machine has been in operation all of that time, without

1-25

SECTION 1.4

67. Assume that lim f (x) = L. Use Theorem 3.1 to prove that x→a 3

lim [ f (x)] = L . Also, show that lim [ f (x)] = L . 3

4

x→a

4

x→a

68. How did you work exercise 67? You probably used Theorem 3.1 to work from lim [ f (x)]2 = L 2 to lim [ f (x)]3 = L 3 , and x→a

x→a

then used lim [ f (x)]3 = L 3 to get lim [ f (x)]4 = L 4 . Going one x→a

x→a

step at a time, we should be able to reach lim [ f (x)]n = L n x→a

for any positive integer n. This is the idea of mathematical induction. Formally, we need to show the result is true for a specific value of n = n 0 [we show n 0 = 2 in the text], then assume the result is true for a general n = k ≥ n 0 . If we show that we can get from the result being true for n = k to the result being true for n = k + 1, we have proved that the result is true for any positive integer n. In one sentence, explain why this is true. Use this technique to prove that lim [ f (x)]n = L n

..

Continuity and Its Consequences

97

76. Suppose a state’s income tax code states that tax liability is 12% on the first $20,000 of taxable earnings and 16% on the remainder. Find constants a and b for the tax function a + 0.12x if x ≤ 20,000 T (x) = b + 0.16(x − 20,000) if x > 20,000 such that lim T (x) = 0 and lim T (x) exists. Why is it x→0+

x→20,000

important for these limits to exist? 77. The greatest integer function is denoted by f (x) = [x] and equals the greatest integer that is less than or equal to x. Thus, [2.3] = 2, [−1.2] = −2 and [3] = 3. In spite of this last fact, show that lim [x] does not exist. x→3

78. Investigate the existence of (a) lim [x], (b) (c) lim [2x], and (d) lim (x − [x]). x→1.5

x→1

lim [x],

x→1.5

x→1

x→a

for any positive integer n. 69. Find all the errors in the following incorrect string of equalities: 1 x 1 = lim 2 = lim x lim 2 = 0 · ? = 0. x→0 x x→0 x→0 x x 70. Find all the errors in the following incorrect string of equalities: lim

x→0

0 sin 2x = = 1. x 0 71. Give an example of functions f and g such that lim [ f (x) + g(x)] exists but lim f (x) and lim g(x) do not exist. lim

x→0

x→0

x→0

x→0

72. Give an example of functions f and g such that lim [ f (x) · g(x)] exists but at least one of lim f (x) and lim g(x) x→0

x→0

x→0

does not exist. 73. If lim f (x) exists and lim g(x) does not exist, is it always true x→a

x→a

that lim [ f (x) + g(x)] does not exist? Explain. x→a

74. Is the following true or false? If lim f (x) does not exist, then x→0

lim

x→0

Compute lim T (x); why is this good? Compute

1.4

sin x = 1. Using graphical x sin 2x , and numerical evidence, conjecture the value of lim x→0 x sin 3x sin π x sin x/2 , lim and lim . In general, conjeclim x→0 x→0 x→0 x x x sin cx for any constant c. Given that ture the value of lim x→0 x sin cx = 1 for any constant c = 0, prove that your conlim x→0 cx jecture is correct. x→0

75. Suppose a state’s income tax code states the tax liability on x dollars of taxable income is given by 0.14x if 0 ≤ x < 10,000 . T (x) = 1500 + 0.21x if 10,000 ≤ x x→0+

1. The value x = 0 is called a zero of multiplicity n (n ≥ 1) f (x) for the function f if lim n exists and is nonzero but x→0 x f (x) = 0. Show that x = 0 is a zero of multiplicity 2 lim x→0 x n−1 2 for x , x = 0 is a zero of multiplicity 3 for x 3 and x = 0 is a zero of multiplicity 4 for x 4 . For polynomials, what does multiplicity describe? The reason the definition is not as straightforward as we might like is so that it can apply to nonpolynomial functions, as well. Find the multiplicity of x = 0 for f (x) = sin x; f (x) = x sin x; f (x) = sin x 2 . If you know that x = 0 is a zero of multiplicity m for f (x) and multiplicity n for g(x), what can you say about the multiplicity of x = 0 for f (x) + g(x)? f (x) · g(x)? f (g(x))? 2. We have conjectured that lim

1 does not exist. Explain. f (x)

why is this bad?

EXPLORATORY EXERCISES

lim

x→10,000

T (x);

CONTINUITY AND ITS CONSEQUENCES When you describe something as continuous, just what do you have in mind? For example, if told that a machine has been in continuous operation for the past 60 hours, most of us would interpret this to mean that the machine has been in operation all of that time, without

98

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..

Limits and Continuity

1-26

any interruption at all, even for a moment. Mathematicians mean much the same thing when we say that a function is continuous. A function is said to be continuous on an interval if its graph on that interval can be drawn without interruption, that is, without lifting your pencil from the paper. It is helpful for us to first try to see what it is about the functions whose graphs are shown in Figures 1.22a–1.22d that makes them discontinuous (i.e., not continuous) at the point x = a. y

y

x

a x

a

FIGURE 1.22b

FIGURE 1.22a

f (a) is defined, but lim f (x) does

f (a) is not defined (the graph has a hole at x = a).

not exist (the graph has a jump at x = a).

x→a

y

y f (a)

a

x a

FIGURE 1.22c lim f (x) exists and f (a) is defined,

FIGURE 1.22d

but lim f (x) = f (a) (the graph has

lim f (x) does not exist (the

x→a

x→a

REMARK 4.1 The definition of continuity all boils down to the one condition in (iii), since conditions (i) and (ii) must hold whenever (iii) is met. Further, this says that a function is continuous at a point exactly when you can compute its limit at that point by simply substituting in.

a hole at x = a).

x

x→a

function “blows up” at x = a).

This suggests the following definition of continuity at a point.

DEFINITION 4.1 A function f is continuous at x = a when (i) f (a) is defined, (ii) lim f (x) exists and x→a

(iii) lim f (x) = f (a).

Otherwise, f is said to be discontinuous at x = a.

x→a

For most purposes, it is best for you to think of the intuitive notion of continuity that we’ve outlined above. Definition 4.1 should then simply follow from your intuitive understanding of the concept.

1-27

SECTION 1.4

..

Continuity and Its Consequences

99

y

EXAMPLE 4.1

Finding Where a Rational Function Is Continuous

Determine where f (x) = 4

y

x2 2x 3 x1

x 2 + 2x − 3 is continuous. x −1

Solution Note that x 2 + 2x − 3 (x − 1)(x + 3) = x −1 x −1 = x + 3, for x = 1.

f (x) =

x 1

Factoring the numerator. Canceling common factors.

This says that the graph of f is a straight line, but with a hole in it at x = 1, as indicated in Figure 1.23. So, f is discontinuous at x = 1, but continuous elsewhere.

FIGURE 1.23 y=

x 2 + 2x − 3 x −1

EXAMPLE 4.2

Removing a Discontinuity

Make the function from example 4.1 continuous everywhere by redefining it at a single point.

REMARK 4.2 You should be careful not to confuse the continuity of a function at a point with its simply being defined there. A function can be defined at a point without being continuous there. (Look back at Figures 1.22b and 1.22c.)

Solution In example 4.1, we saw that the function is discontinuous at x = 1, since it is undefined there. So, suppose we just go ahead and define it, as follows. Let x 2 + 2x − 3 , if x = 1 g(x) = x −1 a, if x = 1, for some real number a. Notice that g(x) is defined for all x and equals f (x) for all x = 1. Here, we have x 2 + 2x − 3 x→1 x −1 = lim (x + 3) = 4.

lim g(x) = lim

x→1

y

x→1

Observe that if we choose a = 4, we now have that lim g(x) = 4 = g(1)

x→1

y g(x)

4

and so, g is continuous at x = 1. Note that the graph of g is the same as the graph of f seen in Figure 1.23, except that we now include the point (1, 4) (see Figure 1.24). Also, note that there’s a very simple way to write g(x). (Think about this.)

x 1

FIGURE 1.24 y = g(x)

When we can remove a discontinuity by redefining the function at that point, we call the discontinuity removable. Not all discontinuities are removable, however. Carefully examine Figures 1.22a–1.22d and convince yourself that the discontinuities in Figures 1.22a and 1.22c are removable, while those in Figures 1.22b and 1.22d are nonremovable. Briefly, a function f has a removable discontinuity at x = a if lim f (x) exists and either f (a) is x→a undefined or lim f (x) = f (a). x→a

EXAMPLE 4.3

Nonremovable Discontinuities

1 1 . Find all discontinuities of f (x) = 2 and g(x) = cos x x

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Limits and Continuity

y

1-28

Solution You should observe from Figure 1.25a (also, construct a table of function values) that lim

x→0

4

1 does not exist. x2

Hence, f is discontinuous at x = 0. Similarly, observe that lim cos(1/x) does not exist, due to the endless oscillation of x→0

2

x

3

cos(1/x) as x approaches 0 (see Figure 1.25b). In both cases, notice that since the limits do not exist, there is no way to redefine either function at x = 0 to make it continuous there.

3

From your experience with the graphs of some common functions, the following result should come as no surprise.

FIGURE 1.25a 1 y= 2 x y

THEOREM 4.1 All polynomials are continuous everywhere. Additionally, sin x, cos x, tan−1 x and e x √ n are continuous everywhere, x is continuous for all x, when n is odd and for x > 0, when n is even. We also have ln x is continuous for x > 0 and sin−1 x and cos−1 x are continuous for −1 < x < 1.

1

x

0.2

0.2

PROOF 1

We have already established (in Theorem 3.2) that for any polynomial p(x) and any real number a,

FIGURE 1.25b

lim p(x) = p(a),

y = cos (1/x)

x→a

from which it follows that p is continuous at x = a. The rest of the theorem follows from Theorem 3.4 in a similar way. From these very basic continuous functions, we can build a large collection of continuous functions, using Theorem 4.2.

THEOREM 4.2 Suppose that f and g are continuous at x = a. Then all of the following are true: (i) ( f ± g) is continuous at x = a, (ii) ( f · g) is continuous at x = a and (iii) ( f /g) is continuous at x = a if g(a) = 0.

Simply put, Theorem 4.2 says that a sum, difference or product of continuous functions is continuous, while the quotient of two continuous functions is continuous at any point at which the denominator is nonzero.

1-29

SECTION 1.4

..

Continuity and Its Consequences

101

PROOF (i) If f and g are continuous at x = a, then lim [ f (x) ± g(x)] = lim f (x) ± lim g(x)

x→a

x→a

x→a

= f (a) ± g(a)

From Theorem 3.1. Since f and g are continuous at a.

= ( f ± g)(a), by the usual rules of limits. Thus, ( f ± g) is also continuous at x = a. Parts (ii) and (iii) are proved in a similar way and are left as exercises. y

EXAMPLE 4.4 150

50 5

x 4 − 3x 2 + 2 . x 2 − 3x − 4 Solution Here, f is a quotient of two polynomial (hence continuous) functions. The graph of the function indicated in Figure 1.26 suggests a vertical asymptote at around x = 4, but doesn’t indicate any other discontinuity. From Theorem 4.2, f will be continuous at all x where the denominator is not zero, that is, where Determine where f is continuous, for f (x) =

100

10

Continuity for a Rational Function

x 5 50

10

100 150

FIGURE 1.26 y=

x 4 − 3x 2 + 2 x 2 − 3x − 4

x 2 − 3x − 4 = (x + 1)(x − 4) = 0. Thus, f is continuous for x = −1, 4. (Think about why you didn’t see anything peculiar about the graph at x = −1.) With the addition of the result in Theorem 4.3, we will have all the basic tools needed to establish the continuity of most elementary functions.

THEOREM 4.3 Suppose that lim g(x) = L and f is continuous at L. Then, x→a lim f (g(x)) = f lim g(x) = f (L). x→a

x→a

A proof of Theorem 4.3 is given in Appendix A. Notice that this says that if f is continuous, then we can bring the limit “inside.” This should make sense, since as x → a, g(x) → L and so, f (g(x)) → f (L), since f is continuous at L.

COROLLARY 4.1 Suppose that g is continuous at a and f is continuous at g(a). Then, the composition f ◦ g is continuous at a.

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Limits and Continuity

1-30

PROOF From Theorem 4.3, we have

lim ( f ◦ g)(x) = lim f (g(x)) = f

x→a

x→a

lim g(x)

x→a

= f (g(a)) = ( f ◦ g)(a).

EXAMPLE 4.5

Since g is continuous at a.

Continuity for a Composite Function

Determine where h(x) = cos(x 2 − 5x + 2) is continuous. Solution Note that h(x) = f (g(x)), where g(x) = x 2 − 5x + 2 and f (x) = cos x. Since both f and g are continuous for all x, h is continuous for all x, by Corollary 4.1.

y

DEFINITION 4.2 If f is continuous at every point on an open interval (a, b), we say that f is continuous on (a, b). Following Figure 1.27, we say that f is continuous on the closed interval [a, b], if f is continuous on the open interval (a, b) and lim f (x) = f (a)

x→a +

x a

b

and

lim f (x) = f (b).

x→b−

Finally, if f is continuous on all of (−∞, ∞), we simply say that f is continuous. (That is, when we don’t specify an interval, we mean continuous everywhere.)

FIGURE 1.27 f continuous on [a, b]

For many functions, it’s a simple matter to determine the intervals on which the function is continuous. We illustrate this in example 4.6. y

EXAMPLE 4.6

Continuity on a Closed Interval

Determine the interval(s) where f is continuous, for f (x) =

x

4 − x2 > 0

2

FIGURE 1.28 y=

√ 4 − x2

4 − x 2.

Solution First, observe that f is defined only for −2 ≤ x ≤ 2. Next, note that f is the composition of two continuous functions and hence, is continuous for all x for which 4 − x 2 > 0. We show a graph of the function in Figure 1.28. Since

2

2

√

for −2 < x < 2, we have that f is continuous for all x in the interval (−2, 2), by Theorem 4.1 and Corollary 4.1. Finally, √ √ we test the endpoints to see that lim− 4 − x 2 = 0 = f (2) and lim + 4 − x 2 = 0 = f (−2), so that f is continuous x→2

x→−2

on the closed interval [−2, 2].

EXAMPLE 4.7

Interval of Continuity for a Logarithm

Determine the interval(s) where f (x) = ln(x − 3) is continuous. Solution It follows from Theorem 4.1 and Corollary 4.1 that f is continuous whenever (x − 3) > 0 (i.e., for x > 3). Thus, f is continuous on the interval (3, ∞).

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SECTION 1.4

..

Continuity and Its Consequences

103

The Internal Revenue Service presides over some of the most despised functions in existence. Look up the current Tax Rate Schedules. In 2002, the first few lines (for single taxpayers) looked like: For taxable amount over $0 $6000 $27,950

but not over $6000 $27,950 $ 67,700

your tax liability is 10% 15% 27%

minus $0 $300 $3654

Where do the numbers $300 and $3654 come from? If we write the tax liability T (x) as a function of the taxable amount x (assuming that x can be any real value and not just a whole dollar amount), we have if 0 < x ≤ 6000 0.10x if 6000 < x ≤ 27,950 T (x) = 0.15x − 300 0.27x − 3654 if 27,950 < x ≤ 67,700. Be sure you understand our translation so far. Note that it is important that this be a continuous function: think of the fairness issues that would arise if it were not!

EXAMPLE 4.8

Continuity of Federal Tax Tables

Verify that the federal tax rate function T (x) is continuous at the “joint” x = 27,950. Then, find a to complete the table. (You will find b and c as exercises.) For taxable amount over

but not over

your tax liability is

minus

$67,700 $141,250 $307,050

$141,250 $307,050 —

30% 35% 38.6%

a b c

Solution For T (x) to be continuous at x = 27,950, we must have lim

x→27,950−

T (x) =

lim

x→27,950+

T (x).

Since both functions 0.15x − 300 and 0.27x − 3654 are continuous, we can compute the one-sided limits by substituting x = 27,950. Thus, lim

x→27,950−

and

lim

x→27,950+

T (x) = 0.15(27,950) − 300 = 3892.50

T (x) = 0.27(27,950) − 3654 = 3892.50.

Since the one-sided limits agree and equal the value of the function at that point, T (x) is continuous at x = 27,950. We leave it as an exercise to establish that T (x) is also continuous at x = 6000. (It’s worth noting that the function could be written with equal signs on all of the inequalities; this would be incorrect if the function were discontinuous.) To complete the table, we choose a to get the one-sided limits at x = 67,700 to match. We have lim

x→67,700−

T (x) = 0.27(67,700) − 3654 = 14,625,

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while

lim

x→67,700+

T (x) = 0.30(67,700) − a = 20,310 − a.

So, we set the one-sided limits equal, to obtain 14,625 = 20,310 − a a = 20,310 − 14,625 = 5685.

or

Theorem 4.4 should seem an obvious consequence of our intuitive definition of continuity.

HISTORICAL NOTES

THEOREM 4.4 (Intermediate Value Theorem)

Karl Weierstrass (1815–1897) A German mathematician who proved the Intermediate Value Theorem and several other fundamental results of the calculus. Weierstrass was known as an excellent teacher whose students circulated his lecture notes throughout Europe, because of their clarity and originality. Also known as a superb fencer, Weierstrass was one of the founders of modern mathematical analysis.

Suppose that f is continuous on the closed interval [a, b] and W is any number between f (a) and f (b). Then, there is a number c ∈ [a, b] for which f (c) = W .

Theorem 4.4 says that if f is continuous on [a, b], then f must take on every value between f (a) and f (b) at least once. That is, a continuous function cannot skip over any numbers between its values at the two endpoints. To do so, the graph would need to leap across the horizontal line y = W , something that continuous functions cannot do (see Figure 1.29a). Of course, a function may take on a given value W more than once (see Figure 1.29b). We must point out that, although these graphs make this result seem reasonable, like any other result, Theorem 4.4 requires proof. The proof is more complicated than you might imagine and we must refer you to an advanced calculus text. y

y f (b)

f (b) W f (c)

yW

a

yW

x c

b

f (a)

a

x c1 c2

c3

b

f (a)

FIGURE 1.29a

FIGURE 1.29b

An illustration of the Intermediate Value Theorem

More than one value of c

In Corollary 4.2, we see an immediate and useful application of the Intermediate Value Theorem.

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Continuity and Its Consequences

105

y

COROLLARY 4.2 f(b)

Suppose that f is continuous on [a, b] and f (a) and f (b) have opposite signs [i.e., f (a) · f (b) < 0]. Then, there is at least one number c ∈ (a, b) for which f (c) = 0. (Recall that c is then a zero of f .) y f (x)

a

x c

b f(a)

Notice that Corollary 4.2 is simply the special case of the Intermediate Value Theorem where W = 0 (see Figure 1.30). The Intermediate Value Theorem and Corollary 4.2 are examples of existence theorems; they tell you that there exists a number c satisfying some condition, but they do not tell you what c is.

The Method of Bisections

FIGURE 1.30 Intermediate Value Theorem where c is a zero of f

In example 4.9, we see how Corollary 4.2 can help us locate the zeros of a function.

EXAMPLE 4.9

Finding Zeros by the Method of Bisections

Find the zeros of f (x) = x 5 + 4x 2 − 9x + 3. y 20 10

2

1

x 1

2

10 20

FIGURE 1.31 y = x 5 + 4x 2 − 9x + 3

Solution If f were a quadratic polynomial, you could certainly find its zeros. However, you don’t have any formulas for finding zeros of polynomials of degree 5. The only alternative is to approximate the zeros. A good starting place would be to draw a graph of y = f (x) like the one in Figure 1.31. There are three zeros visible on the graph. Since f is a polynomial, it is continuous everywhere and so, Corollary 4.2 says that there must be a zero on any interval on which the function changes sign. From the graph, you can see that there must be zeros between −3 and −2, between 0 and 1 and between 1 and 2. We could also conclude this by computing say, f (0) = 3 and f (1) = −1. Although we’ve now found intervals that contain zeros, the question remains as to how we can find the zeros themselves. While a rootfinding program can provide an accurate approximation, the issue here is not so much to get an answer as it is to understand how to find one. We suggest a simple yet effective method, called the method of bisections. For the zero between 0 and 1, a reasonable guess might be the midpoint, 0.5. Since f (0.5) ≈ −0.469 < 0 and f (0) = 3 > 0, there must be a zero between 0 and 0.5. Next, the midpoint of [0, 0.5] is 0.25 and f (0.25) ≈ 1.001 > 0, so that the zero is on the interval (0.25, 0.5). We continue in this way to narrow the interval on which there’s a zero until the interval is sufficiently small so that any point in the interval can serve as an adequate approximation to the actual zero. We do this in the following table. a

b

f(a)

f(b)

Midpoint

f (midpoint)

0 0 0.25 0.375 0.375 0.40625 0.40625 0.40625 0.40625

1 0.5 0.5 0.5 0.4375 0.4375 0.421875 0.4140625 0.41015625

3 3 1.001 0.195 0.195 0.015 0.015 0.015 0.015

−1 −0.469 −0.469 −0.469 −0.156 −0.156 −0.072 −0.029 −0.007

0.5 0.25 0.375 0.4375 0.40625 0.421875 0.4140625 0.41015625 0.408203125

−0.469 1.001 0.195 −0.156 0.015 −0.072 −0.029 −0.007 0.004

If you continue this process through 20 more steps, you ultimately arrive at the approximate zero x = 0.40892288, which is accurate to at least eight decimal places.

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This method of bisections is a tedious process, if you’re working it with pencil and paper. It is interesting because it’s reliable and it’s a simple, yet general method for finding approximate zeros. Computer and calculator rootfinding utilities are very useful, but our purpose here is to provide you with an understanding of how basic rootfinding works. We discuss a more powerful method for finding roots in Chapter 3.

EXERCISES 1.4 WRITING EXERCISES 1. Think about the following “real-life” functions, each of which is a function of the independent variable time: the height of a falling object, the velocity of an object, the amount of money in a bank account, the cholesterol level of a person, the heart rate of a person, the amount of a certain chemical present in a test tube and a machine’s most recent measurement of the cholesterol level of a person. Which of these are continuous functions? For each function you identify as discontinuous, what is the real-life meaning of the discontinuities? 2. Whether a process is continuous or not is not always clear-cut. When you watch television or a movie, the action seems to be continuous. This is an optical illusion, since both movies and television consist of individual “snapshots” that are played back at many frames per second. Where does the illusion of continuous motion come from? Given that the average person blinks several times per minute, is our perception of the world actually continuous? (In what cognitive psychologists call temporal binding, the human brain first decides whether a stimulus is important enough to merit conscious consideration. If so, the brain “predates” the stimulus so that the person correctly identifies when the stimulus actually occurred.) 3. When you sketch the graph of the parabola y = x 2 with pencil or pen, is your sketch (at the molecular level) actually the graph of a continuous function? Is your calculator or computer’s graph actually the graph of a continuous function? On many calculators, you have the option of a connected or disconnected graph. At the pixel level, does a connected graph show the graph of a function? Does a disconnected graph show the graph of a continuous function? Do we ever have problems correctly interpreting a graph due to these limitations? In the exercises in section 1.7, we examine one case where our perception of a computer graph depends on which choice is made.

In exercises 1–6, use the given graph to identify all discontinuities of the functions. y

1. 5

x 5

y

2. 5

x 5

y

3. 5

x 6

4. For each of the graphs in Figures 1.22a–1.22d, describe (with an example) what the formula for f (x) might look like to produce the given discontinuity.

1-35

..

SECTION 1.4

y

4.

107

In exercises 13–24, find all discontinuities of f (x). For each discontinuity that is removable, define a new function that removes the discontinuity.

5

x 5

x −1 x2 − 1 4x 15. f (x) = 2 x +4 17. f (x) = x 2 tan x

4x x2 + x − 2 3x 16. f (x) = 2 x − 2x − 4 18. f (x) = x cot x

13. f (x) =

14. f (x) =

19. f (x) = x ln x 2

20. f (x) = e−4/x sin x 22. f (x) = x 1

21. f (x) =

2x x2

2

if x < 1 if x ≥ 1

3x − 1 if x ≤ −1 23. f (x) = x 2 + 5x if −1 < x < 1 3 3x if x ≥ 1 if x < 0 2x if 0 < x ≤ π 24. f (x) = sin x x − π if x > π

y

5.

Continuity and Its Consequences

5

if x = 0 if x = 0

x 5

In exercises 25–32, determine the intervals on which f (x) is continuous. √ √ 25. f (x) = x + 3 26. f (x) = x 2 − 4 √ 28. f (x) = (x − 1)3/2 27. f (x) = 3 x + 2 1 29. f (x) = sin(x 2 + 2) 30. f (x) = cos x 31. f (x) = ln(x + 1) 32. f (x) = ln(4 − x 2 )

y

6. 5

x 5

In exercises 7–12, explain why each function is discontinuous at the given point by indicating which of the three conditions in Definition 4.1 are not met. x x2 − 1 7. f (x) = at x = 1 8. f (x) = at x = 1 x −1 x −1 1 9. f (x) = sin at x = 0 10. f (x) = e1/x at x = 0 x 2 if x < 2 x if x = 2 at x = 2 11. f (x) = 3 3x − 2 if x > 2 2 x if x < 2 12. f (x) = at x = 2 3x − 2 if x > 2

In exercises 33–35, determine values of a and b that make the given function continuous. 2 sin x if x < 0 33. f (x) = a x if x = 0 b cos x if x > 0 x if x < 0 ae +x1 −1 if 0 ≤ x ≤ 2 34. f (x) = sin 2 2 x − x + b if x > 2 a(tan−1 x + 2) if x < 0 if 0 ≤ x ≤ 3 35. f (x) = 2ebx + 1 ln(x − 2) + x 2 if x > 3 36. Prove Corollary 4.1. 37. Suppose that a state’s income tax code states that the tax liability on x dollars of taxable income is given by if x ≤ 0 0 if 0 < x < 10,000 T (x) = 0.14x c + 0.21x if 10,000 ≤ x. Determine the constant c that makes this function continuous for all x. Give a rationale why such a function should be continuous.

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38. Suppose a state’s income tax code states that tax liability is 12% on the first $20,000 of taxable earnings and 16% on the remainder. Find constants a and b for the tax function 0 if x ≤ 0 if 0 < x ≤ 20,000 T (x) = a + 0.12x b + 0.16(x − 20,000) if x > 20,000 such that T (x) is continuous for all x. 39. In example 4.8, find b and c to complete the table. 40. In example 4.8, show that T (x) is continuous for x = 6000. In exercises 41–46, use the Intermediate Value Theorem to verify that f (x) has a zero in the given interval. Then use the method of bisections to find an interval of length 1/32 that contains the zero. 41. f (x) = x 2 − 7, [2, 3] 42. f (x) = x 3 − 4x − 2, [2, 3]

44. f (x) = x 3 − 4x − 2, [−2, −1] 45. f (x) = cos x − x, [0, 1] 46. f (x) = e x + x, [−1, 0] A function is continuous from the right at x a if lim f (x) f (a). In exercises 47–50, determine whether f (x) x→a

is continuous from the right at x 2. x2 if x < 2 47. f (x) = 3x − 1 if x ≥ 2 2 if x < 2 x if x = 2 48. f (x) = 3 3x − 3 if x > 2 x2 if x ≤ 2 49. f (x) = 3x − 3 if x > 2 50. f (x) =

x2 3x − 2

for some function g(T ). Explain why it is reasonable that f (T ) be continuous. Determine a function g(T ) such that 0 ≤ g(T ) ≤ 100 for 30 ≤ T ≤ 34 and the resulting function f (T ) is continuous. [Hint: It may help to draw a graph first and make g(T ) linear.] x 2 , if x = 0 and g(x) = 2x, show that 54. If f (x) = 4, if x = 0 lim f (g(x)) = f lim g(x) . x→0

55. If you push on a large box resting on the ground, at first nothing will happen because of the static friction force that opposes motion. If you push hard enough, the box will start sliding, although there is again a friction force that opposes the motion. Suppose you are given the following description of the friction force. Up to 100 pounds, friction matches the force you apply to the box. Over 100 pounds, the box will move and the friction force will equal 80 pounds. Sketch a graph of friction as a function of your applied force based on this description. Where is this graph discontinuous? What is significant physically about this point? Do you think the friction force actually ought to be continuous? Modify the graph to make it continuous while still retaining most of the characteristics described. 400 , we have f (−1) > 0 and f (2) < 0. x Does the Intermediate Value Theorem guarantee a zero of f (x) between x = −1 and x = 2? What happens if you try the method of bisections?

56. For f (x) = 2x −

if x < 2 if x > 2

51. Define what it means for a function to be continuous from the left at x = a and determine which of the functions in exercises 47–50 are continuous from the left at x = 2. g(x) 52. Suppose that f (x) = and h(a) = 0. Determine whether h(x) each of the following statements is always true, always false, or maybe true/maybe false. Explain. (a) lim f (x) does not exist. (b) f (x) is discontinuous at x = a.

53. The sex of newborn Mississippi alligators is determined by the temperature of the eggs in the nest. The eggs fail to develop unless the temperature is between 26◦ C and 36◦ C. All eggs between 26◦ C and 30◦ C develop into females, and eggs between 34◦ C and 36◦ C develop into males. The percentage of females decreases from 100% at 30◦ C to 0% at 34◦ C. If f (T ) is the percentage of females developing from an egg at T ◦ C, then 100 if 26 ≤ T ≤ 30 f (T ) = g(T ) if 30 < T < 34 0 if 34 ≤ T ≤ 36,

x→0

43. f (x) = x 3 − 4x − 2, [−1, 0]

1-36

x→a

57. On Monday morning, a saleswoman leaves on a business trip at 7:13 A.M. and arrives at her destination at 2:03 P.M. The following morning, she leaves for home at 7:17 A.M. and arrives at 1:59 P.M. The woman notices that at a particular stoplight along the way, a nearby bank clock changes from 10:32 A.M. to 10:33 A.M. on both days. Therefore, she must have been at the same location at the same time on both days. Her boss doesn’t believe that such an unlikely coincidence could occur. Use the Intermediate Value Theorem to argue that it must be true that at some point on the trip, the saleswoman was at exactly the same place at the same time on both Monday and Tuesday. 58. Suppose you ease your car up to a stop sign at the top of a hill. Your car rolls back a couple of feet and then you drive through

1-37

SECTION 1.4

the intersection. A police officer pulls you over for not coming to a complete stop. Use the Intermediate Value Theorem to argue that there was an instant in time when your car was stopped (in fact, there were at least two). What is the difference between this stopping and the stopping that the police officer wanted to see? 59. Suppose a worker’s salary starts at $40,000 with $2000 raises every 3 months. Graph the salary function s(t); why is it discon2000 tinuous? How does the function f (t) = 40,000 + t (t in 3 months) compare? Why might it be easier to do calculations with f (t) than s(t)? 60. Prove the final two parts of Theorem 4.2. 61. Suppose that f (x) is a continuous function with consecutive zeros at x = a and x = b; that is, f (a) = f (b) = 0 and f (x) = 0 for a < x < b. Further, suppose that f (c) > 0 for some number c between a and b. Use the Intermediate Value Theorem to argue that f (x) > 0 for all a < x < b. 62. Use the method of bisections to estimate the other two zeros in example 4.9. 63. Suppose that f (x) is continuous at x = 0. Prove that lim x f (x) = 0. x→0

64. The converse of exercise 63 is not true. That is, the fact lim x f (x) = 0 does not guarantee that f (x) is continuous at x→0

x = 0. Find a counterexample; that is, find a function f such that lim x f (x) = 0 and f (x) is not continuous at x = 0. x→0

65. If f (x) is continuous at x = a, prove that g(x) = | f (x)| is continuous at x = a.

..

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109

x = 1. For the method of bisections, we guess the midpoint, x = 0.5. Is there any reason to suspect that the solution is actually closer to x = 0 than to x = 1? Using the function values f (0) = −1 and f (1) = 5, devise your own method of guessing the location of the solution. Generalize your method to using f (a) and f (b), where one function value is positive and one is negative. Compare your method to the method of bisections on the problem x 3 + 5x − 1 = 0; for both methods, stop when you are within 0.001 of the solution, x ≈ 0.198437. Which method performed better? Before you get overconfident in your method, compare the two methods again on x 3 + 5x 2 − 1 = 0. Does your method get close on the first try? See if you can determine graphically why your method works better on the first problem. 2. You have probably seen the turntables on which luggage rotates at the airport. Suppose that such a turntable has two long straight parts with a semicircle on each end (see the figure). We will model the left/right movement of the luggage. Suppose the straight part is 40 ft long, extending from x = −20 to x = 20. Assume that our luggage starts at time t = 0 at location x = −20, and that it takes 60 s for the luggage to reach x = 20. Suppose the radius of the circular portion is 5 ft and it takes the luggage 30 s to complete the half-circle. We model the straight-line motion with a linear function x(t) = at + b. Find constants a and b so that x(0) = −20 and x(60) = 20. For the circular motion, we use a cosine (Why is this a good choice?) x(t) = 20 + d · cos (et + f ) for constants d, e and f . The requirements are x(60) = 20 (since the motion is continuous), x(75) = 25 and x(90) = 20. Find values of d, e and f to make this work. Find equations for the position of the luggage along the backstretch and the other semicircle. What would the motion be from then on?

66. Determine whether the converse of exercise 65 is true. That is, if | f (x)| is continuous at x = a, is it necessarily true that f (x) must be continuous at x = a? 67. Let f (x) be a continuous function for x ≥ a and define h(x) = max f (t). Prove that h(x) is continuous for x ≥ a. a≤t≤x

Would this still be true without the assumption that f (x) is continuous? 68. Graph f (x) = tinuities.

sin |x 3 − 3x 2 + 2x| and determine all disconx 3 − 3x 2 + 2x Luggage carousel

EXPLORATORY EXERCISES 1. In the text, we discussed the use of the method of bisections to find an approximate solution of equations such as f(x) = x 3 + 5x −1 = 0. We can start by noticing that f (0) = −1 and f (1) = 5. Since f (x) is continuous, the Intermediate Value Theorem tells us that there is a solution between x = 0 and

3. Determine all x’s for which each function is continuous. 0 if x is irrational , f (x) = x if x is rational 2 x + 3 if x is irrational g(x) = 4x if x is rational and cos 4x if x is irrational . h(x) = sin 4x if x is rational

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LIMITS INVOLVING INFINITY; ASYMPTOTES y

10

f(x)

In this section, we revisit some old limit problems to give more informative answers and examine some related questions.

x x

3

x

3

EXAMPLE 5.1 Examine lim

x→0

f(x) 10

FIGURE 1.32 1 1 = ∞ and lim = −∞ lim x→0+ x x→0− x x

1 x

0.1 0.01 0.001 0.0001 0.00001

10 100 1000 10,000 100,000

x −0.1 −0.01 −0.001 −0.0001 −0.00001

A Simple Limit Revisited

1 . x

Solution Of course, we can draw a graph (see Figure 1.32) and compute a table of function values easily, by hand. (See the tables in the margin.) 1 1 While we say that the limits lim+ and lim− do not exist, the behavior of the x→0 x x→0 x 1 function is clearly quite different for x > 0 than for x < 0. Specifically, as x → 0+ , x 1 increases without bound, while as x → 0− , decreases without bound. To x communicate more about the behavior of the function near x = 0, we write 1 =∞ x

(5.1)

1 = −∞. x

(5.2)

lim+

x→0

and

1 x

lim

x→0−

1 approaches the vertical line x = 0, as x x → 0, as seen in Figure 1.32. When this occurs, we say that the line x = 0 is a vertical asymptote. It is important to note that while the limits (5.1) and (5.2) do not exist, we say that they “equal” ∞ and −∞, respectively, only to be specific as to why they do not exist. Finally, in view of the one-sided limits (5.1) and (5.2), we say that

Graphically, this says that the graph of y =

−10 −100 −1000 −10,000 −100,000

lim

x→0

REMARK 5.1 It may at first seem 1 x does not exist and then to write 1 lim = ∞. Note that since x→0+ x ∞ is not a real number, there is no contradiction here. (When we say that a limit “does not exist,” we are saying that there is no real number L that the function values are approaching.) We say that 1 lim = ∞ to indicate that as x→0+ x + x → 0 , the function values are increasing without bound. contradictory to say that lim

x→0+

EXAMPLE 5.2

1 does not exist. x

A Function Whose One-Sided Limits Are Both Infinite

1 Evaluate lim 2 . x→0 x Solution The graph (in Figure 1.33) seems to indicate a vertical asymptote at x = 0. A table of values is easily constructed by hand (see the accompanying tables). x

1 x2

x

1 x2

0.1 0.01 0.001 0.0001 0.00001

100 10,000 1 × 106 1 × 108 1 × 1010

−0.1 −0.01 −0.001 −0.0001 −0.00001

100 10,000 1 × 106 1 × 108 1 × 1010

1-39

..

SECTION 1.5

y f (x)

lim+

1 =∞ x2

lim

1 = ∞. x2

x→0

4

and

x→0−

Since both one-sided limits agree (i.e., both tend to ∞), we say that

2

lim

x

x→0

x

x

3

3

FIGURE 1.33 lim

111

From this, we can see that

f(x)

x→0

Limits Involving Infinity; Asymptotes

1 =∞ x2

1 = ∞. x2

This one concise statement says that the limit does not exist, but also that f (x) has a vertical asymptote at x = 0 where f (x) → ∞ as x → 0 from either side.

REMARK 5.2 Mathematicians try to convey as much information as possible with as few symbols as 1 1 possible. For instance, we prefer to say lim 2 = ∞ rather than lim 2 does not x→0 x x→0 x exist, since the first statement not only says that the limit does not exist, but also says 1 that 2 increases without bound as x approaches 0, with x > 0 or x < 0. x

y

EXAMPLE 5.3

10

f(x)

Evaluate lim

x→5

5 x 5

x

x 10

5 f(x)

10

1 = ∞ and (x − 5)3 1 lim = −∞ x→5− (x − 5)3

lim

1 . (x − 5)3

Solution In Figure 1.34, we show a graph of the function. From the graph, you should get a pretty clear idea that there’s a vertical asymptote at x = 5 and just how the function is blowing up there (to ∞ from the right side and to −∞ from the left). You can verify this behavior algebraically, by noticing that as x → 5, the denominator approaches 0, while the numerator approaches 1. This says that the fraction grows large in absolute value, without bound as x → 5. Specifically, as x → 5+ ,

FIGURE 1.34 x→5+

A Case Where Infinite One-Sided Limits Disagree

(x − 5)3 → 0

and

(x − 5)3 > 0.

We indicate the sign of each factor by printing a small “+” or “−” sign above or below each one. This enables you to see the signs of the various terms at a glance. In this case, we have +

1 lim = ∞. x→5+ (x − 5)3

Since (x − 5)3 > 0, for x > 5.

+

Likewise, as

x → 5− ,

(x − 5)3 → 0

and

(x − 5)3 < 0.

In this case, we have +

1 lim− = −∞. x→5 (x − 5)3

Since (x − 5)3 < 0, for x < 5.

−

Finally, we say that

1 does not exist, x→5 (x − 5)3 lim

since the one-sided limits are different.

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Learning from the lessons of examples 5.1, 5.2 and 5.3, you should recognize that if the denominator tends to 0 and the numerator does not, then the limit in question does not exist. In this event, we can determine whether the limit tends to ∞ or −∞ by carefully examining the signs of the various factors.

EXAMPLE 5.4 y

Evaluate lim

x→−2

10 f (x) 5 x

x

4

x

Another Case Where Infinite One-Sided Limits Disagree

x +1 . (x − 3)(x + 2)

Solution First, notice from the graph of the function shown in Figure 1.35 that there appears to be a vertical asymptote at x = −2. Further, the function appears to tend to ∞ as x → −2+ , and to −∞ as x → −2− . You can verify this behavior, by observing that −

4

x +1 lim =∞ x→−2+ (x − 3) (x + 2)

5

−

f (x)

10

+

Since (x + 1) < 0, (x − 3) < 0 and (x + 2) > 0, for −2 < x < −1.

−

and

FIGURE 1.35 x +1 does not exist. lim x→−2 (x − 3)(x + 2)

x +1 lim = −∞. x→−2− (x − 3) (x + 2) −

−

Since (x + 1) < 0, (x − 3) < 0 and (x + 2) < 0, for x < −2.

So, we can see that x = −2 is indeed a vertical asymptote and that x +1 does not exist. x→−2 (x − 3)(x + 2) lim

EXAMPLE 5.5 y

A Limit Involving a Trigonometric Function

Evaluate limπ tan x. x→ 2

Solution Notice from the graph of the function shown in Figure 1.36 that there appears π to be a vertical asymptote at x = . 2 You can verify this behavior by observing that +

q

q

p

sin x limπ − tan x = limπ − =∞ cos x x→ 2 x→ 2

x

+

w

+

and

limπ + tan x = limπ +

x→ 2

FIGURE 1.36 y = tan x

x→ 2

sin x = −∞. cos x −

Since sin x > 0 and cos x > 0 π for 0 < x < . 2 Since sin x > 0 and cos x < 0 π < x < π. for 2

π So, we can see that x = is indeed a vertical asymptote and that 2 limπ tan x does not exist. x→ 2

Limits at Infinity We are also interested in examining the limiting behavior of functions as x increases without bound (written x → ∞) or as x decreases without bound (written x → −∞).

1-41

SECTION 1.5

y

Returning to f (x) =

10 f (x)

3

x

3

f(x)

1 1 = 0 and lim =0 x→−∞ x x

lim

x→−∞

1 = 0. x

1 . x

1 Solution We show a graph of y = f (x) in Figure 1.38. Since as x → ±∞, → 0, x we get that 1 lim 2 − =2 x→∞ x 1 and lim 2 − = 2. x→−∞ x

8 6 4 x

Thus, the line y = 2 is a horizontal asymptote.

x f (x) x

2

Finding Horizontal Asymptotes

Look for any horizontal asymptotes of f (x) = 2 −

y

f(x)

Similarly,

EXAMPLE 5.6

FIGURE 1.37 lim

113

Notice that in Figure 1.37, the graph appears to approach the horizontal line y = 0, as x → ∞ and as x → −∞. In this case, we call y = 0 a horizontal asymptote.

10

x→∞

Limits Involving Infinity; Asymptotes

1 1 , we can see that as x → ∞, → 0. In view of this, we write x x 1 lim = 0. x→∞ x

x x

..

2 2

1 , for any positive rational power t, xt 1 as x → ±∞, is largely the same as we observed for f (x) = . x As you can see in Theorem 5.1, the behavior of

FIGURE 1.38

1 = 2 and lim 2 − x→∞ x 1 lim 2 − =2 x→−∞ x

THEOREM 5.1 For any rational number t > 0, lim

x→±∞

1 = 0, xt

where for the case where x → −∞, we assume that t =

REMARK 5.3 All of the usual rules for limits stated in Theorem 3.1 also hold for limits as x → ±∞.

p where q is odd. q

A proof of Theorem 5.1 is given in Appendix A. Be sure that the following argument 1 makes sense to you: for t > 0, as x → ∞, we also have x t → ∞, so that t → 0. x In Theorem 5.2, we see that the behavior of a polynomial at infinity is easy to determine.

THEOREM 5.2 For a polynomial of degree n > 0, pn (x) = an x n + an−1 x n−1 + · · · + a0 , we have ∞, if an > 0 lim pn (x) = . −∞, if an < 0 x→∞

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PROOF lim pn (x) = lim (an x n + an−1 x n−1 + · · · + a0 ) x→∞ an−1 a0 = lim x n an + + ··· + n x→∞ x x = ∞, an−1 a0 if an > 0, since lim an + + · · · + n = an x→∞ x x and lim x n = ∞. The result is proved similarly for an < 0. We have

x→∞

x→∞

Observe that you can make similar statements regarding the value of lim pn (x), but x→−∞

be careful: the answer will change depending on whether n is even or odd. (We leave this as an exercise.) In example 5.7, we again see the need for caution when applying our basic rules for limits (Theorem 3.1), which also apply to limits as x → ∞ or as x → −∞.

y

EXAMPLE 5.7

4 x x

10

10 f (x) 4

A Limit of a Quotient That Is Not the Quotient of the Limits

5x − 7 . 4x + 3 Solution You might be tempted to write lim (5x − 7) 5x − 7 x→∞ lim = x→∞ 4x + 3 lim (4x + 3)

Evaluate lim

x→∞

x→∞

FIGURE 1.39 lim

x→∞

5 5x − 7 = 4x + 3 4

x

5x − 7 4x 3

10 100 1000 10,000 100,000

1 1.223325 1.247315 1.249731 1.249973

This is an incorrect use of Theorem 3.1 (iv), since the limits in the numerator and the denominator do not exist.

∞ = This is incorrect! (5.3) = 1. ∞ The graph in Figure 1.39 and some function values (see the accompanying table) suggest that the conjectured value of 1 is incorrect. Recall that the limit of a quotient is the quotient of the limits only when both limits exist (and the limit in the denominator is nonzero). Since both the limit in the denominator and that in the numerator tend to ∞, the limits do not exist. Further, when a limit looks like ∞ , the actual value of the limit can be anything at ∞ all. For this reason, we call ∞ an indeterminate form, meaning that the value of the ∞ expression cannot be determined solely by noticing that both numerator and denominator tend to ∞. Rule of Thumb: When faced with the indeterminate form ∞ in calculating the ∞ limit of a rational function, divide numerator and denominator by the highest power of x appearing in the denominator. Here, we have Multiply numerator and 5x − 7 5x − 7 (1/x) lim = lim · 1 denominator by . x→∞ 4x + 3 x→∞ 4x + 3 (1/x) x 5 − 7/x 4 + 3/x lim (5 − 7/x)

= lim

x→∞

=

x→∞

lim (4 + 3/x)

Multiply through by

1 . x

By Theorem 3.1 (iv).

x→∞

5 = 1.25, 4 which is consistent with what we observed both graphically and numerically earlier. =

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115

In example 5.8, we apply our rule of thumb to a common limit problem.

EXAMPLE 5.8

Finding Slant Asymptotes

4x + 5 and find any slant asymptotes. x→∞ −6x 2 − 7x Solution As usual, we first examine a graph (see Figure 1.40a). Note that here, the graph appears to tend to −∞ as x → ∞. Further, observe that outside of the interval [−2, 2], the graph looks very much like a straight line. If we look at the graph in a somewhat larger window, this linearity is even more apparent (see Figure 1.40b). 3

Evaluate lim

y

y

20

6

x

6

6

x

20

20

20

6

FIGURE 1.40a 4x 3 + 5 y= −6x 2 − 7x

FIGURE 1.40b y=

Using our rule of thumb, we have 4x 3 + 5 (1/x 2 ) 4x 3 + 5 lim = lim · x→∞ −6x 2 − 7x x→∞ −6x 2 − 7x (1/x 2 ) 4x + 5/x 2 x→∞ −6 − 7/x = −∞, = lim

4x 3 + 5 −6x 2 − 7x

Multiply numerator and 1 denominator by 2 . x Multiply through by

1 . x2

since as x → ∞, the numerator tends to ∞ and the denominator tends to −6. To further explain the behavior seen in Figure 1.40b, we perform a long division. We have 4x 3 + 5 2 7 5 + 49/9x =− x+ + . −6x 2 − 7x 3 9 −6x 2 − 7x Since the third term in this expansion tends to 0 as x → ∞, the function values approach those of the linear function 2 7 − x+ , 3 9 as x → ∞. For this reason, we say that the function has a slant (or oblique) asymptote. That is, instead of approaching a vertical or horizontal line, as happens with vertical or horizontal asymptotes, the graph is approaching the slanted straight line 2 7 y = − x + . (This is the behavior we’re seeing in Figure 1.40b.) 3 9 Limits involving exponential functions are very important in many applications.

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1-44

y

EXAMPLE 5.9 10

Two Limits of an Exponential Function

Evaluate lim− e1/x and lim+ e1/x . x→0

5

10

x

5

5

10

5

x→0

Solution A computer-generated graph is shown in Figure 1.41a. Although it is an unusual looking graph, it appears that the function values are approaching 0, as x approaches 0 from the left and tend to infinity as x approaches 0 from the right. To 1 verify this, recall that lim− = −∞ and lim e x = 0. (See Figure 1.41b for a graph x→−∞ x→0 x of y = e x .) Combining these results, we get

FIGURE 1.41a

lim e1/x = 0.

y = e1/x

x→0−

y

Similarly, lim+ x→0

30

we have

1 = ∞ and lim e x = ∞. (Again, see Figure 1.41b.) In this case, x→∞ x lim e1/x = ∞.

20

x→0+

10 4

As we see in example 5.10, inverse trigonometric functions may have horizontal asymptotes.

x

2

2

4

FIGURE 1.41b

EXAMPLE 5.10

y = ex

Evaluate lim tan−1 x and lim tan−1 x.

y

x→∞

1

4

Two Limits of an Inverse Trigonometric Function

x

2

2

4

1

FIGURE 1.42a y = tan−1 x y

x→−∞

Solution The graph of y = tan−1 x (shown in Figure 1.42a) suggests a horizontal asymptote of about y = −1.5 as x → −∞ and about y = 1.5 as x → ∞. We can be more precise with this, as follows. For lim tan−1 x, we are looking for the angle that θ x→∞ π π must approach, with − < θ < , such that tan θ tends to ∞. Referring to the graph 2 2 π− of y = tan x in Figure 1.42b, we see that tan x tends to ∞ as x approaches . 2 π+ Likewise, tan x tends to −∞ as x approaches − , so that 2 π π lim tan−1 x = and lim tan−1 x = − . x→∞ x→−∞ 2 2 In example 5.11, we consider a model of the size of an animal’s pupils. Recall that in bright light, pupils shrink to reduce the amount of light entering the eye, while in dim light, pupils dilate to allow in more light. (See the chapter introduction.)

−π 2

π 2

x

EXAMPLE 5.11

Finding the Size of an Animal’s Pupils

Suppose that the diameter of an animal’s pupils is given by f (x) mm, where x is the 160x −0.4 + 90 intensity of light on the pupils. If f (x) = , find the diameter of the pupils 4x −0.4 + 15 with (a) minimum light and (b) maximum light. FIGURE 1.42b y = tan x

Solution For part (a), notice that f (0) is undefined, since 0−0.4 indicates a division by 0. We therefore consider the limit of f (x) as x approaches 0, but we compute a

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SECTION 1.5

y

..

Limits Involving Infinity; Asymptotes

117

one-sided limit since x cannot be negative. A computer-generated graph of y = f (x) with 0 ≤ x ≤ 10 is shown in Figure 1.43a. It appears that the y-values approach 20 as x approaches 0. To compute the limit, we multiply numerator and denominator by x 0.4 (to eliminate the negative exponents). We then have

20 15

lim+

10

x→0

160x −0.4 + 90 160x −0.4 + 90 x 0.4 = lim+ · −0.4 x→0 4x + 15 4x −0.4 + 15 x 0.4 = lim+

5

x→0

x 2

4

6

8

10

FIGURE 1.43a y = f (x) y

160 + 90x 0.4 160 = = 40 mm. 4 + 15x 0.4 4

This limit does not seem to match our graph, but notice that Figure 1.43a shows a gap near x = 0. In Figure 1.43b, we have zoomed in so that 0 ≤ x ≤ 0.1. Here, a limit of 40 looks more reasonable. For part (b), we consider the limit as x tends to ∞. From Figure 1.43a, it appears that the graph has a horizontal asymptote at a value close to y = 10. We compute the limit

40

160x −0.4 + 90 90 = = 6 mm. x→∞ 4x −0.4 + 15 15 lim

30

So, the pupils have a limiting size of 6 mm, as the intensity of light tends to ∞.

20

In our final example, we consider the velocity of a falling object.

10 x 0.02 0.04 0.06 0.08 0.1

FIGURE 1.43b

EXAMPLE 5.12

Finding the Limiting Velocity of a Falling Object

The velocity in ft/s of a falling object is modeled by

y = f (x)

√

32 1 − e−2t 32k √ v(t) = − , k 1 + e−2t 32k where k is a constant that depends upon the size and shape of the object and the density of the air. Find the limiting velocity of the object, that is, find lim v(t) and compare limiting t→∞ velocities for skydivers with k = 0.00016 (head first) and k = 0.001 (spread eagle). Solution Observe that the only place√that t appears in the expression for v(t) is in the √ −2t 32k −2t 32k two identical exponential terms: e . Also notice that lim e = 0, since lim e x = 0. We then have x→−∞

t→∞

√ 32 1 − e−2t 32k √ lim v(t) = lim − t→∞ t→∞ k 1 + e−2t 32k 1 − lim e−2t √32k 32 32 32 1 − 0 t→∞ =− √ =− =− ft/s, k k 1+0 k 1 + lim e−2t 32k

t→∞

where the negative sign indicates a downward direction. So, with k = 0.00016, the 32 limiting velocity is − 0.00016 ≈ −447 ft/s (about 300 mph!), and with k = 0.001, the 32 limiting velocity is − 0.001 ≈ −179 ft/s (about 122 mph).

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EXERCISES 1.5 WRITING EXERCISES 1. It may seem odd that we use ∞ in describing limits but do not count ∞ as a real number. Discuss the existence of ∞: is it a number or a concept? 2. In example 5.7, we dealt with the “indeterminate form” ∞ . ∞ Thinking of a limit of ∞ as meaning “getting very large” and a limit of 0 as meaning “getting very close to 0,” explain why the following are indeterminate forms: ∞ , 0 , ∞ − ∞, and ∞ 0 ∞ · 0. Determine what the following non-indeterminate forms represent: ∞ + ∞, −∞ − ∞, ∞ + 0 and 0/∞. 3. On your computer or calculator, graph y = 1/(x − 2) and look for the horizontal asymptote y = 0 and the vertical asymptote x = 2. Most computers will draw a vertical line at x = 2 and will show the graph completely flattening out at y = 0 for large x’s. Is this accurate? misleading? Most computers will compute the locations of points for adjacent x’s and try to connect the points with a line segment. Why might this result in a vertical line at the location of a vertical asymptote?

In exercises 5–24, determine each limit (answer as appropriate, with a number, ∞ , − ∞ or does not exist). −x −2/3 5. lim √ 6. lim (x 2 − 2x − 3) x→2− x→−1− 4 − x2 −x 2x 2 − x + 1 7. lim √ 8. lim x→−∞ x→∞ 4x 2 − 3x − 1 4 + x2 x3 − 2 2x 2 − 1 9. lim 10. lim x→∞ 3x 2 + 4x − 1 x→∞ 4x 3 − 5x − 1 12. lim ln 2x 11. lim ln 2x x→0+

x→∞

13. lim e

−2/x

14. lim e−2/x

x→0+

x→∞

−1

15. lim cot x→∞

17. lim e

16. lim sec−1 x

x

x→∞

2x−1

18. lim e1/x

x→∞

20. lim (e−3x cos 2x)

19. lim sin 2x x→∞

x→∞

ln(2 + e ) ln(1 + e x ) 23. lim e− tan x 3x

21. lim

x→∞

4. Many students learn that asymptotes are lines that the graph gets closer and closer to without ever reaching. This is true for many asymptotes, but not all. Explain why vertical asymptotes are never reached or crossed. Explain why horizontal or slant asymptotes may, in fact, be crossed any number of times; draw one example.

In exercises 1–4, determine each limit (answer as appropriate, with a number, ∞ , − ∞ or does not exist). 1. (a) lim

x→1−

(c) lim

x→1

2. (a)

lim

x→−1−

x→−1

3. (a) lim

x→2−

(c) lim

x→2

x→1+

1 − 2x x2 − 1

1 − 2x x2 − 1

x −4 x 2 − 4x + 4

(b)

lim

x→−1+

1 − 2x x2 − 1

(b) lim

x→2+

x −4 x 2 − 4x + 4

x −4 x 2 − 4x + 4

lim

x→−1

1−x (x + 1)2

1−x (x + 1)2

x→π/2

lim

x→−1+

1−x (x + 1)2

x→∞

24. lim tan−1 (ln x) x→0+

34. f (x) = 3e−1/x

In exercises 35–38, determine all vertical and slant asymptotes. x3 4 − x2 x3 37. y = 2 x +x −4 35. y =

x2 + 1 x −2 x4 38. y = 3 x +2 36. y =

In exercises 39–48, use graphical and numerical evidence to conjecture a value for the indicated limit. x2 x→∞ 2x x 2 − 4x + 7 41. lim x→∞ 2x 2 + x cos x x 3 + 4x + 5 43. lim x→∞ e x/2 39. lim

(b)

22. lim sin(tan−1 x)

In exercises 25–34, determine all horizontal and vertical asymptotes. For each vertical asymptote, determine whether f (x) → ∞ or f (x) → − ∞ on either side of the asymptote. x x 26. f (x) = √ 25. f (x) = √ 4 + x2 4 − x2 x x2 27. f (x) = 28. f (x) = 2 4−x 4 − x2 2 1−x 3x + 1 30. f (x) = 2 29. f (x) = 2 x − 2x − 3 x +x −2 ln(x + 2) 31. f (x) = ln(1 − cos x) 32. f (x) = ln(x 2 + 3x + 3) 33. f (x) = 4 tan−1 x − 1

1 − 2x x2 − 1

x→−1−

(c) lim

(b) lim

1 − 2x x2 − 1

(c) lim

4. (a)

1 − 2x x2 − 1

2

x→0

2x x→∞ x 2 2x 3 + 7x 2 + 1 42. lim x→−∞ x 3 − x sin x

40. lim

44. lim (e x/3 − x 4 ) x→∞

1-47

SECTION 1.5

x − cos (π x) x→−1 x +1 x 47. lim x→0+ cos x − 1

ex − 1 x→0 x ln x 2 48. lim 2 x→0 x

45. lim

46. lim

In exercises 49–52, use graphical and numerical evidence to conjecture the value of the limit. Then, verify your conjecture by finding the limit exactly. 49. lim ( 4x 2 − 2x + 1 − 2x) (Hint: Multiply and divide by the x→∞ √ conjugate expression: 4x 2 − 2x + 1 + 2x and simplify.) 50. lim ( x 2 + 3 − x) (See the hint for exercise 49.) x→∞ 51. lim ( 5x 2 + 4x + 7 − 5x 2 + x + 3) (See the hint for x→∞

exercise 49.) 3 2x 1 x . Hint : e = lim 1 + 52. lim 1 + x→−∞ x→∞ x x 53. Sketch the graph of f (x) = e−x cos x. Identify the horizontal asymptote. Is this asymptote approached in both directions (as x → ∞ and x → −∞)? How many times does the graph cross the horizontal asymptote? 54. Explain why it is reasonable that lim f (x) = lim f (1/x) and x→∞

lim f (x) = lim f (1/x).

x→−∞

x→0+

x→0−

55. In the exercises of section 1.2, we found that lim (1 + x)1/x = x→0+

lim (1 + x)1/x . (It turns out that both limits equal the irra-

x→0−

tional number e.) Use this result and exercise 54 to argue that lim (1 + 1/x)x = lim (1 + 1/x)x . x→∞

x→−∞

56. One of the reasons for saying that infinite limits do not exist is that we would otherwise invalidate Theorem 3.1 in section 1.3. Find examples of functions with infinite limits such that parts (ii) and (iv) of Theorem 3.1 do not hold. 57. Suppose that the length of a small animal t days after birth 300 is h(t) = mm. What is the length of the animal at 1 + 9(0.8)t birth? What is the eventual length of the animal (i.e., the length as t → ∞)? 58. Suppose that the length of a small animal t days after birth 100 mm. What is the length of the animal at is h(t) = 2 + 3(0.4)t birth? What is the eventual length of the animal (i.e., the length as t → ∞)? 59. Suppose that the size of the pupil of a certain animal is given by f (x) (mm), where x is the intensity of the light on the pupil. 80x −0.3 + 60 If f (x) = , find the size of the pupil with no light 2x −0.3 + 5 and the size of the pupil with an infinite amount of light. 60. Repeat exercise 59 with f (x) =

80x −0.3 + 60 . 8x −0.3 + 15

61. Modify the functions in exercises 59 and 60 to find a function f such that lim f (x) = 8 and lim f (x) = 2. x→0+

x→∞

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Limits Involving Infinity; Asymptotes

119

62. After an injection, the concentration of a drug in a muscle varies according to a function of time f (t). Suppose that t is measured in hours and f (t) = e−0.02t − e−0.42t . Find the limit of f (t) both as t → 0 and t → ∞, and interpret both limits in terms of the concentration of the drug. 63. Suppose an object with initial velocity v0 = 0 ft/s and (constant) mass m slugs is accelerated by a constant force F pounds for t seconds. According to Newton’s laws of motion, the object’s speed will be v N = Ft/m. According to Einstein’s √ theory of relativity, the object’s speed will be v E = Fct/ m 2 c2 + F 2 t 2 , where c is the speed of light. Compute lim v N and lim v E . t→∞

t→∞

64. According to Einstein’s theory of relativity, themass of an object traveling at speed v is given by m = m 0 / 1 − v 2 /c2 , where c is the speed of light (about 9.8 × 108 ft/s). Compute lim m and explain why m 0 is called the “rest mass.” Compute v→0

lim m and discuss the implications. (What would happen if

v→c−

you were traveling in a spaceship approaching the speed of light?) How much does the mass of a 192-pound man (m 0 = 6) increase at the speed of 9000 ft/s (about 4 times the speed of sound)? 65. In example 5.12, the velocity t seconds after jump√ of a skydiver 32 1 − e−2t 32k √ . Find the limiting ing is given by v(t) = − k 1 + e−2t 32k velocity with k = 0.00064 and k = 0.00128. By what factor does a skydiver have to change the value of k to cut the limiting velocity in half? 66. Graph the velocity function in exercise 65 with k = 0.00016 (representing a headfirst dive) and estimate how long it takes for the diver to reach a speed equal to 90% of the limiting velocity. Repeat with k = 0.001 (representing a spread-eagle position). 67. Ignoring air resistance, the maximum height reached by a v02 R rocket launched with initial velocity v0 is h = m/s, 19.6R − v02 where R is the radius of the earth. In this exercise, we interpret this as a function of v0 . Explain why the domain of this function must be restricted to v0 ≥ 0. There is an additional restriction. Find the (positive) value ve such that h is undefined. Sketch a possible graph of h with 0 ≤ v0 < ve and discuss the significance of the vertical asymptote at ve . (Explain what would happen to the rocket if it is launched with initial velocity ve .) Explain why ve is called the escape velocity. p(x) with the q(x) degree (largest exponent) of p(x) less than the degree of q(x). Determine the horizontal asymptote of y = f (x).

68. Suppose that f (x) is a rational function f (x) =

p(x) with q(x) the degree of p(x) greater than the degree of q(x). Determine whether y = f (x) has a horizontal asymptote.

69. Suppose that f (x) is a rational function f (x) =

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1-48

p(x) . If q(x) y = f (x) has a horizontal asymptote y = 2, how does the degree of p(x) compare to the degree of q(x)? p(x) Suppose that f (x) is a rational function f (x) = . If q(x) y = f (x) has a slant asymptote y = x + 2, how does the degree of p(x) compare to the degree of q(x)? x2 − 4 Find a quadratic function q(x) such that f (x) = has q(x) one horizontal asymptote y = 2 and two vertical asymptotes x = ±3. x2 − 4 Find a quadratic function q(x) such that f (x) = has q(x) 1 one horizontal asymptote y = − 2 and exactly one vertical asymptote x = 3. x −4 has two horizonFind a function g(x) such that f (x) = g(x) tal asymptotes y = ±1 and no vertical asymptotes.

70. Suppose that f (x) is a rational function f (x) =

71.

72.

73.

74.

In exercises 75–80, label the statement as true or false (not always true) for real numbers a and b. 75. If lim f (x) = a and lim g(x) = b, then x→∞

x→∞

lim [ f (x) + g(x)] = a + b.

x→∞

76. If lim f (x) = a and lim g(x) = b, then lim x→∞

x→∞

x→∞

f (x) a = . g(x) b

77. If lim f (x) = ∞ and lim g(x) = ∞, then x→∞

x→∞

lim [ f (x) − g(x)] = 0.

x→∞

78. If lim f (x) = ∞ and lim g(x) = ∞, then x→∞

x→∞

lim [ f (x) + g(x)] = ∞.

f (x) = 0. x→∞ x→∞ x→∞ g(x) f (x) = 1. 80. If lim f (x) = ∞ and lim g(x) = ∞, then lim x→∞ x→∞ x→∞ g(x) x→∞

79. If lim f (x) = a and lim g(x) = ∞, then lim

In exercises 81 and 82, determine all vertical and horizontal asymptotes. 4x if x < 0 x −4 2 x 81. f (x) = if 0 ≤ x < 4 x −2 −x e if x ≥ 4 x +1 x +3 if x < 0 x 2 − 4x x if 0 ≤ x < 2 82. f (x) = e + 1 2 x − 1 if x ≥ 2 x 2 − 7x + 10 83. Explain why lim (e−at sin t) = 0 for any positive constant a. t→∞

Although this is theoretically true, it is not necessarily useful in practice. The function e−at sin t is a simple model for a

spring-mass system, such as the suspension system on a car. Suppose t is measured in seconds and the car passengers cannot feel any vibrations less than 0.01 (inches). If suspension system A has the vibration function e−t sin t and suspension system B has the vibration function e−t/4 sin t, determine graphically how long it will take before the vibrations damp out, that is, | f (t)| < 0.01. Is the result lim (e−at sin t) = 0 much consolat→∞

tion to the owner of car B? 84. (a) State and prove a result analogous to Theorem 5.2 for lim pn (x), for n odd. x→−∞

(b) State and prove a result analogous to Theorem 5.2 for lim pn (x), for n even. x→−∞

85. It is very difficult to find simple statements in calculus that are always true; this is one reason that a careful development of the theory is so important. You may have heard the simple rule: g(x) , simply set the to find the vertical asymptotes of f (x) = h(x) denominator equal to 0 [i.e., solve h(x) = 0]. Give an example where h(a) = 0 but there is not a vertical asymptote at x = a. 86. In exercise 85, you needed to find an example indicating that the following statement is not (necessarily) true: if h(a) = 0, g(x) has a vertical asymptote at x = a. This is then f (x) = h(x) g(x) has not true, but perhaps its converse is true: if f (x) = h(x) a vertical asymptote at x = a, then h(a) = 0. Is this statement true? What if g and h are polynomials? In exercises 87–90, use numerical evidence to conjecture a decimal representation for the limit. Check your answer with your computer algebra system (CAS); if your answers disagree, which one is correct? 88. lim (ln x)x x→1+ ln x 90. lim x→−∞ x 2

87. lim x 1/ ln x x→0+

89. lim x 1/x x→∞

2 −1

EXPLORATORY EXERCISES 1. Suppose you are shooting a basketball from a (horizontal) distance of L feet, releasing the ball from a location h feet below the basket. To get a perfect swish, it is necessary that the initial velocity v0 and initial release angle θ0 satisfy the equation h

u0 10

√

L

v0 = gL/ 2 cos2 θ0 (tan θ0 − h/L). For a free throw, take L = 15, h = 2 and g = 32 and graph v0 as a function of θ0 .

1-49

SECTION 1.6

What is the significance of the two vertical asymptotes? Explain in physical terms what type of shot corresponds to each vertical asymptote. Estimate the minimum value of v0 (call it vmin ). Explain why it is easier to shoot a ball with a small initial velocity. There is another advantage to this initial velocity. Assume that the basket is 2 ft in diameter and the ball is 1 ft in diameter. For a free throw, L = 15 ft is perfect. What is the maximum horizontal distance the ball could travel and still go in the basket (without bouncing off the backboard)? What is the minimum horizontal distance? Call these numbers L max and L min . Find the angle θ1 corresponding to vmin and L min and the angle θ2 corresponding to vmin and L max . The difference |θ2 − θ1 | is the angular margin of error. Brancazio has shown that the angular margin of error for vmin is larger than for any other initial velocity.

1.6

..

Formal Definition of the Limit

121

2. In applications, it is common to compute lim f (x) to dex→∞

termine the “stability” of the function f (x). Consider the function f (x) = xe−x . As x → ∞, the first factor in f (x) goes to ∞, but the second factor goes to 0. What does the product do when one term is getting smaller and the other term is getting larger? It depends on which one is changing faster. What we want to know is which term “dominates.” Use graphical and numerical evidence to conjecture the value of lim (xe−x ). Which term dominates? In the limit x→∞

lim (x 2 e−x ), which term dominates? Also, try lim (x 5 e−x ).

x→∞

x→∞

Based on your investigation, is it always true that exponentials dominate polynomials? Are you positive? Try to determine which type of function, polynomials or logarithms, dominates.

FORMAL DEFINITION OF THE LIMIT We have now spent many pages discussing various aspects of the computation of limits. This may seem a bit odd, when you realize that we have never actually defined what a limit is. Oh, sure, we have given you an idea of what a limit is, but that’s about all. Once again, we have said that lim f (x) = L ,

x→a

HISTORICAL NOTES Augustin Louis Cauchy (1789–1857) A French mathematician who developed the ε-δ definitions of limit and continuity. Cauchy was one of the most prolific mathematicians in history, making important contributions to number theory, linear algebra, differential equations, astronomy, optics and complex variables. A difficult man to get along with, a colleague wrote, “Cauchy is mad and there is nothing that can be done about him, although right now, he is the only one who knows how mathematics should be done.”

if f (x) gets closer and closer to L as x gets closer and closer to a. So far, we have been quite happy with this somewhat vague, although intuitive, description. In this section, however, we will make this more precise, and you will begin to see how mathematical analysis (that branch of mathematics of which the calculus is the most elementary study) works. Studying more advanced mathematics without an understanding of the precise definition of limit is somewhat akin to studying brain surgery without bothering with all that background work in chemistry and biology. In medicine, it has only been through a careful examination of the microscopic world that a deeper understanding of our own macroscopic world has developed, and good surgeons need to understand what they are doing and why they are doing it. Likewise, in mathematical analysis, it is through an understanding of the microscopic behavior of functions (such as the precise definition of limit) that a deeper understanding of the mathematics will come about. We begin with the careful examination of an elementary example. You should certainly believe that lim (3x + 4) = 10.

x→2

Suppose that you were asked to explain the meaning of this particular limit to a fellow student. You would probably repeat the intuitive explanation we have used so far: that as x gets closer and closer to 2, (3x + 4) gets arbitrarily close to 10. But, exactly what do we mean by close? One answer is that if lim (3x + 4) = 10, we should be able to make x→2

(3x + 4) as close as we like to 10, just by making x sufficiently close to 2. But can we

1-49

SECTION 1.6

What is the significance of the two vertical asymptotes? Explain in physical terms what type of shot corresponds to each vertical asymptote. Estimate the minimum value of v0 (call it vmin ). Explain why it is easier to shoot a ball with a small initial velocity. There is another advantage to this initial velocity. Assume that the basket is 2 ft in diameter and the ball is 1 ft in diameter. For a free throw, L = 15 ft is perfect. What is the maximum horizontal distance the ball could travel and still go in the basket (without bouncing off the backboard)? What is the minimum horizontal distance? Call these numbers L max and L min . Find the angle θ1 corresponding to vmin and L min and the angle θ2 corresponding to vmin and L max . The difference |θ2 − θ1 | is the angular margin of error. Brancazio has shown that the angular margin of error for vmin is larger than for any other initial velocity.

1.6

..

Formal Definition of the Limit

121

2. In applications, it is common to compute lim f (x) to dex→∞

termine the “stability” of the function f (x). Consider the function f (x) = xe−x . As x → ∞, the first factor in f (x) goes to ∞, but the second factor goes to 0. What does the product do when one term is getting smaller and the other term is getting larger? It depends on which one is changing faster. What we want to know is which term “dominates.” Use graphical and numerical evidence to conjecture the value of lim (xe−x ). Which term dominates? In the limit x→∞

lim (x 2 e−x ), which term dominates? Also, try lim (x 5 e−x ).

x→∞

x→∞

Based on your investigation, is it always true that exponentials dominate polynomials? Are you positive? Try to determine which type of function, polynomials or logarithms, dominates.

FORMAL DEFINITION OF THE LIMIT We have now spent many pages discussing various aspects of the computation of limits. This may seem a bit odd, when you realize that we have never actually defined what a limit is. Oh, sure, we have given you an idea of what a limit is, but that’s about all. Once again, we have said that lim f (x) = L ,

x→a

HISTORICAL NOTES Augustin Louis Cauchy (1789–1857) A French mathematician who developed the ε-δ definitions of limit and continuity. Cauchy was one of the most prolific mathematicians in history, making important contributions to number theory, linear algebra, differential equations, astronomy, optics and complex variables. A difficult man to get along with, a colleague wrote, “Cauchy is mad and there is nothing that can be done about him, although right now, he is the only one who knows how mathematics should be done.”

if f (x) gets closer and closer to L as x gets closer and closer to a. So far, we have been quite happy with this somewhat vague, although intuitive, description. In this section, however, we will make this more precise, and you will begin to see how mathematical analysis (that branch of mathematics of which the calculus is the most elementary study) works. Studying more advanced mathematics without an understanding of the precise definition of limit is somewhat akin to studying brain surgery without bothering with all that background work in chemistry and biology. In medicine, it has only been through a careful examination of the microscopic world that a deeper understanding of our own macroscopic world has developed, and good surgeons need to understand what they are doing and why they are doing it. Likewise, in mathematical analysis, it is through an understanding of the microscopic behavior of functions (such as the precise definition of limit) that a deeper understanding of the mathematics will come about. We begin with the careful examination of an elementary example. You should certainly believe that lim (3x + 4) = 10.

x→2

Suppose that you were asked to explain the meaning of this particular limit to a fellow student. You would probably repeat the intuitive explanation we have used so far: that as x gets closer and closer to 2, (3x + 4) gets arbitrarily close to 10. But, exactly what do we mean by close? One answer is that if lim (3x + 4) = 10, we should be able to make x→2

(3x + 4) as close as we like to 10, just by making x sufficiently close to 2. But can we

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1-50

actually do this? For instance, can we force (3x + 4) to be within distance 1 of 10? To see what values of x will guarantee this, we write an inequality that says that (3x + 4) is within 1 unit of 10: |(3x + 4) − 10| < 1. Eliminating the absolute values, we see that this is equivalent to −1 < (3x + 4) − 10 < 1

y y 3x 4

− 1 < 3x − 6 < 1.

or

11 10 9

Since we need to determine how close x must be to 2, we want to isolate x − 2, instead of x. So, dividing by 3, we get − x 2W

2

2W

FIGURE 1.44 1 1 2 − < x < 2 + guarantees 3 3 that |(3x + 4) − 10| < 1

1 1 < x −2< 3 3 |x − 2| <

or

1 . 3

(6.1)

Reversing the steps that lead to inequality (6.1), we see that if x is within distance 13 of 2, then (3x + 4) will be within the specified distance (1) of 10. (See Figure 1.44 for a graphical interpretation of this.) So, does this convince you that you can make (3x + 4) as close as you want to 10? Probably not, but if you used a smaller distance, perhaps you’d be more convinced.

EXAMPLE 6.1

Exploring a Simple Limit

Find the values of x for which (3x + 4) is within distance

1 of 10. 100

Solution We want |(3x + 4) − 10| <

1 . 100

Eliminating the absolute values, we get −

1 1 < (3x + 4) − 10 < 100 100

or

−

1 1 < 3x − 6 < . 100 100

Dividing by 3 yields

−

1 1 < x −2< , 300 300

which is equivalent to

|x − 2| <

1 . 300

So, based on example 6.1, are you now convinced that we can make (3x + 4) as close as desired to 10? All we’ve been able to show is that we can make (3x + 4) pretty close to 10. So, how close do we need to be able to make it? The answer is arbitrarily close, as close as anyone would ever demand. We can show that this is possible by repeating the arguments in example 6.1, this time for an unspecified distance, call it ε (epsilon, where ε > 0).

1-51

SECTION 1.6

..

Formal Definition of the Limit

123

y 10 ´

EXAMPLE 6.2

y 3x 4

Verifying a Limit

Show that we can make (3x + 4) within any specified distance ε of 10 (no matter how small ε is), just by making x sufficiently close to 2.

10 10 ´

Solution The objective is to determine the range of x-values that will guarantee that (3x + 4) stays within ε of 10 (see Figure 1.45 for a sketch of this range). We have |(3x + 4) − 10| < ε. ´

23

2

´

23

x

This is equivalent to or

− ε < 3x − 6 < ε.

Dividing by 3, we get

−

FIGURE 1.45 The range of x-values that keep |(3x + 4) − 10| < ε

− ε < (3x + 4) − 10 < ε

or

ε ε < x −2< 3 3 ε |x − 2| < . 3

ε also implies 3 ε that |(3x + 4) − 10| < ε. This says that as long as x is within distance of 2, (3x + 4) 3 will be within the required distance ε of 10. That is, ε |(3x + 4) − 10| < ε whenever |x − 2| < . 3

Notice that each of the preceding steps is reversible, so that |x − 2| <

Take a moment or two to recognize what we’ve done in example 6.2. By using an unspecified distance, ε, we have verified that we can indeed make (3x + 4) as close to 10 as might be demanded (i.e., arbitrarily close; just name whatever ε > 0 you would like), simply by making x sufficiently close to 2. Further, we have explicitly spelled out what “sufficiently close to 2” means in the context of the present problem. Thus, no matter how close we are asked to make (3x + 4) to 10, we can accomplish this simply by taking x to be in the specified interval. Next, we examine this more precise notion of limit in the case of a function that is not defined at the point in question.

EXAMPLE 6.3

Proving That a Limit Is Correct

2x + 2x − 4 = 6. x→1 x −1 Solution It is easy to use the usual rules of limits to establish this result. It is yet another matter to verify that this is correct using our new and more precise notion of limit. In this case, we want to know how close x must be to 1 to ensure that 2

Prove that lim

f (x) =

2x 2 + 2x − 4 x −1

is within an unspecified distance ε > 0 of 6. First, notice that f is undefined at x = 1. So, we seek a distance δ (delta, δ > 0), such that if x is within distance δ of 1, but x = 1 (i.e., 0 < |x − 1| < δ), then this guarantees that | f (x) − 6| < ε.

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Notice that we have specified that 0 < |x − 1| to ensure that x = 1. Further, | f (x) − 6| < ε is equivalent to −ε <

2x 2 + 2x − 4 − 6 < ε. x −1

Finding a common denominator and subtracting in the middle term, we get −ε <

2x 2 + 2x − 4 − 6(x − 1) <ε x −1

or

−ε <

2x 2 − 4x + 2 < ε. x −1

Since the numerator factors, this is equivalent to y

−ε < y f (x)

6 ´

2(x − 1)2 < ε. x −1

Since x = 1, we can cancel two of the factors of (x − 1) to yield

6

−ε < 2(x − 1) < ε

6 ´

−

or

ε ε < x−1 < , 2 2

Dividing by 2.

which is equivalent to |x − 1| < ε/2. So, taking δ = ε/2 and working backward, we see that requiring x to satisfy x ´ 12

1

0 < |x − 1| < δ =

´ 12

FIGURE 1.46

ε 0 < |x − 1| < guarantees that 2 2x 2 + 2x − 4 6−ε < < 6 + ε. x −1

will guarantee that

ε 2

2 2x + 2x − 4 − 6 < ε. x −1

We illustrate this graphically in Figure 1.46. What we have seen so far motivates us to make the following general definition, illustrated in Figure 1.47.

y

DEFINITION 6.1 (Precise Definition of Limit) y f (x)

For a function f defined in some open interval containing a (but not necessarily at a itself), we say

L ´ L L ´

lim f (x) = L ,

x→a

x

a ad

if given any number ε > 0, there is another number δ > 0, such that 0 < |x − a| < δ guarantees that | f (x) − L| < ε.

ad

FIGURE 1.47 a − δ < x < a + δ guarantees that L − ε < f (x) < L + ε.

Notice that example 6.2 amounts to an illustration of Definition 6.1 for lim (3x + 4). There, we found that δ = ε/3 satisfies the definition.

x→2

1-53

SECTION 1.6

..

Formal Definition of the Limit

125

REMARK 6.1 TODAY IN MATHEMATICS

We want to emphasize that this formal definition of limit is not a new idea. Rather, it is a more precise mathematical statement of the same intuitive notion of limit that we have been using since the beginning of the chapter. Also, we must in all honesty point out that it is rather difficult to explicitly find δ as a function of ε, for all but a few simple examples. Despite this, learning how to work through the definition, even for a small number of problems, will shed considerable light on a deep concept.

Paul Halmos (1916– ) A Hungarian-born mathematician who earned a reputation as one of the best mathematical writers ever. For Halmos, calculus did not come easily, with understanding coming in a flash of inspiration only after a long period of hard work. “I remember standing at the blackboard in Room 213 of the mathematics building with Warren Ambrose and suddenly I understood epsilons. I understood what limits were, and all of that stuff that people had been drilling into me became clear. . . . I could prove the theorems. That afternoon I became a mathematician.’’

Example 6.4, although only slightly more complex than the last several problems, provides an unexpected challenge.

EXAMPLE 6.4

Using the Precise Definition of Limit

Use Definiton 6.1 to prove that lim (x 2 + 1) = 5. x→2

Solution If this limit is correct, then given any ε > 0, there must be a δ > 0 for which 0 < |x − 2| < δ guarantees that |(x 2 + 1) − 5| < ε. Notice that |(x 2 + 1) − 5| = |x 2 − 4|

Factoring the difference

= |x + 2||x − 2|.

of two squares.

(6.2)

Our strategy is to isolate |x − 2| and so, we’ll need to do something with the term |x + 2|. Since we’re interested only in what happens near x = 2, anyway, we will only consider x’s within a distance of 1 from 2, that is, x’s that lie in the interval [1, 3] (so that |x − 2| < 1). Notice that this will be true if we require δ ≤ 1 and |x − 2| < δ. In this case, we have |x + 2| ≤ 5,

Since x ∈ [1, 3].

and so, from (6.2), |(x 2 + 1) − 5| = |x + 2||x − 2| ≤ 5|x − 2|.

(6.3)

Finally, if we require that

y

5|x − 2| < ε, 5 ´

(6.4)

then we will also have from (6.3) that |(x 2 + 1) − 5| ≤ 5|x − 2| < ε.

5

Of course, (6.4) is equivalent to 5 ´

|x − 2| <

y x2 1

x 2d

2

2d

FIGURE 1.48 0 < |x − 2| < δ guarantees that |(x 2 + 1) − 5| < ε.

ε . 5

ε So, in view of this, we now have two restrictions: that |x − 2| < 1 and that |x − 2| < . 5 ε To ensure that both restrictions are met, we choose δ = min 1, i.e., the minimum 5 ε of 1 and . Working backward, we get that for this choice of δ, 5 0 < |x − 2| < δ will guarantee that |(x 2 + 1) − 5| < ε, as desired. We illustrate this in Figure 1.48.

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Exploring the Definition of Limit Graphically As you can see from example 6.4, this business of finding δ’s for a given ε is not easily accomplished. There, we found that even for the comparatively simple case of a quadratic polynomial, the job can be quite a challenge. Unfortunately, there is no procedure that will work for all problems. However, we can explore the definition graphically in the case of more complex functions. First, we reexamine example 6.4 graphically.

EXAMPLE 6.5

Exploring the Precise Definition of Limit Graphically

Explore the precise definition of limit graphically, for lim (x 2 + 1) = 5.

5.5

x→2

ε Solution In example 6.4, we discovered that for δ = min 1, , 5 2 0 < |x − 2| < δ implies that |(x + 1) − 5| < ε.

5.3 y

5.1 4.9 4.7 4.5 1.9

1.95

2 x

2.05

2.1

FIGURE 1.49 y = x2 + 1

This says that (for ε ≤ 5 ) if we draw a graph of y = x 2 + 1 and restrict the x-values to ε ε lie in the interval 2 − , 2 + , then the y-values will lie in the interval (5 − ε, 5 + ε). 5 5 1 Take ε = , for instance. If we draw the graph in the window defined by 2 1 1 2− ≤ x ≤2+ and 4.5 ≤ y ≤ 5.5, then the graph will not run off the top or 10 10 bottom of the screen (see Figure 1.49). Of course, we can draw virtually the same picture for any given value of ε, since we have an explicit formula for finding δ given ε. For most limit problems, we are not so fortunate.

EXAMPLE 6.6

Exploring the Definition of Limit for a Trigonometric Function

1 Graphically find a δ > 0 corresponding to (a) ε = and (b) ε = 0.1 for 2 πx lim sin = 0. x→2 2 2π Solution This limit seems plausible enough. After all, sin = 0 and f (x) = sin x 2 is a continuous function. However, the point is to verify this carefully. Given any ε > 0, we want to find a δ > 0, for which πx 0 < |x − 2| < δ guarantees that sin − 0 < ε. 2 πx Note that since we have no algebra for simplifying sin , we cannot accomplish this 2 symbolically. Instead, we’ll try to graphically find δ’s corresponding to the specific ε’s 1 given. First, for ε = , we would like to find a δ > 0 for which if 0 < |x − 2| < δ, then 2

y 0.5

x 1

1.5

2

2.5

0.5

FIGURE 1.50a y = sin

πx 2

3

−

1 πx 1 < sin −0< . 2 2 2

πx 1 1 Drawing the graph of y = sin with 1 ≤ x ≤ 3 and − ≤ y ≤ , we get 2 2 2 Figure 1.50a. If you trace along a calculator or computer graph, you will notice that the graph stays on the screen (i.e., the y-values stay in the interval [−0.5, 0.5]) for

1-55

SECTION 1.6

y

..

Formal Definition of the Limit

x ∈ [1.666667, 2.333333]. Thus, we have determined experimentally that for ε =

0.5

127

1 , 2

δ = 2.333333 − 2 = 2 − 1.666667 = 0.333333 2–δ

2+δ

2

0.5

FIGURE 1.50b y = sin

πx 2

x

will work. (Of course, any value of δ smaller than 0.333333 will also work.) To illustrate this, we redraw the last graph, but restrict x to lie in the interval [1.67, 2.33] (see Figure 1.50b). In this case, the graph stays in the window over the entire range of displayed x-values. Taking ε = 0.1, we look for an interval of x-values that will πx guarantee that sin stays between −0.1 and 0.1. We redraw the graph from Figure 2 1.50a, with the y-range restricted to the interval [−0.1, 0.1] (see Figure 1.51a). Again, tracing along the graph tells us that the y-values will stay in the desired range for x ∈ [1.936508, 2.063492]. Thus, we have experimentally determined that δ = 2.063492 − 2 = 2 − 1.936508 = 0.063492 will work here. We redraw the graph using the new range of x-values (see Figure 1.51b), since the graph remains in the window for all values of x in the indicated interval. y

y

0.1

0.1

x 1.7

2

2–δ

2.3

0.1

2+δ

2

x

0.1

FIGURE 1.51a πx y = sin 2

FIGURE 1.51b y = sin

πx 2

It is important to recognize that we are not proving that the above limit is correct. To prove this requires us to symbolically find a δ for every ε > 0. The idea here is to use these graphical illustrations to become more familiar with the definition and with what δ and ε represent. x 0.1 0.01 0.001 0.0001

x 2 2x √ x 3 4x 2 1.03711608 1.0037461 1.00037496 1.0000375

EXAMPLE 6.7

Exploring the Definition of Limit Where the Limit Does Not Exist

x 2 + 2x Determine whether or not lim √ = 1. x→0 x 3 + 4x 2 Solution We first construct a table of function values. From the table alone, we might be tempted to conjecture that the limit is 1. However, we would be making a huge error, as we have not considered negative values of x or drawn a graph. This kind of carelessness is dangerous. Figure 1.52a (on the following page) shows the default graph drawn by our computer algebra system. In this graph, the function values do not quite look like they are approaching 1 as x → 0 (at least as x → 0− ). We now investigate the limit graphically for ε = 12 . We need to find a δ > 0 for which 0 < |x| < δ guarantees that 1−

1 1 x 2 + 2x <1+ <√ 2 2 x 3 + 4x 2

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1-56

y 16

y

12

1.5

8 1 4

4

x

2

2

0.5

4

4

x

0.1

0.1

FIGURE 1.52a

FIGURE 1.52b

x 2 + 2x y= √ x 3 + 4x 2

x 2 + 2x y= √ x 3 + 4x 2

x 2 + 2x 1 3 <√ < . 3 2 2 2 x + 4x

or

We try δ = 0.1 to see if this is sufficiently small. So, we set the x-range to the interval [−0.1, 0.1] and the y-range to the interval [0.5, 1.5] and redraw the graph in this window (see Figure 1.52b). Notice that no points are plotted in the window for any x < 0. According to the definition, the y-values must lie in the interval (0.5, 1.5) for all x in the interval (−δ, δ). Further, you can see that δ = 0.1 clearly does not work since x = −0.05 lies in the interval (−δ, δ), but f (−0.05) ≈ −0.981 is not in the interval (0.5, 1.5). You should convince yourself that no matter how small you make δ, there is an x in the interval (−δ, δ) such that f (x) ∈ / (0.5, 1.5). (In fact, notice that for all x’s in the interval (−1, 0), f (x) < 0.) That is, there is no choice of δ that makes the defining inequality true for ε = 12 . Thus, the conjectured limit of 1 is incorrect. You should note here that, while we’ve only shown that the limit is not 1, it’s somewhat more complicated to show that the limit does not exist.

Limits Involving Infinity Recall that we had observed that lim

x→0

written

1 does not exist, but to be more descriptive, we had x2 lim

x→0

1 = ∞. x2

By this statement, we mean that the function increases without bound as x → 0. Just as with our initial intuitive notion of lim f (x) = L, this description is imprecise and needs to be x→a 1 more carefully defined. When we say that 2 increases without bound as x → 0, we mean x 1 that we can make 2 as large as we like, simply by making x sufficiently close to 0. So, x 1 given any large positive number, M, we must be able to make 2 > M, for x sufficiently x close to 0. We measure closeness here the same way as we did before and arrive at the following definition.

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SECTION 1.6

..

Formal Definition of the Limit

129

y

DEFINITION 6.2 For a function f defined in some open interval containing a (but not necessarily at a itself), we say lim f (x) = ∞,

M

x→a

if given any number M > 0, there is another number δ > 0, such that 0 < |x − a| < δ guarantees that f (x) > M. (See Figure 1.53 for a graphical interpretation of this.) x ad

a

ad

FIGURE 1.53 lim f (x) = ∞

Similarly, we had said that if f decreases without bound as x → a, then lim f (x) = −∞. Think of how you would make this more precise and then consider the x→a following definition.

x→a

DEFINITION 6.3

y ad

a

ad x

For a function f defined in some open interval containing a (but not necessarily at a itself), we say lim f (x) = −∞, x→a

if given any number N < 0, there is another number δ > 0, such that 0 < |x − a| < δ guarantees that f (x) < N . (See Figure 1.54 for a graphical interpretation of this.)

N

It’s easy to keep these definitions straight if you think of their meaning. Don’t simply memorize them. FIGURE 1.54 lim f (x) = −∞

x→a

EXAMPLE 6.8

Using the Definition of Limit Where the Limit Is Infinite

1 = ∞. x2 Solution Given any (large) number M > 0, we need to find a distance δ > 0 such that if x is within δ of 0 (but not equal to 0) then

Prove that lim

x→0

1 > M. x2

(6.5)

Since both M and x 2 are positive, (6.5) is equivalent to x2 <

1 . M

√ Taking the square root of both sides and recalling that x 2 = |x|, we get 1 |x| < . M 1 So, for any M > 0, if we take δ = and work backward, we have that 0 < |x − 0| < δ M guarantees that 1 > M, x2 1 as desired. Note that this says, for instance, that for M = 100, 2 > 100, whenever x 1 1 0 < |x| < = . (Verify that this works, as an exercise.) 100 10

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There are two remaining limits that we have yet to place on a careful footing. Before reading on, try to figure out for yourself what appropriate definitions would look like. If we write lim f (x) = L, we mean that as x increases without bound, f (x) gets x→∞ closer and closer to L. That is, we can make f (x) as close to L as we like, by choosing x sufficiently large. More precisely, we have the following definition. y

DEFINITION 6.4 For a function f defined on an interval (a, ∞), for some a > 0, we say

L ´ L L´

lim f (x) = L ,

x→∞

if given any ε > 0, there is a number M > 0 such that x > M guarantees that | f (x) − L| < ε.

x M

(See Figure 1.55 for a graphical interpretation of this.)

FIGURE 1.55 lim f (x) = L

Similarly, we have said that lim f (x) = L means that as x decreases without bound,

x→∞

x→−∞

f (x) gets closer and closer to L. So, we should be able to make f (x) as close to L as desired, just by making x sufficiently large in absolute value and negative. We have the following definition. y

DEFINITION 6.5 L ´ L L´

For a function f defined on an interval (−∞, a), for some a < 0, we say lim f (x) = L ,

x→−∞

if given any ε > 0, there is a number N < 0 such that x < N guarantees that N

| f (x) − L| < ε.

x

(See Figure 1.56 for a graphical interpretation of this.) FIGURE 1.56 lim f (x) = L

x→−∞

We use Definitions 6.4 and 6.5 essentially the same as we do Definitions 6.1–6.3, as we see in example 6.9.

EXAMPLE 6.9 Prove that lim

x→−∞

Using the Definition of Limit Where x Is Becoming Infinite

1 = 0. x

1 Solution Here, we must show that given any ε > 0, we can make within ε of 0, x simply by making x sufficiently large in absolute value and negative. So, we need to determine those x’s for which 1 − 0 < ε x 1 < ε. or (6.6) x

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SECTION 1.6

..

Formal Definition of the Limit

131

Since x < 0, |x| = −x, and so (6.6) becomes

REMARK 6.2 You should take care to note the commonality among the definitions of the five limits we have given. All five deal with a precise description of what it means to be “close.” It is of considerable benefit to work through these definitions until you can provide your own words for each. Don’t just memorize the formal definitions as stated here. Rather, work toward understanding what they mean and come to appreciate the exacting language mathematicians use.

1 < ε. −x Dividing both sides by ε and multiplying by x (remember that x < 0 and ε > 0, so that this will change the direction of the inequality), we get −

1 > x. ε

1 So, if we take N = − and work backward, we have satisfied the definition and thereby ε proved that the limit is correct. We don’t use the limit definitions to prove each and every limit that comes along. Actually, we use them to prove only a few basic limits and to prove the limit theorems that we’ve been using for some time without proof. Further use of these theorems then provides solid justification of new limits. As an illustration, we now prove the rule for a limit of a sum.

THEOREM 6.1 Suppose that for a real number a, lim f (x) = L 1 and lim g(x) = L 2 . Then, x→a

x→a

lim [ f (x) + g(x)] = lim f (x) + lim g(x) = L 1 + L 2 .

x→a

x→a

x→a

PROOF Since lim f (x) = L 1 , we know that given any number ε1 > 0, there is a number δ1 > 0 for x→a which 0 < |x − a| < δ1 guarantees that | f (x) − L 1 | < ε1 .

(6.7)

Likewise, since lim g(x) = L 2 , we know that given any number ε2 > 0, there is a number x→a δ2 > 0 for which 0 < |x − a| < δ2 guarantees that |g(x) − L 2 | < ε2 .

(6.8)

Now, in order to get lim [ f (x) + g(x)] = (L 1 + L 2 ),

x→a

we must show that, given any number ε > 0, there is a number δ > 0 such that 0 < |x − a| < δ guarantees that |[ f (x) + g(x)] − (L 1 + L 2 )| < ε. Notice that |[ f (x) + g(x)] − (L 1 + L 2 )| = |[ f (x) − L 1 ] + [g( x) − L 2 ]| ≤ | f (x) − L 1 | + |g(x) − L 2 |,

(6.9)

by the triangle inequality. Of course, both terms on the right-hand side of (6.9) can be made ε arbitrarily small, from (6.7) and (6.8). In particular, if we take ε1 = ε2 = , then as long as 2 0 < |x − a| < δ1 and 0 < |x − a| < δ2 ,

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we get from (6.7), (6.8) and (6.9) that |[ f (x) + g(x)] − (L 1 + L 2 )| ≤ | f (x) − L 1 | + |g(x) − L 2 | ε ε < + = ε, 2 2 as desired. Of course, this will happen if we take 0 < |x − a| < δ = min{δ1 , δ2 }. The other rules for limits are proven similarly, using the definition of limit. We show these in Appendix A.

EXERCISES 1.6 WRITING EXERCISES 1. In his 1726 masterpiece Mathematical Principles of Natural Philosophy, which introduces many of the fundamentals of calculus, Sir Isaac Newton described the important limit f (a + h) − f (a) (which we study at length in Chapter 2) lim h→0 h as “the limit to which the ratios of quantities decreasing without limit do always converge, and to which they approach nearer than by any given difference, but never go beyond, nor ever reach until the quantities vanish.” If you ever get weary of all the notation that we use in calculus, think of what it would look like in words! Critique Newton’s definition of limit, addressing the following questions in the process. What restrictions do the phrases “never go beyond” and “never reach” put on the limit process? Give an example of a simple limit, not necessarily f (a + h) − f (a) of the form lim , that violates these restrich→0 h tions. Give your own (English language) description of the limit, avoiding restrictions such as Newton’s. Why do mathematicians consider the ε−δ definition simple and elegant? 2. You have computed numerous limits before seeing the definition of limit. Explain how this definition changes and/or improves your understanding of the limit process. 3. Each word in the ε−δ definition is carefully chosen and precisely placed. Describe what is wrong with each of the following slightly incorrect “definitions” (use examples!): (a) There exists ε > 0 such that there exists a δ > 0 such that if 0 < |x − a| < δ, then | f (x) − L| < ε. (b) For all ε > 0 and for all δ > 0, if 0 < |x − a| < δ, then | f (x) − L| < ε. (c) For all δ > 0 there exists ε > 0 such that 0 < |x − a| < δ and | f (x) − L| < ε. 4. In order for the limit to exist, given every ε > 0, we must be able to find a δ > 0 such that the if/then inequalities are true. To prove that the limit does not exist, we must find a particular ε > 0 such that the if/then inequalities are not true for any choice of δ > 0. To understand the logic behind the swapping of the “for every” and “there exists” roles, draw an

analogy with the following situation. Suppose the statement, “Everybody loves somebody” is true. If you wanted to verify the statement, why would you have to talk to every person on earth? But, suppose that the statement is not true. What would you have to do to disprove it? In exercises 1–8, numerically and graphically determine a δ corresponding to (a) ε 0.1 and (b) ε 0.05. Graph the function in the ε − δ window [x-range is (a − δ, a δ) and y-range is (L − ε, L ε)] to verify that your choice works. 1. lim (x 2 + 1) = 1

2. lim (x 2 + 1) = 5

3. lim cos x = 1

4. lim cos x = 0

x→0

x→0

5. lim

x→1

7. lim

x→1

√

x→2

x→π/2

x +3=2

√

6. lim

x→−2

x +2 =3 x2

8. lim

x→2

x +3=1

x +2 =1 x2

In exercises 9–20, symbolically find δ in terms of ε. 9. lim 3x = 0

10. lim 3x = 3

x→0

x→1

11. lim (3x + 2) = 8

12. lim (3x + 2) = 5

13. lim (3 − 4x) = −1

14. lim (3 − 4x) = 7

x +x −2 =3 15. lim x→1 x −1

16. lim

17. lim (x − 1) = 0

18. lim (x 2 − x + 1) = 1

19. lim (x − 1) = 3

20. lim (x 3 + 1) = 1

x→2

x→1

x→1

x→−1

2

2

x→1

x→−1

x2 − 1 = −2 x +1

x→1

2

x→2

x→0

21. Determine a formula for δ in terms of ε for lim (mx + b). (Hint: x→a

Use exercises 9–14.) Does the formula depend on the value of a? Try to explain this answer graphically. 22. Based on exercises 17 and 19, does the value of δ depend on the value of a for lim (x 2 + b)? Try to explain this graphically. x→a

1-61

SECTION 1.6

23. Modify the ε − δ definition to define the one-sided limits lim f (x) and lim f (x). x→a −

x→a +

24. Symbolically find the largest δ corresponding to ε = 0.1 in the definition of lim 1/x = 1. Symbolically find the largest x→1−

δ corresponding to ε = 0.1 in the definition of lim 1/x = 1.

49. f (x) = 50. f (x) =

x→1+

Which δ could be used in the definition of lim 1/x = 1? Briefly x→1

explain. Then prove that lim 1/x = 1. x→1

In exercises 25–30, find a δ corresponding to M 100 or N − 100 (as appropriate) for each limit. 25. lim

x→1+

2 =∞ x −1

27. lim cot x = ∞ x→0+

29. lim √ − x→2

2 4 − x2

=∞

26. lim

x→1−

2 = −∞ x −1

28. lim cot x = −∞ x→π −

30. lim ln x = −∞ x→0+

In exercises 31–36, find an M or N corresponding to ε 0.1 for each limit at infinity. 31. lim

x→∞

33.

x2

x2 − 2 =1 +x +1

x2 + 3 = 0.25 x→−∞ 4x 2 − 4 lim

35. lim e

−2x

x→∞

=0

32. lim

x→∞

34.

x2

x −2 =0 +x +1

3x 2 − 2 =3 x→−∞ x 2 + 1 lim

e +x 36. lim x =1 x→∞ e − x 2

x→∞

2 =0 x3

1 = 0, for k > 0 xk 1 − 3 = −3 41. lim x→∞ x2 + 2

39. lim

x→∞

43. lim

x→−3

−2 = −∞ (x + 3)4

4 =∞ x→5 (x − 5)2

45. lim

38. 40.

lim

3 =0 x3

lim

1 = 0, for k > 0 x 2k

x→−∞

x→−∞

42. lim

1 =0 (x − 7)2

44. lim

3 =∞ (x − 7)2

x→∞

x→7

−6 = −∞ x→−4 (x + 4)6

46. lim

In exercises 47–50, identify a specific ε > 0 for which no δ > 0 exists to satisfy the definition to limit. 2x if x < 1, lim f (x) = 2 47. f (x) = x 2 + 3 if x > 1 x→1 48. f (x) =

Formal Definition of the Limit

133

if x < 1, lim f (x) = 2 if x > 1 x→1

x − 1 if x < 2, lim f (x) = 1 x2 if x > 2 x→2

51. A metal washer of (outer) radius r inches weighs 2r 2 ounces. A company manufactures 2-inch washers for different customers who have different error tolerances. If the customer demands a washer of weight 8 ± ε ounces, what is the error tolerance for the radius? That is, find δ such that a radius of r within the interval (2 − δ, 2 + δ) guarantees a weight within (8 − ε, 8 + ε). 52. A fiberglass company ships its glass as spherical marbles. If the volume of each marble must be within ε of π/6, how close does the radius need to be to 1/2? 53. Prove Theorem 3.1 (i). 54. Prove Theorem 3.1 (ii). 55. Prove the Squeeze Theorem, as stated in Theorem 3.5. 56. Given that lim f (x) = L and lim f (x) = L, prove that x→a −

x→a +

lim f (x) = L.

x→a

57. Prove: if lim f (x) = L, then lim [ f (x) − L] = 0. x→a

x→a

x

In exercises 37–46, prove that the limit is correct using the appropriate definition (assume that k is an integer). 37. lim

2x 5 − x2

..

x 2 − 1 if x < 0, lim f (x) = −2 −x − 2 if x > 0 x→0

58. Prove: if lim [ f (x) − L] = 0, then lim f (x) = L. x→a

x→a

59. In this exercise, we explore the definition of lim x 2 = 4 with x→2 √ ε = 0.1. Show that x 2 − 4 < 0.1 if 2 < x < 4.1. This indicates that δ1 = 0.02484 works for x > 2. Show that √ x 2 − 4 > −0.1 if 3.9 < x < 2. This indicates that δ2 = 0.02515 works for x < 2. For the limit definition, is δ = δ1 or δ = δ2 the correct choice? Briefly explain. √ 60. Generalize exercise 59 to find a δ of the form 4 + ε or √ 4 − ε corresponding to any ε > 0.

EXPLORATORY EXERCISES 1. We hope that working through this section has provided you with extra insight into the limit process. However, we have not yet solved any problems we could not already solve in previous sections. We do so now, while investigating an unusual function. Recall that rational numbers can be written as fractions p/q, where p and q are integers. We will assume that p/q has been simplified by dividing out common factors (e.g., 0 if x is irrational 1/2 and not 2/4). Define f (x) = . 1/q if x = qp is rational We will try to show that lim f (x) exists. Without graphics, x→2/3

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we need a good definition to answer this question. We know that f (2/3) = 1/3, but recall that the limit is independent of the actual function value. We need to think about x’s close to 2/3. If such an x is irrational, f (x) = 0. A simple hypothesis would then be lim f (x) = 0. We’ll try this out for x→2/3

ε = 1/6. We would like to guarantee that | f (x)| < 1/6 whenever 0 < |x − 2/3| < δ. Well, how many x’s have a function value greater than 1/6? The only possible function values are 1/5, 1/4, 1/3, 1/2 and 1. The x’s with function value 1/5 are 1/5, 2/5, 3/5, 4/5 and so on. The closest of these x’s to 2/3

1.7

is 3/5. Find the closest x (not counting x = 2/3) to 2/3 with function value 1/4. Repeat for f (x) = 1/3, f (x) = 1/2 and f (x) = 1. Out of all these closest x’s, how close is the absolute closest? Choose δ to be this number, and argue that if 0 < |x − 2/3| < δ, we are guaranteed that | f (x)| < 1/6. Argue that a similar process can find a δ for any ε. 2. State a definition for “ f (x) is continuous at x = a” using Definition 6.1. Use it to prove that the function in exploratory exercise 1 is continuous at every irrational number and discontinuous at every rational number.

LIMITS AND LOSS-OF-SIGNIFICANCE ERRORS “Pay no attention to that man behind the curtain . . . .” (from The Wizard of Oz) Things are not always what they appear to be. We spend much time learning to distinguish reality from mere appearances. Along the way, we develop a healthy level of skepticism. You may have already come to realize that mathematicians are a skeptical lot. This is of necessity, for you simply can’t accept things at face value. People tend to accept a computer’s answer as a fact not subject to debate. However, when we use a computer (or calculator), we must always keep in mind that these devices perform most computations only approximately. Most of the time, this will cause us no difficulty whatsoever. Modern computational devices generally carry out calculations to a very high degree of accuracy. Occasionally, however, the results of round-off errors in a string of calculations are disastrous. In this section, we briefly investigate these errors and learn how to recognize and avoid some of them. We first consider a relatively tame-looking example.

EXAMPLE 7.1

y 9

A Limit with Unusual Graphical and Numerical Behavior 2

(x 3 + 4) − x 6 . x→∞ x3

Evaluate lim 8

7

x 20,000

60,000

100,000

FIGURE 1.57a 2

y=

(x 3 + 4) − x 6 x3

Solution At first glance, the numerator looks like ∞ − ∞, which is indeterminate, while the denominator tends to ∞. Algebraically, the only reasonable step to take is to multiply out the first term in the numerator. Before we do that, let’s draw a graph and compute some function values. (Different computers and different software will produce somewhat different results, but for large values of x, you should see results similar to those shown here.) In Figure 1.57a, the function appears nearly constant, until it begins oscillating around x = 40,000. Notice that the accompanying table of function values is inconsistent with Figure 1.57a. The last two values in the table may have surprised you. Up until that point, the function values seemed to be settling down to 8.0 very nicely. So, what happened here and what is the correct value of the limit? Obviously, something unusual has occurred

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1-62

we need a good definition to answer this question. We know that f (2/3) = 1/3, but recall that the limit is independent of the actual function value. We need to think about x’s close to 2/3. If such an x is irrational, f (x) = 0. A simple hypothesis would then be lim f (x) = 0. We’ll try this out for x→2/3

ε = 1/6. We would like to guarantee that | f (x)| < 1/6 whenever 0 < |x − 2/3| < δ. Well, how many x’s have a function value greater than 1/6? The only possible function values are 1/5, 1/4, 1/3, 1/2 and 1. The x’s with function value 1/5 are 1/5, 2/5, 3/5, 4/5 and so on. The closest of these x’s to 2/3

1.7

is 3/5. Find the closest x (not counting x = 2/3) to 2/3 with function value 1/4. Repeat for f (x) = 1/3, f (x) = 1/2 and f (x) = 1. Out of all these closest x’s, how close is the absolute closest? Choose δ to be this number, and argue that if 0 < |x − 2/3| < δ, we are guaranteed that | f (x)| < 1/6. Argue that a similar process can find a δ for any ε. 2. State a definition for “ f (x) is continuous at x = a” using Definition 6.1. Use it to prove that the function in exploratory exercise 1 is continuous at every irrational number and discontinuous at every rational number.

LIMITS AND LOSS-OF-SIGNIFICANCE ERRORS “Pay no attention to that man behind the curtain . . . .” (from The Wizard of Oz) Things are not always what they appear to be. We spend much time learning to distinguish reality from mere appearances. Along the way, we develop a healthy level of skepticism. You may have already come to realize that mathematicians are a skeptical lot. This is of necessity, for you simply can’t accept things at face value. People tend to accept a computer’s answer as a fact not subject to debate. However, when we use a computer (or calculator), we must always keep in mind that these devices perform most computations only approximately. Most of the time, this will cause us no difficulty whatsoever. Modern computational devices generally carry out calculations to a very high degree of accuracy. Occasionally, however, the results of round-off errors in a string of calculations are disastrous. In this section, we briefly investigate these errors and learn how to recognize and avoid some of them. We first consider a relatively tame-looking example.

EXAMPLE 7.1

y 9

A Limit with Unusual Graphical and Numerical Behavior 2

(x 3 + 4) − x 6 . x→∞ x3

Evaluate lim 8

7

x 20,000

60,000

100,000

FIGURE 1.57a 2

y=

(x 3 + 4) − x 6 x3

Solution At first glance, the numerator looks like ∞ − ∞, which is indeterminate, while the denominator tends to ∞. Algebraically, the only reasonable step to take is to multiply out the first term in the numerator. Before we do that, let’s draw a graph and compute some function values. (Different computers and different software will produce somewhat different results, but for large values of x, you should see results similar to those shown here.) In Figure 1.57a, the function appears nearly constant, until it begins oscillating around x = 40,000. Notice that the accompanying table of function values is inconsistent with Figure 1.57a. The last two values in the table may have surprised you. Up until that point, the function values seemed to be settling down to 8.0 very nicely. So, what happened here and what is the correct value of the limit? Obviously, something unusual has occurred

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..

Limits and Loss-of-Significance Errors

135

between x = 1 × 104 and x = 1 × 105 . We should look carefully at function values in that interval. A more detailed table is shown below to the right.

y 8.2

Incorrect calculated values

(x 4) − x 6 x3 8.016 8.000016 8.0 8.0 0.0 0.0 3

x

8

7.8 x 20,000

60,000

100,000

FIGURE 1.57b

10 100 1 × 103 1 × 104 1 × 105 1 × 106

2

x 2 × 104 3 × 104 4 × 104 5 × 104

(x 3 4)2 − x 6 x3 8.0 8.14815 7.8125 0

2

y=

(x 3 + 4) − x 6 x3

In Figure 1.57b, we have blown up the graph to enhance the oscillation observed between x = 1 × 104 and x = 1 × 105 . The picture that is emerging is even more confusing. The deeper we look into this limit, the more erratically the function appears to behave. We use the word appears because all of the oscillatory behavior we are seeing is an illusion, created by the finite precision of the computer used to perform the calculations or draw the graph.

Computer Representation of Real Numbers The reason for the unusual behavior seen in example 7.1 boils down to the way in which computers represent real numbers. Without getting into all of the intricacies of computer arithmetic, it suffices to think of computers and calculators as storing real numbers internally in scientific notation. For example, the number 1,234,567 would be stored as 1.234567 × 106 . The number preceding the power of 10 is called the mantissa and the power is called the exponent. Thus, the mantissa here is 1.234567 and the exponent is 6. All computing devices have finite memory and consequently have limitations on the size mantissa and exponent that they can store. (This is called finite precision.) Many calculators carry a 14-digit mantissa and a 3-digit exponent. On a 14-digit computer, this would suggest that the computer would retain only the first 14 digits in the decimal expansion of any given number.

EXAMPLE 7.2

Computer Representation of a Rational Number

1 2 Determine how is stored internally on a 10-digit computer and how is stored internally 3 3 on a 14-digit computer. 1 −1 Solution On a 10-digit computer, is stored internally as 3.333333333 ! " ×10 . On a 3 10 digits

2 −1 14-digit computer, is stored internally as 6.6666666666667 ! " × 10 . 3 14 digits

For most purposes, such finite precision presents no problem. However, we do occasionally come across a disastrous error caused by finite precision. In example 7.3, we subtract two relatively close numbers and examine the resulting catastrophic error.

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EXAMPLE 7.3

A Computer Subtraction of Two “Close” Numbers

Compare the exact value of 18 18 1. 0000000000000 ! " 4 × 10 − 1. 0000000000000 ! " 1 × 10 13 zeros

13 zeros

with the result obtained from a calculator or computer with a 14-digit mantissa. Solution Notice that 18 18 18 1. 0000000000000 ! " 4×10 − 1. 0000000000000 ! " 1×10 = 0. 0000000000000 ! " 3×10 13 zeros

13 zeros

13 zeros

= 30,000.

(7.1)

However, if this calculation is carried out on a computer or calculator with a 14-digit (or smaller) mantissa, both numbers on the left-hand side of (7.1) are stored by the computer as 1 × 1018 and hence, the difference is calculated as 0. Try this calculation for yourself now.

EXAMPLE 7.4

Another Subtraction of Two “Close” Numbers

Compare the exact value of 20 20 1. 0000000000000 ! " 6 × 10 − 1. 0000000000000 ! " 4 × 10 13 zeros

13 zeros

with the result obtained from a calculator or computer with a 14-digit mantissa. Solution Notice that 20 20 20 1.0000000000000 ! " 6 ×10 − 1.0000000000000 ! " 4 ×10 = 0.0000000000000 ! " 2 ×10 13 zeros

13 zeros

13 zeros

= 2,000,000.

However, if this calculation is carried out on a calculator with a 14-digit mantissa, the first number is represented as 1.0000000000001 × 1020 , while the second number is represented by 1.0 × 1020 , due to the finite precision and rounding. The difference between the two values is then computed as 0.0000000000001 × 1020 or 10,000,000, which is, again, a very serious error. In examples 7.3 and 7.4, we witnessed a gross error caused by the subtraction of two numbers whose significant digits are very close to one another. This type of error is called a loss-of-significant-digits error or simply a loss-of-significance error. These are subtle, often disastrous computational errors. Returning now to example 7.1, we will see that it was this type of error that caused the unusual behavior noted.

EXAMPLE 7.5

A Loss-of-Significance Error 2

(x 3 + 4) − x 6 In example 7.1, we considered the function f (x) = . x3 4 Follow the calculation of f (5 × 10 ) one step at a time, as a 14-digit computer would do it.

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..

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137

Solution We have [(5 × 104 )3 + 4]2 − (5 × 104 )6 (5 × 104 )3 (1.25 × 1014 + 4)2 − 1.5625 × 1028 = 1.25 × 1014

f (5 × 104 ) =

=

(125,000,000,000,000 + 4)2 − 1.5625 × 1028 1.25 × 1014

=

(1.25 × 1014 )2 − 1.5625 × 1028 = 0, 1.25 × 1014

REMARK 7.1 If at all possible, avoid subtractions of nearly equal values. Sometimes, this can be accomplished by some algebraic manipulation of the function.

since 125,000,000,000,004 is rounded off to 125,000,000,000,000. Note that the real culprit here was not the rounding of 125,000,000,000,004, but the fact that this was followed by a subtraction of a nearly equal value. Further, note that this is not a problem unique to the numerical computation of limits, but one that occurs in numerical computation, in general.

y

In the case of the function from example 7.5, we can avoid the subtraction and hence, the loss-of-significance error by rewriting the function as follows:

9

2

(x 3 + 4) − x 6 f (x) = x3 6 (x + 8x 3 + 16) − x 6 = x3 3 8x + 16 = , x3

8

7

x 20,000

60,000

100,000

FIGURE 1.58 y=

8x 3 + 16 x3

where we have eliminated the subtraction. Using this new (and equivalent) expression for the function, we can compute a table of function values reliably. Notice, too, that if we redraw the graph in Figure 1.57a using the new expression (see Figure 1.58), we no longer see the oscillation present in Figures 1.57a and 1.57b. From the rewritten expression, we easily obtain 2

8x 16 x3 8.016 8.000016 8.000000016 8.00000000002 8.0 8.0 8.0

(x 3 + 4) − x 6 = 8, x→∞ x3 lim

3

x 10 100 1 × 103 1 × 104 1 × 105 1 × 106 1 × 107

which is consistent with Figure 1.58 and the corrected table of function values. In example 7.6, we examine a loss-of-significance error that occurs for x close to 0.

EXAMPLE 7.6

Loss-of-Significance Involving a Trigonometric Function

1 − cos x 2 . x→0 x4

Evaluate lim

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y

1-66

Solution As usual, we look at a graph (see Figure 1.59) and some function values.

0.5

4

x

2

2

4

1 − cos x x4

2

2

1 − cos x 2 x4

x

1 − cos x 2 x4

0.1 0.01 0.001 0.0001 0.00001

0.499996 0.5 0.5 0.0 0.0

−0.1 −0.01 −0.001 −0.0001 −0.00001

0.499996 0.5 0.5 0.0 0.0

As in example 7.1, note that the function values seem to be approaching 0.5, but then suddenly take a jump down to 0.0. Once again, we are seeing a loss-of-significance error. In this particular case, this occurs because we are subtracting nearly equal values (cos x 2 and 1). We can again avoid the error by eliminating the subtraction. Note that

FIGURE 1.59 y=

x

1 − cos x 2 1 − cos x 2 1 + cos x 2 = · x4 x4 1 + cos x 2 2 2 1 − cos x = 4 x 1 + cos x 2 sin2 x 2 . = 4 x 1 + cos x 2

2

x

sin (x ) x 4 (1 cos x 2 )

±0.1 ±0.01 ±0.001 ±0.0001 ±0.00001

0.499996 0.4999999996 0.5 0.5 0.5

Multiply numerator and denominator by (1 + cos x 2 ). 1 − cos2 (x 2 ) = sin2 (x 2 ).

Since this last (equivalent) expression has no subtraction indicated, we should be able to use it to reliably generate values without the worry of loss-of-significance error. Using this to compute function values, we get the accompanying table. Using the graph and the new table, we conjecture that 1 − cos x 2 1 = . 4 x→0 x 2 lim

We offer one final example where a loss-of-significance error occurs, even though no subtraction is explicitly indicated.

EXAMPLE 7.7

Evaluate lim x[(x 2 + 4)

y

x→−∞

1 108 6 107 2 107

x

1

2

3

FIGURE 1.60 y = x[(x 2 + 4)

1/2

+ x]

A Loss-of-Significance Error Involving a Sum 1/2

+ x].

Solution Initially, you might think that since there is no subtraction (explicitly) indicated, there will be no loss-of-significance error. We first draw a graph (see Figure 1.60) and compute a table of values. #

1/2

x

x (x 2 4)

−100 −1 × 103 −1 × 104 −1 × 105 −1 × 106 −1 × 107 −1 × 108

−1.9998 −1.999998 −2.0 −2.0 −2.0 0.0 0.0

x

$

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SECTION 1.7

..

Limits and Loss-of-Significance Errors

139

You should quickly notice the sudden jump in values in the table and the wild oscillation visible in the graph. Although a subtraction is not explicitly indicated, there 1/2 is indeed a subtraction here, since we have x < 0 and (x 2 + 4) > 0. We can again remedy this with some algebraic manipulation, as follows. #

x (x + 4) 2

1/2

# $ 1/2 $ (x 2 + 4) − x + x = x (x + 4) + x # $ (x 2 + 4)1/2 − x $ # 2 (x + 4) − x 2 = x# $ (x 2 + 4)1/2 − x 4x = # $. (x 2 + 4)1/2 − x $

#

1/2

2

Multiply numerator and denominator by the conjugate.

Simplify the numerator.

We use this last expression to generate a graph in the same window as that used for Figure 1.60 and to generate the accompanying table of values. In Figure 1.61, we can see none of the wild oscillation observed in Figure 1.60, and the graph appears to be a horizontal line. y 1 108 6 107 2 107

x

1

2

3

FIGURE 1.61 y=

4x [(x 2

+ 4)1/2 − x]

4x

x

#

−100 −1 × 103 −1 × 104 −1 × 105 −1 × 106 −1 × 107 −1 × 108

−1.9998 −1.999998 −1.99999998 −1.9999999998 −2.0 −2.0 −2.0

(x 2

4)1/2 − x

$

Further, the values displayed in the table no longer show the sudden jump indicative of a loss-of-significance error. We can now confidently conjecture that lim x[(x 2 + 4)

x→−∞

1/2

+ x] = −2.

BEYOND FORMULAS In examples 7.5–7.7, we demonstrated calculations that suffered from catastrophic lossof-significance errors. In each case, we showed how we could rewrite the expression to avoid this error. We have by no means exhibited a general procedure for recognizing and repairing such errors. Rather, we hope that by seeing a few of these subtle errors, and by seeing how to fix even a limited number of them, you will become a more skeptical and intelligent user of technology.

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Limits and Continuity

1-68

EXERCISES 1.7 WRITING EXERCISES 1. It is probably clear that caution is important in using technology. Equally important is redundancy. This property is sometimes thought to be a negative (wasteful, unnecessary), but its positive role is one of the lessons of this section. By redundancy, we mean investigating a problem using graphical, numerical and symbolic tools. Why is it important to use multiple methods? Answer this from a practical perspective (think of the problems in this section) and a theoretical perspective (if you have learned multiple techniques, do you understand the mathematics better?). 2. The drawback of caution and redundancy is that they take extra time. In computing limits, when should you stop and take extra time to make sure an answer is correct and when is it safe to go on to the next problem? Should you always look at a graph? compute function values? do symbolic work? an ε−δ proof? Prioritize the techniques in this chapter. f (a + h) − f (a) will be very important in h Chapter 2. For a specific function and specific a, we could compute a table of values of the fraction for smaller and smaller values of h. Why should we be wary of loss-of-significance errors?

3 3 11. lim x 4/3 ( x 2 + 1 − x 2 − 1) x→∞

√ √ 3 3 12. lim x 2/3 ( x + 4 − x − 3) x→∞

In exercises 13 and 14, compare the limits to show that small errors can have disastrous effects. x2 + x − 2 x 2 + x − 2.01 and lim x→1 x→1 x −1 x −1 x −2 x −2 14. lim 2 and lim 2 x→2 x − 4 x→2 x − 4.01 13. lim

15. Compare f (x) = sin π x and g(x) = sin 3.14x for x = 1 (radian), x = 10, x = 100 and x = 1000. 16. For exercise 1, follow the calculation of the function for x = 105 as it would proceed for a machine computing with a 10-digit mantissa. Identify where the round-off error occurs.

3. The limit lim

h→0

4. Notice that we rationalized the numerator in example 7.7. The old rule of rationalizing the denominator is another example of rewriting an expression to try to minimize computational errors. Before computers, square roots were very difficult to compute. To see one reason why you might want the square root in the numerator, suppose that dec√ you can get only one 6 imal place of accuracy, so that 3 ≈ 1.7. Compare 1.7 to √63 and then compare 2(1.7) to √63 . Which of the approximations could you do in your head?

In exercises 1–12, (a) use graphics and numerics to conjecture a value of the limit. (b) Find a computer or calculator graph showing a loss-of-significance error. (c) Rewrite the function to avoid the loss-of-significance error. √ √ 1. lim x 4x 2 + 1 − 2x 2. lim x 4x 2 + 1 + 2x x→∞

√ √ √ 3. lim x x + 4 − x + 2 x→∞

5. lim x x→∞

√

x2 + 4 −

√

x2 + 2

x→−∞

4. lim x 2 x→∞

6. lim x x→∞

1 − cos 2x 12x 2

8. lim

1 − cos x 3 x→0 x6

10. lim

7. lim

x→0

9. lim

x→0

√

√

x4 + 8 − x2

x 3 + 8 − x 3/2

1 − cos x x2

1 − cos x 4 x→0 x8

In exercises 17 and 18, compare the exact answer to one obtained by a computer with a six-digit mantissa. 17. (1.000003 − 1.000001) × 107 18. (1.000006 − 1.000001) × 107 19. If you have access to a CAS, test it on the limits of examples 7.1, 7.6 and 7.7. Based on these results, do you think that your CAS does precise calculations or numerical estimates?

EXPLORATORY EXERCISES 1. In this exercise, we look at one aspect of the mathematical study of chaos. First, iterate the function f (x) = x 2 − 2 starting at x0 = 0.5. That is, compute x1 = f (0.5), then x2 = f (x1 ), then x3 = f (x2 ) and so on. Although the sequence of numbers stays bounded, the numbers never repeat (except by the accident of round-off errors). An impressive property of chaotic functions is the butterfly effect (more properly referred to as sensitive dependence on initial conditions). The butterfly effect applies to the chaotic nature of weather and states that the amount of air stirred by a butterfly flapping its wings in Brazil can create or disperse a tornado in Texas a few days later. Therefore, longrange weather prediction is impossible. To illustrate the butterfly effect, iterate f (x) starting at x0 = 0.5 and x0 = 0.51. How many iterations does it take before the iterations are more than 0.1 apart? Try this again with x0 = 0.5 and x0 = 0.501. Repeat this exercise for the function g(x) = x 2 − 1. Even though the functions are almost identical, g(x) is not chaotic and behaves very differently. This represents an important idea in modern medical research called dynamical diseases: a small change in

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one of the constants in a function (e.g., the rate of an electrical signal within the human heart) can produce a dramatic change in the behavior of the system (e.g., the pumping of blood from the ventricles). 2. Just as we are subject to round-off error in using calculations from a computer, so are we subject to errors in a computergenerated graph. After all, the computer has to compute function values before it can decide where to plot points. On your computer or calculator, graph y = sin x 2 (a disconnected graph—or point plot—is preferable). You should see the oscillations you expect from the sine function, but with the oscillations getting faster as x gets larger. Shift your graphing window to the right several times. At some point, the plot will

..

Review Exercises

141

become very messy and almost unreadable. Depending on your technology, you may see patterns in the plot. Are these patterns real or an illusion? To explain what is going on, recall that a computer graph is a finite set of pixels, with each pixel representing one x and one y. Suppose the computer is plotting points at x = 0, x = 0.1, x = 0.2 and so on. The y-values would then be sin 02 , sin 0.12 , sin 0.22 and so on. Investigate what will happen between x = 15 and x = 16. Compute all the points (15, sin 152 ), (15.1, sin 15.12 ) and so on. If you were to graph these points, what pattern would emerge? To explain this pattern, argue that there is approximately half a period of the sine curve missing between each point plotted. Also, investigate what happens between x = 31 and x = 32.

Review Exercises WRITING EXERCISES The following list includes terms that are defined and theorems that are stated in this chapter. For each term or theorem, (1) give a precise definition or statement, (2) state in general terms what it means and (3) describe the types of problems with which it is associated. Secant line One-sided limit Removable discontinuity Vertical asymptote Method of bisections Slope of curve

Limit Continuous Horizontal asymptote Squeeze Theorem Length of line segment

Infinite limit Loss-of-significance error Slant asymptote Intermediate Value Theorem

State whether each statement is true or false and briefly explain why. If the statement is false, try to “fix it” by modifying the given statement to make a new statement that is true. 1. In calculus, problems are often solved by first approximating the solution and then improving the approximation. 2. If f (1.1) = 2.1, f (1.01) = 2.01 and so on, then lim f (x) = 2. 3. lim [ f (x) · g(x)] = [lim f (x)][lim g(x)]

x→a

x→a

In exercises 1 and 2, numerically estimate the slope of y f (x) at x a. 1. f (x) = x 2 − 2x, a = 2

x→a

In exercises 3 and 4, numerically estimate the length of the curve using (a) n 4 and (b) n 8 line segments and evenly spaced x-coordinates. 3. f (x) = sin x, 0 ≤ x ≤

π 4

4. f (x) = x 2 − x, 0 ≤ x ≤ 2

x→1

x→a

lim f (x)

4. lim

x→a

2. f (x) = sin 2x, a = 0

TRUE OR FALSE

x→a

8. Small round-off errors typically have only small effects on a calculation. √ 9. lim f (x) = L if and only if lim f (x) = L.

f (x) x→a = g(x) lim g(x) x→a

5. If f (2) = 1 and f (4) = 2, then there exists an x between 2 and 4 such that f (x) = 0. 6. For any polynomial p(x), lim p(x) = ∞. x→∞

p(x) for polynomials p and q with q(a) = 0, then 7. If f (x) = q(x) f has a vertical asymptote at x = a.

In exercises 5–10, use numerical and graphical evidence to conjecture the value of the limit. tan−1 x 2 x→0 x2

5. lim

x +2 |x + 2| 2 x 9. lim 1 + x→∞ x 7. lim

x→−2

6. lim

x→1

x2 − 1 ln x 2

8. lim (1 + 2x)1/x x→0

10. lim x 2/x x→∞

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Differentiation

2-2

Notice that the speed calculation in m/s is the same calculation we would use for the slope between the points (9.85, 100) and (19.79, 200). The connection between slope and speed (and other quantities of interest) is explored in this chapter.

2.1

TANGENT LINES AND VELOCITY

P Point of release

FIGURE 2.1 Path of rock

A traditional slingshot is essentially a rock on the end of a string, which you rotate around in a circular motion and then release. When you release the string, in which direction will the rock travel? An overhead view of this is illustrated in Figure 2.1. Many people mistakenly believe that the rock will follow a curved path, but Newton’s first law of motion tells us that the path in the horizontal plane is straight. In fact, the rock follows a path along the tangent line to the circle at the point of release. Our aim in this section is to extend the notion of tangent line to more general curves. To make our discussion more concrete, suppose that we want to find the tangent line to the curve y = x 2 + 1 at the point (1, 2) (see Figure 2.2). How could we define this? The tangent line hugs the curve near the point of tangency. In other words, like the tangent line to a circle, this tangent line has the same direction as the curve at the point of tangency. So, if you were standing on the curve at the point of tangency, took a small step and tried to stay on the curve, you would step in the direction of the tangent line. Another way to think of this is to observe that, if we zoom in sufficiently far, the graph appears to approximate that of a straight line. In Figure 2.3, we show the graph of y = x 2 + 1 zoomed in on the small rectangular box indicated in Figure 2.2. (Be aware that the “axes” indicated in Figure 2.3 do not intersect at the origin. We provide them only as a guide as to the scale used to produce the figure.) We now choose two points from the curve—for example, (1, 2) and (3, 10)—and compute the slope of the line joining these two points. Such a line is called a secant line and we denote its slope by m sec : m sec =

10 − 2 = 4. 3−1

An equation of the secant line is then determined by y−2 = 4, x −1 y

y 2.2

12

2.1

8

2.0 4 1.9

(1, 2) 4

x

2

2 4

4

1.8 x 0.92 0.96 1.00 1.04 1.08

FIGURE 2.2

FIGURE 2.3

y = x2 + 1

y = x2 + 1

2-3

SECTION 2.1

147

As can be seen in Figure 2.4a, the secant line doesn’t look very much like a tangent line. Taking the second point a little closer to the point of tangency, say (2, 5), gives the slope of the secant line as

y 12 8

m sec = 4 x

2

Tangent Lines and Velocity

y = 4(x − 1) + 2.

so that

4

..

2

4

4

5−2 =3 2−1

and an equation of the secant line as y = 3(x − 1) + 2. As seen in Figure 2.4b, this looks much more like a tangent line, but it’s still not quite there. Choosing our second point much closer to the point of tangency, say (1.05, 2.1025), should give us an even better approximation to the tangent line. In this case, we have

FIGURE 2.4a Secant line joining (1, 2) and (3, 10)

m sec =

2.1025 − 2 = 2.05 1.05 − 1

and an equation of the secant line is y = 2.05(x − 1) + 2. As can be seen in Figure 2.4c, the secant line looks very much like a tangent line, even when zoomed in quite far, as in Figure 2.4d. We continue this process by computing the slope of the secant line joining (1, 2) and the unspecified point (1 + h, f (1 + h)), for some value of h close to 0. The slope of this secant line is m sec =

f (1 + h) − 2 [(1 + h)2 + 1] − 2 = (1 + h) − 1 h

=

(1 + 2h + h 2 ) − 1 2h + h 2 = h h

Multiply out and cancel.

=

h(2 + h) = 2 + h. h

Factor out common h and cancel.

y

y

y 12

12

8

8

4

4

3

2

1 4

x

2

2

4

4

4

x

2

2

4

4

x 0.6

1.0

1.4

FIGURE 2.4b

FIGURE 2.4c

FIGURE 2.4d

Secant line joining (1, 2) and (2, 5)

Secant line joining (1, 2) and (1.05, 2.1025)

Close-up of secant line

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Differentiation

y

2-4

Notice that as h approaches 0, the slope of the secant line approaches 2, which we define to be the slope of the tangent line.

y f (x)

REMARK 1.1 We should make one more observation before moving on to the general case of tangent lines. Unlike the case for a circle, tangent lines may intersect a curve at more than one point, as indicated in Figure 2.5.

x

FIGURE 2.5

The General Case

Tangent line intersecting a curve at more than one point

To find the slope of the tangent line to y = f (x) at x = a, first pick two points on the curve. One point is the point of tangency, (a, f (a)). Call the x-coordinate of the second point x = a + h, for some small number h; the corresponding y-coordinate is then f (a + h). It is natural to think of h as being positive, as shown in Figure 2.6a, although h can also be negative, as shown in Figure 2.6b. y

y

a

ah

x

ah

x a

FIGURE 2.6a

FIGURE 2.6b

Secant line (h > 0)

Secant line (h < 0)

The slope of the secant line through the points (a, f (a)) and (a + h, f (a + h)) is given by y

m sec = Q

P x

FIGURE 2.7 Secant lines approaching the tangent line at the point P

f (a + h) − f (a) f (a + h) − f (a) = . (a + h) − a h

(1.1)

Notice that the expression in (1.1) (called a difference quotient) gives the slope of the secant line for any second point we might choose (i.e., for any h = 0). Recall that in order to obtain an improved approximation to the tangent line, we zoom in closer and closer toward the point of tangency. This makes the two points closer together, which in turn makes h closer to 0. Just how far should we zoom in? The farther, the better; this means that we want h to approach 0. We illustrate this process in Figure 2.7, where we have plotted a number of secant lines for h > 0. Notice that as the point Q approaches the point P (i.e., as h → 0), the secant line approaches the tangent line at P. We define the slope of the tangent line to be the limit of the slopes of the secant lines in (1.1) as h tends to 0, whenever this limit exists.

2-5

SECTION 2.1

..

Tangent Lines and Velocity

149

DEFINITION 1.1 The slope m tan of the tangent line to y = f (x) at x = a is given by m tan = lim

h→0

f (a + h) − f (a) , h

(1.2)

provided the limit exists. The tangent line is then the line passing through the point (a, f (a)) with slope m tan and so, the point-slope form of the equation of the tangent line is y = m tan (x − a) + f (a).

Equation of tangent line

EXAMPLE 1.1

Finding the Equation of a Tangent Line

Find an equation of the tangent line to y = x 2 + 1 at x = 1. Solution We compute the slope using (1.2): f (1 + h) − f (1) h→0 h [(1 + h)2 + 1] − (1 + 1) lim h→0 h 2 1 + 2h + h + 1 − 2 lim h→0 h 2 2h + h h(2 + h) = lim lim h→0 h→0 h h lim (2 + h) = 2.

m tan = lim y

= 12

=

8

=

4

4

=

x

2

2

4

4

Factor out common h and cancel.

h→0

Notice that the point corresponding to x = 1 is (1, 2) and the line with slope 2 through the point (1, 2) has equation y = 2(x − 1) + 2

FIGURE 2.8 y = x 2 + 1 and the tangent line at x = 1

Multiply out and cancel.

or

y = 2x.

Note how closely this corresponds to the secant lines computed earlier. We show a graph of the function and this tangent line in Figure 2.8.

EXAMPLE 1.2

Tangent Line to the Graph of a Rational Function

Find an equation of the tangent line to y =

2 at x = 2. x

Solution From (1.2), we have m tan

2 −1 f (2 + h) − f (2) 2 + h = lim = lim h→0 h→0 h h 2 − (2 + h) 2−2−h (2 + h) (2 + h) = lim = lim h→0 h→0 h h −h −1 1 = lim = lim =− . h→0 (2 + h)h h→0 2 + h 2

Since f (2 + h) =

2 . 2+h

Add fractions and multiply out.

Cancel h’s.

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The point corresponding to x = 2 is (2, 1), since f (2) = 1. An equation of the tangent line is then 1 y = − (x − 2) + 1. 2 We show a graph of function and this tangent line in Figure 2.9.

y 5 4 3 2

In cases where we cannot (or cannot easily) evaluate the limit for the slope of the tangent line, we can approximate the limit numerically. We illustrate this in example 1.3.

1 x

1 1

1

2

3

4

5

5 x 2

4

5 10

(1, 0)

x −1 x +1

(0.1, −0.8182) (0.01, −0.9802)

y 3

msec

Second Point

0 − (−1) =1 1−0 −0.8182 − (−1) = 1.818 0.1 − 0 −0.9802 − (−1) = 1.98 0.01 − 0

(−0.5, −3) (−0.1, −1.2222) (−0.01, −1.0202)

msec −3 − (−1) = 4.0 −0.5 − 0 −1.2222 − (−1) = 2.22 −0.1 − 0 −1.0202 − (−1) = 2.02 −0.01 − 0

In both columns, as the second point gets closer to (0, −1), the slope of the secant line gets closer to 2. A reasonable estimate of the slope of the curve at the point (0, −1) is then 2.

2 1 1

x −1 is shown in Figure 2.10a. We are interested in the x +1 tangent line at the point (0, −1). We sketch a tangent line in Figure 2.10b, where we have zoomed in to provide better detail. To approximate the slope, we estimate the coordinates of one point on the tangent line other than (0, −1). In Figure 2.10b, it appears that the tangent line passes through the point (1, 1). An estimate of the slope is 1 − (−1) then m tan ≈ = 2. To approximate the slope numerically, we choose several 1−0 points near (0, −1) and compute the slopes of the secant lines. For example, rounding the y-values to four decimal places, we have Second Point

FIGURE 2.10a

2

x −1 at x +1

Solution A graph of y =

10

y=

Graphical and Numerical Approximation of Tangent Lines

x = 0.

y

2

EXAMPLE 1.3

Graphically and numerically approximate the slope of the tangent line to y =

FIGURE 2.9 2 y = and tangent line at (2, 1) x

4

2-6

x 1

1

2 3

FIGURE 2.10b Tangent line

2

Velocity The slopes of tangent lines have many important applications, of which one of the most important is in computing velocity. The term velocity is certainly familiar to you, but can you say precisely what it is? We often describe velocity as a quantity determining the speed and direction of an object, but what exactly is speed? If your car did not have a speedometer, you might determine your speed using the familiar formula distance = rate × time.

(1.3)

Using (1.3), you can find the rate (speed) by simply dividing the distance by the time. However, the rate in (1.3) refers to average speed over a period of time. We are interested in the speed at a specific instant. The following story should indicate the difference. During traffic stops, police officers frequently ask drivers if they know how fast they were going. An overzealous student might answer that during the past, say, 3 years,

2-7

SECTION 2.1

..

Tangent Lines and Velocity

151

2 months, 7 days, 5 hours and 45 minutes, they’ve driven exactly 45,259.7 miles, so that their speed was rate =

distance 45,259.7 miles = ≈ 1.62118 mph. time 27,917.75 hours

Of course, most police officers would not be impressed with this analysis, but, why is it wrong? Certainly there’s nothing wrong with formula (1.3) or the arithmetic. However, it’s reasonable to argue that unless they were in their car during this entire 3-year period, the results are invalid. Suppose that the driver substitutes the following argument instead: “I left home at 6:17 P.M. and by the time you pulled me over at 6:43 P.M., I had driven exactly 17 miles. Therefore, my speed was rate =

17 miles 60 minutes · = 39.2 mph, 26 minutes 1 hour

well under the posted 45-mph speed limit.” While this is a much better estimate of the velocity than the 1.6 mph computed previously, it’s still an average velocity using too long of a time period. More generally, suppose that the function f (t) gives the position at time t of an object moving along a straight line. That is, f (t) gives the displacement (signed distance) from a fixed reference point, so that f (t) < 0 means that the object is located | f (t)| away from the reference point, but in the negative direction. Then, for two times a and b (where a < b), f (b) − f (a) gives the signed distance between positions f (a) and f (b). The average velocity vavg is then given by

vavg =

EXAMPLE 1.4

signed distance f (b) − f (a) = . time b−a

(1.4)

Finding Average Velocity

The position of a car after t minutes driving in a straight line is given by s(t) =

1 2 1 t − t 3, 2 12

0 ≤ t ≤ 4.

Approximate the velocity at time t = 2. Solution Averaging over the 2 minutes from t = 2 to t = 4, we get from (1.4) that 2.666666667 − 1.333333333 s(4) − s(2) ≈ 4−2 2 ≈ 0.666666667 mile/minute ≈ 40 mph. Of course, a 2-minute-long interval is rather long, given that cars can speed up and slow down a great deal in 2 minutes. We get an improved approximation by averaging over just one minute: vavg =

s(3) − s(2) 2.25 − 1.333333333 ≈ 3−2 1 ≈ 0.9166666667 mile/minute ≈ 55 mph.

vavg =

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..

Differentiation

1.0

s(2 h) − s(2) h 0.9166666667

0.1

0.9991666667

0.01

0.9999916667

0.001

0.999999917

0.0001

1.0

0.00001

1.0

h

2-8

While this latest estimate is certainly better than the first one, we can do better. As we make the time interval shorter and shorter, the average velocity should be getting closer and closer to the velocity at the instant t = 2. It stands to reason that, if we compute the average velocity over the time interval [2, 2 + h] and then let h → 0, the resulting average velocities should be getting closer and closer to the velocity at the instant t = 2. s(2 + h) − s(2) s(2 + h) − s(2) vavg = = . We have (2 + h) − 2 h A sequence of these average velocities is displayed in the accompanying table, for h > 0, with similar results if we allow h to be negative. It appears that the average velocity is approaching 1 mile/minute (60 mph), as h → 0. We refer to this limiting value as the instantaneous velocity. This leads us to make the following definition.

NOTE

DEFINITION 1.2

Notice that if (for example) t is measured in seconds and f (t) is measured in feet, then velocity (average or instantaneous) is measured in feet per second (ft/s). The term velocity is always used to refer to instantaneous velocity.

If f (t) represents the position of an object relative to some fixed location at time t as it moves along a straight line, then the instantaneous velocity at time t = a is given by v(a) = lim

h→0

f (a + h) − f (a) f (a + h) − f (a) = lim , h→0 (a + h) − a h

(1.5)

provided the limit exists.

EXAMPLE 1.5

Finding Average and Instantaneous Velocity

Suppose that the height of a falling object t seconds after being dropped from a height of 64 feet is given by f (t) = 64 − 16t 2 feet. Find the average velocity between times t = 1 and t = 2; the average velocity between times t = 1.5 and t = 2; the average velocity between times t = 1.9 and t = 2 and the instantaneous velocity at time t = 2. Solution The average velocity between times t = 1 and t = 2 is vavg =

f (2) − f (1) 64 − 16(2)2 − [64 − 16(1)2 ] = = −48 ft/s. 2−1 1

The average velocity between times t = 1.5 and t = 2 is vavg =

64 − 16(2)2 − [64 − 16(1.5)2 ] f (2) − f (1.5) = = −56 ft/s. 2 − 1.5 0.5

The average velocity between times t = 1.9 and t = 2 is vavg =

f (2) − f (1.9) 64 − 16(2)2 − [64 − 16(1.9)2 ] = = −62.4 ft/s. 2 − 1.9 0.1

2-9

SECTION 2.1

..

Tangent Lines and Velocity

153

The instantaneous velocity is the limit of such average velocities. From (1.5), we have v(2) = lim

h→0

f (2 + h) − f (2) (2 + h) − 2

[64 − 16(2 + h)2 ] − [64 − 16(2)2 ] h→0 h

= lim

[64 − 16(4 + 4h + h 2 )] − [64 − 16(2)2 ] h→0 h

= lim

−64h − 16h 2 −16h(h + 4) = lim h→0 h→0 h h = lim [−16(h + 4)] = −64 ft/s. = lim

Multiply out and cancel.

Factor out common h and cancel.

h→0

Recall that velocity indicates both speed and direction. In this problem, f (t) measures the height above the ground. So, the negative velocity indicates that the object is moving in the negative (or downward) direction. The speed of the object at the 2-second mark is then 64 ft/s. (Speed is simply the absolute value of velocity.) Observe that the formulas for instantaneous velocity (1.5) and for the slope of a tangent line (1.2) are identical. We want to make this connection as strong as possible, by illustrating example 1.5 graphically. We graph the position function f (t) = 64 − 16t 2 for 0 ≤ t ≤ 2. The average velocity between t = 1 and t = 2 corresponds to the slope of the secant line between the points at t = 1 and t = 2. (See Figure 2.11a.) Similarly, the average velocity between t = 1.5 and t = 2 gives the slope of the corresponding secant line. (See Figure 2.11b.) Finally, the instantaneous velocity at time t = 2 corresponds to the slope of the tangent line at t = 2. (See Figure 2.11c.) y

y

y

80

80

80

60

60

60

40

40

40

20

20

20

t 1

2

t

3

1

2

t

3

1

20

20

20

40

40

40

60

60

60

2

FIGURE 2.11a

FIGURE 2.11b

FIGURE 2.11c

Secant line between t = 1 and t =2

Secant line between t = 1.5 and t = 2

Tangent line at t = 2

3

Velocity is a rate (more precisely, the instantaneous rate of change of position with respect to time). We now generalize this notion of instantaneous rate of change. In general,

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the average rate of change of a function f (x) between x = a and x = b (a = b) is given by f (b) − f (a) . b−a The instantaneous rate of change of f (x) at x = a is given by lim

h→0

f (a + h) − f (a) , h

provided the limit exists. The units of the instantaneous rate of change are the units of f divided by (or “per”) the units of x.

EXAMPLE 1.6

Interpreting Rates of Change

If the function f (t) gives the population of a city in millions of people t years after January 1, 2000, interpret each of the following quantities, assuming that they f (2) − f (0) equal the given numbers. (a) = 0.34, (b) f (2) − f (1) = 0.31 and 2 f (2 + h) − f (2) (c) lim = 0.3. h→0 h f (b) − f (a) Solution From the preceding, is the average rate of change of the b−a function f between a and b. Expression (a) tells us that the average rate of change of f between a = 0 and b = 2 is 0.34. Stated in more common language, the city’s population grew at an average rate of 0.34 million people per year between 2000 and 2002. Similarly, expression (b) is the average rate of change between a = 1 and b = 2. That is, the city’s population grew at an average rate of 0.31 million people per year in 2001. Finally, expression (c) gives the instantaneous rate of change of the population at time t = 2. As of January 1, 2002, the city’s population was growing at a rate of 0.3 million people per year. Additional applications of the slope of a tangent line are innumerable. These include the rate of a chemical reaction, the inflation rate in economics and learning growth rates in psychology. Rates of change in nearly any discipline you can name can be thought of as slopes of tangent lines. We explore many of these applications as we progress through the text. You hopefully noticed that we tacked the phrase “provided the limit exists” onto the end of the definitions of the slope of a tangent line, the instantaneous velocity and the instantaneous rate of change. This was important, since these defining limits do not always exist, as we see in example 1.7.

y y x

EXAMPLE 1.7

A Graph with No Tangent Line at a Point

Determine whether there is a tangent line to y = |x| at x = 0. Slope 1

Slope 1 x

FIGURE 2.12 y = |x|

Solution We can look at this problem graphically, numerically and symbolically. The graph is shown in Figure 2.12. Our graphical technique is to zoom in on the point of tangency until the graph appears straight. However, no matter how far we zoom in on (0, 0), the graph continues to look like Figure 2.12. (This is one reason why we left off the scale on Figure 2.12.) From this evidence alone, we would conjecture that the tangent line does not exist. Numerically, the slope of the tangent line is the limit of the

2-11

SECTION 2.1

..

Tangent Lines and Velocity

155

slope of a secant line, as the second point approaches the point of tangency. Observe that the secant line through (0, 0) and (1, 1) has slope 1, as does the secant line through (0, 0) and (0.1, 0.1). In fact, if h is any positive number, the slope of the secant line through (0, 0) and (h, |h|) is 1. However, the secant line through (0, 0) and (−1, 1) has slope −1, as does the secant line through (0, 0) and (h, |h|) for any negative number h. We therefore conjecture that the one-sided limits are different, so that the limit (and also the tangent line) does not exist. To prove this conjecture, we take our cue from the numerical work and look at one-sided limits: if h > 0, then |h| = h, so that lim+

h→0

f (0 + h) − f (0) |h| − 0 h = lim+ = lim+ = 1. h→0 h→0 h h h

On the other hand, if h < 0, then |h| = −h (remember that if h < 0, −h > 0), so that lim−

h→0

f (0 + h) − f (0) |h| − 0 −h = lim− = lim− = −1. h→0 h→0 h h h

Since the one-sided limits are different, we conclude that lim

h→0

f (0 + h) − f (0) does not exist h

and hence, the tangent line does not exist.

EXERCISES 2.1 WRITING EXERCISES 1. What does the phrase “off on a tangent” mean? Relate the common meaning of the phrase to the image of a tangent to a circle (use the slingshot example, if that helps). In what way does the zoomed image of the tangent promote the opposite view of the relationship between a curve and its tangent? 2. In general, the instantaneous velocity of an object cannot be computed directly; the limit process is the only way to compute velocity at an instant. Given this, how does a car’s speedometer compute velocity? (Hint: Look this up in a reference book or on the Internet. An important aspect of the car’s ability to do this seemingly difficult task is that it performs analog calculations. For example, the pitch of a fly’s buzz gives us an analog device for computing the speed of a fly’s wings, since pitch is proportional to speed.) 3. Look in the news media (TV, newspaper, Internet) and find references to at least five different rates. We have defined a rate of change as the limit of the difference quotient of a function. For your five examples, state as precisely as possible what the original function is. Is the rate given quantitatively or qualitatively? If it is given quantitatively, is the rate given as a percentage or a number? In calculus, we usually compute rates (quantitatively) as numbers; is this in line with the standard usage? 4. Sketch the graph of a function that is discontinuous at x = 1. Explain why there is no tangent line at x = 1.

In exercises 1–4, sketch in a plausible tangent line at the given point. (Hint: Mentally zoom in on the point and use the zoomed image of the tangent.) 1.

y

x

p

2.

y

x

3.

at x = π

at x = 0

y

x

at x = 0

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Differentiation

y

4.

2-12

y

8. B

C

A x

D

x

at x = 1

1

In exercises 5 and 6, estimate the slope of the tangent line to the curve at x 1. y

5. 1.5

In exercises 9–12, compute the slope of the secant line between the points at (a) x 1 and x 2, (b) x 2 and x 3, (c) x 1.5 and x 2, (d) x 2 and x 2.5, (e) x 1.9 and x 2, (f) x 2 and x 2.1, and (g) use parts (a)–(f) and other calculations as needed to estimate the slope of the tangent line at x 2. √ 9. f (x) = x 3 − x 10. f (x) = x 2 + 1

1.0

11. f (x) = cos x 2

0.5

In exercises 13–16, use a CAS or graphing calculator. 13. On one graph, sketch the secant lines in exercise 9, parts (a)–(d) and the tangent line in part (g).

x 0.5

1.0

12. f (x) = tan (x/4)

1.5

14. On one graph, sketch the secant lines in exercise 10, parts (a)–(d) and the tangent line in part (g). y

6.

15. Animate the secant lines in exercise 9, parts (a), (c) and (e), converging to the tangent line in part (g).

2 x 1

2

16. Animate the secant lines in exercise 9, parts (b), (d) and (f), converging to the tangent line in part (g).

3

In exercises 17–24, find the equation of the tangent line to y f (x) at x a. Graph y f (x) and the tangent line to verify that you have the correct equation.

2 4

In exercises 7 and 8, list the points A, B, C and D in order of increasing slope of the tangent line.

17. f (x) = x 2 − 2, a = 1

18. f (x) = x 2 − 2, a = 0

19. f (x) = x 2 − 3x, a = −2

20. f (x) = x 3 + x, a = 1

2 ,a =1 x +1 √ 23. f (x) = x + 3, a = −2

x ,a =0 x −1 √ 24. f (x) = x 2 + 1, a = 1

21. f (x) =

22. f (x) =

y

7.

In exercises 25–30, use graphical and numerical evidence to determine whether the tangent line to y f (x) exists at x a. If it does, estimate the slope of the tangent; if not, explain why not.

B A

D C

25. f (x) = |x − 1| at a = 1 x

4x at a = 1 x −1 −2x 2 if x < 0 at a = 0 27. f (x) = x3 if x ≥ 0 26. f (x) =

2-13

SECTION 2.1

29. f (x) = 30. f (x) =

−2x if x < 1 at a = 1 x − 3 if x ≥ 1 x2 − 1 x3 + 1 −2x x 2 − 2x

Tangent Lines and Velocity

157

Water level

28. f (x) =

..

if x < 0 at a = 0 if x ≥ 0 if x < 0 at a = 0 if x > 0

24 hours

Time

In exercises 31–34, the function represents the position in feet of an object at time t seconds. Find the average velocity between (a) t 0 and t 2, (b) t 1 and t 2, (c) t 1.9 and t 2, (d) t 1.99 and t 2, and (e) estimate the instantaneous velocity at t 2.

41. Suppose a hot cup of coffee is left in a room for 2 hours. Sketch a reasonable graph of what the temperature would look like as a function of time. Then sketch a graph of what the rate of change of the temperature would look like.

31. f (t) = 16t 2 + 10 √ 33. f (t) = t 2 + 8t

42. Sketch a graph representing the height of a bungee-jumper. Sketch the graph of the person’s velocity (use + for upward velocity and − for downward velocity).

32. f (t) = 3t 3 + t 34. f (t) = 100 sin(t/4)

In exercises 35 and 36, use the position function f (t) meters to find the velocity at time t a seconds. 35. f (t) = −16t 2 + 5, (a) a = 1; (b) a = 2 √ 36. f (t) = t + 16, (a) a = 0; (b) a = 2 37. The table shows the freezing temperature of water in degrees Celsius at various pressures. Estimate the slope of the tangent line at p = 1 and interpret the result. Estimate the slope of the tangent line at p = 3 and interpret the result. p (atm) ◦ C

0 0

1 −7

2 −20

3 −16

4 −11

38. The table shows the range of a soccer kick launched at 30◦ above the horizontal at various initial speeds. Estimate the slope of the tangent line at v = 50 and interpret the result. Distance (yd) Speed (mph)

19 30

28 40

37 50

47 60

58 70

Elevation

39. The graph shows the elevation of a person on a hike up a mountain as a function of time. When did the hiker reach the top? When was the hiker going the fastest on the way up? When was the hiker going the fastest on the way down? What do you think occurred at places where the graph is level?

4 hours

Time

40. The graph shows the amount of water in a city water tank as a function of time. When was the tank the fullest? the emptiest? When was the tank filling up at the fastest rate? When was the tank emptying at the fastest rate? What time of day do you think the level portion represents?

43. Suppose that f (t) represents the balance in dollars of a bank account t years after January 1, 2000. Interpret each of the folf (4) − f (2) lowing. (a) = 21,034, (b) 2[ f (4) − f (3.5)] = 2 f (4 + h) − f (4) = 30,000. 25,036 and (c) lim h→0 h 44. Suppose that f (m) represents the value of a car that has been driven m thousand miles. Interpret each of the following. f (40) − f (38) = −2103, (b) f (40) − f (39) = −2040 (a) 2 f (40 + h) − f (40) and (c) lim = −2000. h→0 h 45. In using a slingshot, it is important to generate a large angular θ(a + h) − θ (a) , velocity. Angular velocity is defined by lim h→0 h where θ(t) is the angle of rotation at time t. If the angle of a slingshot is θ(t) = 0.4t 2 , what is the angular velocity after three rotations? [Hint: Which value of t (seconds) corresponds to three rotations?] 46. Find the angular velocity of the slingshot in exercise 45 after two rotations. Explain why the third rotation is helpful. 47. Sometimes an incorrect method accidentally produces a correct answer. For quadratic functions (but definitely not most other functions), the average velocity between t = r and t = s equals the average of the velocities at t = r and t = s. To show this, assume that f (t) = at 2 + bt + c is the distance function. Show that the average velocity between t = r and t = s equals a(s + r ) + b. Show that the velocity at t = r is 2ar + b and the velocity at t = s is 2as + b. Finally, show that (2ar + b) + (2as + b) a(s + r ) + b = . 2 48. Find a cubic function [try f (t) = t 3 + · · ·] and numbers r and s such that the average velocity between t = r and t = s is different from the average of the velocities at t = r and t = s.

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f (a + h) − f (a) f (x) − f (a) = lim . (Hint: x→a h x −a

How do reflections normally occur in nature? From water! You would perceive a tree and its reflection in a pool of water. This is the desert mirage!

50. Use the second limit in exercise 49 to recompute the slope in exercises 17 and 19. Which limit do you prefer?

2. You can use a VCR to estimate speed. Most VCRs play at 30 frames per second. So, with a frame-by-frame advance, you can estimate time as the number of frames divided by 30. If you know the distance covered, you can compute the average velocity by dividing distance by time. Try this to estimate how fast you can throw a ball, run 50 yards, hit a tennis ball or whatever speed you find interesting. Some of the possible inaccuracies are explored in exercise 3.

49. Show that lim

h→0

Let h = x − a.)

2 51. A car speeding around a curve in the shape 1 1of y = x (moving from left to right) skids off at the point 2 , 4 . If the car contin ues in a straight path, will it hit a tree located at the point 1, 34 ?

52. For the car in exercise 51, show graphically that there is only 2 one skid point on the curve y = x such that the tangent line passes through the point 1, 34 .

EXPLORATORY EXERCISES 1. Many optical illusions are caused by our brain’s (unconscious) use of the tangent line in determining the positions of objects. Suppose you are in the desert 100 feet from a palm tree. You see a particular spot 10 feet up on the palm tree due to light reflecting from that spot to your eyes. Normally, it is a good approximation to say that the light follows a straight line (top path in the figure).

Two paths of light from tree to person. However, when there is a large temperature difference in the air, light may follow nonlinear paths. If, as in the desert, the air near the ground is much hotter than the air higher up, light will bend as indicated by the bottom path in the figure. Our brains always interpret light coming in straight paths, so you would think the spot on the tree is at y = 10 because of the top path and also at some other y because of the bottom path. If the bottom curve is y = 0.002x 2 − 0.24x + 10, find an equation of the tangent line at x = 100 and show that it crosses the y-axis at y = −10. That is, you would “see” the spot at y = 10 and also at y = −10, a perfect reflection.

Two perceived locations of tree.

3. What is the peak speed for a human being? It has been estimated that Carl Lewis reached a peak speed of 28 mph while winning a gold medal in the 1992 Olympics. Suppose that we have the following data for a sprinter.

Meters 30 40 50 56 58 60

Seconds 3.2 4.2 5.16666 5.76666 5.93333 6.1

Meters 62 64 70 80 90 100

Seconds 6.26666 6.46666 7.06666 8.0 9.0 10.0

We want to estimate peak speed. We could start by computing distance 100 m = = 10 m/s, but this is the average speed time 10 s over the entire race, not the peak speed. Argue that we want to compute average speeds only using adjacent measurements (e.g., 40 and 50 meters, or 50 and 56 meters). Do this for all 11 adjacent pairs and find the largest speed (if you want to convert to mph, divide by 0.447). We will then explore how accurate this estimate might be. Notice that all times are essentially multiples of 1/30, since the data were obtained using the VCR technique in exercise 2. Given this, why is it suspicious that all the distances are whole numbers? To get an idea of how much this might affect your calculations, change some of the distances. For instance, if you change 60 (meters) to 59.8, how much do your average velocity calculations change? One possible way to identify where mistakes have been made is to look at the pattern of average velocities: does it seem reasonable? Would a sprinter speed up and slow down in such a pattern? In places where the pattern seems suspicious, try adjusting the distances and see if you can produce a more realistic pattern. Taking all this into account, try to quantify your error analysis: what is the highest (lowest) the peak speed could be?

2-15

SECTION 2.2

2.2

..

The Derivative

159

THE DERIVATIVE In section 2.1, we investigated two seemingly unrelated concepts: slopes of tangent lines and velocity, both of which are expressed in terms of the same limit. This is an indication of the power of mathematics, that otherwise unrelated notions are described by the same mathematical expression. This particular limit turns out to be so useful that we give it a special name.

DEFINITION 2.1 The derivative of the function f (x) at x = a is defined as f (a + h) − f (a) , (2.1) h provided the limit exists. If the limit exists, we say that f is differentiable at x = a. f (a) = lim

h→0

An alternative form of (2.1) is f (a) = lim

b→a

f (b) − f (a) . b−a

(2.2)

(See exercise 49 in section 2.1.)

EXAMPLE 2.1

Finding the Derivative at a Point

Compute the derivative of f (x) = 3x 3 + 2x − 1 at x = 1. Solution From (2.1), we have f (1 + h) − f (1) h 3(1 + h)3 + 2(1 + h) − 1 − (3 + 2 − 1) = lim h→0 h

f (1) = lim

h→0

3(1 + 3h + 3h 2 + h 3 ) + (2 + 2h) − 1 − 4 h→0 h

Multiply out and cancel.

11h + 9h 2 + 3h 3 h→0 h

Factor out common h and cancel.

= lim = lim

= lim (11 + 9h + 3h 2 ) = 11. h→0

Suppose that in example 2.1 we had also needed to find f (2) and f (3). Must we now repeat the same long limit calculation to find each of f (2) and f (3)? Instead, we compute the derivative without specifying a value for x, leaving us with a function from which we can calculate f (a) for any a, simply by substituting a for x.

EXAMPLE 2.2

Finding the Derivative at an Unspecified Point

Find the derivative of f (x) = 3x 3 + 2x − 1 at an unspecified value of x. Then, evaluate the derivative at x = 1, x = 2 and x = 3.

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2-16

Solution Replacing a with x in the definition of the derivative (2.1), we have f (x + h) − f (x) h→0 h 3(x + h)3 + 2(x + h) − 1 − (3x 3 + 2x − 1) = lim h→0 h

f (x) = lim

3(x 3 + 3x 2 h + 3xh 2 + h 3 ) + (2x + 2h) − 1 − 3x 3 − 2x + 1 h→0 h

= lim

Multiply out and cancel.

9x 2 h + 9xh 2 + 3h 3 + 2h h→0 h = lim (9x 2 + 9xh + 3h 2 + 2)

Factor out common h and cancel.

= lim

h→0

= 9x 2 + 0 + 0 + 2 = 9x 2 + 2. Notice that in this case, we have derived a new function, f (x) = 9x 2 + 2. Simply substituting in for x, we get f (1) = 9 + 2 = 11 (the same as we got in example 2.1!), f (2) = 9(4) + 2 = 38 and f (3) = 9(9) + 2 = 83. Example 2.2 leads us to the following definition.

DEFINITION 2.2 The derivative of f (x) is the function f (x) given by f (x + h) − f (x) , (2.3) h provided the limit exists. The process of computing a derivative is called differentiation. Further, f is differentiable on an interval I if it is differentiable at every point in I . f (x) = lim

h→0

In examples 2.3 and 2.4, observe that the name of the game is to write down the defining limit and then to find some way of evaluating that limit (which initially has the indeterminate form 00 ).

EXAMPLE 2.3

Finding the Derivative of a Simple Rational Function

1 (x = 0), find f (x). x Solution We have

If f (x) =

f (x + h) − f (x) h→0 h

1 1 − x +h x lim h→0 h x − (x + h) x(x + h) lim h→0 h −h lim h→0 hx(x + h) −1 1 lim = − 2, h→0 x(x + h) x

f (x) = lim

=

= = = or f (x) = −x −2 .

Since f (x + h) =

1 . x +h

Add fractions and cancel.

Cancel h’s.

2-17

..

SECTION 2.2

EXAMPLE 2.4 If f (x) =

√

The Derivative

161

The Derivative of the Square Root Function

x (for x ≥ 0), find f (x).

Solution We have f (x + h) − f (x) h √ √ x +h− x = lim h→0 h √ √ √ √ x +h− x x +h+ x = lim √ √ h→0 h x +h+ x

f (x) = lim

h→0

(x + h) − x √ √ x +h+ x h = lim √ √ h→0 h x +h+ x = lim

h→0

h→0

Multiply out and cancel.

h

= lim √

1 x +h+

√

Multiply numerator and denominator by √ √ the conjugate: x + h + x.

Cancel common h’s.

x

1 1 = √ = x −1/2 . 2 2 x Notice that f (x) is defined only for x > 0. y 15 10 5

4

x

2

2

4

The benefits of having a derivative function go well beyond simplifying the computation of a derivative at multiple points. As we’ll see, the derivative function tells us a great deal about the original function. Keep in mind that the value of a derivative at a point is the slope of the tangent line at that point. In Figures 2.13a–2.13c, we have graphed a function along with its tangent lines at three different points. The slope of the tangent line in Figure 2.13a is negative; the slope of the tangent line in Figure 2.13c is positive and the slope of the tangent line in Figure 2.13b is zero. These three tangent lines give us three points on the graph of the derivative function (see Figure 2.13d), by estimating the value of f (x) at the three points. Thus, as x changes, the slope of the tangent line changes and hence f (x) changes.

FIGURE 2.13a m tan < 0 y y

y 4

15

15

10

10

5

5

2

2

x

1

1 2

4

x

2

2

4

4

x

2

2

FIGURE 2.13b

FIGURE 2.13c

m tan = 0

m tan > 0

4

4

FIGURE 2.13d

y = f (x) (three points)

2

162

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CHAPTER 2

Differentiation

2-18

y

Sketching the Graph of f (x ) Given the Graph of f (x )

EXAMPLE 2.5 60

Given the graph of f (x) in Figure 2.14, sketch a plausible graph of f (x).

40 20 4

x

2

2

4

20 40 60

FIGURE 2.14 y = f (x)

y

Solution Rather than worrying about exact values of the slope, we only wish to get the general shape right. As in Figures 2.13a–2.13d, pick a few important points to analyze carefully. You should focus on any discontinuities and any places where the graph of f turns around. The graph levels out at approximately x = −2 and x = 2. At these points, the derivative is 0. As we move from left to right, the graph rises for x < −2, drops for −2 < x < 2 and rises again for x > 2. This means that f (x) > 0 for x < −2, f (x) < 0 for −2 < x < 2 and finally f (x) > 0 for x > 2. We can say even more. As x approaches −2 from the left, observe that the tangent lines get less steep. Therefore, f (x) becomes less positive as x approaches −2 from the left. Moving to the right from x = −2, the graph gets steeper until about x = 0, then gets less steep until it levels out at x = 2. Thus, f (x) gets more negative until x = 0, then less negative until x = 2. Finally, the graph gets steeper as we move to the right from x = 2. Putting this all together, we have the possible graph of f (x) shown in red in Figure 2.15, superimposed on the graph of f (x). The opposite question to that asked in example 2.5 is even more interesting. That is, given the graph of a derivative, what might the graph of the original function look like? We explore this in example 2.6.

60 f ' (x) 40 20 4

x

2

2

4

20 40 f (x)

60

FIGURE 2.15

y = f (x) and y = f (x)

Sketching the Graph of f (x ) Given the Graph of f (x )

EXAMPLE 2.6

Given the graph of f (x) in Figure 2.16, sketch a plausible graph of f (x). Solution Again, do not worry about getting exact values of the function, but rather only the general shape of the graph. Notice from the graph of y = f (x) that f (x) < 0 for x < −2, so that on this interval, the slopes of the tangent lines to y = f (x) are negative and the function is decreasing. On the interval (−2, 1), f (x) > 0, indicating that the tangent lines to the graph of y = f (x) have positive slope and the function is increasing. Further, this says that the graph turns around (i.e., goes from decreasing to increasing) at x = −2. We have drawn a graph exhibiting this behavior in Figure 2.17 y

y 20

20 f (x)

10

10

x

4

2

4

x

4

2

10

10

20

f '(x) 20

FIGURE 2.16 y = f (x)

FIGURE 2.17

4

y = f (x) and a plausible graph of y = f (x)

2-19

SECTION 2.2

..

The Derivative

163

superimposed on the graph of y = f (x). Further, f (x) < 0 on the interval (1, 3), so that the function decreases here. Finally, for x > 3, we have that f (x) > 0, so that the function is increasing here. We show a graph exhibiting all of this behavior in Figure 2.17. We drew the graph of f so that the small “valley” on the right side of the y-axis was not as deep as the one on the left side of the y-axis for a reason. Look carefully at the graph of f (x) and notice that | f (x)| gets much larger on (−2, 1) than on (1, 3). This says that the tangent lines and hence, the graph will be much steeper on the interval (−2, 1) than on (1, 3).

Alternative Derivative Notations

HISTORICAL NOTES Gottfried Leibniz (1646–1716) A German mathematician and philosopher who introduced much of the notation and terminology in calculus and who is credited (together with Sir Isaac Newton) with inventing the calculus. Leibniz was a prodigy who had already received his law degree and published papers on logic and jurisprudence by age 20. A true Renaissance man, Leibniz made important contributions to politics, philosophy, theology, engineering, linguistics, geology, architecture and physics, while earning a reputation as the greatest librarian of his time. Mathematically, he derived many fundamental rules for computing derivatives and helped promote the development of calculus through his extensive communications. The simple and logical notation he invented made calculus accessible to a wide audience and has only been marginally improved upon in the intervening 300 years. He wrote, “In symbols one observes an advantage in discovery which is greatest when they express the exact nature of a thing briefly . . . then indeed the labor of thought is wonderfully diminished.”

We have denoted the derivative function by f (x). There are other commonly used notations, each with advantages and disadvantages. One of the coinventors of the calculus, Gottfried df Leibniz, used the notation (Leibniz notation) for the derivative. If we write y = f (x), dx the following are all alternatives for denoting the derivative: dy df d f (x) = y = = = f (x). dx dx dx d The expression is called a differential operator and tells you to take the derivative of dx whatever expression follows. In section 2.1, we observed that f (x) = |x|, does not have a tangent line at x = 0 (i.e., it is not differentiable at x = 0), although it is continuous everywhere. Thus, there are continuous functions that are not differentiable. You might have already wondered whether the reverse is true. That is, are there differentiable functions that are not continuous? The answer (no) is provided by Theorem 2.1.

THEOREM 2.1 If f (x) is differentiable at x = a, then f (x) is continuous at x = a.

PROOF For f to be continuous at x = a, we need only show that lim f (x) = f (a). We consider x→a f (x) − f (a) lim [ f (x) − f (a)] = lim Multiply and divide by (x − a). (x − a) x→a x→a x −a f (x) − f (a) By Theorem 3.1 (iii) = lim lim (x − a) from section 1.3. x→a x→a x −a = f (a)(0) = 0,

Since f is differentiable at x = a.

where we have used the alternative definition of derivative (2.2) discussed earlier. By Theorem 3.1 in section 1.3, it now follows that 0 = lim [ f (x) − f (a)] = lim f (x) − lim f (a) x→a

x→a

x→a

= lim f (x) − f (a), x→a

which gives us the result. Note that Theorem 2.1 says that if a function is not continuous at a point then it cannot have a derivative at that point. It also turns out that functions are not differentiable at any point where their graph has a “sharp” corner, as is the case for f (x) = |x| at x = 0. (See example 1.7.)

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y

2-20

EXAMPLE 2.7

y f (x)

Show that f (x) =

f(x) 2

4 f(x) 0

Showing That a Function Is Not Differentiable at a Point 4 2x

if x < 2 is not differentiable at x = 2. if x ≥ 2

Solution The graph (see Figure 2.18) indicates a sharp corner at x = 2, so you might expect that the derivative does not exist. To verify this, we investigate the derivative by evaluating one-sided limits. For h > 0, note that (2 + h) > 2 and so, f (2 + h) = 2(2 + h). This gives us

x 2

FIGURE 2.18 lim+

A sharp corner

h→0

y

f (2 + h) − f (2) 2(2 + h) − 4 = lim+ h→0 h h 4 + 2h − 4 = lim+ h→0 h 2h = 2. = lim+ h→0 h

Multiply out and cancel.

Cancel common h’s.

Likewise, if h < 0, (2 + h) < 2 and so, f (2 + h) = 4. Thus, we have lim−

h→0

f (2 + h) − f (2) 4−4 = lim− = 0. h→0 h h

Since the one-sided limits do not agree (0 = 2), f (2) does not exist (i.e., f is not differentiable at x = 2). x

a

FIGURE 2.19a

Figures 2.19a–2.19d show a variety of functions for which f (a) does not exist. In each case, convince yourself that the derivative does not exist.

A jump discontinuity y y

y

a

a

x

a

x

x

FIGURE 2.19b

FIGURE 2.19c

FIGURE 2.19d

A vertical asymptote

A cusp

A vertical tangent line

Numerical Differentiation There are many times in applications when it is not possible or practical to compute derivatives symbolically. This is frequently the case in applications where we have only some data (i.e., a table of values) representing an otherwise unknown function. You will need an understanding of the limit definition to compute reasonable estimates of the derivative.

2-21

SECTION 2.2

EXAMPLE 2.8

..

The Derivative

165

Approximating a Derivative Numerically

√ Numerically estimate the derivative of f (x) = x 2 x 3 + 2 at x = 1.

Solution We are not anxious to struggle through the limit definition for this function. The definition tells us, however, that the derivative at x = 1 is the limit of slopes of secant lines. We compute some of these below: f (1 h) − f (1) h 4.7632 4.3715 4.3342

h 0.1 0.01 0.001

h −0.1 −0.01 −0.001

f (1 h) − f (1) h 3.9396 4.2892 4.3260

Notice that the slopes seem to be converging to approximately 4.33 as h approaches 0. Thus, we make the approximation f (1) ≈ 4.33.

EXAMPLE 2.9

Estimating Velocity Numerically

Suppose that a sprinter reaches the following distances in the given times. Estimate the velocity of the sprinter at the 6-second mark. t(s) f (t) (ft)

5.0 123.7

5.5 141.01

5.8 151.41

5.9 154.90

6.0 158.40

6.1 161.92

6.2 165.42

6.5 175.85

7.0 193.1

Solution The instantaneous velocity is the limit of the average velocity as the time interval shrinks. We first compute the average velocities over the shortest intervals given, from 5.9 to 6.0 and from 6.0 to 6.1.

Time Interval

Average Velocity

Time Interval

Average Velocity

(5.9, 6.0)

35.0 ft/s

(5.5, 6.0)

34.78 ft/s

(6.0, 6.1)

35.2 ft/s

(5.8, 6.0)

34.95 ft/s

(5.9, 6.0)

35.00 ft/s

(6.0, 6.1)

35.20 ft/s

(6.0, 6.2)

35.10 ft/s

(6.0, 6.5)

34.90 ft/s

Since these are the best individual estimates available from the data, we could just split the difference and estimate a velocity of 35.1 ft/s. However, there is useful information in the rest of the data. Based on the accompanying table, we can conjecture that the sprinter was reaching a peak speed at about the 6-second mark. Thus, we might accept the higher estimate of 35.2 ft/s. We should emphasize that there is not a single correct answer to this question, since the data are incomplete (i.e., we know the distance only at fixed times, rather than over a continuum of times).

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BEYOND FORMULAS In sections 2.3–2.8, we derive numerous formulas for computing derivatives. As you learn these formulas, keep in mind the reasons that we are interested in the derivative. Careful studies of the slope of the tangent line to a curve and the velocity of a moving object led us to the same limit, which we named the derivative. In general, the derivative represents the rate of change or the ratio of the change of one quantity to the change in another quantity. The study of change in a quantifiable way led directly to modern science and engineering. If we were limited to studying phenomena with only constant change, how much of the science that you have learned would still exist?

EXERCISES 2.2 WRITING EXERCISES 1. The derivative is important because of its many different uses and interpretations. Describe four aspects of the derivative: graphical (think of tangent lines), symbolic (the derivative function), numerical (approximations) and applications (velocity and others).

In exercises 13–18, match the graphs of the functions on the left with the graphs of their derivatives on the right. 13.

y

2. Mathematicians often use the word “smooth” to describe functions with certain (desirable) properties. Graphically, how are differentiable functions smoother than functions that are continuous but not differentiable, or functions that are not continuous? 3. Briefly describe what the derivative tells you about the original function. In particular, if the derivative is positive at a point, what do you know about the trend of the function at that point? What is different if the derivative is negative at the point?

(a)

y

x x

14.

y

(b)

y

x x

4. Show that the derivative of f (x) = 3x − 5 is f (x) = 3. Explain in terms of slope why this is true. In exercises 1–4, compute f (a) using the limits (2.1) and (2.2). 1. f (x) = 3x + 1, a = 1 3. f (x) =

√ 3x + 1, a = 1

15.

y

(c)

y

2. f (x) = 3x 2 + 1, a = 1 4. f (x) =

x

3 ,a = 2 x +1

x

In exercises 5–12, compute the derivative function f (x) using (2.1) or (2.2). 5. f (x) = 3x 2 + 1

6. f (x) = x 2 − 2x + 1

3 7. f (x) = x +1 √ 9. f (x) = 3x + 1

2 8. f (x) = 2x − 1

11. f (x) = x 3 + 2x − 1

10. f (x) = 2x + 3 12. f (x) = x 4 − 2x 2 + 1

16.

y

(d)

y x

x

2-23

17.

SECTION 2.2

y

y

(e)

..

The Derivative

167

y

22.

x

x

x

18.

y

y

(f)

x x

In exercises 23 and 24, use the given graph of f (x) to sketch a plausible graph of a continuous function f (x). y

23.

In exercises 19–22, use the given graph of f (x) to sketch a graph of f (x). 19.

x

y

y

24.

x

x

20.

25. Graph f (x) = |x| + |x − 2| and identify all x-values at which f (x) is not differentiable.

y

26. Graph f (x) = e−2/(x −x) and identify all x-values at which f (x) is not differentiable. 3

27. Find all real numbers p such that f (0) exists for f (x) = x p . x

28. Prove that if f (x) is differentiable at x = a, then f (a + ch) − f (a) = c f (a). lim h→0 h 29. If

f (x) is differentiable f (x 2 ) − f (a 2 ) . lim x→a x 2 − a2

21.

y

at

x = a = 0,

evaluate

30. Prove that if f (x) is differentiable at x = 0, f (x) ≤ 0 for all x and f (0) = 0, then f (0) = 0.

x

31. The table shows the margin of error in degrees for tennis serves hit at 100 mph with various amounts of topspin (in units of revolutions per second). Estimate the slope of the derivative at x = 60, and interpret it in terms of the benefit of extra spin. (Data adapted from The Physics and Technology of Tennis by Brody, Cross and Lindsey.) Topspin (rps) Margin of error

20 1.8

40 2.4

60 3.1

80 3.9

100 4.6

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32. The table shows the margin of error in degrees for tennis serves hit at 120 mph from various heights. Estimate the slope of the derivative at x = 8.5 and interpret it in terms of hitting a serve from a higher point. (Data adapted from The Physics and Technology of Tennis by Brody, Cross and Lindsey.) Height (ft) Margin of error

7.5 0.3

8.0 0.58

8.5 0.80

9.0 1.04

9.5 1.32

In exercises 33 and 34, use the distances f (t) to estimate the velocity at t 2. 33.

34.

t f (t) t f (t)

1.7 3.1

1.8 3.9

1.7 4.6

1.9 4.8

1.8 5.3

2.0 5.8

1.9 6.1

2.1 6.8

2.0 7.0

2.2 7.7

2.1 7.8

2.3 8.5

2.2 8.6

2.3 9.3

35. The Environmental Protection Agency uses the measurement of ton-MPG to evaluate the power-train efficiency of vehicles. The ton-MPG rating of a vehicle is given by the weight of the vehicle (in tons) multiplied by a rating of the vehicle’s fuel efficiency in miles per gallon. Several years of data for new cars are given in the table. Estimate the rate of change of ton-MPG in (a) 1994 and (b) 2000. Do your estimates imply that cars are becoming more or less efficient? Is the rate of change constant or changing? Year Ton-MPG

1992 44.9

1994 45.7

1996 46.5

1998 47.3

2000 47.7

36. The fuel efficiencies in miles per gallon of cars from 1992 to 2000 are shown in the following table. Estimate the rate of change in MPG in (a) 1994 and (b) 2000. Do your estimates imply that cars are becoming more or less fuel efficient? Comparing your answers to exercise 35, what must be happening to the average weight of cars? If weight had remained constant, what do you expect would have happened to MPG? Year MPG

1992 28.0

1994 28.1

1996 28.3

1998 28.5

2000 28.1

In exercises 37 and 38, use a CAS or graphing calculator. 37. Numerically estimate f (1) for f (x) = x x and verify your answer using a CAS. 38. Numerically estimate f (π) for f (x) = x sin x and verify your answer using a CAS.

In exercises 39 and 40, compute the right-hand derivative f (h) − f (0) D f (0) lim and the left-hand derivative h h→0 f (h) − f (0) D− f (0) lim . h h→0− 2x + 1 if x < 0 39. f (x) = 3x + 1 if x ≥ 0 2 x if x < 0 40. f (x) = 3 x if x ≥ 0 g(x) if x < 0 41. Assume that f (x) = . If f is continuous at k(x) if x ≥ 0 x = 0 and g and k are differentiable at x = 0, prove that D+ f (0) = k (0) and D− f (0) = g (0). Which statement is not true if f has a jump discontinuity at x = 0? 42. Explain why the derivative f (0) exists if and only if the onesided derivatives exist and are equal. 43. If f (x) > 0 for all x, use the tangent line interpretation to argue that f is an increasing function; that is, if a < b, then f (a) < f (b). 44. If f (x) < 0 for all x, use the tangent line interpretation to argue that f is a decreasing function; that is, if a < b, then f (a) > f (b). 45. If f (x) = x 2/3 , show graphically and numerically that f is continuous at x = 0 but f (0) does not exist. 0 if x < 0 , show graphically and numerically 46. If f (x) = 2x if x ≥ 0 that f is continuous at x = 0 but f (0) does not exist. 47. Give an example showing that the following is not true for all functions f : if f (x) ≤ x, then f (x) ≤ 1. 48. Determine whether the following is true for all functions f : if f (0) = 0, f (x) exists for all x and f (x) ≤ x, then f (x) ≤ 1. In exercises 49 and 50, give the units for the derivative function. 49. (a) f (t) represents position, measured in meters, at time t seconds. (b) f (x) represents the demand, in number of items, of a product when the price is x dollars. 50. (a) c(t) represents the amount of a chemical present, in grams, at time t minutes. (b) p(x) represents the mass, in kg, of the first x meters of a pipe. 51. Let f (t) represent the trading value of a stock at time t days. If f (t) < 0, what does that mean about the stock? If you held some shares of this stock, should you sell what you have or buy more?

2-25

SECTION 2.2

..

The Derivative

169

52. Suppose that there are two stocks with trading values f (t) and g(t), where f (t) > g(t) and 0 < f (t) < g (t). Based on this information, which stock should you buy? Briefly explain.

62. Sketch the graph of a function with the following properties: f (−2) = 4, f (0) = −2, f (2) = 1, f (−2) = −2, f (0) = 2 and f (2) = 1.

53. One model for the spread of a disease assumes that at first the disease spreads very slowly, gradually the infection rate increases to a maximum and then the infection rate decreases back to zero, marking the end of the epidemic. If I (t) represents the number of people infected at time t, sketch a graph of both I (t) and I (t), assuming that those who get infected do not recover.

63. A phone company charges one dollar for the first 20 minutes of a call, then 10 cents per minute for the next 60 minutes and 8 cents per minute for each additional minute (or partial minute). Let f (t) be the price in cents of a t-minute phone call, t > 0. Determine f (t) as completely as possible.

54. One model for urban population growth assumes that at first, the population is growing very rapidly, then the growth rate decreases until the population starts decreasing. If P(t) is the population at time t, sketch a graph of both P(t) and P (t). 55. Use the graph to list the following in increasing order: f (1), f (1.5) − f (1) , f (1). f (2) − f (1), 0.5

64. The table shows the percentage of English Premier League soccer players by birth month, where x = 0 represents November, x = 1 represents December and so on. (The data are adapted from John Wesson’s The Science of Soccer.) If these data come from a differentiable function f (x), estimate f (1). Interpret the derivative in terms of the effect of being a month older but in the same grade of school. Month Percent

0 13

1 11

2 9

3 7

4 7

y 10

EXPLORATORY EXERCISES

8

1. Compute the derivative function for x 2 , x 3 and x 4 . Based on your results, identify the pattern and conjecture a general formula √ for the derivative of x n . Test your conjecture on the functions x = x 1/2 and 1/x = x −1 .

6 4 2 3 2 1

x 1

2

3

Exercises 55 and 56 56. Use the graph to list the following in increasing order: f (0), f (0) − f (−0.5) , f (0). f (0) − f (−1), 0.5 In exercises 57–60, the limit equals f (a) for some function f (x) and some constant a. Determine f (x) and a. (1 + h)2 − (1 + h) h→0 h √ 4+h−2 58. lim h→0 h

57. lim

1 1 − 2 + h 2 59. lim h→0 h (h − 1)2 − 1 h→0 h

60. lim

61. Sketch the graph of a function with the following properties: f (0) = 1, f (1) = 0, f (3) = 6, f (0) = 0, f (1) = −1 and f (3) = 4.

2. In Theorem 2.1, it is stated that a differentiable function is guaranteed to be continuous. The converse is not true: continuous functions are not necessarily differentiable (see example 2.7). This fact is carried to an extreme in Weierstrass’ function, to be explored here. First, graph the function f 4 (x) = cos x + 12 cos 3x + 14 cos 9x + 18 cos 27x + 161 cos 81x in the graphing window 0 ≤ x ≤ 2π and −2 ≤ y ≤ 2. Note that the graph appears to have several sharp corners, where a derivative would not exist. Next, graph the function 1 1 f 6 (x) = f 4 (x) + 32 cos 243x + 64 cos 729x. Note that there are even more places where the graph appears to have sharp corners. Explore graphs of f 10 (x), f 14 (x) and so on, with more terms added. Try to give graphical support to the fact that the Weierstrass function f ∞ (x) is continuous for all x but is not differentiable for any x. More graphical evidence comes from the fractal nature of the Weierstrass function: compare the graphs of f 4 (x) with 0 ≤ x ≤ 2π and −2 ≤ y ≤ 2 and f 6 (x) − cos x − 12 cos 3x with 0 ≤ x ≤ 2π and − 12 ≤ y ≤ 12 . 9 Explain why the graphs are identical. Then explain why this indicates that no matter how much you zoom in on a graph of the Weierstrass function, you will continue to see wiggles and corners. That is, you cannot zoom in to find a tangent line. 3. Suppose there is a function F(x) such that F(1) = 1 and F(0) = f 0 , where 0 < f 0 < 1. If F (1) > 1, show graphically that the equation F(x) = x has a solution q where 0 < q < 1. (Hint: Graph y = x and a plausible F(x) and look for

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intersections.) Sketch a graph where F (1) < 1 and there are no solutions to the equation F(x) = x between 0 and 1 (although x = 1 is a solution). Solutions have a connection with the probability of the extinction of animals or family names. Suppose you and your descendants have children according to the following probabilities: f 0 = 0.2 is the probability of having no children, f 1 = 0.3 is the probability of having exactly one child, and f 2 = 0.5 is the probability of having two children. Define F(x) = 0.2 + 0.3x + 0.5x 2 and show that F (1) > 1. Find the solution of F(x) = x between x = 0 and x = 1; this number is the probability that your “line” will go extinct some time into the future. Find nonzero values of f 0 , f 1 and f 2 such that the corresponding F(x) satisfies F (1) < 1 and hence the probability of your line going extinct is 1.

as h approaches 0. Then compute the limit and compare to the derivative f (1) found in example 1.1. For h = 1, h = 0.5 and h = 0.1, compare the actual values of the symmetric difference f (a + h) − f (a) quotient and the usual difference quotient . h In general, which difference quotient provides a better estimate of the derivative? Next, compare the values of the difference quotients with h = 0.5 and h = −0.5 to the derivative f (1). Explain graphically why one is smaller and one is larger. Compare the average of these two difference quotients to the symmetric difference quotient with h = 0.5. Use this result to explain why the symmetric difference quotient might provide a better estimate of the derivative. Next, compute sev4 if x < 2 eral symmetric difference quotients of f (x) = 2x if x ≥ 2 centered at a = 2. Recall that in example 2.7 we showed that the derivative f (2) does not exist. Given this, discuss one major problem with using the symmetric difference quotient to approximate derivatives. Finally, show that if f (a) exists, f (a + h) − f (a − h) = f (a). then lim h→0 2h

4. The symmetric difference quotient of a function f centered at f (a + h) − f (a − h) . If f (x) = x 2 + 1 x = a has the form 2h and a = 1, illustrate the symmetric difference quotient as a slope of a secant line for h = 1 and h = 0.5. Based on your picture, conjecture the limit of the symmetric difference quotient

2.3

COMPUTATION OF DERIVATIVES: THE POWER RULE You have now computed numerous derivatives using the limit definition. In fact, you may have computed enough that you have started taking some shortcuts. In exploratory exercise 1 in section 2.2, we asked you to compile a list of derivatives of basic functions and to generalize. We continue that process in this section.

The Power Rule We first revisit the limit definition of derivative to compute two very simple derivatives.

y yc c

For any constant c,

a

FIGURE 2.20 A horizontal line

d c = 0. dx

(3.1)

x

Notice that (3.1) says that for any constant c, the horizontal line y = c has a tangent line with zero slope. That is, the tangent line to a horizontal line is the same horizontal line (see Figure 2.20). Let f (x) = c, for all x. From the definition in equation (2.3), we have d f (x + h) − f (x) c = f (x) = lim h→0 dx h c−c = lim = lim 0 = 0. h→0 h→0 h

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intersections.) Sketch a graph where F (1) < 1 and there are no solutions to the equation F(x) = x between 0 and 1 (although x = 1 is a solution). Solutions have a connection with the probability of the extinction of animals or family names. Suppose you and your descendants have children according to the following probabilities: f 0 = 0.2 is the probability of having no children, f 1 = 0.3 is the probability of having exactly one child, and f 2 = 0.5 is the probability of having two children. Define F(x) = 0.2 + 0.3x + 0.5x 2 and show that F (1) > 1. Find the solution of F(x) = x between x = 0 and x = 1; this number is the probability that your “line” will go extinct some time into the future. Find nonzero values of f 0 , f 1 and f 2 such that the corresponding F(x) satisfies F (1) < 1 and hence the probability of your line going extinct is 1.

as h approaches 0. Then compute the limit and compare to the derivative f (1) found in example 1.1. For h = 1, h = 0.5 and h = 0.1, compare the actual values of the symmetric difference f (a + h) − f (a) quotient and the usual difference quotient . h In general, which difference quotient provides a better estimate of the derivative? Next, compare the values of the difference quotients with h = 0.5 and h = −0.5 to the derivative f (1). Explain graphically why one is smaller and one is larger. Compare the average of these two difference quotients to the symmetric difference quotient with h = 0.5. Use this result to explain why the symmetric difference quotient might provide a better estimate of the derivative. Next, compute sev4 if x < 2 eral symmetric difference quotients of f (x) = 2x if x ≥ 2 centered at a = 2. Recall that in example 2.7 we showed that the derivative f (2) does not exist. Given this, discuss one major problem with using the symmetric difference quotient to approximate derivatives. Finally, show that if f (a) exists, f (a + h) − f (a − h) = f (a). then lim h→0 2h

4. The symmetric difference quotient of a function f centered at f (a + h) − f (a − h) . If f (x) = x 2 + 1 x = a has the form 2h and a = 1, illustrate the symmetric difference quotient as a slope of a secant line for h = 1 and h = 0.5. Based on your picture, conjecture the limit of the symmetric difference quotient

2.3

COMPUTATION OF DERIVATIVES: THE POWER RULE You have now computed numerous derivatives using the limit definition. In fact, you may have computed enough that you have started taking some shortcuts. In exploratory exercise 1 in section 2.2, we asked you to compile a list of derivatives of basic functions and to generalize. We continue that process in this section.

The Power Rule We first revisit the limit definition of derivative to compute two very simple derivatives.

y yc c

For any constant c,

a

FIGURE 2.20 A horizontal line

d c = 0. dx

(3.1)

x

Notice that (3.1) says that for any constant c, the horizontal line y = c has a tangent line with zero slope. That is, the tangent line to a horizontal line is the same horizontal line (see Figure 2.20). Let f (x) = c, for all x. From the definition in equation (2.3), we have d f (x + h) − f (x) c = f (x) = lim h→0 dx h c−c = lim = lim 0 = 0. h→0 h→0 h

2-27

..

SECTION 2.3

Computation of Derivatives: The Power Rule

171

Similarly, we have

y yx

d x = 1. dx

a

FIGURE 2.21

x

(3.2)

Notice that (3.2) says that the tangent line to the line y = x is a line of slope one (i.e., y = x; see Figure 2.21). This is unsurprising, since intuitively, it should be clear that the tangent line to any line is that same line. We verify this result as follows. Let f (x) = x. From equation (2.3), we have

Tangent line to y = x

d f (x + h) − f (x) x = f (x) = lim h→0 dx h = lim

h→0

(x + h) − x h

h = lim 1 = 1. h→0 h h→0

= lim f (x)

f (x)

1 x x2 x3 x4

0 1 2x 3x 2 4x 3

The table shown in the margin presents a short list of derivatives calculated previously either as examples or in the exercises using the limit definition. Can you identify any pattern to the derivatives shown in the table? There are two features to note. First, the power of x in the derivative is always one less than the power of x in the original function. Second, the coefficient of x in the derivative is the same as the power of x in the original function. Putting these ideas into symbolic form, we make and then prove a reasonable conjecture.

THEOREM 3.1 (Power Rule) For any integer n > 0,

d n x = nx n−1 . dx

PROOF From the limit definition of derivative given in equation (2.3), if f (x) = x n , then d n f (x + h) − f (x) (x + h)n − x n x = f (x) = lim = lim . h→0 h→0 dx h h

(3.3)

To evaluate the limit, we will need to simplify the expression in the numerator. If n is a positive integer, we can multiply out (x + h)n . Recall that (x + h)2 = x 2 + 2xh + h 2 and (x + h)3 = x 3 + 3x 2 h + 3xh 2 + h 3 . More generally, you may recall from the binomial theorem that (x + h)n = x n + nx n−1 h +

n(n − 1) n−2 2 x h + · · · + nxh n−1 + h n . 2

(3.4)

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Substituting (3.4) into (3.3), we get

f (x) = lim

x n + nx n−1 h +

h→0

n(n − 1) n−2 2 x h + · · · + nxh n−1 + h n − x n 2 h

Cancel x n terms.

n(n − 1) n−2 2 x h + · · · + nxh n−1 + h n 2 = lim h→0 h n(n − 1) Factor out h nx n−1 + x n−2 h 1 + · · · + nxh n−2 + h n−1 2 common h = lim h→0 h and cancel. n(n − 1) n−2 1 = lim nx n−1 + x h + · · · + nxh n−2 + h n−1 = nx n−1 , h→0 2 nx n−1 h +

since every term but the first has a factor of h. The power rule is very easy to apply, as we see in example 3.1.

EXAMPLE 3.1

Using the Power Rule

Find the derivative of f (x) = x 8 and g(t) = t 107 . Solution We have f (x) = Similarly,

g (t) =

d 8 x = 8x 8−1 = 8x 7 . dx

d 107 t = 107t 107−1 = 107t 106 . dt

Recall that in section 2.2, we showed that

d 1 1 = − 2. dx x x

(3.5)

Notice that we can rewrite (3.5) as d −1 x = (−1)x −2 . dx

REMARK 3.1 As we will see, the power rule holds for any power of x. We will not be able to prove this fact for some time now, as the proof of Theorem 3.1 does not generalize, since the expansion in equation (3.4) holds only for positive integer exponents. Even so, we will use the rule freely for any power of x. We state this in Theorem 3.2.

That is, the derivative of x −1 follows the same pattern as the power rule that we just stated and proved for positive integer exponents. Likewise, in section 2.2, we used the limit definition to show that d √ 1 x= √ . dx 2 x We can also rewrite (3.6) as

(3.6)

d 1/2 1 x = x −1/2 . dx 2

Here, notice that the derivative of a rational power of x also follows the same pattern as the power rule that we proved for positive integer exponents.

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173

THEOREM 3.2 (General Power Rule) For any real number r ,

d r x = r x r −1 . dx

(3.7)

The power rule is simple to use, as we see in example 3.2.

CAUTION Be careful here to avoid a common error: d −19 = −19x −18 . x dx The power rule says to subtract 1 from the exponent (even if the exponent is negative).

EXAMPLE 3.2

Using the General Power Rule

1 √ 3 , x 2 and x π . x 19 Solution From (3.7), we have

d d −19 1 = = −19x −19−1 = −19x −20 . x d x x 19 dx √ 3 If we rewrite x 2 as a fractional power of x, we can use (3.7) to compute the derivative, as follows. d √ d 2/3 2 2 3 x2 = x = x 2/3−1 = x −1/3 . dx dx 3 3 d π x = π x π −1 . Finally, we have dx

Find the derivative of

Notice that there is the additional conceptual problem in example 3.2 (which we resolve in Chapter 4) of deciding what x π means. Since the exponent isn’t rational, what exactly do we mean when we raise a number to the irrational power π ?

General Derivative Rules The power rule gives us a large class of functions whose derivatives we can quickly compute without using the limit definition. The following rules for combining derivatives further expand the number of derivatives we can compute without resorting to the definition. Keep in mind that a derivative is a limit; the differentiation rules in Theorem 3.3 then follow immediately from the corresponding rules for limits (found in Theorem 3.1 in Chapter 1).

THEOREM 3.3 If f (x) and g(x) are differentiable at x and c is any constant, then d [ f (x) + g(x)] = f (x) + g (x), dx d (ii) [ f (x) − g(x)] = f (x) − g (x) and dx d (iii) [c f (x)] = c f (x). dx (i)

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PROOF We prove only part (i). The proofs of parts (ii) and (iii) are left as exercises. Let k(x) = f (x) + g(x). Then, from the limit definition of the derivative (2.3), we get d k(x + h) − k(x) [ f (x) + g(x)] = k (x) = lim h→0 dx h = lim

h→0

[ f (x + h) + g(x + h)] − [ f (x) + g(x)] h

[ f (x + h) − f (x)] + [g(x + h) − g(x)] h→0 h

= lim = lim

h→0

f (x + h) − f (x) g(x + h) − g(x) + lim h→0 h h

By definition of k(x). Grouping the f terms together and the g terms together. By Theorem 3.1 in Chapter 1. Recognizing the derivatives of f and of g.

= f (x) + g (x).

We illustrate Theorem 3.3 by working through the calculation of a derivative step by step, showing all of the details.

EXAMPLE 3.3

Finding the Derivative of a Sum

√ Find the derivative of f (x) = 2x 6 + 3 x. Solution We have

d d √ (2x 6 ) + 3 x dx dx d d = 2 (x 6 ) + 3 (x 1/2 ) dx dx

1 −1/2 5 x = 2(6x ) + 3 2

By Theorem 3.3 (iii).

3 = 12x 5 + √ . 2 x

Simplifying.

f (x) =

EXAMPLE 3.4

By Theorem 3.3 (i).

By the power rule.

Rewriting a Function before Computing the Derivative

√ 4x 2 − 3x + 2 x Find the derivative of f (x) = . x Solution Note that we don’t yet have any rule for computing the derivative of a quotient. So, we must first rewrite f (x) by dividing out the x in the denominator. We have √ 4x 2 3x 2 x f (x) = − + = 4x − 3 + 2x −1/2 . x x x From Theorem 3.3 and the power rule (3.7), we get

d d −1/2 d 1 −3/2 f (x) = 4 (x) − 3 (1) + 2 (x = 4 − x −3/2 . )=4−0+2 − x dx dx dx 2

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175

y

EXAMPLE 3.5 10

Finding the Equation of the Tangent Line

2 at x = 1. x Solution First, notice that f (x) = 4 − 4x + 2x −1 . From Theorem 3.3 and the power rule, we have Find an equation of the tangent line to f (x) = 4 − 4x +

5 x 1

2

3

5

f (x) = 0 − 4 − 2x −2 = −4 − 2x −2 . At x = 1, the slope of the tangent line is then f (1) = −4 − 2 = −6. The line with slope −6 through the point (1, 2) has equation

10

y − 2 = −6 (x − 1).

FIGURE 2.22 y = f (x) and the tangent line at x = 1

We show a graph of y = f (x) and the tangent line at x = 1 in Figure 2.22.

Higher Order Derivatives One consequence of having the derivative function is that we can compute the derivative of a derivative. It turns out that such higher order derivatives have important applications. Suppose we start with a function f (x) and compute its derivative f (x). We can then compute the derivative of f (x), called the second derivative of f and written f (x). We can then compute the derivative of f (x), called the third derivative of f , written f (x). We can continue to take derivatives indefinitely. Below, we show common notations for the first five derivatives of f [where we assume that y = f (x)]. Order

Prime Notation

Leibniz Notation df dx

1

y = f (x)

2

y = f (x)

d2 f dx2

3

y = f (x)

d3 f dx3

4

y (4) = f (4) (x)

d4 f dx4

5

y (5) = f (5) (x)

d5 f dx5

Computing higher order derivatives is done by simply computing several first derivatives, as we see in example 3.6.

EXAMPLE 3.6

Computing Higher Order Derivatives

If f (x) = 3x 4 − 2x 2 + 1, compute as many derivatives as possible. Solution We have f (x) =

df d = (3x 4 − 2x 2 + 1) = 12x 3 − 4x. dx dx

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2-32

d2 f dx2 d3 f f (x) = dx3 d4 f f (4) (x) = dx4 d5 f f (5) (x) = dx5 and so on. It follows that

Then,

f (x) =

d (12x 3 − 4x) = 36x 2 − 4, dx d = (36x 2 − 4) = 72x, dx d = (72x) = 72, dx d = (72) = 0 dx =

f (n) (x) =

dn f = 0, for n ≥ 5. dxn

Acceleration What information does the second derivative of a function give us? Graphically, we get a property called concavity, which we develop in Chapter 3. One important application of the second derivative is acceleration, which we briefly discuss now. You are probably familiar with the term acceleration, which is the instantaneous rate of change of velocity. Consequently, if the velocity of an object at time t is given by v(t), then the acceleration is dv a(t) = v (t) = . dt

EXAMPLE 3.7

Computing the Acceleration of a Skydiver

Suppose that the height of a skydiver t seconds after jumping from an airplane is given by f (t) = 640 − 20t − 16t 2 feet. Find the person’s acceleration at time t. Solution Since acceleration is the derivative of velocity, we first compute velocity: v(t) = f (t) = 0 − 20 − 32t = −20 − 32t ft/s. Computing the derivative of this function gives us a(t) = v (t) = −32. To finish the problem, we need to determine the units of acceleration. Since the distance here is measured in feet and time is measured in seconds, the units of the velocity are feet per second, so that the units of acceleration are feet per second per second, written ft/s/s, or more commonly ft/s2 (feet per second squared). Our answer says that the velocity changes by −32 ft/s every second. In this case, the speed in the downward (negative) direction increases by 32 ft/s every second due to gravity.

BEYOND FORMULAS The power rule gives us a much-needed shortcut for computing many derivatives. Mathematicians always seek the shortest, most efficient computations. By skipping unnecessary lengthy steps and saving brain power, mathematicians free themselves to tackle complex problems with creativity. It’s important to remember,

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Computation of Derivatives: The Power Rule

177

however, that shortcuts such as the power rule must always be carefully proved. What shortcuts do you use when solving an equation such as 3x 2 − 4 = 5 and what are the intermediate steps that you are skipping?

EXERCISES 2.3 WRITING EXERCISES 1. Explain to a non-calculus-speaking friend how to (mechanically) use the power rule. Decide whether it is better to give separate explanations for positive and negative exponents; integer and noninteger exponents; other special cases. 2. In the 1700s, mathematical “proofs” were, by modern standards, a bit fuzzy and lacked rigor. In 1734, the Irish metaphysician Bishop Berkeley wrote The Analyst to an “infidel mathematician” (thought to be Edmund Halley of Halley’s comet fame). The accepted proof at the time of the power rule may be described as follows. If x is incremented to x + h, then x n is incremented to (x + h)n . It follows that n 2 − n n−2 (x + h)n − x n n−1 = nx hx + + · · · . Now, let the (x + h) − x 2 increment h vanish, and the derivative is nx n−1 . Bishop Berkeley objected to this argument. “But it should seem that the reasoning is not fair or conclusive. For when it is said, ‘let the increments vanish,’ the former supposition that the increments were something, or that there were increments, is destroyed, and yet a consequence of that supposition is retained. Which . . . is a false way of reasoning. Certainly, when we suppose the increments to vanish, we must suppose . . . everything derived from the supposition of their existence to vanish with them.” Do you think Berkeley’s objection is fair? Is it logically acceptable to assume that something exists to draw one conclusion, and then assume that the same thing does not exist to avoid having to accept other consequences? Mathematically speaking, how does the limit avoid Berkeley’s objection of the increment h both existing and not existing? 3. The historical episode in exercise 2 is just one part of an ongoing conflict between people who blindly use mathematical techniques without proof and those who insist on a full proof before permitting anyone to use the technique. To which side are you sympathetic? Defend your position in an essay. Try to anticipate and rebut the other side’s arguments. 4. Explain the first two terms in the expansion (x + h)n = x n + nhx n−1 + · · ·, where n is a positive integer. Think of multiplying out (x + h)(x + h)(x + h) · · · (x + h); how many terms would include x n ? x n−1 ?

In exercises 1–16, find the derivative of each function. 1. f (x) = x 3 − 2x + 1 √ 3. f (t) = 3t 3 − 2 t 3 5. f (x) = − 8x + 1 x 10 7. h(x) = √ − 2x x 9. f (s) = 2s 3/2 − 3s −1/3 √ 11. f (x) = 2 3 x + 3 √ 13. f (x) = x 3x 2 − x 15. f (x) =

3x 2 − 3x + 1 2x

2. f (x) = x 9 − 3x 5 + 4x 2 − 4x √ 4. f (s) = 5 s − 4s 2 + 3 2 6. f (x) = 4 − x 3 + 2 x 3 8. h(x) = 12x − x 2 − √ x 10. f (t) = 3t π − 2t 1.3 √ 3 12. f (x) = 4x − 3 x 2 14. f (x) = (x + 1)(3x 2 − 4) 16. f (x) =

4x 2 − x + 3 √ x

In exercises 17–24, compute the indicated derivative. 17. f (x) for f (x) = x 4 + 3x 2 − 2 18.

√ d2 f for f (x) = x 6 − x dx2

19.

d2 f 3 for f (x) = 2x 4 − √ dx2 x

20. f (t) for f (t) = 4t 2 − 12 +

4 t2

21. f (4) (x) for f (x) = x 4 + 3x 2 − 2 22. f (5) (x) for f (x) = x 10 − 3x 4 + 2x − 1 x2 − x + 1 √ x √ (4) 2 24. f (t) for f (t) = (t − 1)( t + t)

23. f (x) for f (x) =

In exercises 25–28, use the given position function to find the velocity and acceleration functions. 25. s(t) = −16t 2 + 40t + 10 26. s(t) = 12t 3 − 6t − 1 √ 27. s(t) = t + 2t 2 28. s(t) = 10 −

10 t

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In exercises 29–32, the given function represents the height of an object. Compute the velocity and acceleration at time t t0 . Is the object going up or down? Is the speed of the object increasing or decreasing?

y

(c)

29. h(t) = −16t 2 + 40t + 5, t0 = 1

x

30. h(t) = −16t 2 + 40t + 5, t0 = 2 31. h(t) = 10t 2 − 24t, t0 = 2 32. h(t) = 10t 2 − 24t, t0 = 1 In exercises 33–36, find an equation of the tangent line to y f (x) at x a. √ 34. f (x) = x 2 − 2x + 1, a = 2 33. f (x) = 4 x − 2x, a = 4 35. f (x) = x 2 − 2, a = 2

y

42. (a) 10

36. f (x) = 3x + 4, a = 2

In exercises 37 and 38, determine the value(s) of x for which the tangent line to y f (x) is horizontal. Graph the function and determine the graphical significance of each point. 37. f (x) = x 3 − 3x + 1

5

38. f (x) = x 4 − 2x 2 + 2

In exercises 39 and 40, determine the value(s) of x for which the slope of the tangent line to y f (x) does not exist. Graph the function and determine the graphical significance of each point. 39. f (x) = x 2/3

41. (a)

1

2

3

1

2

3

5 10

40. f (x) = x 1/3

In exercises 41 and 42, one curve represents a function f (x) and the other two represent f (x) and f (x). Determine which is which.

x

3 2 1

y

(b) 10

y

5

x

3 2 1 5 x

(b)

10

y

(c)

y

10 5

−3

x

−2

−1

−10

x 1

2

3

2-35

SECTION 2.3

In exercises 43 and 44, find a general formula for the nth derivative f (n) (x). 43. f (x) =

√

44. f (x) =

x

2 x

45. Find a second-degree polynomial (of the form ax 2 + bx + c) such that f (0) = −2, f (0) = 2 and f (0) = 3. 46. Find a second-degree polynomial (of the form ax 2 + bx + c) such that f (0) = 0, f (0) = 5 and f (0) = 1. 47. Find the area of the triangle bounded by x = 0, y = 0 and the tangent line to y = x1 at x = 1. Repeat with the triangle bounded by x = 0, y = 0 and the tangent line to y = x1 at x = 2. Show that you get the same area using the tangent line to y = x1 at any x = a > 0. 48. Show that the result of exercise 47 does not hold for y = x12 . That is, the area of the triangle bounded by x = 0, y = 0 and the tangent line to y = x12 at x = a > 0 does depend on the value of a. 49. Assume that a is a real number, f (x) is differentiable for all x ≥ a and g(x) = max f (t) for x ≥ a. Find g (x) in the cases a≤t≤x

(a) f (x) > 0 and (b) f (x) < 0. 50. Assume that a is a real number, f (x) is differentiable for all x ≥ a and g(x) = min f (t) for x ≥ a. Find g (x) in the cases a≤t≤x

(a) f (x) > 0 and (b) f (x) < 0.

..

Computation of Derivatives: The Power Rule

[Hint: to estimate the second derivative, estimate f (1998) and f (1999) and look for a trend.] t

1996

1997

1998

1999

2000

2001

f (t)

7664.8

8004.5

8347.3

8690.7

9016.8

9039.5

56. Let f (t) equal the average weight of a domestic SUV in year t. Several values are given in the table below. Estimate and interpret f (2000) and f (2000). t f (t)

1985 4055

1990 4189

1995 4353

58. Suppose that the daily output of a manufacturing plant is modeled by Q = 1000K 1/2 L 1/3 , where K is the capital investment in thousands of dollars and L is the size of the labor force in worker-hours. Assume that L stays constant and think of output as a function of capital investment, Q(x) = 1000L 1/3 x 1/2 . Find and interpret Q (40). In exercises 59–62, find a function with the given derivative. 59. f (x) = 4x 3 √

60. f (x) = 5x 4 62. f (x) =

61. f (x) =

52. A rod made of an inhomogeneous material extends from x = 0 to x = 4 meters. The mass of the portion of the rod from x = 0 to x = t is given by m(t) = 3t 2 kg. Compute m (t) and explain why it represents the density of the rod.

64. For f (x) = x|x|, show that lim

54. Suppose the function v(d) represents the average speed in m/s of the world record running time for d meters. For example, if the fastest 200-meter time ever is 19.32 s, then v(200) = 200/19.32 ≈ 10.35. Compare the function f (d) = 26.7d −0.177 to the values of v(d), which you will have to research and compute, for distances ranging from d = 400 to d = 2000. Explain what v (d) would represent. 55. Let f (t) equal the gross domestic product (GDP) in billions of dollars for the United States in year t. Several values are given in the table. Estimate and interpret f (2000) and f (2000).

2000 4619

57. If the position of an object is at time t given by f (t), then f (t) represents velocity and f (t) gives acceleration. By Newton’s second law, acceleration is proportional to the net force on the object (causing it to accelerate). Interpret the third derivative f (t) in terms of force. The term jerk is sometimes applied to f (t). Explain why this is an appropriate term.

51. A public official solemnly proclaims, “We have achieved a reduction in the rate at which the national debt is increasing.” If d(t) represents the national debt at time t years, which derivative of d(t) is being reduced? What can you conclude about the size of d(t) itself?

53. For most land animals, the relationship between leg width w and body length b follows an equation of the form w = cb3/2 for some constant c > 0. Show that if b is large enough, w (b) > 1. Conclude that for larger animals, leg width (necessary for support) increases faster than body length. Why does this put a limitation on the size of land animals?

179

x

1 x2

63. Assume that a is a real number and f (a) exists. Then f (a + h) − 2 f (a) + f (a − h) also exists. Find its value. lim h→0 h2 f (h) − 2 f (0) + f (−h) exists h2 but f (0) does not exist. (That is, the converse of exercise 63 is not true.) h→0

EXPLORATORY EXERCISES 1. A plane is cruising at an altitude of 2 miles at a distance of 10 miles from an airport. Choosing the airport to be at the point (0, 0), the plane starts its descent at the point (10, 2) and lands at the airport. Sketch a graph of a reasonable flight path y = f (x), where y represents altitude and x gives the ground distance from the airport. (Think about it as you draw!) Explain what the derivative f (x) represents. (Hint: It’s not velocity.) Explain why it is important and/or necessary to have f (0) = 0, f (10) = 2, f (0) = 0 and f (10) = 0. The simplest polynomial that can meet these requirements is a cubic polynomial f (x) = ax 3 + bx 2 + cx + d (Note: four requirements, four constants). Find values of the constants a, b, c and d to fit the flight path. [Hint: Start by setting f (0) = 0 and then

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set f (0) = 0. You may want to use your CAS to solve the equations.] Graph the resulting function; does it look right? 2 Suppose that airline regulations prohibit a derivative of 10 or larger. Why might such a regulation exist? Show that the flight path you found is illegal. Argue that in fact all flight paths meeting the four requirements are illegal. Therefore, the descent needs to start farther away than 10 miles. Find a flight path with descent starting at 20 miles away that meets all requirements. 2. We discuss a graphical interpretation of the second derivative in Chapter 3. You can discover the most important aspects of that here. For f (x) = x 4 − 2x 2 − 1, solve the equations f (x) = 0 and f (x) = 0. What do the solutions of the equation f (x) = 0 represent graphically? The solutions of the equation f (x) = 0 are a little harder to interpret. Looking at the graph of f (x) near x = 0, would you say that the graph is curving up or curving down? Compute f (0). Looking at the graph near x = 2 and x = −2, is the graph curving up or down? Compute f (2) and f (−2). Where does the graph change from curving up to curving down and vice versa? Hypothesize a relationship between f (x) and the curving of the graph of y = f (x). Test your hypothesis on a variety of functions. (Try y = x 4 − 4x 3 .) 3. In the enjoyable book Surely You’re Joking Mr. Feynman, physicist Richard Feynman tells the story of a contest he had pitting his brain against the technology of the day (an

2.4

abacus). The contest was to compute the cube root of 1729.03. Feynman came up with 12.002 before the abacus expert gave up. Feynman admits to some luck in the choice of the number 1729.03: he knew that a cubic foot contains 1728 cubic inches. Explain why this told Feynman that the answer is slightly greater than 12. How did he get three digits of accuracy? “I had learned in calculus that for small fractions, the cube root’s excess is one-third of the number’s excess. The excess, 1.03, is only one part in nearly 2000. So all I had to do is find the fraction 1/1728, divide by 3 and multiply by 12.” To see what he did, find an equation of the tangent line to y = x 1/3 at x = 1728 and find the y-coordinate of the tangent line at x = 1729.03. 4. Suppose that you want to find solutions of the equation x 3 − 4x 2 + 2 = 0. Show graphically that there is a solution between x = 0 and x = 1. We will approximate this solution in stages. First, find an equation of the tangent line to y = x 3 − 4x 2 + 2 at x = 1. Then, determine where this tangent line crosses the x-axis. Show graphically that the x-intercept is considerably closer to the solution than is x = 1. Now, repeat the process: for the new x-value, find the equation of the tangent line, determine where it crosses the x-axis and show that this is closer still to the desired solution. This process of using tangent lines to produce continually improved approximations is referred to as Newton’s method. We discuss this in some detail in section 3.1.

THE PRODUCT AND QUOTIENT RULES We have now developed rules for computing the derivatives of a variety of functions, including general formulas for the derivative of a sum or difference of two functions. Given this, you might wonder whether the derivative of a product of two functions is the same as the product of the derivatives. We test this conjecture with a simple example.

Product Rule Consider

d [(x 2 )(x 5 )]. We can compute this derivative by first combining the two terms: dx d d 7 [(x 2 )(x 5 )] = x = 7x 6 . dx dx

Is this derivative the same as the product of the two individual derivatives? Notice that

d 2 d 5 x x = (2x)(5x 4 ) dx dx d = 10x 5 = 7x 6 = [(x 2 )(x 5 )]. (4.1) dx

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set f (0) = 0. You may want to use your CAS to solve the equations.] Graph the resulting function; does it look right? 2 Suppose that airline regulations prohibit a derivative of 10 or larger. Why might such a regulation exist? Show that the flight path you found is illegal. Argue that in fact all flight paths meeting the four requirements are illegal. Therefore, the descent needs to start farther away than 10 miles. Find a flight path with descent starting at 20 miles away that meets all requirements. 2. We discuss a graphical interpretation of the second derivative in Chapter 3. You can discover the most important aspects of that here. For f (x) = x 4 − 2x 2 − 1, solve the equations f (x) = 0 and f (x) = 0. What do the solutions of the equation f (x) = 0 represent graphically? The solutions of the equation f (x) = 0 are a little harder to interpret. Looking at the graph of f (x) near x = 0, would you say that the graph is curving up or curving down? Compute f (0). Looking at the graph near x = 2 and x = −2, is the graph curving up or down? Compute f (2) and f (−2). Where does the graph change from curving up to curving down and vice versa? Hypothesize a relationship between f (x) and the curving of the graph of y = f (x). Test your hypothesis on a variety of functions. (Try y = x 4 − 4x 3 .) 3. In the enjoyable book Surely You’re Joking Mr. Feynman, physicist Richard Feynman tells the story of a contest he had pitting his brain against the technology of the day (an

2.4

abacus). The contest was to compute the cube root of 1729.03. Feynman came up with 12.002 before the abacus expert gave up. Feynman admits to some luck in the choice of the number 1729.03: he knew that a cubic foot contains 1728 cubic inches. Explain why this told Feynman that the answer is slightly greater than 12. How did he get three digits of accuracy? “I had learned in calculus that for small fractions, the cube root’s excess is one-third of the number’s excess. The excess, 1.03, is only one part in nearly 2000. So all I had to do is find the fraction 1/1728, divide by 3 and multiply by 12.” To see what he did, find an equation of the tangent line to y = x 1/3 at x = 1728 and find the y-coordinate of the tangent line at x = 1729.03. 4. Suppose that you want to find solutions of the equation x 3 − 4x 2 + 2 = 0. Show graphically that there is a solution between x = 0 and x = 1. We will approximate this solution in stages. First, find an equation of the tangent line to y = x 3 − 4x 2 + 2 at x = 1. Then, determine where this tangent line crosses the x-axis. Show graphically that the x-intercept is considerably closer to the solution than is x = 1. Now, repeat the process: for the new x-value, find the equation of the tangent line, determine where it crosses the x-axis and show that this is closer still to the desired solution. This process of using tangent lines to produce continually improved approximations is referred to as Newton’s method. We discuss this in some detail in section 3.1.

THE PRODUCT AND QUOTIENT RULES We have now developed rules for computing the derivatives of a variety of functions, including general formulas for the derivative of a sum or difference of two functions. Given this, you might wonder whether the derivative of a product of two functions is the same as the product of the derivatives. We test this conjecture with a simple example.

Product Rule Consider

d [(x 2 )(x 5 )]. We can compute this derivative by first combining the two terms: dx d d 7 [(x 2 )(x 5 )] = x = 7x 6 . dx dx

Is this derivative the same as the product of the two individual derivatives? Notice that

d 2 d 5 x x = (2x)(5x 4 ) dx dx d = 10x 5 = 7x 6 = [(x 2 )(x 5 )]. (4.1) dx

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You can now plainly see from (4.1) that the derivative of a product is not generally the product of the corresponding derivatives. In Theorem 4.1, we state a general rule for computing the derivative of a product of two differentiable functions.

THEOREM 4.1 (Product Rule) Suppose that f and g are differentiable at x. Then d [ f (x)g(x)] = f (x)g(x) + f (x)g (x). dx

(4.2)

PROOF Since we are proving a general rule, we have only the limit definition of derivative to use. For p(x) = f (x)g(x), we have d p(x + h) − p(x) [ f (x)g(x)] = p (x) = lim h→0 dx h = lim

h→0

f (x + h)g(x + h) − f (x)g(x) . h

(4.3)

Notice that the elements of the derivatives of f and g are present [limit, f (x + h), f (x) etc.], but we need to get them into the right form. The trick is to add and subtract f (x)g(x + h) in the numerator. From (4.3), we have p (x) = lim

h→0

f (x + h)g(x + h) − f (x)g(x + h) + f (x)g(x + h) − f (x)g(x) h

Subtract and add f (x)g(x + h).

= lim

h→0

f (x + h)g(x + h) − f (x)g(x + h) f (x)g(x + h) − f (x)g(x) + lim h→0 h h

Break into two pieces.

g(x + h) − g(x) f (x + h) − f (x) g(x + h) + lim f (x) h→0 h h f (x + h) − f (x) g(x + h) − g(x) = lim lim g(x + h) + f (x) lim h→0 h→0 h→0 h h

= lim

h→0

= f (x)g(x) + f (x)g (x).

Recognize the derivative of f and the derivative of g.

Here, we identified the limits of the difference quotients as derivatives. There is also a subtle technical detail in the last step: since g is differentiable at x, recall that it must also be continuous at x, so that g(x + h) → g(x) as h → 0. In example 4.1, notice that the product rule saves us from multiplying out a messy product.

EXAMPLE 4.1

Using the Product Rule

√ 2 Find f (x) if f (x) = (2x 4 − 3x + 5) x 2 − x + . x

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Solution Although we could first multiply out the expression, the product rule will simplify our work:

√ √ d 2 d 2 4 2 4 2 f (x) = (2x − 3x + 5) x − x + + (2x − 3x + 5) x − x+ dx x dx x

√ 2 1 2 = (8x 3 − 3) x 2 − x + + (2x 4 − 3x + 5) 2x − √ − 2 . x x 2 x The product rule usually leaves derivatives in an “unsimplified” form as in example 4.1. Unless you have some particular reason to simplify (or some particularly obvious simplification to make), just leave the results of the product rule alone.

EXAMPLE 4.2

Finding the Equation of the Tangent Line

Find an equation of the tangent line to y = (x 4 − 3x 2 + 2x)(x 3 − 2x + 3) at x = 0. Solution From the product rule, we have y = (4x 3 − 6x + 2)(x 3 − 2x + 3) + (x 4 − 3x 2 + 2x)(3x 2 − 2). Evaluating at x = 0, we have y (0) = (2)(3) + (0)(−2) = 6. The line with slope 6 and passing through the point (0, 0) [why (0, 0)?] has equation y = 6x.

Quotient Rule Given our experience with the product rule, you probably have no expectation that the derivative of a quotient will turn out to be the quotient of the derivatives. Just to be sure, let’s try a simple experiment. Note that

d x5 d 3 = (x ) = 3x 2 , 2 dx x dx

while

d 5

(x ) 5 3 d x5 5x 4 dx 2 . = 1 = x = 3x = d 2 2x 2 dx x2 (x ) dx

Since these are obviously not the same, we know that the derivative of a quotient is generally not the quotient of the corresponding derivatives. Theorem 4.2 provides us with a general rule for computing the derivative of a quotient of two differentiable functions.

THEOREM 4.2 (Quotient Rule) Suppose that f and g are differentiable at x and g(x) = 0. Then d f (x) f (x)g(x) − f (x)g (x) . = d x g(x) [g(x)]2

(4.4)

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PROOF You should be able to guess the structure of the proof. For Q(x) =

f (x) , we have from the g(x)

limit definition of derivative that d Q(x + h) − Q(x) f (x) = Q (x) = lim h→0 d x g(x) h f (x + h) f (x) − g(x + h) g(x) = lim h→0 h f (x + h)g(x) − f (x)g(x + h) g(x + h)g(x) = lim h→0 h f (x + h)g(x) − f (x)g(x + h) . = lim h→0 hg(x + h)g(x)

Add the fractions.

Simplify.

As in the proof of the product rule, we look for the right term to add and subtract in the numerator, so that we can isolate the limit definitions of f (x) and g (x). In this case, we add and subtract f (x)g(x), to get Q (x) = lim

h→0

f (x + h)g(x) − f (x)g(x + h) hg(x + h)g(x)

f (x + h)g(x) − f (x)g(x) + f (x)g(x) − f (x)g(x + h) Subtract and add f (x)g(x). hg(x + h)g(x) g(x + h) − g(x) f (x + h) − f (x) Group first two and last g(x) − f (x) two terms together h h = lim and factor out common h→0 g(x + h)g(x) = lim

h→0

lim

=

h→0

terms.

f (x + h) − f (x) g(x + h) − g(x) g(x) − f (x) lim h→0 h h lim g(x + h)g(x)

h→0

=

f (x)g(x) − f (x)g (x) , [g(x)]2

where we have again recognized the derivatives of f and g and used the fact that g is differentiable to imply that g is continuous, so that lim g(x + h) = g(x).

h→0

Notice that the numerator in the quotient rule looks very much like the product rule, but with a minus sign between the two terms. For this reason, you need to be very careful with the order.

EXAMPLE 4.3

Using the Quotient Rule

Compute the derivative of f (x) =

x2 − 2 . x2 + 1

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Solution Using the quotient rule, we have d 2 d (x − 2) (x 2 + 1) − (x 2 − 2) (x 2 + 1) d x d x f (x) = 2 2 (x + 1) =

2x(x 2 + 1) − (x 2 − 2)(2x) (x 2 + 1)2

=

6x . (x 2 + 1)2

In this case, we rewrote the numerator because it simplified significantly. This often occurs with the quotient rule. Now that we have the quotient rule, we can justify the use of the power rule for negative integer exponents. (Recall that we have been using this rule without proof since section 2.3.)

THEOREM 4.3 (Power Rule) For any integer exponent n,

d n x = nx n−1 . dx

PROOF We have already proved this for positive integer exponents. So, suppose that n < 0 and let M = −n > 0. Then, using the quotient rule, we get

d n 1 d −M d 1 x = x = Since x −M = M . x dx dx dx x M d d (1) x M − (1) (x M ) dx dx = By the quotient rule. (x M )2 =

(0)x M − (1)M x M−1 x 2M

By the power rule, since M > 0.

=

−M x M−1 = −M x M−1−2M x 2M

By the usual rules of exponents.

= (−M)x −M−1 = nx n−1 ,

Since n = −M.

d M x = M x M−1 , since M > 0. dx As we see in example 4.4, it is sometimes preferable to rewrite a function, instead of automatically using the product or quotient rule.

where we have used the fact that

EXAMPLE 4.4

A Case Where the Product and Quotient Rules Are Not Needed

√ 2 Compute the derivative of f (x) = x x + 2 . x

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185

Solution Although it may be tempting to use the product rule for the first term and the quotient rule for the second term, notice that we can rewrite the function and considerably simplify our work. We can combine the two powers of x in the first term and since the second term is a fraction with a constant numerator, we can more simply write it using a negative exponent. We have √ 2 f (x) = x x + 2 = x 3/2 + 2x −2 . x Using the power rule, we have simply f (x) =

3 1/2 x − 4x −3 . 2

Applications You will see important uses of the product and quotient rules throughout your mathematical and scientific studies. We start you off with a couple of simple applications now.

EXAMPLE 4.5

Investigating the Rate of Change of Revenue

Suppose that a product currently sells for $25, with the price increasing at the rate of $2 per year. At this price, consumers will buy 150 thousand items, but the number sold is decreasing at the rate of 8 thousand per year. At what rate is the total revenue changing? Is the total revenue increasing or decreasing? Solution To answer these questions, we need the basic relationship revenue = quantity × price (e.g., if you sell 10 items at $4 each, you earn $40). Since these quantities are changing in time, we write R(t) = Q(t)P(t), where R(t) is revenue, Q(t) is quantity sold and P(t) is the price, all at time t. We don’t have formulas for any of these functions, but from the product rule, we have R (t) = Q (t)P(t) + Q(t)P (t). We have information about each of these terms: the initial price, P(0), is 25 (dollars); the rate of change of the price is P (0) = 2 (dollars per year); the initial quantity, Q(0), is 150 (thousand items) and the rate of change of quantity is Q (0) = −8 (thousand items per year). Note that the negative sign of Q (0) denotes a decrease in Q. Thus, R (0) = (−8)(25) + (150)(2) = 100 thousand dollars per year. Since the rate of change is positive, the revenue is increasing. This may be a surprise since one of the two factors in the equation is decreasing and the rate of decrease of the quantity is more than the rate of increase in the price.

EXAMPLE 4.6

Using the Derivative to Analyze Sports

A golf ball of mass 0.05 kg struck by a golf club of mass m kg with speed 50 m/s will 83m have an initial speed of u(m) = m/s. Show that u (m) > 0 and interpret this m + 0.05 result in golf terms. Compare u (0.15) and u (0.20).

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y

2-42

Solution From the quotient rule, we have u (m) =

60

40

20

m 0.1

0.2

0.3

FIGURE 2.23 u(m) =

83m m + 0.05

83(m + 0.05) − 83m 4.15 = . 2 (m + 0.05) (m + 0.05)2

Both the numerator and denominator are positive, so u (m) > 0. A positive slope for all tangent lines indicates that the graph of u(m) should rise from left to right (see Figure 2.23). Said a different way, u(m) increases as m increases. In golf terms, this says that (all other things being equal) the greater the mass of the club, the greater the velocity of the ball will be. Finally, we compute u (0.15) = 103.75 and u (0.20) = 66.4. This says that the rate of increase in ball speed is much less for the heavier club than for the lighter one. Since heavier clubs can be harder to control, the relatively small increase in ball speed obtained by making the heavy club even heavier may not compensate for the decrease in control.

EXERCISES 2.4 WRITING EXERCISES 1. The product and quotient rules give you the ability to symbolically calculate the derivative of a wide range of functions. However, many calculators and almost every computer algebra system (CAS) can do this work for you. Discuss why you should learn these basic rules anyway. (Keep example 4.5 in mind.) 2. Gottfried Leibniz is recognized (along with Sir Isaac Newton) as a coinventor of calculus. Many of the fundamental methods and (perhaps more importantly) much of the notation of calculus are due to Leibniz. The product rule was worked out by Leibniz in 1675, in the form d(x y) = (d x)y + x(dy). His “proof,” as given in a letter written in 1699, follows. “If we are to differentiate xy we write: (x + d x)(y + dy) − x y = x dy + y d x + d x d y. But here d x d y is to be rejected as incomparably less than x dy + y d x. Thus, in any particular case the error is less than any finite quantity.” Answer Leibniz’ letter with one describing your own “discovery” of the product rule for d(x yz). Use Leibniz’ notation. 3. In example 4.1, we cautioned you against always multiplying out the terms of the derivative. To see one reason for this warning, suppose that you want to find solutions of the equation f (x) = 0. (In fact, we do this routinely in Chapter 3.) Explain why having a factored form of f (x) is very helpful. Discuss the extent to which the product rule gives you a factored form. 4. Many students prefer the product rule to the quotient rule. Many computer algebra systems actually use the product rule to compute the derivative of f (x)[g(x)]−1 instead of using f (x) (see exercise 18 on the next page). the quotient rule on g(x)

Given the simplifications in problems like example 4.3, explain why the quotient rule can be preferable. In exercises 1–16, find the derivative of each function. 1. f (x) = (x 2 + 3)(x 3 − 3x + 1) 2. f (x) = (x 3 − 2x 2 + 5)(x 4 − 3x 2 + 2)

√ 3 3. f (x) = ( x + 3x) 5x 2 − x

3 4. f (x) = (x 3/2 − 4x) x 4 − 2 + 2 x 3x − 2 6. f (x) = 5. f (x) = 5x + 1 √ 3x − 6 x 8. f (x) = 7. f (x) = 5x 2 − 2 (x + 1)(x − 2) 10. f (x) = 9. f (x) = 2 x − 5x + 1 x 2 + 3x − 2 12. f (x) = 11. f (x) = √ x √ 14. f (x) = 13. f (x) = x( 3 x + 3) 15. f (x) = (x 2 − 1)

x 3 + 3x 2 x2 + 2

x 2 + 2x + 5 x 2 − 5x + 1 6x − 2/x √ x2 + x x 2 − 2x x 2 + 5x 2x x2 + 1 5 x2 + 2 3 x x2 − 1 16. f (x) = (x + 2) 2 x +x

17. Write out the product rule for the function f (x)g(x)h(x). (Hint: Group the first two terms together.) Describe the general product rule: for n functions, what is the derivative of the product f 1 (x) f 2 (x) f 3 (x) · · · f n (x)? How many terms are there? What does each term look like?

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SECTION 2.4

18. Use the quotient rule to show that the derivative of [g(x)]−1 is −g (x)[g(x)]−2 . Then use the product rule to compute the derivative of f (x)[g(x)]−1 . In exercises 19 and 20, find the derivative of each function using the general product rule developed in exercise 17. 19. f (x) = x 2/3 (x 2 − 2)(x 3 − x + 1) 20. f (x) = (x + 4)(x 3 − 2x 2 + 1)(3 − 2/x) In exercises 21–24, assume that f and g are differentiable with f (0) − 1, f (1) − 2, f (0) − 1, f (1) 3, g(0) 3, g(1) 1, g (0) − 1 and g (1) − 2. 21. Find an equation of the tangent line to h(x) = f (x)g(x) at (a) x = 1, (b) x = 0. f (x) at 22. Find an equation of the tangent line to h(x) = g(x) (a) x = 1, (b) x = 0. 23. Find an equation of the tangent line to h(x) = x 2 f (x) at (a) x = 1, (b) x = 0. 24. Find an equation of the tangent line to h(x) = (a) x = 1, (b) x = 0.

x2 at g(x)

25. Suppose that for some toy, the quantity sold Q(t) at time t years decreases at a rate of 4%; explain why this translates to Q (t) = −0.04Q(t). Suppose also that the price increases at a rate of 3%; write out a similar equation for P (t) in terms of P(t). The revenue for the toy is R(t) = Q(t)P(t). Substituting the expressions for Q (t) and P (t) into the product rule R (t) = Q (t)P(t) + Q(t)P (t), show that the revenue decreases at a rate of 1%. Explain why this is “obvious.” 26. As in exercise 25, suppose that the quantity sold decreases at a rate of 4%. By what rate must the price be increased to keep the revenue constant? 27. Suppose the price of an object is $20 and 20,000 units are sold. If the price increases at a rate of $1.25 per year and the quantity sold increases at a rate of 2000 per year, at what rate will revenue increase? 28. Suppose the price of an object is $14 and 12,000 units are sold. The company wants to increase the quantity sold by 1200 units per year, while increasing the revenue by $20,000 per year. At what rate would the price have to be increased to reach these goals? 29. A baseball with mass 0.15 kg and speed 45 m/s is struck by a baseball bat of mass m and speed 40 m/s (in the opposite direction of the ball’s motion). After the collision, the ball has 82.5m − 6.75 initial speed u(m) = m/s. Show that u (m) > 0 m + 0.15 and interpret this in baseball terms. Compare u (1) and u (1.2). 30. In exercise 29, if the baseball has mass M kg at speed 45 m/s and the bat has mass 1.05 kg at speed 40 m/s, the ball’s initial

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The Product and Quotient Rules

187

86.625 − 45M m/s. Compute u (M) and M + 1.05 interpret its sign (positive or negative) in baseball terms.

speed is u(M) =

31. In example 4.6, it is reasonable to assume that the speed of the golf club at impact decreases as the mass of the club increases. If, for example, the speed of a club of mass m is v = 8.5/m m/s at impact, then the initial speed of the golf ball 14.11 is u(m) = m/s. Show that u (m) < 0 and interpret m + 0.05 this in golf terms. 32. In example 4.6, if the golf club has mass 0.17 kg and strikes the ball with speed v m/s, the ball has initial speed 0.2822v u(v) = m/s. Compute and interpret the derivative 0.217 u (v). 33. Assume that g(x) is continuous at x = 0 and define f (x) = xg(x). Show that f (x) is differentiable at x = 0. Illustrate the result with g(x) = |x|. 34. Determine whether the result of exercise 33 still holds if x = 0 is replaced with x = a = 0. In exercises 35–40, use the symbolic differentiation feature on your CAS or calculator. 35. Repeat example 4.4 with your CAS. If its answer is not in the same form as ours in the text, explain how the CAS computed its answer. 36. Repeat exercise 15 with your CAS. If its answer is not in the same form as ours in the back of the book, explain how the CAS computed its answer. 37. Use your CAS to sketch the derivative of sin x. What function does this look like? Repeat with sin 2x and sin 3x. Generalize to conjecture the derivative of sin kx for any constant k. 38. Repeat exercise 37 with sin x 2 . To identify the derivative, sketch a curve outlining the tops of the curves of the derivative graph and try to identify the amplitude of the derivative. √ 3x 3 + x 2 39. Find the derivative of f (x) = on your CAS. Comx −3 3 for x > 0 and √ for pare its answer to √ 2 3x + 1 2 3x + 1 x < 0. Explain how to get this answer and your CAS’s answer, if it differs.

2x 2 x2 − x − 2 2x − on 40. Find the derivative of f (x) = x −2 x +1 your CAS. Compare its answer to 2. Explain how to get this answer and your CAS’s answer, if it differs. 41. Suppose that F(x) = f (x)g(x) for infinitely differentiable functions f (x) and g(x) (that is, f (x), f (x), etc. exist for all x). Show that F (x) = f (x)g(x) + 2 f (x)g (x) + f (x)g (x). Compute F (x). Compare F (x) to the binomial formula for (a + b)2 and compare F (x) to the formula for (a + b)3 .

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42. Given that (a + b)4 = a 4 + 4a 3 b + 6a 2 b2 + 4ab3 + b4 , write out a formula for F (4) (x). (See exercise 41.) 43. Use the product rule to show that if g(x) = [ f (x)]2 and f (x) is differentiable, then g (x) = 2 f (x) f (x). This is an example of the chain rule, to be discussed in section 2.5. 44. Use the result from exercise 43 and the product rule to show that if g(x) = [ f (x)]3 and f (x) is differentiable, then g (x) = 3[ f (x)]2 f (x). Hypothesize the correct chain rule for the derivative of [ f (x)]n . 45. The relationship among the pressure P, volume V and temperature T of a gas or liquid is given by van der Waals’ equation

n2a P + 2 (V − nb) = n RT for positive constants a, b, n V and R. Solve the equation for P. Treating T as a constant and V as the variable, find the critical point (Tc , Pc , Vc ) such that P (V ) = P (V ) = 0. (Hint: Don’t solve either equation separately, but substitute the result from one equation into the other.) For temperatures above Tc , the substance can exist only in its gaseous form; below Tc , the substance is a gas or liquid (depending on the pressure and volume). For water, take R = 0.08206 l-atm/mo-K, a = 5.464 12 -atm/mo2 and b = 0.03049 l/mo. Find the highest temperature at which n = 1 mo of water may exist as a liquid. (Note: your answer will be in degrees Kelvin; subtract 273.15 to get degrees Celsius.) 46. The amount of an allosteric enzyme is affected by the presence of an activator. If x is the amount of activator and f is the amount of enzyme, then one model of an allosteric acx 2.7 tivation is f (x) = . Find and interpret lim f (x) and x→0 1 + x 2.7 lim f (x). x→∞

47. For the allosteric enzyme of exercise 46, compute and interpret f (x). 48. Enzyme production can also be inhibited. In this situation, the amount of enzyme as a function of the amount of inhibitor 1 . Find and interpret lim f (x), is modeled by f (x) = x→0 1 + x 2.7 lim f (x) and f (x). x→∞

In exercises 49–52, find the derivative of the expression for an unspecified differentiable function f . √ √ x f (x) 3 49. x f (x) 52. x f (x) 50. 51. x2 f (x)

EXPLORATORY EXERCISES 1. Most cars are rated for fuel efficiency by estimating miles per gallon in city driving (c) and miles per gallon in highway driving (h). The Environmental Protection Agency uses the

2-44

1 as its overall rating of gas us0.55/c + 0.45/ h age (this is called the 55/45 combined mpg). You may want to rewrite the function (find a common denominator in the denominator and simplify) to work this exercise. (a) Think of c as the variable and h as a constant, and show dr that > 0. Interpret this result in terms of gas mileage. dc (b) Think of h as the variable and c as a constant, and show dr that > 0. dh (c) Show that if c = h, then r = c. (d) Show that if c < h, then c < r < h. To do this, assume that c is a constant and c < h. Explain why the results of parts (b) and (c) imply that r > c. Next, show that dr < 0.45. Explain why this result along with the result dh of part (b) implies that r < h. Explain why the results of parts (a)–(d) must be true if the EPA’s combined formula is a reasonable way to average the ratings c and h. To get some sense of how the formula works, take c = 20 and graph r as a function of h. Comment on why the EPA might want to use a function whose graph flattens out as this one does. formula r =

2. In many sports, the collision between a ball and a striking implement is central to the game. Suppose the ball has weight w and velocity v before the collision and the striker (bat, tennis racket, golf club, etc.) has weight W and velocity −V before the collision (the negative indicates the striker is moving in the opposite direction from the ball). The velocity of the ball after W V (1 + c) + v(cW − w) the collision will be u = , where W +w the parameter c, called the coefficient of restitution, represents the “bounciness” of the ball in the collision. Treating W as the independent variable (like x) and the other parameters as constants, compute the derivative and verify that du V (1 + c)w + cvw + vw ≥ 0 since all parameters are = dW (W + w)2 nonnegative. Explain why this implies that if the athlete uses a bigger striker (bigger W ) with all other things equal, the speed of the ball increases. Does this match your intuition? What is doubtful about the assumption of all other things being equal? du du du du Similarly compute and interpret , , and . (Hint: dw dv d V dc c is between 0 and 1 with 0 representing a dead ball and 1 the liveliest ball possible.) 3. Suppose that a soccer player strikes the ball with enough energy that a stationary ball would have initial speed 80 mph. Show that the same energy kick on a ball moving directly to the player at 40 mph will launch the ball at approximately 100 mph. (Use the general collision formula in exploratory exercise 2 with c = 0.5 and assume that the ball’s weight is much less than the soccer player’s weight.) In general, what proportion of the ball’s incoming speed is converted by the kick into extra speed in the opposite direction?

2-45

SECTION 2.5

2.5

..

The Chain Rule

189

THE CHAIN RULE √ Suppose that the function P(t) = 100 + 8t models the population of a city after t years. Then the rate of growth of that population after 2 years is given by P (2). At present, we need the limit definition to compute this derivative. However, observe that √ P(t) is the composition of the two functions f (t) = t and g(t) = 100 + 8t, so that P(t) = f (g(t)). Also, notice that both f (t) and g (t) are easily computed using existing derivative rules. We now develop a general rule for the derivative of a composition of two functions. The following simple examples will help us to identify the form of the chain rule. Notice that from the product rule d d [(x 2 + 1)2 ] = [(x 2 + 1)(x 2 + 1)] dx dx = 2x(x 2 + 1) + (x 2 + 1)2x = 2(x 2 + 1)2x. Of course, we can write this as 4x(x 2 + 1), but the unsimplified form helps us to understand the form of the chain rule. Using this result and the product rule, notice that d d [(x 2 + 1)3 ] = [(x 2 + 1)(x 2 + 1)2 ] dx dx = 2x(x 2 + 1)2 + (x 2 + 1)2(x 2 + 1)2x = 3(x 2 + 1)2 2x. We leave it as a straightforward exercise to extend this result to d [(x 2 + 1)4 ] = 4(x 2 + 1)3 2x. dx You should observe that, in each case, we have brought the exponent down, lowered the power by one and then multiplied by 2x, the derivative of x 2 + 1. Notice that we can write (x 2 + 1)4 as the composite function f (g(x)) = (x 2 + 1)4 , where g(x) = x 2 + 1 and f (x) = x 4 . Finally, observe that the derivative of the composite function is d d [ f (g(x))] = [(x 2 + 1)4 ] = 4(x 2 + 1)3 2x = f (g(x))g (x). dx dx This is an example of the chain rule, which has the following general form.

THEOREM 5.1 (Chain Rule) If g is differentiable at x and f is differentiable at g(x), then d [ f (g(x))] = f (g(x)) g (x). dx

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PROOF At this point, we can prove only the special case where g (x) = 0. Let F(x) = f (g(x)). Then, d F(x + h) − F(x) [ f (g(x))] = F (x) = lim h→0 dx h f (g(x + h)) − f (g(x)) = lim h→0 h = lim

h→0

Since F(x) = f (g(x)).

f (g(x + h)) − f (g(x)) g(x + h) − g(x) h g(x + h) − g(x)

f (g(x + h)) − f (g(x)) g(x + h) − g(x) = lim lim h→0 h→0 g(x + h) − g(x) h =

lim

g(x+h)→g(x)

Multiply numerator and denominator by g(x + h) − g(x). Regroup terms.

f (g(x + h)) − f (g(x)) g(x + h) − g(x) lim h→0 g(x + h) − g(x) h

= f (g(x))g (x), where the next to the last line is valid since as h → 0, g(x + h) → g(x), by the continuity of g. (Recall that since g is differentiable, it is also continuous.) You will be asked in the exercises to fill in some of the gaps in this argument. In particular, you should identify why we need g (x) = 0 in this proof. It is often helpful to think of the chain rule in Leibniz notation. If y = f (u) and u = g(x), then y = f (g(x)) and the chain rule says that dy dy du = . dx du d x

REMARK 5.1 The chain rule should make sense intuitively as follows. dy as the We think of dx (instantaneous) rate of change dy as the of y with respect to x, du (instantaneous) rate of change du as of y with respect to u and dx the (instantaneous) rate of change of u with respect to x. dy = 2 (i.e., y is So, if du changing at twice the rate of u) du = 5 (i.e., u is changing and dx at five times the rate of x), it should make sense that y is changing at 2 × 5 = 10 times dy = 10, the rate of x. That is, dx which is precisely what equation (5.1) says.

EXAMPLE 5.1

(5.1)

Using the Chain Rule

Differentiate y = (x + x − 1)5 . 3

Solution For u = x 3 + x − 1, note that y = u 5 . From (5.1), we have dy dy du d 5 du = = (u ) dx du d x du dx

Since y = u 5 .

d 3 (x + x − 1) dx = 5(x 3 + x − 1)4 (3x 2 + 1).

= 5u 4

It is helpful to think of the chain rule in terms of inside functions and outside functions. For the composition f (g(x)), f is referred to as the outside function and g is referred to as the inside function. The chain rule derivative f (g(x))g (x) can then be viewed as the derivative of the outside function times the derivative of the inside function. In example 5.1, the inside function is x 3 + x − 1 (the expression √ inside the parenof 100 + 8t as comtheses) and the outside function is u 5 . In example 5.2, we think √ posed of the inside function 100 + 8t and the outside function u. Some careful thought about the pieces of a composition of functions will help you use the chain rule effectively.

2-47

SECTION 2.5

EXAMPLE 5.2 Find

..

The Chain Rule

191

Using the Chain Rule on a Radical Function

d √ ( 100 + 8t). dt

Solution Let u = 100 + 8t and note that

√

100 + 8t = u 1/2 . Then, from (5.1),

d √ d 1 du ( 100 + 8t) = (u 1/2 ) = u −1/2 dt dt 2 dt 1 4 d = √ (100 + 8t) = √ . 2 100 + 8t dt 100 + 8t Notice that here, the derivative of the inside is the derivative of the expression under the square root sign. You are now in a position to calculate the derivative of a very large number of functions, by using the chain rule in combination with other differentiation rules.

EXAMPLE 5.3

Derivatives Involving Chain Rules and Other Rules

√ Compute the derivative of f (x) = x 3 4x + 1, g(x) =

8x 8 and h(x) = 3 . (x 3 + 1)2 (x + 1)2

Solution Notice the differences in these three functions. The first function f (x) is a product of two functions, g(x) is a quotient of two functions and h(x) is a constant divided by a function. This tells us to use the product rule for f (x), the quotient rule for g(x) and simply the chain rule for h(x). For the first function, we have √ d √ d 3√ x 4x + 1 = 3x 2 4x + 1 + x 3 4x + 1 dx dx √ 1 d = 3x 2 4x + 1 + x 3 (4x + 1)−1/2 (4x + 1) 2 d x

f (x) =

= 3x

2

√

By the product rule.

By the chain rule.

derivative of the inside

4x + 1 + 2x (4x + 1) 3

−1/2

.

Simplifying.

Next, we have 8(x 3 + 1)2 − 8x d [(x 3 + 1)2 ] 8x dx = (x 3 + 1)2 (x 3 + 1)4 d 3 3 2 3 8(x + 1) − 8x 2(x + 1) (x + 1) d x

d g (x) = dx

=

derivative of the inside

(x 3 + 1)4

=

8(x 3 + 1)2 − 16x(x 3 + 1)3x 2 (x 3 + 1)4

=

8(x 3 + 1) − 48x 3 8 − 40x 3 = . (x 3 + 1)3 (x 3 + 1)3

By the quotient rule.

By the chain rule.

Simplification.

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2-48

For h(x), notice that instead of using the quotient rule, it is simpler to rewrite the function as h(x) = 8(x 3 + 1)−2 . Then h (x) =

d d [8(x 3 + 1)−2 ] = −16(x 3 + 1)−3 (x 3 + 1) = −16(x 3 + 1)−3 (3x 2 ) dx d x derivative of the inside −3

= −48x (x + 1) . 2

3

In example 5.4, we apply the chain rule to a composition of a function with a composition of functions.

EXAMPLE 5.4

A Derivative Involving Multiple Chain Rules

√ Find the derivative of f (x) = ( x 2 + 4 − 3x 2 )3/2 .

Solution We have 1/2 d 3 2 f (x) = x + 4 − 3x 2 x 2 + 4 − 3x 2 2 dx

1/2 3 1 2 d = x 2 + 4 − 3x 2 (x + 4)−1/2 (x 2 + 4) − 6x 2 2 dx

1/2 1 3 2 2 2 −1/2 (x + 4) = x + 4 − 3x (2x) − 6x 2 2 1/2 3 2 x(x 2 + 4)−1/2 − 6x . = x + 4 − 3x 2 2

By the chain rule.

By the chain rule.

Simplification.

We now use the chain rule to compute the derivative of an inverse function in terms of the original function. Recall that we write g(x) = f −1 (x) if g( f (x)) = x for all x in the domain of f and f (g(x)) = x for all x in the domain of g. From this last equation, assuming that f and g are differentiable, it follows that d d [ f (g(x))] = (x). dx dx From the chain rule, we now have f (g(x))g (x) = 1. Solving this for g (x), we get

g (x) =

1 , f (g(x))

assuming we don’t divide by zero. We now state the result as Theorem 5.2.

THEOREM 5.2 If f is differentiable at all x and has an inverse function g(x) = f −1 (x), then 1 g (x) = , f (g(x)) provided f (g(x)) = 0. As we see in example 5.5, in order to use Theorem 5.2, we must be able to compute values of the inverse function.

2-49

SECTION 2.5

EXAMPLE 5.5

..

The Chain Rule

193

The Derivative of an Inverse Function

Given that the function f (x) = x 5 + 3x 3 + 2x + 1 has an inverse function g(x), compute g (7). y 8 6 4 2 −3

−2 −1

x −2

1

2

3

−4 −6 −8

TODAY IN MATHEMATICS Fan Chung (1949– ) A Taiwanese mathematician with a highly successful career in American industry. She says, “As an undergraduate in Taiwan, I was surrounded by good friends and many women mathematicians. . . . A large part of education is learning from your peers, not just the professors.” Collaboration has been a hallmark of her career. “Finding the right problem is often the main part of the work in establishing the connection. Frequently a good problem from someone else will give you a push in the right direction and the next thing you know, you have another good problem.”

FIGURE 2.24 y = x 5 + 3x 3 + 2x + 1

Solution First, notice from Figure 2.24 that f appears to be one-to-one and so, will have an inverse. From Theorem 5.2, we have g (7) =

1 f (g(7))

.

(5.2)

It’s easy to compute f (x) = 5x 4 + 9x 2 + 2, but to use Theorem 5.2 we also need the value of g(7). If we write x = g(7), then x = f −1 (7), so that f (x) = 7. In general, solving the equation f (x) = 7 may be beyond our algebraic abilities. (Try solving x 5 + 3x 3 + 2x + 1 = 7 to see what we mean.) By trial and error, however, it is not hard to see that f (1) = 7, so that g(7) = 1. [Keep in mind that for inverse functions, f (x) = y and g(y) = x are equivalent statements.] Returning to equation (5.2), we now have g (7) =

1 1 = . f (1) 16

Notice that our solution in example 5.5 is dependent on finding an x for which f (x) = 7. This particular example was workable by trial and error, but finding most values other than g(7) would have been quite difficult or impossible to solve exactly.

BEYOND FORMULAS If you think that the method used in example 5.5 is roundabout, then you have the right idea. The chain rule in particular and calculus in general give us methods for determining quantities that are not directly computable. In the case of example 5.5, we use the properties of one function to determine properties of another function. The key to our ability to do this is understanding the theory behind the chain rule. Can you think of situations in your life where your understanding of your friends’ personalities helped you convince them, in a roundabout way, to do something that at first they didn’t want to do?

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2-50

EXERCISES 2.5 WRITING EXERCISES 1. If Fred can run 10 mph and Greg can run twice as fast as Fred, how fast can Greg run? The answer is obvious for most people. Formulate this simple problem as a chain rule calculation and conclude that the chain rule (in this context) is obvious.

In exercises 23 and 24, find an equation of the tangent line to y f (x) at x a. √ 23. f (x) = x 2 + 16, a = 3

2. The biggest challenge in computing √ the derivatives of (x 2 + 4)(x 3 − x + 1), (x 2 + 4) x 3 − x + 1 and √ x 2 + 4 x 3 − x + 1 is knowing which rule (product, chain etc.) to use when. Discuss how you know which rule to use when. (Hint: Think of the order in which you would perform operations to compute the value of each function for a specific choice of x.)

24. f (x) =

3. One simple implication of the chain rule is: if g(x) = f (x − a), then g (x) = f (x − a). Explain this derivative graphically: how does g(x) compare to f (x) graphically and why do the slopes of the tangent lines relate as the formula indicates? 4. Another simple implication of the chain rule is: if h(x) = f (2x), then h (x) = 2 f (2x). Explain this derivative graphically: how does h(x) compare to f (x) graphically and why do the slopes of the tangent lines relate as the formula indicates? In exercises 1–4, find the derivative with and without using the chain rule. 1. f (x) = (x 3 − 1)2

2. f (x) = (x 2 + 2x + 1)2

3. f (x) = (x + 1)

4. f (x) = (2x + 1)4

2

3

In exercises 5–22, find the derivative of each function. √ 6. f (x) = (x 3 + x − 1)3 5. f (x) = x 2 + 4 √ √ 7. f (x) = x 5 x 3 + 2 8. f (x) = (x 3 + 2) x 5 x3 x2 + 4 9. f (x) = 2 10. f (x) = 2 (x + 4) (x 3 )2 5

6

11. f (x) = √ x2 + 4 √ 13. f (x) = ( x + 3)4/3 √ 15. f (x) = ( x 3 + 2 + 2x)−2 x 17. f (x) = √ x2 + 1 x 19. f (x) = x2 + 1 21. f (x) = x x 4 + 2x 4 3

22. f (x) =

(x 3 + 4) 12. f (x) = 8 √ 14. f (x) = x(x 4/3 + 3) 16. f (x) = 4x 2 + (8 − x 2 )2 (x 2 − 1)2 x2 + 1 √ 20. f (x) = (x 2 + 1)( x + 1)3

18. f (x) =

x2

6 , a = −2 +4

In exercises 25 and 26, use the position function to find the velocity at time t 2. (Assume units of meters and seconds.) 25. s(t) =

√ t2 + 8

60t 26. s(t) = √ t2 + 1

In exercises 27 and 28, compute f (x), f (x) and f (4) (x), and identify a pattern for the nth derivative f (n) (x). 27. f (x) =

√ 2x + 1

28. f (x) =

In exercises 29–32, use the table of values to estimate the derivative of h(x) f (g(x)) or k(x) g( f (x)). x f (x) g(x) 29. h (1)

−3 −2 6

−2 −1 4

−1 0 2

0 −1 2

30. k (1)

1 −2 4

3x 2 + 2 x 3 + 4/x 4 √ (x 3 − 4) x 2 + 2

2 −3 6

31. k (3)

3 −2 4

4 0 2

5 2 1

32. h (3)

In exercises 33 and 34, use the relevant information to compute the derivative for h(x) f (g(x)). 33. h (1), where f (1) = 3, g(1) = 2, f (1) = 4, f (2) = 3, g (1) = −2 and g (3) = 5 34. h (2), where f (2) = 1, g(2) = 3, f (2) = −1, f (3) = −3, g (1) = 2 and g (2) = 4 In exercises 35–40, f (x) has an inverse g(x). Use Theorem 5.2 to find g (a). 35. f (x) = x 3 + 4x − 1, a = −1 36. f (x) = x 3 + 2x + 1, a = −2 37. f (x) = x 5 + 3x 3 + x, a = 5

8 x +2

2 x +1

38. f (x) = x 5 + 4x − 2, a = −2 √ 39. f (x) = x 3 + 2x + 4, a = 2 √ 40. f (x) = x 5 + 4x 3 + 3x + 1, a = 3

2-51

SECTION 2.5

In exercises 41–44, find a function g(x) such that g (x) f (x). 41. f (x) = (x + 3) (2x) 2

43. f (x) = √

42. f (x) = x (x + 4)

2

2

x

44. f (x) =

x2 + 1

3

..

The Chain Rule

195

y 12

2/3

x (x 2 + 1)2

10 8

45. A function f (x) is an even function if f (−x) = f (x) for all x and is an odd function if f (−x) = − f (x) for all x. Prove that the derivative of an even function is odd and the derivative of an odd function is even.

6 4 2

In exercises 46–49, find the derivative of the expression for an unspecified differentiable function f (x). 46. f (x 2 )

√ 47. f ( x)

48.

4 f (x) + 1

49.

1 1 + [ f (x)]2

50. If the graph of a differentiable function f (x) is symmetric about the line x = a, what can you say about the symmetry of the graph of f (x)? In exercises 51–54, use the given graphs to estimate the derivative. y 10

x

1

2

3

y = c(x) 51. 52. 53. 54.

f (2), where f (x) = b(a(x)) f (0), where f (x) = a(b(x)) f (−1), where f (x) = c(a(x)) f (1), where f (x) = b(c(x))

√ 55. Determine all values of x such that f (x) = 3 x 3 − 3x 2 + 2x is not differentiable. Describe the graphical property that prevents the derivative from existing. 56. Determine all values of x for which f (x) = |2x| + |x − 4| + |x + 4| is not differentiable. Describe the graphical property that prevents the derivative from existing. 57. Which steps in our outline of the proof of the chain rule are not well documented? Where do we use the assumption that g (x) = 0?

8 6 4 2 4

2

EXPLORATORY EXERCISES x 2

4

2

1 T df . Compute the derivative , 2L p dT where we think of T as the independent variable and treat p and L as constants. Interpret this derivative in terms of a guitarist tightening or loosening the string to “tune” it. Compute df the derivative and interpret it in terms of a guitarist playing dL notes by pressing the string against a fret.

at the frequency f =

y = a(x)

y 10 5

4

1. A guitar string of length L, density p and tension T will vibrate

x

2

2 5 10

y = b(x)

4

2. Newton’s second law of motion is F = ma, where m is the mass of the object that undergoes an acceleration a due to an applied force F. This law is accurate at low speeds. At high speeds, we use the corresponding formula from Einstein’s v(t) d , where v(t) theory of relativity, F = m dt 1 − v 2 (t)/c2 is the velocity function and c is the speed of light. Compute v(t) d . What has to be “ignored” to simplify dt 1 − v 2 (t)/c2 this expression to the acceleration a = v (t) in Newton’s second law?

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2.6

..

Differentiation

2-52

DERIVATIVES OF TRIGONOMETRIC FUNCTIONS

Displacement u(t) Equilibrium position

FIGURE 2.25 Spring-mass system

Springs are essential components of many mechanical systems, including shock absorbing systems for cars, stereo equipment and other sensitive devices. Imagine a weight hanging from a spring suspended from the ceiling (see Figure 2.25). If we set the weight in motion (e.g., by tapping down on it), the weight will bounce up and down in ever-shortening strokes until it eventually is again at rest (equilibrium). For short periods of time, this motion is nearly periodic. Suppose we measure the vertical displacement of the weight from its natural resting (equilibrium) position (see Figure 2.25). When we pull the weight down, its vertical displacement is negative. The weight then swings up to where the displacement is positive, swings down to a negative displacement and so on. The only functions we’ve experienced that exhibit this kind of behavior are the sine and cosine functions. We calculate the derivatives of these and the other trigonometric functions in this section. We can learn a lot about the derivatives of sin x and cos x from their graphs. From the graph of y = sin x in Figure 2.26a, notice the horizontal tangents at x = −3π/2, −π/2, π/2 and 3π/2. At these x-values, the derivative must equal 0. The tangent lines have positive slope for−2π < x < −3π/2, negative slope for −3π/2 < x < −π/2 and so on. For each interval on which the derivative is positive (or negative), the graph appears to be steepest in the middle of the interval: for example, from x = −π/2, the graph gets steeper until about x = 0 and then gets less steep until leveling out at x = π/2. A sketch of the derivative graph should then look like the graph in Figure 2.26b, which looks like the graph of y = cos x. We show here that this conjecture is, in fact, correct. In the exercises, you are asked to perform a similar graphical analysis to conjecture that the derivative of f (x) = cos x equals −sin x. y

y

1

1

x w

q

q

w

2p

1

x

p

p

2p

1

FIGURE 2.26a

FIGURE 2.26b

y = sin x

The derivative of f (x) = sin x

Before we move to the calculation of the derivatives of the six trigonometric functions, we first consider a few limits involving trigonometric functions. (We refer to these results as lemmas—minor theorems that lead up to some more significant result.) You will see shortly why we must consider these first.

LEMMA 6.1 lim sin θ = 0.

θ →0

2-53

SECTION 2.6

y

..

Derivatives of Trigonometric Functions

197

This result certainly seems reasonable, especially when we consider the graph of y = sin x. In fact, we have been using this for some time now, having stated this (without proof) as part of Theorem 3.4 in section 1.3. We now prove the result. u

1

PROOF

sin u u

x

For 0 < θ <

π , consider Figure 2.27. From the figure, observe that 2 0 ≤ sin θ ≤ θ.

(6.1)

It is a simple matter to see that lim 0 = 0 = lim+ θ.

θ →0+

FIGURE 2.27

θ →0

(6.2)

From the Squeeze Theorem (see section 1.3), it now follows from (6.1) and (6.2) that

Definition of sin θ

lim sin θ = 0,

θ →0+

also. Similarly, you can show that lim sin θ = 0.

θ →0−

This is left as an exercise. Since both one-sided limits are the same, it follows that lim sin θ = 0.

θ →0

y

LEMMA 6.2

Q(1, tan u)

lim cos θ = 1.

P(cos u, sin u)

θ →0

(0, 1)

The proof of this result is straightforward and follows from Lemma 6.1 and the Pythagorean Theorem. We leave this as an exercise. The following result was conjectured to be true (based on a graph and some computations) when we first examined limits in section 1.2. We can now prove the result.

u x

O

R(1, 0)

LEMMA 6.3

FIGURE 2.28a

sin θ = 1. θ →0 θ lim

y

PROOF

1

π . 2 Referring to Figure 2.28a, observe that the area of the circular sector OPR is larger than the area of the triangle OPR, but smaller than the area of the triangle OQR. That is,

Assume 0 < θ < u

x

0 < Area OPR < Area sector OPR < Area OQR.

(6.3)

You can see from Figure 2.28b that FIGURE 2.28b A circular sector

Area sector OPR = π (radius)2 (fraction of circle included) θ θ = π (12 ) = . 2π 2

(6.4)

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CHAPTER 2

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Differentiation

2-54

1 1 (base) (height) = (1) sin θ 2 2

(6.5)

1 (1) tan θ. 2

(6.6)

1 θ 1 sin θ < < tan θ. 2 2 2

(6.7)

Area OPR =

Also,

Area OQR =

and

Thus, from (6.3), (6.4), (6.5) and (6.6), we have 0< If we divide (6.7) by affected), we get

1 2

sin θ (note that this is positive, so that the inequalities are not 1<

θ tan θ 1 < = . sin θ sin θ cos θ

Taking reciprocals (again, everything here is positive), we find 1>

sin θ > cos θ. θ

(6.8)

The inequality (6.8) also holds if − π2 < θ < 0. (This is left as an exercise.) Finally, note that lim cos θ = 1 = lim 1.

θ →0

θ →0

Thus, it follows from (6.8) and the Squeeze Theorem that sin θ =1 θ →0 θ lim

also.

We need one additional limit result before we tackle the derivatives of the trigonometric functions.

LEMMA 6.4 lim

θ →0

1 − cos θ = 0. θ

y

Before we prove this, we want to make this conjecture seem reasonable. We draw a 1 − cos x graph of y = in Figure 2.29. The tables of function values that follow should x prove equally convincing.

0.8 0.4

4

x

2

2

4

0.8

FIGURE 2.29 1 − cos x y= x

0.1

1 − cos x x 0.04996

−0.1

1 − cos x x −0.04996

0.01

0.00499996

−0.01

−0.00499996

0.001

0.0005

−0.001

−0.0005

0.0001

0.00005

−0.0001

−0.00005

x

x

Now that we have strong evidence for the conjecture, we prove the lemma.

2-55

SECTION 2.6

..

Derivatives of Trigonometric Functions

199

PROOF

1 − cos θ 1 − cos θ 1 + cos θ lim = lim θ→0 θ →0 θ θ 1 + cos θ 1 − cos2 θ θ →0 θ(1 + cos θ)

= lim

Multiply out numerator and denominator.

sin2 θ θ →0 θ(1 + cos θ)

sin θ sin θ = lim θ →0 θ 1 + cos θ

sin θ sin θ = lim lim θ →0 θ θ →0 1 + cos θ

0 = (1) = 0, 1+1 = lim

Multiply numerator and denominator by 1 + cos θ.

Since sin2 θ + cos2 θ = 1.

Split up terms, since both limits exist.

as conjectured. We are finally in a position to compute the derivatives of the sine and cosine functions. The derivatives of the other trigonometric functions will then follow by the quotient rule.

THEOREM 6.1 d sin x = cos x. dx

PROOF From the limit definition of derivative, for f (x) = sin x, we have d sin (x + h) − sin (x) sin x = f (x) = lim h→0 dx h sin x cos h + sin h cos x − sin x = lim h→0 h sin x cos h − sin x sin h cos x + lim = lim h→0 h→0 h h cos h − 1 sin h + (cos x) lim = (sin x) lim h→0 h→0 h h = (sin x)(0) + (cos x)(1) = cos x, from Lemmas 6.3 and 6.4.

THEOREM 6.2 d cos x = − sin x. dx

Trig identity: sin (α + β) = sin α cos β + sin β cos α. Grouping terms with sin x and terms with sin h separately. Factoring sin x from the first term and cos x from the second term.

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The proof of Theorem 6.2 is left as an exercise. For the remaining four trigonometric functions, we can use the quotient rule in conjunction with the derivatives of sin x and cos x.

THEOREM 6.3 d tan x = sec2 x. dx

PROOF We have from the quotient rule that

d d sin x tan x = dx d x cos x d d (sin x)(cos x) − (sin x) (cos x) dx = dx (cos x)2 cos x(cos x) − sin x(−sin x) = (cos x)2 =

cos2 x + sin2 x 1 = = sec2 x, (cos x)2 (cos x)2

where we have used the quotient rule and the preceding results on the derivatives of sin x and cos x (Theorems 6.1 and 6.2). The derivatives of the remaining trigonometric functions are left as exercises. The derivatives of all six trigonometric functions are summarized below.

d sin x = cos x dx

d cos x = −sin x dx

d tan x = sec2 x dx

d cot x = −csc2 x dx

d sec x = sec x tan x dx

d csc x = −csc x cot x dx

Example 6.1 shows where the product rule is necessary.

EXAMPLE 6.1

A Derivative That Requires the Product Rule

Find the derivative of f (x) = x 5 cos x. Solution From the product rule, we have d 5 d d 5 (x cos x) = (x ) cos x + x 5 (cos x) dx dx dx = 5x 4 cos x − x 5 sin x.

2-57

SECTION 2.6

EXAMPLE 6.2

..

Derivatives of Trigonometric Functions

201

Computing Some Routine Derivatives

Compute the derivatives of (a) f (x) = sin2 x and (b) g(x) = 4 tan x − 5 csc x. Solution For (a), we first rewrite the function as f (x) = (sin x)2 and use the chain rule. We have d f (x) = (2 sin x) (sin x) = 2 sin x cos x. d x derivative of the inside

g (x) = 4 sec2 x + 5 csc x cot x.

For (b), we have

You must be very careful to distinguish between similar notations with very different meanings, as you see in example 6.3.

EXAMPLE 6.3

The Derivatives of Some Similar Trigonometric Functions

Compute the derivative of f (x) = cos x 3 , g(x) = cos3 x and h(x) = cos 3x. Solution Note the differences in these three functions. Using the implied parentheses we normally do not bother to include, we have f (x) = cos(x 3 ), g(x) = (cos x)3 and h(x) = cos (3x). For the first function, we have f (x) =

d d 3 cos (x 3 ) = −sin(x 3 ) (x ) = −sin(x 3 )(3x 2 ) = −3x 2 sin x 3 . dx d x derivative of the inside

Next, we have g (x) =

d d (cos x)3 = 3(cos x)2 (cos x) dx d x derivative of the inside

= 3(cos x)2 (−sin x) = −3 sin x cos2 x. Finally, we have h (x) =

d d (cos 3x) = −sin (3x) (3x) = −sin (3x)(3) = −3 sin 3x. dx d x derivative of the inside

By combining our trigonometric rules with the product, quotient and chain rules, we can now differentiate many complicated functions.

EXAMPLE 6.4

A Derivative Involving the Chain Rule and the Quotient Rule

Find the derivative of f (x) = sin Solution We have f (x) = cos

2x . x +1

d 2x 2x x + 1 dx x + 1 derivative of the inside

2(x + 1) − 2x(1) 2x x +1 (x + 1)2

2x 2 = cos . x + 1 (x + 1)2 = cos

By the chain rule.

By the quotient rule.

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Applications The trigonometric functions arise quite naturally in the solution of numerous physical problems of interest. For instance, it can be shown that the vertical displacement of a weight suspended from a spring, in the absence of damping (i.e., when resistance to the motion, such as air resistance, is negligible), is given by u(t) = a cos (ωt) + b sin (ωt), Displacement u(t)

where ω is the frequency, t is time and a and b are constants. (See Figure 2.30 for a depiction of such a spring-mass system.) Equilibrium position

EXAMPLE 6.5 FIGURE 2.30 Spring-mass system

Analysis of a Spring-Mass System

Suppose that u(t) measures the displacement (measured in inches) of a weight suspended from a spring t seconds after it is released and that u(t) = 4 cos t. Find the velocity at any time t and determine the maximum velocity. Solution Since u(t) represents position (displacement), the velocity is given by u (t). We have u (t) = 4(−sin t) = −4 sin t, where u (t) is measured in inches per second. Of course, sin t oscillates between −1 and 1 and hence, the largest that u (t) can be is −4(−1) = 4 inches per second. This occurs when sin t = −1, that is, at t = 3π/2, t = 7π/2 and so on. Notice that at these times u(t) = 0, so that the weight is moving fastest when it is passing through its resting position.

1 farad

EXAMPLE 6.6 1 henry

Analysis of a Simple Electrical Circuit

The diagram in Figure 2.31 shows a simple electrical circuit. If the capacitance is 1 (farad), the inductance is 1 (henry) and the impressed voltage is 2t 2 (volts) at time t, then a model for the total charge Q(t) in the circuit at time t is Q(t) = 2 sin t + 2t 2 − 4 (coulombs).

~ 2t 2 volts

FIGURE 2.31 A simple circuit

The current is defined to be the rate of change of the charge with respect to time and so is given by dQ (amperes). dt Compare the current at times t = 0 and t = 1. I (t) =

Solution In general, the current is given by dQ = 2 cos t + 4t (amperes). dt Notice that at time t = 0, I (0) = 2 (amperes). At time t = 1, I (t) =

I (1) = 2 cos 1 + 4 ≈ 5.08 (amperes). This represents an increase of I (1) − I (0) 3.08 (100)% ≈ (100)% = 154%. I (0) 2

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EXAMPLE 6.7

..

Derivatives of Trigonometric Functions

203

Finding the Equation of the Tangent Line

Find an equation of the tangent line to y = 3 tan x − 2 csc x π at x = . 3 Solution The derivative is

y 10

y = 3 sec2 x − 2(−csc x cot x) = 3 sec2 x + 2 csc x cot x.

5 x 0.4

0.8

1.2

5 10

FIGURE 2.32 y = 3 tan x − 2 csc x and the π tangent line at x = 3

π , we have 3

π 40 1 4 2 y = 3(2)2 + 2 √ ≈ 13.33333. = 12 + = √ 3 3 3 3 3

40 4 π √ The tangent line with slope and point of tangency ,3 3 − √ has equation 3 3 3

√ 40 4 π y= x− +3 3− √ . 3 3 3 At x =

We show a graph of the function and the tangent line in Figure 2.32.

EXERCISES 2.6 WRITING EXERCISES 1. Most people draw sine curves that are very steep and rounded. Given the results of this section, discuss the actual shape of the sine curve. Starting at (0, 0), how steep should the graph be drawn? What is the steepest the graph should be drawn anywhere? In which regions is the graph almost straight and where does it curve a lot? 2. In many common physics and engineering applications, the term sin x makes calculations difficult. A common simplification is to replace sin x with x, accompanied by the justification “sin x approximately equals x for small angles.” Discuss this approximation in terms of the tangent line to y = sin x at x = 0. How small is the “small angle” for which the approximation is good? The tangent line to y = cos x at x = 0 is simply y = 1, but the simplification “cos x approximately equals 1 for small angles” is almost never used. Why would this approximation be less useful than sin x ≈ x? 1. Use a graphical analysis as in the text to argue that the derivative of cos x is −sin x. d 2. In the proof of (sin x) = cos x, where do we use the asdx sumption that x is in radians? If you have access to a CAS, find out what the derivative of sin x ◦ is; explain where the π/180 came from.

In exercises 3–20, find the derivative of each function. 3. f (x) = 4 sin x − x

4. f (x) = x 2 + 2 cos2 x

5. f (x) = tan3 x − csc4 x

6. f (x) = 4 sec x 2 − 3 cot x

7. f (x) = x cos 5x 2 9. f (x) = sin(tan x 2 ) 11. f (x) =

sin x 2 x2

13. f (t) = sin t sec t 1 sin 4x 17. f (x) = 2 sin x cos x √ 19. f (x) = tan x 2 + 1 15. f (x) =

8. f (x) = 4x 2 − 3 tan x 10. f (x) = sin2 x + 2 x2 csc4 x √ 14. f (t) = cos t sec t

12. f (x) =

16. f (x) = x 2 sec2 3x 18. f (x) = 4 sin2 x + 4 cos2 x 20. f (x) = 4x 2 sin x sec 3x

In exercises 21–24, use your CAS or graphing calculator. 21. Repeat exercise 17 with your CAS. If its answer is not in the same form as ours in the back of the book, explain how the CAS computed its answer. 22. Repeat exercise 18 with your CAS. If its answer is not 0, explain how the CAS computed its answer.

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23. Find the derivative of f (x) = 2 sin2 x + cos 2x on your CAS. Compare its answer to 0. Explain how to get this answer and your CAS’s answer, if it differs. tan x on your CAS. Compare its sin x answer to sec x tan x. Explain how to get this answer and your CAS’s answer, if it differs.

24. Find the derivative of f (x) =

In exercises 25–28, find an equation of the tangent line to y f (x) at x a. π 25. f (x) = sin 4x, a = 8 26. f (x) = tan 3x, a = 0 π 27. f (x) = cos x, a = 2 π 28. f (x) = x sin x, a = 2

41. Use the identity cos (x + h) = cos x cos h − sin x sin h to prove Theorem 6.2. 42. Use the quotient rule to derive formulas for the derivatives of cot x, sec x and csc x. 43. Use the basic limits lim

x→0

find the following limits: sin 3x x

(b) lim

cos x − 1 x→0 5x

(d) lim

(a) lim

x→0

t→0

x→0

find the following limits: (a) lim t→0

2t sin t

sin t 4t

sin x 2 x→0 x 2

(c) lim

44. Use the basic limits lim

In exercises 29–32, use the position function to find the velocity at time t t0 . Assume units of feet and seconds.

sin x cos x − 1 = 1 and lim = 0 to x→0 x x

sin x cos x − 1 = 1 and lim = 0 to x→0 x x cos x 2 − 1 x→0 x2

(b) lim

sin 6x tan 2x (d) lim x→0 sin 5x x sin x if x = 0 show that f is continuous and 45. For f (x) = x 1 if x = 0 differentiable for all x. (Hint: Focus on x = 0.) (c) lim

x→0

29. s(t) = t 2 − sin 2t, t0 = 0 30. s(t) = t cos(t 2 + π ), t0 = 0 cos t , t0 = π t 32. s(t) = 4 + 3 sin t, t0 = π 31. s(t) =

33. A spring hanging from the ceiling vibrates up and down. Its vertical position at time t is given by f (t) = 4 sin 3t. Find the velocity of the spring at time t. What is the spring’s maximum speed? What is its location when it reaches its maximum speed?

46. For the function of exercise 45, show that the derivative f (x) is continuous. (In this case, we say that the function f is C 1 .) 47. Show that the function of exercise 45 is C 2 . (That is, f (x) exists and is continuous for all x.)

34. In exercise 33, for what time values is the velocity 0? What is the location of the spring when its velocity is 0? When does the spring change directions?

48. As in exercise 46, show that 3 x sin x1 if x = 0 f (x) = is C 1 . 0 if x = 0

In exercises 35 and 36, refer to example 6.6.

49. Sketch a graph of y = sin x and its tangent line at x = 0. Try to determine how many times they intersect by zooming in on the graph (but don’t spend too much time on this). Show that for f (x) = sin x, f (x) < 1 for 0 < x < 1. Explain why this implies that sin x < x for 0 < x < 1. Use a similar argument to show that sin x > x for −1 < x < 0. Explain why y = sin x intersects y = x at only one point.

35. If the total charge in an electrical circuit at time t is given by Q(t) = 3 sin 2t + t + 4 coulombs, compare the current at times t = 0 and t = 1. 36. If the total charge in an electrical circuit at time t is given by Q(t) = 4 cos 4t − 3t + 1 coulombs, compare the current at times t = 0 and t = 1. 37. For f (x) = sin x, find f

(75)

(x) and f

(150)

(x).

38. For f (x) = cos x, find f (77) (x) and f (120) (x). 39. For Lemma 6.1, show that lim sin θ = 0. θ→0−

40. Use Lemma 6.1 and the identity cos2 θ + sin2 θ = 1 to prove Lemma 6.2.

50. For different positive values of k, determine how many times y = sin kx intersects y = x. In particular, what is the largest value of k for which there is only one intersection? Try to determine the largest value of k for which there are three intersections. 51. On a graphing calculator, graph y = sin x using the following range of x-values: [−50, 50], [−60, 60] and so on. You should find graphs similar to the following.

2-61

SECTION 2.6

y

y

x

..

Derivatives of Trigonometric Functions

205

identify the pattern. (Hint: Write the coefficients— a, b, etc.— in the form 1/n! for some integer n.) We will take a longer look at these approximations, called Taylor polynomials, in Chapter 8. x

y

x

Briefly explain why the graph seems to change so much. In particular, is the calculator showing all of the graph or are there large portions missing?

3. When a ball bounces, we often think of the bounce occurring instantaneously, as in the accompanying figure. The sharp corner in the graph at the point of impact does not take into account that the ball actually compresses and maintains contact with the ground for a brief period of time. As shown in John Wesson’s The Science of Soccer, the amount s that the ball is compressed satisfies the equation s (t) = − cp s(t), where c is m the circumference of the ball, p is the pressure of air in the ball and m is the mass of the ball. Assume that the ball hits the ground at time 0 with vertical speed v m/s. Then s(0) = 0 and s (0) = v. Show that s(t) = vk sin kt satisfies the three condi cp s(t), s(0) = 0 and s (0) = v with k = . tions s (t) = − cp m m Use the properties of the sine function to show that the duration of the bounce is πk seconds and find the maximum compression. For a soccer ball with c = 0.7 m, p = 0.86 × 105 N/m2 , v = 15 m/s, radius R = 0.112 m and m = 0.43 kg, compute the duration of the bounce and the maximum compression. y 62.5 50

EXPLORATORY EXERCISES

1

x 2 sin x if x = 0 has several un0 if x = 0 usual properties. Show that f is continuous and differentiable at x = 0. However, f (x) is discontinuous at x = 0. To see this, 1 1 show that f (x) = −1 for x = ,x = and so on. Then 2π 4π 1 1 and so on. Explain show that f (x) = 1 for x = , x = π 3π why this proves that f (x) cannot be continuous at x = 0.

1. The function f (x) =

2. We have seen that the approximation sin x ≈ x for small x derives from the tangent line to y = sin x at x = 0. We can also sin x = 1. Explain think of it as arising from the result lim x→0 x why the limit implies sin x ≈ x for small x. How can we get a better approximation? Instead of using the tangent line, we can try a quadratic function. To see what to multiply x 2 by, nusin x − x . Then sin x − x ≈ ax 2 merically compute a = lim x→0 x2 or sin x ≈ x + ax 2 for small x. How about a cubic sin x − (x + ax 2 ) . Then approximation? Compute b = lim x→0 x3 sin x − (x + ax 2 ) ≈ bx 3 or sin x ≈ x + ax 2 + bx 3 for small x. cos x − 1 Starting with lim = 0, find a cubic approximation x→0 x of the cosine function. If you have a CAS, take the approximations for sine and cosine out to a seventh-order polynomial and

37.5 25 12.5 0

x 0

1

2

3

4

An idealized bounce y

x

A ball being compressed Putting together the physics for before, during and after the bounce, we obtain the height of the center of mass of

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a ball of radius R: −4.9t 2 − vt + R R − vk sin kt h(t) = −4.9(t − πk )2 + v(t − πk ) + R

2.7

π k

DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS y

30

20

10

4

if t < 0 if 0 ≤ t ≤ if t > πk .

Determine whether h(t) is continuous for all t and sketch a reasonable graph of this function to replace the figure shown here.

x

2

2

4

FIGURE 2.33a y = 2x y 20

Exponential and logarithmic functions are among the most common functions encountered in applications. We begin with a simple application in business. Suppose that you have an investment that doubles in value every year. If you start with $100, then the value of the investment after 1 year is $100(2) or $200. After 2 years, its value is $100(2)(2) = $400 and after 3 years, its value is $100(23 ) = $800. In general, after t years, the value of your investment is $100(2t ). Since the value doubles every year, you might describe the rate of return on your investment as 100% (usually called the annual percentage yield or APY). To a calculus student, the term rate should suggest the derivative. We first consider f (x) = a x for some constant a > 1 (called the base). Recall from our discussion in Chapter 0 that the graph will look something like the graph of f (x) = 2x , shown in Figure 2.33a. Observe that as you look from left to right, the graph rises. Thus, the slopes of the tangent lines and so, too, the values of the derivative, are always positive. Also, the farther to the right you look, the steeper the graph is and so, the more positive the derivative is. Further, to the left of the origin, the tangent lines are nearly horizontal and hence, the derivative is positive but close to zero. The sketch of y = f (x) shown in Figure 2.33b is consistent with all of the above information. (Use your computer algebra system to generate a series of graphs of y = a x for various values of a > 0, along with the graphs of the corresponding derivatives and you may detect a pattern.) In particular, notice that the sketch of the derivative closely resembles the graph of the function itself.

10

Derivatives of the Exponential Functions 4

2

x 2

4

FIGURE 2.33b The derivative of f (x) = 2x

Let’s start by using the usual limit definition to try to compute the derivative of f (x) = a x , for a > 1. We have f (x + h) − f (x) a x+h − a x = lim h→0 h→0 h h x h x a a −a = lim From the usual rules of exponents. h→0 h ah − 1 . Factor out the common term of a x . = a x lim h→0 h

f (x) = lim

(7.1)

Unfortunately, we have, at present, no means of computing the limit in (7.1). Nonetheless, assuming the limit exists, (7.1) says that d x a = (constant) a x . dx

(7.2)

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a ball of radius R: −4.9t 2 − vt + R R − vk sin kt h(t) = −4.9(t − πk )2 + v(t − πk ) + R

2.7

π k

DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS y

30

20

10

4

if t < 0 if 0 ≤ t ≤ if t > πk .

Determine whether h(t) is continuous for all t and sketch a reasonable graph of this function to replace the figure shown here.

x

2

2

4

FIGURE 2.33a y = 2x y 20

Exponential and logarithmic functions are among the most common functions encountered in applications. We begin with a simple application in business. Suppose that you have an investment that doubles in value every year. If you start with $100, then the value of the investment after 1 year is $100(2) or $200. After 2 years, its value is $100(2)(2) = $400 and after 3 years, its value is $100(23 ) = $800. In general, after t years, the value of your investment is $100(2t ). Since the value doubles every year, you might describe the rate of return on your investment as 100% (usually called the annual percentage yield or APY). To a calculus student, the term rate should suggest the derivative. We first consider f (x) = a x for some constant a > 1 (called the base). Recall from our discussion in Chapter 0 that the graph will look something like the graph of f (x) = 2x , shown in Figure 2.33a. Observe that as you look from left to right, the graph rises. Thus, the slopes of the tangent lines and so, too, the values of the derivative, are always positive. Also, the farther to the right you look, the steeper the graph is and so, the more positive the derivative is. Further, to the left of the origin, the tangent lines are nearly horizontal and hence, the derivative is positive but close to zero. The sketch of y = f (x) shown in Figure 2.33b is consistent with all of the above information. (Use your computer algebra system to generate a series of graphs of y = a x for various values of a > 0, along with the graphs of the corresponding derivatives and you may detect a pattern.) In particular, notice that the sketch of the derivative closely resembles the graph of the function itself.

10

Derivatives of the Exponential Functions 4

2

x 2

4

FIGURE 2.33b The derivative of f (x) = 2x

Let’s start by using the usual limit definition to try to compute the derivative of f (x) = a x , for a > 1. We have f (x + h) − f (x) a x+h − a x = lim h→0 h→0 h h x h x a a −a = lim From the usual rules of exponents. h→0 h ah − 1 . Factor out the common term of a x . = a x lim h→0 h

f (x) = lim

(7.1)

Unfortunately, we have, at present, no means of computing the limit in (7.1). Nonetheless, assuming the limit exists, (7.1) says that d x a = (constant) a x . dx

(7.2)

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Derivatives of Exponential and Logarithmic Functions

207

Further, (7.2) is consistent with what we observed graphically in Figures 2.33a and 2.33b. The question that we now face is: does ah − 1 h→0 h lim

exist for all (or any) values of a > 1? We explore this limit numerically in the following table for a = 2. h 0.01 0.0001 0.000001 0.0000001

2h − 1 h 0.6955550 0.6931712 0.6931474 0.6931470

h −0.01 −0.0001 −0.000001 −0.0000001

2h − 1 h 0.6907505 0.6931232 0.6931469 0.6931472

The numerical evidence suggests that the limit in question exists and that 2h − 1 ≈ 0.693147. h→0 h lim

In the same way, the numerical evidence suggests that 3h − 1 ≈ 1.098612. h→0 h lim

The approximate values of these limits are largely unremarkable until you observe that ln 2 ≈ 0.693147

and

ln 3 ≈ 1.098612.

You’ll find similar results if you consider the limit in (7.1) for other values of a > 1. (Try estimating several of these for yourself.) This suggests that for a > 1,

y 30

d x a = a x ln a. dx 20

10

4

x

2

2

FIGURE 2.34a y = 2−x

4

Next, the exponential function f (x) = a x for 0 < a < 1, for instance, consider x f (x) = 12 . Notice that we may write such a function as f (x) = b−x , where b = a1 > 1. For example, for b = 12 , we have x 1 f (x) = = 2−x . 2 The graph of this type of exponential looks something like the graph of y = 2−x shown in Figure 2.34a. Assuming that the chain rule gives us

d x 2 = 2x ln 2, dx

d 1 x d −x = 2 dx 2 dx d (−x) dx = 2−x (−ln 2) x

1 1 = ln , 2 2

= 2−x ln 2

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y

2-64

where we have used the fact that

1 ln = ln(2−1 ) = −ln 2. 2

x

4

2

4

10

A sketch of the derivative is shown in Figure 2.34b. This suggests the following general result.

20

THEOREM 7.1 FIGURE 2.34b

The derivative of f (x) = 2−x

For any constant a > 0,

d x a = a x ln a. dx

(7.3)

The proof of this result is complete except for the precise evaluation of the limit in (7.1). Unfortunately, we will not be in a position to complete this work until Chapter 4. For the moment, you should be content with the strong numerical, graphical and (nearly complete) algebraic arguments supporting this conjecture. You should notice that (7.3) is consistent with what we observed in Figures 2.33a, 2.33b, 2.34a and 2.34b. In particular, recall that for 0 < a < 1, ln a < 0. This accounts for why the graph of the derivative for 0 < a < 1 resembles the reflection of the graph of the original function through the x-axis.

EXAMPLE 7.1

Finding the Rate of Change of an Investment

If the value of a 100-dollar investment doubles every year, its value after t years is given by v(t) = 100 2t . Find the instantaneous percentage rate of change of the worth. Solution The instantaneous rate of change is the derivative v (t) = 100 2t ln 2. The relative rate of change is then v (t) 100 2t ln 2 = ln 2 ≈ 0.693. = v(t) 100 2t The percentage change is then about 69.3% per year. This is surprising to most people. A percentage rate of 69.3% will double your investment each year if it is compounded “continuously.” We explore the notion of continuous compounding of interest further in exercise 37. The most commonly used base (by far) is the (naturally occurring) irrational number e. You should immediately notice the significance of this choice. Since ln e = 1, the derivative of f (x) = e x is simply d x e = e x ln e = e x . dx We now have the following result.

THEOREM 7.2 d x e = ex . dx

2-65

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..

Derivatives of Exponential and Logarithmic Functions

209

PROOF As we have just seen, the proof of this formula follows immediately from Theorem 7.1. You will probably agree that this is the easiest derivative formula to remember. In section 2.6, we looked at a simple model of the oscillations of a weight hanging from a spring. Using the cosine function, we had a model that produced the oscillations of the spring, but did not reproduce the physical reality of the spring eventually coming to rest. Coupling the trigonometric functions with the exponential functions produces a more realistic model.

Displacement u(t) Equilibrium position

EXAMPLE 7.2

Finding the Velocity of a Hanging Weight

If we build damping (i.e., resistance to the motion due to friction, for instance) into our model spring-mass system (see Figure 2.35), the vertical displacement at time t of a weight hanging from a spring can be described by

FIGURE 2.35 Spring-mass system

u(t) = Aeαt cos (ωt) + Beαt sin (ωt),

u 1.2

where A, B, α and ω are constants. For each of

1

(a) u(t) = e−t cos t

0.8

and

(b) v(t) = e−t/6 cos 4t,

0.6

sketch a graph of the motion of the weight and find its velocity at any time t.

0.4 0.2 t 5

0.2

10

FIGURE 2.36a u(t) = e−t cos t

Solution Figure 2.36a displays a graph of u(t) = e−t cos t. Notice that it seems to briefly oscillate and then quickly come to rest at u = 0. You should observe that this is precisely the kind of behavior you expect from your car’s suspension system (the spring-mass system most familiar to you) when you hit a bump in the road. If your car’s suspension system needs repair, you might get the behavior shown in Figure 2.36b, which is a graph of v(t) = e−t/6 cos (4t). The velocity of the weight is given by the derivative. From the product rule, we get

y

d d −t (e ) cos t + e−t (cos t) dt dt d = e−t (−t) cos t − e−t sin t dt

u (t) =

1 0.8 0.6 0.4 0.2

= −e−t (cos t + sin t)

x −0.2 −0.4 −0.6 −0.8 −1

1

3

5

and

FIGURE 2.36b y=e

−t/6

cos (4t)

d d −t/6 (e ) cos (4t) + e−t/6 [cos (4t)] dt dt

d t d = e−t/6 − cos (4t) + e−t/6 [− sin (4t)] (4t) dt 6 dt

v (t) =

1 = − e−t/6 cos (4t) − 4e−t/6 sin (4t). 6 At first glance, the chain rule for exponential functions may look a bit different, in part because the “inside” of the exponential function e f (x) is the exponent f (x). Be careful not to change the exponent when computing the derivative of an exponential function.

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EXAMPLE 7.3

The Chain Rule with Exponential Functions 2

2

Find the derivative of f (x) = 3e x , g(x) = xe2/x and h(x) = 32x . Solution From the chain rule, we have d 2 2 2 (x ) = 3e x (2x) = 6xe x . dx Using the product rule and the chain rule, we get

2 2/x 2/x d g (x) = (1)e + xe dx x

2 = e2/x + xe2/x − 2 x f (x) = 3e x

2

e2/x x = e2/x (1 − 2/x). = e2/x − 2

h (x) = 32x ln 3 2

Finally, we have

d (2x 2 ) dx

2

= 32x ln 3 (4x) 2

= 4(ln 3) x32x .

Derivative of the Natural Logarithm The natural logarithm function ln x is closely connected with exponential functions. We’ve already seen it arise as a part of the general exponential derivative formula (7.3). Recall from Chapter 0 that the graph of the natural logarithm looks like the one shown in Figure 2.37a. The function is defined only for x > 0 and as you look to the right, the graph always rises. Thus, the slopes of the tangent lines and hence, also the values of the derivative are always positive. Further, as x → ∞, the slopes of the tangent lines become less positive and seem to approach 0. On the other hand, as x approaches 0 from the right, the graph gets steeper and steeper and hence, the derivative gets larger and larger, without bound. The graph shown in Figure 2.37b is consistent with all of these observations. Do you recognize it? y

y 3

5

2

4

1

3 x 2

4

1

6

2 1

2

x

3

1

2

3

4

5

FIGURE 2.37a

FIGURE 2.37b

y = ln x

The derivative of f (x) = ln x

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211

Using the definition of derivative, we get the following for f (x) = ln x: f (x) = lim

h→0

f (x + h) − f (x) ln (x + h) − ln x = lim . h→0 h h

Unfortunately, we don’t yet know how to evaluate this limit or even whether it exists. (We’ll develop all of this in Chapter 4.) Our only option at present is to investigate this limit numerically. In the following tables, we compute some values for x = 2.

h 0.1 0.001 0.00001 0.0000001

ln (2 h) − ln 2 h 0.4879016 0.4998750 0.4999988 0.5000002

h −0.1 −0.001 −0.00001 −0.0000001

ln (2 h) − ln 2 h 0.5129329 0.5001250 0.5000013 0.5000000

This leads us to the approximation ln (2 + h) − ln 2 1 ≈ . h→0 h 2

f (2) = lim

Similarly, we compute some values for x = 3.

h 0.1 0.001 0.00001 0.0000001

ln (3 h) − ln 3 h 0.3278982 0.3332778 0.3333328 0.3333333

h −0.1 −0.001 −0.00001 −0.0000001

ln (3 h) − ln 3 h 0.3390155 0.3333889 0.3333339 0.3333333

This leads us to the approximation ln (3 + h) − ln 3 1 ≈ . h→0 h 3

f (3) = lim

You should verify the approximations f (4) ≈ 14 , f (5) ≈ 15 and so on. Look back one more time at Figure 2.37b. Do you recognize the function now? You should convince yourself that the result stated in Theorem 7.3 makes sense.

THEOREM 7.3 For x > 0,

d 1 (ln x) = . dx x

(7.4)

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PROOF The proof of this follows directly from Theorems 5.2 and 7.2. Recall that y = ln x if and only if e y = x. From Theorem 5.2, we have for f (x) = ln x and g(x) = e x f (x) =

1 g (y)

=

1 1 = , y e x

as desired.

EXAMPLE 7.4

Derivatives of Logarithms

Find the derivative of f (x) = x ln x, g(x) = ln x 3 and h(x) = ln(x 2 + 1). Solution Using the product rule, we get

1 f (x) = (1) ln x + x = ln x + 1. x

We could certainly use the chain rule to differentiate g(x). However, using the properties of logarithms, recall that we can rewrite g(x) = ln x 3 = 3 ln x and using (7.4), we get

3 1 d g (x) = 3 (ln x) = 3 = . dx x x Using the chain rule for h(x), we get h (x) =

EXAMPLE 7.5

x2

1 1 2x d 2 (x + 1) = 2 (2x) = 2 . + 1 dx x +1 x +1

Finding the Maximum Concentration of a Chemical

The concentration x of a certain chemical after t seconds of an autocatalytic reaction is 10 . Show that x (t) > 0 and use this information to determine given by x(t) = −20t 9e +1 that the concentration of the chemical never exceeds 10. Solution Before computing the derivative, look carefully at the function x(t). The independent variable is t and the only term involving t is in the denominator. So, we don’t need to use the quotient rule. Instead, first rewrite the function as x(t) = 10(9e−20t + 1)−1 and use the chain rule. We get

x 10

d (9e−20t + 1) dt + 1)−2 (−180e−20t )

x (t) = −10(9e−20t + 1)−2

8 6

= −10(9e−20t

4

= 1800e−20t (9e−20t + 1)−2

2

=

1800e−20t . (9e−20t + 1)2

t 0.2

0.4

0.6

0.8

1.0

FIGURE 2.38 Chemical concentration

Notice that since e−20t > 0 for all t, both the numerator and denominator are positive, so that x (t) > 0. Since all of the tangent lines have positive slope, the graph of y = x(t) rises from left to right, as shown in Figure 2.38.

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213

Since the concentration increases for all time, the concentration is always less than the limiting value lim x(t), which is easily computed to be t→∞

lim

10

t→∞ 9e−20t

+1

=

10 = 10. 0+1

Logarithmic Differentiation A clever technique called logarithmic differentiation uses the rules of logarithms to help find derivatives of certain functions for which we don’t presently have derivative formulas. For instance, note that the function f (x) = x x is not a power function because the exponent is not a constant and is not an exponential function because the base is not constant. In example 7.6, we show how to take advantage of the properties of logarithms to find the derivative of such a function.

EXAMPLE 7.6

Logarithmic Differentiation

Find the derivative of f (x) = x x , for x > 0. Solution As already noted, none of our existing derivative rules apply. We begin by taking the natural logarithm of both sides of the equation f (x) = x x . We have ln [ f (x)] = ln (x x ) = x ln x, from the usual properties of logarithms. We now differentiate both sides of this last equation. Using the chain rule on the left side and the product rule on the right side, we get 1 1 f (x) = (1) ln x + x f (x) x or Solving for f (x), we get

f (x) = ln x + 1. f (x) f (x) = (ln x + 1) f (x).

Substituting f (x) = x x gives us f (x) = (ln x + 1)x x .

EXERCISES 2.7 WRITING EXERCISES 1. The graph of f (x) = e x curves upward in the interval from x = −1 to x = 1. Interpreting f (x) = e x as the slopes of tangent lines and noting that the larger x is, the larger e x is, explain why the graph curves upward. For larger values of x, the graph of f (x) = e x appears to shoot straight up with no curve. Using the tangent line, determine whether this is correct or just an optical illusion.

2. The graph of f (x) = ln x appears to get flatter as x gets larger. Interpret the derivative f (x) = x1 as the slopes of tangent lines to determine whether this is correct or just an optical illusion. 3. Graphically compare and contrast the functions x 2 , x 3 , x 4 and e x for x > 0. Sketch the graphs for large x (and very large

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y’s) and compare the relative growth rates of the functions. In general, how does the exponential function compare to polynomials? 4. Graphically compare and contrast the functions x 1/2 , x 1/3 , x 1/4 and ln x for x > 1. Sketch the graphs for large x and compare the relative growth √ rates of the functions. In general, how does ln x compare to n x?

relative increase of worth in one year. Find the APY for the following interest rates: (a) 5% (d) 100 ln 2%

2. f (x) = e2x cos 4x

3. f (x) = x + 2 5. f (x) = 2e

4x+1

7. f (x) = (1/3)

x2

9. f (x) = 4−3x+1 e4x x

4. f (x) = x4

3x

6. f (x) = (1/e)

x

−x 2

8. f (x) = 4

10. f (x) = (1/2)1−x

13. f (x) = ln 2x

x e6x √ 14. f (x) = ln 8x

15. f (x) = ln (x 3 + 3x)

16. f (x) = x 3 ln x

17. f (x) = ln (cos x)

18. f (x) = esin 2x

19. f (x) = sin [ln (cos x 3 )] √ ln x 2 21. f (x) = x

20. f (x) = ln (sin x 2 )

11. f (x) =

23. f (x) = ln (sec x + tan x)

(b) 10%

In exercises 39–44, use logarithmic differentiation to find the derivative.

In exercises 1–24, find the derivative of the function.

x

(c) 20%

38. Determine the interest rate needed to obtain an APY of (a) 100%

1. f (x) = x 3 e x

(b) 10% (e) 100%

12. f (x) =

ex 2x √ 3 24. f (x) = e2x x 3 22. f (x) =

In exercises 25–30, find an equation of the tangent line to y f (x) at x 1. 25. f (x) = 3e x

26. f (x) = 2e x−1

27. f (x) = 3x

28. f (x) = 2x

29. f (x) = x 2 ln x

30. f (x) = 2 ln x 3

In exercises 31–34, the value of an investment at time t is given by v(t). Find the instantaneous percentage rate of change. 31. v(t) = 100 3t

32. v(t) = 100 4t

33. v(t) = 100 et

34. v(t) = 100 e−t

35. A bacterial population starts at 200 and triples every day. Find a formula for the population after t days and find the percentage rate of change in population.

2

39. f (x) = x sin x

40. f (x) = x 4−x

41. f (x) = (sin x)x

42. f (x) = (x 2 )4x

43. f (x) = x ln x

44. f (x) = x

√

x

45. The motion of a spring is described by f (t) = e−t cos t. Compute the velocity at time t. Graph the velocity function. When is the velocity zero? What is the position of the spring when the velocity is zero? 46. The motion of a spring is described by f (t) = e−2t sin 3t. Compute the velocity at time t. Graph the velocity function. When is the velocity zero? What is the position of the spring when the velocity is zero? 47. In exercise 45, graphically estimate the value of t > 0 at which the maximum velocity is reached. 48. In exercise 46, graphically estimate the value of t > 0 at which the maximum velocity is reached. In exercises 49–52, involve the hyperbolic sine and hyperbolic e x e− x e x − e− x and cosh x . cosine functions: sinh x 2 2 d d 49. Show that (sinh x) = cosh x and (cosh x) = sinh x. dx dx 50. Find the derivative of the hyperbolic tangent function: sinh x tanh x = . cosh x 51. Show that both sinh x and cosh x have the property that f (x) = f (x). f (x) = sinh (cos x)

52. Find the derivative of (a) (b) f (x) = cosh (x 2 ) − sinh (x 2 ).

and

53. Find the value of a such that the tangent to ln x at x = a is a line through the origin. 54. Find the value of a such that the tangent to e x at x = a is a line through the origin. Compare the slopes of the lines in exercises 53 and 54.

36. A bacterial population starts at 500 and doubles every four days. Find a formula for the population after t days and find the percentage rate of change in population.

In exercises 55–58, use a CAS or graphing calculator.

37. An investment of A dollars receiving 100r percent (per year) interest compounded continuously will be worth f (t) = Aer t dollars after t years. APY can be defined as [ f (1) − A]/A, the

55. Find the derivative of f (x) = eln x on your CAS. Compare its answer to 2x. Explain how to get this answer and your CAS’s answer, if it differs.

2

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SECTION 2.7 2

56. Find the derivative of f (x) = eln (−x ) on your CAS. The correct answer is that it does not exist. Explain how to get this answer and your CAS’s answer, if it differs. √ 57. Find the derivative of f (x) = ln 4e3x on your CAS. Compare 3 its answer to 2 . Explain how to get this answer and your CAS’s answer, if it differs. 4x

e on your CAS. Com58. Find the derivative of f (x) = ln x2 pare its answer to 4 − 2/x. Explain how to get this answer and your CAS’s answer, if it differs. 59. Numerically estimate the limit in (7.1) for a = 3 and compare your answer to ln 3. 60. Numerically estimate the limit in (7.1) for a = your answer to ln 13 .

1 3

and compare

61. The concentration of a certain chemical after t seconds of an 6 autocatalytic reaction is given by x(t) = −8t . Show that 2e + 1 x (t) > 0 and use this information to determine that the concentration of the chemical never exceeds 6. 62. The concentration of a certain chemical after t seconds of an 10 . Show that autocatalytic reaction is given by x(t) = −10t 9e +2 x (t) > 0 and use this information to determine that the concentration of the chemical never exceeds 5. 63. The Pad´e approximation of e x is the function of the form a + bx for which the values of f (0), f (0) and f (0) f (x) = 1 + cx match the corresponding values of e x . Show that these values all equal 1 and find the values of a, b and c that make f (0) = 1, f (0) = 1 and f (0) = 1. Compare the graphs of f (x) and e x . 64. In a World Almanac or Internet resource, look up the population of the United States by decade for as many years as are available. If there are not columns indicating growth by decade, both numerically and by percentage, compute these yourself. (A spreadsheet is helpful here.) The United States has had periods of both linear and exponential growth. Explain why linear growth corresponds to a constant numerical increase; during which decades has the numerical growth been (approximately) constant? Explain why exponential growth corresponds to a constant percentage growth. During which decades has the percentage growth been (approximately) constant? 65. In statistics, the function f (x) = e−x /2 is used to analyze random quantities that have a bell-shaped distribution. Solutions of the equation f (x) = 0 give statisticians a measure of the variability of the random variable. Find all solutions. 2

66. Repeat exercise 65 for the function f (x) = e−x /8 . Comparing the graphs of the two functions, explain why you would say that this distribution is more spread out than that of exercise 65. 2

67. Repeat exercise 65 for the general 2 2 f (x) = e−(x−m) /2c , where m and c are constants.

function

..

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215

68. In exercise 67, find the solution to the equation f (x) = 0. This value is known as the mode (or average) of the distribution.

EXPLORATORY EXERCISES Ax n for positive constants A, n θn + xn and θ are used to model a variety of chemical and biological processes. In this exercise, we explore a technique for determin ing the values of the constants. First, show that

f (x) > 0 for f (x)/A x > 0 and lim f (x) = A. Second, if u = ln x→∞ 1 − f (x)/A and v = ln x, show that u is a linear function of v. To see why these facts are important, consider the following data collected in a study of the binding of oxygen to hemoglobin. Here, x is the percentage of oxygen in the air and y is the percentage of hemoglobin saturated with oxygen.

1. The Hill functions f (x) =

x y

1 2

2 13

3 32

4 52

5 67

6 77

7 84

8 88

9 91

Plot these data points. As you already showed, Hill functions are increasing and level off at a horizontal asymptote. Explain why it would be reasonable to try to find a Hill function to match these data. Use the limiting value of f (x) to explain why in this data set, A = 100. Next, for each x-y pair, compute u and v as defined above. Plot the u-v points and show that they are (almost exactly) linear. Find the slope and use this to determine the values of n and θ. 2. In Chapter 0, we defined e as the limit e = lim (1 + 1/n)n . n→∞

In this section, e is given as the value of the base a such that d x a = a x . There are a variety of other interesting properties dx of this important number. We discover one here. Graph the functions x 2 and 2x . For x > 0, how many times do they intersect? Graph the functions x 3 and 3x . For x > 0, how many times do they intersect? Try the pair x 2.5 and 2.5x and the pair x 4 and 4x . What would be a reasonable conjecture for the number of intersections (x > 0) of the functions x a and a x ? Explain why x = a is always a solution. Under what circumstances is there another solution less than a? greater than a? By trial and error, verify that e is the value of a at which the “other” solution changes. e−1/x 3. For n = 1 and n = 2, investigate lim n numerically and x→0 x e−1/x graphically. Conjecture the value of lim n for any posix→0 x tive integer n and use your conjecture for the remainder of the 0 if x ≤ 0 , show that f is differexercise. For f (x) = e−1/x if x > 0 entiable at each x and that f (x) is continuous for all x. Then show that f (0) exists and compare the work needed to show that f (x) is continuous at x = 0 and to show that f (0) exists.

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IMPLICIT DIFFERENTIATION AND INVERSE TRIGONOMETRIC FUNCTIONS y

Compare the following two equations describing familiar curves:

2

y = x 2 + 3 (parabola) and

x 2 + y 2 = 4 (circle).

x

2

2

2

FIGURE 2.39 The tangent line √ at the point (1, − 3)

The first equation defines y as a function of x explicitly, since for each x, the equation gives an explicit formula y = f (x) for finding the corresponding value of y. On the other hand, the second equation does not define a function, since the circle in Figure 2.39 doesn’t pass the However, √ vertical line test. √ you can solve for y and find at least two functions y = 4 − x 2 and y = − 4 − x 2 that are defined implicitly by the equation x 2 + y 2 = 4. 2 2 Suppose √ that we want to find the slope of the tangent line to the circle x + y = 4 at the point 1, − 3 (see Figure 2.39). We can think of the circle as the graph of two semicircles, √ √ √ in the point (1, − 3), we use y = 4 − x 2 and y = − 4 − x 2 . Since we are interested √ the equation describing the bottom semicircle, y = − 4 − x 2 to compute the derivative 1 x y (x) = − √ (−2x) = √ . 2 4 − x2 4 − x2 √ 1 So, the slope of the tangent line at the point 1, − 3 is then y (1) = √ . 3 This calculation was not especially challenging, although we will soon see an easier way to do it. However, it’s not always possible to explicitly solve for a function defined implicitly by a given equation. For example, van der Waals’ equation relating the pressure P, volume V and temperature T of a gas has the form

an 2 P + 2 (V − nb) = nRT, (8.1) V where a, n, b and R are constants. Notice the difficulty in solving this for V as a function of P. If we want the derivative dV , we will need a method for computing the derivative dP directly from the implicit representation given in (8.1). Consider each of the following calculations: d 3 (x + 4)2 = 2(x 3 + 4)(3x 2 ), dx d (sin x − 3x)2 = 2(sin x − 3x)(cos x − 3) dx and

d (tan x + 2)2 = 2(tan x + 2) sec2 x. dx

Notice that each of these calculations has the form d [y(x)]2 = 2[y(x)]y (x), dx for some choice of the function y(x). This last equation is simply an expression of the chain rule. We can use this notion to find the derivatives of functions defined implicitly by equations.

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217

We first return to the simple case of the circle x 2 + y 2 = 4. Assuming this equation defines one or more differentiable functions of x: y = y(x), the equation is x 2 + [y(x)]2 = 4.

(8.2)

Differentiating both sides of equation (8.2) with respect to x, we obtain d 2 d x + [y(x)]2 = (4). dx dx From the chain rule,

d [y(x)]2 = 2y(x)y (x), as above and so, we have dx 2x + 2y(x)y (x) = 0.

(8.3)

Subtracting 2x from both sides of (8.3) gives us 2y(x)y (x) = −2x and dividing by 2y(x) (assuming this is not zero), we have solved for y (x): y (x) =

−2x −x = . 2y(x) y(x)

of both x and y. To get the slope at the point Notice √that here y (x) is expressed in terms √ (1, − 3), substitute x = 1 and y = − 3. We have −x −1 1 y (1) = = √ =√ . y(x) x=1 − 3 3

Notice that this is the same as we had found earlier by first solving for y explicitly and then differentiating. This process of differentiating both sides of an equation with respect to x and then solving for y (x) is called implicit differentiation. When faced with an equation implicitly defining one or more differentiable functions y = y(x), differentiate both sides with respect to x, being careful to recognize that differentiating any function of y will require the chain rule: d g(y) = g (y)y (x). dx Then, gather any terms with a factor of y (x) on one side of the equation, with the remaining terms on the other side of the equation and solve for y (x). We illustrate this process in the examples that follow.

EXAMPLE 8.1

Finding a Slope Implicitly

Find y (x) for x + y − 2y = 3. Then, find the slope of the tangent line at the point (2, 1). 2

3

Solution Since we can’t (easily) solve for y in terms of x explicitly, we compute the derivative implicitly. Differentiating both sides with respect to x, we get d 2 d (x + y 3 − 2y) = (3) dx dx and so,

2x + 3y 2 y (x) − 2y (x) = 0.

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To solve for y (x), simply write all terms involving y (x) on one side of the equation and all other terms on the other side. We have 3y 2 y (x) − 2y (x) = −2x

Subtracting 2x from both sides.

and hence, after factoring, we have (3y 2 − 2)y (x) = −2x. Solving for y (x), we get

y

y (x) =

3 2

2

x 2

−2x . 3y 2 − 2

Dividing by (3y 2 − 2).

Substituting x = 2 and y = 1, we find that the slope of the tangent line at the point (2, 1) is

(2, 1)

1 4

Factoring y (x) from both terms on the left side.

y (2) =

4

1 2

−4 = −4. 3−2

The equation of the tangent line is then

3

y − 1 = −4(x − 2).

FIGURE 2.40

We have plotted a graph of the equation and the tangent line in Figure 2.40 using the implicit plot mode of our computer algebra system.

Tangent line at (2, 1)

EXAMPLE 8.2

Finding a Tangent Line by Implicit Differentiation

Find y (x) for x 2 y 2 − 2x = 4 − 4y. Then, find an equation of the tangent line at the point (2, −2). Solution Differentiating both sides with respect to x, we get d 2 2 d (x y − 2x) = (4 − 4y). dx dx Since the first term is the product of x 2 and y 2 , we must use the product rule. We get 2x y 2 + x 2 (2y)y (x) − 2 = 0 − 4y (x). Grouping the terms with y (x) on one side, we get (2x 2 y + 4)y (x) = 2 − 2x y 2 ,

y 2

so that x

3

2

4

2 4

y (x) =

Substituting x = 2 and y = −2, we get the slope of the tangent line, y (2) =

2 − 16 7 = . −16 + 4 6

Finally, an equation of the tangent line is given by y+2=

FIGURE 2.41 Tangent line at (2, −2)

2 − 2x y 2 . 2x 2 y + 4

7 (x − 2). 6

We have plotted the curve and the tangent line at (2, −2) in Figure 2.41 using the implicit plot mode of our computer algebra system.

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Implicit Differentiation and Inverse Trigonometric Functions

219

You can use implicit differentiation to find a needed derivative from virtually any equation you can write down. We illustrate this next for an application.

EXAMPLE 8.3

Rate of Change of Volume with Respect to Pressure

Suppose that van der Waals’ equation for a specific gas is

5 P + 2 (V − 0.03) = 9.7. V

(8.4)

Thinking of the volume V as a function of pressure P, use implicit differentiation to find dV the derivative at the point (5, 1). dP Solution Differentiating both sides of (8.4) with respect to P, we have d d [(P + 5V −2 )(V − 0.03)] = (9.7). dP dP From the product rule and the chain rule, we get

dV −3 dV 1 − 10V (V − 0.03) + (P + 5V −2 ) = 0. dP dP

V 6

Grouping the terms containing

4

dV , we get dP

[−10V −3 (V − 0.03) + P + 5V −2 ] 2

dV 0.03 − V . = dP −10V −3 (V − 0.03) + P + 5V −2

so that P 2

4

6

FIGURE 2.42 Graph of van der Waals’ equation and the tangent line at the point (5, 1)

dV = 0.03 − V, dP

We now have V (5) =

−0.97 97 0.03 − 1 = =− . −10(1)(0.97) + 5 + 5(1) 0.3 30

(The units are in terms of volume per unit pressure.) We show a graph of van der Waals’ equation, along with the tangent line to the graph at the point (5, 1) in Figure 2.42. Of course, since we can find one derivative implicitly, we can also find second and higher order derivatives implicitly. In example 8.4, notice that you can choose which equation to use for the second derivative. A smart choice can save you time and effort.

EXAMPLE 8.4

Finding a Second Derivative Implicitly

Find y (x) implicitly for y 2 + 2e−x y = 6. Then find the value of y at the point (0, 2). Solution As always, start by differentiating both sides of the equation with respect to x. We have d 2 d (y + 2e−x y ) = (6). dx dx By the chain rule, we have 2yy (x) + 2e−x y [−y − x y (x)] = 0.

(8.5)

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Notice that we don’t need to solve this for y (x). Dividing out the common factor of 2 and differentiating again, we get y (x)y (x) + yy (x) − e−x y [−y − x y (x)][y + x y (x)] − e−x y [y (x) + y (x) + x y (x)] = 0.

y 2.4 2.2 2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 3 2 1

Grouping all the terms involving y (x) on one side of the equation gives us yy (x) − xe−x y y (x) = −[y (x)]2 − e−x y [y + x y (x)]2 + 2e−x y y (x). Factoring out the y (x) on the left, we get (y − xe−x y )y (x) = −[y (x)]2 − e−x y [y + x y (x)]2 + 2e−x y y (x), so that x 1

2

3

FIGURE 2.43 y 2 + 2e−x y = 6

y (x) =

−[y (x)]2 − e−x y [y + x y (x)]2 + 2e−x y y (x) . y − xe−x y

(8.6)

Notice that (8.6) gives us a (rather messy) formula for y (x) in terms of x, y and y (x). If we need to have y (x) in terms of x and y only, we can solve (8.5) for y (x) and substitute into (8.6). However, we don’t need to do this to find y (0). Instead, first substitute x = 0 and y = 2 into (8.5) to get 4y (0) + 2(−2) = 0, from which we conclude that y (0) = 1. Then substitute x = 0, y = 2 and y (0) = 1 into (8.6) to get y (0) =

−1 − (2)2 + 2 3 =− . 2 2

See Figure 2.43 for a graph of y 2 + 2e−x y = 6 near the point (0, 2).

TODAY IN MATHEMATICS Dusa McDuff (1945– ) A British mathematician who has won prestigious awards for her research in multidimensional geometry. McDuff was inspired by Russian mathematician Israel Gelfand. “Gelfand amazed me by talking of mathematics as if it were poetry. . . . I had always thought of mathematics as being much more straightforward: a formula is a formula, and an algebra is an algebra, but Gelfand found hedgehogs lurking in the rows of his spectral sequences!’’ McDuff has made important contributions to undergraduate teaching and Women in Science and Engineering, lectures around the world and has coauthored several research monographs.

Recall that, up to this point, we have proved the power rule d r x = r x r −1 dx only for integer exponents (see Theorems 3.1 and 4.3), although we have been freely using this result for any real exponent, r . Now that we have developed implicit differentiation, however, we have the tools we need to prove the power rule for the case of any rational exponent.

THEOREM 8.1 For any rational exponent, r,

d r x = r x r −1 . dx

PROOF Suppose that r is any rational number. Then r =

p , for some integers p and q. Let q

y = x r = x p/q .

(8.7)

Then, raising both sides of equation (8.7) to the qth power, we get yq = x p .

(8.8)

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Differentiating both sides of equation (8.8) with respect to x, we get d q d p (y ) = (x ). dx dx From the chain rule, we have Solving for

qy q−1

dy = px p−1 . dx

dy , we have dx dy px p−1 px p−1 = = dx qy q−1 q(x p/q )q−1 px p−1 p = x p−1− p+ p/q q x p− p/q q p = x p/q−1 = r x r −1 , q =

Since y = x p/q .

Using the usual rules of exponents.

Since

p = r. q

as desired.

Derivatives of the Inverse Trigonometric Functions The inverse trigonometric functions are useful in any number of applications and are essential for solving equations. We now develop derivative rules for these functions. Recall from our discussion in Chapter 0 that you must pay very close attention to the domains and ranges for these functions. In particular, the inverse sine (or function is defined by ! arcsine) π π" restricting the domain of the sine function to the interval − , . Specifically, we had 2 2 π π y = sin−1 x if and only if sin y = x and − ≤ y ≤ . 2 2 Differentiating the equation sin y = x implicitly, we have d d sin y = x dx dx dy cos y = 1. dx

and so, Solving this for

dy , we find (for cos y = 0) that dx dy 1 = . dx cos y

This is not entirely satisfactory, though, since this gives us the derivative in terms of y. Notice that for − π2 ≤ y ≤ π2 , cos y ≥ 0 and hence, cos y = 1 − sin2 y = 1 − x 2 . dy 1 1 = =√ , dx cos y 1 − x2

This leaves us with for −1 < x < 1. That is,

d 1 sin−1 x = √ , dx 1 − x2

for −1 < x < 1.

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Alternatively, we can derive this formula using Theorem 5.2 in section 2.5. Similarly, we can show that d −1 , cos−1 x = √ dx 1 − x2 To find

for −1 < x < 1.

d tan−1 x, recall that we have dx y = tan−1 x if and only if tan y = x

and

−

π π

Using implicit differentiation, we then have d d tan y = x dx dx (sec2 y)

and so, We solve this for

dy = 1. dx

dy , to obtain dx dy 1 = dx sec2 y 1 = 1 + tan2 y 1 . = 1 + x2

That is,

d 1 . tan−1 x = dx 1 + x2

The derivatives of the remaining inverse trigonometric functions are left as exercises. The derivatives of all six inverse trigonometric functions are summarized here.

d sin−1 x dx d cos−1 x dx d tan−1 x dx d cot−1 x dx d sec−1 x dx

= √ = √ = = =

d csc−1 x = dx

1 1 − x2 −1

,

for −1 < x < 1

, for −1 < x < 1 1 − x2 1 1 + x2 −1 1 + x2 1 , for |x| > 1 √ |x| x 2 − 1 −1 , for |x| > 1 √ |x| x 2 − 1

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EXAMPLE 8.5

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223

Finding the Derivative of an Inverse Trigonometric Function

Compute the derivative of (a) cos−1 (3x 2 ), (b) (sec−1 x)2 and (c) tan−1 (x 3 ). Solution From the chain rule, we have (a)

d d −1 cos−1 (3x 2 ) = (3x 2 ) 2 2 dx 1 − (3x ) d x = √

(b)

and (c)

1 − 9x 4

.

d d (sec−1 x)2 = 2(sec−1 x) (sec−1 x) dx dx 1 = 2(sec−1 x) √ |x| x 2 − 1 d d 3 1 [tan−1 (x 3 )] = (x ) dx 1 + (x 3 )2 d x =

EXAMPLE 8.6

−6x

3x 2 . 1 + x6

Modeling the Rate of Change of a Ballplayer’s Gaze

One of the guiding principles of most sports is to “keep your eye on the ball.” In baseball, a batter stands 2 feet from home plate as a pitch is thrown with a velocity of 130 ft/s (about 90 mph). At what rate does the batter’s angle of gaze need to change to follow the ball as it crosses home plate? Solution First, look at the triangle shown in Figure 2.44. We denote the distance from the ball to home plate by d and the angle of gaze by θ . Since the distance is changing with time, we write d = d(t). The velocity of 130 ft/s means that d (t) = −130. [Why Overhead view

d

θ 2

FIGURE 2.44 A ballplayer’s gaze

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would d (t) be negative?] From Figure 2.44, notice that d(t) θ(t) = tan−1 . 2 The rate of change of the angle is then θ (t) =

1+

=

d (t) 2 2 d(t)

1

2

2d (t) radians/second. 4 + [d(t)]2

When d(t) = 0 (i.e., when the ball is crossing home plate), the rate of change is then 2(−130) = −65 radians/second. 4 One problem with this is that most humans can accurately track objects only at the rate of about 3 radians/second. Keeping your eye on the ball in this case is thus physically impossible. (See Watts and Bahill, Keep Your Eye on the Ball.) θ (t) =

BEYOND FORMULAS Implicit differentiation allows us to find the derivative of functions even when we don’t have a formula for the function. This remarkable result means that if we have any equation for the relationship between two quantities, we can find the rate of change of one with respect to the other. Here is a case where mathematics requires creative thinking beyond formula memorization. In what other situations have you seen the need for creativity in mathematics?

EXERCISES 2.8 WRITING EXERCISES 1. For implicit differentiation, we assume that y is a function of x: we write y(x) to remind ourselves of this. However, for the 2 circle x√ + y 2 = 1, it is not true that y is a function of x. Since y = ± 1 − x 2 , there are actually (at least) two functions of x defined implicitly. Explain why this is not really a contradiction; that is, explain exactly what we are assuming when we do implicit differentiation. 2. To perform implicit differentiation on an equation such as x 2 y 2 + 3 = x, we start by differentiating all terms. We get 2x y 2 + x 2 (2y)y = 1. Many students learn the rules this way: take “regular” derivatives of all terms, and tack on a y every time you take a y-derivative. Explain why this works, and rephrase the rule in a more accurate and understandable form. 3. In implicit differentiation, the derivative is typically a function of both x and y; for example, on the circle x 2 + y 2 = r 2 , we have y = −x/y. If we take the derivative −x/y and plug in

any x and y, will it always be the slope of a tangent line? That is, are there any requirements on which x’s and y’s we can plug in? 4. In each example in this section, after we differentiated the given equation, we were able to rewrite the resulting equation in the form f (x, y)y (x) = g(x, y) for some functions f (x, y) and g(x, y). Explain why this can always be done; that is, why doesn’t the chain rule ever produce a term like [y (x)]2 1 or ? y (x) In exercises 1–4, compute the slope of the tangent line at the given point both explicitly (first solve for y as a function of x) and implicitly. 1. x 2 + 4y 2 = 8 at (2, 1) √ √ 2. x 3 y − 4 x = x 2 y at (2, 2)

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3. y − 3x 2 y = cos x at (0, 1) 4. y 2 + 2x y + 4 = 0 at (−2, 2) In exercises 5–16, find the derivative y (x) implicitly. 5. x 2 y 2 + 3y = 4x √ 7. x y − 4y 2 = 12 9.

6. 3x y 3 − 4x = 10y 2 8. sin x y = x 2 − 3

x +3 = 4x + y 2 y

10. 3x + y 3 − 4y = 10x 2

2

11. e x y − e y = x √ 13. x + y − 4x 2 = y

12. xe y − 3y sin x = 1 14. cos y − y 2 = 8

15. e4y − ln y = 2x

16. e x y − 3y = x 2 + 1

2

In exercises 17–22, find an equation of the tangent line at the given point. If you have a CAS that will graph implicit curves, sketch the curve and the tangent line. 17. x 2 − 4y 3 = 0 at (2, 1)

18. x 2 y 2 = 4x at (1, 2)

19. x y = 4y at (2, 1)

20. x y = −3x y at (−1, −3) √ 22. x 4 = 8(x 2 − y 2 ) at (2, − 2)

2 2

21. x 4 = 4(x 2 − y 2 ) at (1,

3 2

√ 3 ) 2

In exercises 23 and 24, find the locations of all horizontal and vertical tangents. 23. x 2 + y 3 − 3y = 4

24. x y 2 − 2y = 2

In exercises 25–28, find the second derivative y (x). 25. x 2 y 2 + 3x − 4y = 5 27. y = x − 6x + 4 cos y 2

3

26. x 2/3 + y 2/3 = 4 28. e

xy

+ 2y − 3x = sin y

In exercises 29–38, find the derivative of the given function. √ 29. f (x) = tan−1 x 30. f (x) = sin−1 (x 3 + 1) 31. f (x) = tan−1 (cos x) 33. f (x) = 4 sec (x 4 ) −1

35. f (x) = etan 37. f (x) =

x

tan−1 x x2 + 1

32. f (x) = 4 sec−1 (x 4 ) √ 34. f (x) = 2 + tan−1 x 36. f (x) =

x2 cot−1 x

38. f (x) = sin−1 (sin x)

39. In example 8.1, it is easy to find a y-value for x = 2, but other y-values are not so easy to find. Try solving for y if x = 1.9. Use the tangent line found in example 8.1 to estimate a y-value. Repeat for x = 2.1. 40. Use the tangent line found in example 8.2 to estimate a y-value corresponding to x = 1.9; x = 2.1.

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41. For elliptic curves, there are nice ways of finding points with rational coordinates (see Ezra Brown’s article “Three Fermat Trails to Elliptic Curves” in the May 2000 College Mathematics Journal for more information). If you have access to an implicit plotter, graph the elliptic curve defined by y 2 = x 3 − 6x + 9. Show that the points (−3, 0) and (0, 3) are on the curve. Find the line through these two points and show that the line intersects the curve in another point with rational (in this case, integer) coordinates. 42. For the elliptic curve y 2 = x 3 − 6x + 4, show that the point (−1, 3) is on the curve. Find the tangent line to the curve at this point and show that it intersects the curve at another point with rational coordinates. e x − e−x e x + e−x and coth x = x both x −x e +e e − e−x have inverses on the appropriate domains and ranges. Show 1 d 1 d tanh−1 x = coth−1 x = and . Identhat dx 1 − x2 dx 1 − x2 tify the set of x-values for which each formula is valid. Is it fair to say that the derivatives are equal?

43. The functions tanh x =

44. Use a CAS to plot the set of points for which (cos x)2 + (sin y)2 = 1. Determine whether the segments plotted are straight or not. 45. Find and simplify the derivative of sin−1 x + cos−1 x. Use the result to write out an equation relating sin−1 x and cos−1 x.

x 46. Find and simplify the derivative of sin−1 √ . Use the x 2 + 1

x −1 and result to write out an equation relating sin √ x2 + 1 −1 tan x. 47. Use implicit differentiation to find y (x) for x 2 y − 2y = 4. Based on this equation, why would you expect to find vertical √ tangents at x = ± 2 and horizontal tangents at y = 0? Show that there are no points for these values. To see what’s going on, solve the original equation for √ y and sketch the graph. Describe what’s happening at x = ± 2 and y = 0. 48. Show that any curve of the form x y = c for some constant c intersects any curve of the form x 2 − y 2 = k for some constant k at right angles (that is, the tangent lines to the curves at the intersection points are perpendicular). In this case, we say that the families of curves are orthogonal. In exercises 49–52, show that the families of curves are orthogonal (see exercise 48). c 49. y = and y 2 = x 2 + k x 50. x 2 + y 2 = cx and x 2 + y 2 = ky 51. y = cx 3 and x 2 + 3y 2 = k 52. y = cx 4 and x 2 + 4y 2 = k

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53. Based on exercises 51 and 52, make a conjecture for a family of functions that is orthogonal to y = cx n . Show that your conjecture is correct. Are there any values of n that must be excluded? 54. Suppose that a circle of radius r and center (0, c) is inscribed in the parabola y = x 2 . At the point of tangency, the slopes must be the same. Find the slope of the circle implicitly and show that at the point of tangency, y = c − 12 . Then use the equations of the circle and parabola to show that c = r 2 + 14 . y

x

55. In example 8.6, it was shown that by the time the baseball reached home plate, the rate of rotation of the player’s gaze (θ ) was too fast for humans to track. Given a maximum rotational rate of θ = −3 radians per second, find d such that θ = −3. That is, find how close to the plate a player can track the ball. In a major league setting, the player must start swinging by the time the pitch is halfway (30 ) to home plate. How does this correspond to the distance at which the player loses track of the ball? 56. Suppose the pitching speed d in example 8.6 is different. Then θ will be different and the value of d for which θ = −3 will change. Find d as a function of d for d ranging from 30 ft/s (slowpitch softball) to 140 ft/s (major league fastball), and sketch the graph. 57. Suppose a painting hangs on a wall. The frame extends from 6 feet to 8 feet above the floor. A person stands x feet from the wall and views the painting, with a viewing angle A formed by the ray from the person’s eye (5 feet above the floor) to the top of the frame and the ray from the person’s eye to the bottom of the frame. Find the value of x that maximizes the viewing angle A.

2.9

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58. What changes in exercise 57 if the person’s eyes are 6 feet above the floor?

EXPLORATORY EXERCISES 1. Suppose a slingshot (see section 2.1) rotates counterclockwise along the circle x 2 + y 2 = 9 and the rock is released at the point (2.9, 0.77). If the rock travels 300 feet, where does it land? [Hint: Find the tangent line at (2.9, 0.77), and find the point (x, y) on that line such that the distance is (x − 2.9)2 + (y − 0.77)2 = 300.] 2. A landowner’s property line runs along the path y = 6 − x. The landowner wants to run an irrigation ditch from a reservoir bounded by the ellipse 4x 2 + 9y 2 = 36. The landowner wants to build the shortest ditch possible from the reservoir to the closest point on the property line. We explore how to find the best path. Sketch the line and ellipse, and draw in a tangent line to the ellipse that is parallel to the property line. Argue that the ditch should start at the point of tangency and run perpendicular to the two lines. We start by identifying the point on the right side of the ellipse with tangent line parallel to y = 6 − x. Find the slope of the tangent line to the ellipse at (x, y) and set it equal to −1. Solve for x and substitute into the equation of the ellipse. Solve for y and you have the point on the ellipse at which to start the ditch. Find an equation of the (normal) line through this point perpendicular to y = 6 − x and find the intersection of the normal line and y = 6 − x. This point is where the ditch ends. 3. In this exercise, you will design a movie theater with all seats having an equal view of the screen. Suppose the screen extends vertically from 10 feet to 30 feet above the floor. The first row of seats is 15 feet from the screen. Your task is to determine a function h(x) such that if seats x feet from the screen are raised h(x) feet above floor level, then the angle from the bottom of the screen to the viewer to the top of the screen will be the same as for a viewer sitting in the first row. You will be able to accomplish this only for a limited range of x-values. Beyond the maximum such x, find the height that maximizes the viewing angle. [Hint: Write the angle as a difference of inverse tan a − tan b tangents and use the formula tan (a − b) = .] 1 + tan a tan b

THE MEAN VALUE THEOREM In this section, we present the Mean Value Theorem, which is so significant that we will be deriving new ideas from it for many chapters to come. Before considering the main result, we look at a special case, called Rolle’s Theorem. The idea behind Rolle’s Theorem is really quite simple. For any function f that is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) and where f (a) = f (b), there must be at least one point between x = a and x = b where the

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53. Based on exercises 51 and 52, make a conjecture for a family of functions that is orthogonal to y = cx n . Show that your conjecture is correct. Are there any values of n that must be excluded? 54. Suppose that a circle of radius r and center (0, c) is inscribed in the parabola y = x 2 . At the point of tangency, the slopes must be the same. Find the slope of the circle implicitly and show that at the point of tangency, y = c − 12 . Then use the equations of the circle and parabola to show that c = r 2 + 14 . y

x

55. In example 8.6, it was shown that by the time the baseball reached home plate, the rate of rotation of the player’s gaze (θ ) was too fast for humans to track. Given a maximum rotational rate of θ = −3 radians per second, find d such that θ = −3. That is, find how close to the plate a player can track the ball. In a major league setting, the player must start swinging by the time the pitch is halfway (30 ) to home plate. How does this correspond to the distance at which the player loses track of the ball? 56. Suppose the pitching speed d in example 8.6 is different. Then θ will be different and the value of d for which θ = −3 will change. Find d as a function of d for d ranging from 30 ft/s (slowpitch softball) to 140 ft/s (major league fastball), and sketch the graph. 57. Suppose a painting hangs on a wall. The frame extends from 6 feet to 8 feet above the floor. A person stands x feet from the wall and views the painting, with a viewing angle A formed by the ray from the person’s eye (5 feet above the floor) to the top of the frame and the ray from the person’s eye to the bottom of the frame. Find the value of x that maximizes the viewing angle A.

2.9

2-82

58. What changes in exercise 57 if the person’s eyes are 6 feet above the floor?

EXPLORATORY EXERCISES 1. Suppose a slingshot (see section 2.1) rotates counterclockwise along the circle x 2 + y 2 = 9 and the rock is released at the point (2.9, 0.77). If the rock travels 300 feet, where does it land? [Hint: Find the tangent line at (2.9, 0.77), and find the point (x, y) on that line such that the distance is (x − 2.9)2 + (y − 0.77)2 = 300.] 2. A landowner’s property line runs along the path y = 6 − x. The landowner wants to run an irrigation ditch from a reservoir bounded by the ellipse 4x 2 + 9y 2 = 36. The landowner wants to build the shortest ditch possible from the reservoir to the closest point on the property line. We explore how to find the best path. Sketch the line and ellipse, and draw in a tangent line to the ellipse that is parallel to the property line. Argue that the ditch should start at the point of tangency and run perpendicular to the two lines. We start by identifying the point on the right side of the ellipse with tangent line parallel to y = 6 − x. Find the slope of the tangent line to the ellipse at (x, y) and set it equal to −1. Solve for x and substitute into the equation of the ellipse. Solve for y and you have the point on the ellipse at which to start the ditch. Find an equation of the (normal) line through this point perpendicular to y = 6 − x and find the intersection of the normal line and y = 6 − x. This point is where the ditch ends. 3. In this exercise, you will design a movie theater with all seats having an equal view of the screen. Suppose the screen extends vertically from 10 feet to 30 feet above the floor. The first row of seats is 15 feet from the screen. Your task is to determine a function h(x) such that if seats x feet from the screen are raised h(x) feet above floor level, then the angle from the bottom of the screen to the viewer to the top of the screen will be the same as for a viewer sitting in the first row. You will be able to accomplish this only for a limited range of x-values. Beyond the maximum such x, find the height that maximizes the viewing angle. [Hint: Write the angle as a difference of inverse tan a − tan b tangents and use the formula tan (a − b) = .] 1 + tan a tan b

THE MEAN VALUE THEOREM In this section, we present the Mean Value Theorem, which is so significant that we will be deriving new ideas from it for many chapters to come. Before considering the main result, we look at a special case, called Rolle’s Theorem. The idea behind Rolle’s Theorem is really quite simple. For any function f that is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) and where f (a) = f (b), there must be at least one point between x = a and x = b where the

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The Mean Value Theorem

227

tangent line to y = f (x) is horizontal. In Figures 2.45a to 2.45c, we draw a number of graphs satisfying the above criteria. Notice that each one has at least one point where there is a horizontal tangent line. Draw your own graphs, to convince yourself that, under these circumstances, it’s not possible to connect the two points (a, f (a)) and (b, f (b)) without having at least one horizontal tangent line. Note that since f (x) = 0 at a horizontal tangent, this says that there is at least one point c in (a, b), for which f (c) = 0 (see Figures 2.45a to 2.45c). y

y

y

c2 x a

c

a

x

b

a

c

x

c1

b

b

FIGURE 2.45a

FIGURE 2.45b

FIGURE 2.45c

Graph initially rising

Graph initially falling

Graph with two horizontal tangents

THEOREM 9.1 (Rolle’s Theorem)

HISTORICAL NOTES Michel Rolle (1652–1719) A French mathematician who proved Rolle’s Theorem for polynomials. Rolle came from a poor background, being largely self-taught and struggling through a variety of jobs including assistant attorney, scribe and elementary school teacher. He was a vigorous member of the French Academy of Sciences, arguing against such luminaries as Descartes that if a < b then −b < −a (so, for instance, −2 < −1). Oddly, Rolle was known as an opponent of the newly developed calculus, calling it a “collection of ingenious fallacies.”

Suppose that f is continuous on the interval [a, b], differentiable on the interval (a, b) and f (a) = f (b). Then there is a number c ∈ (a, b) such that f (c) = 0.

SKETCH OF PROOF We present the main ideas of the proof from a graphical perspective. First, note that if f (x) is constant on [a, b], then f (x) = 0 for all x’s between a and b. On the other hand, if f (x) is not constant on [a, b], then, as you look from left to right, the graph must at some point start to either rise or fall (see Figures 2.46a and 2.46b). For the case where the graph starts to rise, notice that in order to return to the level at which it started, it will need to turn around at some point and start to fall. (Think about it this way: if you start to climb up a mountain—so that your altitude rises—if you are to get back down to where you started, you will need to turn around at some point—where your altitude starts to fall.) So, there is at least one point where the graph turns around, changing from rising to falling (see Figure 2.46a). Likewise, in the case where the graph first starts to fall, the graph y

y f (c) 0

(a, f (a))

(a, f (a))

(b, f(b))

(b, f (b))

f (c) 0 x

c

x c

FIGURE 2.46a

FIGURE 2.46b

Graph rises and turns around to fall back to where it started.

Graph falls and then turns around to rise back to where it started.

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must turn around from falling to rising (see Figure 2.46b). We name this point x = c. Since we know that f (c) exists, we have that either f (c) > 0, f (c) < 0 or f (c) = 0. We want to argue that f (c) = 0, as Figures 2.46a and 2.46b suggest. To establish this, it is easier to show that it is not true that f (c) > 0 or f (c) < 0. If it were true that f (c) > 0, then from the alternative definition of the derivative given in equation (2.2) in section 2.2, we have f (c) = lim

x→c

f (x) − f (c) > 0. x −c

This says that for every x sufficiently close to c, f (x) − f (c) > 0. x −c

(9.1)

In particular, for the case where the graph first rises, if x − c > 0 (i.e., x > c), this says that f (x) − f (c) > 0 or f (x) > f (c), which can’t happen for every x > c (with x sufficiently close to c) if the graph has turned around at c and started to fall. From this, we conclude that it can’t be true that f (c) > 0. Similarly, we can show that it is not true that f (c) < 0. Therefore, f (c) = 0, as desired. The proof for the case where the graph first falls is nearly identical and is left to the reader. We now give an illustration of the conclusion of Rolle’s Theorem.

EXAMPLE 9.1

An Illustration of Rolle’s Theorem

Find a value of c satisfying the conclusion of Rolle’s Theorem for f (x) = x 3 − 3x 2 + 2x + 2 on the interval [0, 1]. Solution First, we verify that the hypotheses of the theorem are satisfied: f is differentiable and continuous for all x [since f (x) is a polynomial and all polynomials are continuous and differentiable everywhere]. Also, f (0) = f (1) = 2. We have f (x) = 3x 2 − 6x + 2. We now look for values of c such that f (c) = 3c2 − 6c + 2 = 0. √ By the quadratic formula, we get c = 1 + 13 3 ≈ 1.5774 [not in the interval (0, 1)] and √ c = 1 − 13 3 ≈ 0.42265 ∈ (0, 1).

REMARK 9.1 We want to emphasize that example 9.1 is merely an illustration of Rolle’s Theorem. Finding the number(s) c satisfying the conclusion of Rolle’s Theorem is not the point of our discussion. Rather, Rolle’s Theorem is of interest to us primarily because we use it to prove one of the fundamental results of elementary calculus, the Mean Value Theorem.

Although Rolle’s Theorem is a simple result, we can use it to derive numerous properties of functions. For example, we are often interested in finding the zeros of a function f (that is, solutions of the equation f (x) = 0). In practice, it is often difficult to determine how many zeros a given function has. Rolle’s Theorem can be of help here.

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THEOREM 9.2 If f is continuous on the interval [a, b], differentiable on the interval (a, b) and f (x) = 0 has two solutions in [a, b], then f (x) = 0 has at least one solution in (a, b).

PROOF This is just a special case of Rolle’s Theorem. Identify the two zeros of f (x) as x = s and x = t, where s < t. Since f (s) = f (t), Rolle’s Theorem guarantees that there is a number c such that s < c < t (and hence a < c < b) where f (c) = 0. We can easily generalize the result of Theorem 9.2, as in the following theorem.

THEOREM 9.3 For any integer n > 0, if f is continuous on the interval [a, b] and differentiable on the interval (a, b) and f (x) = 0 has n solutions in [a, b], then f (x) = 0 has at least (n − 1) solutions in (a, b).

PROOF From Theorem 9.2, between every pair of solutions of f (x) = 0 is at least one solution of f (x) = 0. In this case, there are (n − 1) consecutive pairs of solutions of f (x) = 0 and so, the result follows. We can use Theorems 9.2 and 9.3 to investigate the number of zeros a given function has. (Here, we consider only real zeros of a function and not complex zeros.) y

EXAMPLE 9.2 10

Prove that x 3 + 4x + 1 = 0 has exactly one solution.

5

2

x

1

Determining the Number of Zeros of a Function

1 5 10

FIGURE 2.47 y = x 3 + 4x + 1

2

Solution The graph shown in Figure 2.47 makes the result seem reasonable: we can see one solution, but how can we be sure there are no others outside of the displayed window? Notice that if f (x) = x 3 + 4x + 1, then f (x) = 3x 2 + 4 > 0 for all x. By Theorem 9.2, if f (x) = 0 had two solutions, then f (x) = 0 would have at least one solution. However, since f (x) = 0 for all x, it can’t be true that f (x) = 0 has two (or more) solutions. Therefore, f (x) = 0 has exactly one solution. We now generalize Rolle’s Theorem to one of the most significant results of elementary calculus.

THEOREM 9.4 (Mean Value Theorem) Suppose that f is continuous on the interval [a, b] and differentiable on the interval (a, b). Then there exists a number c ∈ (a, b) such that f (c) =

f (b) − f (a) . b−a

(9.2)

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PROOF

NOTE Note that in the special case where f (a) = f (b), (9.2) simplifies to the conclusion of Rolle’s Theorem, that f (c) = 0.

Note that the hypotheses are identical to those of Rolle’s Theorem, except that there is no f (b) − f (a) assumption about the values of f at the endpoints. The expression is the slope b−a of the secant line connecting the endpoints, (a, f (a)) and (b, f (b)). The theorem states that there is a line tangent to the curve at some point x = c in (a, b) that has the same slope as (and hence, is parallel to) the secant line (see Figures 2.48 and 2.49). If you tilt your head so that the line segment looks horizontal, Figure 2.49 will look like a figure for Rolle’s Theorem (Figures 2.46a and 2.46b). The proof is to “tilt” the function and then apply Rolle’s Theorem. y

y m

y f (x)

f (b) f (a) ba

x

x a

y f (x)

m f (c)

a

b

c

b

FIGURE 2.48

FIGURE 2.49

Secant line

Mean Value Theorem

The equation of the secant line through the endpoints is y − f (a) = m(x − a), where

m=

f (b) − f (a) . b−a

Define the “tilted” function g to be the difference between f and the function whose graph is the secant line: g(x) = f (x) − [m(x − a) + f (a)].

(9.3)

Note that g is continuous on [a, b] and differentiable on (a, b), since f is. Further, g(a) = f (a) − [0 + f (a)] = 0 and

g(b) = f (b) − [m(b − a) + f (a)] = f (b) − [ f (b) − f (a) + f (a)] = 0.

Since m =

f (b) − f (a) . b−a

Since g(a) = g(b), we have by Rolle’s Theorem that there exists a number c in the interval (a, b) such that g (c) = 0. Differentiating (9.3), we get 0 = g (c) = f (c) − m. Finally, solving (9.4) for f (c) gives us f (c) = m = as desired.

f (b) − f (a) , b−a

(9.4)

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Before we demonstrate some of the power of the Mean Value Theorem, we first briefly illustrate its conclusion.

EXAMPLE 9.3

An Illustration of the Mean Value Theorem

Find a value of c satisfying the conclusion of the Mean Value Theorem for f (x) = x 3 − x 2 − x + 1 on the interval [0, 2]. Solution Notice that f is continuous on [0, 2] and differentiable on (0, 2). The Mean Value Theorem then says that there is a number c in (0, 2) for which f (c) =

y 3

f (2) − f (0) 3−1 = = 1. 2−0 2−0

To find this number c, we set

2

f (c) = 3c2 − 2c − 1 = 1

1

3c2 − 2c − 2 = 0. √ 1± 7 From the quadratic formula, we get c = . In this case, only one of these, 3 √ 1+ 7 c= , is in the interval (0, 2). In Figure 2.50, we show the graphs of y = f (x), 3 the secant line joining the endpoints of the portion of the curve on the interval [0, 2] and √ 1+ 7 the tangent line at x = . 3 or

x 1 1 2

FIGURE 2.50 Mean Value Theorem

2

The illustration in example 9.3, where we found the number c whose existence is guaranteed by the Mean Value Theorem, is not the point of the theorem. In fact, these c’s usually remain unknown. The significance of the Mean Value Theorem is that it relates a difference of function values to the difference of the corresponding x-values, as in equation (9.5) below. Note that if we take the conclusion of the Mean Value Theorem (9.2) and multiply both sides by the quantity (b − a), we get f (b) − f (a) = f (c)(b − a).

(9.5)

As it turns out, many of the most important results in the calculus (including one known as the Fundamental Theorem of Calculus) follow from the Mean Value Theorem. For now, we derive a result essential to our work in Chapter 4. The question concerns how many functions share the same derivative. Recall that for any constant c, d (c) = 0. dx A question that you probably haven’t thought to ask is: Are there any other functions whose derivative is zero? The answer is no, as we see in Theorem 9.5.

THEOREM 9.5 Suppose that f (x) = 0 for all x in some open interval I . Then, f (x) is constant on I .

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PROOF Pick any two numbers, say a and b, in I , with a < b. Since f is differentiable in I and (a, b) ⊂ I , f is continuous on [a, b] and differentiable on (a, b). By the Mean Value Theorem, we know that f (b) − f (a) = f (c), b−a

(9.6)

for some number c ∈ (a, b) ⊂ I . Since, f (x) = 0 for all x ∈ I , f (c) = 0 and it follows from (9.6) that f (b) − f (a) = 0

or

f (b) = f (a).

Since a and b were arbitrary points in I , this says that f is constant on I , as desired. A question closely related to Theorem 9.5 is the following. We know, for example, that d 2 (x + 2) = 2x. dx But are there any other functions with the same derivative? You should quickly come up with several. For instance, x 2 + 3 and x 2 − 4 also have the derivative 2x. In fact, d 2 (x + c) = 2x, dx for any constant c. Are there any other functions, though, with the derivative 2x? Corollary 9.1 says that there are no such functions.

COROLLARY 9.1 Suppose that g (x) = f (x) for all x in some open interval I . Then, for some constant c, g(x) = f (x) + c, for all x ∈ I.

PROOF Define h(x) = g(x) − f (x). Then h (x) = g (x) − f (x) = 0 for all x in I . From Theorem 9.5, h(x) = c, for some constant c. The result then follows immediately from the definition of h(x). We see in Chapter 4 that Corollary 9.1 has significant implications when we try to reverse the process of differentiation (called antidifferentiation). We take a look ahead to this in example 9.4.

EXAMPLE 9.4

Finding Every Function with a Given Derivative

Find all functions that have a derivative equal to 3x 2 + 1. Solution We first write down (from our experience with derivatives) one function with the correct derivative: x 3 + x. Then, Corollary 9.1 tells us that any other function with the same derivative differs by at most a constant. So, every function whose derivative equals 3x 2 + 1 has the form x 3 + x + c, for some constant c.

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As our final example, we demonstrate how the Mean Value Theorem can be used to establish a useful inequality.

EXAMPLE 9.5

Proving an Inequality for sin x |sin a| ≤ |a| for all a.

Prove that

Solution First, note that f (x) = sin x is continuous and differentiable on any interval and that for any a, |sin a| = |sin a − sin 0|, since sin 0 = 0. From the Mean Value Theorem, we have that (for a = 0) sin a − sin 0 = f (c) = cos c, (9.7) a−0 for some number c between a and 0. Notice that if we multiply both sides of (9.7) by a and take absolute values, we get |sin a| = |sin a − sin 0| = |cos c| |a − 0| = |cos c| |a|.

(9.8)

But, |cos c| ≤ 1, for all real numbers c and so, from (9.8), we have |sin a| = |cos c| |a| ≤ (1) |a| = |a|, as desired.

BEYOND FORMULAS The Mean Value Theorem is subtle, but its implications are far-reaching. Although the illustration in Figure 2.49 makes the result seem obvious, the consequences of the Mean Value Theorem, such as example 9.4, are powerful and not at all obvious. For example, most of the rest of the calculus developed in this book depends on the Mean Value Theorem either directly or indirectly. A thorough understanding of the theory of calculus can lead you to important conclusions, particularly when the problems are beyond what your intuition alone can handle. What other theorems have you learned that continue to provide insight beyond their original context?

EXERCISES 2.9 WRITING EXERCISES 1. Notice that for both Rolle’s Theorem and the Mean Value Theorem, we have assumed that the function is continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Recall that if a function is differentiable at x = a, then it is also continuous at x = a. Therefore, if we had assumed that the function was differentiable on the closed interval, we would not have had to mention continuity. We do not do so because the assumption of being differentiable at the endpoints is not

necessary. The “ethics” of the statement of a theorem is to include only assumptions that are absolutely necessary. Discuss the virtues of this tradition. Is this common practice in our social dealings, such as financial obligations or personal gossip? 2. One of the results in this section is that if f (x) = g (x), then g(x) = f (x) + c for some constant c. Explain this result graphically.

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3. Explain the result of Corollary 9.1 in terms of position and velocity functions. That is, if two objects have the same velocity functions, what can you say about the relative positions of the two objects?

26. Two runners start a race at time 0. At some time t = a, one runner has pulled ahead, but the other runner has taken the lead by time t = b. Prove that at some time t = c > 0, the runners were going exactly the same speed.

4. As we mentioned, you can derive Rolle’s Theorem from the Mean Value Theorem simply by setting f (a) = f (b). Given this, it may seem odd that Rolle’s Theorem rates its own name and portion of the book. To explain why we do this, discuss ways in which Rolle’s Theorem is easier to understand than the Mean Value Theorem.

27. If f and g are differentiable functions on the interval [a, b] with f (a) = g(a) and f (b) = g(b), prove that at some point in the interval [a, b], f and g have parallel tangent lines.

In exercises 1–6, check the hypotheses of Rolle’s Theorem and the Mean Value Theorem and find a value of c that makes the appropriate conclusion true. Illustrate the conclusion with a graph.

28. Prove that the result of exercise 27 still holds if the assumptions f (a) = g(a) and f (b) = g(b) are relaxed to requiring f (b) − f (a) = g(b) − g(a). In exercises 29–34, find all functions g such that g (x) f (x). 29. f (x) = x 2 31. f (x) = 1/x 2

30. f (x) = 9x 4 √ 32. f (x) = x 34. f (x) = cos x

1. f (x) = x 2 + 1, [−2, 2]

2. f (x) = x 2 + 1, [0, 2]

33. f (x) = sin x

3. f (x) = x 3 + x 2 , [0, 1]

4. f (x) = x 3 + x 2 , [−1, 1]

5. f (x) = sin x, [0, π/2]

6. f (x) = sin x, [−π, 0]

In exercises 35–38, explain why it is not valid to use the Mean Value Theorem. When the hypotheses are not true, the theorem does not tell you anything about the truth of the conclusion. In three of the four cases, show that there is no value of c that makes the conclusion of the theorem true. In the fourth case, find the value of c. 1 1 35. f (x) = , [−1, 1] 36. f (x) = 2 , [−1, 2] x x 37. f (x) = tan x, [0, π] 38. f (x) = x 1/3 , [−1, 1]

7. If f (x) > 0 for all x, prove that f (x) is an increasing function: that is, if a < b, then f (a) < f (b). 8. If f (x) < 0 for all x, prove that f (x) is a decreasing function: that is, if a < b, then f (a) > f (b). In exercises 9–16, determine whether the function is increasing, decreasing or neither. 9. f (x) = x 3 + 5x + 1 10. f (x) = x 5 + 3x 3 − 1 11. f (x) = −x 3 − 3x + 1

12. f (x) = x 4 + 2x 2 + 1

13. f (x) = e x

14. f (x) = e−x

15. f (x) = ln x

16. f (x) = ln x 2

17. Prove that x 3 + 5x + 1 = 0 has exactly one solution. 18. Prove that x 3 + 4x − 3 = 0 has exactly one solution. 19. Prove that x 4 + 3x 2 − 2 = 0 has exactly two solutions. 20. Prove that x 4 + 6x 2 − 1 = 0 has exactly two solutions. 21. Prove that x 3 + ax + b = 0 has exactly one solution for a > 0. 22. Prove that x 4 + ax 2 − b = 0 (a > 0, b > 0) has exactly two solutions.

39. Assume that f is a differentiable function such that f (0) = f (0) = 0 and f (0) > 0. Argue that there exists a positive constant a > 0 such that f (x) > 0 for all x in the interval (0, a). Can anything be concluded about f (x) for negative x’s? 40. Show that for any real numbers u and v, | cos u − cos v| ≤ |u − v|. (Hint: Use the Mean Value Theorem.) 41. Prove that | sin a| < |a| for all a = 0 and use the result to show that the only solution to the equation sin x = x is x = 0. What happens if you try to find all intersections with a graphing calculator? 42. Use the Mean Value Theorem to show that |tan−1 a| < |a| for all a = 0 and use this inequality to find all solutions of the equation tan−1 x = x. 43. Prove that |x| < |sin−1 x| for 0 < |x| < 1.

23. Prove that x 5 + ax 3 + bx + c = 0 has exactly one solution for a > 0, b > 0.

44. Prove that |x| ≤ |tan x| for |x| <

24. Prove that a third-degree (cubic) polynomial has at most three zeros. (You may use the quadratic formula.)

45. For f (x) =

25. Suppose that s(t) gives the position of an object at time t. If s is differentiable on the interval [a, b], prove that at some time t = c, the instantaneous velocity at t = c equals the average velocity between times t = a and t = b.

π . 2

2x if x ≤ 0 show that f is continuous on 2x − 4 if x > 0 the interval (0, 2), differentiable on the interval (0, 2) and has f (0) = f (2). Show that there does not exist a value of c such that f (c) = 0. Which hypothesis of Rolle’s Theorem is not satisfied?

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46. Assume that f is a differentiable function such that f (0) = f (0) = 0. Show by example that it is not necessarily true that f (x) = 0 for all x. Find the flaw in the following bogus “proof.” Using the Mean Value Theorem with a = x f (x) − f (0) . Since f (0) = 0 and b = 0, we have f (c) = x −0 f (x) so that f (x) = 0. and f (c) = 0, we have 0 = x

EXPLORATORY EXERCISES 1. In section 2.1, we gave an example of computing the velocity of a moving car. The point of the story was that computing instantaneous velocity requires us to compute a limit. However, we left an interesting question unanswered. If you have an average velocity of 60 mph over 1 hour and the speed limit is 65 mph, you are unable to prove that you never exceeded the speed limit. What is the longest time interval over which you can average 60 mph and still guarantee no speeding? We can use the Mean Value Theorem to answer the question after clearing up a couple of preliminary questions. First, argue that we need to know the maximum acceleration of a car and the maximum positive acceleration may differ from the maximum negative acceleration. Based on your experience, what is the fastest your car could accelerate (speed up)? What is the fastest your car could decelerate (slow down)? Back up your estimates with some real data (e.g., my car goes from 0 to 60 in 15 seconds). Call the larger number A (use units of mph per second). Next, argue that if acceleration (the derivative of velocity) is constant, then the velocity function is linear. Therefore, if the velocity varies from 55 mph to 65 mph at constant acceleration, the average

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velocity will be 60 mph. Now, apply the Mean Value Theorem to the velocity function v(t) on a time interval [0, T ], where the velocity changes from 55 mph to 65 mph at constant accel65 − 55 v(T ) − v(0) eration A: v (c) = becomes A = and so T −0 T −0 T = 10/A. For how long is the guarantee good? 2. Suppose that a pollutant is dumped into a lake at the rate of p (t) = t 2 − t + 4 tons per month. The amount of pollutant dumped into the lake in the first two months is A = p(2) − p(0). Using c = 1 (the midpoint of the interval), estimate A by applying the Mean Value Theorem to p(t) on the interval [0, 2]. To get a better estimate, apply the Mean Value Theorem to the intervals [0, 1/2], [1/2, 1], [1, 3/2] and [3/2, 2]. [Hint: A = p(1/2) − p(0) + p(1) − p(1/2) + p(3/2) − p(1) + p(2) − p(3/2).] If you have access to a CAS, get better estimates by dividing the interval [0, 2] into more and more pieces and try to conjecture the limit of the estimates. You will be well on your way to understanding Chapter 4 on integration. 3. A result known as the Cauchy Mean Value Theorem states that if f and g are differentiable on the interval (a, b) and continuous on [a, b], then there exists a number c with a < c < b and [ f (b) − f (a)]g (c) = [g(b) − g(a)] f (c). Find all flaws in the following invalid attempt to prove the result, and then find a correct proof. Invalid attempt: The hypotheses of the Mean Value Theorem are satisfied by both functions, so there exists a number c with a < c < b f (b) − f (a) g(b) − g(a) and f (c) = and g (c) = . Then b−a b−a g(b) − g(a) f (b) − f (a) = and thus b−a = f (c) g (c) [ f (b) − f (a)]g (c) = [g(b) − g(a)] f (c).

Review Exercises TRUE OR FALSE

WRITING EXERCISES The following list includes terms that are defined and theorems that are stated in this chapter. For each term or theorem, (1) give a precise definition or statement, (2) state in general terms what it means and (3) describe the types of problems with which it is associated. Tangent line Derivative Product rule Implicit differentiation

Velocity Power rule Quotient rule Mean Value Theorem

Average velocity Acceleration Chain rule Rolle’s Theorem

State the derivative of each function: sin x, cos x, tan x, cot x, sec x, csc x, sin−1 x, cos−1 x, tan−1 x, −1 cot x, sec−1 x, csc−1 x, e x , b x , ln x, logb x

State whether each statement is true or false and briefly explain why. If the statement is false, try to “fix it” by modifying the given statement to make a new statement that is true. 1. If a function is continuous at x = a, then it has a tangent line at x = a. 2. The average velocity between t = a and t = b is the average of the velocities at t = a and t = b. 3. The derivative of a function gives its slope. 4. Given the graph of f (x), you can construct the graph of f (x).

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Jupiter and an asteroid is a third system. The clusters of asteroids at the L4 and L5 Lagrange points of the Sun-Jupiter system are called Trojan asteroids. For a given system, the locations of the five Lagrange points can be determined by solving equations. As you will see in the section 3.1 exercises, the equation for the location of SOHO is a difficult fifth-order polynomial equation. For a fifth-order equation, we usually are forced to gather graphical and numerical evidence to approximate solutions. The graphing and analysis of complicated functions and the solution of equations involving these functions are the emphases of this chapter.

3.1

LINEAR APPROXIMATIONS AND NEWTON’S METHOD There are two distinctly different tasks for which you use scientific calculators. First, they perform basic arithmetic operations much faster than any of us could. We all know how to multiply 1024 by 1673, but it is time-consuming to carry out this calculation with pencil and paper. For such problems, calculators are a tremendous convenience. More significantly, we also use calculators to compute values of transcendental functions such as sine, cosine, tangent, exponentials and logarithms. In this case, the calculator is much more than a mere convenience. How would you calculate sin(1.2345678) without a calculator? Don’t worry if you don’t know how to do this. The problem is that the sine function is not algebraic. That is, there is no formula for sin x involving only the arithmetic operations. So, how does your calculator “know” that sin(1.2345678) ≈ 0.9440056953? In short, it doesn’t know this at all. Rather, the calculator has a built-in program that generates approximate values of the sine and other transcendental functions. In this section, we develop a simple approximation method. Although somewhat crude, it points the way toward more sophisticated approximation techniques to follow later in the text.

Linear Approximations Suppose we wanted to find an approximation for f (x1 ), where f (x1 ) is unknown, but where f (x0 ) is known for some x0 “close” to x1 . For instance, the value of cos(1) is unknown, but we do know that cos(π/3) = 12 (exactly) and π/3 ≈ 1.047 is “close” to 1. While we could use 12 as an approximation to cos(1), we can do better. Referring to Figure 3.1, notice that if x1 is “close” to x0 and we follow the tangent line at x = x0 to the point corresponding to x = x1 , then the y-coordinate of that point (y1 ) should be “close” to the y-coordinate of the point on the curve y = f (x) [i.e., f (x1 )]. Since the slope of the tangent line to y = f (x) at x = x0 is f (x0 ), the equation of the tangent line to y = f (x) at x = x0 is found from y − f (x0 ) . x − x0

(1.1)

y = f (x0 ) + f (x0 )(x − x0 ).

(1.2)

m tan = f (x0 ) = Solving equation (1.1) for y gives us

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243

y y f (x) f (x 1) y f (x0 ) f (x0)(x x 0) y1 f (x 0)

x x0

x1

FIGURE 3.1 Linear approximation of f (x1 )

Notice that (1.2) is the equation of the tangent line to the graph of y = f (x) at x = x0 . We give the linear function defined by this equation a name, as follows.

DEFINITION 1.1 The linear (or tangent line) approximation of f (x) at x = x0 is the function L(x) = f (x0 ) + f (x0 )(x − x0 ).

Observe that the y-coordinate y1 of the point on the tangent line corresponding to x = x1 is simply found by substituting x = x1 in equation (1.2). This gives us y1 = f (x0 ) + f (x0 )(x1 − x0 ).

(1.3)

We define the increments x and y by x = x1 − x0 and

y = f (x1 ) − f (x0 ).

Using this notation, equation (1.3) gives us the approximation f (x1 ) ≈ y1 = f (x0 ) + f (x0 )x.

(1.4)

We illustrate this in Figure 3.2. We sometimes rewrite (1.4) by subtracting f (x0 ) from both sides, to yield y = f (x1 ) − f (x0 ) ≈ f (x0 ) x = dy,

(1.5)

where dy = f (x0 )x is called the differential of y. When using this notation, we also define d x, the differential of x, by d x = x, so that by (1.5), dy = f (x0 ) d x.

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y y f (x) f (x 1)

y1

y f (x0 ) f (x0)(x x 0)

y dy

f (x 0) x x x0

x1

FIGURE 3.2 Increments and differentials

We can use linear approximations to produce approximate values of transcendental functions, as in example 1.1.

EXAMPLE 1.1

Finding a Linear Approximation

Find the linear approximation to f (x) = cos x at x0 = π/3 and use it to approximate cos(1). y

p u

FIGURE 3.3 y = cos x and its linear approximation at x0 = π/3

Solution From Definition 1.1, the linear approximation is defined as L(x) = f (x0 ) + f (x0 )(x − x0 ). Here, x0 = π/3, f (x) = cos x and f (x) = − sin x. So, we have √ π π π 1 π 3 L(x) = cos − sin x− = − x− . 3 3 3 2 2 3 x

In Figure 3.3, we show a graph of y = cos x and the linear approximation to cos x for x0 = π/3. Notice that the linear approximation (i.e., the tangent line at x0 = π/3) stays close to the graph of y = cos x only for x close to π/3. In fact, for x < 0 or x > π , the linear approximation is obviously quite poor. It is typical of linear approximations (tangent lines) to stay close to the curve only nearby the point of tangency. Observe that we chose x0 = π3 since π3 is the value closest to 1 at which we know the value of the cosine exactly. Finally, an estimate of cos(1) is √ 1 3 π L(1) = − 1− ≈ 0.5409. 2 2 3 Your calculator gives you cos(1) ≈ 0.5403 and so, we have found a fairly good approximation to the desired value. In example 1.2, we derive a useful approximation to sin x, valid for x close to 0. This approximation is often used in applications in physics and engineering to simplify equations involving sin x.

EXAMPLE 1.2

Linear Approximation of sin x

Find the linear approximation of f (x) = sin x, for x close to 0.

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245

Solution Here, f (x) = cos x, so that from Definition 1.1, we have

y

sin x ≈ L(x) = f (0) + f (0) (x − 0) = sin 0 + cos 0 (x) = x.

1

x

1

..

1 1

FIGURE 3.4 y = sin x and y = x

This says that for x close to 0, sin x ≈ x. We illustrate this in Figure 3.4, where we show graphs of both y = sin x and y = x. Observe from Figure 3.4 that the graph of y = x stays close to the graph of y = sin x only in the vicinity of x = 0. Thus, the approximation sin x ≈ x is valid only for x close to 0. Also note that the farther x gets from 0, the worse the approximation becomes. This becomes even more apparent in example 1.3, where we also illustrate the use of the increments x and y.

EXAMPLE 1.3

Linear Approximation to Some Cube Roots

√ √ √ 8.02, 3 8.07, 3 8.15 and 3 25.2. √ Solution Here we are approximating values of the function f (x) = 3 x = x 1/3 . So, f (x) = 13 x −2/3 . The closest number to any of 8.02, 8.07 or 8.15 whose cube root we know exactly is 8. So, we write

Use a linear approximation to approximate

√ 3

f (8.02) = f (8) + [ f (8.02) − f (8)]

Add and substract f (8).

= f (8) + y.

(1.6)

From (1.5), we have y ≈ dy = f (8)x 1 −2/3 1 = 8 . (8.02 − 8) = 3 600

Since x = 8.02 − 8.

(1.7)

Using (1.6) and (1.7), we get f (8.02) ≈ f (8) + dy = 2 + while your calculator accurately returns

√ 3

1 ≈ 2.0016667, 600

8.02 ≈ 2.0016653. Similarly, we get

1 f (8.07) ≈ f (8) + 8−2/3 (8.07 − 8) ≈ 2.0058333 3 1 f (8.15) ≈ f (8) + 8−2/3 (8.15 − 8) ≈ 2.0125, 3 √ √ while your calculator√returns 3 8.07 ≈ 2.005816 and 3 8.15 ≈ 2.012423, respectively. To approximate 3 25.2, observe that 8 is not the closest number to 25.2 whose cube root we know exactly. Since 25.2 is much closer to 27 than to 8, we write and

f (25.2) = f (27) + y ≈ f (27) + dy = 3 + dy. In this case, 1 1 dy = f (27)x = 27−2/3 (25.2 − 27) = 3 3

and we have

f (25.2) ≈ 3 + dy = 3 −

1 1 (−1.8) = − 9 15

1 ≈ 2.9333333, 15

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y

3-6

compared to the value of 2.931794, produced by your calculator. It is important to recognize here that the farther the value of x gets from the point of tangency, the worse the approximation tends to be. You can see this clearly in Figure 3.5.

2

x 8

FIGURE 3.5 √ y = 3 x and the linear approximation at x0 = 8

x f (x)

6 84

10 60

Our first three examples were intended to familiarize you with the technique and to give you a feel for how good (or bad) linear approximations tend to be. In example 1.4, there is no exact answer to compare with the approximation. Our use of the linear approximation here is referred to as linear interpolation.

EXAMPLE 1.4

Using a Linear Approximation to Perform Linear Interpolation

The price of an item affects consumer demand for that item. Suppose that based on market research, a company estimates that f (x) thousand small cameras can be sold at the price of $x, as given in the accompanying table. Estimate the number of cameras that can be sold at $7.

14 32

Solution The closest x-value to x = 7 in the table is x = 6. [In other words, this is the closest value of x at which we know the value of f (x).] The linear approximation of f (x) at x = 6 would look like L(x) = f (6) + f (6)(x − 6). From the table, we know that f (6) = 84, but we do not know f (6). Further, we can’t compute f (x), since we don’t have a formula for f (x). The best we can do with the given data is to approximate the derivative by

Number of cameras sold

y

80

f (6) ≈

60 40

f (10) − f (6) 60 − 84 = = −6. 10 − 6 4

The linear approximation is then L(x) ≈ 84 − 6(x − 6).

20 0

x 0

5 7 10 15 Price of cameras

FIGURE 3.6 Linear interpolation

Using this, we estimate that the number of cameras sold at x = 7 would be L(7) ≈ 84 − 6 = 78 thousand. That is, we would expect to sell approximately 78 thousand cameras at a price of $7. We show a graphical interpretation of this in Figure 3.6, where the straight line is the linear approximation (in this case, the secant line joining the first two data points).

Newton’s Method We now return to the question of finding zeros of a function. In section 1.4, we introduced the method of bisections as one procedure for finding zeros of a continuous function. Here, we explore a method that is usually much more efficient than bisections. We are again looking for values of x such that f (x) = 0. These values are called roots of the equation f (x) = 0 or zeros of the function f . It’s easy to find the zeros of f (x) = ax 2 + bx + c, but how would you find zeros of f (x) = tan x − x?

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y

y 5

y f (x) x

3

3

x 5

x0

x1

x2

FIGURE 3.7

FIGURE 3.8

y = tan x − x

Newton’s method

This function is not algebraic and there are no formulas available for finding the zeros. Even so, we can clearly see zeros in Figure 3.7. (In fact, there are infinitely many of them.) The question is, how are we to find them? In general, to find approximate solutions to f (x) = 0, we first make an initial guess, denoted x0 , of the location of a solution. Since the tangent line to y = f (x) at x = x0 tends to hug the curve, we follow the tangent line to where it intersects the x-axis (see Figure 3.8). This appears to provide an improved approximation to the zero. The equation of the tangent line to y = f (x) at x = x0 is given by the linear approximation at x0 [see equation (1.2)], y = f (x0 ) + f (x0 )(x − x0 ).

(1.8)

We denote the x-intercept of the tangent line by x1 [found by setting y = 0 in (1.8)]. We then have

HISTORICAL NOTES Sir Isaac Newton (1642–1727) An English mathematician and scientist known as the co-inventor of calculus. In a 2-year period from 1665 to 1667, Newton made major discoveries in several areas of calculus, as well as optics and the law of gravitation. Newton’s mathematical results were not published in a timely fashion. Instead, techniques such as Newton’s method were quietly introduced as useful tools in his scientific papers. Newton’s Mathematical Principles of Natural Philosophy is widely regarded as one of the greatest achievements of the human mind.

0 = f (x0 ) + f (x0 )(x1 − x0 ) and, solving this for x1 , we get x1 = x0 −

f (x0 ) . f (x0 )

If we repeat this process, using x1 as our new guess, we should produce a further improved approximation, x2 = x1 −

f (x1 ) f (x1 )

and so on (see Figure 3.8). In this way, we generate a sequence of successive approximations determined by

xn+1 = xn −

f (xn ) , for n = 0, 1, 2, 3, . . . . f (xn )

(1.9)

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This procedure is called the Newton-Raphson method, or simply Newton’s method. If Figure 3.8 is any indication, xn should get closer and closer to a zero as n increases. Newton’s method is generally a very fast, accurate method for approximating the zeros of a function, as we illustrate with example 1.5.

EXAMPLE 1.5

Using Newton’s Method to Approximate a Zero

Find a zero of f (x) = x 5 − x + 1. y

3

2

x

1

1

Solution From Figure 3.9, it appears as if the only zero of f is located between x = −2 and x = −1. We further observe that f (−1) = 1 > 0 and f (−2) = −29 < 0. Since f is continuous, the Intermediate Value Theorem (Theorem 4.4 in section 1.4) says that f must have a zero on the interval (−2, −1). Because the zero appears to be closer to x = −1, we make the initial guess x0 = −1. Finally, f (x) = 5x 4 − 1 and so, Newton’s method gives us xn+1 = xn −

3

= xn −

FIGURE 3.9 y = x −x +1 5

f (xn ) f (xn ) xn5 − xn + 1 , 5xn4 − 1

n = 0, 1, 2, . . . .

Using the initial guess x0 = −1, we get x1 = −1 − = −1 −

(−1)5 − (−1) + 1 5(−1)4 − 1 1 5 =− . 4 4

5 Likewise, from x1 = − , we get the improved approximation 4 5 5 5 − − +1 − 5 4 4 x2 = − − 4 5 4 5 − −1 4 ≈ −1.178459394 and so on. We find that

x3 ≈ −1.167537389, x4 ≈ −1.167304083

and

x5 ≈ −1.167303978 ≈ x6 .

Since x5 = x6 , we will make no further progress by calculating additional steps. As a final check on the accuracy of our approximation, we compute f (x6 ) ≈ 1 × 10−13 . Since this is very close to zero, we say that x6 ≈ −1.167303978 is an approximate zero of f . You can bring Newton’s method to bear on a variety of approximation problems. As we illustrate in example 1.6, you may first need to rephrase the problem as a rootfinding problem.

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EXAMPLE 1.6

Linear Approximations and Newton’s Method

249

Using Newton’s Method to Approximate a Cube Root

Use Newton’s method to approximate y

.. √ 3

7.

Solution Recall that we can use a linear approximation to do this. On the other hand, Newton’s method is used to solve equations of the form√f (x) = 0. We can rewrite the current problem in this form, as follows. Suppose x = 3 7. Then, x 3 = 7, which can be rewritten as

30

f (x) = x 3 − 7 = 0.

x 2

Here, f (x) = 3x 2 and we obtain an initial guess from a graph of y = f (x) (see Figure 3.10). Notice that there is a zero near x = 2 and so we take x0 = 2. Newton’s method then yields

30

FIGURE 3.10

x1 = 2 −

y = x3 − 7

23 − 7 23 = ≈ 1.916666667. 2 3(2 ) 12

Continuing this process, we have x2 ≈ 1.912938458 and Further,

x3 ≈ 1.912931183 ≈ x4 . f (x4 ) ≈ 1 × 10−13

and so, x4 is an approximate zero of f . This also says that √ 3 7 ≈ 1.912931183, √ which compares very favorably with the value of 3 7 produced by your calculator.

REMARK 1.1 Although it seemed to be very efficient in examples 1.5 and 1.6, Newton’s method does not always work. We urge you to make sure that the values coming from the method are getting progressively closer and closer together (zeroing in, we hope, on the desired solution). Don’t stop until you’ve reached the limits of accuracy of your computing device. Also, be sure to compute the value of the function at the suspected approximate zero. If the function value is not close to zero, do not accept the value as an approximate zero.

As we illustrate in example 1.7, Newton’s method needs a good initial guess to find an accurate approximation.

y 8

EXAMPLE 1.7 x 2

3

8

FIGURE 3.11 y = x 3 − 3x 2 + x − 1

The Effect of a Bad Guess on Newton’s Method

Use Newton’s method to find an approximate zero of f (x) = x 3 − 3x 2 + x − 1. Solution From the graph in Figure 3.11, there appears to be a zero on the interval (2, 3). Using the (not particularly good) initial guess x0 = 1, we get x1 = 0, x2 = 1, x3 = 0 and so on. Try this for yourself. Newton’s method is sensitive to the initial guess and x0 = 1 is just a bad initial guess. If we had instead started with the improved initial guess x0 = 2, Newton’s method would have quickly converged to the approximate zero 2.769292354. (Again, try this for yourself.)

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3-10

As we see in example 1.7, making a good initial guess is essential with Newton’s method. However, this alone will not guarantee rapid convergence. By rapid convergence, we mean that it takes only a few iterations to obtain an accurate approximation.

n

xn

1 2 3 4

−9.5 −65.9 −2302 −2,654,301

5

−3.5 × 1012

EXAMPLE 1.8

6

−6.2 × 1024

Use Newton’s method with (a) x0 = −2, (b) x0 = −1 and (c) x0 = 0 to try to locate the (x − 1)2 zero of f (x) = 2 . x +1 Solution Of course, there’s no mystery here: f has only one zero, located at x = 1. However, watch what happens when we use Newton’s method with the specified guesses. (a) If we take x0 = −2, and apply Newton’s method, we obtain the values in the table found in the margin. Obviously, the successive iterations are blowing up with this initial guess. To see why, look at Figure 3.12, which shows the graphs of both y = f (x) and the tangent line at x = −2. If you follow the tangent line to where it intersects the x-axis, you will be going away from the zero (far away). Since all of the tangent lines for x ≤ −2 have positive slope [compute f (x) to see why this is true], each subsequent step takes you farther from the zero. (b) If we use the improved initial guess x0 = −1, we cannot even compute x1 . In this case, f (x0 ) = 0 and so, Newton’s method fails. Graphically, this means that the tangent line to y = f (x) at x = −1 is horizontal (see Figure 3.13), so that the tangent line never intersects the x-axis. (c) With the even better initial guess x0 = 0, we obtain the successive approximations in the following table.

Newton’s method iterations for x0 = −2 y

1

x

2

FIGURE 3.12 (x − 1)2 and the tangent line y= 2 x +1 at x = −2

Unusually Slow Convergence for Newton’s Method

y

1

1

x

FIGURE 3.13 y=

(x − 1) and the tangent line x2 + 1 at x = −1

n

xn

n

xn

1 2 3 4 5 6

0.5 0.70833 0.83653 0.912179 0.95425 0.976614

7 8 9 10 11 12

0.9881719 0.9940512 0.9970168 0.9985062 0.9992525 0.9996261

Newton’s method iterations for x0 = 0

2

Finally, we happened upon an initial guess for which Newton’s method converges to the root x = 1. What is unusual here is that the successive approximations shown in the table are converging to 1 much more slowly than in previous examples. By comparison, note that in example 1.5, the iterations stop changing at x5 . Here, x5 is not particularly close to the desired zero of f (x). In fact, in this example, x12 is not as close to the zero as x5 is in example 1.5. We look further into this type of behavior in the exercises. Despite the minor problems experienced in examples 1.7 and 1.8, you should view Newton’s method as a generally reliable and efficient method of locating zeros approximately. Just use a bit of caution and common sense. If the successive approximations are converging to some value that does not appear consistent with the graph, then you need to scrutinize your results more carefully and perhaps try some other initial guesses.

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BEYOND FORMULAS Approximations are at the heart of calculus. To find the slope of a tangent line, for example, we start by approximating the tangent line with secant lines. The fact that there are numerous simple derivative formulas to help us compute exact slopes is an unexpected bonus. In this section, the tangent line is thought of as an approximation of a curve and is used to approximate solutions of equations for which algebra fails. Although this doesn’t provide us the luxury of simply computing an exact answer, we can make the approximation as accurate as we like and so, for most practical purposes, we can “solve” the equation. Think about a situation where you need the time of day. How often do you need the exact time?

EXERCISES 3.1 WRITING EXERCISES 1. Briefly explain in terms of tangent lines why the approximation in example 1.3 gets worse as x gets farther from 8. 2. We constructed a variety of linear approximations in this section. Approximations can be “good” approximations or “bad” approximations. Explain why it can be said that y = x is a good approximation to y = sin x near x = 0 but y = 1 is not a good approximation to y = cos x near x = 0. (Hint: Look at the graphs of y = sin x and y = x on the same axes, then do the same with y = cos x and y = 1.) 3. In example 1.6, we mentioned that you might think of using a linear approximation instead of Newton’s method.√Discuss the relationship between a linear approximation to 3 7 and a √ 3 Newton’s method approximation to 7. (Hint: Compare the first step of Newton’s method to a linear approximation.) 4. Explain why Newton’s method fails computationally if f (x0 ) = 0. In terms of tangent lines intersecting the x-axis, explain why having f (x0 ) = 0 is a problem.

In exercises 1–6, find the linear approximation to f (x) at x x0 . Graph the function and its linear approximation. √ 2. f (x) = (x + 1)1/3 , x0 = 0 1. f (x) = x, x0 = 1 √ 4. f (x) = 2/x, x0 = 1 3. f (x) = 2x + 9, x0 = 0 5. f (x) = sin 3x, x0 = 0

6. f (x) = sin x, x0 = π

7. (a) Find the linear approximation at x = 0 to each of f (x) = (x + 1)2 , g(x) = 1 + sin(2x) and h(x) = e2x . Compare your results. (b) Graph each function in part (a) together with its linear approximation derived in part (a). Which function has the closest fit with its linear approximation?

8. (a) Find the linear approximation at x = 0 to each of f (x) = e x − e−x sin x, g(x) = tan−1 x and h(x) = sinh x = . 2 Compare your results. (b) Graph each function in part (a) together with its linear approximation derived in part (a). Which function has the closest fit with its linear approximation? In exercises 9 and 10, use linear approximations to estimate the quantity. √ √ √ 9. (a) 4 16.04 (b) 4 16.08 (c) 4 16.16 10. (a) sin (0.1) (b) sin (1.0) (c) sin 94 11. For exercise 9, compute the errors (the absolute value of the difference between the exact values and the linear approximations). 12. √ Thinking of exercises 9a–9c as numbers of the form 4 16 + x, denote the errors as e(x) (where x = 0.04, x = 0.08 and x = 0.16). Based on these three computations, determine a constant c such that e(x) ≈ c(x)2 . In exercises 13–16, use linear interpolation to estimate the desired quantity. 13. A company estimates that f (x) thousand software games can be sold at the price of $x as given in the table. x f (x)

20 18

30 14

40 12

Estimate the number of games that can be sold at (a) $24 and (b) $36.

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14. A vending company estimates that f (x) cans of soft drink can be sold in a day if the temperature is x ◦ F as given in the table. x f (x)

60 84

80 120

100 168

Estimate the number of cans that can be sold at (a) 72◦ and (b) 94◦ . 15. An animation director enters the position f (t) of a character’s head after t frames of the movie as given in the table.

3-12

In exercises 21–24, use Newton’s method with the given x0 to (a) compute x1 and x2 by hand and (b) use a computer or calculator to find the root to at least five decimal places of accuracy. 21. x 3 + 3x 2 − 1 = 0, x0 = 1 22. x 3 + 4x 2 − x − 1 = 0, x0 = −1 23. x 4 − 3x 2 + 1 = 0, x0 = 1 24. x 4 − 3x 2 + 1 = 0, x0 = −1 In exercises 25–32, use Newton’s method to find an approximate root (accurate to six decimal places). Sketch the graph and explain how you determined your initial guess. 25. x 3 + 4x 2 − 3x + 1 = 0

26. x 4 − 4x 3 + x 2 − 1 = 0

27. x + 3x + x − 1 = 0 29. sin x = x 2 − 1

28. cos x − x = 0 30. cos x 2 = x √ 32. e−x = x

5

t f (t)

200 128

220 142

240 136

31. e x = −x

If the computer software uses interpolation to determine the intermediate positions, determine the position of the head at frame numbers (a) 208 and (b) 232. 16. A sensor measures the position f (t) of a particle t microseconds after a collision as given in the table. t f (t)

5 8

10 14

33. Show that Newton’s method applied to x 2 − c = 0 (where c > 0 is some constant) produces the √ iterative scheme xn+1 = 12 (xn + c/xn ) for approximating c. This scheme has been known for over 2000 years. To understand why it works, √ suppose that your initial guess√(x0 ) for c is a little too small. How would c/x0 compare to c? Explain why the√average of x0 and c/x0 would give a better approximation to c. 34. Show that Newton’s method applied to x n − c = 0 (where n and c are positive constants) produces the iterative √ scheme xn+1 = n1 [(n − 1)xn + cxn1−n ] for approximating n c.

15 18

Estimate the position of the particle at times (a) t = 8 and (b) t = 12. 17. Given the graph of y = f (x), draw in the tangent lines used in Newton’s method to determine x1 and x2 after starting at x0 = 2. Which of the zeros will Newton’s method converge to? y

2

2

3

x 2

18. Repeat exercise 17 with x0 = −2 and x0 = 0.4. 19. What would happen to Newton’s method in exercise 17 if you had a starting value of x0 = 0? 20. Consider the use of Newton’s method in exercise 17 with x0 = 0.2 and x0 = 10. Obviously, x0 = 0.2 is much closer to a zero of the function, but which initial guess would work better in Newton’s method? Explain.

In exercises 35–40, use Newton’s method [state the function f (x) you use] to estimate the given number. (Hint: See exercises 33 and 34.) √ √ √ √ 35. 11 36. 23 37. 3 11 38. 3 23 √ √ 40. 46 24 39. 44 24 In exercises 41–46, Newton’s method fails. Explain why the method fails and, if possible, find a root by correcting the problem. 41. 4x 3 − 7x 2 + 1 = 0, x0 = 0 42. 4x 3 − 7x 2 + 1 = 0, x0 = 1 43. x 2 + 1 = 0, x0 = 0 44. x 2 + 1 = 0, x0 = 1 4x 2 − 8x + 1 = 0, x0 = −1 45. 4x 2 − 3x − 7 1/3 x +1 46. = 0, x0 = 0.5 x −2 47. Use Newton’s method with (a) x0 = 1.2 and (b) x0 = 2.2 to find a zero of f (x) = x 3 − 5x 2 + 8x − 4. Discuss the difference in the rates of convergence in each case. 48. Use Newton’s method with (a) x0 = 0.2 and (b) x0 = 3.0 to find a zero of f (x) = x sin x. Discuss the difference in the rates of convergence in each case. 49. Use Newton’s method with (a) x0 = −1.1 and (b) x0 = 2.1 to find a zero of f (x) = x 3 − 3x − 2. Discuss the difference in the rates of convergence in each case.

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50. Factor the polynomials in exercises 47 and 49. Find a relationship between the factored polynomial and the rate at which Newton’s method converges to a zero. Explain how the function in exercise 48, which does not factor, fits into this relationship. (Note: The relationship will be explored further in exploratory exercise 2.)

61. The spruce budworm is an enemy of the balsam fir tree. In one model of the interaction between these organisms, possible long-term populations of the budworm are solutions of the equation r (1 − x/k) = x/(1 + x 2 ), for positive constants r and k (see Murray’s Mathematical Biology). Find all positive solutions of the equation with r = 0.5 and k = 7.

In exercises 51–54, find the linear approximation at x 0 to show that the following commonly used approximations are valid for “small” x. Compare the approximate and exact values for x 0.01, x 0.1 and x 1. √ 51. tan x ≈ x 52. 1 + x ≈ 1 + 12 x √ 54. e x ≈ 1 + x 53. 4 + x ≈ 2 + 14 x

62. Repeat exercise 61 with r = 0.5 and k = 7.5. For a small change in the environmental constant k (from 7 to 7.5), how did the solution change from exercise 61 to exercise 62? The largest solution corresponds to an “infestation” of the spruce budworm.

55. Use a computer algebra system (CAS) to determine the range of x’s in exercise 51 for which the approximation is accurate to within 0.01. That is, find x such that | tan x − x| < 0.01. 56. Use a CAS to determine the range of x’s in exercise 54 for which the approximation is accurate to within 0.01. That is, find x such that |e x − (1 + x)| < 0.01. 57. A water wave of length L meters in water of depth d meters has velocity v satisfying the equation v2 =

4.9L e2π d/L − e−2π d/L . π e2π d/L + e−2π d/L

Treating L as a constant and thinking of v 2 as a function f (d), use a linear approximation to show that f (d) ≈ 9.8d for small values of d. That √is, for small depths, the velocity of the wave is approximately 9.8d and is independent of the wavelength L. 58. Planck’s law states that the energy density of blackbody radiation of wavelength x is given by f (x) =

8π hcx −5 . ehc/(kT x) − 1

Use the linear approximation in exercise 54 to show that f (x) ≈ 8πkT /x 4 , which is known as the Rayleigh-Jeans law. 59. Suppose that a species reproduces as follows: with probability p0 , an organism has no offspring; with probability p1 , an organism has one offspring; with probability p2 , an organism has two offspring and so on. The probability that the species goes extinct is given by the smallest nonnegative solution of the equation p0 + p1 x + p2 x 2 + · · · = x (see Sigmund’s Games of Life). Find the positive solutions of the equations 0.1 + 0.2x + 0.3x 2 + 0.4x 3 = x and 0.4 + 0.3x + 0.2x 2 + 0.1x 3 = x. Explain in terms of species going extinct why the first equation has a smaller solution than the second. 60. In Einstein’s theory of relativity, the length of an object depends on its velocity. If L 0 is the length of the object at rest, v is the object’s velocity and c is the speed of light, the Lorentz contraction formula for the length of the object is L = L 0 1 − v 2 /c2 . Treating L as a function of v, find the linear approximation of L at v = 0.

63. Newton’s theory of gravitation states that the weight of a person at elevation x feet above sea level is W (x) = P R 2 /(R + x)2 , where P is the person’s weight at sea level and R is the radius of the earth (approximately 20,900,000 feet). Find the linear approximation of W (x) at x = 0. Use the linear approximation to estimate the elevation required to reduce the weight of a 120-pound person by 1%. 64. One important aspect of Einstein’s theory of relativity is that mass is not constant. For a person with mass m 0 at rest, the mass will equal m = m 0 / 1 − v 2 /c2 at velocity v (where c is the speed of light). Thinking of m as a function of v, find the linear approximation of m(v) at v = 0. Use the linear approximation to show that mass is essentially constant for small velocities. 65. In the figure, a beam is clamped at its left end and simply supported at its right end. A force P (called an axial load) is applied at the right end.

P

If enough force is applied, the beam will buckle. Obviously, it is important for engineers to be able to compute this buckling load. For certain types of beams, the buckling√load √ is the smallest positive solution of the equation tan √x = √x. The √ shape of √ the buckled √ beam is given by y = L − L x − L cos L x + sin L x, where L is the buckling load. Find L and the shape of the buckled beam. 66. The spectral radiancy S of an ideal radiator at constant temperature can be thought of as a function S( f ) of the radiant frequency f . The function S( f ) attains its maximum when 3e−c f + c f − 3 = 0 for the constant c = 10−13 . Use Newton’s method to approximate the solution. 67. In the diagram (on the following page), a hockey player is D feet from the net on the central axis of the rink. The goalie blocks off a segment of width w and stands d feet from the net. The shooting angle to the left of the goalie is given by 3(1 − d/D) − w/2 −1 φ = tan . Use a linear approximation D−d −1 . of tan x at x = 0 to show that if d = 0, then φ ≈ 3−w/2 D Based on this, describe how φ changes if there is an increase in (a) w or (b) D.

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3-14

D

w

Exercise 67

the iterates√converge to 0. (Note: For complex numbers, |a + bi| = a 2 + b2 .) Color the pixel at the point (a, b) black if the iterates converge to 0 and white if not. You can change the ranges of a and b and the step size to “zoom in” on interesting regions. The accompanying pictures show the basin of attraction (in black) for x = 1. In the first figure, we display the region with −1.5 ≤ x ≤ 3.5. In the second figure, we have zoomed in to the portion for 0.26 ≤ x ≤ 0.56. The third shows an even tighter zoom: 0.5 ≤ x ≤ 0.555.

68. The shooter in exercise 67 is assumed to be in the center of the ice. Suppose that the line from the shooter to the center of the goal makes an angle of θ with center line. For the goalie to completely block the goal, he must stand d feet away from the net where d = D(1 − w/6 cos θ ). Show that for small angles, d ≈ D(1 − w/6). In exercises 69 and 70, we explore the convergence of Newton’s method for f (x) x 3 − 3x 2 2x. 69. The zeros of f are x = 0, x = 1 and x = 2. Determine which of the three zeros Newton’s method iterates converge to for (a) x0 = 0.1, (b) x0 = 1.1 and (c) x0 = 2.1. 70. Determine which of the three zeros Newton’s method iterates converge to for (a) x0 = 0.54, (b) x0 = 0.55 and (c) x0 = 0.56.

EXPLORATORY EXERCISES 1. In this exercise, you will extend the work of exercises 69 and 70. First, a definition: The basin of attraction of a zero is the set of starting values x0 for which Newton’s method iterates converge to the zero. As exercises 69 and 70 indicate, the basin boundaries are more complicated than you might expect. For example, you have seen that the interval [0.54, 0.56] contains points in all three basins of attraction. Show that the same is true of the interval [0.552, 0.553]. The picture gets even more interesting when you use complex numbers. These √ are numbers of the form a + bi where i = −1. The remainder of the exercise requires a CAS or calculator that is programmable and performs calculations with complex numbers. First, try Newton’s method with starting point x0 = 1 + i. The formula is exactly the same! Use your computer to show x 3 − 3x02 + 2x0 that x1 = x0 − 0 2 = 1 + 12 i. Then verify that 3x0 − 6x0 + 2 1 x2 = 1 + 17 i and x3 = 1 + 182 i. It certainly appears that the iterates are converging to the zero x = 1. Now, for some programming: set up a double loop with the parameter a running from 0 to 2 in steps of 0.02 and b running from −1 to 1 in steps of 0.02. Within the double loop, set x0 = a + bi and compute 10 Newton’s method iterates. If x10 is close to 0, say |x10 − 0| < 0.1, then we can conjecture that

2. Another important question involving Newton’s method is how fast it converges to a given zero. Intuitively, we can distinguish between the rate of convergence for f (x) = x 2 − 1 (with x0 = 1.1) and that for g(x) = x 2 − 2x + 1 (with x0 = 1.1). But how can we measure this? One method is to take successive approximations xn−1 and xn and compute the difference n = xn − xn−1 . To discover the importance of this quantity, run Newton’s method with x0 = 1.5 and then compute the ratios 3 /2 , 4 /3 , 5 /4 and so on, for each of the following functions: F1 (x) F2 (x) F3 (x) F4 (x)

= = = =

(x (x (x (x

− 1)(x + 2)3 = x 4 + 5x 3 + 6x 2 − 4x − 8, − 1)2 (x + 2)2 = x 4 + 2x 3 − 3x 2 − 4x + 4, − 1)3 (x + 2) = x 4 − x 3 − 3x 2 + 5x − 2 and − 1)4 = x 4 − 4x 3 + 6x 2 − 4x + 1.

n+1 . n If the limit exists and is nonzero, we say that Newton’s method converges linearly. How does r relate to your intuitive sense of how fast the method converges? For f (x) = (x − 1)4 , In each case, conjecture a value for the limit r = lim

n→∞

3-15

SECTION 3.2

we say that the zero x = 1 has multiplicity 4. For f (x) = (x − 1)3 (x + 2), x = 1 has multiplicity 3 and so on. How does r relate to the multiplicity of the zero? Based on this analysis, why did Newton’s method converge faster for f (x) = x 2 − 1 than for g(x) = x 2 − 2x + 1? Finally, use Newton’s method to compute the rate r and hypothesize the multiplicity of the zero x = 0 for f (x) = x sin x and g(x) = x sin x 2 . 3. This exercise looks at a special case of the three-body problem, in which there is a large object A of mass m A , a much smaller object B of mass m B m A and an object C of negligible mass. (Here, a b means that a is much smaller than b.) Assume that object B orbits in a circular path around the common center of mass. There are five circular orbits for object C that maintain constant relative positions of the three objects. These are called Lagrange points L 1 , L 2 , L 3 , L 4 and L 5 , as shown in the figure. L4

B

A L3

L1

L2

L5

3.2

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Indeterminate Forms and L’Hˆ opital’s Rule

255

To derive equations for the Lagrange points, set up a coordinate system with object A at the origin and object B at the point (1, 0). Then L 1 is at the point (x1 , 0), where x1 is the solution of (1 + k)x 5 − (3k + 2)x 4 + (3k + 1)x 3 − x 2 + 2x − 1 = 0; L 2 is at the point (x2 , 0), where x2 is the solution of (1 + k)x 5 − (3k + 2)x 4 + (3k + 1)x 3 − (2k + 1)x 2 + 2x − 1 = 0 and L 3 is at the point (−x3 , 0), where x3 is the solution of (1 + k)x 5 + (3k + 2)x 4 + (3k + 1)x 3 − x 2 − 2x − 1 = 0, mB . Use Newton’s method to find approximate mA solutions of the following.

where k =

(a) Find L 1 for the Earth-Sun system with k = 0.000002. This point has an uninterrupted view of the sun and is the location of the solar observatory SOHO. (b) Find L 2 for the Earth-Sun system with k = 0.000002. This is the future location of NASA’s Microwave Anistropy Probe. (c) Find L 3 for the Earth-Sun system with k = 0.000002. This point is invisible from the Earth and is the location of Planet X in many science fiction stories. (d) Find L 1 for the Moon-Earth system with k = 0.01229. This point has been suggested as a good location for a space station to help colonize the moon. (e) The points L 4 and L 5 form equilateral triangles with objects A and B. Explain why means that polar coordinates this for L 4 are (r, θ) = 1, π6 . Find (x, y)-coordinates for L 4 and L 5 . In the Jupiter-Sun system, these are locations of numerous Trojan asteroids.

ˆ INDETERMINATE FORMS AND L’H OPITAL’S RULE In this section, we reconsider the problem of computing limits. You have frequently seen limits of the form lim

x→a

f (x) , g(x)

where lim f (x) = lim g(x) = 0 or where lim f (x) = lim g(x) = ∞ (or −∞). Recall that x→a x→a x→a x→a from either of these forms 00 or ∞ , called indeterminate forms , we cannot determine ∞ the value of the limit, or even whether the limit exists. For instance, note that lim

x2 − 1 (x − 1)(x + 1) x +1 2 = lim = lim = = 2, x→1 x→1 x −1 x −1 1 1

lim

x −1 x −1 1 1 = lim = lim = 2 x→1 (x − 1)(x + 1) x→1 x + 1 x −1 2

x→1

x→1

3-15

SECTION 3.2

we say that the zero x = 1 has multiplicity 4. For f (x) = (x − 1)3 (x + 2), x = 1 has multiplicity 3 and so on. How does r relate to the multiplicity of the zero? Based on this analysis, why did Newton’s method converge faster for f (x) = x 2 − 1 than for g(x) = x 2 − 2x + 1? Finally, use Newton’s method to compute the rate r and hypothesize the multiplicity of the zero x = 0 for f (x) = x sin x and g(x) = x sin x 2 . 3. This exercise looks at a special case of the three-body problem, in which there is a large object A of mass m A , a much smaller object B of mass m B m A and an object C of negligible mass. (Here, a b means that a is much smaller than b.) Assume that object B orbits in a circular path around the common center of mass. There are five circular orbits for object C that maintain constant relative positions of the three objects. These are called Lagrange points L 1 , L 2 , L 3 , L 4 and L 5 , as shown in the figure. L4

B

A L3

L1

L2

L5

3.2

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Indeterminate Forms and L’Hˆ opital’s Rule

255

To derive equations for the Lagrange points, set up a coordinate system with object A at the origin and object B at the point (1, 0). Then L 1 is at the point (x1 , 0), where x1 is the solution of (1 + k)x 5 − (3k + 2)x 4 + (3k + 1)x 3 − x 2 + 2x − 1 = 0; L 2 is at the point (x2 , 0), where x2 is the solution of (1 + k)x 5 − (3k + 2)x 4 + (3k + 1)x 3 − (2k + 1)x 2 + 2x − 1 = 0 and L 3 is at the point (−x3 , 0), where x3 is the solution of (1 + k)x 5 + (3k + 2)x 4 + (3k + 1)x 3 − x 2 − 2x − 1 = 0, mB . Use Newton’s method to find approximate mA solutions of the following.

where k =

(a) Find L 1 for the Earth-Sun system with k = 0.000002. This point has an uninterrupted view of the sun and is the location of the solar observatory SOHO. (b) Find L 2 for the Earth-Sun system with k = 0.000002. This is the future location of NASA’s Microwave Anistropy Probe. (c) Find L 3 for the Earth-Sun system with k = 0.000002. This point is invisible from the Earth and is the location of Planet X in many science fiction stories. (d) Find L 1 for the Moon-Earth system with k = 0.01229. This point has been suggested as a good location for a space station to help colonize the moon. (e) The points L 4 and L 5 form equilateral triangles with objects A and B. Explain why means that polar coordinates this for L 4 are (r, θ) = 1, π6 . Find (x, y)-coordinates for L 4 and L 5 . In the Jupiter-Sun system, these are locations of numerous Trojan asteroids.

ˆ INDETERMINATE FORMS AND L’H OPITAL’S RULE In this section, we reconsider the problem of computing limits. You have frequently seen limits of the form lim

x→a

f (x) , g(x)

where lim f (x) = lim g(x) = 0 or where lim f (x) = lim g(x) = ∞ (or −∞). Recall that x→a x→a x→a x→a from either of these forms 00 or ∞ , called indeterminate forms , we cannot determine ∞ the value of the limit, or even whether the limit exists. For instance, note that lim

x2 − 1 (x − 1)(x + 1) x +1 2 = lim = lim = = 2, x→1 x→1 x −1 x −1 1 1

lim

x −1 x −1 1 1 = lim = lim = 2 x→1 (x − 1)(x + 1) x→1 x + 1 x −1 2

x→1

x→1

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and

lim

x→1

3-16

x2

x −1 x −1 1 = lim = lim , which does not exist, 2 x→1 x→1 − 2x + 1 x −1 (x − 1)

even though all three limits initially look like 00 . The lesson here is that the expression 00 is mathematically meaningless. It indicates only that both the numerator and denominator tend to zero and that we’ll need to dig deeper to find the value of the limit. Similarly, each of the following limits has the indeterminate form ∞ : ∞ 1 1 1 + 3 x2 + 1 0 x3 x x = lim lim = lim = = 0, x→∞ x 3 + 5 x→∞ x→∞ 5 1 1 1+ 3 (x 3 + 5) 3 x x

(x 2 + 1)

CAUTION will frequently write 00 or We ∞ next to an expression, for ∞ instance, x −1 0 lim 2 . x→1 x − 1 0 We use this shorthand to indicate that the limit has the indicated indeterminate form. This notation does not mean that the value of the limit is 00 . You should take care to avoid writing lim f (x) = 00 or ∞ , ∞ x→a

as these are meaningless expressions.

1 5 (x + 5) x+ 2 x3 + 5 x2 x = lim lim = lim =∞ x→∞ x 2 + 1 x→∞ x→∞ 1 1 1 + (x 2 + 1) x2 x2

3

and

1 (2x + 3x − 5) 2+ 2x 2 + 3x − 5 x2 = lim lim 2 = lim x→∞ x + 4x − 11 x→∞ x→∞ 1 1+ (x 2 + 4x − 11) x2 2

5 3 − 2 2 x x = = 2. 4 11 1 − 2 x x

So, as with limits of the form 00 , if a limit has the form ∞ , we must dig deeper to determine the ∞ value of the limit or whether the limit even exists. Unfortunately, limits with indeterminate forms are frequently more difficult than those just given. For instance, back at section sin x 2.6, where we struggled with the limit lim , ultimately resolving it only with an x→0 x intricate geometric argument. This limit has the indeterminate form 00 , but there is no way to manipulate the numerator or denominator to simplify the expression. In the case of f (x) lim , where lim f (x) = lim g(x) = 0, we can use linear approximations to suggest a x→c g(x) x→c x→c solution, as follows. If both functions are differentiable at x = c, then they are also continuous at x = c, so that f (c) = lim f (x) = 0 and g(c) = lim g(x) = 0. We now have the linear approximations x→c

x→c

f (x) ≈ f (c) + f (c)(x − c) = f (c)(x − c) g(x) ≈ g(c) + g (c)(x − c) = g (c)(x − c),

and

since f (c) = 0 and g(c) = 0. As we have seen, the approximation should improve as x approaches c, so we would expect that if the limits exist, lim

x→c

f (c) f (x) f (c)(x − c) f (c) = lim = lim = , x→c g (c)(x − c) x→c g (c) g(x) g (c)

assuming that g (c) = 0. Note that if f (x) and g (x) are continuous at x = c and g (c) = 0, f (c) f (x) then = lim . This suggests the following result. x→c g (x) g (c)

3-17

SECTION 3.2

..

Indeterminate Forms and L’Hˆ opital’s Rule

257

THEOREM 2.1 (L’Hˆ opital’s Rule)

HISTORICAL NOTES Guillaume de l’Hˆ opital (1661–1704) A French mathematician who first published the result now known as l’Hˆ opital’s Rule. Born into nobility, l’Hˆ opital was taught calculus by the brilliant mathematician Johann Bernoulli. A competent mathematician, l’Hˆ opital is best known as the author of the first calculus textbook. L’Hˆ opital was a friend and patron of many of the top mathematicians of the seventeenth century.

Suppose that f and g are differentiable on the interval (a, b), except possibly at some fixed point c ∈ (a, b) and that g (x) = 0 on (a, b), except possibly at c. f (x) 0 ∞ Suppose further that lim has the indeterminate form or and that x→c g(x) 0 ∞ f (x) lim = L (or ±∞). Then, x→c g (x) f (x) f (x) lim = lim . x→c g(x) x→c g (x)

PROOF Here, we prove only the 00 case where f, f , g and g are all continuous on all of (a, b) and g (c) = 0, while leaving the more intricate general 00 case for Appendix A. First, recall the alternative form of the definition of derivative (found in section 2.2): f (c) = lim

x→c

f (x) − f (c) . x −c

Working backward, we have by continuity that f (x) f (c) lim = = x→c g (x) g (c)

f (x) − f (c) f (x) − f (c) f (x) − f (c) x −c x −c = lim = lim . g(x) − g(c) x→c g(x) − g(c) x→c g(x) − g(c) lim x→c x −c x −c

lim

x→c

Further, since f and g are continuous at x = c, we have that f (c) = lim f (x) = 0 x→c

and

g(c) = lim g(x) = 0. x→c

It now follows that lim

x→c

f (x) f (x) − f (c) f (x) = lim = lim , x→c g(x) − g(c) x→c g(x) g (x)

which is what we wanted. We leave the proof for the

∞ ∞

case to more advanced texts.

REMARK 2.1 f (x) is replaced with any of the g(x) f (x) f (x) f (x) f (x) limits lim+ , lim , lim or lim . (In each case, we must make x→−∞ g(x) x→c g(x) x→c− g(x) x→∞ g(x) appropriate adjustments to the hypotheses.)

The conclusion of Theorem 2.1 also holds if lim

x→c

EXAMPLE 2.1 Evaluate lim

x→0

The Indeterminate Form

0 0

1 − cos x . sin x

Solution This has the indeterminate form 00 , and both (1 − cos x) and sin x are continuous and differentiable everywhere. Further, ddx sin x = cos x = 0 in some

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interval containing x = 0. (Can you determine one such interval?) From the graph of 1 − cos x seen in Figure 3.14, it appears that f (x) → 0, as x → 0. We can f (x) = sin x confirm this with l’Hˆopital’s Rule, as follows: d (1 − cos x) 0 1 − cos x sin x dx lim = lim = = 0. = lim d x→0 x→0 x→0 cos x sin x 1 (sin x) dx

y 3 2 1 x

2

1

3-18

2

1 2

L’Hˆopital’s Rule is equally easy to apply with limits of the form

3

EXAMPLE 2.2

FIGURE 3.14 y=

1 − cos x sin x

The Indeterminate Form

∞ . ∞

∞ ∞

ex . x→∞ x Solution This has the form ∞ and from the graph in Figure 3.15, it appears that the ∞ function grows larger and larger, without bound, as x → ∞. Applying l’Hˆopital’s Rule confirms our suspicions, as d x (e ) ex ex dx lim = lim = ∞. = lim x→∞ x x→∞ d x→∞ 1 (x) dx

Evaluate lim

y 30

20

10

For some limits, you may need to apply l’Hˆopital’s Rule repeatedly. Just be careful to verify the hypotheses at each step. x 1

2

3

4

5

EXAMPLE 2.3 FIGURE 3.15 y=

ex x

4

6

0.6

0.4

0.2

x 8

FIGURE 3.16 y=

x2 ex

x2 . x→∞ e x Solution First, note that this limit has the form ∞ . From the graph in Figure 3.16, it ∞ seems that the function tends to 0 as x → ∞ (and does so very rapidly, at that). Applying l’Hˆopital’s Rule twice, we get d 2 (x ) x2 2x ∞ dx lim x = lim = lim x x→∞ e x→∞ d x→∞ e ∞ (e x ) dx d (2x) 2 dx = lim = lim x = 0, x→∞ d x→∞ e (e x ) dx as expected.

Evaluate lim

y

2

A Limit Requiring Two Applications of L’Hˆ opital’s Rule

10

REMARK 2.2 A very common error is to apply l’Hˆopital’s Rule indiscriminately, without first checking that the limit has the indeterminate form 00 or ∞ . Students also sometimes ∞ incorrectly compute the derivative of the quotient, rather than the quotient of the derivatives. Be very careful here.

3-19

SECTION 3.2

EXAMPLE 2.4 y

..

Indeterminate Forms and L’Hˆ opital’s Rule

259

An Erroneous Use of L’Hˆ opital’s Rule

Find the mistake in the string of equalities x2 2x 2 2 = lim x = lim x = = 2. x→0 e x − 1 x→0 e x→0 e 1 lim

x 3

3

FIGURE 3.17 y=

Solution From the graph in Figure 3.17, we can see that the limit is approximately 0, x2 so 2 appears to be incorrect. The first limit, lim x , has the form 00 and the functions x→0 e − 1 f (x) = x 2 and g(x) = e x − 1 satisfy the hypotheses of l’Hˆopital’s Rule. Therefore, the x2 2x 2x 0 first equality, lim x = lim x , holds. However, notice that lim x = = 0 and x→0 e − 1 x→0 e x→0 e 1 l’Hˆopital’s Rule does not apply here. The correct evaluation is then

x2 ex − 1

lim

x→0 e x

x2 2x 0 = lim x = = 0. − 1 x→0 e 1

Sometimes an application of l’Hˆopital’s Rule must be followed by some simplification, as we see in example 2.5.

EXAMPLE 2.5 y

Evaluate lim+ x→0

Simplification of the Indeterminate Form

∞ ∞

ln x . csc x

0.4 0.2 x 0.4

0.8

1.2

0.2 0.4

FIGURE 3.18 y=

ln x csc x

Solution First, notice that this limit has the form ∞ . From the graph in Figure 3.18, it ∞ appears that the function tends to 0 as x → 0+ . Applying l’Hˆopital’s Rule, we have d 1 (ln x) ∞ ln x dx x lim+ = lim+ . = lim+ d x→0 csc x x→0 x→0 −csc x cot x ∞ (csc x) dx This last limit still has the indeterminate form ∞ , but rather than apply l’Hˆopital’s Rule ∞ again, observe that we can rewrite the expression. (Do this wherever possible when a limit expression gets too complicated.) We have 1 sin x ln x x lim = lim+ = lim+ − tan x = (−1)(0) = 0, x→0+ csc x x→0 −csc x cot x x→0 x as expected, where we have used the fact (established in section 2.6) that sin x lim = 1. x→0 x (You can also establish this by using l’Hˆopital’s Rule.) Notice that if we had simply blasted away with further applications of l’Hˆopital’s Rule to lim+ would never have resolved the limit. (Why not?)

x→0

1 x

−csc x cot x

, we

Other Indeterminate Forms There are five additional indeterminate forms to consider: ∞ − ∞, 0 · ∞, 00 , 1∞ and ∞0 . Look closely at each of these to see why they are indeterminate. When evaluating a limit of this type, the objective is to somehow reduce it to one of the indeterminate forms 00 or ∞ , at which point we can apply l’Hˆopital’s Rule. ∞

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3-20

y

EXAMPLE 2.6

x

3 2

2

3

Evaluate lim

x→0

The Indeterminate Form ∞ − ∞

1 1 − 4 . x2 x

Solution First, notice that the limit has the form (∞ − ∞). From the graph of the function in Figure 3.19, it appears that the function values tend to −∞ as x → 0. If we add the fractions, we get 2 1 x −1 1 lim = lim = −∞, − x→0 x 2 x→0 x4 x4

5

10

FIGURE 3.19

as conjectured, where we have resolved the limit without using l’Hˆopital’s Rule, which does not apply here. (Why not?)

1 1 y= 2 − 4 x x

EXAMPLE 2.7

The Indeterminate Form ∞ − ∞

1 1 Evaluate lim − . x→0 ln (x + 1) x

y 1.0 0.8 0.6 0.4 0.2 x

1

1

2

FIGURE 3.20 y=

1 1 − ln (x + 1) x

3

Solution In this case, the limit has the form (∞ − ∞). From the graph in Figure 3.20, it appears that the limit is somewhere around 0.5. If we add the fractions, we get a form to which we can apply l’Hˆopital’s Rule. We have 1 1 0 x − ln (x + 1) lim − = lim x→0 ln (x + 1) x→0 ln (x + 1)x x 0 d [x − ln (x + 1)] dx = lim By l’Hˆopital’s Rule. d x→0 [ln (x + 1)x] dx 1 1− 0 x +1 = lim . x→0 1 0 x + ln (x + 1)(1) x +1 Rather than apply l’Hˆopital’s Rule to this last expression, we first simplify the expression, by multiplying top and bottom by (x + 1). We now have 1 1− 1 1 x +1 x +1 lim − = lim x→0 ln (x + 1) x→0 1 x x +1 x + ln (x + 1)(1) x +1 (x + 1) − 1 0 = lim x→0 x + (x + 1) ln (x + 1) 0 d (x) dx = lim By l’Hˆopital’s Rule. x→0 d [x + (x + 1) ln (x + 1)] dx 1 1 = , = lim 1 x→0 2 1 + (1) ln (x + 1) + (x + 1) (x + 1) which is consistent with Figure 3.20.

3-21

..

SECTION 3.2

EXAMPLE 2.8

Evaluate lim

x→∞

y 0.4 0.3 0.2 0.1 x 20

40

60

80

100

FIGURE 3.21 y=

1 ln x x

Indeterminate Forms and L’Hˆ opital’s Rule

261

The Indeterminate Form 0 · ∞

1 ln x . x

Solution This limit has the indeterminate form (0 · ∞). From the graph in Figure 3.21, it appears that the function is decreasing very slowly toward 0 as x → ∞. It’s easy to rewrite this in the form ∞ , from which we can use l’Hˆopital’s Rule. Note that ∞ 1 ∞ ln x lim ln x = lim x→∞ x x→∞ x ∞ d ln x dx = lim By l’Hˆopital’s Rule. x→∞ d x dx 1 0 x = lim = = 0. x→∞ 1 1 Note: If lim [ f (x)]g(x) has one of the indeterminate forms 00 , ∞0 or 1∞ , then, letting x→c

y = [ f (x)]g(x) , we have for f (x) > 0 that ln y = ln[ f (x)]g(x) = g(x) ln [ f (x)], so that lim ln y = lim {g(x) ln [ f (x)]} will have the indeterminate form 0 · ∞, which we x→c x→c can deal with as in example 2.8.

EXAMPLE 2.9

The Indeterminate Form 1∞

1

Evaluate lim+ x x−1 . x→1

Solution First, note that this limit has the indeterminate form (1∞ ). From the graph in 1 Figure 3.22, it appears that the limit is somewhere around 3. We define y = x x−1 , so that

y 20

1

ln y = ln x x−1 =

15

1 ln x. x −1

We now consider the limit

10

1 ln x (∞ · 0) x→1 x→1 x − 1 0 ln x = lim+ x→1 x − 1 0 d (ln x) x −1 dx = lim+ = 1. By l’Hˆopital’s Rule. = lim+ d x→1 x→1 1 (x − 1) dx Be careful; we have found that lim+ ln y = 1, but this is not the original limit. We want lim+ ln y = lim+

5 x 0.5

1.0

1.5

FIGURE 3.22 1

y = x x−1

2.0

x→1

lim y = lim+ eln y = e1 ,

x→1+

which is consistent with Figure 3.22.

x→1

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y

3-22

Often, the computation of a limit in this form requires several applications of l’Hˆopital’s Rule. Just be careful (in particular, verify the hypotheses at every step) and do not lose sight of the original problem.

1.0 0.9

EXAMPLE 2.10

0.8

The Indeterminate Form 00

Evaluate lim+ (sin x)x . x→0

0.7 x 0.2

0.4

0.6

0.8

Solution This limit has the indeterminate form (00 ). In Figure 3.23, it appears that the limit is somewhere around 1. We let y = (sin x)x , so that ln y = ln (sin x)x = x ln (sin x).

1.0

FIGURE 3.23 y = (sin x)x

Now consider the limit lim ln y = lim+ ln (sin x)x = lim+ [x ln (sin x)] (0 · ∞) x→0 x→0 ln (sin x) ∞ = lim+ 1 x→0 ∞ x d [ln (sin x)] dx = lim+ By l’Hˆopital’s Rule. d −1 x→0 (x ) dx ∞ (sin x)−1 cos x . = lim+ −2 x→0 −x ∞

x→0+

TODAY IN MATHEMATICS Vaughan Jones (1952– ) A New Zealand mathematician whose work has connected apparently disjoint areas of mathematics. He was awarded the Fields Medal in 1990 for mathematics that was described by peers as ‘astonishing’. One of his major accomplishments is a discovery in knot theory that has given biologists insight into the replication of DNA. A strong supporter of science and mathematics education in New Zealand, Jones’ “style of working is informal, and one which encourages the free and open interchange of ideas . . . His openness and generosity in this regard have been in the best tradition and spirit of mathematics.” His ideas have “served as a rich source of ideas for the work of others.”

As we have seen earlier, we should rewrite the expression before proceeding. Here, we multiply top and bottom by x 2 sin x to get (sin x)−1 cos x x 2 sin x lim ln y = lim+ x→0+ x→0 −x −2 x 2 sin x 2 −x cos x 0 = lim+ x→0 sin x 0 d (−x 2 cos x) dx = lim+ By l’Hˆopital’s Rule. d x→0 (sin x) dx −2x cos x + x 2 sin x 0 = lim+ = = 0. x→0 cos x 1 Again, we have not yet found the original limit. However, lim y = lim+ eln y = e0 = 1,

x→0+

x→0

which is consistent with Figure 3.23.

EXAMPLE 2.11

The Indeterminate Form ∞0

Evaluate lim (x + 1)2/x . x→∞

Solution This limit has the indeterminate form (∞0 ). From the graph in Figure 3.24, it appears that the function tends to a limit around 1 as x → ∞. We let y = (x + 1)2/x

3-23

..

SECTION 3.2

y

and consider

lim ln y = lim ln (x + 1)

4

x→∞

3

2/x

x→∞

2 ln (x + 1) x→∞ x

= lim

2

x 40

60

80

= lim

x→∞

∞ ∞

100

FIGURE 3.24 y = (x + 1)2/x

We now have that

263

2 ln (x + 1) x

d [2 ln (x + 1)] 2(x + 1)−1 dx = lim = lim x→∞ x→∞ d 1 x dx 2 = lim = 0. x→∞ x + 1

1 20

Indeterminate Forms and L’Hˆ opital’s Rule

(0 · ∞)

By l’Hˆopital’s Rule.

lim y = lim eln y = e0 = 1,

x→∞

x→∞

as expected.

BEYOND FORMULAS On several occasions, we have had the benefit of rewriting a function or an equation into a more convenient form. L’Hˆopital’s Rule gives us a way to rewrite limits that are indeterminate. As we have seen, sometimes just the act of rewriting a function as a single fraction lets us determine the limit. What are other examples where an equation is made easier by rewriting, perhaps with a trigonometric identity or exponential rule?

EXERCISES 3.2 WRITING EXERCISES 1. L’Hˆopital’s Rule states that, in certain situations, the ratios of function values approach the same limits as the ratios of corresponding derivatives (rates of change). Graphically, this may f (x) be hard to understand. To get a handle on this, consider g(x) where both f (x) = ax + b and g(x) = cx + d are linear funcf (x) tions. Explain why the value of lim should depend on x→∞ g(x) the relative sizes of the slopes of the lines; that is, it should be f (x) equal to lim . x→∞ g (x) 2. Think of a limit of 0 as actually meaning “getting very small” and a limit of ∞ as meaning “getting very large.” Discuss whether the following limit forms are indeterminate or not and explain your answer: ∞ − ∞, 10 , 0 · ∞, ∞ · ∞, ∞0 , 0∞ and 00 . 3. A friend is struggling with l’Hˆopital’s Rule. When asked to work a problem, your friend says, “First, I plug in for x and

get 0 over 0. Then I use the quotient rule to take the derivative. Then I plug x back in.” Explain to your friend what the mistake is and how to correct it. 4. Suppose that two runners begin a race from the starting line, with one runner initially going twice as fast as the other. If f (t) and g(t) represent the positions of the runners at time t ≥ 0, explain why we can assume that f (0) = g(0) = 0 and f (t) lim = 2. Explain in terms of the runners’ positions why + t→0 g (t) f (t) = 2. l’Hˆopital’s Rule holds: that is, lim t→0 g(t) In exercises 1–38, find the indicated limits. x +2 x2 − 4 1. lim 2 2. lim 2 x→−2 x − 4 x→2 x − 3x + 2 3x 2 + 2 x +1 3. lim 2 4. lim 2 x→∞ x − 4 x→−∞ x + 4x + 3

264

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Applications of Differentiation

e2x − 1 x→0 x tan−1 x lim x→0 sin x sin 2x lim x→π sin x sin x − x lim x→0 x3 √ x −1 lim x→1 x − 1 x3 lim x x→∞ e x cos x − sin x lim x→0 x sin2 x sin π x lim x→1 x − 1 ln x lim x→∞ x 2 lim xe−x

sin x e3x − 1 sin x lim x→0 sin−1 x cos−1 x lim 2 x→−1 x − 1 tan x − x lim x→0 x3 ln x lim x→1 x − 1 ex lim 4 x→∞ x 1 2 lim cot x − 2 x→0 x e x−1 − 1 lim x→1 x 2 − 1 ln x lim √ x→∞ x lim x sin (1/x)

5. lim

6. lim

7.

8.

9. 11. 13. 15. 17. 19. 21. 23.

x→∞

ln (ln x) 25. lim x→1 ln x

10. 12. 14. 16. 18. 20. 22. 24.

ln x cot x 31. lim ( x 2 + 1 − x)

29. lim

x→0+

1 x 33. lim 1 + x→∞ x 1 x 35. lim √ − x→0+ x +1 x x 37. lim (1/x)

x→0+

32. lim (ln x − x) x→∞

√ 2

x + 1 x −4

34. lim

x→∞ x − 2

√ 5−x −2 36. lim √ x→1 10 − x − 3 38. lim (cos x)1/x x→0+

In exercises 39 and 40, find the error(s) in the incorrect calculations. cos x −sin x −cos x 1 39. lim 2 = lim = lim =− x→0 x x→0 x→0 2x 2 2 1 ex − 1 ex ex = lim = = lim 40. lim x→0 x→0 2x x→0 2 x2 2 41. Find all errors in the string x2 x2 2x 2 = lim = lim = lim x→0 ln x 2 x→0 2 ln x x→0 2/x x→0 −2/x 2 = lim (−x 2 ) = 0. lim

x→0

Then, determine the correct value of the limit. 42. Find all errors in the string lim

44. 45.

46.

47.

48.

x→0

x→0+

x→0

x→0

x→∞

sin (sin x) sin x sin x 28. lim √ x→0+ x √ x 30. lim x→0+ ln x

sin x cos x −sin x = lim = 0. = lim x→0 2x x→0 x2 2

Then, determine the correct value of the limit.

sin 3x 3x , cancel sin to get lim , then cancel x→0 2x sin 2x 3 x’s to get 2 . This answer is correct. Is either of the steps used valid? Use linear approximations to argue that the first step is likely to give a correct answer. sin nx Evaluate lim for nonzero constants n and m. x→0 sin mx sin x 2 sin x (a) Compute lim and compare your result to lim . x→0 x 2 x→0 x 2 1 − cos x and compare your result to (b) Compute lim x→0 x4 1 − cos x . lim x→0 x2 sin x 3 Use your results from exercise 45 to evaluate lim and x→0 x 3 1 − cos x 3 lim without doing any calculations. x→0 x6 sin kx 2 has the indeterminate form 00 and then Show that lim x→0 x2 evaluate the limit (where k is some real number). What is the range of values that a limit of the indeterminate form 00 can have? cot kx 2 has the indeterminate form ∞ and then Show that lim ∞ x→0 csc x 2 evaluate the limit (where k is some real number). What is the range of values that a limit of the indeterminate form ∞ can ∞ have?

43. Starting with lim

x→0

26. lim

27. lim x ln x

x→∞

3-24

In exercises 49 and 50, determine which function “dominates,” where we say that the function f (x) dominates the function g(x) as x → ∞ if lim f (x) lim g(x) ∞ and either x→∞

lim

x→∞

x→∞

f (x) g(x) ∞ or lim 0. x→∞ f (x) g(x)

49. e x or x n (n = any positive integer) 50. ln x or x p (for any number p > 0) ecx − 1 for any constant c. 51. Evaluate lim x→0 x tan cx − cx 52. Evaluate lim for any constant c. x→0 x3 f (x) f (x 2 ) = L , what can be said about lim ? Explain 53. If lim x→0 g(x) x→0 g(x 2 ) f (x) = L for a = 0, 1 does not tell why knowing that lim x→a g(x) 2 f (x ) you anything about lim . x→a g(x 2 ) f (x 2 ) 54. Give an example of functions f and g for which lim x→0 g(x 2 ) f (x) exists, but lim does not exist. x→0 g(x) 55. In section 1.2, we briefly discussed the position of a baseball thrown with the unusual knuckleball pitch. The left/right position (in feet) of a ball thrown with spin rate ω and a particular grip at time t seconds is f (ω) = (2.5/ω)t − (2.5/4ω2 ) sin 4ωt. Treating t as a constant and ω as the variable (change to x if you

3-25

SECTION 3.3

like), show that lim f (ω) = 0 for any value of t. (Hint: Find ω→0

a common denominator and use l’Hˆopital’s Rule.) Conclude that this pitch does not move left or right at all. 56. In this exercise, we look at a knuckleball thrown with a different grip than that of exercise 55. The left or right position (in feet) of a ball thrown with spin rate ω and this new grip at time t seconds is f (ω) = (2.5/4ω2 ) − (2.5/4ω2 ) sin (4ωt + π/2). Treating t as a constant and ω as the variable (change to x if you like), find lim f (ω). Your answer should depend on t. By ω→0

graphing this function of t, you can see the path of the pitch (use a domain of 0 ≤ t ≤ 0.68). Describe this pitch. 57. Find functions f such that lim f (x) has the indeterminate x→∞

form ∞ , but where the limit (a) does not exist; (b) equals 0; ∞ (c) equals 3 and (d) equals −4. 58. Find functions f such that lim f (x) has the indeterminate x→∞

form ∞ − ∞, but where the limit (a) does not exist; (b) equals 0 and (c) equals 2. 59. In the figure shown here, a section of the unit circle is determined by angle θ . Region 1 is the triangle ABC. Region 2 is bounded by the line segments AB and BC and the arc of the circle. As the angle θ decreases, the difference between the two regions decreases, also. You might expect that the areas of the regions become nearly equal, in which case the ratio of the areas approaches 1. To see what really happens, show that the area of region 1 divided by the area of region 2 equals sin θ − 12 sin 2θ (1 − cos θ ) sin θ = and find the limit of this θ − cos θ sin θ θ − 12 sin 2θ expression as θ → 0. Surprise! y 1

A

θ

C B

Exercise 59

3.3

1

x

..

Maximum and Minimum Values

265

60. The size of an animal’s pupils expand and contract depending 160x −0.4 + 90 on the amount of light available. Let f (x) = be 8x −0.4 + 10 the size in mm of the pupils at light intensity x. Find lim f (x) x→0+

and lim f (x), and argue that these represent the largest and x→∞

smallest possible sizes of the pupils, respectively.

EXPLORATORY EXERCISES 1. In this exercise, you take a quick look at what we call Taylor sin x = 1. Briefly series in Chapter 8. Start with the limit lim x→0 x explain why this means that for x close to 0, sin x ≈ x. Graph y = sin x and y = x to see why this is true. If you look far enough away from x = 0, the graph of y = sin x eventually curves noticeably. Let’s see if we can find polynomials of higher order to match this curving. Show that sin x − x lim = 0. This means that sin x − x ≈ 0 or (again) x→0 x2 sin x − x 1 = − . This says that if x is sin x ≈ x. Show that lim x→0 x3 6 1 1 close to 0, then sin x − x ≈ − x 3 or sin x ≈ x − x 3 . Graph 6 6 these two functions to see how well they match up. To consin x − (x − x 3 /6) sin x − f (x) tinue, compute lim and lim 4 x→0 x→0 x x5 for the appropriate approximation f (x). At this point, look at the pattern of terms you have (Hint: 6 = 3! and 120 = 5!). Using this pattern, approximate sin x with an 11th-degree polynomial and graph the two functions. 2. A zero of a function f (x) is a solution of the equation f (x) = 0. Clearly, not all zeros are created equal. For example, x = 1 is a zero of f (x) = x − 1, but in some ways it seems that x = 1 should count as two zeros of f (x) = (x − 1)2 . To quantify this, we say that x = 1 is a zero of multiplicity 2 of f (x) = (x − 1)2 . The precise definition is: x = c is a zero of f (x) exists and multiplicity n of f (x) if f (c) = 0 and lim x→c (x − c)n is nonzero. Thus, x = 0 is a zero of multiplicity 2 of x sin x x sin x sin x = 1. Find the multiplicity = lim since lim x→0 x→0 x x2 of each zero of the following functions: x 2 sin x, x sin x 2 , x 4 sin x 3 , (x − 1) ln x, ln (x − 1)2 , e x − 1 and cos x − 1.

MAXIMUM AND MINIMUM VALUES To remain competitive in today’s global economy, businesses need to minimize waste and maximize the return on their investment. In the extremely competitive personal computer industry, companies must continually evaluate how low they can afford to set their prices and still earn a profit. With this backdrop, it should be apparent that it is increasingly important to

3-25

SECTION 3.3

like), show that lim f (ω) = 0 for any value of t. (Hint: Find ω→0

a common denominator and use l’Hˆopital’s Rule.) Conclude that this pitch does not move left or right at all. 56. In this exercise, we look at a knuckleball thrown with a different grip than that of exercise 55. The left or right position (in feet) of a ball thrown with spin rate ω and this new grip at time t seconds is f (ω) = (2.5/4ω2 ) − (2.5/4ω2 ) sin (4ωt + π/2). Treating t as a constant and ω as the variable (change to x if you like), find lim f (ω). Your answer should depend on t. By ω→0

graphing this function of t, you can see the path of the pitch (use a domain of 0 ≤ t ≤ 0.68). Describe this pitch. 57. Find functions f such that lim f (x) has the indeterminate x→∞

form ∞ , but where the limit (a) does not exist; (b) equals 0; ∞ (c) equals 3 and (d) equals −4. 58. Find functions f such that lim f (x) has the indeterminate x→∞

form ∞ − ∞, but where the limit (a) does not exist; (b) equals 0 and (c) equals 2. 59. In the figure shown here, a section of the unit circle is determined by angle θ . Region 1 is the triangle ABC. Region 2 is bounded by the line segments AB and BC and the arc of the circle. As the angle θ decreases, the difference between the two regions decreases, also. You might expect that the areas of the regions become nearly equal, in which case the ratio of the areas approaches 1. To see what really happens, show that the area of region 1 divided by the area of region 2 equals sin θ − 12 sin 2θ (1 − cos θ ) sin θ = and find the limit of this θ − cos θ sin θ θ − 12 sin 2θ expression as θ → 0. Surprise! y 1

A

θ

C B

Exercise 59

3.3

1

x

..

Maximum and Minimum Values

265

60. The size of an animal’s pupils expand and contract depending 160x −0.4 + 90 on the amount of light available. Let f (x) = be 8x −0.4 + 10 the size in mm of the pupils at light intensity x. Find lim f (x) x→0+

and lim f (x), and argue that these represent the largest and x→∞

smallest possible sizes of the pupils, respectively.

EXPLORATORY EXERCISES 1. In this exercise, you take a quick look at what we call Taylor sin x = 1. Briefly series in Chapter 8. Start with the limit lim x→0 x explain why this means that for x close to 0, sin x ≈ x. Graph y = sin x and y = x to see why this is true. If you look far enough away from x = 0, the graph of y = sin x eventually curves noticeably. Let’s see if we can find polynomials of higher order to match this curving. Show that sin x − x lim = 0. This means that sin x − x ≈ 0 or (again) x→0 x2 sin x − x 1 = − . This says that if x is sin x ≈ x. Show that lim x→0 x3 6 1 1 close to 0, then sin x − x ≈ − x 3 or sin x ≈ x − x 3 . Graph 6 6 these two functions to see how well they match up. To consin x − (x − x 3 /6) sin x − f (x) tinue, compute lim and lim 4 x→0 x→0 x x5 for the appropriate approximation f (x). At this point, look at the pattern of terms you have (Hint: 6 = 3! and 120 = 5!). Using this pattern, approximate sin x with an 11th-degree polynomial and graph the two functions. 2. A zero of a function f (x) is a solution of the equation f (x) = 0. Clearly, not all zeros are created equal. For example, x = 1 is a zero of f (x) = x − 1, but in some ways it seems that x = 1 should count as two zeros of f (x) = (x − 1)2 . To quantify this, we say that x = 1 is a zero of multiplicity 2 of f (x) = (x − 1)2 . The precise definition is: x = c is a zero of f (x) exists and multiplicity n of f (x) if f (c) = 0 and lim x→c (x − c)n is nonzero. Thus, x = 0 is a zero of multiplicity 2 of x sin x x sin x sin x = 1. Find the multiplicity = lim since lim x→0 x→0 x x2 of each zero of the following functions: x 2 sin x, x sin x 2 , x 4 sin x 3 , (x − 1) ln x, ln (x − 1)2 , e x − 1 and cos x − 1.

MAXIMUM AND MINIMUM VALUES To remain competitive in today’s global economy, businesses need to minimize waste and maximize the return on their investment. In the extremely competitive personal computer industry, companies must continually evaluate how low they can afford to set their prices and still earn a profit. With this backdrop, it should be apparent that it is increasingly important to

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3-26

use mathematical methods to maximize and minimize various quantities of interest. In this section, we investigate the notion of maximum and minimum from a purely mathematical standpoint. In section 3.7, we examine how to apply these notions to problems of an applied nature. We begin by giving careful mathematical definitions of some familiar terms.

DEFINITION 3.1 For a function f defined on a set S of real numbers and a number c ∈ S, (i) f (c) is the absolute maximum of f on S if f (c) ≥ f (x) for all x ∈ S and (ii) f (c) is the absolute minimum of f on S if f (c) ≤ f (x) for all x ∈ S. An absolute maximum or an absolute minimum is referred to as an absolute extremum. If a function has more than one extremum, we refer to these as extrema (the plural form of extremum).

The first question you might ask is whether every function has an absolute maximum and an absolute minimum. The answer is no, as we can see from Figures 3.25a and 3.25b. y y

Has no absolute maximum

Absolute maximum f (c)

c

x c

Absolute minimum

f (c)

FIGURE 3.25a

EXAMPLE 3.1

x

Has no absolute minimum

FIGURE 3.25b

Absolute Maximum and Minimum Values

(a) Locate any absolute extrema of f (x) = x 2 − 9 on the interval (−∞, ∞). (b) Locate any absolute extrema of f (x) = x 2 − 9 on the interval (−3, 3). (c) Locate any absolute extrema of f (x) = x 2 − 9 on the interval [−3, 3]. Solution (a) In Figure 3.26, notice that f has an absolute minimum value of f (0) = −9, but has no absolute maximum value. (b) In Figure 3.27a, we see that f has an absolute minimum value of f (0) = −9, but still has no absolute maximum value. Your initial reaction might be to say that f has an absolute maximum of 0, but f (x) = 0 for any x ∈ (−3, 3), since this is an open interval and hence, does not include the endpoints −3 and 3.

3-27

SECTION 3.3

Maximum and Minimum Values

y

No absolute maximum

y

..

y No absolute maximum 3

3

267

x

Absolute maximum f (3) f (3) 0 3

3

x

x

3

3 f (0) 9 (Absolute minimum) f(0) 9 (Absolute minimum)

FIGURE 3.26

f(0) 9 (Absolute minimum)

FIGURE 3.27a

FIGURE 3.27b

y = x 2 − 9 on (−3, 3)

y = x 2 − 9 on [−3, 3]

(c) In this case, the endpoints 3 and −3 are in the interval [−3, 3]. Here, f assumes its absolute maximum at two points: f (3) = f (−3) = 0 (see Figure 3.27b).

y = x 2 − 9 on (−∞, ∞)

We have seen that a function may or may not have absolute extrema, depending on the interval on which we’re looking. In example 3.1, the function failed to have an absolute maximum, except on the closed, bounded interval [−3, 3]. This provides some clues, and example 3.2 provides another piece of the puzzle.

EXAMPLE 3.2 y

A Function with No Absolute Maximum or Minimum

Locate any absolute extrema of f (x) = 1/x, on the interval [−3, 3]. Solution From the graph in Figure 3.28, f clearly fails to have either an absolute maximum or an absolute minimum on [−3, 3]. The following table of values for f (x) for x close to 0 suggests the same conclusion.

2 3

x

x 1 0.1 0.01 0.001 0.0001 0.00001 0.000001

3

FIGURE 3.28 y = 1/x

1/x 1 10 100 1000 10,000 100,000 1,000,000

x −1 −0.1 −0.01 −0.001 −0.0001 −0.00001 −0.000001

1/x −1 −10 −100 −1000 −10,000 −100,000 −1,000,000

The most obvious difference between the functions in examples 3.1 and 3.2 is that f (x) = 1/x is discontinuous at a point in the interval [−3, 3]. We offer the following theorem without proof.

THEOREM 3.1 (Extreme Value Theorem) A continuous function f defined on a closed, bounded interval [a, b] attains both an absolute maximum and an absolute minimum on that interval.

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3-28

While you do not need to have a continuous function or a closed interval to have an absolute extremum, Theorem 3.1 says that continuous functions are guaranteed to have an absolute maximum and an absolute minimum on a closed, bounded interval. In example 3.3, we revisit the function from example 3.2, but look on a different interval. y

EXAMPLE 3.3

Finding Absolute Extrema of a Continuous Function

Find the absolute extrema of f (x) = 1/x on the interval [1, 3].

1

x 1

3

Solution Notice that on the interval [1, 3], f is continuous. Consequently, the Extreme Value Theorem guarantees that f has both an absolute maximum and an absolute minimum on [1, 3]. Judging from the graph in Figure 3.29, it appears that f (x) reaches its maximum value of 1 at x = 1 and its minimum value of 1/3 at x = 3.

FIGURE 3.29 y = 1/x on [1, 3]

Our objective is to determine how to locate the absolute extrema of a given function. Before we do this, we need to consider an additional type of extremum.

DEFINITION 3.2 (i) f (c) is a local maximum of f if f (c) ≥ f (x) for all x in some open interval containing c. (ii) f (c) is a local minimum of f if f (c) ≤ f (x) for all x in some open interval containing c. In either case, we call f (c) a local extremum of f . Local maxima and minima (the plural forms of maximum and minimum, respectively) are sometimes referred to as relative maxima and minima, respectively. Notice from Figure 3.30 that each local extremum seems to occur either at a point where the tangent line is horizontal [i.e., where f (x) = 0], at a point where the tangent line is vertical [where f (x) is undefined] or at a corner [again, where f (x) is undefined]. We can see this behavior quite clearly in examples 3.4 and 3.5. y Local maximum [ f (d) is undefined] Local maximum [ f (b) 0]

a

c b

x d

Local minimum [ f (a) 0] Local minimum [ f (c) is undefined]

FIGURE 3.30 Local extrema

3-29

SECTION 3.3

..

Maximum and Minimum Values

269

y

EXAMPLE 3.4

A Function with a Zero Derivative at a Local Maximum

Locate any local extrema for f (x) = 9 − x 2 and describe the behavior of the derivative at the local extremum.

5 x

2

2

Solution We can see from Figure 3.31 that there is a local maximum at x = 0. Further, note that f (x) = −2x and so, f (0) = 0. Note that this says that the tangent line to y = f (x) at x = 0 is horizontal, as indicated in Figure 3.31.

10

EXAMPLE 3.5 FIGURE 3.31 y = 9 − x 2 and the tangent line at x = 0 y

A Function with an Undefined Derivative at a Local Minimum

Locate any local extrema for f (x) = |x| and describe the behavior of the derivative at the local extremum. Solution We can see from Figure 3.32 that there is a local minimum at x = 0. As we have noted in section 2.1, the graph has a corner at x = 0 and hence, f (0) is undefined. [See example 1.7 in Chapter 2.]

3

x 2

FIGURE 3.32

The graphs shown in Figures 3.30–3.32 are not unusual. Here is a small challenge: spend a little time now drawing graphs of functions with local extrema. It should not take long to convince yourself that local extrema occur only at points where the derivative is either zero or undefined. Because of this, we give these points a special name.

y = |x|

DEFINITION 3.3 A number c in the domain of a function f is called a critical number of f if f (c) = 0 or f (c) is undefined.

HISTORICAL NOTES Pierre de Fermat (1601–1665) A French mathematician who discovered many important results, including the theorem named for him. Fermat was a lawyer and member of the Toulouse supreme court, with mathematics as a hobby. The “Prince of Amateurs” left an unusual legacy by writing in the margin of a book that he had discovered a wonderful proof of a clever result, but that the margin of the book was too small to hold the proof. Fermat’s Last Theorem confounded many of the world’s best mathematicians for more than 300 years before being proved by Andrew Wiles in 1995.

It turns out that our earlier observation regarding the location of extrema is correct. That is, local extrema occur only at points where the derivative is zero or undefined. We state this formally in Theorem 3.2.

THEOREM 3.2 (Fermat’s Theorem) Suppose that f (c) is a local extremum (local maximum or local minimum). Then c must be a critical number of f .

PROOF Suppose that f is differentiable at x = c. (If not, c is a critical number of f and we are done.) Suppose further that f (c) = 0. Then, either f (c) > 0 or f (c) < 0. If f (c) > 0, we have by the definition of derivative that f (c + h) − f (c) f (c) = lim > 0. h→0 h So, for all h sufficiently small, f (c + h) − f (c) > 0. (3.1) h

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3-30

For h > 0, (3.1) says that

f (c + h) − f (c) > 0 f (c + h) > f (c).

and so, Thus, f (c) is not a local maximum. For h < 0, (3.1) says that

TODAY IN MATHEMATICS Andrew Wiles (1953– ) A British mathematician who in 1995 published a proof of Fermat’s Last Theorem, the most famous unsolved problem of the 20th century. Fermat’s Last Theorem states that there is no integer solution x , y and z of the equation x n + y n = zn for integers n > 2. Wiles had wanted to prove the theorem since reading about it as a 10-year-old. After more than ten years as a successful research mathematician, Wiles isolated himself from colleagues for seven years as he developed the mathematics needed for his proof. “I realised that talking to people casually about Fermat was impossible because it generated too much interest. You cannot focus yourself for years unless you have this kind of undivided concentration which too many spectators would destroy.” The last step of his proof came, after a year of intense work on this one step, as “this incredible revelation” that was “so indescribably beautiful, it was so simple and elegant.”

f (c + h) − f (c) < 0 and so,

f (c + h) < f (c).

Thus, f (c) is not a local minimum. Since we had assumed that f (c) was a local extremum, this is a contradiction. This rules out the possibility that f (c) > 0. Similarly, if f (c) < 0, we obtain the same contradiction. This is left as an exercise. The only remaining possibility is to have f (c) = 0 and this proves the theorem.

We can use Fermat’s Theorem and calculator- or computer-generated graphs to find local extrema, as in examples 3.6 and 3.7.

EXAMPLE 3.6

Finding Local Extrema of a Polynomial

Find the critical numbers and local extrema of f (x) = 2x 3 − 3x 2 − 12x + 5.

y

20

1

x 2

20

FIGURE 3.33 y = 2x 3 − 3x 2 − 12x + 5

Solution Here,

f (x) = 6x 2 − 6x − 12 = 6(x 2 − x − 2) = 6(x − 2)(x + 1).

Thus, f has two critical numbers, x = −1 and x = 2. Notice from the graph in Figure 3.33 that these correspond to the locations of a local maximum and a local minimum, respectively.

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..

Maximum and Minimum Values

271

y

EXAMPLE 3.7

An Extremum at a Point Where the Derivative Is Undefined

Find the critical numbers and local extrema of f (x) = (3x + 1)2/3 .

4

Solution Here, we have f (x) = x

2

2 2 (3x + 1)−1/3 (3) = . 3 (3x + 1)1/3

Of course, f (x) = 0 for all x, but f (x) is undefined at x = − 13 . Be sure to note that − 13 is in the domain of f . Thus, x = − 13 is the only critical number of f . From the graph in Figure 3.34, we see that this corresponds to the location of a local minimum (also the absolute minimum). If you use your graphing utility to try to produce a graph of y = f (x), you may get only half of the graph displayed in Figure 3.34. The reason is that the algorithms used by most calculators and many computers will return a complex number (or an error) when asked to compute certain fractional powers of negative numbers. While this annoying shortcoming presents only occasional difficulties, we mention this here only so that you are aware that technology has limitations.

2

FIGURE 3.34 y = (3x + 1)2/3

REMARK 3.1 Fermat’s Theorem says that local extrema can occur only at critical numbers. This does not say that there is a local extremum at every critical number. In fact, this is false, as we illustrate in examples 3.8 and 3.9.

EXAMPLE 3.8

A Horizontal Tangent at a Point That Is Not a Local Extremum

Find the critical numbers and local extrema of f (x) = x 3 .

y

Solution It should be clear from Figure 3.35 that f has no local extrema. However, f (x) = 3x 2 = 0 for x = 0 (the only critical number of f ). In this case, f has a horizontal tangent line at x = 0, but does not have a local extremum there.

2 x

2

EXAMPLE 3.9

2 2

A Vertical Tangent at a Point That Is Not a Local Extremum

Find the critical numbers and local extrema of f (x) = x 1/3 . Solution As in example 3.8, f has no local extrema (see Figure 3.36). Here, f (x) = 13 x −2/3 and so, f has a critical number at x = 0 (in this case the derivative is undefined at x = 0). However, f does not have a local extremum at x = 0.

FIGURE 3.35 y = x3 y

You should always check that a given value is in the domain of the function before declaring it a critical number, as in example 3.10.

2 x

2

2 2

FIGURE 3.36 y = x 1/3

EXAMPLE 3.10

Finding Critical Numbers of a Rational Function

2x 2 . x +2 Solution You should note that the domain of f consists of all real numbers other than x = −2. Here, we have

Find all the critical numbers of f (x) =

4x(x + 2) − 2x 2 (1) (x + 2)2 2x(x + 4) = . (x + 2)2

f (x) =

From the quotient rule.

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3-32

Notice that f (x) = 0 for x = 0, −4 and f (x) is undefined for x = −2. However, −2 is not in the domain of f and consequently, the only critical numbers are x = 0 and x = −4.

REMARK 3.2

We have observed that local extrema occur only at critical numbers and that continuous functions must have an absolute maximum and an absolute minimum on a closed, bounded interval. However, we haven’t yet really been able to say how to find these extrema. Theorem 3.3 is particularly useful.

When we use the terms maximum, minimum or extremum without specifying absolute or local, we will always be referring to absolute extrema.

THEOREM 3.3 Suppose that f is continuous on the closed interval [a, b]. Then, the absolute extrema of f must occur at an endpoint (a or b) or at a critical number.

PROOF First, recall that by the Extreme Value Theorem, f will attain its maximum and minimum values on [a, b], since f is continuous. Let f (c) be an absolute extremum. If c is not an endpoint (i.e., c = a and c = b), then c must be in the open interval (a, b). Thus, f (c) must be a local extremum, also. Finally, by Fermat’s Theorem, c must be a critical number, since local extrema occur only at critical numbers.

REMARK 3.3 Theorem 3.3 gives us a simple procedure for finding the absolute extrema of a continuous function on a closed, bounded interval: 1. Find all critical numbers in the interval and compute function values at these points. 2. Compute function values at the endpoints. 3. The largest function value is the absolute maximum and the smallest function value is the absolute minimum. We illustrate Theorem 3.3 for the case of a polynomial function in example 3.11.

EXAMPLE 3.11

Finding Absolute Extrema on a Closed Interval

Find the absolute extrema of f (x) = 2x 3 − 3x 2 − 12x + 5 on the interval [−2, 4].

y

Solution From the graph in Figure 3.37, the maximum appears to be at the endpoint x = 4, while the minimum appears to be at a local minimum near x = 2. In example 3.6, we found that the critical numbers of f are x = −1 and x = 2. Further, both of these are in the interval [−2, 4]. So, we compare the values at the endpoints:

40

20

f (−2) = 1 x

2

2

4

20

FIGURE 3.37 y = 2x 3 − 3x 2 − 12x + 5

and

f (4) = 37,

and the values at the critical numbers: f (−1) = 12 and f (2) = −15. Since f is continuous on [−2, 4], Theorem 3.3 says that the absolute extrema must be among these four values. Thus, f (4) = 37 is the absolute maximum and f (2) = −15 is the absolute minimum. Note that these values are consistent with what we see in the graph in Figure 3.37.

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..

Maximum and Minimum Values

273

Of course, most real problems of interest are unlikely to result in derivatives with integer zeros. Consider the following somewhat less user-friendly example. y

EXAMPLE 3.12

Finding Extrema for a Function with Fractional Exponents

Find the absolute extrema of f (x) = 4x 5/4 − 8x 1/4 on the interval [0, 4].

10

x 2

4

Solution First, we draw a graph of the function to get an idea of where the extrema are located (see Figure 3.38). From the graph, it appears that the maximum occurs at the endpoint x = 4 and the minimum near x = 12 . Next, observe that 5x − 2 . x 3/4

Thus, the critical numbers are x = 25 since f 25 = 0 and x = 0 (since f (0) is undefined and 0 is in the domain of f ). We now need only compare f (0) = 0, f (4) ≈ 11.3137 and f 25 ≈ −5.0897.

5

f (x) = 5x 1/4 − 2x −3/4 =

FIGURE 3.38 y = 4x 5/4 − 8x 1/4

So, the absolute maximum is f (4) ≈ 11.3137 and the absolute minimum is f 25 ≈ −5.0897, which is consistent with what we expected from Figure 3.38.

y 8

In practice, the critical numbers are not always as easy to find as they were in examples 3.11 and 3.12. In example 3.13, it is not even known how many critical numbers there are. We can, however, estimate the number and locations of these from a careful analysis of computer-generated graphs. x

2

3

EXAMPLE 3.13

Finding Absolute Extrema Approximately

Find the absolute extrema of f (x) = x 3 − 5x + 3 sin x 2 on the interval [ −2, 2.5]. 4

Solution We first draw a graph to get an idea of where the extrema will be located (see Figure 3.39). From the graph, we can see that the maximum seems to occur near x = −1, while the minimum seems to occur near x = 2. Next, we compute

FIGURE 3.39 y = f (x) = x 3 − 5x + 3 sin x 2

f (x) = 3x 2 − 5 + 6x cos x 2 .

y

20

x

2

3 10

a ≈ −1.26410884789,

FIGURE 3.40

y = f (x) = 3x − 5 + 6x cos x 2

Unlike examples 3.11 and 3.12, there is no algebra we can use to find the zeros of f . Our only alternative is to find the zeros approximately. You can do this by using Newton’s method to solve f (x) = 0. (You can also use any other rootfinding method built into your calculator or computer.) First, we’ll need adequate initial guesses. We obtain these from the graph of y = f (x) found in Figure 3.40. From the graph, it appears that there are four zeros of f (x) on the interval in question, located near x = −1.3, 0.7, 1.2 and 2.0. Further, referring back to Figure 3.39, these four zeros correspond with the four local extrema seen in the graph of y = f (x). We now apply Newton’s method to solve f (x) = 0, using the preceding four values as our initial guesses. This leads us to four approximate critical numbers of f on the interval [−2, 2.5]. We have

2

c ≈ 1.2266828947 and

b ≈ 0.674471354085, d ≈ 2.01830371473.

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3-34

We now need only compare the values of f at the endpoints and the approximate critical numbers: f (a) ≈ 7.3, f (d) ≈ −4.3,

f (b) ≈ −1.7, f (−2) ≈ −0.3

f (c) ≈ −1.3 f (2.5) ≈ 3.0.

and

Thus, the absolute maximum is approximately f (−1.26410884789) ≈ 7.3 and the absolute minimum is approximately f (2.01830371473) ≈ −4.3. It is important (especially in light of how much of our work here was approximate and graphical) to verify that the approximate extrema correspond with what we expect from the graph of y = f (x). Since these correspond closely, we have great confidence in their accuracy. We have now seen how to locate the absolute extrema of a continuous function on a closed interval. In section 3.4, we see how to find local extrema.

BEYOND FORMULAS The Extreme Value Theorem is an important but subtle result. Think of it this way. If the hypotheses of the theorem are met, you will never waste your time looking for the maximum of a function that does not have a maximum. That is, the problem is always solvable. The technique described in Remark 3.3 always works. If you are asked to find a novel with a certain plot, does it help to know that there actually is such a novel?

EXERCISES 3.3 WRITING EXERCISES 1. Using one or more graphs, explain why the Extreme Value Theorem is true. Is the conclusion true if we drop the hypothesis that f is a continuous function? Is the conclusion true if we drop the hypothesis that the interval is closed? 2. Using one or more graphs, argue that Fermat’s Theorem is true. Discuss how Fermat’s Theorem is used. Restate the theorem in your own words to make its use clearer. 3. Suppose that f (t) represents your elevation after t hours on a mountain hike. If you stop to rest, explain why f (t) = 0. Discuss the circumstances under which you would be at a local maximum, local minimum or neither.

converse always true? Sometimes true? Apply this logic to both the Extreme Value Theorem and Fermat’s Theorem: state the converse and decide whether it is sometimes true or always true.

In exercises 1–4, use the graph to locate the absolute extrema (if they exist) of the function on the given interval. 1. f (x) =

1 on (a) (−∞, ∞), (b) [−1, 1] and (c) (0, 1) x2 − 1 y 10

4. Mathematically, an if/then statement is usually strictly onedirectional. When we say “If A, then B” it is generally not the case that “If B, then A” is also true: when both are true, we say “A if and only if B,” which is abbreviated to “A iff B.” Unfortunately, common English usage is not always this precise. This occasionally causes a problem interpreting a mathematical theorem. To get this straight, consider the statement, “If you wrote a best-selling book, then you made a lot of money.” Is this true? How does this differ from its converse, “If you made a lot of money, then you wrote a best-selling book.” Is the

5

4

x

2

2 5 10

4

3-35

SECTION 3.3

2. f (x) =

x2 on (a) (−∞, ∞), (b) [−1, 1] and (c) (0, 1) (x − 1)2 y 10 8

..

Maximum and Minimum Values

In exercises 11–30, find all critical numbers by hand. If available, use graphing technology to determine whether the critical number represents a local maximum, local minimum or neither. 11. f (x) = x 4 − 3x 3 + 2

12. f (x) = x 4 + 6x 2 − 2

13. f (x) = x 3/4 − 4x 1/4

14. f (x) = (x 2/5 − 3x 1/5 )2 √ 16. f (x) = 3 sin x + cos x

15. f (x) = sin x cos x, [0, 2π] 6

x2 − 2 x +2 x 19. f (x) = 2 x +1

4 2 4

x

2

2

4

3. f (x) = sin x on (a) (−∞, ∞), (b) 0, π4 and (c) π4 , 3π4 y 1 0.5

10

x

5

5

10

0.5

x2 − x + 4 x −1 3x 20. f (x) = 2 x −1

17. f (x) =

18. f (x) =

21. f (x) = 12 (e x + e−x )

22. f (x) = xe−2x

23. f (x) = x 4/3 + 4x 1/3 + 4x −2/3 √ 25. f (x) = 2x x + 1

24. f (x) = x 7/3 − 28x 1/3 √ 26. f (x) = x/ x 2 + 1

27. f (x) = |x 2 − 1| √ 28. f (x) = 3 x 3 − 3x 2 + 2x x 2 + 2x − 1 if x < 0 29. f (x) = x 2 − 4x + 3 if x ≥ 0 sin x if −π < x < π 30. f (x) = − tan x if |x| ≥ π In exercises 31–38, find the absolute extrema of the given function on each indicated interval.

1

4. f (x) = x 3 − 3x + 1 on (a) (−∞, ∞), (b) [−2, 2] and (c) (0, 2) y

35. f (x) = e−x on (a) [0, 2] and (b) [−3, 2] 2

x 2

4

5 10

36. f (x) = x 2 e−4x on (a) [−2, 0] and (b) [0, 4] 3x 2 37. f (x) = on (a) [−2, 2] and (b) [2, 8] x −3 38. f (x) = tan−1 (x 2 ) on (a) [0, 1] and (b) [−3, 4] In exercises 39–44, numerically estimate the absolute extrema of the given function on the indicated intervals.

In exercises 5–10, find all critical numbers by hand. Use your knowledge of the type of graph (i.e., parabola or cubic) to determine whether the critical number represents a local maximum, local minimum or neither. 5. f (x) = x 2 + 5x − 1

6. f (x) = −x 2 + 4x + 2

7. f (x) = x 3 − 3x + 1

8. f (x) = −x 3 + 6x 2 + 2

9. f (x) = x 3 − 3x 2 + 6x

32. f (x) = x 4 − 8x 2 + 2 on (a) [−3, 1] and (b) [−1, 3]

34. f (x) = sin x + cos x on (a) [0, 2π] and (b) [π/2, π ]

5

2

31. f (x) = x 3 − 3x + 1 on (a) [0, 2] and (b) [−3, 2]

33. f (x) = x 2/3 on (a) [−4, −2] and (b) [−1, 3]

10

4

275

10. f (x) = x 3 − 3x 2 + 3x

39. f (x) = x 4 − 3x 2 + 2x + 1 on (a) [−1, 1] and (b) [−3, 2] 40. f (x) = x 6 − 3x 4 − 2x + 1 on (a) [−1, 1] and (b) [−2, 2] 41. f (x) = x 2 − 3x cos x on (a) [−2, 1] and (b) [−5, 0] 42. f (x) = xecos 2x on (a) [−2, 2] and (b) [2, 5]

43. f (x) = x sin x + 3 on (a) −π , π2 and (b) [0, 2π ] 2 44. f (x) = x 2 + e x on (a) [0, 1] and (b) [−2, 2]

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45. Repeat exercises 31–38, except instead of finding extrema on the closed interval, find the extrema on the open interval, if they exist. 46. Briefly outline a procedure for finding extrema on an open interval (a, b), a procedure for the half-open interval (a, b] and a procedure for the half-open interval [a, b). 47. Sketch a graph of a function f such that the absolute maximum of f (x) on the interval [−2, 2] equals 3 and the absolute minimum does not exist. 48. Sketch a graph of a continuous function f such that the absolute maximum of f (x) on the interval (−2, 2) does not exist and the absolute minimum equals 2. 49. Sketch a graph of a continuous function f such that the absolute maximum of f (x) on the interval (−2, 2) equals 4 and the absolute minimum equals 2.

3-36

59. If two soccer teams each score goals at a rate of r goals per minute, the probability that n goals will be scored in t minutes is n 1 P = (rn!t) e−r t . Take r = 25 . Show that for a 90-minute game, P is maximized with n = 3. Briefly explain why this makes sense. 60. In the situation of exercise 59, find t to maximize the probability that exactly 1 goal has been scored. Briefly explain why your answer makes sense. 61. If f is differentiable on the interval [a, b] and f (a) < 0 < f (b), prove that there is a c with a < c < b for which f (c) = 0. (Hint: Use the Extreme Value Theorem and Fermat’s Theorem.) 62. Sketch a graph showing that y = f (x) = x 2 + 1 and y = g(x) = ln x do not intersect. Find x to minimize f (x) − g(x). At this value of x, show that the tangent lines to y = f (x) and y = g(x) are parallel. Explain graphically why it makes sense that the tangent lines are parallel. x2 for x > 0 and determine x2 + 1 where the graph is steepest. (That is, find where the slope is a maximum.)

50. Sketch a graph of a function f such that the absolute maximum of f (x) on the interval [−2, 2] does not exist and the absolute minimum does not exist.

63. Sketch a graph of f (x) =

51. Give an example showing that the following statement is false (not always true): between any two local minima of f (x) there is a local maximum.

64. Sketch a graph of f (x) = e−x and determine where the graph is steepest. (Note: This is an important problem in probability theory.)

52. If you have won three out of four matches against someone, does that mean that the probability that you will win the next one is 34 ? In general, if you have a probability p of winning each match, the probability of winning m out of n matches n! p m (1 − p)n−m . Find p to maximize f . is f ( p) = (n − m)! m! This value of p is called the maximum likelihood estimator of the probability. Briefly explain why your answer makes sense. 53. In this exercise, we will explore the family of functions f (x) = x 3 + cx + 1, where c is constant. How many and what types of local extrema are there? (Your answer will depend on the value of c.) Assuming that this family is indicative of all cubic functions, list all types of cubic functions. 54. Prove that any fourth-order polynomial must have at least one local extremum and can have a maximum of three local extrema. Based on this information, sketch several possible graphs of fourth-order polynomials. 55. Show that f (x) = x 3 + bx 2 + cx + d has both a local maximum and a local minimum if c < 0.

2

65. A section of roller coaster is in the shape of y = x 5 − 4x 3 − x + 10, where x is between −2 and 2. Find all local extrema and explain what portions of the roller coaster they represent. Find the location of the steepest part of the roller coaster. 66. Suppose a large computer file is sent over the Internet. If the probability that it reaches its destination without any errors is x, then the probability that an error is made is 1 − x. The field of information theory studies such situations. An important quantity is entropy (a measure of unpredictability), defined by H = −x ln x − (1 − x) ln(1 − x), for 0 < x < 1. Find the value of x that maximizes this quantity. Explain why this probability would maximize the measure of unpredictability of errors. (Hint: If x = 0 or x = 1, are errors unpredictable?) 67. Researchers in a number of fields (including population biology, economics and the study of animal tumors) make use of −t the Gompertz growth curve, W (t) = ae−be . As t → ∞, show that W (t) → a and W (t) → 0. Find the maximum growth rate.

57. For the family of functions f (x) = x 4 + cx 2 + 1, find all local extrema (your answer will depend on the value of the constant c).

68. The rate R of an enzymatic reaction as a function of the sub[S]Rm strate concentration [S] is given by R = , where Rm K m + [S] and K m are constants. K m is called the Michaelis constant and Rm is referred to as the maximum reaction rate. Show that Rm is not a proper maximum in that the reaction rate can never be equal to Rm .

58. For the family of functions f (x) = x 4 + cx 3 + 1, find all local extrema. (Your answer will depend on the value of the constant c.)

69. Suppose a painting hangs on a wall as in the figure. The frame extends from 6 feet to 8 feet above the floor. A person whose eyes are 5 feet above the ground stands x feet from the wall

56. In exercise 55, show that the sum of the critical numbers is − 2b3 .

3-37

SECTION 3.4

and views the painting, with a viewing angle A formed by the ray from the person’s eye to the top of the frame and the ray from the person’s eye to the bottom of the bottom of the frame. Find the value of x that maximizes the viewing angle A. 8'

6' A

5'

..

Increasing and Decreasing Functions

277

and h = 2x so V = 2π y 2 x. Call the rod measurement z. By the Pythagorean Theorem, x 2 + (2y)2 = z 2 . Kepler’s mystery was how to compute V given only z. The key observation made by Kepler was that Austrian wine casks were made with the same height-to-diameter ratio (for us, x/y). Let t = x/y 2 and show that z 2 /y 2 = t 2 + 4. Use this to replace y in the 2 2 volume formula. Then replace x with z + 4y . Show that 2π z 3 t . In this formula, t is a constant, so the vintV = (4 + t 2 )3/2 ner could measure z and quickly estimate the volume. We haven’t told you yet what t equals. Kepler assumed that the vintners would have made a smart choice for this ratio. Find the value of t that maximizes the volume for a given z. This is, in fact, the ratio used in the construction of Austrian wine casks!

x

70. What changes in exercise 69 if the person’s eyes are 6 feet above the ground? z

EXPLORATORY EXERCISES 1. Explore the graphs of e−x , xe−x , x 2 e−x and x 3 e−x . Find all local extrema and use l’Hˆopital’s Rule to determine the behavior as x → ∞. You can think of the graph of x n e−x as showing the results of a tug-of-war: x n → ∞ as x → ∞ but e−x → 0 as x → ∞. Describe the graph of x n e−x in terms of this tug-of-war. 2. Johannes Kepler (1571–1630) is best known as an astronomer, especially for his three laws of planetary motion. However, his discoveries were primarily due to his brilliance as a mathematician. While serving in Austrian Emperor Matthew I’s court, Kepler observed the ability of Austrian vintners to quickly and mysteriously compute the capacities of a variety of wine casks. Each cask (barrel) had a hole in the middle of its side (see Figure a). The vintner would insert a rod in the hole until it hit the far corner and then announce the volume. Kepler first analyzed the problem for a cylindrical barrel (see Figure b). The volume of a cylinder is V = πr 2 h. In Figure b, r = y

3.4

FIGURE a

z

2y 2x

FIGURE b 3. Suppose that a hockey player is shooting at a 6-foot-wide net from a distance of d feet away from the goal line and 4 feet to the side of the center line. (a) Find the distance d that maximizes the shooting angle. (b) Repeat part (a) with the shooter 2 feet to the side of the center line. Explain why the answer is so different. (c) Repeat part (a) with the goalie blocking all but the far 2 feet of the goal.

INCREASING AND DECREASING FUNCTIONS In section 3.3, we determined that local extrema occur only at critical numbers. However, not all critical numbers correspond to local extrema. In this section, we see how to determine which critical numbers correspond to local extrema. At the same time, we’ll learn more about the connection between the derivative and graphing.

3-37

SECTION 3.4

and views the painting, with a viewing angle A formed by the ray from the person’s eye to the top of the frame and the ray from the person’s eye to the bottom of the bottom of the frame. Find the value of x that maximizes the viewing angle A. 8'

6' A

5'

..

Increasing and Decreasing Functions

277

and h = 2x so V = 2π y 2 x. Call the rod measurement z. By the Pythagorean Theorem, x 2 + (2y)2 = z 2 . Kepler’s mystery was how to compute V given only z. The key observation made by Kepler was that Austrian wine casks were made with the same height-to-diameter ratio (for us, x/y). Let t = x/y 2 and show that z 2 /y 2 = t 2 + 4. Use this to replace y in the 2 2 volume formula. Then replace x with z + 4y . Show that 2π z 3 t . In this formula, t is a constant, so the vintV = (4 + t 2 )3/2 ner could measure z and quickly estimate the volume. We haven’t told you yet what t equals. Kepler assumed that the vintners would have made a smart choice for this ratio. Find the value of t that maximizes the volume for a given z. This is, in fact, the ratio used in the construction of Austrian wine casks!

x

70. What changes in exercise 69 if the person’s eyes are 6 feet above the ground? z

EXPLORATORY EXERCISES 1. Explore the graphs of e−x , xe−x , x 2 e−x and x 3 e−x . Find all local extrema and use l’Hˆopital’s Rule to determine the behavior as x → ∞. You can think of the graph of x n e−x as showing the results of a tug-of-war: x n → ∞ as x → ∞ but e−x → 0 as x → ∞. Describe the graph of x n e−x in terms of this tug-of-war. 2. Johannes Kepler (1571–1630) is best known as an astronomer, especially for his three laws of planetary motion. However, his discoveries were primarily due to his brilliance as a mathematician. While serving in Austrian Emperor Matthew I’s court, Kepler observed the ability of Austrian vintners to quickly and mysteriously compute the capacities of a variety of wine casks. Each cask (barrel) had a hole in the middle of its side (see Figure a). The vintner would insert a rod in the hole until it hit the far corner and then announce the volume. Kepler first analyzed the problem for a cylindrical barrel (see Figure b). The volume of a cylinder is V = πr 2 h. In Figure b, r = y

3.4

FIGURE a

z

2y 2x

FIGURE b 3. Suppose that a hockey player is shooting at a 6-foot-wide net from a distance of d feet away from the goal line and 4 feet to the side of the center line. (a) Find the distance d that maximizes the shooting angle. (b) Repeat part (a) with the shooter 2 feet to the side of the center line. Explain why the answer is so different. (c) Repeat part (a) with the goalie blocking all but the far 2 feet of the goal.

INCREASING AND DECREASING FUNCTIONS In section 3.3, we determined that local extrema occur only at critical numbers. However, not all critical numbers correspond to local extrema. In this section, we see how to determine which critical numbers correspond to local extrema. At the same time, we’ll learn more about the connection between the derivative and graphing.

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Salary

3-38

We are all familiar with the terms increasing and decreasing. If your employer informs you that your salary will be increasing steadily over the term of your employment, you have in mind that as time goes on, your salary will rise something like Figure 3.41. If you take out a loan to purchase a car, once you start paying back the loan, your indebtedness will decrease over time. If you plotted your debt against time, the graph might look something like Figure 3.42. We now carefully define these notions. Notice that Definition 4.1 is merely a formal statement of something you already understand. Time

DEFINITION 4.1

FIGURE 3.41

A function f is (strictly) increasing on an interval I if for every x1 , x2 ∈ I with x1 < x2 , f (x1 ) < f (x2 ) [i.e., f (x) gets larger as x gets larger]. A function f is (strictly) decreasing on the interval I if for every x1 , x2 ∈ I with x1 < x2 , f (x1 ) > f (x2 ) [i.e., f (x) gets smaller as x gets larger].

Increasing salary

Debt

Time

FIGURE 3.42 Decreasing debt

Why do we bother with such an obvious definition? Of course, anyone can look at a graph of a function and immediately see where that function is increasing and decreasing. The real challenge is to determine where a function is increasing and decreasing, given only a mathematical formula for the function. For example, can you determine where f (x) = x 2 sin x is increasing and decreasing, without looking at a graph? Look carefully at Figure 3.43 to see if you can notice what happens at every point at which the function is increasing or decreasing. f increasing (tangent lines have positive slope)

y

y f (x)

x

f decreasing (tangent lines have negative slope)

FIGURE 3.43 Increasing and decreasing

Observe that on intervals where the tangent lines have positive slope, f is increasing, while on intervals where the tangent lines have negative slope, f is decreasing. Of course, the slope of the tangent line at a point is given by the value of the derivative at that point. So, whether a function is increasing or decreasing on an interval seems to be connected to the sign of its derivative on that interval. This conjecture, although it’s based on only a single picture, sounds like a theorem and it is.

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Increasing and Decreasing Functions

279

THEOREM 4.1 Suppose that f is differentiable on an interval I . (i) If f (x) > 0 for all x ∈ I , then f is increasing on I . (ii) If f (x) < 0 for all x ∈ I , then f is decreasing on I .

PROOF (i) Pick any two points x1 , x2 ∈ I , with x1 < x2 . Applying the Mean Value Theorem (Theorem 9.4 in section 2.9) to f on the interval (x1 , x2 ), we get f (x2 ) − f (x1 ) = f (c), x2 − x1

(4.1)

for some number c ∈ (x1 , x2 ). (Why can we apply the Mean Value Theorem here?) By hypothesis, f (c) > 0 and since x1 < x2 (so that x2 − x1 > 0), we have from (4.1) that 0 < f (x2 ) − f (x1 ) f (x1 ) < f (x2 ).

or

(4.2)

Since (4.2) holds for all x1 < x2 , f is increasing on I . The proof of (ii) is nearly identical and is left as an exercise.

What You See May Not Be What You Get One aim here and in sections 3.5 and 3.6 is to learn how to draw representative graphs of functions (i.e., graphs that display all of the significant features of a function: where it is increasing or decreasing, any extrema, asymptotes and two features we’ll introduce in section 3.5, concavity and inflection points). When we draw a graph, we are drawing in a particular viewing window (i.e., a particular range of x- and y-values). In the case of computer- or calculator-generated graphs, the window is often chosen by the machine. So, how do we know when significant features are hidden outside of a given window? Further, how do we determine the precise locations of features that we can see in a given window? As we’ll discover, the only way we can resolve these questions is with some calculus.

EXAMPLE 4.1

y

Draw a graph of f (x) = 2x 3 + 9x 2 − 24x − 10 showing all local extrema.

10

x

10

Drawing a Graph

10 10

FIGURE 3.44 y = 2x 3 + 9x 2 − 24x − 10

Solution Many graphing calculators use the default window defined by −10 ≤ x ≤ 10 and −10 ≤ y ≤ 10. Using this window, the graph of y = f (x) looks like that displayed in Figure 3.44, although the three segments shown are not particularly revealing. Instead of blindly manipulating the window in the hope that a reasonable graph will magically appear, we stop briefly to determine where the function is increasing and decreasing. First, observe that f (x) = 6x 2 + 18x − 24 = 6(x 2 + 3x − 4) = 6(x − 1)(x + 4).

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..

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Note that the critical numbers (1 and −4) are the only possible locations for local extrema. We can see where the two factors and consequently the derivative are positive and negative from the number lines displayed in the margin. From this, note that

6(x 1)

1

(x 4)

4

0

0

4

6(x 1)(x 4)

f (x) > 0 on (−∞, −4) ∪ (1, ∞)

f increasing.

f (x) < 0 on (−4, 1).

f decreasing.

1

and

y

100

x

8

4 50

FIGURE 3.45a y = 2x 3 + 9x 2 − 24x − 10 y f(x)

f '(x)

For convenience, we have placed arrows indicating where the function is increasing and decreasing beneath the last number line. In Figure 3.45a, we redraw the graph in the window defined by −8 ≤ x ≤ 4 and −50 ≤ y ≤ 125. Here, we have selected the y-range so that the critical points (−4, 102) and (1, −23) are displayed. Since f is increasing on all of (−∞, −4), we know that the function is still increasing to the left of the portion displayed in Figure 3.45a. Likewise, since f is increasing on all of (1, ∞), we know that the function continues to increase to the right of the displayed portion. In Figure 3.45b, we have plotted both y = f (x) (shown in blue) and y = f (x) (shown in red). Notice the connection between the two graphs. When f (x) > 0, f is increasing; when f (x) < 0, f is decreasing and also notice what happens to f (x) at the local extrema of f . (We’ll say more about this shortly.)

100

x

8

3-40

4 50

You may be tempted to think that you can draw graphs by machine and with a little fiddling with the graphing window, get a reasonable looking graph. Unfortunately, this frequently isn’t enough. For instance, while it’s clear that the graph in Figure 3.44 is incomplete, the initial graph in example 4.2 has a familiar shape and may look reasonable, but it is incorrect. The calculus tells you what features you should expect to see in a graph. Without it, you’re simply taking a shot in the dark.

FIGURE 3.45b

y = f (x) and y = f (x)

EXAMPLE 4.2

Uncovering Hidden Behavior in a Graph

Graph f (x) = 3x 4 + 40x 3 − 0.06x 2 − 1.2x showing all local extrema.

0

0.1

0

0

(x 10)

10

0 10

0

0.1

f (x) = 12x 3 + 120x 2 − 0.12x − 1.2 = 12(x 2 − 0.01)(x + 10)

(x 0.1)

0.1

12(x 0.1)

Solution We first show the default graph drawn by our computer algebra system (see Figure 3.46a). We show a common default graphing calculator graph in Figure 3.46b. You can certainly make Figure 3.46b look more like Figure 3.46a by adjusting the window some. But with some calculus, you can discover features of the graph that are hidden in both graphs. First, notice that

0 0.1

= 12(x − 0.1)(x + 0.1)(x + 10).

f'(x)

We show number lines for the three factors in the margin. Observe that f (x)

> 0 on (−10, −0.1) ∪ (0.1, ∞) < 0 on (−∞, −10) ∪ (−0.1, 0.1).

f increasing. f decreasing.

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SECTION 3.4

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Increasing and Decreasing Functions

281

y

6000

y 10

3000

x

4

4 3000

y

10,000

x

15

5

10,000

FIGURE 3.47a The global behavior of f (x) = 3x 4 + 40x 3 − 0.06x 2 − 1.2x

x

10

10 10

FIGURE 3.46a

FIGURE 3.46b

Default CAS graph of y = 3x 4 + 40x 3 − 0.06x 2 − 1.2x

Default calculator graph of y = 3x 4 + 40x 3 − 0.06x 2 − 1.2x

This says that neither of the machine-generated graphs seen in Figures 3.46a or 3.46b is adequate, as the behavior on (−∞, −10) ∪ (−0.1, 0.1) cannot be seen in either graph. As it turns out, no single graph captures all of the behavior of this function. However, by increasing the range of x-values to the interval [−15, 5], we get the graph seen in Figure 3.47a. This shows the big picture, what we refer to as the global behavior of the function. Here, you can see the local minimum at x = −10, which was missing in our earlier graphs, but the behavior for values of x close to zero is not clear. To see this, we need a separate graph, restricted to a smaller range of x-values, as seen in Figure 3.47b. Notice that here, we can see the behavior of the function for x close to zero quite clearly. In particular, the local maximum at x = −0.1 and the local minimum at x = 0.1 are clearly visible. We often say that a graph such as Figure 3.47b shows the local behavior of the function. In Figures 3.48a and 3.48b, we show graphs indicating the global and local behavior of f (x) (in blue) and f (x) (in red) on the same set of axes. Pay particular attention to the behavior of f (x) in the vicinity of local extrema of f (x). y

y

y

f '(x)

0.4

f (x)

0.4

10,000

x

0.3

0.3

f '(x) x

0.3

0.3

x

15

5

10,000 0.4

FIGURE 3.48a

y = f (x) and y = f (x) (global behavior)

FIGURE 3.47b Local behavior of f (x) = 3x 4 + 40x 3 − 0.06x 2 − 1.2x

f (x) 1.2

FIGURE 3.48b

y = f (x) and y = f (x) (local behavior)

You may have already noticed a connection between local extrema and the intervals on which a function is increasing and decreasing. We state this in Theorem 4.2.

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y

THEOREM 4.2 (First Derivative Test) Local maximum f (x) 0 f increasing

Suppose that f is continuous on the interval [a, b] and c ∈ (a, b) is a critical number. f (x) 0 f decreasing

(i) If f (x) > 0 for all x ∈ (a, c) and f (x) < 0 for all x ∈ (c, b) (i.e., f changes from increasing to decreasing at c), then f (c) is a local maximum. (ii) If f (x) < 0 for all x ∈ (a, c) and f (x) > 0 for all x ∈ (c, b) (i.e., f changes from decreasing to increasing at c), then f (c) is a local minimum. (iii) If f (x) has the same sign on (a, c) and (c, b), then f (c) is not a local extremum.

x

c

It’s easiest to think of this result graphically. If f is increasing to the left of a critical number and decreasing to the right, then there must be a local maximum at the critical number (see Figure 3.49a). Likewise, if f is decreasing to the left of a critical number and increasing to the right, then there must be a local minimum at the critical number (see Figure 3.49b). This suggests a proof of the theorem; the job of writing out all of the details is left as an exercise.

FIGURE 3.49a Local maximum y

c

x

EXAMPLE 4.3

Finding Local Extrema Using the First Derivative Test

Find the local extrema of the function from example 4.1, f (x) = 2x 3 + 9x 2 − 24x − 10. f (x) 0 f decreasing

f (x) 0 f increasing

Solution We had found in example 4.1 that

f (x) Local minimum

FIGURE 3.49b

> 0, on (−∞, −4) ∪ (1, ∞)

f increasing.

< 0, on (−4, 1).

f decreasing.

It now follows from the First Derivative Test that f has a local maximum located at x = −4 and a local minimum located at x = 1.

Local minimum

Theorem 4.2 works equally well for a function with critical points where the derivative is undefined.

EXAMPLE 4.4

Finding Local Extrema of a Function with Fractional Exponents

Find the local extrema of f (x) = x 5/3 − 3x 2/3 . f (x) =

Solution We have

5 2/3 2 x −3 x −1/3 3 3

5x − 6 , 3x 1/3 so that the critical numbers are 65 [ f 65 = 0] and 0 [ f (0) is undefined]. Again drawing number lines for the factors, we determine where f is increasing and decreasing. Here, we have placed an above the 0 on the number line for f (x) to indicate that f (x) is not defined at x = 0. From this, we can see at a glance where f is positive and negative: =

0

(5x 6)

6/5

0

3x1/3

0

0

0 6/5

f (x)

> 0, on (−∞, 0) ∪ f (x) < 0, on 0, 65 .

6 5

,∞

f increasing. f decreasing.

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SECTION 3.4

..

Increasing and Decreasing Functions

Consequently, f has a local maximum at x = 0 and a local minimum at x = 65 . These local extrema are both clearly visible in the graph in Figure 3.50.

y

1 x 2

EXAMPLE 4.5

Finding Local Extrema Approximately

Find the local extrema of f (x) = x 4 + 4x 3 − 5x 2 − 31x + 29 and draw a graph. Solution A graph of y = f (x) using the most common graphing calculator default window appears in Figure 3.51. Without further analysis, we do not know whether this graph shows all of the significant behavior of the function. [Note that some fourth-degree polynomials (e.g., f (x) = x 4 ) have graphs that look very much like the one in Figure 3.51.] First, we compute

FIGURE 3.50 y = x 5/3 − 3x 2/3

f (x) = 4x 3 + 12x 2 − 10x − 31.

y 10

x

10

10

However, this derivative does not easily factor. A graph of y = f (x) (see Figure 3.52) reveals three zeros, one near each of x = −3, −1.5 and 1.5. Since a cubic polynomial has at most three zeros, there are no others. Using Newton’s method or some other rootfinding method [applied to f (x)], we can find approximations to the three zeros of f . We get a ≈ −2.96008, b ≈ −1.63816 and c ≈ 1.59824. From Figure 3.52, we can see that

10

and FIGURE 3.51 f (x) = x 4 + 4x 3 − 5x 2 − 31x + 29

f (x) > 0 on (a, b) ∪ (c, ∞)

f increasing.

f (x) < 0 on (−∞, a) ∪ (b, c).

f decreasing.

You can quickly read off the local extrema: a local minimum at a ≈ −2.96008, a local maximum at b ≈ −1.63816 and a local minimum at c ≈ 1.59824. Since only the local minimum at x = c is visible in the graph in Figure 3.51, this graph is clearly not representative of the behavior of the function. By narrowing the range of displayed x-values and widening the range of displayed y-values, we obtain the far more useful graph seen in Figure 3.53. You should convince yourself, using the preceding analysis, that the local minimum at x = c ≈ 1.59824 is also the absolute minimum.

y

y

100

100

50 a

b

c

4

x 4

50

FIGURE 3.52

f (x) = 4x 3 + 12x 2 − 10x − 31

c 4

a

b

x 4

50

FIGURE 3.53 f (x) = x 4 + 4x 3 − 5x 2 − 31x + 29

283

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3-44

EXERCISES 3.4 WRITING EXERCISES 1. Suppose that f (0) = 2 and f is an increasing function. To sketch the graph of y = f (x), you could start by plotting the point (0, 2). Filling in the graph to the left, would you move your pencil up or down? How does this fit with the definition of increasing? 2. Suppose you travel east on an east-west interstate highway. You reach your destination, stay a while and then return home. Explain the First Derivative Test in terms of your velocities (positive and negative) on this trip. 3. Suppose that you have a differentiable function f with two critical numbers. Your computer has shown you a graph that looks like the one in the figure. y

In exercises 11–20, find (by hand) all critical numbers and use the First Derivative Test to classify each as the location of a local maximum, local minimum or neither. 11. y = x 4 + 4x 3 − 2

12. y = x 5 − 5x 2 + 1

13. y = xe−2x

14. y = x 2 e−x 16. y = sin−1 1 −

15. y = tan−1 (x 2 ) x 1 + x3 √ 19. y = x 3 + 3x 2

17. y =

23. y =

4

4

x 1 + x4

20. y = x 4/3 + 4x 1/3

x2 −1

x2

x −1

22. y =

x2

x2 − 4x + 3

24. y =

x 1 − x4

26. y =

x2 + 2 (x + 1)2

25. y = √

x

In exercises 21–26, find (by hand) all asymptotes and extrema, and sketch a graph. 21. y =

10

18. y =

1 x2

x x2 + 1

x2

In exercises 27–34, find the x-coordinates of all extrema and sketch graphs showing global and local behavior of the function.

10

27. y = x 3 − 13x 2 − 10x + 1 Discuss the possibility that this is a representative graph: that is, is it possible that there are any important points not shown in this window? 4. Suppose that the function in exercise 3 has three critical numbers. Explain why the graph is not a representative graph. Explain how you would change the graphing window to show the rest of the graph.

28. y = x 3 + 15x 2 − 70x + 2 29. y = x 4 − 15x 3 − 2x 2 + 40x − 2 30. y = x 4 − 16x 3 − 0.1x 2 + 0.5x − 1 31. y = x 5 − 200x 3 + 605x − 2 32. y = x 4 − 0.5x 3 − 0.02x 2 + 0.02x + 1 33. y = (x 2 + x + 0.45)e−2x

In exercises 1–10, find (by hand) the intervals where the function is increasing and decreasing. Use this information to sketch a graph. 1. y = x 3 − 3x + 2

2. y = x 3 + 2x 2 + 1

3. y = x 4 − 8x 2 + 1

4. y = x 3 − 3x 2 − 9x + 1

5. y = (x + 1)2/3

6. y = (x − 1)1/3

7. y = sin x + cos x

8. y = sin2 x

9. y = e

x 2 −1

10. y = ln(x − 1) 2

34. y = x 5 ln 8x 2 In exercises 35–38, sketch a graph of a function with the given properties. 35. f (0) = 1, f (2) = 5, f (x) < 0 for x < 0 and x > 2, f (x) > 0 for 0 < x < 2. 36. f (−1) = 1, f (2) = 5, f (x) < 0 for x < −1 and x > 2, f (x) > 0 for −1 < x < 2, f (−1) = 0, f (2) does not exist. 37. f (3) = 0, f (x) < 0 for x < 0 and x > 3, f (x) > 0 for 0 < x < 3, f (3) = 0, f (0) and f (0) do not exist.

3-45

SECTION 3.4

38. f (1) = 0, lim f (x) = 2, f (x) < 0 for x < 1, f (x) > 0 for

x→∞

x > 1, f (1) = 0. In exercises 39–42, estimate critical numbers and sketch graphs showing both global and local behavior. 39. y =

x − 30 x4 − 1

40. y =

x2 − 8 x4 − 1

41. y =

x + 60 x2 + 1

42. y =

x − 60 x2 − 1

43. For f (x) =

x + 2x 2 sin 1/x if x = 0 0

if x = 0 show that f (0) > 0,

but that f is not increasing in any interval around 0. Explain why this does not contradict Theorem 4.1. 44. For f (x) = x 3 , show that f is increasing in any interval around 0, but f (0) = 0. Explain why this does not contradict Theorem 4.1. 45. Prove Theorem 4.2 (the First Derivative Test). 46. Give a graphical argument that if f (a) = g(a) and f (x) > g (x) for all x > a, then f (x) > g(x) for all x > a. Use the Mean Value Theorem to prove it. In exercises 47–50, use the result of exercise 46 to verify the inequality. √ 1 47. 2 x > 3 − for x > 1 48. x > sin x for x > 0 x x 50. x − 1 > ln x for x > 1 49. e > x + 1 for x > 0 51. Give an example showing that the following statement is false. If f (0) = 4 and f (x) is a decreasing function, then the equation f (x) = 0 has exactly one solution. 52. Assume that f is an increasing function with inverse function f −1 . Show that f −1 is also an increasing function.

..

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285

function with local extrema so close together that they’re difficult to see. For instance, suppose you want local extrema at x = −0.1 and x = 0.1. Explain why you could start with f (x) = (x − 0.1)(x + 0.1) = x 2 − 0.01. Look for a function whose derivative is as given. Graph your function to see if the extrema are “hidden.” Next, construct a polynomial of degree 4 with two extrema very near x = 1 and another near x = 0. 60. Suppose that f and g are differentiable functions and x = c is a critical number of both functions. Either prove (if it is true) or disprove (with a counterexample) that the composition f ◦ g also has a critical number at x = c. 61. Show that f (x) = x 3 + bx 2 + cx + d is an increasing function if b2 ≤ 3c. 62. Find a condition on the coefficients b and c similar to exercise 61 that guarantees that f (x) = x 5 + bx 3 + cx + d is an increasing function. 63. The table shows the coefficient of friction µ of ice as a function of temperature. The lower µ is, the more “slippery” the ice is. Estimate µ (C) at (a) C = −10 and (b) C = −6. If skating warms the ice, does it get easier or harder to skate? Briefly explain. ◦

C −12 −10 −8 −6 −4 −2 µ 0.0048 0.0045 0.0043 0.0045 0.0048 0.0055

64. For a college football field with the dimensions shown, the angle θ for kicking a field goal from a (horizontal) distance of x feet from the goal post is given by θ(x) = tan−1 (29.25/x) − tan−1 (10.75/x). Show that t f (t) = 2 is increasing for a > t and use this fact to show a + t2 that θ(x) is a decreasing function for x ≥ 30. Announcers often say that for a short field goal (50 ≤ x ≤ 60), a team can improve the angle by backing up 5 yards with a penalty. Is this true?

53. State the domain for sin−1 x and determine where it is increasing and decreasing. 54. State the domain for sin−1 π2 tan−1 x and determine where it is increasing and decreasing. 55. If f and g are both increasing functions, is it true that f (g(x)) is also increasing? Either prove that it is true or give an example that proves it false. 56. If f and g are both increasing functions with f (5) = 0, find the maximum and minimum of the following values: g(1), g(4), g( f (1)), g( f (4)). 57. Suppose that the√total sales of a product after t months is given by s(t) = t + 4 thousand dollars. Compute and interpret s (t). 58. In exercise 57, show that s (t) > 0 for all t > 0. Explain in business terms why it is impossible to have s (t) < 0. 59. In this exercise, you will play the role of professor and construct a tricky graphing exercise. The first goal is to find a

18.5'

40' x

EXPLORATORY EXERCISES 1. In this exercise, we look at the ability of fireflies to synchronize their flashes. (To see a remarkable demonstration of this ability, see David Attenborough’s video series Trials of Life.) Let the function f represent an individual firefly’s

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3-46

other hand, if T (t) is large enough, then T (t) < 0 and the T-cell population will decrease. For what value of T (t) is T (t) = 0? Even though this population would remain stable, explain why this would be bad news for the infected human.

rhythm, so that the firefly flashes whenever f (t) equals an integer. Let e(t) represent the rhythm of a neighboring firefly, where again e(t) = n, for some integer n, whenever the neighbor flashes. One model of the interaction between fireflies is f (t) = ω + A sin [e(t) − f (t)] for constants ω and A. If the fireflies are synchronized [e(t) = f (t)], then f (t) = ω, so the fireflies flash every 1/ω time units. Assume that the difference between e(t) and f (t) is less than π. Show that if f (t) < e(t), then f (t) > ω. Explain why this means that the individual firefly is speeding up its flash to match its neighbor. Similarly, discuss what happens if f (t) > e(t).

3. In a sport like soccer or hockey where ties are possible, the probability that the stronger team wins depends in an interesting way on the number of goals scored. Suppose that at any point, the probability that team A scores the next goal is p, where 0 < p < 1. If 2 goals are scored, a 1-1 tie could result from team A scoring first (probability p) and then team B tieing the score (probability 1 − p), or vice versa. The probability of a tie in a 2-goal game is then 2 p(1 − p). Similarly, the probability of a 2-2 tie in a 4-goal game is 4·3 2 p (1 − p)2 , the probability of a 3-3 tie in a 6-goal game 2·1

2. The HIV virus attacks specialized T-cells that trigger the human immune system response to a foreign substance. If T (t) is the population of uninfected T-cells at time t (days) and V (t) is the population of infectious HIV in the bloodstream, a model that has been used to study AIDS is given by the following differential equation that describes the rate at which the population of T cells changes. 1 T (t)V (t) T (t) = 10 1 + − 0.02T (t) + 0.01 1 + V (t) 100 + V (t)

is 63 ·· 52 ·· 41 p 3 (1 − p)3 and so on. As the number of goals increases, does the probability of a tie increase or decrease? < 4 for x > 0 and To find out, first show that (2x+2)(2x+1) (x+1)2 x(1 − x) ≤ 14 for 0 ≤ x ≤ 1. Use these inequalities to show that the probability of a tie decreases as the (even) number of goals increases. In a 1-goal game, the probability that team A wins is p. In a 2-goal game, the probability that team A wins is p 2 . In a 3-goal game, the probability that team A wins is p 3 + 3 p 2 (1 − p). In a 4-goal game, the probability that team A wins is p 4 + 4 p 3 (1 − p). In a 5-goal game, the probability that team A wins is p 5 + 5 p 4 (1 − p) + 52 ·· 41 p 3 (1 − p)2 . Explore the extent to which the probability that team A wins increases as the number of goals increases.

− 0.000024 T (t)V (t). If there is no HIV present [that is, V (t) = 0] and T (t) = 1000, show that T (t) = 0. Explain why this means that the T-cell count will remain constant at 1000 (cells per cubic mm). Now, suppose that V (t) = 100. Show that if T (t) is small enough, then T (t) > 0 and the T-cell population will increase. On the

3.5

CONCAVITY AND THE SECOND DERIVATIVE TEST In section 3.4, we saw how to determine where a function is increasing and decreasing and how this relates to drawing a graph of the function. First, recognize that simply knowing where a function increases and decreases is not sufficient to draw a good graph. In Figures 3.54a and 3.54b, we show two very different shapes of increasing functions joining the same two points. y

y

x a

b

x a

b

FIGURE 3.54a

FIGURE 3.54b

Increasing function

Increasing function

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3-46

other hand, if T (t) is large enough, then T (t) < 0 and the T-cell population will decrease. For what value of T (t) is T (t) = 0? Even though this population would remain stable, explain why this would be bad news for the infected human.

rhythm, so that the firefly flashes whenever f (t) equals an integer. Let e(t) represent the rhythm of a neighboring firefly, where again e(t) = n, for some integer n, whenever the neighbor flashes. One model of the interaction between fireflies is f (t) = ω + A sin [e(t) − f (t)] for constants ω and A. If the fireflies are synchronized [e(t) = f (t)], then f (t) = ω, so the fireflies flash every 1/ω time units. Assume that the difference between e(t) and f (t) is less than π. Show that if f (t) < e(t), then f (t) > ω. Explain why this means that the individual firefly is speeding up its flash to match its neighbor. Similarly, discuss what happens if f (t) > e(t).

3. In a sport like soccer or hockey where ties are possible, the probability that the stronger team wins depends in an interesting way on the number of goals scored. Suppose that at any point, the probability that team A scores the next goal is p, where 0 < p < 1. If 2 goals are scored, a 1-1 tie could result from team A scoring first (probability p) and then team B tieing the score (probability 1 − p), or vice versa. The probability of a tie in a 2-goal game is then 2 p(1 − p). Similarly, the probability of a 2-2 tie in a 4-goal game is 4·3 2 p (1 − p)2 , the probability of a 3-3 tie in a 6-goal game 2·1

2. The HIV virus attacks specialized T-cells that trigger the human immune system response to a foreign substance. If T (t) is the population of uninfected T-cells at time t (days) and V (t) is the population of infectious HIV in the bloodstream, a model that has been used to study AIDS is given by the following differential equation that describes the rate at which the population of T cells changes. 1 T (t)V (t) T (t) = 10 1 + − 0.02T (t) + 0.01 1 + V (t) 100 + V (t)

is 63 ·· 52 ·· 41 p 3 (1 − p)3 and so on. As the number of goals increases, does the probability of a tie increase or decrease? < 4 for x > 0 and To find out, first show that (2x+2)(2x+1) (x+1)2 x(1 − x) ≤ 14 for 0 ≤ x ≤ 1. Use these inequalities to show that the probability of a tie decreases as the (even) number of goals increases. In a 1-goal game, the probability that team A wins is p. In a 2-goal game, the probability that team A wins is p 2 . In a 3-goal game, the probability that team A wins is p 3 + 3 p 2 (1 − p). In a 4-goal game, the probability that team A wins is p 4 + 4 p 3 (1 − p). In a 5-goal game, the probability that team A wins is p 5 + 5 p 4 (1 − p) + 52 ·· 41 p 3 (1 − p)2 . Explore the extent to which the probability that team A wins increases as the number of goals increases.

− 0.000024 T (t)V (t). If there is no HIV present [that is, V (t) = 0] and T (t) = 1000, show that T (t) = 0. Explain why this means that the T-cell count will remain constant at 1000 (cells per cubic mm). Now, suppose that V (t) = 100. Show that if T (t) is small enough, then T (t) > 0 and the T-cell population will increase. On the

3.5

CONCAVITY AND THE SECOND DERIVATIVE TEST In section 3.4, we saw how to determine where a function is increasing and decreasing and how this relates to drawing a graph of the function. First, recognize that simply knowing where a function increases and decreases is not sufficient to draw a good graph. In Figures 3.54a and 3.54b, we show two very different shapes of increasing functions joining the same two points. y

y

x a

b

x a

b

FIGURE 3.54a

FIGURE 3.54b

Increasing function

Increasing function

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287

Given that a curve joins two particular points and is increasing, we need further information to determine which of the two shapes shown (if either) we should draw. Realize that this is an important distinction to make. For example, suppose that Figure 3.54a or 3.54b depicts the balance in your bank account. Both indicate a balance that is growing. However, the rate of growth in Figure 3.54a, is increasing, while the rate of growth depicted in Figure 3.54b is decreasing. Which would you want to have describe your bank balance? Why? Figures 3.55a and 3.55b are the same as Figures 3.54a and 3.54b, respectively, but with a few tangent lines drawn in. y

y

x a

x

b

a

b

FIGURE 3.55a

FIGURE 3.55b

Concave up, increasing

Concave down, increasing

Although all of the tangent lines have positive slope [since f (x) > 0], the slopes of the tangent lines in Figure 3.55a are increasing, while those in Figure 3.55b are decreasing. We call the graph in Figure 3.55a concave up and the graph in Figure 3.55b concave down. The situation is similar for decreasing functions. In Figures 3.56a and 3.56b, we show two different shapes of decreasing functions. The one shown in Figure 3.56a is concave up (slopes of tangent lines increasing) and the one shown in Figure 3.56b is concave down (slopes of tangent lines decreasing). We summarize this in Definition 5.1. y

y

x a

b

x a

b

FIGURE 3.56a

FIGURE 3.56b

Concave up, decreasing

Concave down, decreasing

DEFINITION 5.1 For a function f that is differentiable on an interval I , the graph of f is (i) concave up on I if f is increasing on I or (ii) concave down on I if f is decreasing on I .

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3-48

Note that you can tell when f is increasing or decreasing from the derivative of f (i.e., f ). Theorem 5.1 connects concavity with what we already know about increasing and decreasing functions. The proof is a straightforward application of Theorem 4.1 to Definition 5.1.

THEOREM 5.1 Suppose that f exists on an interval I . (i) If f (x) > 0 on I , then the graph of f is concave up on I . (ii) If f (x) < 0 on I , then the graph of f is concave down on I .

EXAMPLE 5.1

Determining Concavity

Determine where the graph of f (x) = 2x 3 + 9x 2 − 24x − 10 is concave up and concave down, and draw a graph showing all significant features of the function.

y 100

Solution Here, we have Inflection point x

4

4 50

and from our work in example 4.3, we have > 0 on (−∞, −4) ∪ (1, ∞) f (x) < 0 on (−4, 1).

Further, we have f (x) = 12x + 18

FIGURE 3.57 y = 2x + 9x − 24x − 10 3

2

NOTES If (c, f (c)) is an inflection point, then either f (c) = 0 or f (c) is undefined. So, finding all points where f (x) is zero or is undefined gives you all possible candidates for inflection points. But beware: not all points where f (x) is zero or undefined correspond to inflection points.

0

0

3

0

0

3

0 0

0 3

< 0, for x <

− 32 .

f decreasing.

Concave up. Concave down.

Using all of this information, we areable to draw the graph shown in Figure 3.57. Notice that at the point − 32 , f − 32 , the graph changes from concave down to concave up. Such points are called inflection points, which we define more precisely in Definition 5.2.

DEFINITION 5.2 Suppose that f is continuous on the interval (a, b) and that the graph changes concavity at a point c ∈ (a, b) (i.e., the graph is concave down on one side of c and concave up on the other). Then, the point (c, f (c)) is called an inflection point of f .

EXAMPLE 5.2

( x 3 )

Determine where the graph of f (x) = x 4 − 6x 2 + 1 is concave up and concave down, find any inflection points and draw a graph showing all significant features.

( x 3 )

3

> 0, for x > − 32

f increasing.

4x

0

f (x) = 6x 2 + 18x − 24

f (x)

Determining Concavity and Locating Inflection Points

Solution Here, we have f (x) = 4x 3 − 12x = 4x(x 2 − 3) √ √ = 4x(x − 3)(x + 3).

3-49

SECTION 3.5

0

0

(x 1)

1

0

0

1

0

3

0

f (x)

0

f (x) = 12x 2 − 12 = 12(x − 1)(x + 1).

f'(x)

3

0

0

289

1

Next, we have

Concavity and the Second Derivative Test

We have drawn number lines for the factors of f (x) in the margin. From this, we can see that √ √ > 0, on (− 3, 0) ∪ ( 3, ∞) f increasing. √ √ f (x) < 0, on (−∞, − 3) ∪ (0, 3). f decreasing.

12(x 1)

1

..

1

We have drawn number lines for the two factors in the margin. From this, we can see that

0

f''(x)

1

f (x)

> 0, on (−∞, −1) ∪ (1, ∞)

Concave up.

< 0, on (−1, 1).

Concave down.

y 10 5 x

3

3

For convenience, we have indicated the concavity below the bottom number line, with small concave up and concave down segments. Finally, observe that since the graph changes concavity at x = −1 and x = 1, there are inflection points located at (−1, −4) and (1, −4). Using all of this information, we are able to draw the graph shown in Figure 3.58. For your convenience, we have reproduced the number lines for f (x) and f (x) together above the figure. As we see in example 5.3, having f (x) = 0 does not imply the existence of an inflection point.

5 10

EXAMPLE 5.3 FIGURE 3.58

Determine the concavity of f (x) = x 4 and locate any inflection points.

y = x 4 − 6x 2 + 1

Solution There’s nothing tricky about this function. We have f (x) = 4x 3 and f (x) = 12x 2 . Since f (x) > 0 for x > 0 and f (x) < 0 for x < 0, we know that f is increasing for x > 0 and decreasing for x < 0. Further, f (x) > 0 for all x = 0, while f (0) = 0. So, the graph is concave up for x = 0. Further, even though f (0) = 0, there is no inflection point at x = 0. We show a graph of the function in Figure 3.59.

y

4

We now explore a connection between second derivatives and extrema. Suppose that f (c) = 0 and that the graph of f is concave down in some open interval containing c. Then, near x = c, the graph looks like that in Figure 3.60a and hence, f (c) is a local maximum. Likewise, if f (c) = 0 and the graph of f is concave up in some open interval containing c, then near x = c, the graph looks like that in Figure 3.60b and hence, f (c) is a local minimum.

2

2

A Graph with No Inflection Points

x

1

1

2

y

y

FIGURE 3.59

f (c) 0

y = x4

f (c) 0

f (c) 0 c

f (c) 0 x

c

FIGURE 3.60a

FIGURE 3.60b

Local maximum

Local minimum

x

290

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3-50

We state this more precisely in Theorem 5.2.

THEOREM 5.2 (Second Derivative Test) Suppose that f is continuous on the interval (a, b) and f (c) = 0, for some number c ∈ (a, b). (i) If f (c) < 0, then f (c) is a local maximum. (ii) If f (c) > 0, then f (c) is a local minimum.

We leave a formal proof of this theorem as an exercise. When applying the theorem, simply think about Figures 3.60a and 3.60b.

EXAMPLE 5.4

Using the Second Derivative Test to Find Extrema

Use the Second Derivative Test to find the local extrema of f (x) = x 4 − 8x 2 + 10.

y

Solution Here, 20

f (x) = 4x 3 − 16x = 4x(x 2 − 4) = 4x(x − 2)(x + 2). Thus, the critical numbers are x = 0, 2 and −2. We also have f (x) = 12x 2 − 16

4

x

2

2

4

and so,

f (0) = −16 < 0, f (−2) = 32 > 0

10

and

FIGURE 3.61

f (2) = 32 > 0.

So, by the Second Derivative Test, f (0) is a local maximum and f (−2) and f (2) are local minima. We show a graph of y = f (x) in Figure 3.61.

y = x 4 − 8x 2 + 10

REMARK 5.1 If f (c) = 0 or f (c) is undefined, the Second Derivative Test yields no conclusion. That is, f (c) may be a local maximum, a local minimum or neither. In this event, we must rely solely on first derivative information (i.e., the First Derivative Test) to determine whether f (c) is a local extremum. We illustrate this with example 5.5.

y 30

4

x

2

2

30

FIGURE 3.62a y = x3

4

EXAMPLE 5.5

Functions for Which the Second Derivative Test Is Inconclusive

Use the Second Derivative Test to try to classify any local extrema for (a) f (x) = x 3 , (b) g(x) = x 4 and (c) h(x) = −x 4 . Solution (a) Note that f (x) = 3x 2 and f (x) = 6x. So, the only critical number is x = 0 and f (0) = 0, also. We leave it as an exercise to show that the point (0, 0) is not a local extremum (see Figure 3.62a).

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291

y

y

2

x

1

1

2

4 2 2 4 2

x

1

1

2

FIGURE 3.62b

FIGURE 3.62c

y = x4

y = −x 4

(b) We have g (x) = 4x 3 and g (x) = 12x 2 . Again, the only critical number is x = 0 and g (0) = 0. In this case, though, g (x) < 0 for x < 0 and g (x) > 0 for x > 0. So, by the First Derivative Test, (0, 0) is a local minimum (see Figure 3.62b). (c) Finally, we have h (x) = −4x 3 and h (x) = −12x 2 . Once again, the only critical number is x = 0, h (0) = 0 and we leave it as an exercise to show that (0, 0) is a local maximum for h (see Figure 3.62c). We can use first and second derivative information to help produce a meaningful graph of a function, as in example 5.6.

EXAMPLE 5.6

Drawing a Graph of a Rational Function

25 , showing all significant features. x Solution The domain of f consists of all real numbers other than x = 0. Then,

Draw a graph of f (x) = x +

25 x 2 − 25 = 2 x x2 (x − 5)(x + 5) = . x2

f (x) = 1 −

Add the fractions.

So, the only two critical numbers are x = −5 and x = 5. (Why is x = 0 not a critical number?) Looking at the three factors in f (x), we get the number lines shown in the margin. Thus, f (x)

0

(x 5)

> 0, on (−∞, −5) ∪ (5, ∞)

f increasing.

< 0, on (−5, 0) ∪ (0, 5).

f decreasing.

5

0

(x 5)

5

0

0

5

0

0 5

f (x) =

50 x3

> 0, on (0, ∞)

Concave up.

< 0, on (−∞, 0).

Concave down.

x2

0

Further,

f (x)

Be careful here. There is no inflection point on the graph, even though the graph is concave up on one side of x = 0 and concave down on the other. (Why not?) We can now use either the First Derivative Test or the Second Derivative Test to determine the

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3-52

f (5) =

local extrema. Since

f (−5) = −

and

y 20

50 >0 125 50 < 0, 125

there is a local minimum at x = 5 and a local maximum at x = −5 by the Second Derivative Test. Finally, before we can draw a representative graph, we need to know what happens to the graph near x = 0, since 0 is not in the domain of f. We have

10

x

15 10 5

5

10

lim f (x) = lim+

15

x→0+

x→0

10

20

FIGURE 3.63 y=x+

lim f (x) = lim−

and

x→0−

x→0

25 x+ x

25 x+ x

=∞

= −∞,

so that there is a vertical asymptote at x = 0. Putting together all of this information, we get the graph shown in Figure 3.63.

25 x

In example 5.6, we computed lim+ f (x) and lim− f (x) to uncover the behavior of the x→0

x→0

function near x = 0, since x = 0 was not in the domain of f. In example 5.7, we’ll see that since x = −2 is not in the domain of f (although it is in the domain of f ), we must compute lim + f (x) and lim − f (x) to uncover the behavior of the tangent lines near x = −2. x→−2

x→−2

EXAMPLE 5.7

A Function with a Vertical Tangent Line at an Inflection Point

Draw a graph of f (x) = (x + 2)1/5 + 4, showing all significant features. Solution First, notice that the domain of f is the entire real line. We also have f (x) =

So, f is increasing everywhere, except at x = −2 [the only critical number, where f (−2) is undefined]. This also says that f has no local extrema. Further, > 0, on (−∞, −2) Concave up. 4 −9/5 f (x) = − (x + 2) 25 < 0, on (−2, ∞). Concave down.

y 6

So, there is an inflection point at x = −2. In this case, f (x) is undefined at x = −2. Since −2 is in the domain of f , but not in the domain of f , we consider

4

1 lim − f (x) = lim − (x + 2)−4/5 = ∞ x→−2 x→−2 5

2

4 3 2 1

1 (x + 2)−4/5 > 0, for x = −2. 5

x 1

FIGURE 3.64 y = (x + 2)1/5 + 4

2

and

1 lim f (x) = lim + (x + 2)−4/5 = ∞. x→−2 5

x→−2+

This says that the graph has a vertical tangent line at x = −2. Putting all of this information together, we get the graph shown in Figure 3.64.

3-53

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..

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293

EXERCISES 3.5 WRITING EXERCISES 1. It is often said that a graph is concave up if it “holds water.” This is certainly true for parabolas like y = x 2 , but is it true for graphs like y = 1/x 2 ? It is always helpful to put a difficult concept into everyday language, but the danger is in oversimplification. Do you think that “holds water” is helpful or can it be confusing? Give your own description of concave up, using everyday language. (Hint: One popular image involves smiles and frowns.)

In exercises 15–26, determine all significant features by hand and sketch a graph.

19. f (x) = sin x + cos x

20. f (x) = e−x sin x

2. Find a reference book with the population of the United States since 1800. From 1800 to 1900, the numerical increase by decade increased. Argue that this means that the population curve is concave up. From 1960 to 1990, the numerical increase by decade has been approximately constant. Argue that this means that the population curve is near a point where the curve is neither concave up nor concave down. Why does this not necessarily mean that we are at an inflection point? Argue that we should hope, in order to avoid overpopulation, that it is indeed an inflection point.

21. f (x) = x 3/4 − 4x 1/4

22. f (x) = x 2/3 − 4x 1/3

23. f (x) = x|x|

24. f (x) = x 2 |x| √ x 26. f (x) = √ 1+ x

3. The goal of investing in the stock market is to buy low and sell high. But, how can you tell whether a price has peaked or not? Once a stock price goes down, you can see that it was at a peak, but then it’s too late to do anything about it! Concavity can help. Suppose a stock price is increasing and the price curve is concave up. Why would you suspect that it will continue to rise? Is this a good time to buy? Now, suppose the price is increasing but the curve is concave down. Why should you be preparing to sell? Finally, suppose the price is decreasing. If the curve is concave up, should you buy or sell? What if the curve is concave down? 4. Suppose that f (t) is the amount of money in your bank account at time t. Explain in terms of spending and saving what would cause f (t) to be decreasing and concave down; increasing and concave up; decreasing and concave up.

In exercises 1–8, determine the intervals where the graph of the given function is concave up and concave down. 1. f (x) = x 3 − 3x 2 + 4x − 1

2. f (x) = x 4 − 6x 2 + 2x + 3

3. f (x) = x + 1/x

4. f (x) = x + 3(1 − x)1/3

5. f (x) = sin x − cos x

6. f (x) = tan−1 (x 2 )

7. f (x) = x 4/3 + 4x 1/3

8. f (x) = xe−4x

15. f (x) = (x 2 + 1)2/3 17. f (x) =

10. f (x) = x 4 + 4x 2 + 1 2 12. f (x) = e−x 2 x −1 14. f (x) = x

x x2 − 9

18. f (x) =

25. f (x) = x 1/5 (x + 1)

x x +2

In exercises 27–36, determine all significant features (approximately if necessary) and sketch a graph. 27. f (x) = x 4 − 26x 3 + x 28. f (x) = 2x 4 − 11x 3 + 17x 2 √ 29. f (x) = 3 2x 2 − 1 √ 30. f (x) = x 3 + 1 31. f (x) = x 4 − 16x 3 + 42x 2 − 39.6x + 14 32. f (x) = x 4 + 32x 3 − 0.02x 2 − 0.8x √ 33. f (x) = x x 2 − 4 1 35. f (x) = tan−1 x2 − 1

34. f (x) = √

2x x2 + 4

36. f (x) = e−2x cos x

In exercises 37–40, estimate the intervals where the function is concave up and concave down. (Hint: Estimate where the slope is increasing and decreasing.) y

37. 20

10

In exercises 9–14, find all critical numbers and use the Second Derivative Test to determine all local extrema. 9. f (x) = x 4 + 4x 3 − 1 11. f (x) = xe−x x 2 − 5x + 4 13. f (x) = x

16. f (x) = x ln x

2

−3 −2

x 2

3

294

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Applications of Differentiation

45. Show that any cubic f (x) = ax 3 + bx 2 + cx + d has one inflection point. Find conditions on the coefficients a−e that guarantee that the quartic f (x) = ax 4 + bx 3 + cx 2 + d x + e has two inflection points.

y

38. 1

−4

46. If f and g are functions with two derivatives for all x, f (0) = g(0) = f (0) = g (0) = 0, f (0) > 0 and g (0) < 0, state as completely as possible what can be said about whether f (x) > g(x) or f (x) < g(x).

x

−2

2

4

In exercises 47 and 48, estimate the intervals of increase and decrease, the locations of local extrema, intervals of concavity and locations of inflection points.

−1

y

39.

3-54

y

47.

10

8

5

6 4

x

−2

2

4

2

−5 −3 −2

− 10

x

−1

1

2

3

−2

y

40.

y

48.

10 4

5

−3

x

−1

1

2

3

x

−1

−5

1

2

3

−4

−10

In exercises 41–44, sketch a graph with the given properties. 41. f (0) = 0, f (x) > 0 for x < −1 and −1 < x < 1, f (x) < 0 for x > 1, f (x) > 0 for x < −1, 0 < x < 1 and x > 1, f (x) < 0 for −1 < x < 0 42. f (0) = 2, f (x) > 0 for all x, f (0) = 1, f (x) > 0 for x < 0, f (x) < 0 for x > 0 43. f (0) = 0, f (−1) = −1, f (1) = 1, f (x) > 0 for x < −1 and 0 < x < 1, f (x) < 0 for −1 < x < 0 and x > 1, f (x) < 0 for x < 0 and x > 0 44. f (1) = 0, f (x) < 0 for x < 1, f (x) < 0 for x < 1 and x > 1

f (x) > 0 for x > 1,

49. Repeat exercises 47 and 48 if the given graph is of f (x) instead of f (x). 50. Repeat exercises 47 and 48 if the given graph is of f (x) instead of f (x). 51. Suppose that w(t) is the depth of water in a city’s water reservoir at time t. Which would be better news at time t = 0, w (0) = 0.05 or w (0) = −0.05, or would you need to know the value of w (0) to determine which is better? 52. Suppose that T (t) is a sick person’s temperature at time t. Which would be better news at time t, T (0) = 2 or T (0) = −2, or would you need to know the value of T (0) and T (0) to determine which is better?

3-55

SECTION 3.5

53. Suppose that a company that spends $x thousand on advertising sells $s(x) of merchandise, where s(x) = −3x 3 + 270x 2 − 3600x + 18,000. Find the value of x that maximizes the rate of change of sales. (Hint: Read the question carefully!) Find the inflection point and explain why in advertising terms this is the “point of diminishing returns.” 54. The number of units Q that a worker has produced in a day is related to the number of hours t since the work day began. Suppose that Q(t) = −t 3 + 6t 2 + 12t. Explain why Q (t) is a measure of the efficiency of the worker at time t. Find the time at which the worker’s efficiency is a maximum. Explain why it is reasonable to call the inflection point the “point of diminishing returns.” 55. Suppose that it costs a company C(x) = 0.01x + 40x + 3600 dollars to manufacture x units of a product. For this cost funcC(x) tion, the average cost function is C(x) = . Find the x value of x that minimizes the average cost. The cost function can be related to the efficiency of the production process. Explain why a cost function that is concave down indicates better efficiency than a cost function that is concave up. 2

56. The antiarrhythmic drug lidocaine slowly decays after entering the bloodstream. The plasma concentration t minutes after administering the drug can be modeled by c(t) = 92.8 −0.129e−t/6.55 + 0.218e−t/65.7 − 0.089e−t/13.3 .

..

Concavity and the Second Derivative Test

295

points and asymptotes (for x ≥ 0) for each function and comparing to the graph. Each of these functions is called a Hill function. 58. Two functions with similar properties to the Hill functions of 1 exercise 57 are g1 (x) = tan−1 x and g2 (x) = . As 1 + 99e−x/2 in exercise 57, test these functions for use as a model of the hemoglobin data. 59. Give an example of a function showing that the following statement is false. If the graph of y = f (x) is concave down for all x, the equation f (x) = 0 has at least one solution. 60. Determine whether the following statement is true or false. If f (0) = 1, f (x) exists for all x and the graph of y = f (x) is concave down for all x, the equation f (x) = 0 has at least one solution. 61. A basic principle of physics is that light follows the path of minimum time. Assuming that the speed of light in the earth’s atmosphere decreases as altitude decreases, argue that the path that light follows is concave down. Explain why this means that the setting sun appears higher in the sky than it really is.

Use a CAS to estimate the time of maximum concentration and inflection point (t > 0). Suppose that f (t) represents the concentration of another drug. If the graph of f (t) has a similar shape and the same maximum point as the graph of c(t) but the inflection point occurs at a larger value of t, would this drug be more or less effective than lidocaine? Briefly explain. 57. The plot shows the relationship between the specific partial pressure of oxygen (pO2 , measured in mm Hg) and the saturation level of hemoglobin (y = 1 would mean that no more oxygen can bind). 1.0

0.98

EXPLORATORY EXERCISES

0.9 0.8

Saturation (%)

62. Prove Theorem 5.2 (the Second Derivative Test). (Hint: Think about what the definition of f (c) says when f (c) > 0 or f (c) < 0.)

0.75

0.7 0.6 0.50

0.5 0.4 0.3 0.2 0.1

27

40

103

0.0 0

10

20

30

40

50

60

70

80

90

100

110

120

pO2 (mm Hg)

x x3 or f 2 (x) = 3 is 27 + x 27 + x 3 a better model for these data by finding extrema, inflection Determine whether f 1 (x) =

1. The linear approximation that we defined in section 3.1 is the line having the same location and the same slope as the function being approximated. Since two points determine a line, two requirements (point, slope) are all that a linear function can satisfy. However, a quadratic function can satisfy three requirements, since three points determine a parabola (and there are three constants in a general quadratic function ax 2 + bx + c). Suppose we want to define a quadratic approximation to f (x) at x = a. Building on the linear approximation, the general form is g(x) = f (a) + f (a)(x − a) + c(x − a)2 for some constant c to be determined. In this way, show that g(a) = f (a) and g (a) = f (a). That is, g(x) has the right position and slope at x = a. The third requirement is that g(x) have the right

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concavity at x = a, so that g (a) = f (a). Find the constant c that makes this true. Then, find such a quadratic approximation for each of the functions sin x, cos x and e x at x = 0. In each case, graph the original function, linear approximation and quadratic approximation, and describe how close the approximations are to the original functions. 2. In this exercise, we explore a basic problem in genetics. Suppose that a species reproduces according to the following probabilities: p0 is the probability of having no children, p1 is the probability of having one offspring, p2 is the probability of having two offspring, . . . , pn is the probability of having n offspring and n is the largest number of offspring possible. Explain why for each i, we have 0 ≤ pi ≤ 1 and p0 + p1 + p2 + · · · + pn = 1. We define the function

3.6

F(x) = p0 + p1 x + p2 x 2 + · · · + pn x n . The smallest nonnegative solution of the equation F(x) = x for 0 ≤ x ≤ 1 represents the probability that the species becomes extinct. Show graphically that if p0 > 0 and F (1) > 1, then there is a solution of F(x) = x with 0 < x < 1. Thus, there is a positive probability of survival. However, if p0 > 0 and F (1) < 1, show that there are no solutions of F(x) = x with 0 < x < 1. (Hint: First show that F is increasing and concave up.) x +c x2 − 1 as possible. In particular, find the values of c for which there are two critical points (or one critical point, or no critical points) and identify any extrema. Similarly, determine how the existence or not of inflection points depends on the value of c.

3. Give as complete a description of the graph of f (x) =

OVERVIEW OF CURVE SKETCHING Graphing calculators and computer algebra systems are powerful tools in the study or application of mathematics. However, they do not actually draw graphs. What they do is plot points (albeit lots of them) and then connect the points as smoothly as possible. While this is very helpful, it often leaves something to be desired. The problem boils down to knowing the window in which you should draw a given graph and how many points you plot in that window. The only way to know how to choose this is to use the calculus to determine the properties of the graph that you are interested in seeing. We have already made this point a number of times. We begin this section by summarizing the various tests that you should perform on a function when trying to draw a graph of y = f (x). r Domain: You should always determine the domain of f first. r Vertical Asymptotes: For any isolated point not in the domain of f, check the r r

r r r

limiting value of the function as x approaches that point, to see if there is a vertical asymptote or a jump or removable discontinuity at that point. First Derivative Information: Determine where f is increasing and decreasing, and find any local extrema. Vertical Tangent Lines: At any isolated point not in the domain of f , but in the domain of f , check the limiting values of f (x), to determine whether there is a vertical tangent line at that point. Second Derivative Information: Determine where the graph is concave up and concave down, and locate any inflection points. Horizontal Asymptotes: Check the limit of f (x) as x → ∞ and as x → −∞. Intercepts: Locate x- and y-intercepts, if any. If this can’t be done exactly, then do so approximately (e.g., using Newton’s method). We start with a very straightforward example.

EXAMPLE 6.1

Drawing a Graph of a Polynomial

Draw a graph of f (x) = x 4 + 6x 3 + 12x 2 + 8x + 1, showing all significant features.

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concavity at x = a, so that g (a) = f (a). Find the constant c that makes this true. Then, find such a quadratic approximation for each of the functions sin x, cos x and e x at x = 0. In each case, graph the original function, linear approximation and quadratic approximation, and describe how close the approximations are to the original functions. 2. In this exercise, we explore a basic problem in genetics. Suppose that a species reproduces according to the following probabilities: p0 is the probability of having no children, p1 is the probability of having one offspring, p2 is the probability of having two offspring, . . . , pn is the probability of having n offspring and n is the largest number of offspring possible. Explain why for each i, we have 0 ≤ pi ≤ 1 and p0 + p1 + p2 + · · · + pn = 1. We define the function

3.6

F(x) = p0 + p1 x + p2 x 2 + · · · + pn x n . The smallest nonnegative solution of the equation F(x) = x for 0 ≤ x ≤ 1 represents the probability that the species becomes extinct. Show graphically that if p0 > 0 and F (1) > 1, then there is a solution of F(x) = x with 0 < x < 1. Thus, there is a positive probability of survival. However, if p0 > 0 and F (1) < 1, show that there are no solutions of F(x) = x with 0 < x < 1. (Hint: First show that F is increasing and concave up.) x +c x2 − 1 as possible. In particular, find the values of c for which there are two critical points (or one critical point, or no critical points) and identify any extrema. Similarly, determine how the existence or not of inflection points depends on the value of c.

3. Give as complete a description of the graph of f (x) =

OVERVIEW OF CURVE SKETCHING Graphing calculators and computer algebra systems are powerful tools in the study or application of mathematics. However, they do not actually draw graphs. What they do is plot points (albeit lots of them) and then connect the points as smoothly as possible. While this is very helpful, it often leaves something to be desired. The problem boils down to knowing the window in which you should draw a given graph and how many points you plot in that window. The only way to know how to choose this is to use the calculus to determine the properties of the graph that you are interested in seeing. We have already made this point a number of times. We begin this section by summarizing the various tests that you should perform on a function when trying to draw a graph of y = f (x). r Domain: You should always determine the domain of f first. r Vertical Asymptotes: For any isolated point not in the domain of f, check the r r

r r r

limiting value of the function as x approaches that point, to see if there is a vertical asymptote or a jump or removable discontinuity at that point. First Derivative Information: Determine where f is increasing and decreasing, and find any local extrema. Vertical Tangent Lines: At any isolated point not in the domain of f , but in the domain of f , check the limiting values of f (x), to determine whether there is a vertical tangent line at that point. Second Derivative Information: Determine where the graph is concave up and concave down, and locate any inflection points. Horizontal Asymptotes: Check the limit of f (x) as x → ∞ and as x → −∞. Intercepts: Locate x- and y-intercepts, if any. If this can’t be done exactly, then do so approximately (e.g., using Newton’s method). We start with a very straightforward example.

EXAMPLE 6.1

Drawing a Graph of a Polynomial

Draw a graph of f (x) = x 4 + 6x 3 + 12x 2 + 8x + 1, showing all significant features.

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Overview of Curve Sketching

297

y y

1600 10

1200 800

− 4 −3 −2 −1

x

−10

400 x 1

2

3

10 −10

4

FIGURE 3.65a

FIGURE 3.65b

y = x 4 + 6x 3 + 12x 2 + 8x + 1 (one view)

y = x 4 + 6x 3 + 12x 2 + 8x + 1 (standard calculator view)

Solution Computer algebra systems and graphing calculators usually do one of two things to determine the window in which they will display a graph. One method is to compute a set number of function values over a given standard range of x-values. The y-range is then chosen so that all of the calculated points can be displayed. This might result in a graph that looks like the one in Figure 3.65a. Another method is to draw a graph in a fixed, default window. For instance, most graphing calculators use the default window defined by −10 ≤ x ≤ 10

and

−10 ≤ y ≤ 10.

Using this window, we get the graph shown in Figure 3.65b. Of course, these two graphs are very different. Without the calculus, it’s difficult to tell which, if either, of these is truly representative of the behavior of f . Some analysis will clear up the situation. First, note that the domain of f is the entire real line. Further, since f (x) is a polynomial, it doesn’t have any vertical or horizontal asymptotes. Next, note that

0

f (x) = 4x 3 + 18x 2 + 24x + 8 = 2(2x + 1)(x + 2)2 .

2(2x 1)

Q

0

(x 2)2

2

0

0

2

f (x)

Q

This also tells us that there is a local minimum at x = − 12 and that there are no local maxima. Next, we have

0

12(x 2)

2

0

(x 1)

1

0 2

f (x)

1

0

2

0

0 2

0

f (x)

Q

0 1

Drawing number lines for the individual factors of f (x), we have that > 0, on − 12 , −∞ f increasing. f (x) < 0, on (−∞, −2) ∪ −2, − 12 . f decreasing.

f (x)

f (x) = 12x 2 + 36x + 24 = 12(x + 2)(x + 1). Drawing number lines for the factors of f (x), we have > 0, on (−∞, −2) ∪ (−1, ∞) f (x) < 0, on (−2, −1).

Concave up. Concave down.

From this, we see that there are inflection points at x = −2 and at x = −1. Finally, to find the x-intercepts, we need to solve f (x) = 0 approximately. Doing this (we leave the details as an exercise: use Newton’s method or your calculator’s solver), we find that there are two x-intercepts: x = −1 (exactly) and x ≈ −0.160713. Notice that the significant x-values that we have identified are x = −2, x = −1 and x = − 12 . Computing the corresponding y-values from y = f (x), we get the points (−2, 1), (−1, 0) and − 12 , − 11 . We summarize the first and second derivative 16

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y

3-58

information in the number lines in the margin. In Figure 3.66, we include all of these important points by setting the x-range to be −4 ≤ x ≤ 1 and the y-range to be −2 ≤ y ≤ 8.

8 6

In example 6.2, we examine a function that has local extrema, inflection points and both vertical and horizontal asymptotes.

4 2 −4

−3

−2

x

−1

1 −2

Drawing a Graph of a Rational Function

x2 − 3 , showing all significant features. x3 Solution The default graph drawn by our computer algebra system appears in Figure 3.67a. Notice that this doesn’t seem to be a particularly useful graph, since very little is visible (or at least distinguishable from the axes). The graph drawn using the most common graphing calculator default window is seen in Figure 3.67b. This is arguably an improvement over Figure 3.67a, but does this graph convey all that it could about the function (e.g., about local extrema, inflection points, etc.)? We can answer this question only after we do some calculus. We follow the outline given at the beginning of the section. First, observe that the domain of f includes all real numbers x = 0. Since x = 0 is an isolated point not in the domain of f, we scrutinize the limiting behavior of f as x approaches 0. We have

Draw a graph of f (x) =

FIGURE 3.66 y = x 4 + 6x 3 + 12x 2 + 8x + 1

y 3e + 24 1e + 24 −4

EXAMPLE 6.2

x 4

−1e + 24 −3e + 24

−

2 3 = −∞ lim+ f (x) = lim+ x − 3 x→0 x→0 x

FIGURE 3.67a

+

x2 − 3 x3

y=

−

2 3 = ∞. lim− f (x) = lim− x − 3 x→0 x→0 x

and

From (6.1) and (6.2), we see that the graph has a vertical asymptote at x = 0. Next, we look for whatever information the first derivative will yield. We have

10

x

−10

10

x2 − 3 y= x3 +

0

−

(3 − x)

3 +

(3 + x)

−3 +

0

x4

0 −

0 3 −3

+

× 0

+

0 3

−

2x(x 3 ) − (x 2 − 3)(3x 2 ) (x 3 )2

x 2 [2x 2 − 3(x 2 − 3)] x6 9 − x2 = x4 (3 − x)(3 + x) = . x4

FIGURE 3.67b

+

f (x) = =

−10

0

(6.2)

−

y

−

(6.1)

f'(x)

Quotient rule.

Factor out x 2 .

Combine terms.

Factor difference of two squares.

Looking at the individual factors in f (x), we have the number lines shown in the margin. Thus, f increasing. > 0, on (−3, 0) ∪ (0, 3) f (x) (6.3) < 0, on (−∞, −3) ∪ (3, ∞). f decreasing. Note that this says that f has a local minimum at x = −3 and a local maximum at x = 3.

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Overview of Curve Sketching

299

Next, we look at f (x) =

−2x(x 4 ) − (9 − x 2 )(4x 3 ) (x 4 )2

−2x 3 [x 2 + (9 − x 2 )(2)] x8 −2(18 − x 2 ) = x5 √ √ 2(x − 18)(x + 18) = . x5 =

0

2 ( x 18 )

18

0

0

x5

0

0 18

f (x)

0

f (x)

18

0

Factor out −2x 3 .

Combine terms.

Factor difference of two squares.

Looking at the individual factors in f (x), we obtain the number lines shown in the margin. Thus, we have

( x 18 )

18

Quotient rule.

√ √ > 0, on (− 18, 0) ∪ ( 18, ∞) √ √ < 0, on (−∞, − 18) ∪ (0, 18).

Concave up.

(6.4)

Concave down.

√ This says that there are inflection points at x = ± 18. (Why is there no inflection point at x = 0?) To determine the limiting behavior as x → ±∞, we consider x2 − 3 x→∞ x3 1 3 − 3 = 0. = lim x→∞ x x

lim f (x) = lim

x→∞

0

3

0

0 18

0

f (x)

3

0

f (x)

18

0

0.4

x 5 0.4

FIGURE 3.68 y=

lim f (x) = 0.

x→−∞

x2 − 3 x3

10

(6.6)

So, the line y = 0 is a horizontal asymptote both as x → ∞ and as x → −∞. Finally, the x-intercepts are where 0 = f (x) =

y

5

Likewise, we have

(6.5)

x2 − 3 , x3

√ that is, at x = ± 3. Notice that there are no y-intercepts, since x = 0 is not in the domain of the function. We now have all of the information that we need to draw a representative graph. With some experimentation, you can set the x- and y-ranges so that most of the significant features of the graph (i.e., vertical and horizontal asymptotes, local extrema, inflection points, etc.) are displayed, as in Figure 3.68, which is consistent with all of the information that we accumulated about the function in (6.1)–(6.6). Although the existence of the inflection points is clearly indicated by the change in concavity, their precise location is as yet a bit fuzzy in this graph. Notice, however, that both vertical and horizontal asymptotes and the local extrema are clearly indicated, something that cannot be said about either of Figures 3.67a or 3.67b. In example 6.3, there are multiple vertical asymptotes, only one extremum and no inflection points.

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3-60

y

EXAMPLE 6.3

A Graph with Two Vertical Asymptotes

200

Draw a graph of f (x) =

150 100 50 x

4

4

x2 showing all significant features. x2 − 4

Solution The default graph produced by our computer algebra system is seen in Figure 3.69a, while the default graph drawn by most graphing calculators looks like the graph seen in Figure 3.69b. Notice that the domain of f includes all x except x = ±2 (since the denominator is zero at x = ±2). Figure 3.69b suggests that there are vertical asymptotes at x = ±2, but let’s establish this carefully. We have

50

+

x2 x2 lim+ 2 = lim+ = ∞. (x − 2) (x + 2) x→2 x − 4 x→2

FIGURE 3.69a y=

+

2

x x2 − 4

Similarly, we get

y

lim−

x→2

5 10

5

x2 = −∞, x2 − 4

and

lim −

x→−2

x 5

5

lim +

x→−2

x2 = −∞ x2 − 4

x2 = ∞. x2 − 4

(6.8) (6.9)

Thus, there are vertical asymptotes at x = ±2. Next, we have

10

f (x) = FIGURE 3.69b y=

(6.7)

+

2x(x 2 − 4) − x 2 (2x) (x 2 − 4)

2

=

−8x (x 2 − 4)2

.

Since the denominator is positive for x = ±2, it is a simple matter to see that > 0, on (−∞, −2) ∪ (−2, 0) f increasing. f (x) < 0, on (0, 2) ∪ (2, ∞). f decreasing.

x2 x2 − 4

(6.10)

In particular, notice that the only critical number is x = 0 (since x = −2, 2 are not in the domain of f ). Thus, the only local extremum is the local maximum located at x = 0. Next, we have

0

−8(x 2 − 4)2 + (8x)2(x 2 − 4)1 (2x) (x 2 − 4)4

Quotient rule.

=

8(x 2 − 4)[−(x 2 − 4) + 4x 2 ] (x 2 − 4)4

Factor out 8(x 2 − 4).

=

8(3x 2 + 4) (x 2 − 4)3

Combine terms.

=

8(3x 2 + 4) . (x − 2)3 (x + 2)3

f (x) = (x 2)3

2

0

(x 2)3

2

2

2

2

f (x)

2

0

0

2

2

f (x)

f (x)

Factor difference of two squares.

Since the numerator is positive for all x, we need only consider the terms in the denominator, as seen in the margin. We then have > 0, on (−∞, −2) ∪ (2, ∞) Concave up. f (x) (6.11) < 0, on (−2, 2). Concave down. However, since x = 2, −2 are not in the domain of f , there are no inflection points. It is

3-61

SECTION 3.6

y

Overview of Curve Sketching

301

an easy exercise to verify that

6

lim

x2 =1 x2 − 4

(6.12)

lim

x2 = 1. x2 − 4

(6.13)

x→∞

4 2 x

6 4

..

4

2

and

x→−∞

6

From (6.12) and (6.13), we have that y = 1 is a horizontal asymptote, both as x → ∞ and as x → −∞. Finally, we observe that the only x-intercept is at x = 0. We summarize the information in (6.7)–(6.13) in the graph seen in Figure 3.70.

4 6

FIGURE 3.70

In example 6.4, we need to use computer-generated graphs, as well as a rootfinding method to determine the behavior of the function.

x2 y= 2 x −4

EXAMPLE 6.4

y 10

Graphing Where the Domain and Extrema Must Be Approximated

1 showing all significant features. x 3 + 3x 2 + 3x + 3 Solution The default graph drawn by most graphing calculators and computer algebra systems looks something like the one shown in Figure 3.71. As we have already seen, we can only determine all the significant features by doing some calculus. Since f is a rational function, it is defined for all x, except for where the denominator is zero, that is, where

Draw a graph of f (x) = x

10

10 10

FIGURE 3.71 y=

1 x 3 + 3x 2 + 3x + 3

x 3 + 3x 2 + 3x + 3 = 0. If you don’t see how to factor the expression to find the zeros exactly, you must rely on approximate methods. First, to get an idea of where the zero(s) might be, draw a graph of the cubic (see Figure 3.72). The graph does not need to be elaborate, merely detailed enough to get an idea of where and how many zeros there are. In the present case, we see that there is only one zero, around x = −2. We can verify that this is the only zero, since

y

20 10 4

2

d 3 (x + 3x 2 + 3x + 3) = 3x 2 + 6x + 3 = 3(x + 1)2 ≥ 0. dx

x 2 10 20

FIGURE 3.72

Since the derivative is never negative, observe that the function cannot decrease to cross the x-axis a second time. You can get the approximate zero x = a ≈ −2.25992 using Newton’s method or your calculator’s solver. We can use the graph in Figure 3.72 to help us compute the limits

y = x 3 + 3x 2 + 3x + 3

+

lim f (x) = lim+

x→a +

x→a

1 =∞ x 3 + 3x 2 + 3x + 3

(6.14)

+

+

and

lim f (x) = lim−

x→a −

x→a

1 = −∞. x + 3x + 3x + 3 3

2

(6.15)

−

From (6.14) and (6.15), f has a vertical asymptote at x = a. Turning to the derivative

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information, we have f (x) = −(x 3 + 3x 2 + 3x + 3)−2 (3x 2 + 6x + 3) (x + 1)2 = −3 (x 3 + 3x 2 + 3x + 3)2 2 x +1 = −3 x 3 + 3x 2 + 3x + 3 < 0, for x = a or −1

(6.16)

and f (−1) = 0. Thus, f is decreasing for x < a and x > a. Also, notice that the only critical number is x = −1, but since f is decreasing everywhere except at x = a, there are no local extrema. Turning to the second derivative, we get x +1 1(x 3 + 3x 2 + 3x + 3) − (x + 1)(3x 2 + 6x + 3) f (x) = −6 x 3 + 3x 2 + 3x + 3 (x 3 + 3x 2 + 3x + 3)2 = =

(x 3

−6(x + 1) (−2x 3 − 6x 2 − 6x) + 3x 2 + 3x + 3)3

12x(x + 1)(x 2 + 3x + 3) . (x 3 + 3x 2 + 3x + 3)3

Since (x 2 + 3x + 3) > 0 for all x (why is that?), we need not consider this factor. Considering the remaining factors, we have the number lines shown here.

0

12x

0

0

(x 1)

1

0

(x 3 3x 2 3x 3)

a 2.2599…

0 1

a

0

f (x)

0

Thus, we have that

f (x) y 2 1 3

1 1

x 1

> 0, on (a, −1) ∪ (0, ∞) < 0, on (−∞, a) ∪ (−1, 0)

2

3

FIGURE 3.73 1 x 3 + 3x 2 + 3x + 3

Concave up.

(6.17)

Concave down.

It now follows that there are inflection points at x = 0 and at x = −1. Notice that in Figure 3.71, the concavity information is not very clear and the inflection points are difficult to discern. We note the obvious fact that the function is never zero and hence, there are no x-intercepts. Finally, we consider the limits lim

1 =0 x 3 + 3x 2 + 3x + 3

(6.18)

lim

1 = 0. x 3 + 3x 2 + 3x + 3

(6.19)

x→∞

2

y=

and

x→−∞

Using all of the information in (6.14)–(6.19), we draw the graph seen in Figure 3.73. Here, we can clearly see the vertical and horizontal asymptotes, the inflection points and the fact that the function is decreasing across its entire domain.

3-63

SECTION 3.6

..

Overview of Curve Sketching

303

In example 6.5, we consider the graph of a transcendental function with a vertical asymptote.

EXAMPLE 6.5

Graphing Where Some Features Are Difficult to See

Draw a graph of f (x) = e1/x showing all significant features. Solution The default graph produced by our computer algebra system is not particularly helpful (see Figure 3.74a). The default graph produced by most graphing calculators (see Figure 3.74b) is certainly better, but we can’t be sure this is adequate without further analysis. First, notice that the domain of f is (−∞, 0) ∪ (0, ∞). Thus, we consider

y 5e08

lim e1/x = ∞,

2.5e08

x→0+

(6.20)

since 1/x → ∞ as x → 0+ . Also, since 1/x → −∞ as x → 0− (and et → 0 as t → −∞), we have 4

x

2

2

4

lim e1/x = 0.

x→0−

FIGURE 3.74a

(6.21)

From (6.20) and (6.21), there is a vertical asymptote at x = 0, but an unusual one, in that f (x) → ∞ on one side of 0 and f (x) → 0 on the other side. Next, 1 1/x d f (x) = e dx x −1 < 0, for all x = 0, (6.22) = e1/x x2

y = e1/x y 10

x

10

10 10

FIGURE 3.74b y = e1/x

since e1/x > 0, for all x = 0. From (6.22), we have that f is decreasing for all x = 0. We also have −1 −1 2 1/x f (x) = e1/x + e 2 2 x x x3 1 1 + 2x 2 1/x 1/x =e + 3 =e x4 x x4 < 0, on −∞, − 12 Concave down. (6.23) 1 > 0, on − 2 , 0 ∪ (0, ∞). Concave up. Since x = 0 is not in the domain of f , the only inflection point is at x = − 12 . Next, note that lim e1/x = 1,

x→∞

(6.24)

since 1/x → 0 as x → ∞ and et → 1 as t → 0. Likewise, lim e1/x = 1.

x→−∞

From (6.24) and (6.25), y = 1 is a horizontal asymptote, both as x → ∞ and as x → −∞. Finally, since e1/x > 0,

(6.25)

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y 5 4 3 2 1 4

x

2

2

4

3-64

for all x = 0, there are no x-intercepts. It is worthwhile noting that all of these features of the graph were discernible from Figure 3.74b, except for the inflection point at x = − 12 . Note that in almost any graph you draw, it is difficult to see all of the features of the function. This happens because the inflection point − 12 , e−2 or (−0.5, 0.135335 . . .) is so close to the x-axis. Since the horizontal asymptote is the line y = 1, it is difficult to see both of these features on the same graph (without drawing the graph on a very large piece of paper). We settle for the graph seen in Figure 3.75, which shows 1 all of the features except the inflection point and the concavity on the interval − 2 , 0 . To clearly see the behavior near the inflection point, we draw a graph that is zoomed-in on the area of the inflection point (see Figure 3.76). Here, while we have resolved the problem of the concavity near x = 0 and the inflection point, we have lost the details of the “big picture.”

FIGURE 3.75 y = e1/x

In our final example, we consider the graph of a function that is the sum of a trigonometric function and a polynomial.

y 0.3

EXAMPLE 6.6 0.2

Graphing the Sum of a Polynomial and a Trigonometric Function

Draw a graph of f (x) = cos x − x, showing all significant features. y

0.1 4 0.8 0.6 0.4 0.2

y

x 0.2

2

10

FIGURE 3.76 y = e1/x

4

5

x

2

2

4 10

2

5

x 5

5

10

10

4

FIGURE 3.77a

FIGURE 3.77b

y = cos x − x

y = cos x − x

Solution The default graph provided by our computer algebra system can be seen in Figure 3.77a. The graph produced by most graphing calculators looks like that in Figure 3.77b. As always, we will use the calculus to determine the behavior of the function more precisely. First, notice that the domain of f is the entire real line. Consequently, there are no vertical asymptotes. Next, we have f (x) = − sin x − 1 ≤ 0,

for all x.

(6.26)

Further, f (x) = 0 if and only if sin x = −1. So, there are critical numbers (here, these are all locations of horizontal tangent lines), but since f (x) does not change sign, there are no local extrema. Even so, it is still of interest to find the locations of the horizontal tangent lines. Recall that sin x = −1

for x =

3π 2

3-65

SECTION 3.6

x=

and more generally, for

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Overview of Curve Sketching

305

3π + 2nπ, 2

for any integer n. Next, we see that f (x) = − cos x and on the interval [0, 2π ], we have π 3π ∪ , 2π > 0, on 0, 2 2 cos x π 3π < 0, on , . 2 2

So,

π 3π ∪ , 2π < 0, on 0, 2 2 f (x) = − cos x π 3π > 0, on , . 2 2

Concave down.

(6.27) Concave up.

Outside of [0, 2π ], f (x) simply repeats this pattern. In particular, this says that the graph has infinitely many inflection points, located at odd multiples of π/2. To determine the behavior as x → ±∞, we examine the limits lim (cos x − x) = −∞

(6.28)

lim (cos x − x) = ∞,

(6.29)

x→∞

and

y

since −1 ≤ cos x ≤ 1, for all x and since lim x = ∞. x→∞ Finally, to determine the x-intercept(s), we need to solve

15 10 5 15 10 5 5

x→−∞

f (x) = cos x − x = 0.

x 5

10

10 15

FIGURE 3.78 y = cos x − x

15

This can’t be solved exactly, however. Since f (x) ≤ 0 for all x and Figures 3.77a and 3.77b show a zero around x = 1, there is only one zero and we must approximate this. (Use Newton’s method or your calculator’s solver.) We get x ≈ 0.739085 as an approximation to the only x-intercept. Assembling all of the information in (6.26)–(6.29), we can draw the graph seen in Figure 3.78. Notice that Figure 3.77b shows the behavior just as clearly as Figure 3.78, but for a smaller range of x- and y-values. Which of these is more “representative” is open to discussion.

BEYOND FORMULAS The main characteristic of the examples in sections 3.4–3.6 is the interplay between graphing and equation solving. To analyze the graph of a function, you will go back and forth several times between solving equations (for critical numbers and inflection points and so on) and identifying graphical features of interest. Even if you have access to graphing technology, the equation solving may lead you to uncover hidden features of the graph. What types of graphical features can sometimes be hidden?

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EXERCISES 3.6 WRITING EXERCISES 1. We have talked about sketching representative graphs, but it is often impossible to draw a graph correctly to scale that shows all of the properties we might be interested in. For example, try to generate a computer or calculator graph that shows all three local extrema of x 4 − 25x 3 − 2x 2 + 80x − 3. When two extrema have y-coordinates of approximately −60 and 50, it takes a very large graph to also show a point with y = −40,000! If an accurate graph cannot show all the points of interest, perhaps a freehand sketch like the one shown below is needed. y

4 x 7. f (x) = x ln x √ 9. f (x) = x 2 + 1 5. f (x) = x +

x2 − 1 x 8. f (x) = x ln x 2 √ 10. f (x) = 2x − 1 6. f (x) =

4x x2 − x + 1 √ 13. f (x) = 3 x 3 − 3x 2 + 2x

4x 2 x2 − x + 1 √ 14. f (x) = x 3 − 3x 2 + 2x

15. f (x) = x 5 − 5x

16. f (x) = x 3 −

17. f (x) = e−2/x

18. f (x) = e1/x

11. f (x) =

12. f (x) =

2

3 x 400

In exercises 19–32, determine all significant features (approximately if necessary) and sketch a graph. 19. f (x) = (x 3 − 3x 2 + 2x)2/3 x

20. f (x) = x 6 − 10x 5 − 7x 4 + 80x 3 + 12x 2 − 192x x2 + 1 3x 2 − 1 5x 23. f (x) = 3 x −x +1 25. f (x) = x 2 x 2 − 9

2x 2 x3 + 1 4x 24. f (x) = 2 x +x +1 3 26. f (x) = 2x 2 − 1

27. f (x) = e−2x sin x

28. f (x) = sin x −

21. f (x) =

There is no scale shown on the graph because we have distorted different portions of the graph in an attempt to show all of the interesting points. Discuss the relative merits of an “honest” graph with a consistent scale but not showing all the points of interest versus a caricature graph that distorts the scale but does show all the points of interest. 2. While studying for a test, a friend of yours says that a graph is not allowed to intersect an asymptote. While it is often the case that graphs don’t intersect asymptotes, there is definitely not any rule against it. Explain why graphs can intersect a horizontal asymptote any number of times (Hint: Look at the graph of e−x sin x), but can’t pass through a vertical asymptote. 3. Explain why polynomials never have vertical or horizontal asymptotes. 4. Explain how the graph of f (x) = cos x − x in example 6.6 relates to the graphs of y = cos x and y = −x. Based on this discussion, explain how to sketch the graph of y = x + sin x.

In exercises 1–18, graph the function and completely discuss the graph as in example 6.2. 1. f (x) = x 3 − 3x 2 + 3x

2. f (x) = x 4 − 3x 2 + 2x

3. f (x) = x 5 − 2x 3 + 1

4. f (x) = sin x − cos x

22. f (x) =

29. f (x) = x 4 − 16x 3 + 42x 2 − 39.6x + 14

1 sin 2x 2

30. f (x) = x 4 + 32x 3 − 0.02x 2 − 0.8x √ 1 25 − 50 x 2 + 0.25 −1 32. f (x) = tan 31. f (x) = x x2 − 1 In exercises 33–38, the “family of functions” contains a parameter c. The value of c affects the properties of the functions. Determine what differences, if any, there are for c being zero, positive or negative. Then determine what the graph would look like for very large positive c’s and for very large negative c’s. 33. f (x) = x 4 + cx 2 x2 35. f (x) = 2 x + c2 37. f (x) = sin(cx)

34. f (x) = x 4 + cx 2 + x 36. f (x) = e−x /c √ 38. f (x) = x 2 c2 − x 2 2

39. In a variety of applications, researchers model a phenomenon whose graph starts at the origin, rises to a single maximum and then drops off to a horizontal asymptote of y = 0. For example, the probability density function of events such as the time from conception to birth of an animal and the amount of time surviving after contracting a fatal disease might have these properties. Show that the family of functions xe−bx has these properties for all positive constants b. What effect does

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SECTION 3.6

b have on the location of the maximum? In the case of the time since conception, what would b represent? In the case of survival time, what would b represent? 40. The “FM” in FM radio stands for frequency modulation, a method of transmitting information encoded in a radio wave by modulating (or varying) the frequency. A basic example of such a modulated wave is f (x) = cos (10x + 2 cos x). Use computer-generated graphs of f (x), f (x) and f (x) to try to locate all local extrema of f (x). p(x) , where p(x) q(x) and q(x) are polynomials. Is it true that all rational functions have vertical asymptotes? Is it true that all rational functions have horizontal asymptotes?

41. A rational function is a function of the form

42. It can be useful to identify asymptotes other than vertical and horizontal. For example, the parabola y = x 2 is an asymptote of f (x) if lim [ f (x) − x 2 ] = 0 and/or lim [ f (x) − x 2 ] = 0. x→∞

x→−∞ 4

x − x2 + 1 . Graph x2 − 1 y = f (x) and zoom out until the graph looks like a parabola. (Note: The effect of zooming out is to emphasize large values of x.)

Show that x is an asymptote of f (x) = 2

A function f has a slant asymptote y mx b (m 0) if lim [ f (x) − (mx b)] 0 and/or lim [ f (x) − (mx b)] 0. x→∞

x→− ∞

In exercises 43–48, find the slant asymptote. (Use long division to rewrite the function.) Then, graph the function and its asymptote on the same axes. 3x 2 − 1 x x 3 − 2x 2 + 1 45. f (x) = x2 4 x 47. f (x) = 3 x +1 43. f (x) =

3x 2 − 1 x −1 x3 − 1 46. f (x) = 2 x −1 x4 − 1 48. f (x) = 3 x +x 44. f (x) =

In exercises 49–52, find a function whose graph has the given asymptotes. 49. x = 1, x = 2 and y = 3

50. x = −1, x = 1 and y = 0

51. x = −1, x = 1, y = −2 and y = 2 52. x = 1, y = 2 and x = 3 53. Find all extrema and inflection points, and sketch the graphs e x + e−x e x − e−x and y = cosh x = . of y = sinh x = 2 2 54. On the same axes as the graphs of exercise 53, sketch in the graphs of y = 12 e x and y = 12 e−x . Explain why these graphs serve as an envelope for the graphs in exercise 53. (Hint: As x → ±∞, what happens to e x and e−x ?)

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307

EXPLORATORY EXERCISES 1. For each function, find a polynomial p(x) such that lim [ f (x) − p(x)] = 0. x→∞

(a)

x4 x +1

(b)

x5 − 1 x +1

(c)

x6 − 2 x +1

Show by zooming out that f (x) and p(x) look similar for large x. The first term of a polynomial is the term with the highest power (e.g., x 3 is the first term of x 3 − 3x + 1). Can you zoom out enough to make the graph of f (x) look like the first term of its polynomial asymptote? State a very quick rule enabling you to look at a rational function and determine the first term of its polynomial asymptote (if one exists). 2. One of the natural enemies of the balsam fir tree is the spruce budworm, which attacks the leaves of the fir tree in devastating outbreaks. Define N (t) to be the number of worms on a particular tree at time t. A mathematical model of the population dynamics of the worm must include a term to indicate the worm’s death rate due to its predators (e.g., birds). The form B[N (t)]2 of this term is often taken to be 2 for positive conA + [N (t)]2 2 2x 2 x2 x , , stants A and B. Graph the functions 2 2 4 + x 1 + x 9 + x2 3x 2 and for x > 0. Based on these graphs, discuss why 1 + x2 B[N (t)]2 is a plausible model for the death rate by predaA2 + [N (t)]2 tion. What role do the constants A and B play? The possible stable population levels for the spruce budworms are determined by intersections of the graphs of y = r (1 − x/k) and x y= . Here, x = N /A, r is proportional to the birthrate 1 + x2 of the budworms and k is determined by the amount of food available to the budworms. Note that y = r (1 − x/k) is a line with y-intercept r and x-intercept k. How many solutions are x there to the equation r (1 − x/k) = ? (Hint: The answer 1 + x2 depends on the values of r and k.) One current theory is that outbreaks are caused in situations where there are three solutions and the population of budworms jumps from a small population to a large population. 3. Suppose that f (x) is a function with two derivatives and that f (a) = f (a) = 0 but f (a) = 0 for some number a. Show that f (x) has a local extremum at x = a. Next, suppose that f (x) is a function with three derivatives and that f (a) = f (a) = f (a) = 0 but f (a) = 0 for some number a. Show that f (x) does not have a local extremum at x = a. Generalize your work to the case where f (k) (a) = 0 for k = 0, 1, . . . , n − 1, but f (n) (a) = 0, keeping in mind that there are different conclusions depending on whether n is odd or even. Use this result to determine whether f (x) = x sin x 2 or g(x) = x 2 sin(x 2 ) has a local extremum at x = 0.

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3-68

OPTIMIZATION Everywhere in business and industry today, we see people struggling to minimize waste and maximize productivity. We are now in a position to bring the power of the calculus to bear on problems involving finding a maximum or a minimum. This section contains a number of illustrations of such problems. Pay close attention to how we solve the problems. We start by giving a few general guidelines. (Notice that we said guidelines and not rules.) Do not memorize them, but rather keep these in mind as you work through the section. r If there’s a picture to draw, draw it! Don’t try to visualize how things look in your

head. Put a picture down on paper and label it.

r Determine what the variables are and how they are related. r Decide what quantity needs to be maximized or minimized. r Write an expression for the quantity to be maximized or minimized in terms of only

one variable. To do this, you may need to solve for any other variables in terms of this one variable. r Determine the minimum and maximum allowable values (if any) of the variable you’re using. r Solve the problem. (Be sure to answer the question that is asked.) We begin with a simple example where the goal is to accomplish what businesses face every day: getting the most from limited resources.

EXAMPLE 7.1

Constructing a Rectangular Garden of Maximum Area

You have 40 (linear) feet of fencing with which to enclose a rectangular space for a garden. Find the largest area that can be enclosed with this much fencing and the dimensions of the corresponding garden.

OR

Solution First, note that there are lots of possibilities. We could enclose a plot that is very long but narrow, or one that is very wide but not very long (see Figure 3.79). How are we to decide which configuration is optimal? We first draw a picture and label it appropriately (see Figure 3.80). The variables for a rectangular plot are length and width, which we name x and y, respectively. We want to find the dimensions of the largest possible area, that is, maximize FIGURE 3.79 Possible plots

A = x y. Notice immediately that this function has two variables and so, cannot be dealt with via the means we have available. However, there is another piece of information that we can use here. If we want the maximum area, then all of the fencing must be used. This says that the perimeter of the resulting fence must be 40 and hence,

y

40 = perimeter = 2x + 2y. x

FIGURE 3.80 Rectangular plot

(7.1)

Notice that we can use (7.1) to solve for one variable (either one) in terms of the other. We have 2y = 40 − 2x

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..

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309

y = 20 − x.

and hence, Substituting for y, we get that

A = x y = x(20 − x). So, our job is to find the maximum value of the function A(x) = x(20 − x). y

Hold it! You may want to multiply out this expression. Unless you have a clear picture of how it will affect your later work, leave the function alone! Before we maximize A(x), we need to determine the interval in which x must lie. Since x is a distance, we must have 0 ≤ x. Further, since the perimeter is 40 , we must have x ≤ 20. (Why don’t we have x ≤ 40?) So, we want to find the maximum value of A(x) on the closed interval [0, 20]. This is now a simple problem. As a check on what a reasonable answer should be, we draw a graph of y = A(x) (see Figure 3.81) on the interval [0, 20]. The maximum value appears to occur around x = 10. Now, let’s analyze the problem carefully. We have

100 80 60 40 20 x 5

10

15

FIGURE 3.81 y = x(20 − x)

20

A (x) = 1(20 − x) + x(−1) = 20 − 2x = 2(10 − x). So, the only critical number is x = 10 and this is in the interval under consideration. Recall that the maximum and minimum values of a continuous function on a closed and bounded interval must occur at either the endpoints or a critical number. This says that we need only compare the function values A(0) = 0,

A(20) = 0

and

A(10) = 100.

Thus, the maximum area that can be enclosed with 40 of fencing is 100 ft2 . We also want the dimensions of the plot. (This result is only of theoretical value if we don’t know how to construct the rectangle with the maximum area.) We have that x = 10 and y = 20 − x = 10. That is, the rectangle of perimeter 40 with maximum area is a square 10 on a side. A more general problem that you can now solve is to show that (given a fixed perimeter) the rectangle of maximum area is a square. This is virtually identical to example 7.1 and is left as an exercise. It’s worth noting here that this more general problem is one that cannot be solved by simply using a calculator to draw a graph. You’ll need to use some calculus here. Manufacturing companies routinely make countless decisions that affect the efficiency of their production processes. One decision that is surprisingly important is how to economically package products for shipping. Example 7.2 provides a simple illustration of this problem.

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18

18

EXAMPLE 7.2

3-70

Constructing a Box of Maximum Volume

A square sheet of cardboard 18 on a side is made into an open box (i.e., there’s no top), by cutting squares of equal size out of each corner (see Figure 3.82a) and folding up the sides along the dotted lines (see Figure 3.82b). Find the dimensions of the box with the maximum volume.

18 2x

x

Solution Recall that the volume of a rectangular parallelepiped (a box) is given by V = l × w × h.

18 2x x

From Figure 3.82b, we can see that the height is h = x, while the length and width are l = w = 18 − 2x. Thus, we can write the volume in terms of the one variable x as

FIGURE 3.82a A sheet of cardboard

V = V (x) = (18 − 2x)2 (x) = 4x(9 − x)2 .

x 18 2x 18 2x

Once again, don’t multiply this out, just out of habit. Notice that since x is a distance, we have x ≥ 0. Further, we have x ≤ 9, since cutting squares of side 9 out of each corner will cut up the entire sheet of cardboard. Thus, we are faced with finding the absolute maximum of the continuous function V (x) = 4x(9 − x)2

FIGURE 3.82b

on the closed interval 0 ≤ x ≤ 9. This should be a simple matter. The graph of y = V (x) on the interval [0, 9] is seen in Figure 3.83. From the graph, the maximum volume seems to be somewhat over 400 and seems to occur around x = 3. Now, we solve the problem precisely. We have

Rectangular box y 500 400

V (x) = 4(9 − x)2 + 4x(2)(9 − x)(−1)

300

= 4(9 − x)[(9 − x) − 2x]

200

= 4(9 − x)(9 − 3x).

100 x 2

4

6

FIGURE 3.83 y = 4x(9 − x)2

8

Product rule and chain rule. Factor out 4(9 − x).

So, V has two critical numbers, 3 and 9, and these are both in the interval under consideration. We now need only compare the value of the function at the endpoints and the critical numbers. We have V (0) = 0,

V (9) = 0

and

V (3) = 432.

Obviously, the maximum possible volume is 432 cubic inches. We can achieve this volume if we cut squares of side 3 out of each corner. You should note that this corresponds with what we expected from the graph of y = V (x) in Figure 3.83. Finally, observe that the dimensions of this optimal box are 12 long by 12 wide by 3 deep. When a new building is built, it must be connected to existing telephone and power cables, water and sewer lines and paved roads. If the cables, water or sewer lines or road are curved, then it may not be obvious how to make the shortest (i.e., least expensive) connection possible. In examples 7.3 and 7.4, we consider the general problem of finding the shortest distance from a point to a curve.

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..

Optimization

311

y

EXAMPLE 7.3 9

Finding the Closest Point on a Parabola

(3, 9)

Find the point on the parabola y = 9 − x 2 closest to the point (3, 9) (see Figure 3.84).

(x, y)

Solution Using the usual distance formula, we find that the distance between the point (3, 9) and any point (x, y) is d = (x − 3)2 + (y − 9)2 .

y 9 x2

x

4

4

FIGURE 3.84

If the point (x, y) is on the parabola, note that its coordinates satisfy the equation y = 9 − x 2 and so, we can write the distance in terms of the single variable x as follows d(x) =

y = 9 − x2

= =

(x − 3)2 + [(9 − x 2 ) − 9]2 (x − 3)2 + (−x 2 )2 (x − 3)2 + x 4 .

Although we can certainly solve the problem in its present form, we can simplify our work by observing that d(x) is minimized if and only if the quantity under the square root is minimized. (We leave it as an exercise to show why this is true.) So, instead of minimizing d(x) directly, we minimize the square of d(x):

y 80 60

f (x) = [d(x)]2 = (x − 3)2 + x 4

40 20 x 1

2

FIGURE 3.85 y = (x − 3)2 + x 4

instead. Notice from Figure 3.84 that any point on the parabola to the left of the y-axis is farther away from (3, 9) than is the point (0, 9). Likewise, any point on the parabola below the x-axis is farther from (3, 9) than is the point (3, 0). So, it suffices to look for the closest point with

3

0 ≤ x ≤ 3. See Figure 3.85 for a graph of y = f (x) over the interval of interest. Observe that the minimum value of f (the square of the distance) seems to be around 5 and seems to occur near x = 1. We have f (x) = 2(x − 3)1 + 4x 3 = 4x 3 + 2x − 6. Notice that f (x) factors. [One way to see this is to recognize that x = 1 is a zero of f (x), which makes (x − 1) a factor.] We have f (x) = 2(x − 1)(2x 2 + 2x + 3). So, x = 1 is a critical number. In fact, it’s the only critical number, since (2x 2 + 2x + 3) has no zeros. (Why not?) We now need only compare the value of f at the endpoints and the critical number. We have f (0) = 9,

f (3) = 81

and

f (1) = 5.

Thus, the minimum value of f√(x) is 5. This says that the minimum distance from the point (3, 9) to the parabola is 5 and the closest point on the parabola is (1, 8). Again, notice that this corresponds with what we expected from the graph of y = f (x).

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y

3-72

Example 7.4 is very similar to example 7.3, except that we need to use approximate methods to find the critical number.

(5, 11) 9 (x, y) y 9 x2

EXAMPLE 7.4

Finding Minimum Distance Approximately

Find the point on the parabola y = 9 − x 2 closest to the point (5, 11) (see Figure 3.86). x

4

4

FIGURE 3.86

Solution As in example 7.3, we want to minimize the distance from a fixed point [in this case, the point (5, 11)] to a point (x, y) on the parabola. Using the distance formula, the distance from any point (x, y) on the parabola to the point (5, 11) is d=

y = 9 − x2

= y

=

(x − 5)2 + (y − 11)2 (x − 5)2 + [(9 − x 2 ) − 11]2 (x − 5)2 + (x 2 + 2)2 .

600

Again, it is equivalent (and simpler) to minimize the quantity under the square root: 400

f (x) = [d(x)]2 = (x − 5)2 + (x 2 + 2)2 .

200

As in example 7.3, we can see from Figure 3.86 that any point on the parabola to the left of the y-axis is farther from (5, 11) than is (0, 9). Likewise, any point on the parabola to the right of x = 5 is farther from (5, 11) than is (5, −16). Thus, we minimize f (x) for 0 ≤ x ≤ 5. In Figure 3.87 we see a graph of y = f (x) on the interval of interest. The minimum value of f seems to occur around x = 1. We can make this more precise as follows:

x 1

2

3

4

5

FIGURE 3.87 y = f (x) = [d(x)]2

f (x) = 2(x − 5) + 2(x 2 + 2)(2x) = 4x 3 + 10x − 10.

y

Unlike in example 7.3, the expression for f (x) has no obvious factorization. Our only choice then is to find zeros of f (x) approximately. First, we draw a graph of y = f (x) on the interval of interest (see Figure 3.88). The only zero appears to be slightly less than 1. Using x0 = 1 as an initial guess in Newton’s method (applied to f (x) = 0) or using your calculator’s solver, you should get the approximate root xc ≈ 0.79728. We now compare function values:

300

200

100

x 1

2

3

4

FIGURE 3.88 y = f (x)

f (0) = 29,

f (5) = 729

and

f (xc ) ≈ 24.6.

5

Thus, the minimum distance from (5, 11) to the parabola is approximately √ 24.6 ≈ 4.96 and the closest point on the parabola is located at approximately (0.79728, 8.364).

Notice that in both Figures 3.84 and 3.86, the shortest path appears to be perpendicular to the tangent line to the curve at the point where the path intersects the curve. We leave it as an exercise to prove that this is always the case. This observation is an important geometric principle that applies to many problems of this type.

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313

REMARK 7.1 At this point you might be tempted to forgo the comparison of function values at the endpoints and at the critical numbers. After all, in all of the examples we have seen so far, the desired maximizer or minimizer (i.e., the point at which the maximum or minimum occurred) was the only critical number in the interval under consideration. You might just suspect that if there is only one critical number, it will correspond to the maximizer or minimizer for which you are searching. Unfortunately, this is not always the case. In 1945, two prominent aeronautical engineers derived a function to model the range of an aircraft. Their intention was to use this function to discover how to maximize the range. They found a critical number of this function (corresponding to distributing virtually all of the plane’s weight in the wings) and reasoned that it gave the maximum range. The result was the famous “Flying Wing” aircraft. Some years later, it was argued that the critical number they found corresponded to a local minimum of the range function. In the engineers’ defense, they did not have easy, accurate computational power at their fingertips, as we do today. Remarkably, this design strongly resembles the modern B-2 Stealth bomber. This story came out as controversy brewed over the production of the B-2 (see Science, 244, pp. 650–651, May 12, 1989; also see the Monthly of the Mathematical Association of America, October, 1993, pp. 737–738). The moral should be crystal clear: check the function values at the critical numbers and at the endpoints. Do not simply assume (even by virtue of having only one critical number) that a given critical number corresponds to the extremum you are seeking.

Next, we consider an optimization problem that cannot be restricted to a closed interval. We will use the fact that for a continuous function, a single local extremum must be an absolute extremum. (Think about why this is true.)

EXAMPLE 7.5

A soda can is to hold 12 fluid ounces. Find the dimensions that will minimize the amount of material used in its construction, assuming that the thickness of the material is uniform (i.e., the thickness of the aluminum is the same everywhere in the can).

r

h

FIGURE 3.89 Soda can

Designing a Soda Can That Uses a Minimum Amount of Material

Solution First, we draw and label a picture of a typical soda can (see Figure 3.89). Here we have drawn a right circular cylinder of height h and radius r . Assuming uniform thickness of the aluminum, notice that we minimize the amount of material by minimizing the surface area of the can. We have area = area of top + area of bottom + curved surface area = 2πr 2 + 2πr h.

(7.2)

We can eliminate one of the variables by using the fact that the volume (using 1 fluid ounce ≈ 1.80469 in.3 ) must be 12 fluid ounces ≈ 12 fl oz × 1.80469

in.3 = 21.65628 in.3 . fl oz

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3-74

Further, the volume of a right circular cylinder is vol = πr 2 h and so, y

h=

vol 21.65628 ≈ . πr 2 πr 2

(7.3)

Thus, from (7.2) and (7.3), the surface area is approximately 21.65628 A(r ) = 2πr 2 + 2πr πr 2 21.65628 = 2π r 2 + . πr

150 100 50 x 1

2

3

4

5

FIGURE 3.90 y = A(r )

6

So, our job is to minimize A(r ), but here, there is no closed and bounded interval of allowable values. In fact, all we can say is that r > 0. We can have r as large or small as you can imagine, simply by taking h to be correspondingly small or large, respectively. That is, we must find the absolute minimum of A(r ) on the open and unbounded interval (0, ∞). To get an idea of what a plausible answer might be, we graph y = A(r ) (see Figure 3.90). There appears to be a local minimum (slightly less than 50) located between r = 1 and r = 2. Next, we compute d 21.65628 A (r ) = 2π r 2 + dr πr 21.65628 = 2π 2r − πr 2 2πr 3 − 21.65628 = 2π . πr 2 Notice that the only critical numbers are those for which the numerator of the fraction is zero: 0 = 2πr 3 − 21.65628. r3 =

This occurs if and only if

21.65628 2π

and hence, the only critical number is r = rc =

3

21.65628 ≈ 1.510548. 2π

Further, notice that for 0 < r < rc , A (r ) < 0 and for rc < r, A (r ) > 0. That is, A(r ) is decreasing on the interval (0, rc ) and increasing on the interval (rc , ∞). Thus, A(r ) has not only a local minimum, but also an absolute minimum at r = rc . Notice, too, that this corresponds with what we expected from the graph of y = A(r ) in Figure 3.90. This says that the can that uses a minimum of material has radius rc ≈ 1.510548 and height h=

21.65628 ≈ 3.0211. πrc2

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315

Note that the optimal can from example 7.5 is “square,” in the sense that the height (h) equals the diameter (2r ). Also, we should observe that example 7.5 is not completely realistic. A standard 12-ounce soda can has a radius of about 1.156 . You should review example 7.5 to find any unrealistic assumptions we made. We study the problem of designing a soda can further in the exercises. In our final example, we consider a problem where most of the work must be done numerically and graphically.

EXAMPLE 7.6

Minimizing the Cost of Highway Construction

The state wants to build a new stretch of highway to link an existing bridge with a turnpike interchange, located 8 miles to the east and 8 miles to the south of the bridge. There is a 5-mile-wide stretch of marshland adjacent to the bridge that must be crossed (see Figure 3.91). Given that the highway costs $10 million per mile to build over the marsh and only $7 million to build over dry land, how far to the east of the bridge should the highway be when it crosses out of the marsh?

Bridge

5

Marsh x

8x 3 Interchange

FIGURE 3.91 A new highway

Solution You might guess that the highway should cut directly across the marsh, so as to minimize the amount built over marshland. We will use the calculus to decide this question. We let x represent the distance in question (see Figure 3.91). Then, the interchange lies (8 − x) miles to the east of the point where the highway leaves the marsh. Thus, the total cost (in millions of dollars) is

y

cost = 10(distance across marsh) + 7(distance across dry land ).

120

Using the Pythagorean Theorem on the two right triangles seen in Figure 3.91, we get the cost function C(x) = 10 x 2 + 25 + 7 (8 − x)2 + 9.

110

100 x 2

4

6

FIGURE 3.92 y = C(x)

8

Observe from Figure 3.91 that we must have 0 ≤ x ≤ 8. So, we have the routine problem of minimizing a continuous function C(x) over the closed and bounded interval [0, 8]. Or is it really that routine? First, we draw a graph of y = C(x) on the interval in question to get an idea of a plausible answer (see Figure 3.92). From the graph, the minimum appears to be slightly less than 100 and occurs around x = 4.

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y

3-76

We now compare d 2 10 x + 25 + 7 (8 − x)2 + 9 dx 7 = 5(x 2 + 25)−1/2 (2x) + [(8 − x)2 + 9]−1/2 (2)(8 − x)1 (−1) 2 10x 7(8 − x) = √ − . x 2 + 25 (8 − x)2 + 9

10

C (x) = 5 x 2

4

6

8

5 10

FIGURE 3.93 y = C (x)

First, note that the only critical numbers are where C (x) = 0. (Why?) The only way to find these is to approximate them. From the graph of y = C (x) seen in Figure 3.93, the only zero of C (x) on the interval [0, 8] appears to be between x = 3 and x = 4. We approximate this zero numerically (e.g., with bisections or your calculator’s solver), to obtain the approximate critical number xc ≈ 3.560052. Now, we need only compare the value of C(x) at the endpoints and at this one critical number: C(0) ≈ $109.8 million, C(8) ≈ $115.3 million and

C(xc ) ≈ $98.9 million.

So, by using a little calculus, we can save the taxpayers more than $10 million over cutting directly across the marsh and more than $16 million over cutting diagonally across the marsh (not a bad reward for a few minutes of work). The examples that we’ve presented in this section together with the exercises should give you the basis for solving a wide range of applied optimization problems. When solving these problems, be careful to draw good pictures, as well as graphs of the functions involved. Make sure that the answer you obtain computationally is consistent with what you expect from the graphs. If not, further analysis is required to see what you have missed. Also, make sure that the solution makes physical sense, when appropriate. All of these multiple checks on your work will reduce the likelihood of error.

EXERCISES 3.7 WRITING EXERCISES 1. Suppose some friends complain to you that they can’t work any of the problems in this section. When you ask to see their work, they say that they couldn’t even get started. In the text, we have emphasized sketching a picture and defining variables. Part of the benefit of this is to help you get started writing something (anything) down. Do you think this advice helps? What do you think is the most difficult aspect of these problems? Give your friends the best advice you can.

2. We have neglected one important aspect of optimization problems, an aspect that might be called “common sense.” For example, suppose you are finding the optimal dimensions for a fence and the √mathematical solution is to build a square fence of length 10 5 feet on each side. At the meeting with the carpenter who is going √to build the fence, what length fence do you order? Why is 10 5 probably not √ the best way to express the length? We can approximate 10 5 ≈ 22.36. What would you

3-77

tell the carpenter? Suppose the carpenter accepts measurements down to the inch. Assuming that the building constraint was that the perimeter of the fence could not exceed a certain figure, why should you truncate to 22 4 instead of rounding up to 22 5 ? √ 3. In example 7.3, we stated that d(x) = f (x) is minimized √ by exactly the same x-value(s) as f (x). Use the fact that x is an increasing function to explain why this is true. 4. Suppose that f (x) is a continuous function with a single critical number and f (x) has a local minimum at that critical number. Explain why f (x) also has an absolute minimum at the critical number. 1. Give an example showing that f (x) and sin ( f (x)) need not be minimized by the same x-values. 2. True or false: e f (x) is minimized by exactly the same x-value(s) as f (x). 3. A three-sided fence is to be built next to a straight section of river, which forms the fourth side of a rectangular region. The enclosed area is to equal 1800 ft2 . Find the minimum perimeter and the dimensions of the corresponding enclosure. 4. A three-sided fence is to be built next to a straight section of river, which forms the fourth side of a rectangular region. There is 96 feet of fencing available. Find the maximum enclosed area and the dimensions of the corresponding enclosure. 5. A two-pen corral is to be built. The outline of the corral forms two identical adjoining rectangles. If there is 120 ft of fencing available, what dimensions of the corral will maximize the enclosed area? 6. A showroom for a department store is to be rectangular with walls on three sides, 6-ft door openings on the two facing sides and a 10-ft door opening on the remaining wall. The showroom is to have 800 ft2 of floor space. What dimensions will minimize the length of wall used? 7. Show that the rectangle of maximum area for a given perimeter P is always a square. 8. Show that the rectangle of minimum perimeter for a given area A is always a square. 9. Find the point on the curve y = x 2 closest to the point (0, 1). 10. Find the point on the curve y = x 2 closest to the point (3, 4). 11. Find the point on the curve y = cos x closest to the point (0, 0). 12. Find the point on the curve y = cos x closest to the point (1, 1). 13. In exercises 9 and 10, find the slope of the line through the given point and the closest point on the given curve. Show that in each case, this line is perpendicular to the tangent line to the curve at the given point. 14. Sketch the graph of some function y = f (x) and mark a point not on the curve. Explain why the result of exercise 13

SECTION 3.7

..

Optimization

317

is true. (Hint: Pick a point for which the joining line is not perpendicular and explain why you can get closer.) 15. A box with no top is to be built by taking a 6 -by-10 sheet of cardboard and cutting x-in. squares out of each corner and folding up the sides. Find the value of x that maximizes the volume of the box. 16. A box with no top is to be built by taking a 12 -by-16 sheet of cardboard and cutting x-in. squares out of each corner and folding up the sides. Find the value of x that maximizes the volume of the box. 17. A water line runs east-west. A town wants to connect two new housing developments to the line by running lines from a single point on the existing line to the two developments. One development is 3 miles south of the existing line; the other development is 4 miles south of the existing line and 5 miles east of the first development. Find the place on the existing line to make the connection to minimize the total length of new line. 18. A company needs to run an oil pipeline from an oil rig 25 miles out to sea to a storage tank that is 5 miles inland. The shoreline runs east-west and the tank is 8 miles east of the rig. Assume it costs $50 thousand per mile to construct the pipeline under water and $20 thousand per mile to construct the pipeline on land. The pipeline will be built in a straight line from the rig to a selected point on the shoreline, then in a straight line to the storage tank. What point on the shoreline should be selected to minimize the total cost of the pipeline? 19. A city wants to build a new section of highway to link an existing bridge with an existing highway interchange, which lies 8 miles to the east and 10 miles to the south of the bridge. The first 4 miles south of the bridge is marshland. Assume that the highway costs $5 million per mile over marsh and $2 million per mile over dry land. The highway will be built in a straight line from the bridge to the edge of the marsh, then in a straight line to the existing interchange. At what point should the highway emerge from the marsh in order to minimize the total cost of the new highway? How much is saved over building the new highway in a straight line from the bridge to the interchange? (Hint: Use similar triangles to find the point on the boundary corresponding to a straight path and evaluate your cost function at that point.) 20. After construction has begun on the highway in exercise 19, the cost per mile over marshland is reestimated at $6 million. Find the point on the marsh/dry land boundary that would minimize the total cost of the highway with the new cost function. If the construction is too far along to change paths, how much extra cost is there in using the path from exercise 19? 21. After construction has begun on the highway in exercise 19, the cost per mile over dry land is reestimated at $3 million. Find the point on the marsh/dry land boundary that would minimize the total cost of the highway with the new cost function. If the construction is too far along to change paths, how much extra cost is there in using the path from exercise 19?

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3-78

22. In an endurance contest, contestants 2 miles at sea need to reach a location 2 miles inland and 3 miles east (the shoreline runs east-west). Assume a contestant can swim 4 mph and run 10 mph. To what point on the shoreline should the person swim to minimize the total time? Compare the amount of time spent in the water and the amount of time spent on land. 23. Suppose that light travels from point A to point B as shown in the figure. (Recall that light always follows the path that minimizes time.) Assume that the velocity of light above the boundary line is v1 and the velocity of light below the boundary is v2 . Show that the total time to get from point A to point B is T (x) =

√ 1 + x2 1 + (2 − x)2 + . v1 v2

Write out the equation T (x) = 0, replace the square roots using the sines of the angles in the figure and derive Snell’s sin θ1 v1 Law = . sin θ2 v2 A

25. A soda can is to hold 12 fluid ounces. Suppose that the bottom and top are twice as thick as the sides. Find the dimensions of the can that minimize the amount of material used. (Hint: Instead of minimizing surface area, minimize the cost, which is proportional to the product of the thickness and the area.) 26. Following example 7.5, we mentioned that real soda cans have a radius of about 1.156 . Show that this radius minimizes the cost if the top and bottom are 2.23 times as thick as the sides. 27. The human cough is intended to increase the flow of air to the lungs, by dislodging any particles blocking the windpipe and changing the radius of the pipe. Suppose a windpipe under no pressure has radius r0 . The velocity of air through the windpipe at radius r is approximately V (r ) = cr 2 (r0 − r ) for some constant c. Find the radius that maximizes the velocity of air through the windpipe. Does this mean the windpipe expands or contracts? 28. To supply blood to all parts of the body, the human artery system must branch repeatedly. Suppose an artery of radius r branches off from an artery of radius R (R > r ) at an angle θ . The energy lost due to friction is approximately E(θ) =

1

u1

2x x

csc θ 1 − cot θ + . r4 R4

Find the value of θ that minimizes the energy loss. 29. In an electronic device, individual circuits may serve many purposes. In some cases, the flow of electricity must be controlled by reducing the power instead of amplifying it. In the circuit shown here, a voltage V volts and resistance R ohms are given. We want to determine the size of the remaining resistor (x ohms). The power absorbed by the circuit is

u2 1

B

p(x) =

Exercise 23 24. Suppose that light reflects off a mirror to get from point A to point B as indicated in the figure. Assuming a constant velocity of light, we can minimize time by minimizing the distance traveled. Find the point on the mirror that minimizes the distance traveled. Show that the angles in the figure are equal (the angle of incidence equals the angle of reflection).

V 2x . (R + x)2

Find the value of x that maximizes the power absorbed. R V

x

A

B 2 u1

u2 4x

x

Exercise 24

1

30. In an AC circuit with voltage V (t) = v sin 2π ft, a voltmeter actually shows the average (root-mean-square) voltage of √ v/ 2. If the frequency is f = 60 (Hz) and the meter registers 115 volts, find the maximum voltage reached. [Hint: This is “obvious” if you determine v and think about the graph of V (t).] 31. A Norman window has the outline of a semicircle on top of a rectangle, as shown in the figure. Suppose there is 8 + π feet of wood trim available. Discuss why a window designer might want to maximize the area of the window. Find the dimensions

3-79

SECTION 3.7

of the rectangle (and, hence, the semicircle) that will maximize the area of the window.

..

Optimization

319

42. Suppose that Q(t) represents the output of a worker, as in exercise 41. If T is the length of a workday, then the graph of Q(t) should be increasing for 0 ≤ t ≤ T . Suppose that the graph of Q(t) has a single inflection point for 0 ≤ t ≤ T , called the point of diminishing returns. Show that the worker’s efficiency is maximized at the point of diminishing returns. 43. Suppose that group tickets to a concert are priced at $40 per ticket if 20 tickets are ordered, but cost $1 per ticket less for each extra ticket ordered, up to a maximum of 50 tickets. (For example, if 22 tickets are ordered, the price is $38 per ticket.) Find the number of tickets that maximizes the total cost of the tickets.

32. Suppose a wire 2 ft long is to be cut into two pieces, each of which will be formed into a square. Find the size of each piece to maximize the total area of the two squares. 33. An advertisement consists of a rectangular printed region plus 1-in. margins on the sides and 2-in. margins at top and bottom. If the area of the printed region is to be 92 in.2 , find the dimensions of the printed region and overall advertisement that minimize the total area. 34. An advertisement consists of a rectangular printed region plus 1-in. margins on the sides and 1.5-in. margins at top and bottom. If the total area of the advertisement is to be 120 in.2 , what dimensions should the advertisement be to maximize the area of the printed region? 35. A hallway of width a = 5 ft meets a hallway of width b = 4 ft at a right angle. Find the length of the longest ladder that could be carried around the corner. (Hint: Express the length of the ladder as a function of the angle θ in the figure.) u

b

44. In exercise 43, if management wanted the solution to be 50 (that is, ordering the maximum number of tickets produces the maximum cost), how much should the price be discounted for extra tickets ordered? 45. In sports where balls are thrown or hit, the ball often finishes at a different height than it starts. Examples include a downhill golf shot and a basketball shot. In the diagram, a ball is released at an angle θ and finishes at an angle β above the horizontal (for downhill trajectories, β would be negative). Neglecting air resistance and spin, the horizontal range is given by R=

2v 2 cos2 θ (tan θ − tan β) g

if the initial velocity is v and g is the gravitational constant. In the following cases, find θ to maximize R (treat v and g as constants): (a) β = 10◦ , (b) β = 0◦ and (c) β = −10◦ . Verify that θ = 45◦ + β ◦ /2 maximizes the range.

u

b

a

36. In exercise 35, show that the maximum ladder length for general a and b equals (a 2/3 + b2/3 )3/2 . 37. In exercise 35, suppose that a = 5 and the ladder is 8 ft long. Find the minimum value of b such that the ladder can turn the corner. 38. Solve exercise 37 for a general a and ladder length L. 39. A company’s revenue for selling x (thousand) items is given 35x − x 2 . Find the value of x that maximizes the by R(x) = 2 x + 35 revenue and find the maximum revenue. 40. The function in exercise 39 has a special form. For any positive cx − x 2 constant c, find x to maximize R(x) = 2 . x +c 41. In t hours, a worker makes Q(t) = −t 3 + 12t 2 + 60t items. Graph Q (t) and explain why it can be interpreted as the efficiency of the worker. Find the time at which the worker’s efficiency is maximum.

46. For your favorite sport in which it is important to throw or hit a ball a long way, explain the result of exercise 45 in the language of your sport. 47. A ball is thrown from s = b to s = a (where a < b) with initial speed v0 . Assuming that air resistance is proportional to speed, the time it takes the ball to reach s = a is b−a 1 , T = − ln 1 − c c v0 where c is a constant of proportionality. A baseball player is 300 ft from home plate and throws a ball directly toward home plate with an initial speed of 125 ft/s. Suppose that c = 0.1. How long does it take the ball to reach home plate? Another player standing x feet from home plate has the option of catching the ball and then, after a delay of 0.1 s, relaying the ball

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toward home plate with an initial speed of 125 ft/s. Find x to minimize the total time for the ball to reach home plate. Is the straight throw or the relay faster? What, if anything, changes if the delay is 0.2 s instead of 0.1 s?

3-80

imperfections in the dimensions of the cask would have little effect on the volume. Explain why V (y) = 0 means that small variations in y would convert to small errors in the volume V .

48. For the situation in exercise 47, for what length delay is it equally fast to have a relay and not have a relay? Do you think that you could catch and throw a ball in such a short time? Why do you think it is considered important to have a relay option in baseball?

z 2x

FIGURE a

49. Repeat exercises 47 and 48 if the second player throws the ball with initial speed 100 ft/s. 50. For a delay of 0.1 s in exercise 47, find the value of the initial speed of the second player’s throw for which it is equally fast to have a relay and not have a relay. x2 y2 + 2 = 1 defines an ellipse with 2 a b −a ≤ x ≤ a and −b ≤ y ≤ b. The area enclosed by the ellipse equals πab. Find the maximum area of a rectangle inscribed in the ellipse (that is, a rectangle with sides parallel to the x-axis and y-axis and vertices on the ellipse). Show that the ratio of the maximum inscribed area to the area of the ellipse to the area of the circumscribed rectangle is 1 : π2 : 2.

z

53. Find the maximum area of an isosceles triangle of given perimeter p. [Hint: Use Heron’s formula for the area of a √ triangle of sides a, b and c : A = s(s − a)(s − b)(s − c), where s = 12 (a + b + c).]

EXPLORATORY EXERCISES 1. In exploratory exercise 2 in section 3.3, you did a preliminary investigation of Kepler’s wine cask √ problem. You showed that a height-to-diameter ratio (x/y) of 2 for a cylindrical barrel will maximize the volume (see Figure a). However, real wine casks are bowed out (like beer kegs). Kepler continued his investigation of wine cask construction by approximating a cask with the straight-sided barrel in Figure b. It can be shown (we told you Kepler was good!) that the√volume of this barrel is V = 23 π[y 2 + (w − y)2 + y(w − y)] z 2 − w 2 . Treating w and z as constants, show that V (y) = 0 if y = w/2. Recall that such a critical point can correspond to a maximum or minimum of V (y), but it also could correspond to something else (e.g., an inflection point). To discover which one we have here, redraw Figure b to scale (show the correct relationship between 2y and w). In physical terms (think about increasing and decreasing y), argue that this critical point is neither a maximum nor minimum. Interestingly enough, such a nonextreme critical point would have a definite advantage to the Austrian vintners. Recall that their goal was to convert the measurement z into an estimate of the volume. The vintners would hope that small

w

2y

FIGURE b

51. The equation

52. Show that the maximum volume enclosed by a right circular cylinder inscribed in a sphere equals √13 times the volume of the sphere.

2y

2. The following problem is fictitious, but involves the kind of ambiguity that can make technical jobs challenging. The Band Candy Company decides to liquidate one of its candies. The company has 600,000 bags in inventory that it wants to sell. The candy had cost 35 cents per bag to manufacture and originally sold for 90 cents per bag. A marketing study indicates that if the candy is priced at p cents per bag, approximately Q( p) = − p 2 + 40 p + 250 thousand bags will be sold. Your task as consultant is to recommend the best selling price for the candy. As such, you should do the following: (a) find p to maximize Q( p); (b) find p to maximize 10 p Q( p), which is the actual revenue brought in by selling the candy. Then form your opinion, based on your evaluation of the relative importance of getting rid of as much candy as possible and making the most money possible. 3. A wonderful article by Timothy J. Pennings in the May 2003 issue of The College Mathematics Journal asks the question, “Do Dogs Know Calculus?” A slightly simplified version of exercises 17–22 can be solved in a very general form. Suppose that a ball is thrown from point A on the edge of the water and lands at point B, which is x meters into the water and z meters downshore from point A. (See the diagram.) At what point C should a dog enter the water to minimize the time to reach the ball? Assume that the dog’s running speed is r m/s and the dog’s swimming speed is s m/s. Find y as a function of x to minimize the time to reach the ball. Show that the answer is independent of z! Explain why your solution is invalid if r ≤ s and explain what the dog should do in this case. Dr. Pennings’ dog Elvis was clocked at r = 6.4 m/s and s = 0.9 m/s. Show that Elvis should follow the rule y = 0.144x. In fact, Elvis’ actual entry points are very close to these values for a variety of throws! B x A

C z

y

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3.8

..

Related Rates

321

RELATED RATES In this section, we present a group of problems known as related rates problems. The common thread in each problem is an equation relating two or more quantities that are all changing with time. In each case, we will use the chain rule to find derivatives of all terms in the equation (much as we did in section 2.8 with implicit differentiation). The differentiated equation allows us to determine how different derivatives (rates) are related.

EXAMPLE 8.1

A Related Rates Problem

An oil tanker has an accident and oil pours out at the rate of 150 gallons per minute. 1 Suppose that the oil spreads onto the water in a circle at a thickness of 10 (see Figure 3.94). Given that 1 ft3 equals 7.5 gallons, determine the rate at which the radius of the spill is increasing when the radius reaches 500 feet. FIGURE 3.94

Solution Since the area of a circle of radius r is πr 2 , the volume of oil is given by

Oil spill

V = (depth)(area) = since the depth is

1 10

=

1 πr 2 , 120

1 120

ft. Both volume and radius are functions of time, so π V (t) = [r (t)]2 . 120 Differentiating both sides of the equation with respect to t, we get π V (t) = 2r (t)r (t). 120 We are given a radius of 500 feet. The volume increases at a rate of 150 gallons per minute, or 150 = 20 ft3 /min. Substituting in V (t) = 20 and r = 500, we have 7.5 π 2(500)r (t). 120 Finally, solving for r (t), we find that the radius is increasing at the rate of 2.4 ≈ 0.76394 feet per minute. π 20 =

Although the details change from problem to problem, the general pattern of solution is the same for all related rates problems. Looking back, you should be able to identify each of the following steps in example 8.1.

1. 2. 3. 4. 5.

Make a simple sketch, if appropriate. Set up an equation relating all of the relevant quantities. Differentiate (implicitly) both sides of the equation with respect to time (t). Substitute in values for all known quantities and derivatives. Solve for the remaining rate.

EXAMPLE 8.2

A Sliding Ladder

A 10-foot ladder leans against the side of a building. If the top of the ladder begins to slide down the wall at the rate of 2 ft/sec, how fast is the bottom of the ladder sliding away from the wall when the top of the ladder is 8 feet off the ground?

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y

10 x

FIGURE 3.95

3-82

Solution First, we make a sketch of the problem, as seen in Figure 3.95. We have denoted the height of the top of the ladder as y and the distance from the wall to the bottom of the ladder as x. Since the ladder is sliding down the wall at the rate of 2 ft/sec, dy we must have that = −2. (Note the minus sign here.) Observe that both x and y are dt functions of time, t. We can relate the variables by observing that, since the ladder is 10 feet long, the Pythagorean Theorem gives us

Sliding ladder

[x(t)]2 + [y(t)]2 = 100. Differentiating both sides of this equation with respect to time gives us d d (100) = [x(t)]2 + [y(t)]2 dt dt = 2x(t)x (t) + 2y(t)y (t).

0=

We can solve for x (t), to obtain x (t) = −

y(t) y (t). x(t)

To make use of this, we need values for x(t), y(t) and y (t) at the point in question. Since we know that the height above ground of the top of the ladder is 8 feet, we have y = 8 and from the Pythagorean Theorem, we get 100 = x 2 + 82 , so that x = 6. We now have that at the point in question, x (t) = −

y(t) 8 8 y (t) = − (−2) = . x(t) 6 3

So, the bottom of the ladder is sliding away from the building at the rate of

EXAMPLE 8.3 y

50

x 40

FIGURE 3.96 Cars approaching an intersection

8 3

ft/sec.

Another Related Rates Problem

A car is traveling at 50 mph due south at a point 12 mile north of an intersection. A police car is traveling at 40 mph due west at a point 14 mile east of the same intersection. At that instant, the radar in the police car measures the rate at which the distance between the two cars is changing. What does the radar gun register? Solution First, we sketch a picture and denote the vertical distance of the first car from the center of the intersection y and the horizontal distance of the police car x dx (see Figure 3.96). Notice that this says that = −40, since the police car is moving dt dy in the direction of the negative x-axis and = −50, since the other car is moving in dt the direction of the negative y-axis. From the Pythagorean Theorem, the distance between the two cars is d = x 2 + y 2 . Since all quantities are changing with time, our equation is d(t) = [x(t)]2 + [y(t)]2 = {[x(t)]2 + [y(t)]2 }1/2 .

3-83

SECTION 3.8

..

Related Rates

323

Differentiating both sides with respect to t, we have by the chain rule that 1 {[x(t)]2 + [y(t)]2 }−1/2 2[x(t)x (t) + y(t)y (t)] 2 x(t)x (t) + y(t)y (t) = . [x(t)]2 + [y(t)]2

d (t) =

Substituting in x(t) = 14 , x (t) = −40, y(t) = d (t) =

1 (−40) 4

1 4

1 2

+ 12 (−50) +

1 16

and y (t) = −50, we have −140 = √ ≈ −62.6, 5

so that the radar gun registers 62.6 mph. Note that this is a poor estimate of the car’s actual speed. For this reason, police nearly always take radar measurements from a stationary position. In some problems, the variables are not related by a geometric formula, in which case you will not need to follow the first two steps of our outline. In example 8.4, the third step is complicated by the lack of a given value for one of the rates of change.

EXAMPLE 8.4

Estimating a Rate of Change in Economics

A small company estimates that when it spends x thousand dollars for advertising in a year, its annual sales will be described by s = 60 − 40e−0.05x thousand dollars. The four most recent annual advertising totals are given in the following table. Year Dollars

1 14,500

2 16,000

3 18,000

4 20,000

Estimate the current (year 4) value of x (t) and the current rate of change of sales. Solution From the table, we see that the recent trend is for advertising to increase by $2000 per year. A good estimate is then x (4) ≈ 2. Starting with the sales equation s(t) = 60 − 40e−0.05x(t) , we use the chain rule to obtain s (t) = −40e−0.05x(t) [−0.05x (t)] = 2x (t)e−0.05x(t) . Using our estimate that x (4) ≈ 2 and since x(4) = 20, we get s (4) ≈ 2(2)e−1 ≈ 1.472. Thus, sales are increasing at the rate of approximately $1472 per year. Notice that example 8.5 is similar to example 8.6 of section 2.8.

EXAMPLE 8.5

Tracking a Fast Jet

A spectator at an air show is trying to follow the flight of a jet. The jet follows a straight path in front of the observer at 540 mph. At its closest approach, the jet passes 600 feet in front of the person. Find the maximum rate of change of the angle between the spectator’s line of sight and a line perpendicular to the flight path, as the jet flies by.

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y

3-84

Solution Place the spectator at the origin (0, 0) and the jet’s path left to right on the line y = 600, and call the angle between the positive y-axis and the line of sight θ (see Figure 3.97). If we measure distance in feet and time in seconds, we first need to convert the jet’s speed to feet per second. We have 1 h ft 540 mi 5280 mi = 792 fts . = 540 mi h h 3600 s

Path of plane 600

θ

Observer

x

FIGURE 3.97 Path of jet

From triangle trigonometry (see Figure 3.97), an equation relating the angle θ with x x and y is tan θ = . Be careful with this; since we are measuring θ from the vertical, this y equation may not be what you expect. Since all quantities are changing with time, we have x(t) tan θ (t) = . y(t) Differentiating both sides with respect to time, we have [sec2 θ (t)] θ (t) =

x (t)y(t) − x(t)y (t) . [y(t)]2

With the jet moving left to right along the line y = 600, we have x (t) = 792, y(t) = 600 and y (t) = 0. Substituting these quantities, we have [sec2 θ(t)] θ (t) =

792(600) = 1.32. 6002

Solving for the rate of change θ (t), we get θ (t) =

1.32 = 1.32 cos2 θ(t). sec2 θ(t)

Observe that the rate of change is a maximum when cos2 θ(t) is a maximum. Since the maximum of the cosine function is 1, the maximum value of cos2 θ (t) is 1, occurring when θ = 0. We conclude that the maximum rate of angle change is 1.32 radians/second. This occurs when θ = 0, that is, when the jet reaches its closest point to the observer. (Think about this; it should match your intuition!) Since humans can track objects at up to about 3 radians/second, this means that we can visually follow even a fast jet at a very small distance.

EXERCISES 3.8 WRITING EXERCISES 1. As you read examples 8.1–8.3, to what extent do you find the pictures helpful? In particular, would it be clear what x and y represent in example 8.3 without a sketch? Also, in example 8.3 explain why the derivatives x (t), y (t) and d (t) are all negative. Does the sketch help in this explanation?

1 ft3 equals 7.5 gallons, determine the rate at which the radius of the spill is increasing when the radius reaches (a) 100 ft and (b) 200 ft. Explain why the rate decreases as the radius increases.

2. In example 8.4, the increase in advertising dollars from year 1 to year 2 was $1500. Explain why this amount is not especially relevant to the approximation of s (4).

2. Oil spills out of a tanker at the rate of 90 gallons per minute. The oil spreads in a circle with a thickness of 18 . Determine the rate at which the radius of the spill is increasing when the radius reaches 100 feet.

1. Oil spills out of a tanker at the rate of 120 gallons per minute. The oil spreads in a circle with a thickness of 14 . Given that

3. Oil spills out of a tanker at the rate of g gallons per minute. The oil spreads in a circle with a thickness of 14 . Given that the radius of the spill is increasing at a rate of

3-85

SECTION 3.8

..

Related Rates

325

0.6 ft/min when the radius equals 100 feet, determine the value of g.

15. Rework example 8.3 if the police car is not moving. Does this make the radar gun’s measurement more accurate?

4. In exercises 1–3 and example 8.1, if the thickness of the oil is doubled, how does the rate of increase of the radius change?

16. Show that the radar gun of example 8.3 gives the correct speed if the police car is located at the origin.

5. Assume that the infected area of an injury is circular. If the radius of the infected area is 3 mm and growing at a rate of 1 mm/hr, at what rate is the infected area increasing? 6. For the injury of exercise 5, find the rate of increase of the infected area when the radius reaches 6 mm. Explain in commonsense terms why this rate is larger than that of exercise 5. 7. Suppose that a raindrop evaporates in such a way that it maintains a spherical shape. Given that the volume of a sphere of radius r is V = 43 πr 3 and its surface area is A = 4πr 2 , if the radius changes in time, show that V = Ar . If the rate of evaporation (V ) is proportional to the surface area, show that the radius changes at a constant rate. 8. Suppose a forest fire spreads in a circle with radius changing at a rate of 5 feet per minute. When the radius reaches 200 feet, at what rate is the area of the burning region increasing? 9. A 10-foot ladder leans against the side of a building as in example 8.2. If the bottom of the ladder is pulled away from the wall at the rate of 3 ft/s and the ladder remains in contact with the wall, find the rate at which the top of the ladder is dropping when the bottom is 6 feet from the wall. 10. In exercise 9, find the rate at which the angle between the ladder and the horizontal is changing when the bottom of the ladder is 6 feet from the wall. 11. Two buildings of height 20 feet and 40 feet, respectively, are 60 feet apart. Suppose that the intensity of light at a point between the buildings is proportional to the angle θ in the figure. If a person is moving from right to left at 4 ft/s, at what rate is θ changing when the person is exactly halfway between the two buildings?

20'

40'

θ

17. Show that the radar gun of example 8.3 gives the correct speed if the police car is at x = 12 moving at a speed of √ ( 2 − 1) 50 mph. 18. Find a position and speed for which the radar gun of example 8.3 has a slower reading than the actual speed. 19. Suppose that the average yearly cost per item for producing x items of a business product is C(x) = 10 + 100 . If the curx rent production is x = 10 and production is increasing at a rate of 2 items per year, find the rate of change of the average cost. 20. Suppose that the average yearly cost per item for producing x items of a business product is C(x) = 12 + 94 . The three most x recent yearly production figures are given in the table. Year Prod. (x)

0 8.2

1 8.8

2 9.4

Estimate the value of x (2) and the current (year 2) rate of change of the average cost. 21. For a small company spending $x thousand per year in advertising, suppose that annual sales in thousands of dollars equal s = 60 − 40e−0.05x . The three most recent yearly advertising figures are given in the table. Year Adver.

0 16,000

1 18,000

2 20,000

Estimate the value of x (2) and the current (year 2) rate of change of sales. 22. For a small company spending $x thousand per year in advertising, suppose that annual sales in thousands of dollars equal s = 80 − 20e−0.04x . If the current advertising budget is x = 40 and the budget is increasing at a rate of $1500 per year, find the rate of change of sales.

60'

12. Find the location in exercise 11 where the angle θ is maximum. 13. A plane is located x = 40 miles (horizontally) away from an airport at an altitude of h miles. Radar at the airport detects that the distance s(t) between the plane and airport is changing at the rate of s (t) = −240 mph. If the plane flies toward the airport at the constant altitude h = 4, what is the speed |x (t)| of the airplane? 14. Repeat exercise 13 with a height of 6 miles. Based on your answers, how important is it to know the actual height of the airplane?

23. A baseball player stands 2 feet from home plate and watches a pitch fly by. In the diagram, x is the distance from the ball to home plate and θ is the angle indicating the direction of the player’s gaze. Find the rate θ at which his eyes must move to watch a fastball with x (t) = −130 ft/s as it crosses home plate at x = 0. x

Plate u

2 Player

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Applications of Differentiation

24. In the situation of exercise 23, humans can maintain focus only when θ ≤ 3 (see Watts and Bahill’s book Keep Your Eye on the Ball). Find the fastest pitch that you could actually watch cross home plate. 25. A camera tracks the launch of a vertically ascending spacecraft. The camera is located at ground level 2 miles from the launchpad. If the spacecraft is 3 miles up and traveling at 0.2 mile per second, at what rate is the camera angle (measured from the horizontal) changing? 26. Repeat exercise 25 for the spacecraft at 1 mile up (assume the same velocity). Which rate is higher? Explain in commonsense terms why it is larger. 27. Suppose a 6-ft-tall person is 12 ft away from an 18-ft-tall lamppost (see the figure). If the person is moving away from the lamppost at a rate of 2 ft/s, at what rate is the length of the x +s s shadow changing? Hint: Show that = . 18 6

3-86

constant rate of 5 m3 /s, at what rate is the height of the pile increasing when the height is 2 meters? 33. The frequency at which a guitar string vibrates (which determines the pitch of the note we hear) is related to the tension T to which the string is tightened, the density ρ of the string and the effective length L of the string by the equation 1 T . By running his finger along a string, a guitarist 2L ρ can change L by changing the distance between the bridge f =

T = 220 ft/s so ρ that the units of f are Hertz (cycles per second). If the guitarist’s hand slides so that L (t) = −4, find f (t). At this rate, how long will it take to raise the pitch one octave (that is, double f )? and his finger. Suppose that L =

1 2

ft and

18 ft

6 ft

x

s

Exercise 27 28. Rework exercise 27 if the person is 6 ft away from the lamppost and is walking toward the lamppost at a rate of 3 ft/s. 29. Boyle’s law for a gas at constant temperature is PV = c, where P is pressure, V is volume and c is a constant. Assume that both P and V are functions of time. Show that P (t)/V (t) = −c/V 2 . 30. In exercise 29, solve for P as a function of V . Treating V as an independent variable, compute P (V ). Compare P (V ) and P (t)/V (t) from exercise 29. 31. A dock is 6 feet above water. Suppose you stand on the edge of the dock and pull a rope attached to a boat at the constant rate of 2 ft/s. Assume that the boat remains at water level. At what speed is the boat approaching the dock when it is 20 feet from the dock? 10 feet from the dock? Isn’t it surprising that the boat’s speed is not constant? 32. Sand is poured into a conical pile with the height of the pile equalling the diameter of the pile. If the sand is poured at a

34. Suppose that you are blowing up a balloon by adding air at the rate of 1 ft3 /s. If the balloon maintains a spherical shape, the volume and radius are related by V = 43 πr 3 . Compare the rate at which the radius is changing when r = 0.01 ft versus when r = 0.1 ft. Discuss how this matches the experience of a person blowing up a balloon. 35. Water is being pumped into a spherical tank of radius 60 feet at the constant rate of 10 ft3 /s. Find the rate at which the radius of the top level of water in the tank changes when the tank is half full. 36. For the water tank in exercise 35, find the height at which the height of the water in the tank changes at the same rate as the radius. 37. Sand is dumped such that the shape of the sandpile remains a cone with height equal to twice the radius. If the sand is dumped at the constant rate of 20 ft3 /s, find the rate at which the radius is increasing when the height reaches 6 feet. 38. Repeat exercise 37 for a sandpile for which the edge of the sandpile forms an angle of 45◦ with the horizontal. 39. To start skating, you must angle your foot and push off the ice. Alain Hach´e’s The Physics of Hockey derives the relationship between the skate angle θ, the sideways stride distance s, the stroke period 2s T and the forward speed v of the skater, with θ = tan−1 vT . For T = 1 second, s = 60 cm and an acceleration of 1 m/s2 , find the rate of change of the angle θ when the

3-87

SECTION 3.9

skater reaches (a) 1 m/s and (b) 2 m/s. Interpret the sign and size of θ in terms of skating technique. v

s

θ

push

..

Rates of Change in Economics and the Sciences

327

are analyzed here. In the diagram below, a camera follows an object directly from left to right. If the camera is at the origin, the object moves with speed 1 m/s and the line of motion is at y = c, find an expression for θ as a function of the position of the object. In the diagram to the right, the camera looks down into a parabolic mirror and indirectly views the object. If the mirror has polar coordinates (in this case, the angle θ 1 − sin θ is measured from the horizontal) equation r = and 2 cos2 θ x = r cos θ, find an expression for θ as a function of the position of the object. Compare values of θ at x = 0 and other x-values. If a large value of θ causes the image to blur, which camera system is better? Does the distance y = c affect your preference? (x, y)

EXPLORATORY EXERCISES

(x, y)

1. A thin rectangular advertising sign rotates clockwise at the rate of 6 revolutions per minute. The sign is 4 feet wide and an observer stands 40 feet away. Find the rate of change of the viewing angle θ as a function of the sign angle A. At which position of the sign is the rate of change maximum?

θ

θ

3. A particle moves down a ramp subject only to the force of gravity. Let y0 be the maximum height of the particle. Then conservation of energy gives A θ

2. Vision has proved to be the biggest challenge for building functional robots. Robot vision can either be designed to mimic human vision or follow a different design. Two possibilities

3.9

1 2 mv + mgy = mgy0 2 (a) From the definition v(t) = [x (t)]2 + [y (t)]2 , conclude that |y (t)| ≤ |v(t)|. (b) Show that |v (t)| ≤ g. (c) What shape must the ramp have to get equality in part (b)? Briefly explain in physical terms why g is the maximum value of |v (t)|.

RATES OF CHANGE IN ECONOMICS AND THE SCIENCES It has often been said that mathematics is the language of nature. Today, the concepts of calculus are being applied in virtually every field of human endeavor. The applications in this section represent but a small sampling of some elementary uses of the derivative. These are not all of the uses of the derivative nor are they necessarily the most important uses, but rather, represent some interesting applications in a variety of fields. Recall that the derivative of a function gives the instantaneous rate of change of that function. So, when you see the word rate, you should be thinking derivative. You can hardly pick up a newspaper without finding reference to some rates (e.g., inflation rate, interest rate, unemployment rate, etc.). These can be thought of as derivatives. There are also many quantities with which you are familiar, but that you might not recognize as rates of change. Our first example, which comes from economics, is of this type.

3-87

SECTION 3.9

skater reaches (a) 1 m/s and (b) 2 m/s. Interpret the sign and size of θ in terms of skating technique. v

s

θ

push

..

Rates of Change in Economics and the Sciences

327

are analyzed here. In the diagram below, a camera follows an object directly from left to right. If the camera is at the origin, the object moves with speed 1 m/s and the line of motion is at y = c, find an expression for θ as a function of the position of the object. In the diagram to the right, the camera looks down into a parabolic mirror and indirectly views the object. If the mirror has polar coordinates (in this case, the angle θ 1 − sin θ is measured from the horizontal) equation r = and 2 cos2 θ x = r cos θ, find an expression for θ as a function of the position of the object. Compare values of θ at x = 0 and other x-values. If a large value of θ causes the image to blur, which camera system is better? Does the distance y = c affect your preference? (x, y)

EXPLORATORY EXERCISES

(x, y)

1. A thin rectangular advertising sign rotates clockwise at the rate of 6 revolutions per minute. The sign is 4 feet wide and an observer stands 40 feet away. Find the rate of change of the viewing angle θ as a function of the sign angle A. At which position of the sign is the rate of change maximum?

θ

θ

3. A particle moves down a ramp subject only to the force of gravity. Let y0 be the maximum height of the particle. Then conservation of energy gives A θ

2. Vision has proved to be the biggest challenge for building functional robots. Robot vision can either be designed to mimic human vision or follow a different design. Two possibilities

3.9

1 2 mv + mgy = mgy0 2 (a) From the definition v(t) = [x (t)]2 + [y (t)]2 , conclude that |y (t)| ≤ |v(t)|. (b) Show that |v (t)| ≤ g. (c) What shape must the ramp have to get equality in part (b)? Briefly explain in physical terms why g is the maximum value of |v (t)|.

RATES OF CHANGE IN ECONOMICS AND THE SCIENCES It has often been said that mathematics is the language of nature. Today, the concepts of calculus are being applied in virtually every field of human endeavor. The applications in this section represent but a small sampling of some elementary uses of the derivative. These are not all of the uses of the derivative nor are they necessarily the most important uses, but rather, represent some interesting applications in a variety of fields. Recall that the derivative of a function gives the instantaneous rate of change of that function. So, when you see the word rate, you should be thinking derivative. You can hardly pick up a newspaper without finding reference to some rates (e.g., inflation rate, interest rate, unemployment rate, etc.). These can be thought of as derivatives. There are also many quantities with which you are familiar, but that you might not recognize as rates of change. Our first example, which comes from economics, is of this type.

328

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3-88

In economics, the term marginal is used to indicate a rate. Thus, marginal cost is the derivative of the cost function, marginal profit is the derivative of the profit function and so on. We introduce marginal cost in some detail here, with further applications given in the exercises. Suppose that you are manufacturing an item, where your start-up costs are $4000 and productions costs are $2 per item. The total cost of producing x items would then be 4000 + 2x. Of course, the assumption that the cost per item is constant is unrealistic. Efficient mass-production techniques could reduce the cost per item, but machine maintenance, labor, plant expansion and other factors could drive costs up as production (x) increases. In example 9.1, a quadratic cost function is used to take into account some of these extra factors. In practice, you would find a cost function by making some observations of the cost of producing a number of different quantities and then fitting the data to the graph of a known function. (This is one way in which the calculus is brought to bear on real-world problems.) When the cost per item is not constant, an important question for managers to answer is how much it will cost to increase production. This is the idea behind marginal cost.

EXAMPLE 9.1

Analyzing the Marginal Cost of Producing a Commercial Product

Suppose that C(x) = 0.02x 2 + 2x + 4000 is the total cost (in dollars) for a company to produce x units of a certain product. Compute the marginal cost at x = 100 and compare this to the actual cost of producing the 100th unit. Solution The marginal cost function is the derivative of the cost function: C (x) = 0.04x + 2 and so, the marginal cost at x = 100 is C (100) = 4 + 2 = 6 dollars per unit. On the other hand, the actual cost of producing item number 100 would be C(100) − C(99). (Why?) We have C(100) − C(99) = 200 + 200 + 4000 − (196.02 + 198 + 4000) = 4400 − 4394.02 = 5.98 dollars. Note that this is very close to the marginal cost of $6. Also notice that the marginal cost is easier to compute. Another quantity that businesses use to analyze production is average cost. You can easily remember the formula for average cost by thinking of an example. If it costs a total of $120 to produce 12 items, then the average cost would be $10 $ 120 per item. In 12 general, the total cost is given by C(x) and the number of items by x, so average cost is defined by C(x) =

C(x) . x

A business manager would want to know the level of production that minimizes average cost.

3-89

SECTION 3.9

EXAMPLE 9.2

..

Rates of Change in Economics and the Sciences

329

Minimizing the Average Cost of Producing a Commercial Product C(x) = 0.02x 2 + 2x + 4000

Suppose that

is t