Expand your learning experience with the
Tools for Enriching Calculus CD-ROM The Tools for Enriching Calculus CD-ROM is the ideal complement to Essential Calculus. This innovative learning tool uses a discovery and exploratory approach to help you explore calculus in new ways. Visuals and Modules on the CD-ROM provide geometric visualizations and graphical applications to enrich your understanding of major concepts. Exercises and examples, built from the content in the applets, take a discovery approach, allowing you to explore open-ended questions about the way certain mathematical objects behave. The CD-ROM’s simulation modules include audio explanations of the concept, along with exercises, examples, and instructions. Tools for Enriching Calculus also contains Homework Hints for representative exercises from the text (indicated in blue in the text).
Hours of interactive video instruction!
Interactive Video Skillbuilder CD-ROM The Interactive Video Skillbuilder CD-ROM contains more than eight hours of video instruction.The problems worked during each video lesson are shown first so that you can try working them before watching the solution.To help you evaluate your progress, each section of the text contains a ten-question web quiz (the results of which can be e-mailed to your instructor), and each chapter contains a chapter test, with answers to every problem. Icons in the text direct you to examples that are worked out on the CD-ROM. If you would like to purchase these resources, visit www.cengage.com/highered
REFERENCE PAGES
Cut here and keep for reference
ALGEBRA
G E O M E T RY
ARITHMETIC OPERATIONS
GEOMETRIC FORMULAS
a c ad bc b d bd a b a d ad c b c bc d
ab c ab ac a c ac b b b
Formulas for area A, circumference C, and volume V: Triangle
Circle
Sector of Circle
A 12 bh 12 ab sin
A r 2 C 2 r
A 12 r 2 s r in radians
a
EXPONENTS AND RADICALS xm x mn xn 1 xn n x
x m x n x mn x x m n
mn
n
x y
xyn x n y n
n
n n n xy s xs y s
r
s
¨
b
r
xn yn
Sphere V 43 r 3 A 4 r 2
n n x mn s x m (s x )m
n x 1n s x
r
h
¨
n x x s n y sy
Cylinder V r 2h
Cone V 13 r 2h
r r
h
h
FACTORING SPECIAL POLYNOMIALS
r
x 2 y 2 x yx y x 3 y 3 x yx 2 xy y 2 x 3 y 3 x yx 2 xy y 2
DISTANCE AND MIDPOINT FORMULAS BINOMIAL THEOREM
Distance between P1x1, y1 and P2x 2, y2:
x y2 x 2 2xy y 2
x y2 x 2 2xy y 2
d sx 2 x12 y2 y12
x y3 x 3 3x 2 y 3xy 2 y 3 x y3 x 3 3x 2 y 3xy 2 y 3 x yn x n nx n1y
nn 1 n2 2 x y 2
Midpoint of P1 P2 :
n nk k x y nxy n1 y n k
x1 x 2 y1 y2 , 2 2
LINES
n nn 1 n k 1 where k 1 2 3 k
Slope of line through P1x1, y1 and P2x 2, y2: m
QUADRATIC FORMULA If ax 2 bx c 0, then x
b sb 2 4ac . 2a
y2 y1 x 2 x1
Point-slope equation of line through P1x1, y1 with slope m: y y1 mx x1
INEQUALITIES AND ABSOLUTE VALUE If a b and b c, then a c.
Slope-intercept equation of line with slope m and y-intercept b:
If a b, then a c b c.
y mx b
If a b and c 0, then ca cb. If a b and c 0, then ca cb. If a 0, then
x a x a x a
means
x a or
CIRCLES
x a
Equation of the circle with center h, k and radius r:
means a x a means
x a or
x h2 y k2 r 2
x a 1
REFERENCE PAGES T R I G O N O M E T RY ANGLE MEASUREMENT
FUNDAMENTAL IDENTITIES
radians 180 1
rad 180
180
1 rad
s
r r
in radians
RIGHT ANGLE TRIGONOMETRY
cos tan
hyp csc opp
adj hyp
sec
opp adj
cot
1 sin
sec
1 cos
tan
sin cos
cot
cos sin
cot
1 tan
sin 2 cos 2 1
¨
s r
opp sin hyp
csc
hyp
hyp adj
opp
¨ adj
1 tan 2 sec 2
1 cot 2 csc 2
sin sin
cos cos
tan tan
sin
cos 2
tan
cot 2
adj opp
cos
sin 2
TRIGONOMETRIC FUNCTIONS sin
y r
csc
cos
x r
sec
r x
tan
y x
cot
x y
y
sin A sin B sin C a b c
(x, y)
C c
¨
THE LAW OF COSINES x
y y=sin x
y
b
a 2 b 2 c 2 2bc cos A b 2 a 2 c 2 2ac cos B c 2 a 2 b 2 2ab cos C
y=tan x
A
y=cos x
1
1 π
a
r
GRAPHS OF THE TRIGONOMETRIC FUNCTIONS y
B
THE LAW OF SINES
r y
2π
ADDITION AND SUBTRACTION FORMULAS
2π x
_1
π
2π x
sinx y sin x cos y cos x sin y
x
π
sinx y sin x cos y cos x sin y
_1
cosx y cos x cos y sin x sin y y
y=csc x
y
y=sec x
y
cosx y cos x cos y sin x sin y
y=cot x
1
1 π
2π x
π
π
2π x
2π x
tanx y
tan x tan y 1 tan x tan y
tanx y
tan x tan y 1 tan x tan y
_1
_1
DOUBLE-ANGLE FORMULAS sin 2x 2 sin x cos x
TRIGONOMETRIC FUNCTIONS OF IMPORTANT ANGLES
radians
sin
cos
tan
0 30 45 60 90
0 6 4 3 2
0 12 s22 s32 1
1 s32 s22 12 0
0 s33 1 s3 —
cos 2x cos 2x sin 2x 2 cos 2x 1 1 2 sin 2x tan 2x
2 tan x 1 tan2x
HALF-ANGLE FORMULAS sin 2x
2
1 cos 2x 2
cos 2x
1 cos 2x 2
ESSENTIAL CALCULUS EARLY TRANSCENDENTALS JAMES STEWART McMaster University and University of Toronto
Australia . Brazil . Japan . Korea . Mexico . Singapore . Spain . United Kingdom . United States
Essential Calculus: Early Transcendentals James Stewart Publisher: Bob Pirtle Assistant Editor: Stacy Green Editorial Assistant: Magnolia Molcan Technology Project Manager: Earl Perry Senior Marketing Manager: Karin Sandberg Marketing Communications Manager: Darlene Amidon-Brent Project Manager, Editorial Production: Cheryll Linthicum Creative Director: Rob Hugel Senior Art Director: Vernon T. Boes
© 2011 Brooks/Cole, Cengage Learning ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without the prior written permission of the publisher. For product information and technology assistance, contact us at Cengage Learning Customer & Sales Support, 1-800-354-9706 For permission to use material from this text or product, submit all requests online at www.cengage.com/permissions Further permissions questions can be emailed to
[email protected]
Senior Print/Media Buyer: Rebecca Cross Permissions Editor: Bob Kaueser
Library of Congress Control Number: 2009939223
Production Service: TECH· arts
ISBN-13: 978-0-538-49739-8
Text Designer: Stephanie Kuhns, Kathi Townes
ISBN-10: 0-538-49739-4
Copy Editor: Kathi Townes Illustrators: Brian Betsill, Stephanie Kuhns Cover Designer: William Stanton Cover Image: Terry Why/IndexStock Imagery Compositor: TECH· arts
Brooks/Cole 10 Davis Drive Belmont, CA 94002-3098 USA Cengage Learning is a leading provider of customized learning solutions with office locations around the globe, including Singapore, the United Kingdom, Australia, Mexico, Brazil and Japan. Locate your local office at: www.cengage.com/global Cengage Learning products are represented in Canada by Nelson Education, Ltd. To learn more about Brooks/Cole, visit www.cengage.com/brookscole Purchase any of our products at your local college store or at our preferred online store www.ichapters.com
Printed in Canada 1 2 3 4 5 6 7 8 9 1 0 1 3 1 2 11 1 0 0 9
K08T09
Trademarks Derive is a registered trademark of Soft Warehouse, Inc. Maple is a registered trademark of Waterloo Maple, Inc. Mathematica is a registered trademark of Wolfram Research, Inc. Tools for Enriching is a trademark used herein under license.
CONTENTS
1
FUNCTIONS AND LIMITS 1.1 1.2 1.3 1.4 1.5 1.6
2
DERIVATIVES 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8
3
1
Functions and Their Representations 1 A Catalog of Essential Functions 10 The Limit of a Function 24 Calculating Limits 35 Continuity 45 Limits Involving Infinity 56 Review 69
73 Derivatives and Rates of Change 73 The Derivative as a Function 83 Basic Differentiation Formulas 94 The Product and Quotient Rules 106 The Chain Rule 113 Implicit Differentiation 121 Related Rates 127 Linear Approximations and Differentials Review 138
133
INVERSE FUNCTIONS: Exponential, Logarithmic, and Inverse Trigonometric Functions 142 3.1 3.2 3.3 3.4 3.5 3.6 3.7
Exponential Functions 142 Inverse Functions and Logarithms 148 Derivatives of Logarithmic and Exponential Functions Exponential Growth and Decay 167 Inverse Trigonometric Functions 175 Hyperbolic Functions 181 Indeterminate Forms and L’Hospital’s Rule 187 Review 195
160
iii
iv
■
CONTENTS
4
APPLICATIONS OF DIFFERENTIATION 4.1 4.2 4.3 4.4 4.5 4.6 4.7
5
INTEGRALS 5.1 5.2 5.3 5.4 5.5
6
7
251 Areas and Distances 251 The Definite Integral 262 Evaluating Definite Integrals 274 The Fundamental Theorem of Calculus The Substitution Rule 293 Review 300
SERIES
304
357
Areas between Curves 357 Volumes 362 Volumes by Cylindrical Shells 373 Arc Length 378 Applications to Physics and Engineering Differential Equations 397 Review 407
410 8.1 8.2
284
Integration by Parts 304 Trigonometric Integrals and Substitutions 310 Partial Fractions 320 Integration with Tables and Computer Algebra Systems Approximate Integration 333 Improper Integrals 345 Review 354
APPLICATIONS OF INTEGRATION 7.1 7.2 7.3 7.4 7.5 7.6
8
Maximum and Minimum Values 198 The Mean Value Theorem 205 Derivatives and the Shapes of Graphs 211 Curve Sketching 220 Optimization Problems 226 Newton’s Method 236 Antiderivatives 241 Review 247
TECHNIQUES OF INTEGRATION 6.1 6.2 6.3 6.4 6.5 6.6
198
Sequences 410 Series 420
384
328
CONTENTS
8.3 8.4 8.5 8.6 8.7 8.8
9
PARAMETRIC EQUATIONS AND POLAR COORDINATES 9.1 9.2
9.3 9.4 9.5
10
The Integral and Comparison Tests 429 Other Convergence Tests 437 Power Series 447 Representing Functions as Power Series 452 Taylor and Maclaurin Series 458 Applications of Taylor Polynomials 471 Review 479
Parametric Curves 482 Calculus with Parametric Curves 488 Polar Coordinates 496 Areas and Lengths in Polar Coordinates 504 Conic Sections in Polar Coordinates 509 Review 515
VECTORS AND THE GEOMETRY OF SPACE
517
10.1 Three-Dimensional Coordinate Systems 10.2 Vectors
517
522
10.3 The Dot Product 10.4 10.5 10.6 10.7 10.8 10.9
11
530 The Cross Product 537 Equations of Lines and Planes 545 Cylinders and Quadric Surfaces 553 Vector Functions and Space Curves 559 Arc Length and Curvature 570 Motion in Space: Velocity and Acceleration Review 587
PARTIAL DERIVATIVES
591
11.1 Functions of Several Variables 11.2 Limits and Continuity
11.5 11.6 11.7 11.8
591
601 609 Tangent Planes and Linear Approximations 617 The Chain Rule 625 Directional Derivatives and the Gradient Vector 633 Maximum and Minimum Values 644 Lagrange Multipliers 652 Review 659
11.3 Partial Derivatives 11.4
578
482
■
v
vi
■
CONTENTS
12
MULTIPLE INTEGRALS 12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8
13
Double Integrals over Rectangles 663 Double Integrals over General Regions 674 Double Integrals in Polar Coordinates 682 Applications of Double Integrals 688 Triple Integrals 693 Triple Integrals in Cylindrical Coordinates 703 Triple Integrals in Spherical Coordinates 707 Change of Variables in Multiple Integrals 713 Review 722
VECTOR CALCULUS 13.1 13.2 13.3 13.4 13.5 13.6 13.7 13.8 13.9
APPENDIXES A B C D E
INDEX
A83
663
725
Vector Fields 725 Line Integrals 731 The Fundamental Theorem for Line Integrals Green’s Theorem 751 Curl and Divergence 757 Parametric Surfaces and Their Areas 765 Surface Integrals 775 Stokes’ Theorem 786 The Divergence Theorem 791 Review 797
A1 Trigonometry A1 Proofs A10 Sigma Notation A26 The Logarithm Defined as an Integral Answers to Odd-Numbered Exercises
A31 A39
742
PREFACE
This book is a response to those instructors who feel that calculus textbooks are too big. In writing the book I asked myself: What is essential for a three-semester calculus course for scientists and engineers? The book is about two-thirds the size of my other calculus books (Calculus, Fifth Edition and Calculus, Early Transcendentals, Fifth Edition) and yet it contains almost all of the same topics. I have achieved relative brevity mainly by condensing the exposition and by putting some of the features on the website www.stewartcalculus.com. Here, in more detail are some of the ways I have reduced the bulk: ■ I have organized topics in an efficient way and rewritten some sections with briefer exposition. ■ The design saves space. In particular, chapter opening spreads and photographs have been eliminated. ■ The number of examples is slightly reduced. Additional examples are provided online. ■ The number of exercises is somewhat reduced, though most instructors will find that there are plenty. In addition, instructors have access to the archived problems on the website. ■ Although I think projects can be a very valuable experience for students, I have removed them from the book and placed them on the website. ■ A discussion of the principles of problem solving and a collection of challenging problems for each chapter have been moved to the web. Despite the reduced size of the book, there is still a modern flavor: Conceptual understanding and technology are not neglected, though they are not as prominent as in my other books.
CONTENT This book treats the exponential, logarithmic, and inverse trigonometric functions early, in Chapter 3. Those who wish to cover such functions later, with the logarithm defined as an integral, should look at my book titled simply Essential Calculus. CHAPTER 1 FUNCTIONS AND LIMITS After a brief review of the basic functions, limits and continuity are introduced, including limits of trigonometric functions, limits involving infinity, and precise definitions. ■
DERIVATIVES The material on derivatives is covered in two sections in order to give students time to get used to the idea of a derivative as a function. The CHAPTER 2
■
vii
viii
■
PREFACE
formulas for the derivatives of the sine and cosine functions are derived in the section on basic differentiation formulas. Exercises explore the meanings of derivatives in various contexts. CHAPTER 3
■
INVERSE FUNCTIONS: EXPONENTIAL, LOGARITHMIC, AND INVERSE TRIGONOMETRIC FUNCTIONS
Exponential functions are defined first and the number e is defined as a limit. Logarithms are then defined as inverse functions. Applications to exponential growth and decay follow. Inverse trigonometric functions and hyperbolic functions are also covered here. L’Hospital’s Rule is included in this chapter because limits of transcendental functions so often require it. APPLICATIONS OF DIFFERENTIATION The basic facts concerning extreme values and shapes of curves are deduced from the Mean Value Theorem. The section on curve sketching includes a brief treatment of graphing with technology. The section on optimization problems contains a brief discussion of applications to business and economics.
CHAPTER 4
■
CHAPTER 5 INTEGRALS The area problem and the distance problem serve to motivate the definite integral, with sigma notation introduced as needed. (Full coverage of sigma notation is provided in Appendix C.) A quite general definition of the definite integral (with unequal subintervals) is given initially before regular partitions are employed. Emphasis is placed on explaining the meanings of integrals in various contexts and on estimating their values from graphs and tables. ■
TECHNIQUES OF INTEGRATION All the standard methods are covered, as well as computer algebra systems, numerical methods, and improper integrals. CHAPTER 6
■
APPLICATIONS OF INTEGRATION General methods are emphasized. The goal is for students to be able to divide a quantity into small pieces, estimate with Riemann sums, and recognize the limit as an integral. The chapter concludes with an introduction to differential equations, including separable equations and direction fields.
CHAPTER 7
■
SERIES The convergence tests have intuitive justifications as well as formal proofs. The emphasis is on Taylor series and polynomials and their applications to physics. Error estimates include those based on Taylor’s Formula (with Lagrange’s form of the remainder term) and those from graphing devices.
CHAPTER 8
■
CHAPTER 9 PARAMETRIC EQUATIONS AND POLAR COORDINATES This chapter introduces parametric and polar curves and applies the methods of calculus to them. A brief treatment of conic sections in polar coordinates prepares the way for Kepler’s Laws in Chapter 10. ■
VECTORS AND THE GEOMETRY OF SPACE In addition to the material on vectors, dot and cross products, lines, planes, and surfaces, this chapter covers vectorvalued functions, length and curvature of space curves, and velocity and acceleration along space curves, culminating in Kepler’s laws. CHAPTER 10
■
CHAPTER 11 PARTIAL DERIVATIVES In view of the fact that many students have difficulty forming mental pictures of the concepts of this chapter, I’ve placed a special emphasis on graphics to elucidate such ideas as graphs, contour maps, directional derivatives, gradients, and Lagrange multipliers. ■
CHAPTER 12 MULTIPLE INTEGRALS Cylindrical and spherical coordinates are introduced in the context of evaluating triple integrals. ■
PREFACE
■
ix
VECTOR CALCULUS The similarities among the Fundamental Theorem for line integrals, Green’s Theorem, Stokes’ Theorem, and the Divergence Theorem are emphasized. CHAPTER 13
■
WEBSITE The website www.stewartcalculus.com includes the following. ■
Review of Algebra, Analytic Geometry, and Conic Sections
■
Additional Examples
■
Projects
■
Archived Problems (drill exercises that have appeared in previous editions of my other books), together with their solutions
■
Challenge Problems
■
Complex Numbers
■
Graphing Calculators and Computers
■
Lies My Calculator and Computer Told Me
■
Additional Topics (complete with exercise sets): Principles of Problem Solving, Strategy for Integration, Strategy for Testing Series, Fourier Series, Area of a Surface of Revolution, Linear Differential Equations, SecondOrder Linear Differential Equations, Nonhomogeneous Linear Equations, Applications of Second-Order Differential Equations, Using Series to Solve Differential Equations, Complex Numbers, Rotation of Axes
■
Links, for particular topics, to outside web resources
■
History of Mathematics, with links to the better historical websites
ACKNOWLEDGMENTS I thank the following reviewers for their thoughtful comments. Ulrich Albrecht, Auburn University Christopher Butler, Case Western Reserve University Joe Fisher, University of Cincinnati John Goulet, Worchester Polytechnic Institute Irvin Hentzel, Iowa State University Joel Irish, University of Southern Maine Mary Nelson, University of Colorado, Boulder Ed Slaminka, Auburn University Li (Jason) Zhongshan, Georgia State University I also thank Marv Riedesel for accuracy in proofreading and Dan Clegg for detailed discussions on how to achieve brevity. In addition, I thank Kathi Townes, Stephanie
x
■
PREFACE
Kuhns, Jenny Turney, and Brian Betsill of TECHarts for their production services and the following Brooks/Cole staff: Cheryll Linthicum, editorial production project manager; Vernon Boes, art director; Karin Sandberg and Darlene Amidon-Brent, marketing team; Earl Perry, technology project manager; Stacy Green, assistant editor; Magnolia Molcan, editorial assistant; Bob Kauser, permissions editor; Rebecca Cross, print/media buyer; and William Stanton, cover designer. They have all done an outstanding job. The idea for this book came from my editor, Bob Pirtle, who had been hearing of the desire for a much shorter calculus text from numerous instructors. I thank him for encouraging me to pursue this idea and for his advice and assistance whenever I needed it. JA M E S S T E WA RT
PREFACE
ANCILLARIES FOR INSTRUCTORS
ANCILLARIES FOR STUDENTS
COMPLETE SOLUTIONS MANUAL ISBN 0495014303
STUDENT SOLUTIONS MANUAL ISBN 049501429X
The Complete Solutions Manual provides worked-out solutions to all of the problems in the text.
The Student Solutions Manual provides completely worked-out solutions to all odd-numbered exercises within the text, giving students a way to check their answers and ensure that they took the correct steps to arrive at an answer.
SOLUTIONS BUILDER CD ISBN 0495106925
This CD is an electronic version of the complete solutions manual. It provides instructors with an efficient method for creating solution sets to homework or exams. Instructors can easily view, select, and save solution sets that can then be printed or posted. TOOLS FOR ENRICHING CALCULUS ISBN 0495107638
TEC contains Visuals and Modules for use as classroom demonstrations. Exercises for each Module allow instructors to make assignments based on the classroom demonstration. TEC also includes Homework Hints for representative exercises. Students can benefit from this additional help when instructors assign these exercises.
■
INTERACTIVE VIDEO SKILLBUILDER CD ISBN 0495113719
The Interactive Video Skillbuilder CD-ROM contains more than eight hours of instruction. To help students evaluate their progress, each section contains a tenquestion web quiz (the results of which can be e-mailed to the instructor) and each chapter contains a chapter test, with the answer to each problem on each test. TOOLS FOR ENRICHING CALCULUS ISBN 0495107638
TEC provides a laboratory environment in which students can enrich their understanding by revisiting and exploring selected topics. TEC also includes Homework Hints for representative exercises.
Ancillaries for students are available for purchase at www.cengage.com
xi
TO THE STUDENT
Reading a calculus textbook is different from reading a newspaper or a novel, or even a physics book. Don’t be discouraged if you have to read a passage more than once in order to understand it. You should have pencil and paper and calculator at hand to sketch a diagram or make a calculation. Some students start by trying their homework problems and read the text only if they get stuck on an exercise. I suggest that a far better plan is to read and understand a section of the text before attempting the exercises. In particular, you should look at the definitions to see the exact meanings of the terms. And before you read each example, I suggest that you cover up the solution and try solving the problem yourself. You’ll get a lot more from looking at the solution if you do so. Part of the aim of this course is to train you to think logically. Learn to write the solutions of the exercises in a connected, step-by-step fashion with explanatory sentences—not just a string of disconnected equations or formulas. The answers to the odd-numbered exercises appear at the back of the book, in Appendix E. Some exercises ask for a verbal explanation or interpretation or description. In such cases there is no single correct way of expressing the answer, so don’t worry that you haven’t found the definitive answer. In addition, there are often several different forms in which to express a numerical or algebraic answer, so if your answer differs from mine, don’t immediately assume you’re wrong. For example, if the answer given in the back of the book is s2 1 and you obtain 1(1 s2 ), then you’re right and rationalizing the denominator will show that the answers are equivalent. The icon ; indicates an exercise that definitely requires the use of either a graphing calculator or a computer with graphing software. But that doesn’t mean that graphing devices can’t be used to check your work on the other exercises as well. The symbol CAS is reserved for problems in which the full resources of a computer algebra system (like Derive, Maple, Mathematica, or the TI-89/92) are required. You will also encounter the symbol | , which warns you against committing an error. I have placed this symbol in the margin in situations where I have observed that a large proportion of my students tend to make the same mistake. xii
The CD-ROM Tools for Enriching™ Calculus is referred to by means of the symbol . It directs you to Visuals and Modules in which you can explore aspects of calculus for which the computer is particularly useful. TEC also provides Homework Hints for representative exercises that are indicated by printing the exercise number in blue: 43. . These homework hints ask you questions that allow you to make progress toward a solution without actually giving you the answer. You need to pursue each hint in an active manner with pencil and paper to work out the details. If a particular hint doesn’t enable you to solve the problem, you can click to reveal the next hint. (See the front endsheet for information on how to purchase this and other useful tools.) The Interactive Video Skillbuilder CD-ROM contains videos of instructors explaining two or three of the examples in every section of the text. (The symbol V has been placed beside these examples in the text.) Also on the CD is a video in which I offer advice on how to succeed in your calculus course. I also want to draw your attention to the website www.stewartcalculus.com. There you will find an Algebra Review (in case your precalculus skills are weak) as well as Additional Examples, Challenging Problems, Projects, Lies My Calculator and Computer Told Me (explaining why calculators sometimes give the wrong answer), History of Mathematics, Additional Topics, chapter quizzes, and links to outside resources. I recommend that you keep this book for reference purposes after you finish the course. Because you will likely forget some of the specific details of calculus, the book will serve as a useful reminder when you need to use calculus in subsequent courses. And, because this book contains more material than can be covered in any one course, it can also serve as a valuable resource for a working scientist or engineer. Calculus is an exciting subject, justly considered to be one of the greatest achievements of the human intellect. I hope you will discover that it is not only useful but also intrinsically beautiful. JA M E S S T E WA RT
1
FUNCTIONS AND LIMITS Calculus is fundamentally different from the mathematics that you have studied previously. Calculus is less static and more dynamic. It is concerned with change and motion; it deals with quantities that approach other quantities. So in this first chapter we begin our study of calculus by investigating how the values of functions change and approach limits.
1.1
Year
Population (millions)
1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000
1650 1750 1860 2070 2300 2560 3040 3710 4450 5280 6080
FUNCTIONS AND THEIR REPRESENTATONS Functions arise whenever one quantity depends on another. Consider the following four situations. A. The area A of a circle depends on the radius r of the circle. The rule that connects r and A is given by the equation A r 2. With each positive number r there is associated one value of A, and we say that A is a function of r. B. The human population of the world P depends on the time t. The table gives estimates of the world population Pt at time t, for certain years. For instance, P1950 2,560,000,000 But for each value of the time t there is a corresponding value of P, and we say that P is a function of t. C. The cost C of mailing a first-class letter depends on the weight w of the letter. Although there is no simple formula that connects w and C, the post office has a rule for determining C when w is known. D. The vertical acceleration a of the ground as measured by a seismograph during an earthquake is a function of the elapsed time t. Figure 1 shows a graph generated by seismic activity during the Northridge earthquake that shook Los Angeles in 1994. For a given value of t, the graph provides a corresponding value of a. a {cm/s@} 100
50
5
FIGURE 1
Vertical ground acceleration during the Northridge earthquake
10
15
20
25
30
t (seconds)
_50 Calif. Dept. of Mines and Geology
Each of these examples describes a rule whereby, given a number (r, t, w, or t), another number ( A, P, C, or a) is assigned. In each case we say that the second number is a function of the first number. 1
2
■
CHAPTER 1
FUNCTIONS AND LIMITS
A function f is a rule that assigns to each element x in a set A exactly one element, called f x, in a set B.
x (input)
f
ƒ (output)
FIGURE 2
Machine diagram for a function ƒ
x
ƒ a
f(a)
f
A
We usually consider functions for which the sets A and B are sets of real numbers. The set A is called the domain of the function. The number f x is the value of f at x and is read “ f of x.” The range of f is the set of all possible values of f x as x varies throughout the domain. A symbol that represents an arbitrary number in the domain of a function f is called an independent variable. A symbol that represents a number in the range of f is called a dependent variable. In Example A, for instance, r is the independent variable and A is the dependent variable. It’s helpful to think of a function as a machine (see Figure 2). If x is in the domain of the function f, then when x enters the machine, it’s accepted as an input and the machine produces an output f x according to the rule of the function. Thus we can think of the domain as the set of all possible inputs and the range as the set of all possible outputs. Another way to picture a function is by an arrow diagram as in Figure 3. Each arrow connects an element of A to an element of B. The arrow indicates that f x is associated with x, f a is associated with a, and so on. The most common method for visualizing a function is its graph. If f is a function with domain A, then its graph is the set of ordered pairs
x, f x
B
x A
(Notice that these are input-output pairs.) In other words, the graph of f consists of all points x, y in the coordinate plane such that y f x and x is in the domain of f . The graph of a function f gives us a useful picture of the behavior or “life history” of a function. Since the y-coordinate of any point x, y on the graph is y f x, we can read the value of f x from the graph as being the height of the graph above the point x. (See Figure 4.) The graph of f also allows us to picture the domain of f on the x-axis and its range on the y-axis as in Figure 5.
FIGURE 3
Arrow diagram for ƒ
y
y
{ x, ƒ}
y ⫽ ƒ(x)
range
ƒ f (2) f(1) 0
1
2
x
x
x
0
domain FIGURE 4
FIGURE 5
y
EXAMPLE 1 The graph of a function f is shown in Figure 6.
(a) Find the values of f 1 and f 5. (b) What are the domain and range of f ? 1 0
FIGURE 6
SOLUTION 1
x
(a) We see from Figure 6 that the point 1, 3 lies on the graph of f , so the value of f at 1 is f 1 3. (In other words, the point on the graph that lies above x 1 is 3 units above the x-axis.) When x 5, the graph lies about 0.7 unit below the x-axis, so we estimate that f 5 0.7.
SECTION 1.1
■ The notation for intervals is given on Reference Page 3. The Reference Pages are located at the front and back of the book.
FUNCTIONS AND THEIR REPRESENTATIONS
■
3
(b) We see that f x is defined when 0 x 7, so the domain of f is the closed interval 0, 7 . Notice that f takes on all values from 2 to 4, so the range of f is
y
2 y 4 2, 4
■
REPRESENTATIONS OF FUNCTIONS
There are four possible ways to represent a function: ■ ■
verbally (by a description in words) numerically (by a table of values)
■ ■
visually (by a graph) algebraically (by an explicit formula)
If a single function can be represented in all four ways, it is often useful to go from one representation to another to gain additional insight into the function. But certain functions are described more naturally by one method than by another. With this in mind, let’s reexamine the four situations that we considered at the beginning of this section. A. The most useful representation of the area of a circle as a function of its radius
is probably the algebraic formula Ar r 2, though it is possible to compile a table of values or to sketch a graph (half a parabola). Because a circle has to have a positive radius, the domain is r r 0 0, , and the range is also 0, .
Year
Population (millions)
1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000
1650 1750 1860 2070 2300 2560 3040 3710 4450 5280 6080
B. We are given a description of the function in words: Pt is the human population
of the world at time t. The table of values of world population provides a convenient representation of this function. If we plot these values, we get the graph (called a scatter plot) in Figure 7. It too is a useful representation; the graph allows us to absorb all the data at once. What about a formula? Of course, it’s impossible to devise an explicit formula that gives the exact human population Pt at any time t. But it is possible to find an expression for a function that approximates Pt. In fact, we could use a graphing calculator with exponential regression capabilities to obtain the approximation Pt f t 0.008079266 1.013731t and Figure 8 shows that it is a reasonably good “fit.” The function f is called a mathematical model for population growth. In other words, it is a function with an explicit formula that approximates the behavior of our given function. We will see, however, that the ideas of calculus can be applied to a table of values; an explicit formula is not necessary.
P
P
6x10'
6x10'
1900
1920
1940
1960
1980
2000 t
FIGURE 7 Scatter plot of data points for population growth
1900
1920
1940
1960
1980
2000 t
FIGURE 8 Graph of a mathematical model for population growth
4
■
CHAPTER 1
FUNCTIONS AND LIMITS
■ A function defined by a table of values is called a tabular function.
w (ounces)
Cw (dollars)
0w 1 1w 2 2w 3 3w 4 4w 5
0.39 0.63 0.87 1.11 1.35
12 w 13
3.27
The function P is typical of the functions that arise whenever we attempt to apply calculus to the real world. We start with a verbal description of a function. Then we may be able to construct a table of values of the function, perhaps from instrument readings in a scientific experiment. Even though we don’t have complete knowledge of the values of the function, we will see throughout the book that it is still possible to perform the operations of calculus on such a function. C. Again the function is described in words: Cw is the cost of mailing a first-class letter with weight w. The rule that the US Postal Service used as of 2006 is as follows: The cost is 39 cents for up to one ounce, plus 24 cents for each successive ounce up to 13 ounces. The table of values shown in the margin is the most convenient representation for this function, though it is possible to sketch a graph (see Example 6). D. The graph shown in Figure 1 is the most natural representation of the vertical acceleration function at. It’s true that a table of values could be compiled, and it is even possible to devise an approximate formula. But everything a geologist needs to know—amplitudes and patterns—can be seen easily from the graph. (The same is true for the patterns seen in electrocardiograms of heart patients and polygraphs for lie-detection.) In the next example we sketch the graph of a function that is defined verbally. EXAMPLE 2 When you turn on a hot-water faucet, the temperature T of the water
depends on how long the water has been running. Draw a rough graph of T as a function of the time t that has elapsed since the faucet was turned on.
T
0
FIGURE 9
t
SOLUTION The initial temperature of the running water is close to room temperature because the water has been sitting in the pipes. When the water from the hotwater tank starts flowing from the faucet, T increases quickly. In the next phase, T is constant at the temperature of the heated water in the tank. When the tank is drained, T decreases to the temperature of the water supply. This enables us to make the rough sketch of T as a function of t in Figure 9. ■
EXAMPLE 3 Find the domain of each function.
(a) f x sx 2
(b) tx
1 x x 2
SOLUTION If a function is given by a formula and the domain is not stated explicitly, the convention is that the domain is the set of all numbers for which the formula makes sense and defines a real number. ■
(a) Because the square root of a negative number is not defined (as a real number), the domain of f consists of all values of x such that x 2 0. This is equivalent to x 2, so the domain is the interval 2, . (b) Since 1 1 tx 2 x x xx 1 and division by 0 is not allowed, we see that tx is not defined when x 0 or x 1. Thus the domain of t is x x 0, x 1 , which could also be written in interval notation as , 0 0, 1 1, . ■
The graph of a function is a curve in the xy-plane. But the question arises: Which curves in the xy-plane are graphs of functions? This is answered by the following test. THE VERTICAL LINE TEST A curve in the xy-plane is the graph of a function of
x if and only if no vertical line intersects the curve more than once.
SECTION 1.1
■
FUNCTIONS AND THEIR REPRESENTATIONS
5
The reason for the truth of the Vertical Line Test can be seen in Figure 10. If each vertical line x a intersects a curve only once, at a, b, then exactly one functional value is defined by f a b. But if a line x a intersects the curve twice, at a, b and a, c, then the curve can’t represent a function because a function can’t assign two different values to a. y
y
x=a
(a, c)
x=a
(a, b) (a, b) x
a
0
a
0
x
FIGURE 10
PIECEWISE DEFINED FUNCTIONS
The functions in the following three examples are defined by different formulas in different parts of their domains. V EXAMPLE 4
A function f is defined by f x
1 x if x 1 x2 if x 1
Evaluate f 0, f 1, and f 2 and sketch the graph. SOLUTION Remember that a function is a rule. For this particular function the rule is the following: First look at the value of the input x. If it happens that x 1, then the value of f x is 1 x. On the other hand, if x 1, then the value of f x is x 2.
Since 0 1, we have f 0 1 0 1. Since 1 1, we have f 1 1 1 0.
y
Since 2 1, we have f 2 2 2 4.
1
1
x
FIGURE 11
How do we draw the graph of f ? We observe that if x 1, then f x 1 x, so the part of the graph of f that lies to the left of the vertical line x 1 must coincide with the line y 1 x, which has slope 1 and y-intercept 1. If x 1, then f x x 2, so the part of the graph of f that lies to the right of the line x 1 must coincide with the graph of y x 2, which is a parabola. This enables us to sketch the graph in Figure l1. The solid dot indicates that the point 1, 0 is included on the graph; the open dot indicates that the point 1, 1 is excluded from the graph. ■ The next example of a piecewise defined function is the absolute value function. Recall that the absolute value of a number a, denoted by a , is the distance from a to 0 on the real number line. Distances are always positive or 0, so we have
www.stewartcalculus.com For a more extensive review of absolute values, click on Review of Algebra. ■
For example,
3 3
3 3
a 0 0 0
for every number a
s2 1 s2 1
3 3
6
■
CHAPTER 1
FUNCTIONS AND LIMITS
In general, we have
a a a a
if a 0 if a 0
(Remember that if a is negative, then a is positive.)
EXAMPLE 5 Sketch the graph of the absolute value function f x x .
y
SOLUTION From the preceding discussion we know that
y=| x |
x 0
x
x x
if x 0 if x 0
Using the same method as in Example 4, we see that the graph of f coincides with the line y x to the right of the y-axis and coincides with the line y x to the left of the y-axis (see Figure 12). ■
FIGURE 12
EXAMPLE 6 In Example C at the beginning of this section we considered the cost
Cw of mailing a first-class letter with weight w. In effect, this is a piecewise defined function because, from the table of values, we have
C 1
0
1
FIGURE 13
2
3
4
5
0.39 0.63 Cw 0.87 1.11
w
if if if if
0w 1w 2w 3w
1 2 3 4
The graph is shown in Figure 13. You can see why functions similar to this one are called step functions—they jump from one value to the next. ■ SYMMETRY
If a function f satisfies f x f x for every number x in its domain, then f is called an even function. For instance, the function f x x 2 is even because f x x2 x 2 f x The geometric significance of an even function is that its graph is symmetric with respect to the y-axis (see Figure 14). This means that if we have plotted the graph of y
y
f(_x)
ƒ _x
0
x
FIGURE 14 An even function
_x x
ƒ
0 x
FIGURE 15 An odd function
x
SECTION 1.1
f
_1
■
7
f for x 0, we obtain the entire graph simply by reflecting this portion about the y-axis. If f satisfies f x f x for every number x in its domain, then f is called an odd function. For example, the function f x x 3 is odd because
y 1
FUNCTIONS AND THEIR REPRESENTATIONS
x
1
f x x3 x 3 f x
_1
The graph of an odd function is symmetric about the origin (see Figure 15 on page 6). If we already have the graph of f for x 0, we can obtain the entire graph by rotating this portion through 180 about the origin.
(a) y 1
V EXAMPLE 7 Determine whether each of the following functions is even, odd, or neither even nor odd. (a) f x x 5 x (b) tx 1 x 4 (c) hx 2x x 2
g 1
SOLUTION
x
f x x5 x 15x 5 x
(a)
x 5 x x 5 x f x
(b)
Therefore, f is an odd function.
y
tx 1 x4 1 x 4 tx
(b)
h
1
So t is even. x
1
(c)
hx 2x x2 2x x 2
Since hx hx and hx hx, we conclude that h is neither even nor odd.
■
(c)
The graphs of the functions in Example 7 are shown in Figure 16. Notice that the graph of h is symmetric neither about the y-axis nor about the origin.
FIGURE 16
INCREASING AND DECREASING FUNCTIONS y
B
The graph shown in Figure 17 rises from A to B, falls from B to C, and rises again from C to D. The function f is said to be increasing on the interval a, b , decreasing on b, c , and increasing again on c, d . Notice that if x 1 and x 2 are any two numbers between a and b with x 1 x 2, then f x 1 f x 2 . We use this as the defining property of an increasing function.
D
y=ƒ C f(x™) A
f(x¡)
A function f is called increasing on an interval I if 0 a x¡
FIGURE 17
x™
b
c
d
x
f x 1 f x 2
whenever x 1 x 2 in I
It is called decreasing on I if f x 1 f x 2
whenever x 1 x 2 in I
8
■
CHAPTER 1
FUNCTIONS AND LIMITS
In the definition of an increasing function it is important to realize that the inequality f x 1 f x 2 must be satisfied for every pair of numbers x 1 and x 2 in I with x 1 x 2. You can see from Figure 18 that the function f x x 2 is decreasing on the interval , 0 and increasing on the interval 0, . y
y=≈
1.1
EXERCISES
1. The graph of a function f is given.
(a) (b) (c) (d) (e) (f )
x
0
FIGURE 18
■ Determine whether the curve is the graph of a function of x. If it is, state the domain and range of the function.
3–6
State the value of f 1. Estimate the value of f 2. For what values of x is f x 2? Estimate the values of x such that f x 0. State the domain and range of f . On what interval is f increasing?
y
3.
y
4.
1
1 0
0
x
1
1
x
1
x
y
1 0
y
5.
y
6.
x
1
1
1 0
1
0
x
2. The graphs of f and t are given.
(a) (b) (c) (d) (e) (f )
State the values of f 4 and t3. For what values of x is f x tx? Estimate the solution of the equation f x 1. On what interval is f decreasing? State the domain and range of f. State the domain and range of t.
■
■
■
■
■
■
■
■
■
■
7. The graph shown gives the weight of a certain person as a
function of age. Describe in words how this person’s weight varies over time. What do you think happened when this person was 30 years old?
y 200
g f
Weight (pounds)
2
0
2
x
■
150 100 50 0
10
20 30 40
50
60 70
Age (years)
SECTION 1.1
8. The graph shown gives a salesman’s distance from his home
as a function of time on a certain day. Describe in words what the graph indicates about his travels on this day.
8 AM
■ Evaluate the difference quotient for the given function. Simplify your answer.
f 3 h f 3 h
f a h f a h
20. f x x 3,
10
NOON
2
4
6 PM
Time (hours)
water, and then let the glass sit on a table. Describe how the temperature of the water changes as time passes. Then sketch a rough graph of the temperature of the water as a function of the elapsed time. 10. Sketch a rough graph of the number of hours of daylight as
a function of the time of year. 11. Sketch a rough graph of the outdoor temperature as a func-
tion of time during a typical spring day. 12. Sketch a rough graph of the market value of a new car as a
function of time for a period of 20 years. Assume the car is well maintained. 13. Sketch the graph of the amount of a particular brand of cof-
fee sold by a store as a function of the price of the coffee.
1 , x
22. f x
x3 , x1
15. A homeowner mows the lawn every Wednesday afternoon.
Sketch a rough graph of the height of the grass as a function of time over the course of a four-week period. 16. A jet takes off from an airport and lands an hour later at
another airport, 400 miles away. If t represents the time in minutes since the plane has left the terminal, let xt be the horizontal distance traveled and yt be the altitude of the plane. (a) Sketch a possible graph of xt. (b) Sketch a possible graph of yt. (c) Sketch a possible graph of the ground speed. (d) Sketch a possible graph of the vertical velocity. 17. If f x 3x 2 x 2, find f 2, f 2, f a, f a,
■
23–27
■
■
Vr 43 r 3 . Find a function that represents the amount of air required to inflate the balloon from a radius of r inches to a radius of r 1 inches.
■
■
■
■
■
■
■
■
■
■
■
■
Find the domain of the function. x 3x 1
24. f x
5x 4 x 2 3x 2
3 t 25. f t st s
26. tu su s4 u 27. hx ■
■
1 4 x 2 5x s ■
■
■
■
■
28. Find the domain and range and sketch the graph of the
function hx s4 x 2 . 29– 40
■
Find the domain and sketch the graph of the function.
29. f x 5
30. Fx 2 x 3
31. f t t 2 6t
32. Ht
33. tx sx 5
34. Fx 2x 1
35. Gx 37. f x 38. f x 39. f x
f a 1, 2 f a, f 2a, f a 2 , [ f a] 2, and f a h.
18. A spherical balloon with radius r inches has volume
f x f 1 x1
23. f x
14. You place a frozen pie in an oven and bake it for an
hour. Then you take it out and let it cool before eating it. Describe how the temperature of the pie changes as time passes. Then sketch a rough graph of the temperature of the pie as a function of time.
f x f a xa
21. f x
■
9. You put some ice cubes in a glass, fill the glass with cold
9
19–22
19. f x 4 3x x 2 ,
Distance from home (miles)
■
FUNCTIONS AND THEIR REPRESENTATIONS
40. f x
■
■
1
3x x x
x2 1x 3 12 x 2x 5
36.
4 t2 2t
x tx
x2
if x 0 if x 0 if x 2 if x 2
x 2 if x 1 x2 if x 1
1 if x 1 3x 2 if x 1 7 2x if x 1
■
■
■
■
■
■
■
■
■
10
■
CHAPTER 1
FUNCTIONS AND LIMITS
■ Find an expression for the function whose graph is the given curve.
stairs. Give two other examples of step functions that arise in everyday life.
41– 44
41. The line segment joining the points 2, 1 and 4, 6 42. The line segment joining the points 3, 2 and 6, 3
53–54 ■ Graphs of f and t are shown. Decide whether each function is even, odd, or neither. Explain your reasoning.
43. The bottom half of the parabola x y 12 0
53.
54.
y
y
g
44. The top half of the circle x 1 y 1 2
2
f
f ■
■
45– 49
■
■
■
■
■
■
■
■
■
■
Find a formula for the described function and state its
x
x g
domain. 45. A rectangle has perimeter 20 m. Express the area of the
rectangle as a function of the length of one of its sides. ■
■
■
■
■
■
■
■
■
■
■
46. A rectangle has area 16 m2. Express the perimeter of the
rectangle as a function of the length of one of its sides.
55. (a) If the point 5, 3 is on the graph of an even function,
what other point must also be on the graph? (b) If the point 5, 3 is on the graph of an odd function, what other point must also be on the graph?
47. Express the area of an equilateral triangle as a function of
the length of a side. 48. Express the surface area of a cube as a function of its
56. A function f has domain 5, 5 and a portion of its graph
volume.
is shown. (a) Complete the graph of f if it is known that f is even. (b) Complete the graph of f if it is known that f is odd.
3
49. An open rectangular box with volume 2 m has a square
base. Express the surface area of the box as a function of the length of a side of the base. ■
■
■
■
■
■
■
■
■
■
y
■
50. A taxi company charges two dollars for the first mile (or
part of a mile) and 20 cents for each succeeding tenth of a mile (or part). Express the cost C (in dollars) of a ride as a function of the distance x traveled (in miles) for 0 x 2, and sketch the graph of this function.
0
_5
x
5
51. In a certain country, income tax is assessed as follows.
There is no tax on income up to $10,000. Any income over $10,000 is taxed at a rate of 10%, up to an income of $20,000. Any income over $20,000 is taxed at 15%. (a) Sketch the graph of the tax rate R as a function of the income I. (b) How much tax is assessed on an income of $14,000? On $26,000? (c) Sketch the graph of the total assessed tax T as a function of the income I. 52. The functions in Example 6 and Exercises 50 and 51(a)
are called step functions because their graphs look like
1.2
57–62 ■ Determine whether f is even, odd, or neither. If you have a graphing calculator, use it to check your answer visually.
x2 x 1
57. f x
x x 1
58. f x
59. f x
x x1
60. f x x x
2
61. f x 1 3x 2 x 4 ■
■
■
4
■
■
62. f x 1 3x 3 x 5 ■
■
■
■
■
■
A CATALOG OF ESSENTIAL FUNCTIONS In solving calculus problems you will find that it is helpful to be familiar with the graphs of some commonly occurring functions. These same basic functions are often used to model real-world phenomena, so we begin with a discussion of mathematical modeling. We also review briefly how to transform these functions by shifting, stretching, and reflecting their graphs as well as how to combine pairs of functions by the standard arithmetic operations and by composition.
SECTION 1.2
A CATALOG OF ESSENTIAL FUNCTIONS
■
11
MATHEMATICAL MODELING
A mathematical model is a mathematical description (often by means of a function or an equation) of a real-world phenomenon such as the size of a population, the demand for a product, the speed of a falling object, the concentration of a product in a chemical reaction, the life expectancy of a person at birth, or the cost of emission reductions. The purpose of the model is to understand the phenomenon and perhaps to make predictions about future behavior. Figure 1 illustrates the process of mathematical modeling. Given a real-world problem, our first task is to formulate a mathematical model by identifying and naming the independent and dependent variables and making assumptions that simplify the phenomenon enough to make it mathematically tractable. We use our knowledge of the physical situation and our mathematical skills to obtain equations that relate the variables. In situations where there is no physical law to guide us, we may need to collect data (either from a library or the Internet or by conducting our own experiments) and examine the data in the form of a table in order to discern patterns. From this numerical representation of a function we may wish to obtain a graphical representation by plotting the data. The graph might even suggest a suitable algebraic formula in some cases. Real-world problem
Formulate
Mathematical model
Solve
Mathematical conclusions
Interpret
Real-world predictions
Test FIGURE 1 The modeling process
The second stage is to apply the mathematics that we know (such as the calculus that will be developed throughout this book) to the mathematical model that we have formulated in order to derive mathematical conclusions. Then, in the third stage, we take those mathematical conclusions and interpret them as information about the original real-world phenomenon by way of offering explanations or making predictions. The final step is to test our predictions by checking against new real data. If the predictions don’t compare well with reality, we need to refine our model or to formulate a new model and start the cycle again. A mathematical model is never a completely accurate representation of a physical situation—it is an idealization. A good model simplifies reality enough to permit mathematical calculations but is accurate enough to provide valuable conclusions. It is important to realize the limitations of the model. In the end, Mother Nature has the final say. There are many different types of functions that can be used to model relationships observed in the real world. In what follows, we discuss the behavior and graphs of these functions and give examples of situations appropriately modeled by such functions. ■
www.stewartcalculus.com To review the coordinate geometry of lines, click on Review of Analytic Geometry. ■
LINEAR MODELS
When we say that y is a linear function of x, we mean that the graph of the function is a line, so we can use the slope-intercept form of the equation of a line to write a formula for the function as y f x mx b where m is the slope of the line and b is the y-intercept.
12
■
CHAPTER 1
FUNCTIONS AND LIMITS
A characteristic feature of linear functions is that they grow at a constant rate. For instance, Figure 2 shows a graph of the linear function f x 3x 2 and a table of sample values. Notice that whenever x increases by 0.1, the value of f x increases by 0.3. So f x increases three times as fast as x. Thus the slope of the graph y 3x 2, namely 3, can be interpreted as the rate of change of y with respect to x. y
y=3x-2
0
x
_2
x
f x 3x 2
1.0 1.1 1.2 1.3 1.4 1.5
1.0 1.3 1.6 1.9 2.2 2.5
FIGURE 2
V EXAMPLE 1
(a) As dry air moves upward, it expands and cools. If the ground temperature is 20 C and the temperature at a height of 1 km is 10 C, express the temperature T (in °C) as a function of the height h (in kilometers), assuming that a linear model is appropriate. (b) Draw the graph of the function in part (a). What does the slope represent? (c) What is the temperature at a height of 2.5 km? SOLUTION
(a) Because we are assuming that T is a linear function of h, we can write T mh b We are given that T 20 when h 0, so 20 m 0 b b In other words, the y-intercept is b 20. We are also given that T 10 when h 1, so 10 m 1 20
T
The slope of the line is therefore m 10 20 10 and the required linear function is T 10h 20
20
T=_10h+20 10
0
1
FIGURE 3
3
h
(b) The graph is sketched in Figure 3. The slope is m 10 Ckm, and this represents the rate of change of temperature with respect to height. (c) At a height of h 2.5 km, the temperature is T 102.5 20 5 C
■
SECTION 1.2
■
A CATALOG OF ESSENTIAL FUNCTIONS
■
13
POLYNOMIALS
A function P is called a polynomial if Px a n x n a n1 x n1 a 2 x 2 a 1 x a 0 where n is a nonnegative integer and the numbers a 0 , a 1, a 2 , . . . , a n are constants called the coefficients of the polynomial. The domain of any polynomial is ⺢ , . If the leading coefficient a n 0, then the degree of the polynomial is n. For example, the function Px 2x 6 x 4 25 x 3 s2 is a polynomial of degree 6. A polynomial of degree 1 is of the form Px mx b and so it is a linear function. A polynomial of degree 2 is of the form Px ax 2 bx c and is called a quadratic function. Its graph is always a parabola obtained by shifting the parabola y ax 2. The parabola opens upward if a 0 and downward if a 0. (See Figure 4.) y
y
2 2
x
1 0
FIGURE 4
The graphs of quadratic functions are parabolas.
1
x
(b) y=_2≈+3x+1
(a) y=≈+x+1
A polynomial of degree 3 is of the form Px ax 3 bx 2 cx d
a0
and is called a cubic function. Figure 5 shows the graph of a cubic function in part (a) and graphs of polynomials of degrees 4 and 5 in parts (b) and (c). We will see later why the graphs have these shapes. y
y
1
2
0
FIGURE 5
y 20 1
1
(a) y=˛-x+1
x
x
(b) y=x$-3≈+x
1
x
(c) y=3x%-25˛+60x
Polynomials are commonly used to model various quantities that occur in the natural and social sciences. For instance, in Chapter 2 we will explain why economists often use a polynomial Px to represent the cost of producing x units of a commodity.
14
■
CHAPTER 1
FUNCTIONS AND LIMITS
■
POWER FUNCTIONS
A function of the form f x x a, where a is a constant, is called a power function. We consider several cases. (i) a n, where n is a positive integer
The graphs of f x x n for n 1, 2, 3, 4, and 5 are shown in Figure 6. (These are polynomials with only one term.) You are familiar with the shape of the graphs of y x (a line through the origin with slope 1) and y x 2 (a parabola). y
y=x
0
1
x
0
y=x #
y
1
1
FIGURE 6
y=≈
y
x
0
1
x
y=x%
y
1
1
1
y=x$
y
1
0
1
x
0
x
1
Graphs of ƒ=x n for n=1, 2, 3, 4, 5
The general shape of the graph of f x x n depends on whether n is even or odd. If n is even, then f x x n is an even function and its graph is similar to the parabola y x 2. If n is odd, then f x x n is an odd function and its graph is similar to that of y x 3. Notice from Figure 7, however, that as n increases, the graph of y x n becomes flatter near 0 and steeper when x 1. (If x is small, then x 2 is smaller, x 3 is even smaller, x 4 is smaller still, and so on.)
y
y
y=x $ y=x ^
y=x # y=≈
(_1, 1)
FIGURE 7
Families of power functions
(1, 1) y=x %
(1, 1)
x
0
(_1, _1) x
0
(ii) a 1n, where n is a positive integer n The function f x x 1n s x is a root function. For n 2 it is the square root function f x sx , whose domain is 0, and whose graph is the upper half of the parabola x y 2. [See Figure 8(a).] For other even values of n, the graph of n ys x is similar to that of y sx . For n 3 we have the cube root function
y
y
(1, 1) 0
(1, 1) x
0
FIGURE 8
Graphs of root functions
x (a) ƒ=œ„
x (b) ƒ=Œ„
x
SECTION 1.2
A CATALOG OF ESSENTIAL FUNCTIONS
15
■
3 f x s x whose domain is ⺢ (recall that every real number has a cube root) and n whose graph is shown in Figure 8(b). The graph of y s x for n odd n 3 is 3 similar to that of y sx .
(iii) a 1
y
The graph of the reciprocal function f x x 1 1x is shown in Figure 9. Its graph has the equation y 1x, or xy 1, and is a hyperbola with the coordinate axes as its asymptotes. This function arises in physics and chemistry in connection with Boyle’s Law, which says that, when the temperature is constant, the volume V of a gas is inversely proportional to the pressure P:
y=Δ 1 0
x
1
V FIGURE 9
C P
where C is a constant. Thus the graph of V as a function of P has the same general shape as the right half of Figure 9.
The reciprocal function
■
y
RATIONAL FUNCTIONS
A rational function f is a ratio of two polynomials: 20
f x
0
x
2
Px Qx
where P and Q are polynomials. The domain consists of all values of x such that Qx 0. A simple example of a rational function is the function f x 1x, whose domain is x x 0 ; this is the reciprocal function graphed in Figure 9. The function
f x
FIGURE 10
ƒ=
2x$-≈+1 ≈-4
is a rational function with domain x ■
2x 4 x 2 1 x2 4
x 2 . Its graph is shown in Figure 10.
TRIGONOMETRIC FUNCTIONS
Trigonometry and the trigonometric functions are reviewed on Reference Page 2 and also in Appendix A. In calculus the convention is that radian measure is always used (except when otherwise indicated). For example, when we use the function f x sin x, it is understood that sin x means the sine of the angle whose radian measure is x. Thus the graphs of the sine and cosine functions are as shown in Figure 11. y _ _π
π 2
y 3π 2
1 0 _1
π 2
π
_π 2π
5π 2
3π
_
π 2
π 0
x _1
(a) ƒ=sin x FIGURE 11
1 π 2
3π 3π 2
2π
5π 2
x
(b) ©=cos x
Notice that for both the sine and cosine functions the domain is , and the range is the closed interval 1, 1 . Thus, for all values of x, we have 1 sin x 1
1 cos x 1
16
■
CHAPTER 1
FUNCTIONS AND LIMITS
or, in terms of absolute values,
sin x 1
cos x 1
Also, the zeros of the sine function occur at the integer multiples of ; that is, sin x 0
when
x n
n an integer
An important property of the sine and cosine functions is that they are periodic functions and have period 2. This means that, for all values of x, sinx 2 sin x
y
The periodic nature of these functions makes them suitable for modeling repetitive phenomena such as tides, vibrating springs, and sound waves. The tangent function is related to the sine and cosine functions by the equation
1 _
cosx 2 cos x
0
3π _π π _ 2 2
π 2
3π 2
π
x
tan x
sin x cos x
and its graph is shown in Figure 12. It is undefined whenever cos x 0, that is, when x 2, 32, . . . . Its range is , . Notice that the tangent function has period :
FIGURE 12
y=tan x
tanx tan x y
0
The remaining three trigonometric functions (cosecant, secant, and cotangent) are the reciprocals of the sine, cosine, and tangent functions. Their graphs are shown in Appendix A.
y
1
1 0
x
1
(a) y=2®
■
x
1
(b) y=(0.5)®
FIGURE 13 y
y=log™ x y=log£ x
1
0
1
for all x
x
EXPONENTIAL FUNCTIONS AND LOGARITHMS
The exponential functions are the functions of the form f x a x, where the base a is a positive constant. The graphs of y 2 x and y 0.5 x are shown in Figure 13. In both cases the domain is , and the range is 0, . Exponential functions will be studied in detail in Section 3.1, and we will see that they are useful for modeling many natural phenomena, such as population growth (if a 1) and radioactive decay (if a 1. The logarithmic functions f x log a x, where the base a is a positive constant, are the inverse functions of the exponential functions. They will be studied in Section 3.2. Figure 14 shows the graphs of four logarithmic functions with various bases. In each case the domain is 0, , the range is , , and the function increases slowly when x 1.
y=log∞ x y=log¡¸ x TRANSFORMATIONS OF FUNCTIONS
FIGURE 14
By applying certain transformations to the graph of a given function we can obtain the graphs of certain related functions. This will give us the ability to sketch the graphs of many functions quickly by hand. It will also enable us to write equations for given
SECTION 1.2
■
A CATALOG OF ESSENTIAL FUNCTIONS
17
graphs. Let’s first consider translations. If c is a positive number, then the graph of y f x c is just the graph of y f x shifted upward a distance of c units (because each y-coordinate is increased by the same number c). Likewise, if tx f x c, where c 0, then the value of t at x is the same as the value of f at x c (c units to the left of x). Therefore, the graph of y f x c is just the graph of y f x shifted c units to the right.
■ Figure 15 illustrates these shifts by showing how the graph of y x 3 2 1 is obtained from the graph of the parabola y x 2 : Shift 3 units to the left and 1 unit upward.
y y=≈
VERTICAL AND HORIZONTAL SHIFTS Suppose c 0. To obtain the graph of (_3, 1) _3
1 0
x
FIGURE 15
y f x c, shift the graph of y f x a distance c units upward y f x c, shift the graph of y f x a distance c units downward y f x c, shift the graph of y f x a distance c units to the right y f x c, shift the graph of y f x a distance c units to the left
y=(x+3)@+1
Now let’s consider the stretching and reflecting transformations. If c 1, then the graph of y cf x is the graph of y f x stretched by a factor of c in the vertical direction (because each y-coordinate is multiplied by the same number c). The graph of y f x is the graph of y f x reflected about the x-axis because the point x, y is replaced by the point x, y. The following chart also incorporates the results of other stretching, compressing, and reflecting transformations. VERTICAL AND HORIZONTAL STRETCHING AND REFLECTING
Suppose c 1. To obtain the graph of y cf x, stretch the graph of y f x vertically by a factor of c y 1cf x, compress the graph of y f x vertically by a factor of c y f cx, compress the graph of y f x horizontally by a factor of c y f xc, stretch the graph of y f x horizontally by a factor of c y f x, reflect the graph of y f x about the x-axis y f x, reflect the graph of y f x about the y-axis
Figure 16 illustrates these stretching transformations when applied to the cosine function with c 2. For instance, in order to get the graph of y 2 cos x we multiply the y-coordinate of each point on the graph of y cos x by 2. This means that the graph of y cos x gets stretched vertically by a factor of 2. y
y=2 cos x
y
2
y=cos x
2
1 2
1
1 0
y= cos x 1
x
y=cos 1 x 2
0
x
y=cos x FIGURE 16
y=cos 2x
18
■
CHAPTER 1
FUNCTIONS AND LIMITS
V EXAMPLE 2 Given the graph of y sx , use transformations to graph y sx 2, y sx 2 , y sx , y 2sx , and y sx .
SOLUTION The graph of the square root function y sx , obtained from Figure 8(a), is shown in Figure 17(a). In the other parts of the figure we sketch y sx 2 by shifting 2 units downward, y sx 2 by shifting 2 units to the right, y sx by reflecting about the x-axis, y 2sx by stretching vertically by a factor of 2, and y sx by reflecting about the y-axis. y
y
y
y
y
y
1 0
1
x
x
0
0
x
2
x
0
x
0
0
x
_2
(a) y=œ„x
(b) y=œ„-2 x
(c) y=œ„„„„ x-2
(d) y=_ œ„x
(f ) y=œ„„ _x
(e) y=2 œ„x
■
FIGURE 17
EXAMPLE 3 Sketch the graph of the function y 1 sin x. SOLUTION To obtain the graph of y 1 sin x, we start with y sin x. We reflect about the x -axis to get the graph y sin x and then we shift 1 unit upward to get y 1 sin x. (See Figure 18.) y
y 2
y=sin x 1
y=1-sin x
1 0
π 2
π
x
0
π 2
π
3π 2
2π
x
■
FIGURE 18
COMBINATIONS OF FUNCTIONS
Two functions f and t can be combined to form new functions f t, f t, ft, and ft in a manner similar to the way we add, subtract, multiply, and divide real numbers. The sum and difference functions are defined by f tx f x tx
f tx f x tx
If the domain of f is A and the domain of t is B, then the domain of f t is the intersection A B because both f x and tx have to be defined. For example, the domain of f x sx is A 0, and the domain of tx s2 x is B , 2 , so the domain of f tx sx s2 x is A B 0, 2 . Similarly, the product and quotient functions are defined by ftx f xtx
f f x x t tx
SECTION 1.2
A CATALOG OF ESSENTIAL FUNCTIONS
■
19
The domain of ft is A B, but we can’t divide by 0 and so the domain of ft is x A B tx 0 . For instance, if f x x 2 and tx x 1, then the domain of the rational function ftx x 2x 1 is x x 1 , or , 1 1, . There is another way of combining two functions to get a new function. For example, suppose that y f u su and u tx x 2 1. Since y is a function of u and u is, in turn, a function of x, it follows that y is ultimately a function of x. We compute this by substitution:
y f u f tx f x 2 1 sx 2 1 The procedure is called composition because the new function is composed of the two given functions f and t. In general, given any two functions f and t, we start with a number x in the domain of t and find its image tx. If this number tx is in the domain of f , then we can calculate the value of f tx. The result is a new function hx f tx obtained by substituting t into f . It is called the composition (or composite) of f and t and is denoted by f ⴰ t (“f circle t”). DEFINITION Given two functions f and t, the composite function f ⴰ t (also called the composition of f and t) is defined by
f ⴰ tx f tx The domain of f ⴰ t is the set of all x in the domain of t such that tx is in the domain of f . In other words, f ⴰ tx is defined whenever both tx and f tx are defined. Figure 19 shows how to picture f ⴰ t in terms of machines.
FIGURE 19
g
x (input)
The f • g machine is composed of the g machine (first) and then the f machine.
g(x)
f
f { ©} (output)
f•g
EXAMPLE 4 If f x x 2 and tx x 3, find the composite functions f ⴰ t
and t ⴰ f .
SOLUTION We have
f ⴰ tx f tx f x 3 x 32 t ⴰ f x t f x tx 2 x 2 3
|
■
NOTE You can see from Example 4 that, in general, f ⴰ t t ⴰ f . Remember, the notation f ⴰ t means that the function t is applied first and then f is applied second. In Example 4, f ⴰ t is the function that first subtracts 3 and then squares; t ⴰ f is the function that first squares and then subtracts 3.
20
■
CHAPTER 1
FUNCTIONS AND LIMITS
V EXAMPLE 5
domain. (a) f ⴰ t
If f x sx and tx s2 x , find each function and its (b) t ⴰ f
(c) f ⴰ f
(d) t ⴰ t
SOLUTION
(a)
4 f ⴰ tx f tx f (s2 x ) ss2 x s 2x
The domain of f ⴰ t is x
x
x 2 , 2 .
t ⴰ f x t f x t(sx ) s2 sx
(b) If 0 a b, then a 2 b 2.
2 x 0
For sx to be defined we must have x 0. For s2 sx to be defined we must have 2 sx 0, that is, sx 2, or x 4. Thus we have 0 x 4, so the domain of t ⴰ f is the closed interval 0, 4 . 4 f ⴰ f x f f x f (sx ) ssx s x
(c)
The domain of f ⴰ f is 0, . t ⴰ tx ttx t(s2 x ) s2 s2 x
(d)
This expression is defined when both 2 x 0 and 2 s2 x 0. The first inequality means x 2, and the second is equivalent to s2 x 2, or 2 x 4, or x 2. Thus, 2 x 2, so the domain of t ⴰ t is the closed interval 2, 2 .
■
It is possible to take the composition of three or more functions. For instance, the composite function f ⴰ t ⴰ h is found by first applying h, then t, and then f as follows: f ⴰ t ⴰ hx f thx So far we have used composition to build complicated functions from simpler ones. But in calculus it is often useful to be able to decompose a complicated function into simpler ones, as in the following example. EXAMPLE 6 Given Fx cos2x 9, find functions f , t, and h such that
F f ⴰ t ⴰ h.
SOLUTION Since Fx cosx 9 2, the formula for F says: First add 9, then
take the cosine of the result, and finally square. So we let hx x 9
tx cos x
f x x 2
Then f ⴰ t ⴰ hx f thx f tx 9 f cosx 9 cosx 9 2 Fx
■
SECTION 1.2
1.2
slope 2 and sketch several members of the family. (b) Find an equation for the family of linear functions such that f 2 1 and sketch several members of the family. (c) Which function belongs to both families? 2. What do all members of the family of linear functions
f x 1 mx 3 have in common? Sketch several members of the family. 3. What do all members of the family of linear functions
f x c x have in common? Sketch several members of the family. 4. Find expressions for the quadratic functions whose graphs
are shown. y
y (_2, 2)
f
(0, 1) (4, 2)
0
x
g 3
■
21
EXERCISES
1. (a) Find an equation for the family of linear functions with
0
A CATALOG OF ESSENTIAL FUNCTIONS
x
(1, _2.5)
5. Find an expression for a cubic function f if f 1 6 and
f 1 f 0 f 2 0.
6. Some scientists believe that the average surface temperature
of the world has been rising steadily. They have modeled the temperature by the linear function T 0.02t 8.50, where T is temperature in C and t represents years since 1900. (a) What do the slope and T -intercept represent? (b) Use the equation to predict the average global surface temperature in 2100. 7. If the recommended adult dosage for a drug is D (in mg),
then to determine the appropriate dosage c for a child of age a, pharmacists use the equation c 0.0417Da 1. Suppose the dosage for an adult is 200 mg. (a) Find the slope of the graph of c. What does it represent? (b) What is the dosage for a newborn? 8. The manager of a weekend flea market knows from past
experience that if he charges x dollars for a rental space at the flea market, then the number y of spaces he can rent is given by the equation y 200 4x. (a) Sketch a graph of this linear function. (Remember that the rental charge per space and the number of spaces rented can’t be negative quantities.) (b) What do the slope, the y-intercept, and the x-intercept of the graph represent? 9. The relationship between the Fahrenheit F and Celsius
C temperature scales is given by the linear function F 95 C 32. (a) Sketch a graph of this function.
(b) What is the slope of the graph and what does it represent? What is the F-intercept and what does it represent? 10. Jason leaves Detroit at 2:00 PM and drives at a constant
speed west along I-96. He passes Ann Arbor, 40 mi from Detroit, at 2:50 PM. (a) Express the distance traveled in terms of the time elapsed. (b) Draw the graph of the equation in part (a). (c) What is the slope of this line? What does it represent? 11. Biologists have noticed that the chirping rate of crickets of
a certain species is related to temperature, and the relationship appears to be very nearly linear. A cricket produces 113 chirps per minute at 70 F and 173 chirps per minute at 80 F. (a) Find a linear equation that models the temperature T as a function of the number of chirps per minute N. (b) What is the slope of the graph? What does it represent? (c) If the crickets are chirping at 150 chirps per minute, estimate the temperature. 12. The manager of a furniture factory finds that it costs $2200
to manufacture 100 chairs in one day and $4800 to produce 300 chairs in one day. (a) Express the cost as a function of the number of chairs produced, assuming that it is linear. Then sketch the graph. (b) What is the slope of the graph and what does it represent? (c) What is the y-intercept of the graph and what does it represent? 13. At the surface of the ocean, the water pressure is the same
as the air pressure above the water, 15 lbin2. Below the surface, the water pressure increases by 4.34 lbin2 for every 10 ft of descent. (a) Express the water pressure as a function of the depth below the ocean surface. (b) At what depth is the pressure 100 lbin2 ? 14. The monthly cost of driving a car depends on the number of
miles driven. Lynn found that in May it cost her $380 to drive 480 mi and in June it cost her $460 to drive 800 mi. (a) Express the monthly cost C as a function of the distance driven d, assuming that a linear relationship gives a suitable model. (b) Use part (a) to predict the cost of driving 1500 miles per month. (c) Draw the graph of the linear function. What does the slope represent? (d) What does the y-intercept represent? (e) Why does a linear function give a suitable model in this situation?
22
■
CHAPTER 1
FUNCTIONS AND LIMITS
15. Suppose the graph of f is given. Write equations for the
19. The graph of f is given. Use it to graph the following
graphs that are obtained from the graph of f as follows. (a) Shift 3 units upward. (b) Shift 3 units downward. (c) Shift 3 units to the right. (d) Shift 3 units to the left. (e) Reflect about the x-axis. (f ) Reflect about the y-axis. (g) Stretch vertically by a factor of 3. (h) Shrink vertically by a factor of 3. 16. Explain how the following graphs are obtained from the
graph of y f x. (a) y 5 f x (b) y f x 5 (c) y f x (d) y 5 f x (e) y f 5x (f ) y 5 f x 3
its graph and give reasons for your choices. (a) y f x 4 (b) y f x 3 (c) y 13 f x (d) y f x 4 (e) y 2 f x 6 y 6
3
!
_3
%
y 1 0
x
1
20. (a) How is the graph of y 2 sin x related to the graph of
y sin x ? Use your answer and Figure 18(a) to sketch the graph of y 2 sin x. (b) How is the graph of y 1 sx related to the graph of y sx ? Use your answer and Figure 17(a) to sketch the graph of y 1 sx .
■ Graph the function by hand, not by plotting points, but by starting with the graph of one of the standard functions and then applying the appropriate transformations.
f
0
21. y x 3
22. y 1 x 2
23. y x 12
24. y x 2 4x 3
25. y 1 2 cos x
26. y 4 sin 3x
27. y sin x2
28. y
29. y sx 3
30. y x 24 3
31. y 2 x 2 8x
3 x1 32. y 1 s
1
33. y
#
$ _6
(b) y f ( 12 x) (d) y f x
21–34
17. The graph of y f x is given. Match each equation with
@
functions. (a) y f 2x (c) y f x
■
3
6
x
■
35–36
2 x1
34. y
■
■
■
■
35. f x x 3 2x 2,
■
18. The graph of f is given. Draw the graphs of the following
■
■
■
■
■
(c) y 2 f x
(d) y f x 3
38. f x 1 x ,
tx 1x
3
40. f x 1 3x, 41. f x x
■
■
■
■
■
1 , x
tx 5x 2 3x 2 tx
42. f x s2x 3 , x
■
tx 1 sx
39. f x sin x,
1
■
■ Find the functions (a) f ⴰ t, (b) t ⴰ f , (c) f ⴰ f , and (d) t ⴰ t and their domains.
tx 2x 1
0
■
37– 42
37. f x x 2 1,
1
■
tx s1 x
(b) y f x 4
y
■
tx 3x 2 1
functions. (a) y f x 4
1 2
1 tan x 4 4
Find f t, f t, f t, and ft and state their domains.
36. f x s1 x ,
_3
■
1 x4
■
■
■
■
x1 x2
tx x 2 1 ■
■
■
■
■
■
■
SECTION 1.2
43– 44
■
Find f ⴰ t ⴰ h.
44. f x ■
■
2 , x1 ■
tx cos x,
■
■
23
55. A stone is dropped into a lake, creating a circular ripple that
tx x 2 2,
43. f x sx 1,
A CATALOG OF ESSENTIAL FUNCTIONS
■
■
hx x 3
hx sx 3 ■
■
■
■
■
travels outward at a speed of 60 cms. (a) Express the radius r of this circle as a function of the time t (in seconds). (b) If A is the area of this circle as a function of the radius, find A ⴰ r and interpret it. 56. An airplane is flying at a speed of 350 mih at an altitude
45– 48
■
Express the function in the form f ⴰ t.
45. Fx x 1
46. Fx sin( sx )
47. ut scos t
48. ut
2
■
■
49–51
10
■
■
■
■
■
■
tan t 1 tan t
■
■
■
■
57. The Heaviside function H is defined by
Express the function in the form f ⴰ t ⴰ h.
49. Hx 1 3
50. Hx s2 x
x2
8
Ht
51. Hx sec4 (sx ) ■
■
■
■
■
■
■
■
■
■
■
52. Use the table to evaluate each expression.
(a) f t1 (d) t t1
(b) t f 1 (e) t ⴰ f 3
(c) f f 1 (f ) f ⴰ t6
x
1
2
3
4
5
6
f x
3
1
4
2
2
5
tx
6
3
2
1
2
3
53. Use the given graphs of f and t to evaluate each expression,
or explain why it is undefined. (a) f t2 (b) t f 0 (d) t ⴰ f 6 (e) t ⴰ t2
(c) f ⴰ t0 (f ) f ⴰ f 4
y
g
f
2
0
2
of one mile and passes directly over a radar station at time t 0. (a) Express the horizontal distance d (in miles) that the plane has flown as a function of t. (b) Express the distance s between the plane and the radar station as a function of d. (c) Use composition to express s as a function of t.
x
0 1
if t 0 if t 0
It is used in the study of electric circuits to represent the sudden surge of electric current, or voltage, when a switch is instantaneously turned on. (a) Sketch the graph of the Heaviside function. (b) Sketch the graph of the voltage Vt in a circuit if the switch is turned on at time t 0 and 120 volts are applied instantaneously to the circuit. Write a formula for Vt in terms of Ht. (c) Sketch the graph of the voltage Vt in a circuit if the switch is turned on at time t 5 seconds and 240 volts are applied instantaneously to the circuit. Write a formula for Vt in terms of Ht. (Note that starting at t 5 corresponds to a translation.) 58. The Heaviside function defined in Exercise 57 can also be
used to define the ramp function y ctHt, which represents a gradual increase in voltage or current in a circuit. (a) Sketch the graph of the ramp function y tHt. (b) Sketch the graph of the voltage Vt in a circuit if the switch is turned on at time t 0 and the voltage is gradually increased to 120 volts over a 60-second time interval. Write a formula for Vt in terms of Ht for t 60. (c) Sketch the graph of the voltage Vt in a circuit if the switch is turned on at time t 7 seconds and the voltage is gradually increased to 100 volts over a period of 25 seconds. Write a formula for Vt in terms of Ht for t 32. 59. Let f and t be linear functions with equations
54. A spherical balloon is being inflated and the radius of the
balloon is increasing at a rate of 2 cms. (a) Express the radius r of the balloon as a function of the time t (in seconds). (b) If V is the volume of the balloon as a function of the radius, find V ⴰ r and interpret it.
f x m1 x b1 and tx m 2 x b 2. Is f ⴰ t also a linear function? If so, what is the slope of its graph?
60. If you invest x dollars at 4% interest compounded annually,
then the amount Ax of the investment after one year is Ax 1.04x. Find A ⴰ A, A ⴰ A ⴰ A, and A ⴰ A ⴰ A ⴰ A. What do these compositions represent? Find a formula for the composition of n copies of A.
24
■
CHAPTER 1
FUNCTIONS AND LIMITS
63. (a) Suppose f and t are even functions. What can you say
61. (a) If tx 2x 1 and hx 4x 2 4x 7, find a
function f such that f ⴰ t h. (Think about what operations you would have to perform on the formula for t to end up with the formula for h.) (b) If f x 3x 5 and hx 3x 2 3x 2, find a function t such that f ⴰ t h.
62. If f x x 4 and hx 4x 1, find a function t such
that t ⴰ f h.
about f t and f t ? (b) What if f and t are both odd?
64. Suppose f is even and t is odd. What can you say about f t ? 65. Suppose t is an even function and let h f ⴰ t. Is h always
an even function? 66. Suppose t is an odd function and let h f ⴰ t. Is h always
an odd function? What if f is odd? What if f is even?
1.3
THE LIMIT OF A FUNCTION Our aim in this section is to explore the meaning of the limit of a function. We begin by showing how the idea of a limit arises when we try to find the velocity of a falling ball. V EXAMPLE 1 Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450 m above the ground. Find the velocity of the ball after 5 seconds.
SOLUTION Through experiments carried out four centuries ago, Galileo discovered that the distance fallen by any freely falling body is proportional to the square of the time it has been falling. (This model for free fall neglects air resistance.) If the distance fallen after t seconds is denoted by st and measured in meters, then Galileo’s law is expressed by the equation
st 4.9t 2 The difficulty in finding the velocity after 5 s is that we are dealing with a single instant of time t 5, so no time interval is involved. However, we can approximate the desired quantity by computing the average velocity over the brief time interval of a tenth of a second from t 5 to t 5.1: average velocity
Time interval
Average velocity (ms)
5 t 6 5 t 5.1 5 t 5.05 5 t 5.01 5 t 5.001
53.9 49.49 49.245 49.049 49.0049
change in position time elapsed
s5.1 s5 0.1
4.95.12 4.952 49.49 ms 0.1
The table shows the results of similar calculations of the average velocity over successively smaller time periods. It appears that as we shorten the time period, the average velocity is becoming closer to 49 ms. The instantaneous velocity when t 5 is defined to be the limiting value of these average velocities over shorter and shorter time periods that start at t 5. Thus the (instantaneous) velocity after 5 s is v 49 ms
■
SECTION 1.3
THE LIMIT OF A FUNCTION
■
25
INTUITIVE DEFINITION OF A LIMIT
Let’s investigate the behavior of the function f defined by f x x 2 x 2 for values of x near 2. The following table gives values of f x for values of x close to 2, but not equal to 2. y
ƒ approaches 4.
y=≈-x+2
4
0
2
As x approaches 2, FIGURE 1
x
f x
x
f x
1.0 1.5 1.8 1.9 1.95 1.99 1.995 1.999
2.000000 2.750000 3.440000 3.710000 3.852500 3.970100 3.985025 3.997001
3.0 2.5 2.2 2.1 2.05 2.01 2.005 2.001
8.000000 5.750000 4.640000 4.310000 4.152500 4.030100 4.015025 4.003001
x
From the table and the graph of f (a parabola) shown in Figure 1 we see that when x is close to 2 (on either side of 2), f x is close to 4. In fact, it appears that we can make the values of f x as close as we like to 4 by taking x sufficiently close to 2. We express this by saying “the limit of the function f x x 2 x 2 as x approaches 2 is equal to 4.” The notation for this is lim x 2 x 2 4 x l2
In general, we use the following notation. 1 DEFINITION
We write lim f x L
xla
and say
“the limit of f x, as x approaches a, equals L”
if we can make the values of f x arbitrarily close to L (as close to L as we like) by taking x to be sufficiently close to a (on either side of a) but not equal to a. Roughly speaking, this says that the values of f x tend to get closer and closer to the number L as x gets closer and closer to the number a (from either side of a) but x a. An alternative notation for lim f x L
xla
is
f x l L
as
xla
which is usually read “ f x approaches L as x approaches a.” Notice the phrase “but x a” in the definition of limit. This means that in finding the limit of f x as x approaches a, we never consider x a. In fact, f x need not even be defined when x a. The only thing that matters is how f is defined near a.
26
■
CHAPTER 1
FUNCTIONS AND LIMITS
Figure 2 shows the graphs of three functions. Note that in part (c), f a is not defined and in part (b), f a L. But in each case, regardless of what happens at a, it is true that lim x l a f x L. y
y
y
L
L
L
0
a
0
x
a
(a)
0
x
(b)
x
a
(c)
FIGURE 2 lim ƒ=L in all three cases x a
EXAMPLE 2 Guess the value of lim x l1
x1
f x
0.5 0.9 0.99 0.999 0.9999
0.666667 0.526316 0.502513 0.500250 0.500025
x1 . x2 1
SOLUTION Notice that the function f x x 1x 2 1 is not defined when
x 1, but that doesn’t matter because the definition of lim x l a f x says that we consider values of x that are close to a but not equal to a. The tables at the left give values of f x (correct to six decimal places) for values of x that approach 1 (but are not equal to 1). On the basis of the values in the tables, we make the guess that lim x l1
x1
f x
1.5 1.1 1.01 1.001 1.0001
0.400000 0.476190 0.497512 0.499750 0.499975
x1 0.5 x2 1
■
Example 2 is illustrated by the graph of f in Figure 3. Now let’s change f slightly by giving it the value 2 when x 1 and calling the resulting function t :
t(x)
x1 x2 1
if x 1
2
if x 1
This new function t still has the same limit as x approaches 1. (See Figure 4.) y
y 2
y=
x-1 ≈-1
y=©
0.5
0
FIGURE 3
0.5
1
x
0
FIGURE 4
1
x
SECTION 1.3
EXAMPLE 3 Estimate the value of lim tl0
THE LIMIT OF A FUNCTION
■
27
st 2 9 3 . t2
SOLUTION The table lists values of the function for several values of t near 0.
t
st 2 9 3 t2
1.0 0.5 0.1 0.05 0.01
0.16228 0.16553 0.16662 0.16666 0.16667
As t approaches 0, the values of the function seem to approach 0.1666666 . . . and so we guess that lim
st 2 9 3 t2
t
tl0
1 st 2 9 3 2 t 6
■
In Example 3 what would have happened if we had taken even smaller values of t? The table in the margin shows the results from one calculator; you can see that something strange seems to be happening. If you try these calculations on your own calculator you might get different values, but eventually you will get the value 0 if you make t sufficiently small. Does this mean 1 that the answer is really 0 instead of 6? No, the value of the limit is 16 , as we will show | in the next section. The problem is that the calculator gave false values because st 2 9 is very close to 3 when t is small. (In fact, when t is sufficiently small, a cal■ www.stewartcalculus.com culator’s value for st 2 9 is 3.000. . . to as many digits as the calculator is capable For a further explanation of why calof carrying.) culators sometimes give false values, Something similar happens when we try to graph the function 0.0005 0.0001 0.00005 0.00001
0.16800 0.20000 0.00000 0.00000
click on Lies My Calculator and Computer Told Me. In particular, see the section called The Perils of Subtraction.
f t
st 2 9 3 t2
of Example 3 on a graphing calculator or computer. Parts (a) and (b) of Figure 5 show quite accurate graphs of f , and when we use the trace mode (if available) we can estimate easily that the limit is about 16. But if we zoom in too much, as in parts (c) and (d), then we get inaccurate graphs, again because of problems with subtraction.
0.2
0.2
0.1
0.1
(a) _5, 5 by _0.1, 0.3 FIGURE 5
(b) _0.1, 0.1 by _0.1, 0.3
(c) _10–^, 10–^ by _0.1, 0.3
(d) _10–&, 10–& by _ 0.1, 0.3
28
■
CHAPTER 1
FUNCTIONS AND LIMITS
sin x . x SOLUTION The function f x sin xx is not defined when x 0. Using a calculator (and remembering that, if x ⺢, sin x means the sine of the angle whose radian measure is x), we construct the table of values correct to eight decimal places. From the table at the left and the graph in Figure 6 we guess that V EXAMPLE 4
x
sin x x
1.0 0.5 0.4 0.3 0.2 0.1 0.05 0.01 0.005 0.001
0.84147098 0.95885108 0.97354586 0.98506736 0.99334665 0.99833417 0.99958339 0.99998333 0.99999583 0.99999983
Guess the value of lim
xl0
lim
xl0
sin x 1 x
This guess is in fact correct, as will be proved in the next section using a geometric argument. y
_1
FIGURE 6
V EXAMPLE 5 COMPUTER ALGEBRA SYSTEMS Computer algebra systems (CAS) have commands that compute limits. In order to avoid the types of pitfalls demonstrated in Examples 3 and 5, they don’t find limits by numerical experimentation. Instead, they use more sophisticated techniques such as computing infinite series. If you have access to a CAS, use the limit command to compute the limits in the examples of this section and to check your answers in the exercises of this chapter. ■
Investigate lim sin xl0
1
y=
0
1
sin x x
x
■
. x
SOLUTION Again the function f x sinx is undefined at 0. Evaluating the
function for some small values of x, we get f 1 sin 0
f ( 12 ) sin 2 0
f ( 13) sin 3 0
f ( 14 ) sin 4 0
f 0.1 sin 10 0
f 0.01 sin 100 0
Similarly, f 0.001 f 0.0001 0. On the basis of this information we might be tempted to guess that lim sin 0 xl0 x
| but this time our guess is wrong. Note that although f 1n sin n 0 for any integer n, it is also true that f x 1 for infinitely many values of x that approach 0. The graph of f is given in Figure 7. y
y=sin(π/x)
1
_1 1
_1
FIGURE 7
x
SECTION 1.3
THE LIMIT OF A FUNCTION
■
29
The dashed lines near the y-axis indicate that the values of sinx oscillate between 1 and 1 infinitely often as x approaches 0. (Use a graphing device to graph f and zoom in toward the origin several times. What do you observe?) Since the values of f x do not approach a fixed number as x approaches 0, lim sin
xl0
|
x
The Heaviside function H is defined by
Ht y
0
FIGURE 8
t
■
Examples 3 and 5 illustrate some of the pitfalls in guessing the value of a limit. It is easy to guess the wrong value if we use inappropriate values of x, but it is difficult to know when to stop calculating values. And, as the discussion after Example 3 shows, sometimes calculators and computers give the wrong values. In the next section, however, we will develop foolproof methods for calculating limits. V EXAMPLE 6
1
does not exist
0 1
if t 0 if t 0
[This function is named after the electrical engineer Oliver Heaviside (1850–1925) and can be used to describe an electric current that is switched on at time t 0.] Its graph is shown in Figure 8. As t approaches 0 from the left, Ht approaches 0. As t approaches 0 from the right, Ht approaches 1. There is no single number that Ht approaches as t approaches 0. Therefore, lim t l 0 Ht does not exist. ■
ONE-SIDED LIMITS
We noticed in Example 6 that Ht approaches 0 as t approaches 0 from the left and Ht approaches 1 as t approaches 0 from the right. We indicate this situation symbolically by writing lim Ht 0
t l0
and
lim Ht 1
t l0
The symbol “t l 0 ” indicates that we consider only values of t that are less than 0. Likewise, “t l 0 ” indicates that we consider only values of t that are greater than 0.
2 DEFINITION
We write lim f x L
x la
and say the left-hand limit of f x as x approaches a [or the limit of f x as x approaches a from the left] is equal to L if we can make the values of f x arbitrarily close to L by taking x to be sufficiently close to a and x less than a.
30
■
CHAPTER 1
FUNCTIONS AND LIMITS
Notice that Definition 2 differs from Definition 1 only in that we require x to be less than a. Similarly, if we require that x be greater than a, we get “the right-hand limit of f x as x approaches a is equal to L” and we write lim f x L
x l a
Thus, the symbol “x l a” means that we consider only x a. These definitions are illustrated in Figure 9. y
y
L
ƒ 0
x
a
0
x
a
(a) lim ƒ=L
FIGURE 9
ƒ
L
x
x
(b) lim ƒ=L
x a_
x a+
By comparing Definition l with the definitions of one-sided limits, we see that the following is true.
3
3
y=©
lim f x L
x la
(a) lim tx
(b) lim tx
(c) lim tx
(d) lim tx
(e) lim tx
(f ) lim tx
xl2
xl5
1
FIGURE 10
if and only if
and
lim f x L
x la
V EXAMPLE 7 The graph of a function t is shown in Figure 10. Use it to state the values (if they exist) of the following:
y 4
0
lim f x L
xla
1
2
3
4
5
x
xl2
xl2
xl5
xl5
SOLUTION From the graph we see that the values of tx approach 3 as x
approaches 2 from the left, but they approach 1 as x approaches 2 from the right. Therefore (a) lim tx 3
and
xl2
(b) lim tx 1 xl2
(c) Since the left and right limits are different, we conclude from (3) that lim x l 2 tx does not exist. The graph also shows that (d) lim tx 2
and
xl5
(e) lim tx 2 xl5
(f ) This time the left and right limits are the same and so, by (3), we have lim tx 2
xl5
Despite this fact, notice that t5 2.
■
SECTION 1.3
EXAMPLE 8 Find lim
xl0
THE LIMIT OF A FUNCTION
■
31
1 if it exists. x2
SOLUTION As x becomes close to 0, x 2 also becomes close to 0, and 1x 2 becomes
very large. (See the following table.) In fact, it appears from the graph of the function f x 1x 2 shown in Figure 11 that the values of f x can be made arbitrarily large by taking x close enough to 0. Thus the values of f x do not approach a number, so lim x l 0 1x 2 does not exist.
x
1 x2
1 0.5 0.2 0.1 0.05 0.01 0.001
1 4 25 100 400 10,000 1,000,000
y
y=
1 ≈
x
0
■
FIGURE 11
PRECISE DEFINITION OF A LIMIT
Definition 1 is appropriate for an intuitive understanding of limits, but for deeper understanding and rigorous proofs we need to be more precise. We want to express, in a quantitative manner, that f x can be made arbitrarily close to L by taking x to be sufficiently close to a (but x a. This means that f x can be made to lie within any preassigned distance from L (traditionally denoted by , the Greek letter epsilon) by requiring that x be within a specified distance (the Greek letter delta) from a . That is, f x L when x a and x a. Notice that we can stipulate that x a by writing 0 x a . The resulting precise definition of a limit is as follows.
4 DEFINITION Let f be a function defined on some open interval that contains the number a, except possibly at a itself. Then we say that the limit of f x as x approaches a is L, and we write
lim f x L
xla
if for every number 0 there is a corresponding number 0 such that if
In Module 1.3/1.6 you can explore the precise definition of a limit both graphically and numerically.
0 xa
then
f x L
Definition 4 is illustrated in Figures 12 –14. If a number 0 is given, then we draw the horizontal lines y L and y L and the graph of f . (See Figure 12.) If limx l a f x L, then we can find a number 0 such that if we restrict x to lie in the interval a , a and take x a, then the curve y f x lies
32
■
CHAPTER 1
FUNCTIONS AND LIMITS
between the lines y L and y L . (See Figure 13.) You can see that if such a has been found, then any smaller will also work. y
y
y
y=ƒ
L+∑ y=L+∑
ƒ is in here
∑ ∑
L
y=L+∑
y=L-∑
y=L+∑
∑ ∑
L
y=L-∑
y=L-∑ L-∑
0
0
x
a
a-∂
0
x
a
a+∂
x
a
a-∂
a+∂
when x is in here (x≠ a) FIGURE 12
FIGURE 13
FIGURE 14
It’s important to realize that the process illustrated in Figures 12 and 13 must work for every positive number , no matter how small it is chosen. Figure 14 shows that if a smaller is chosen, then a smaller may be required. In proving limit statements it may be helpful to think of the definition of limit as a challenge. First it challenges you with a number . Then you must be able to produce a suitable . You have to be able to do this for every 0, not just a particular . Prove that lim 4x 5 7.
V EXAMPLE 9
Figure 15 shows the geometry behind Example 9. ■
y
and L 7, we need to find a number such that
then 4x 5 7 But 4x 5 7 4x 12 4x 3 4 x 3 . Therefore, we want: if 0 x 3 then 4 x 3 0 x3
if
y=4x-5
7+∑
x l3
SOLUTION Let be a given positive number. According to Definition 4 with a 3
7
7-∑
We can choose to be 4 because
0
4
then
4 x3 4
x
3
3-∂
0 x3
if
4
Therefore, by the definition of a limit,
3+∂
lim 4x 5 7
FIGURE 15
■
x l3
For a left-hand limit we restrict x so that x a , so in Definition 4 we replace 0 x a by a x a . Similarly, for a right-hand limit we use a x a .
y
y=œ„ x
V EXAMPLE 10
y=∑
∑
Prove that lim sx 0. x l0
SOLUTION Let be a given positive number. We want to find a number such that
if 0
∂=∑@
FIGURE 16
x
0x
then
sx 0
that is
sx
But sx &? x 2 . So if we choose 2 and 0 x 2, then sx . (See Figure 16.) This shows that sx l 0 as x l 0.
■
SECTION 1.3
1.3
(g) t2
height in feet t seconds later is given by y 40t 16t 2. (a) Find the average velocity for the time period beginning when t 2 and lasting (i) 0.5 second (ii) 0.1 second (iii) 0.05 second (iv) 0.01 second (b) Estimate the instantaneous velocity when t 2.
(h) lim tt tl4
y 4 2
2. If an arrow is shot upward on the moon with a velocity of 2
58 ms, its height in meters t seconds later is given by h 58t 0.83t 2. (a) Find the average velocity over the given time intervals: (i) [1, 2] (ii) [1, 1.5] (iii) [1, 1.1] (iv) [1, 1.01] (v) [1, 1.001] (b) Estimate the instantaneous velocity when t 1.
determine the values of a for which lim x l a f x exists:
2x f x x x 12
if it exists. If it does not exist, explain why. (a) lim f x (b) lim f x (c) lim f x xl1
xl1
xl1
(e) f 5
■ Sketch the graph of an example of a function f that satisfies all of the given conditions.
7. lim f x 2,
x l 1
lim f x 2,
8. lim f x 1,
x l 0
xl1
4
lim f x 1,
2
4
9. lim f x 4,
f 3 3,
each quantity, if it exists. If it does not exist, explain why. (a) lim f x (b) lim f x (c) lim f x xl3
xl3
xl3
(e) f 3
■
■
lim f x 3,
x l 4
f 4 1 ■
■
■
■
■
x 2 2x x l2 x x 2 x 2.5, 2.1, 2.05, 2.01, 2.005, 2.001, 1.9, 1.95, 1.99, 1.995, 1.999
11. lim
2
0
2
4
x
each quantity, if it exists. If it does not exist, explain why. (a) lim tt (b) lim tt (c) lim tt tl0
(e) lim tt tl2
tl0
( f ) lim tt tl2
2
x 2 2x xl 1 x x 2 x 0, 0.5, 0.9, 0.95, 0.99, 0.999, 2, 1.5, 1.1, 1.01, 1.001
12. lim
5. For the function t whose graph is given, state the value of
tl2
lim f x 3,
x l 4
■
■
■ Guess the value of the limit (if it exists) by evaluating the function at the given numbers (correct to six decimal places).
4
(d) lim tt
f 2 1
xl1
■
lim f x 2,
x l 2
11–14
y
tl0
lim f x 2,
10. lim f x 3,
f 1 1,
lim f x 0,
x l 2
f 0 is undefined
x l 3
xl3
x
4. For the function f whose graph is given, state the value of xl0
f 2 1,
x l 2
2
f 1 2
lim f x 1,
xl0
(d) lim f x
if x 1 if 1 x 1 if x 1
7–10
y
0
t
4
6. Sketch the graph of the following function and use it to
3. Use the given graph of f to state the value of each quantity,
xl5
33
EXERCISES
1. If a ball is thrown into the air with a velocity of 40 fts, its
(d) lim f x
■
THE LIMIT OF A FUNCTION
2
sin x x tan x x 1, 0.5, 0.2, 0.1, 0.05, 0.01
13. lim
xl0
■
34
■
CHAPTER 1
FUNCTIONS AND LIMITS
sx 4 , x 17, 16.5, 16.1, 16.05, 16.01, x 16 15, 15.5, 15.9, 15.95, 15.99
23. Use the given graph of f x 1x to find a number such
14. lim
x l 16
that
x 2
if ■
■
■
■
■
■
■
■
■
■
■
■
■
■
x 1 x10 1 ■
9 5 x x
18. lim
xl0
■
■
■
■
■
0.5 0.3 x
0 ■
■
ing in toward the point where the graph crosses the y-axis, estimate the value of lim x l 0 f x. (b) Check your answer in part (a) by evaluating f x for values of x that approach 0.
that
x 1
if
x
then
2
y
6x 2x x
xl0
y=≈
1 0.5
by graphing the function y 6 2 x. State your answer correct to two decimal places. (b) Check your answer in part (a) by evaluating f x for values of x that approach 0. x
x
21. (a) Evaluate the function f x x 2 2 x1000 for x 1,
0.8, 0.6, 0.4, 0.2, 0.1, and 0.05, and guess the value of 2x lim x xl0 1000 2
(b) Evaluate f x for x 0.04, 0.02, 0.01, 0.005, 0.003, and 0.001. Guess again. 22. (a) Evaluate hx tan x xx for x 1, 0.5, 0.1, 0.05, 3
0.01, and 0.005.
tan x x . x3 (c) Evaluate hx for successively smaller values of x until you finally reach 0 values for hx. Are you still confident that your guess in part (b) is correct? Explain why you eventually obtained 0 values. (In Section 3.7 a method for evaluating the limit will be explained.) (d) Graph the function h in the viewing rectangle 1, 1 by 0, 1 . Then zoom in toward the point where the graph crosses the y-axis to estimate the limit of hx as x approaches 0. Continue to zoom in until you observe distortions in the graph of h. Compare with the results of part (c). (b) Guess the value of lim
xl0
1 21
1.5
lim
x
10 3
24. Use the given graph of f x x 2 to find a number such
; 20. (a) Estimate the value of
2
10 7
■
; 19. (a) By graphing the function f x tan 4xx and zoom-
;
1 y= x
1
tan 3x 16. lim x l 0 tan 5x
6
xl1
1 0.5 0.2 x
0.7
sx 4 2 15. lim xl0 x 17. lim
y
Use a table of values to estimate the value of the limit. If you have a graphing device, use it to confirm your result graphically.
15–18
then
0
?
1
x
?
; 25. Use a graph to find a number such that
x 2
if
then
s4x 1 3 0.5
; 26. Use a graph to find a number such that if
x
6
then
| sin x | 0.1 1 2
27. A machinist is required to manufacture a circular metal disk
with area 1000 cm2. (a) What radius produces such a disk? (b) If the machinist is allowed an error tolerance of 5 cm2 in the area of the disk, how close to the ideal radius in part (a) must the machinist control the radius? (c) In terms of the , definition of limx l a f x L , what is x ? What is f x ? What is a? What is L ? What value of is given? What is the corresponding value of ?
; 28. A crystal growth furnace is used in research to determine
how best to manufacture crystals used in electronic components for the space shuttle. For proper growth of the crystal, the temperature must be controlled accurately by adjusting the input power. Suppose the relationship is given by Tw 0.1w 2 2.155w 20
SECTION 1.4
39. lim x 2 0
where T is the temperature in degrees Celsius and w is the power input in watts. (a) How much power is needed to maintain the temperature at 200 C ? (b) If the temperature is allowed to vary from 200 C by up to 1 C , what range of wattage is allowed for the input power? (c) In terms of the , definition of limx l a f x L, what is x ? What is f x ? What is a? What is L ? What value of is given? What is the corresponding value of ?
31. lim 1 4x 13
32. lim 7 3x 5
xl1
x l3
■
■
33– 44
■
■
33. lim x l3
35. lim
■
■
■
■
■
CAS
■
34. lim
3x 4 5
xl6
7
37. lim x a
■
■
■
■
■
■
■
■
■
45. (a) For the limit lim x l 1 x 3 x 1 3, use a graph to
find a value of that corresponds to 0.4. (b) By using a computer algebra system to solve the cubic equation x 3 x 1 3 , find the largest possible value of that works for any given 0. (c) Put 0.4 in your answer to part (b) and compare with your answer to part (a).
■
x 9 3 4 2
46. If H is the Heaviside function defined in Example 6, prove,
x 2 x 12 7 36. lim x l3 x3
using Definition 4, that lim t l 0 Ht does not exist. [Hint: Use an indirect proof as follows. Suppose that the limit is L. Take 21 in the definition of a limit and try to arrive at a contradiction.]
xla
1.4
■
38. lim c c
xla
[Hint: If x 3 1 , what can you say about x 4 ?]
Prove the statement using the , definition of limit.
x 3 5 5
x l5
■
[Hint: Write x 2 9 x 3 x 3 .
x l3
xl4
■
xl9
44. lim x 2 x 4 8
( 12 x 3) 2 x l2
30. lim
4 9x 0 42. lim s
Show that if x 3 1 , then x 3 7 . If you let be the smaller of the numbers 1 and 7 , show that this works.]
■
29. lim 2x 3 5
43. lim x 2 9
Prove the statement using the , definition of limit and illustrate with a diagram like Figure 15. 29–32
xl0
41. lim x 0
xl3
35
40. lim x 3 0
xl0
xl0
■
CALCULATING LIMITS
CALCULATING LIMITS In Section 1.3 we used calculators and graphs to guess the values of limits, but we saw that such methods don’t always lead to the correct answer. In this section we use the following properties of limits, called the Limit Laws, to calculate limits. LIMIT LAWS Suppose that c is a constant and the limits
lim f x
xla
and
exist. Then 1. lim f x tx lim f x lim tx xla
xla
xla
2. lim f x tx lim f x lim tx xla
xla
xla
3. lim cf x c lim f x xla
xla
4. lim f xtx lim f x lim tx xla
xla
5. lim
lim f x f x xla tx lim tx
xla
xla
xla
if lim tx 0 xla
lim tx
xla
36
■
CHAPTER 1
FUNCTIONS AND LIMITS
Sum Law Difference Law Constant Multiple Law
These five laws can be stated verbally as follows: 1. The limit of a sum is the sum of the limits. 2. The limit of a difference is the difference of the limits. 3. The limit of a constant times a function is the constant times the limit of the function.
Product Law
4. The limit of a product is the product of the limits.
Quotient Law
5. The limit of a quotient is the quotient of the limits (provided that the limit of
the denominator is not 0). It is easy to believe that these properties are true. For instance, if f x is close to L and tx is close to M, it is reasonable to conclude that f x tx is close to L M. This gives us an intuitive basis for believing that Law 1 is true. All of these laws can be proved using the precise definition of a limit. (See Appendix B.) If we use the Product Law repeatedly with tx f x, we obtain the following law. Power Law
6. lim f x n lim f x x la
[
x la
n
]
where n is a positive integer
In applying these six limit laws, we need to use two special limits: 7. lim c c
8. lim x a
xla
xla
These limits are obvious from an intuitive point of view (state them in words or draw graphs of y c and y x), but they can be proved from the precise definition. (See Exercises 37 and 38 in Section 1.3.) If we now put f x x in Law 6 and use Law 8, we get another useful special limit. 9. lim x n a n
where n is a positive integer
xla
A similar limit holds for roots as follows. n n 10. lim s x s a
where n is a positive integer
xla
(If n is even, we assume that a 0.)
More generally, we have the following law. Root Law
n 11. lim s f x)
x la
f x) s lim x la n
where n is a positive integer
[If n is even, we assume that lim f x 0.] x la
SECTION 1.4
NEWTON AND LIMITS Isaac Newton was born on Christmas Day in 1642, the year of Galileo’s death. When he entered Cambridge University in 1661 Newton didn’t know much mathematics, but he learned quickly by reading Euclid and Descartes and by attending the lectures of Isaac Barrow. Cambridge was closed because of the plague in 1665 and 1666, and Newton returned home to reflect on what he had learned. Those two years were amazingly productive for at that time he made four of his major discoveries: (1) his representation of functions as sums of infinite series, including the binomial theorem; (2) his work on differential and integral calculus; (3) his laws of motion and law of universal gravitation; and (4) his prism experiments on the nature of light and color. Because of a fear of controversy and criticism, he was reluctant to publish his discoveries and it wasn’t until 1687, at the urging of the astronomer Halley, that Newton published Principia Mathematica. In this work, the greatest scientific treatise ever written, Newton set forth his version of calculus and used it to investigate mechanics, fluid dynamics, and wave motion, and to explain the motion of planets and comets. The beginnings of calculus are found in the calculations of areas and volumes by ancient Greek scholars such as Eudoxus and Archimedes. Although aspects of the idea of a limit are implicit in their “method of exhaustion,” Eudoxus and Archimedes never explicitly formulated the concept of a limit. Likewise, mathematicians such as Cavalieri, Fermat, and Barrow, the immediate precursors of Newton in the development of calculus, did not actually use limits. It was Isaac Newton who was the first to talk explicitly about limits. He explained that the main idea behind limits is that quantities “approach nearer than by any given difference.” Newton stated that the limit was the basic concept in calculus, but it was left to later mathematicians like Cauchy to clarify his ideas about limits. ■
CALCULATING LIMITS
■
37
EXAMPLE 1 Evaluate the following limits and justify each step.
(a) lim 2x 2 3x 4
(b) lim
x l5
x l 2
x 3 2x 2 1 5 3x
SOLUTION
lim 2x 2 3x 4 lim 2x 2 lim 3x lim 4
(a)
x l5
x l5
x l5
x l5
(by Laws 2 and 1)
2 lim x 2 3 lim x lim 4
(by 3)
25 2 35 4
(by 9, 8, and 7)
x l5
x l5
x l5
39 (b) We start by using Law 5, but its use is fully justified only at the final stage when we see that the limits of the numerator and denominator exist and the limit of the denominator is not 0. lim
x l2
lim x 3 2x 2 1 x 3 2x 2 1 x l2 5 3x lim 5 3x
(by Law 5)
x l2
lim x 3 2 lim x 2 lim 1
x l2
x l2
x l2
x l2
lim 5 3 lim x
23 222 1 5 32
(by 1, 2, and 3)
x l2
(by 9, 8, and 7)
1 11
■
NOTE If we let f x 2x 2 3x 4, then f 5 39. In other words, we would
have gotten the correct answer in Example 1(a) by substituting 5 for x. Similarly, direct substitution provides the correct answer in part (b). The functions in Example 1 are a polynomial and a rational function, respectively, and similar use of the Limit Laws proves that direct substitution always works for such functions (see Exercises 49 and 50). We state this fact as follows. DIRECT SUBSTITUTION PROPERTY If f is a polynomial or a rational function
and a is in the domain of f , then lim f x f a x la
y
P(cos ¨, sin ¨) 1 ¨ 0
FIGURE 1
(1, 0)
x
The trigonometric functions also enjoy the Direct Substitution Property. We know from the definitions of sin and cos that the coordinates of the point P in Figure 1 are cos , sin . As l 0, we see that P approaches the point 1, 0 and so cos l 1 and sin l 0. Thus 1
lim cos 1
l0
lim sin 0
l0
Since cos 0 1 and sin 0 0, the equations in (1) assert that the cosine and sine
38
■
CHAPTER 1
FUNCTIONS AND LIMITS
■ Another way to establish the limits in (1) is to use the inequality sin (for 0 ), which is proved on page 42.
functions satisfy the Direct Substitution Property at 0. The addition formulas for cosine and sine can then be used to deduce that these functions satisfy the Direct Substitution Property everywhere (see Exercises 51 and 52). In other words, for any real number a, lim sin sin a lim cos cos a la
la
This enables us to evaluate certain limits quite simply. For example, lim x cos x lim x lim cos x cos
(
xl
)(
xl
xl
)
Functions with the Direct Substitution Property are called continuous at a and will be studied in Section 1.5. However, not all limits can be evaluated by direct substitution, as the following examples show. EXAMPLE 2 Find lim
xl1
x2 1 . x1
SOLUTION Let f x x 2 1x 1. We can’t find the limit by substituting
x 1 because f 1 isn’t defined. Nor can we apply the Quotient Law, because the limit of the denominator is 0. Instead, we need to do some preliminary algebra. We factor the numerator as a difference of squares: x2 1 x 1x 1 x1 x1 The numerator and denominator have a common factor of x 1. When we take the limit as x approaches 1, we have x 1 and so x 1 0. Therefore, we can cancel the common factor and compute the limit as follows: lim
xl1
x2 1 x 1x 1 lim x l 1 x1 x1 lim x 1 xl1
112
■
NOTE In Example 2 we were able to compute the limit by replacing the given function f x x 2 1x 1 by a simpler function, tx x 1, with the same limit. This is valid because f x tx except when x 1, and in computing a limit as x approaches 1 we don’t consider what happens when x is actually equal to 1. In general, we have the following useful fact.
If f x tx when x a, then lim f x lim tx, provided the limits exist. xla
xla
EXAMPLE 3 Find lim tx where x l1
tx
x 1 if x 1 if x 1
SECTION 1.4
y
y=ƒ
3 2
xl1
1
2
3
x
y=©
V EXAMPLE 4
Evaluate lim
hl0
SOLUTION If we define
Fh
1 1
2
3
x
■
xl1
3 h2 9 . h
2
0
39
Note that the values of the functions in Examples 2 and 3 are identical except when x 1 (see Figure 2) and so they have the same limit as x approaches 1.
y 3
■
SOLUTION Here t is defined at x 1 and t1 , but the value of a limit as x approaches 1 does not depend on the value of the function at 1. Since tx x 1 for x 1, we have lim tx lim x 1 2
1 0
CALCULATING LIMITS
3 h2 9 h
then, as in Example 2, we can’t compute lim h l 0 Fh by letting h 0 since F0 is undefined. But if we simplify Fh algebraically, we find that
FIGURE 2
The graphs of the functions f (from Example 2) and g (from Example 3)
Fh
9 6h h 2 9 6h h 2 6h h h
(Recall that we consider only h 0 when letting h approach 0.) Thus lim
hl0
EXAMPLE 5 Find lim tl0
3 h2 9 lim 6 h 6 hl0 h
■
st 2 9 3 . t2
SOLUTION We can’t apply the Quotient Law immediately, since the limit of the denominator is 0. Here the preliminary algebra consists of rationalizing the numerator:
lim tl0
st 2 9 3 st 2 9 3 st 2 9 3 lim tl0 t2 t2 st 2 9 3 lim
t 2 9 9 t2 lim 2 2 2 t l 0 t (st 9 3) t (st 9 3)
lim
1 st 9 3
tl0
tl0
2
2
1 1 1 s lim t 9 3 3 3 6 2
tl0
This calculation confirms the guess that we made in Example 3 in Section 1.3.
■
Some limits are best calculated by first finding the left- and right-hand limits. The following theorem is a reminder of what we discovered in Section 1.3. It says that a two-sided limit exists if and only if both of the one-sided limits exist and are equal.
2
THEOREM
lim f x L
xla
if and only if
lim f x L lim f x
xla
x la
40
■
CHAPTER 1
FUNCTIONS AND LIMITS
When computing one-sided limits, we use the fact that the Limit Laws also hold for one-sided limits.
EXAMPLE 6 Show that lim x 0. xl0
SOLUTION Recall that
x
if x 0 if x 0
x x
Since x x for x 0, we have
The result of Example 6 looks plausible from Figure 3. ■
lim x lim x 0
x l 0
y
y=| x |
xl0
For x 0 we have x x and so
lim x lim x 0
x l 0
0
x
Therefore, by Theorem 2,
lim x 0
FIGURE 3
xl0
V EXAMPLE 7
SOLUTION
y | x|
y= x
xl0
lim
x
lim
x
xl0
x
x
xl0
1 0
x does not exist.
Prove that lim
x l 0
■
x
x
lim
x lim 1 1 xl0 x
lim
x lim 1 1 xl0 x
x l 0
xl0
_1
Since the right- and left-hand limits are different, it follows from Theorem 2 that lim x l 0 x x does not exist. The graph of the function f x x x is shown in Figure 4 and supports the one-sided limits that we found. ■
FIGURE 4 ■ Other notations for x are x and ⎣ x⎦. The greatest integer function is sometimes called the floor function.
EXAMPLE 8 The greatest integer function is defined by x the largest integer
y
that is less than or equal to x. (For instance, 4 4, 4.8 4, 3, s2 1, 12 1.) Show that lim x l3 x does not exist.
4
SOLUTION The graph of the greatest integer function is shown in Figure 5. Since
3
x 3 for 3 x 4, we have y=[ x]
2
lim x lim 3 3
x l3
1 0
1
2
3
4
5
x
x l3
Since x 2 for 2 x 3, we have lim x lim 2 2
x l3
FIGURE 5
Greatest integer function
x l3
Because these one-sided limits are not equal, lim x l3 x does not exist by Theorem 2.
■
SECTION 1.4
CALCULATING LIMITS
■
41
The next two theorems give two additional properties of limits. Their proofs can be found in Appendix B. 3 THEOREM If f x tx when x is near a (except possibly at a) and the limits of f and t both exist as x approaches a, then
lim f x lim tx
xla
4
xla
THE SQUEEZE THEOREM If f x tx hx when x is near a (except
possibly at a) and lim f x lim hx L
y
xla
h g
xla
lim tx L
then
xla
L
f 0
x
a
FIGURE 6
The Squeeze Theorem, which is sometimes called the Sandwich Theorem or the Pinching Theorem, is illustrated by Figure 6. It says that if tx is squeezed between f x and hx near a, and if f and h have the same limit L at a, then t is forced to have the same limit L at a. 1 0. x SOLUTION First note that we cannot use V EXAMPLE 9
Show that lim x 2 sin xl0
lim x 2 sin
xl0
1 1 lim x 2 lim sin xl0 xl0 x x
because lim x l 0 sin1x does not exist (see Example 5 in Section 1.3). However, since 1 sin y
1
1 x
we have, as illustrated by Figure 7,
y=≈
x 2 x 2 sin
1
x2 x
x
0
We know that y=_≈ FIGURE 7
y=≈ sin(1/x)
lim x 2 0
xl0
and
lim x 2 0
xl0
Taking f x x 2, tx x 2 sin1x, and hx x 2 in the Squeeze Theorem, we obtain 1 lim x 2 sin 0 xl0 ■ x
42
■
CHAPTER 1
FUNCTIONS AND LIMITS
In Example 4 in Section 1.3 we made the guess, on the basis of numerical and graphical evidence, that
lim
5
D
l0
We can prove Equation 5 with help from the Squeeze Theorem. Assume first that lies between 0 and 2. Figure 8(a) shows a sector of a circle with center O, central angle , and radius 1. BC is drawn perpendicular to OA. By the definition of radian measure, we have arc AB . Also, BC OB sin sin . From the diagram we see that
B
¨
sin
Therefore 1
C
A
(a)
sin 1
so
Let the tangent lines at A and B intersect at E. You can see from Figure 8(b) that the circumference of a circle is smaller than the length of a circumscribed polygon, and so arc AB AE EB . Thus
B E A
O
BC AB arc AB
E
O
sin 1
arc AB AE EB AE ED
AD OA tan tan (b) FIGURE 8
(In Appendix B the inequality tan is proved directly from the definition of the length of an arc without resorting to geometric intuition as we did here.) Therefore, we have
sin cos
and so
cos
sin 1
We know that lim l 0 1 1 and lim l 0 cos 1, so by the Squeeze Theorem, we have lim
l 0
sin 1
But the function sin is an even function, so its right and left limits must be equal. Hence, we have lim
l0
so we have proved Equation 5.
EXAMPLE 10 Find lim
xl0
sin 7x . 4x
sin 1
SECTION 1.4
CALCULATING LIMITS
■
43
SOLUTION In order to apply Equation 5, we first rewrite the function by multiplying and dividing by 7:
sin 7x 7 4x 4
Note that sin 7x 7 sin x.
sin 7x 7x
Notice that as x l 0, we have 7x l 0, and so, by Equation 5 with 7x, sin 7x sin7x lim 1 7x l 0 7x 7x
lim
xl0
Thus
lim
xl0
sin 7x 7 lim xl0 4 4x
EXAMPLE 11 Evaluate lim
l0
sin 7x 7x
7 sin 7x 7 7 lim 1 4 x l 0 7x 4 4
■
cos 1 .
SOLUTION ■ We multiply numerator and denominator by cos 1 in order to put the function in a form in which we can use the limits we know.
lim
l0
cos 1 lim l0 lim
l0
sin 2 lim l0 cos 1
lim
l0
1
1.4
x la
lim tx 0
lim hx 8
x la
(b) lim f x 2
3 hx (c) lim s
(d) lim
1 f x
(f ) lim
tx f x
(e) lim x la
(g) lim x la
f x hx f x tx
sin sin cos 1
sin sin lim l 0 cos 1
0 11
0
(by Equation 5)
■
limit, if it exists. If the limit does not exist, explain why. y
x la
(a) lim f x hx
xla
l0
cos2 1 cos 1
2. The graphs of f and t are given. Use them to evaluate each
find the limits that exist. If the limit does not exist, explain why. x la
lim
EXERCISES
1. Given that
lim f x 3
cos 1 cos 1 cos 1
y=ƒ
x la
(h) lim x la
2 f x hx f x
y=©
1 1
x la
x la
y
x
1
0
1
(a) lim f x tx
(b) lim f x tx
(c) lim f xtx
(d) lim
(e) lim x 3f x
(f ) lim s3 f x
x l2
x l0
x l2
x l1
x l 1
x l1
f x tx
x
44
■
CHAPTER 1
FUNCTIONS AND LIMITS
■ Evaluate the limit and justify each step by indicating the appropriate Limit Law(s).
3–9
3. lim 3x 4 2x 2 x 1
4. lim t 2 13t 35
3 5. lim (1 s x )2 6x 2 x 3
6. lim su 4 3u 6
x l 2
xl8
7. lim x l1
9. ■
1 3x 1 4x 2 3x 4
f x
t l 1
3
8. lim
xl0
cos 4 x 5 2x 3
; 27. Use the Squeeze Theorem to show that
lim sin ■
■
■
■
■
■
■
■
■
lim x l 0 x 2 cos 20 x 0. Illustrate by graphing the functions f x x 2, tx x 2 cos 20 x, and hx x 2 on the same screen.
■
10. (a) What is wrong with the following equation?
; 28. Use the Squeeze Theorem to show that
x2 x 6 x3 x2
lim sx 3 x 2 sin x l0
lim x l2
x2 x 6 lim x 3 x l2 x2
29. If 4x 9 f x x 2 4x 7 for x 0, find
lim x l 4 f x.
is correct.
11. lim x l2
30. If 2x tx x 4 x 2 2 for all x, evaluate lim x l 1 tx.
Evaluate the limit, if it exists.
■
x2 x 6 x2
31. Prove that lim x 4 cos
12. lim
x l4
x2 x 6 13. lim x l2 x2 t2 9 2t 2 7t 3
16. lim
17. lim
4 h 16 h
18. lim
x l1
2
hl0
19. lim
x l2
h l0
x2 x3 8
20. lim
■
24. lim tl0
■
x l0
■ Find the limit, if it exists. If the limit does not exist, explain why.
33–36
x 2 4x x 2 3x 4
33. lim (2x x 3 xl3
35. lim x l0
x 2 2x 1 x4 1
■
■
■
■
1 1 2 t t t ■
■
■
■
x l0
)
x l6
■
■
36. lim x l0
■
■
■
x tx 1 x 2 x1
■
34. lim
1 1 x x
37. Let
■
2x 12 x6
1 1 x x ■
■
■
if x 1 if 1 x 1 if x 1
(a) Evaluate each of the following limits, if it exists. ■
; 25. (a) Estimate the value of lim
2 0. x
32. Prove that lim sx 1 sin2 2x 0 .
3 h1 3 1 22. lim hl0 h
1 1 4 x 23. lim x l 4 4 x ■
x 2 5x 4 x 2 3x 4
s1 h 1 h
x l1
sx 2 3 21. lim x l7 x7
■
x l0
x 2 4x 14. lim 2 x l 4 x 3x 4
15. lim
t l3
0 x
Illustrate by graphing the functions f, t, and h (in the notation of the Squeeze Theorem) on the same screen.
(b) In view of part (a), explain why the equation
11–24
s3 x s3 x
to estimate the value of lim x l 0 f x to two decimal places. (b) Use a table of values of f x to estimate the limit to four decimal places. (c) Use the Limit Laws to find the exact value of the limit.
u l2
l2 ■
; 26. (a) Use a graph of
x s1 3x 1
by graphing the function f x x(s1 3x 1). (b) Make a table of values of f x for x close to 0 and guess the value of the limit. (c) Use the Limit Laws to prove that your guess is correct.
(i) lim tx x l1
(ii) lim tx
(iii) lim tx
(v) lim tx
(vi) lim tx
x l1
(iv) lim tx x l1
x l 1
(b) Sketch the graph of t. 38. Let Fx
(a) Find
x2 1 . x1
(i) lim Fx x l1
(ii) lim Fx x l1
x l0
x l 1
■
SECTION 1.5
we need to show that lim x l a sin x sin a for every real number a. If we let h x a, then x a h and x l a &? h l 0 . So an equivalent statement is that
39. (a) If the symbol denotes the greatest integer function
defined in Example 8, evaluate (i) lim x (ii) lim x
(iii) lim x
x l 2
lim sina h sin a
x l 2.4
h l0
(b) If n is an integer, evaluate (i) lim x (ii) lim x x ln
Use (1) to show that this is true.
xln
(c) For what values of a does lim x l a x exist?
52. Prove that cosine has the Direct Substitution Property.
40. Let f x x x.
53. Show by means of an example that lim x l a f x tx
may exist even though neither limx l a f x nor limx l a tx exists.
(a) Sketch the graph of f. (b) If n is an integer, evaluate (i) lim f x (ii) lim f x x ln
54. Show by means of an example that limx l a f xtx may
x ln
(c) For what values of a does lim x l a f x exist?
exist even though neither lim x l a f x nor limx l a tx exists.
41. If f x x x , show that lim x l 2 f x exists but is
55. Is there a number a such that
not equal to f 2.
42. In the theory of relativity, the Lorentz contraction formula
lim
x l2
L L 0 s1 v 2c 2
43. lim
xl0
45. lim tl0
47. lim
l0
■
■
■
56. The figure shows a fixed circle C1 with equation
x 12 y 2 1 and a shrinking circle C2 with radius r and center the origin. P is the point 0, r, Q is the upper point of intersection of the two circles, and R is the point of intersection of the line PQ and the x-axis. What happens to R as C2 shrinks, that is, as r l 0 ?
Find the limit. sin 3x x
44. lim
xl0
sin 4x sin 6x
y
2
tan 6t sin 2t
46. lim tl0
■
sin 3t t2
P
Q
C™
sin tan ■
3x 2 ax a 3 x2 x 2
exists? If so, find the value of a and the value of the limit.
expresses the length L of an object as a function of its velocity v with respect to an observer, where L 0 is the length of the object at rest and c is the speed of light. Find lim v l c L and interpret the result. Why is a left-hand limit necessary? 43– 48
45
■
51. To prove that sine has the Direct Substitution Property
(b) Does lim x l 1 Fx exist? (c) Sketch the graph of F.
x l2
CONTINUITY
48. lim x cot x xl0
■
■
■
■
■
■
■
■
49. If p is a polynomial, show that lim xl a px pa.
■
0
R
x
C¡
50. If r is a rational function, use Exercise 49 to show that
lim x l a rx ra for every number a in the domain of r.
1.5
CONTINUITY We noticed in Section 1.4 that the limit of a function as x approaches a can often be found simply by calculating the value of the function at a. Functions with this property are called continuous at a. We will see that the mathematical definition of continuity corresponds closely with the meaning of the word continuity in everyday language. (A continuous process is one that takes place gradually, without interruption or abrupt change.)
46
■
CHAPTER 1
FUNCTIONS AND LIMITS
1 DEFINITION
A function f is continuous at a number a if lim f x f a x la
■ As illustrated in Figure 1, if f is continuous, then the points x, f x on the graph of f approach the point a, f a on the graph. So there is no gap in the curve.
y
ƒ approaches f(a).
f(a)
x
a
As x approaches a, FIGURE 1
y
0
1. f a is defined (that is, a is in the domain of f ) 2. lim f x exists x la
3. lim f x f a x la
y=ƒ
0
Notice that Definition l implicitly requires three things if f is continuous at a:
The definition says that f is continuous at a if f x approaches f a as x approaches a. Thus a continuous function f has the property that a small change in x produces only a small change in f x. In fact, the change in f x can be kept as small as we please by keeping the change in x sufficiently small. If f is defined near a (in other words, f is defined on an open interval containing a, except perhaps at a), we say that f is discontinuous at a (or f has a discontinuity at a) if f is not continuous at a. Physical phenomena are usually continuous. For instance, the displacement or velocity of a vehicle varies continuously with time, as does a person’s height. But discontinuities do occur in such situations as electric currents. [See Example 6 in Section 1.3, where the Heaviside function is discontinuous at 0 because lim t l 0 Ht does not exist.] Geometrically, you can think of a function that is continuous at every number in an interval as a function whose graph has no break in it. The graph can be drawn without removing your pen from the paper. EXAMPLE 1 Figure 2 shows the graph of a function f. At which numbers is f discontinuous? Why?
1
2
3
4
5
x
SOLUTION It looks as if there is a discontinuity when a 1 because the graph has a break there. The official reason that f is discontinuous at 1 is that f 1 is not defined. The graph also has a break when a 3, but the reason for the discontinuity is different. Here, f 3 is defined, but lim x l3 f x does not exist (because the left and right limits are different). So f is discontinuous at 3. What about a 5? Here, f 5 is defined and lim x l5 f x exists (because the left and right limits are the same). But
FIGURE 2
lim f x f 5
xl5
■
So f is discontinuous at 5.
Now let’s see how to detect discontinuities when a function is defined by a formula. V EXAMPLE 2
x2 x 2 (a) f x x2 (c) f x
Where are each of the following functions discontinuous?
x2 x 2 x2 1
(b) f x if x 2 if x 2
1 x2 1
(d) f x x
if x 0 if x 0
SECTION 1.5
■
CONTINUITY
47
SOLUTION
(a) Notice that f 2 is not defined, so f is discontinuous at 2. Later we’ll see why f is continuous at all other numbers. (b) Here f 0 1 is defined but lim f x lim
xl0
xl0
1 x2
does not exist. (See Example 8 in Section 1.3.) So f is discontinuous at 0. (c) Here f 2 1 is defined and lim f x lim x l2
x l2
x2 x 2 x 2x 1 lim lim x 1 3 x l2 x l2 x2 x2
exists. But lim f x f 2 x l2
so f is not continuous at 2. (d) The greatest integer function f x x has discontinuities at all of the integers because lim x ln x does not exist if n is an integer. (See Example 8 and Exercise 39 in Section 1.4.) ■ Figure 3 shows the graphs of the functions in Example 2. In each case the graph can’t be drawn without lifting the pen from the paper because a hole or break or jump occurs in the graph. The kind of discontinuity illustrated in parts (a) and (c) is called removable because we could remove the discontinuity by redefining f at just the single number 2. [The function tx x 1 is continuous.] The discontinuity in part (b) is called an infinite discontinuity. The discontinuities in part (d) are called jump discontinuities because the function “jumps” from one value to another. y
y
y
y
1
1
1
1
0
(a) ƒ=
1
2
≈-x-2 x-2
x
0
1 if x≠0 (b) ƒ= ≈ 1 if x=0
0
x
(c) ƒ=
1
2
x
≈-x-2 if x≠2 x-2 1 if x=2
0
1
2
3
(d) ƒ=[ x ]
FIGURE 3 Graphs of the functions in Example 2
2 DEFINITION
A function f is continuous from the right at a number a if lim f x f a
x la
and f is continuous from the left at a if lim f x f a
x la
x
48
■
CHAPTER 1
FUNCTIONS AND LIMITS
EXAMPLE 3 At each integer n, the function f x x [see Figure 3(d)] is continu-
ous from the right but discontinuous from the left because lim f x lim x n f n
x ln
x ln
lim f x lim x n 1 f n
but
x ln
x ln
■
3 DEFINITION A function f is continuous on an interval if it is continuous at every number in the interval. (If f is defined only on one side of an endpoint of the interval, we understand continuous at the endpoint to mean continuous from the right or continuous from the left.)
EXAMPLE 4 Show that the function f x 1 s1 x 2 is continuous on the
interval 1, 1 .
SOLUTION If 1 a 1, then using the Limit Laws, we have
lim f x lim (1 s1 x 2 )
xla
xla
1 lim s1 x 2
(by Laws 2 and 7)
xla
1 s lim 1 x 2
(by 11)
xla
1 s1 a 2
(by 2, 7, and 9)
f a Thus, by Definition l, f is continuous at a if 1 a 1. Similar calculations show that
y
ƒ=1-œ„„„„„ 1-≈
lim f x 1 f 1
1
-1
FIGURE 4
0
x l1
1
x
and
lim f x 1 f 1
x l1
so f is continuous from the right at 1 and continuous from the left at 1. Therefore, according to Definition 3, f is continuous on 1, 1 . The graph of f is sketched in Figure 4. It is the lower half of the circle x 2 y 12 1
■
Instead of always using Definitions 1, 2, and 3 to verify the continuity of a function as we did in Example 4, it is often convenient to use the next theorem, which shows how to build up complicated continuous functions from simple ones. 4 THEOREM If f and t are continuous at a and c is a constant, then the following functions are also continuous at a : 1. f t 2. f t 3. cf f 4. ft 5. if ta 0 t
SECTION 1.5
CONTINUITY
■
49
PROOF Each of the five parts of this theorem follows from the corresponding Limit Law in Section 1.4. For instance, we give the proof of part 1. Since f and t are continuous at a, we have
lim f x f a
lim tx ta
and
xla
xla
Therefore lim f tx lim f x tx
xla
xla
lim f x lim tx xla
xla
(by Law 1)
f a ta f ta This shows that f t is continuous at a.
■
It follows from Theorem 4 and Definition 3 that if f and t are continuous on an interval, then so are the functions f t, f t, cf, ft, and (if t is never 0) ft. The following theorem was stated in Section 1.4 as the Direct Substitution Property. 5 THEOREM
(a) Any polynomial is continuous everywhere; that is, it is continuous on ⺢ , . (b) Any rational function is continuous wherever it is defined; that is, it is continuous on its domain.
PROOF
(a) A polynomial is a function of the form Px cn x n cn1 x n1 c1 x c0 where c0 , c1, . . . , cn are constants. We know that lim c0 c0
xla
and
lim x m a m
xla
(by Law 7)
m 1, 2, . . . , n
(by 9)
This equation is precisely the statement that the function f x x m is a continuous function. Thus, by part 3 of Theorem 4, the function tx cx m is continuous. Since P is a sum of functions of this form and a constant function, it follows from part 1 of Theorem 4 that P is continuous. (b) A rational function is a function of the form f x
Px Qx
where P and Q are polynomials. The domain of f is D x ⺢ Qx 0 . We know from part (a) that P and Q are continuous everywhere. Thus, by part 5 of ■ Theorem 4, f is continuous at every number in D.
50
■
CHAPTER 1
FUNCTIONS AND LIMITS
As an illustration of Theorem 5, observe that the volume of a sphere varies continuously with its radius because the formula Vr 43 r 3 shows that V is a polynomial function of r. Likewise, if a ball is thrown vertically into the air with a velocity of 50 fts, then the height of the ball in feet t seconds later is given by the formula h 50t 16t 2. Again this is a polynomial function, so the height is a continuous function of the elapsed time. Knowledge of which functions are continuous enables us to evaluate some limits very quickly, as the following example shows. Compare it with Example 1(b) in Section 1.4. EXAMPLE 5 Find lim
x l 2
x 3 2x 2 1 . 5 3x
SOLUTION The function
f x
x 3 2x 2 1 5 3x
is rational, so by Theorem 5 it is continuous on its domain, which is {x Therefore lim
x l2
23 222 1 1 5 32 11
lim sin sin a
la
1 _
π 2
0
π 2
π
3π 2
x
y=tan x
lim cos cos a
la
In other words, the sine and cosine functions are continuous everywhere. It follows from part 5 of Theorem 4 that tan x
FIGURE 5
■
It turns out that most of the familiar functions are continuous at every number in their domains. For instance, Limit Law 10 (page 36) is exactly the statement that root functions are continuous. From the appearance of the graphs of the sine and cosine functions (Figure 11 in Section 1.2), we would certainly guess that they are continuous. And in Section 1.4 we showed that
y
3π _π
5 3
x 3 2x 2 1 lim f x f 2 x l2 5 3x
_ 2
x }.
sin x cos x
is continuous except where cos x 0. This happens when x is an odd integer multiple of 2, so y tan x has infinite discontinuities when x 2, 32, 52, and so on (see Figure 5).
6 THEOREM The following types of functions are continuous at every number in their domains: polynomials, rational functions, root functions, trigonometric functions
SECTION 1.5
CONTINUITY
■
51
EXAMPLE 6 On what intervals is each function continuous?
(a) f x x 100 2x 37 75 (c) hx sx
(b) tx
x1 x1 2 x1 x 1
x 2 2x 17 x2 1
SOLUTION
(a) f is a polynomial, so it is continuous on , by Theorem 5(a). (b) t is a rational function, so by Theorem 5(b) it is continuous on its domain, which is D x x 2 1 0 x x 1 . Thus, t is continuous on the intervals , 1, 1, 1, and 1, . (c) We can write hx Fx Gx Hx, where
Fx sx
Gx
x1 x1
Hx
x1 x2 1
F is continuous on 0, by Theorem 6. G is a rational function, so it is continuous everywhere except when x 1 0, that is, x 1. H is also a rational function, but its denominator is never 0, so H is continuous everywhere. Thus, by parts 1 and 2 of Theorem 4, h is continuous on the intervals 0, 1 and 1, . ■ Another way of combining continuous functions f and t to get a new continuous function is to form the composite function f ⴰ t. This fact is a consequence of the following theorem. ■ This theorem says that a limit symbol can be moved through a function symbol if the function is continuous and the limit exists. In other words, the order of these two symbols can be reversed.
7 THEOREM
If f is continuous at b and lim tx b, then lim f tx f b. x la
In other words,
lim f tx f lim tx
(
xla
xla
x la
)
Intuitively, Theorem 7 is reasonable because if x is close to a, then tx is close to b, and since f is continuous at b, if tx is close to b, then f tx is close to f b. A proof of Theorem 7 is given in Appendix B. 8 THEOREM If t is continuous at a and f is continuous at ta, then the composite function f ⴰ t given by f ⴰ tx f tx is continuous at a.
This theorem is often expressed informally by saying “a continuous function of a continuous function is a continuous function.” PROOF Since t is continuous at a, we have
lim tx ta
xla
Since f is continuous at b ta, we can apply Theorem 7 to obtain lim f tx f ta
xla
which is precisely the statement that the function hx f tx is continuous at a; that is, f ⴰ t is continuous at a. ■
52
■
CHAPTER 1
FUNCTIONS AND LIMITS
Where are the following functions continuous? 1 (a) hx sinx 2 (b) Fx sx 2 7 4 V EXAMPLE 7
SOLUTION
(a) We have hx f tx, where tx x 2
and
f x sin x
Now t is continuous on ⺢ since it is a polynomial, and f is also continuous everywhere by Theorem 6. Thus, h f ⴰ t is continuous on ⺢ by Theorem 8. (b) Notice that F can be broken up as the composition of four continuous functions: Ffⴰtⴰhⴰk where
f x
1 x
or
tx x 4
Fx f thkx hx sx
kx x 2 7
We know that each of these functions is continuous on its domain (by Theorems 5 and 6), so by Theorem 8, F is continuous on its domain, which is
{ x ⺢ sx 2 7
4} x
x 3 , 3 3, 3 3,
■
An important property of continuous functions is expressed by the following theorem, whose proof is found in more advanced books on calculus. Suppose that f is continuous on the closed interval a, b and let N be any number between f a and f b, where f a f b. Then there exists a number c in a, b such that f c N . 9 THE INTERMEDIATE VALUE THEOREM
The Intermediate Value Theorem states that a continuous function takes on every intermediate value between the function values f a and f b. It is illustrated by Figure 6. Note that the value N can be taken on once [as in part (a)] or more than once [as in part (b)]. y
y
f(b)
f(b)
y=ƒ
N N
y=ƒ
f(a) 0
FIGURE 6
a
f(a)
c
(a)
b
x
0
a c¡
c™
c£
b
x
(b)
If we think of a continuous function as a function whose graph has no hole or break, then it is easy to believe that the Intermediate Value Theorem is true. In geometric terms it says that if any horizontal line y N is given between y f a and
SECTION 1.5
y=ƒ y=N
N f(b) 0
a
b
■
53
y f b as in Figure 7, then the graph of f can’t jump over the line. It must intersect y N somewhere. It is important that the function f in Theorem 9 be continuous. The Intermediate Value Theorem is not true in general for discontinuous functions (see Exercise 34). One use of the Intermediate Value Theorem is in locating roots of equations as in the following example.
y f(a)
CONTINUITY
x
V EXAMPLE 8
Show that there is a root of the equation
FIGURE 7
4x 3 6x 2 3x 2 0 between 1 and 2. SOLUTION Let f x 4x 3 6x 2 3x 2. We are looking for a solution of the
given equation, that is, a number c between 1 and 2 such that f c 0. Therefore, we take a 1, b 2, and N 0 in Theorem 9. We have f 1 4 6 3 2 1 0 f 2 32 24 6 2 12 0
and
Thus f 1 0 f 2; that is, N 0 is a number between f 1 and f 2. Now f is continuous since it is a polynomial, so the Intermediate Value Theorem says there is a number c between 1 and 2 such that f c 0. In other words, the equation 4x 3 6x 2 3x 2 0 has at least one root c in the interval 1, 2. In fact, we can locate a root more precisely by using the Intermediate Value Theorem again. Since f 1.2 0.128 0
f 1.3 0.548 0
and
a root must lie between 1.2 and 1.3. A calculator gives, by trial and error, f 1.22 0.007008 0
f 1.23 0.056068 0
and
so a root lies in the interval 1.22, 1.23.
■
We can use a graphing calculator or computer to illustrate the use of the Intermediate Value Theorem in Example 8. Figure 8 shows the graph of f in the viewing rectangle 1, 3 by 3, 3 and you can see that the graph crosses the x-axis between 1 and 2. Figure 9 shows the result of zooming in to the viewing rectangle 1.2, 1.3 by 0.2, 0.2 . 3
0.2
3
_1
_3
FIGURE 8
1.2
_0.2
FIGURE 9
1.3
54
■
CHAPTER 1
FUNCTIONS AND LIMITS
In fact, the Intermediate Value Theorem plays a role in the very way these graphing devices work. A computer calculates a finite number of points on the graph and turns on the pixels that contain these calculated points. It assumes that the function is continuous and takes on all the intermediate values between two consecutive points. The computer therefore connects the pixels by turning on the intermediate pixels.
1.5
EXERCISES
1. Write an equation that expresses the fact that a function f
(b) Discuss the discontinuities of this function and their significance to someone who parks in the lot.
is continuous at the number 4. 2. If f is continuous on , , what can you say about its
8. Explain why each function is continuous or discontinuous.
(a) The temperature at a specific location as a function of time (b) The temperature at a specific time as a function of the distance due west from New York City (c) The altitude above sea level as a function of the distance due west from New York City (d) The cost of a taxi ride as a function of the distance traveled (e) The current in the circuit for the lights in a room as a function of time
graph? 3. (a) From the graph of f , state the numbers at which f is
discontinuous and explain why. (b) For each of the numbers stated in part (a), determine whether f is continuous from the right, or from the left, or neither. y
9. If f and t are continuous functions with f 3 5 and
lim x l 3 2 f x tx 4, find t3.
_4
0
_2
2
4
x
6
■ Use the definition of continuity and the properties of limits to show that the function is continuous at the given number a.
10 –11
10. f x x 2 s7 x , 11. f x x 2x ,
4. From the graph of t, state the intervals on which t is
■
continuous.
■
■
■
a4
a 1
3 4
■
■
■
■
■
■
■
■
12. Use the definition of continuity and the properties of
y
limits to show that the function f x x s16 x 2 is continuous on the interval 4, 4 . ■ Explain why the function is discontinuous at a 1. Sketch the graph of the function.
13–16 _4
_2
2
4
6
8
x
13. f x
5. Sketch the graph of a function that is continuous every-
where except at x 3 and is continuous from the left at 3.
14. f x
6. Sketch the graph of a function that has a jump discontinuity
at x 2 and a removable discontinuity at x 4, but is continuous elsewhere. 7. A parking lot charges $3 for the first hour (or part of an
hour) and $2 for each succeeding hour (or part), up to a daily maximum of $10. (a) Sketch a graph of the cost of parking at this lot as a function of the time parked there.
15. f x
1 x 12
1 x1 2
if x 1
1 x2 1x
if x 1 if x 1
x2 x 16. f x x 2 1 1 ■
■
■
■
if x 1
if x 1 if x 1 ■
■
■
■
■
■
■
■
SECTION 1.5
3 18. Gx s x 1 x 3
19. Rx x 2 s2 x 1
20. hx
21. Fx sx sin x ■
■
■
■
tx
■
■
■
■
■
■
■
■ Locate the discontinuities of the function and illustrate by graphing.
■
■
■
25–26
■
25. lim x l4
■
28. f x ■
■
■
■
■
■
26. lim sinx sin x ■
■
■
■
■
■
■
if x 4 if x 4
sin x cos x
■
■
■
■
■
■
■
■
is discontinuous. At which of these points is f continuous from the right, from the left, or neither? Sketch the graph of f . 30. The gravitational force exerted by the Earth on a unit mass
at a distance r from the center of the planet is if r R if r R
where M is the mass of the Earth, R is its radius, and G is the gravitational constant. Is F a continuous function of r ? 31. For what value of the constant c is the function f continu-
ous on , ? f x
3 sx 9x
a9
that f c 1000.
■
x 2 if x 0 if 0 x 1 f x 2x 2 2 x if x 1
Fr
(d) f x
35. If f x x 2 10 sin x, show that there is a number c such ■
36. Use the Intermediate Value Theorem to prove that there is a
positive number c such that c 2 2. (This proves the existence of the number s2 .)
29. Find the numbers at which the function
GMr R3 GM r2
a 4
0.25 and that f 0 1 and f 1 3. Let N 2. Sketch two possible graphs of f , one showing that f might not satisfy the conclusion of the Intermediate Value Theorem and one showing that f might still satisfy the conclusion of the Intermediate Value Theorem (even though it doesn’t satisfy the hypothesis).
■
if x 1 if x 1
x2 sx
x 64 x4
34. Suppose that a function f is continuous on [0, 1] except at
x l
■
(c) f x ■
Show that f is continuous on , .
27. f x
■
■
5 sx s5 x ■
■
■
if x 4 if x 4
3
Use continuity to evaluate the limit.
■
27–28
24. y tan sx
■
x2 c2 cx 20
tinuity at a ? If the discontinuity is removable, find a function t that agrees with f for x a and is continuous at a. x 2 2x 8 (a) f x a 2 x2 x7 (b) f x a7 x7
; 23–24
1 23. y 1 sin x
33. Which of the following functions f has a removable discon-
sin x x1
22. Fx sincossin x ■
55
32. Find the constant c that makes t continuous on , .
■ Explain, using Theorems 4, 5, 6, and 8, why the function is continuous at every number in its domain. State the domain.
17–22
x 17. Fx 2 x 5x 6
■
CONTINUITY
cx 2 2x if x 2 x 3 cx if x 2
■ Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval.
37– 40
37. x 4 x 3 0, 39. cos x x, ■
■
■
1, 2
3 38. s x 1 x,
0, 1 ■
40. tan x 2x, ■
■
■
■
■
0, 1 0, 1.4
■
■
■
41– 42 ■ (a) Prove that the equation has at least one real root. (b) Use your calculator to find an interval of length 0.01 that contains a root. 41. cos x x 3 ■
■
■
42. x 5 x 2 2x 3 0 ■
■
■
■
■
■
■
■
■
; 43– 44
■ (a) Prove that the equation has at least one real root. (b) Use your graphing device to find the root correct to three decimal places.
43. x 5 x 2 4 0 ■
■
■
■
44. sx 5 ■
■
■
■
■
1 x3 ■
■
45. Is there a number that is exactly 1 more than its cube?
■
56
■
CHAPTER 1
FUNCTIONS AND LIMITS
46. (a) Show that the absolute value function Fx x is
47. A Tibetan monk leaves the monastery at 7:00 AM and
takes his usual path to the top of the mountain, arriving at 7:00 PM. The following morning, he starts at 7:00 AM at the top and takes the same path back, arriving at the monastery at 7:00 PM. Use the Intermediate Value Theorem to show that there is a point on the path that the monk will cross at exactly the same time of day on both days.
continuous everywhere. (b) Prove that if f is a continuous function on an interval, then so is f . (c) Is the converse of the statement in part (b) also true? In other words, if f is continuous, does it follow that f is continuous? If so, prove it. If not, find a counterexample.
1.6
LIMITS INVOLVING INFINITY In this section we investigate the global behavior of functions and, in particular, whether their graphs approach asymptotes, vertical or horizontal. INFINITE LIMITS
1 0.5 0.2 0.1 0.05 0.01 0.001
In Example 8 in Section 1.3 we concluded that
1 x2
x
lim x l0
1 4 25 100 400 10,000 1,000,000
1 x2
by observing, from the table of values and the graph of y 1x 2 in Figure 1, that the values of 1x 2 can be made arbitrarily large by taking x close enough to 0. Thus the values of f x do not approach a number, so lim x l 0 1x 2 does not exist. To indicate this kind of behavior we use the notation lim x l0
y
does not exist
1 x2
| This does not mean that we are regarding as a number. Nor does it mean that the y=
limit exists. It simply expresses the particular way in which the limit does not exist: 1x 2 can be made as large as we like by taking x close enough to 0. In general, we write symbolically
1 ≈
lim f x x la
0
x
FIGURE 1
to indicate that the values of f x become larger and larger (or “increase without bound”) as x approaches a. 1 DEFINITION
■ A more precise version of Definition 1 is given at the end of this section.
The notation lim f x x la
means that the values of f x can be made arbitrarily large (as large as we please) by taking x sufficiently close to a (on either side of a) but not equal to a. Another notation for lim x l a f x is f x l
as
xla
SECTION 1.6
y=ƒ
a
x
x=a
57
or
“ f x becomes infinite as x approaches a”
or
“ f x increases without bound as x approaches a ”
This definition is illustrated graphically in Figure 2. Similarly, as shown in Figure 3,
FIGURE 2
lim f x
lim ƒ=`
x la
x a
■ When we say that a number is “large negative,” we mean that it is negative but its magnitude (absolute value) is large.
y
means that the values of f x are as large negative as we like for all values of x that are sufficiently close to a, but not equal to a. The symbol lim x l a f x can be read as “the limit of f x, as x approaches a, is negative infinity” or “ f x decreases without bound as x approaches a.” As an example we have
x=a
lim x l0
0
a
x
1 x2
Similar definitions can be given for the one-sided infinite limits
y=ƒ
lim f x
lim f x
x la
x la
lim f x
lim f x
x la
FIGURE 3 x a
y
y
a
0
(a) lim ƒ=` a_
x la
remembering that “x l a” means that we consider only values of x that are less than a, and similarly “x l a” means that we consider only x a. Illustrations of these four cases are given in Figure 4.
lim ƒ=_`
x
■
Again, the symbol is not a number, but the expression lim x l a f x is often read as “the limit of f x, as x approaches a, is infinity”
y
0
LIMITS INVOLVING INFINITY
x
y
a
0
x
(b) lim ƒ=` x
a+
y
a
0
(c) lim ƒ=_` x
a
0
x
x
(d) lim ƒ=_`
a_
x
a+
FIGURE 4 2 DEFINITION The line x a is called a vertical asymptote of the curve y f x if at least one of the following statements is true:
lim f x x la
lim f x x la
lim f x
x la
lim f x
x la
lim f x
x la
lim f x
x la
For instance, the y-axis is a vertical asymptote of the curve y 1x 2 because lim x l 0 1x 2 . In Figure 4 the line x a is a vertical asymptote in each of the four cases shown.
58
■
CHAPTER 1
FUNCTIONS AND LIMITS
EXAMPLE 1 Find lim x l3
2x 2x and lim . x l3 x 3 x3
SOLUTION If x is close to 3 but larger than 3, then the denominator x 3 is a small positive number and 2x is close to 6. So the quotient 2xx 3 is a large positive number. Thus, intuitively, we see that
y
y=
2x x-3
lim
x l3
5
Likewise, if x is close to 3 but smaller than 3, then x 3 is a small negative number but 2x is still a positive number (close to 6). So 2xx 3 is a numerically large negative number. Thus
x
0
2x x3
x=3
lim
x l3
2x x3
The graph of the curve y 2xx 3 is given in Figure 5. The line x 3 is a vertical asymptote.
FIGURE 5
■
EXAMPLE 2 Find the vertical asymptotes of f x tan x. SOLUTION Because
y
tan x 1 3π _π
_ 2
_
π 2
0
π 2
π
3π 2
x
sin x cos x
there are potential vertical asymptotes where cos x 0. In fact, since cos x l 0 as x l 2 and cos x l 0 as x l 2, whereas sin x is positive (and not near 0) when x is near 2, we have lim tan x
x l2
lim tan x
and
x l2
This shows that the line x 2 is a vertical asymptote. Similar reasoning shows that the lines x 2n 12, where n is an integer, are all vertical asymptotes of f x tan x. The graph in Figure 6 confirms this. ■
FIGURE 6
y=tan x
LIMITS AT INFINITY x
f x
0 1 2 3 4 5 10 50 100 1000
1 0 0.600000 0.800000 0.882353 0.923077 0.980198 0.999200 0.999800 0.999998
In computing infinite limits, we let x approach a number and the result was that the values of y became arbitrarily large (positive or negative). Here we let x become arbitrarily large (positive or negative) and see what happens to y. Let’s begin by investigating the behavior of the function f defined by f x
as x becomes large. The table at the left gives values of this function correct to six decimal places, and the graph of f has been drawn by a computer in Figure 7. y
y=1
0
FIGURE 7
x2 1 x2 1
1
y=
≈-1 ≈+1
x
SECTION 1.6
LIMITS INVOLVING INFINITY
59
■
As x grows larger and larger you can see that the values of f x get closer and closer to 1. In fact, it seems that we can make the values of f x as close as we like to 1 by taking x sufficiently large. This situation is expressed symbolically by writing x2 1 1 x2 1
lim
x l
In general, we use the notation lim f x L
x l
to indicate that the values of f x approach L as x becomes larger and larger. 3 DEFINITION
Let f be a function defined on some interval a, . Then lim f x L
x l
means that the values of f x can be made as close to L as we like by taking x sufficiently large. Another notation for lim x l f x L is f x l L
as
xl
The symbol does not represent a number. Nonetheless, the expression lim f x L x l is often read as “the limit of f x, as x approaches infinity, is L” or
“the limit of f x, as x becomes infinite, is L”
or
“the limit of f x, as x increases without bound, is L”
The meaning of such phrases is given by Definition 3. A more precise definition, similar to the , definition of Section 1.3, is given at the end of this section. Geometric illustrations of Definition 3 are shown in Figure 8. Notice that there are many ways for the graph of f to approach the line y L (which is called a horizontal asymptote) as we look to the far right of each graph. y
y
y=L
y
y=L
y=ƒ
y=ƒ y=ƒ
y=L 0
x
FIGURE 8
Examples illustrating lim ƒ=L x `
0
0
x
x
Referring back to Figure 7, we see that for numerically large negative values of x, the values of f x are close to 1. By letting x decrease through negative values without bound, we can make f x as close to 1 as we like. This is expressed by writing lim
x l
x2 1 1 x2 1
60
■
CHAPTER 1
FUNCTIONS AND LIMITS
y
In general, as shown in Figure 9, the notation y=ƒ
lim f x L
x l
means that the values of f x can be made arbitrarily close to L by taking x sufficiently large negative. Again, the symbol does not represent a number, but the expression lim x l f x L is often read as
y=L 0
x y
“the limit of f x, as x approaches negative infinity, is L”
y=ƒ
4 DEFINITION The line y L is called a horizontal asymptote of the curve y f x if either
y=L
0
lim f x L
x
or
x l
lim f x L
x l
FIGURE 9
Examples illustrating lim ƒ=L
For instance, the curve illustrated in Figure 7 has the line y 1 as a horizontal asymptote because
x _`
lim
x l
x2 1 1 x2 1
The curve y f x sketched in Figure 10 has both y 1 and y 2 as horizontal asymptotes because lim f x 1
xl
and
lim f x 2
x l
y 2
y=2
0
y=_1
y=ƒ x
_1
FIGURE 10
EXAMPLE 3 Find the infinite limits, limits at infinity, and asymptotes for the funcy
tion f whose graph is shown in Figure 11. SOLUTION We see that the values of f x become large as x l 1 from both sides,
so lim f x
2
x l1
0
2
x
Notice that f x becomes large negative as x approaches 2 from the left, but large positive as x approaches 2 from the right. So lim f x
x l2
FIGURE 11
and
lim f x
x l2
Thus, both of the lines x 1 and x 2 are vertical asymptotes.
SECTION 1.6
LIMITS INVOLVING INFINITY
■
61
As x becomes large, it appears that f x approaches 4. But as x decreases through negative values, f x approaches 2. So lim f x 4
lim f x 2
and
x l
x l
This means that both y 4 and y 2 are horizontal asymptotes. EXAMPLE 4 Find lim
x l
■
1 1 and lim . x l x x
SOLUTION Observe that when x is large, 1x is small. For instance,
1 0.01 100
1 0.0001 10,000
In fact, by taking x large enough, we can make 1x as close to 0 as we please. Therefore, according to Definition 3, we have
y
y=Δ
lim
x l
0
x
lim
lim x `
1 1 =0, lim =0 x x _` x
1 0 x
Similar reasoning shows that when x is large negative, 1x is small negative, so we also have
x l
FIGURE 12
1 0.000001 1,000,000
1 0 x
It follows that the line y 0 (the x-axis) is a horizontal asymptote of the curve y 1x. (This is an equilateral hyperbola; see Figure 12.)
■
Most of the Limit Laws that were given in Section 1.4 also hold for limits at infinity. It can be proved that the Limit Laws listed in Section 1.4 (with the exception of Laws 9 and 10) are also valid if “x l a” is replaced by “x l ” or “ x l .” In particular, if we combine Law 6 with the results of Example 4 we obtain the following important rule for calculating limits.
5
If n is a positive integer, then lim
x l
V EXAMPLE 5
1 0 xn
lim
x l
1 0 xn
Evaluate lim
x l
3x 2 x 2 5x 2 4x 1
SOLUTION As x becomes large, both numerator and denominator become large, so it isn’t obvious what happens to their ratio. We need to do some preliminary algebra. To evaluate the limit at infinity of any rational function, we first divide both the numerator and denominator by the highest power of x that occurs in the denomi-
62
■
CHAPTER 1
FUNCTIONS AND LIMITS
nator. (We may assume that x 0, since we are interested only in large values of x.) In this case the highest power of x is x 2, and so, using the Limit Laws, we have 3x 2 x 2 1 2 3 2 3x x 2 x2 x x lim lim lim x l 5x 2 4x 1 x l 5x 2 4x 1 x l 4 1 5 2 x2 x x 2
■ Figure 13 illustrates Example 5 by showing how the graph of the given rational function approaches the horizontal asymptote y 53 .
y
y=0.6
1 2 2 x x
lim 5
4 1 2 x x
x l
0
1
x
lim 3
x l
1 2 lim x l x 1 lim 5 4 lim lim x l x l x x l lim 3 lim
x l
FIGURE 13
y=
3≈-x-2 5≈+4x+1
x l
300 500
3 5
1 x2 1 x2
[by (5)]
A similar calculation shows that the limit as x l is also 35.
■
EXAMPLE 6 Compute lim (sx 2 1 x). x l
SOLUTION Because both sx 2 1 and x are large when x is large, it’s difficult to We can think of the given function as having a denominator of 1. ■
see what happens to their difference, so we use algebra to rewrite the function. We first multiply numerator and denominator by the conjugate radical: lim (sx 2 1 x) lim (sx 2 1 x)
x l
x l
lim
x l
x 2 1 x 2 1 lim 2 1 x 2 1 x x l sx sx
The Squeeze Theorem could be used to show that this limit is 0. But an easier method is to divide numerator and denominator by x. Doing this and remembering that x sx 2 for x 0, we obtain 1 1 x lim (sx 2 1 x) lim lim x l x l sx 2 1 x x l sx 2 1 x x
y
y=œ„„„„„-x ≈+1 1
lim 0
FIGURE 14
1
sx 2 1 x sx 2 1 x
x l
x
Figure 14 illustrates this result.
1
1 x 1 1 x2
0 0 s1 0 1 ■
SECTION 1.6
EXAMPLE 7 Evaluate lim sin xl
LIMITS INVOLVING INFINITY
■
63
1 . x
SOLUTION If we let t 1x, then t l 0 as x l . Therefore
1 lim sin t 0 tl0 x
lim sin
xl
■
(See Exercise 55.) EXAMPLE 8 Evaluate lim sin x. x l
SOLUTION As x increases, the values of sin x oscillate between 1 and 1 infinitely often. Thus lim x l sin x does not exist. ■ INFINITE LIMITS AT INFINITY
lim f x
The notation
x l
is used to indicate that the values of f x become large as x becomes large. Similar meanings are attached to the following symbols: lim f x
lim f x
x l
x l
lim f x
x l
EXAMPLE 9 Find lim x 3 and lim x 3. xl
y
x l
SOLUTION When x becomes large, x 3 also becomes large. For instance,
10 3 1000
y=˛
0
x
100 3 1,000,000
1000 3 1,000,000,000
In fact, we can make x 3 as big as we like by taking x large enough. Therefore, we can write lim x 3 xl
Similarly, when x is large negative, so is x 3. Thus lim x 3
x l
FIGURE 15
lim x#=`, lim x#=_` x `
x _`
These limit statements can also be seen from the graph of y x 3 in Figure 15.
■
EXAMPLE 10 Find lim x 2 x. x l
| SOLUTION It would be wrong to write lim x 2 x lim x 2 lim x
x l
x l
x l
The Limit Laws can’t be applied to infinite limits because is not a number ( can’t be defined). However, we can write lim x 2 x lim xx 1
x l
x l
because both x and x 1 become arbitrarily large.
■
64
■
CHAPTER 1
FUNCTIONS AND LIMITS
EXAMPLE 11 Find lim
x l
x2 x . 3x
SOLUTION We divide numerator and denominator by x (the highest power of x that
occurs in the denominator): lim
x l
x2 x x1 lim x l 3 3x 1 x
because x 1 l and 3x 1 l 1 as x l .
■
PRECISE DEFINITIONS
The following is a precise version of Definition 1. 6 DEFINITION Let f be a function defined on some open interval that contains the number a, except possibly at a itself. Then
lim f x
xla
means that for every positive number M there is a positive number such that
y
y=M
M
0
x
a
a-∂ FIGURE 16
a+∂
0 xa
if
then
f x M
This says that the values of f x can be made arbitrarily large (larger than any given number M ) by taking x close enough to a (within a distance , where depends on M , but with x a). A geometric illustration is shown in Figure 16. Given any horizontal line y M , we can find a number 0 such that if we restrict x to lie in the interval a , a but x a, then the curve y f x lies above the line y M . You can see that if a larger M is chosen, then a smaller may be required. 1 . x2 SOLUTION Let M be a given positive number. According to Definition 6, we need to find a number such that V EXAMPLE 12
Use Definition 6 to prove that lim
xl0
0 x
if
But x 2 1M &? if
1 M x2
then
that is
x 1sM . We can choose 1sM
0 x
1 s
then
x2
1 M
because
1 1 2 M x2
Therefore, by Definition 6, lim
xl0
1 x2
■
SECTION 1.6
LIMITS INVOLVING INFINITY
■
65
Similarly, limx l a f x means that for every negative number N there is a positive number such that if 0 x a , then f x N. Definition 3 can be stated precisely as follows.
7 DEFINITION
Let f be a function defined on some interval a, . Then lim f x L
xl
means that for every 0 there is a corresponding number N such that xN
if
Module 1.3/1.6 illustrates Definition 7 graphically and numerically.
f x L
then
In words, this says that the values of f x can be made arbitrarily close to L (within a distance , where is any positive number) by taking x sufficiently large (larger than N , where N depends on ). Graphically it says that by choosing x large enough (larger than some number N ) we can make the graph of f lie between the given horizontal lines y L and y L as in Figure 17. This must be true no matter how small we choose . Figure 18 shows that if a smaller value of is chosen, then a larger value of N may be required. y
y=ƒ
y=L +∑ ∑ L ∑ y=L -∑
ƒ is in here
0
x
N
FIGURE 17
lim ƒ=L
when x is in here
x `
y
y=ƒ y=L+∑
L
y=L-∑ 0
N
FIGURE 18
x
lim ƒ=L x `
Similarly, limx l f x L means that for every 0 there is a corresponding number N such that if x N, then f x L .
1 0. x
EXAMPLE 13 Use Definition 7 to prove that lim
xl
SOLUTION Given 0, we want to find N such that
if
xN
then
1 0 x
66
■
CHAPTER 1
FUNCTIONS AND LIMITS
In computing the limit we may assume that x 0. Then 1x &? x 1. Let’s choose N 1. So xN
if
1
then
1 1 0 x x
Therefore, by Definition 7, lim
xl
1 0 x
Figure 19 illustrates the proof by showing some values of and the corresponding values of N . y
y
y
∑=1 ∑=0.2 0
x
N=1
∑=0.1
0
N=5
x
0
N=10
x
FIGURE 19
Finally we note that an infinite limit at infinity can be defined as follows. The geometric illustration is given in Figure 20.
y
y=M
8 DEFINITION
M
Let f be a function defined on some interval a, . Then lim f x
xl
0
x
N
FIGURE 20
means that for every positive number M there is a corresponding positive number N such that if xN then f x M
lim ƒ=` x `
Similar definitions apply when the symbol is replaced by .
1.6
EXERCISES
1. For the function f whose graph is given, state the following.
(a) lim f x
(b)
(c) lim f x
(d) lim f x
x l2
x l1
y
lim f x
x l1
x l
(e) lim f x x l
(f ) The equations of the asymptotes
1 1
x
SECTION 1.6
2. For the function t whose graph is given, state the following.
(a) lim tx
(b) lim tx
(c) lim tx
(d) lim tx
(e) lim tx
(f ) The equations of the asymptotes
x l
asymptotes of the curve y
x l0
x l2
x
2
13. 3– 8 ■ Sketch the graph of an example of a function f that satisfies all of the given conditions.
lim f x 0,
x l0
lim f x ,
x l
lim f x ,
x l2
lim f x ,
x l
lim f x 0,
x l
16. lim cot x x l
19. lim
x 3 5x 2x x 2 4
20. lim
21. lim
4u 4 5 u 22u 2 1
22. lim
ul
lim f x
x l0
xl5
18. lim
x l
5. lim f x , x l0
lim
6 x5
14. lim
2x x 12
x l2
x l
lim f x 1
x2 x3
sec x
lim f x 1,
x l0
lim
x l3
x l1
17. 4. lim f x ,
x
Find the limit.
■
15. lim
f is odd
x l
2 x
to estimate the value of lim x l f x correct to two decimal places. (b) Use a table of values of f x to estimate the limit to four decimal places. 13–31
f 1 1,
f x 1
1
3. f 0 0,
x3 x 2x 1 3
; 12. (a) Use a graph of
y
0
67
; 11. Use a graph to estimate all the vertical and horizontal
x l
x l3
■
LIMITS INVOLVING INFINITY
x l
3
3x 5 x4
t l
2
x l
t2 2 t t2 1 3
x2 s9x 2 1
23. lim (s9x 2 x 3x) x l
6. lim f x ,
lim f x 3,
lim f x 3
24. lim (sx 2 ax sx 2 bx
)
25. lim cos x
26. lim
sin 2x x2
lim f x 3
27. lim ( x sx )
28. lim
x 3 2x 3 5 2x 2
8. lim f x ,
29. lim x x
30. lim x 2 x 4
x l2
7. f 0 3,
x l
lim f x 4,
x l0
lim f x ,
x l
lim f x 2,
x l0
lim f x ,
x l 4
x l
x l
x l
lim f x ,
x l 4
xl
x l
lim f x 2,
xl3
■
■
■
x l
■
■
■
■
f 0 0, ■
■
■
■
x2 lim x x l 2 by evaluating the function f x x 22 x for x 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 50, and 100. Then use a graph of f to support your guess. 1 1 and lim 3 x l1 x 1 x 1 (a) by evaluating f x 1x 3 1 for values of x that approach 1 from the left and from the right, (b) by reasoning as in Example 1, and (c) from a graph of f .
10. Determine lim x l1
4
f is even
; 9. Guess the value of the limit
;
x l
xl
5
xl
x l
■
xx x 1 x2 x4 3
31. lim
xl
■
■
■
5
■
■
■
■
■
■
■
■
; 32. (a) Graph the function f x
s2x 2 1 3x 5
How many horizontal and vertical asymptotes do you observe? Use the graph to estimate the values of the limits
3
lim
x l
s2x 2 1 3x 5
and
lim
x l
s2x 2 1 3x 5
(b) By calculating values of f x, give numerical estimates of the limits in part (a).
■
68
■
CHAPTER 1
FUNCTIONS AND LIMITS
(b) Two functions are said to have the same end behavior if their ratio approaches 1 as x l . Show that P and Q have the same end behavior.
(c) Calculate the exact values of the limits in part (a). Did you get the same value or different values for these two limits? [In view of your answer to part (a), you might have to check your calculation for the second limit.]
; 33–34
41. Let P and Q be polynomials. Find
Find the horizontal and vertical asymptotes of each curve. Check your work by graphing the curve and estimating the asymptotes. ■
33. y ■
■
2x 2 x 1 x2 x 2 ■
■
34. Fx ■
■
■
■
lim
xl
x9 s4x 2 3x 2 ■
■
■
Px Qx
if the degree of P is (a) less than the degree of Q and (b) greater than the degree of Q. ■
42. Make a rough sketch of the curve y x n (n an integer)
; 35. (a) Estimate the value of
for the following five cases: (i) n 0 (ii) n 0, n odd (iii) n 0, n even (iv) n 0, n odd (v) n 0, n even Then use these sketches to find the following limits. (a) lim x n (b) lim x n
lim (sx 2 x 1 x)
x l
by graphing the function f x sx 2 x 1 x. (b) Use a table of values of f x to guess the value of the limit. (c) Prove that your guess is correct.
x l0
x l0
(c) lim x n
(d) lim x n
x l
; 36. (a) Use a graph of
x l
43. Find lim x l f x if, for all x 5,
f x s3x 2 8x 6 s3x 2 3x 1
4x 1 4x 2 3x f x x x2
to estimate the value of lim x l f x to one decimal place. (b) Use a table of values of f x to estimate the limit to four decimal places. (c) Find the exact value of the limit.
44. In the theory of relativity, the mass of a particle with velocity v is
m
; 37. Estimate the horizontal asymptote of the function f x
3x 3 500x 2 x 500x 2 100x 2000
where m 0 is the mass of the particle at rest and c is the speed of light. What happens as v l c?
3
by graphing f for 10 x 10. Then calculate the equation of the asymptote by evaluating the limit. How do you explain the discrepancy?
45. (a) A tank contains 5000 L of pure water. Brine that con-
tains 30 g of salt per liter of water is pumped into the tank at a rate of 25 Lmin. Show that the concentration of salt t minutes later (in grams per liter) is
38. Find a formula for a function that has vertical asymptotes
x 1 and x 3 and horizontal asymptote y 1.
Ct
39. Find a formula for a function f that satisfies the following
conditions: lim f x 0, x l
lim f x ,
x l3
lim f x , x l0
30t 200 t
(b) What happens to the concentration as t l ?
f 2 0,
lim f x
4x 2 5x 2. x l 2x 2 1 (b) By graphing the function in part (a) and the line y 1.9 on a common screen, find a number N such that
46. (a) Show that lim
x l3
; 40. By the end behavior of a function we mean the behavior of
its values as x l and as x l . (a) Describe and compare the end behavior of the functions Px 3x 5 5x 3 2x
m0 s1 v 2c 2
Qx 3x 5
by graphing both functions in the viewing rectangles 2, 2 by 2, 2 and 10, 10 by 10,000, 10,000 .
;
4x 2 5x 1.9 2x 2 1
when
What if 1.9 is replaced by 1.99?
xN
CHAPTER 1
47. How close to 3 do we have to take x so that
x l3
49. Prove that lim x l 1
lim
xl
1 . x 34
(b) Taking n 2 in (5), we have the statement lim
xl
s4x 2 1 2 x1
54. Prove, using Definition 8, that lim x 3 . xl
55. Prove that
lim f x lim f 1t
; 51. Use a graph to find a number N such that
6x 2 5x 3 3 0.2 2x2 1
1
xl
xN
REVIEW
2. Discuss four ways of representing a function. Illustrate your
3. (a) What is an even function? How can you tell if a func-
tion is even by looking at its graph? (b) What is an odd function? How can you tell if a function is odd by looking at its graph? 4. What is a mathematical model? 5. Give an example of each type of function.
(b) Power function (d) Quadratic function (f ) Rational function
6. Sketch by hand, on the same axes, the graphs of the follow-
(b) tx x 2 (d) jx x 4
7. Draw, by hand, a rough sketch of the graph of each function.
8. Suppose that f has domain A and t has domain B.
(a) What is the domain of f t ? (b) What is the domain of f t ? (c) What is the domain of ft ?
9. How is the composite function f ⴰ t defined? What is its
domain?
discussion with examples.
(a) y sin x (c) y 2 x (e) y x
tl0
CONCEPT CHECK
(b) What is the graph of a function? (c) How can you tell whether a given curve is the graph of a function?
ing functions. (a) f x x (c) hx x 3
x l
if these limits exist.
1. (a) What is a function? What are its domain and range?
(a) Linear function (c) Exponential function (e) Polynomial of degree 5
tl0
lim f x lim f 1t
and whenever
1 0 x2
Prove this directly using Definition 7.
illustrate Definition 7 by finding values of N that correspond to 0.5 and 0.1.
2x 1 sx 1
53. (a) How large do we have to take x so that 1x 2 0.0001?
; 50. For the limit lim
69
illustrate Definition 8 by finding a value of N that corresponds to M 100.
5 . x 1 3
xl
■
; 52. For the limit
1 10,000 x 34 48. Prove, using Definition 6, that lim
REVIEW
(b) y tan x (d) y 1x (f ) y sx
10. Suppose the graph of f is given. Write an equation for
each of the graphs that are obtained from the graph of f as follows. (a) Shift 2 units upward. (b) Shift 2 units downward. (c) Shift 2 units to the right. (d) Shift 2 units to the left. (e) Reflect about the x-axis. (f ) Reflect about the y-axis. (g) Stretch vertically by a factor of 2. (h) Shrink vertically by a factor of 2. (i) Stretch horizontally by a factor of 2. ( j) Shrink horizontally by a factor of 2. 11. Explain what each of the following means and illustrate
with a sketch. (a) lim f x L
(b) lim f x L
(c) lim f x L
(d) lim f x
x la
x la
(e) lim f x L x l
x la
x la
70
■
CHAPTER 1
FUNCTIONS AND LIMITS
12. Describe several ways in which a limit can fail to exist.
(b) What does it mean for f to be continuous on the interval , ? What can you say about the graph of such a function?
Illustrate with sketches. 13. State the following Limit Laws.
(a) (c) (e) (g)
Sum Law Constant Multiple Law Quotient Law Root Law
(b) Difference Law (d) Product Law (f ) Power Law
16. What does the Intermediate Value Theorem say? 17. (a) What does it mean to say that the line x a is a vertical
asymptote of the curve y f x? Draw curves to illustrate the various possibilities. (b) What does it mean to say that the line y L is a horizontal asymptote of the curve y f x? Draw curves to illustrate the various possibilities.
14. What does the Squeeze Theorem say? 15. (a) What does it mean for f to be continuous at a?
T R U E - FA L S E Q U I Z Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.
11. If p is a polynomial, then lim x l b px pb. 12. If lim x l 0 f x and lim x l 0 tx , then
lim x l 0 f x tx 0.
1. If f is a function, then f s t f s f t. 2. If f s f t, then s t.
13. A function can have two different horizontal asymptotes.
3. A vertical line intersects the graph of a function at most
14. If f has domain 0, and has no horizontal asymptote,
then lim x l f x or lim x l f x .
once. 4. If f and t are functions, then f ⴰ t t ⴰ f .
2x 8 5. lim x l4 x4 x4
x l1
lim x 2 6x 7
x 6x 7 x l1 x 2 5x 6 lim x 2 5x 6 2
6. lim
2x 8 lim lim x l4 x 4 x l4 x 4
x l1
15. If the line x 1 is a vertical asymptote of y f x, then f
is not defined at 1. 16. If f 1 0 and f 3 0, then there exists a number c
between 1 and 3 such that f c 0.
17. If f is continuous at 5 and f 5 2 and f 4 3, then
lim x l 2 f 4x 2 11 2.
lim x 3
7. lim x l1
x3 x l1 x 2 2x 4 lim x 2 2x 4
18. If f is continuous on 1, 1 and f 1 4 and f 1 3,
then there exists a number r such that r 1 and f r .
x l1
8. If lim x l 5 f x 2 and lim x l 5 tx 0, then
limx l 5 f xtx does not exist.
19. Let f be a function such that lim x l 0 f x 6. Then
9. If lim x l5 f x 0 and lim x l 5 tx 0, then
lim x l 5 f xtx does not exist.
10. If lim x l 6 f xtx exists, then the limit must be f 6t6.
there exists a number such that if 0 x , then f x 6 1.
20. If f x 1 for all x and lim x l 0 f x exists, then
lim x l 0 f x 1.
EXERCISES 1. Let f be the function whose graph is given.
(a) (b) (c) (d) (e) (f )
Estimate the value of f 2. Estimate the values of x such that f x 3. State the domain of f. State the range of f. On what interval is f increasing? Is f even, odd, or neither even nor odd? Explain.
y
f 1 1
x
CHAPTER 1
2. Determine whether each curve is the graph of a function
■
71
16. Find an expression for the function whose graph consists of
the line segment from the point 2, 2 to the point 1, 0 together with the top half of the circle with center the origin and radius 1.
of x. If it is, state the domain and range of the function. y y (a) (b) 2
REVIEW
2
17. If f x sx and tx sin x, find the functions (a) f ⴰ t,
(b) t ⴰ f , (c) f ⴰ f , (d) t ⴰ t, and their domains.
0
x
1
0
x
1
18. Express the function Fx 1sx sx as a composition
of three functions.
; 19. Use graphs to discover what members of the family of func3–6
Find the domain and range of the function.
■
3. f x s4 3x 2
4. tx 1x 1
5. y 1 sin x
6. y tan 2x
■
tions f x sin n x have in common, where n is a positive integer. How do they differ? What happens to the graphs as n becomes large?
■
■
■
■
■
■
■
■
20. A small-appliance manufacturer finds that it costs $9000 to ■
■
■
7. Suppose that the graph of f is given. Describe how the
graphs of the following functions can be obtained from the graph of f. (a) y f x 8 (b) y f x 8 (c) y 1 2 f x (d) y f x 2 2 (e) y f x (f ) y 3 f x 8. The graph of f is given. Draw the graphs of the following
functions. (a) y f x 8 (c) y 2 f x
produce 1000 toaster ovens a week and $12,000 to produce 1500 toaster ovens a week. (a) Express the cost as a function of the number of toaster ovens produced, assuming that it is linear. Then sketch the graph. (b) What is the slope of the graph and what does it represent? (c) What is the y-intercept of the graph and what does it represent? 21. The graph of f is given.
(a) Find each limit, or explain why it does not exist. (i) lim f x (ii) lim f x
(b) y f x (d) y 12 f x 1
x l2
x l3
(iii) lim f x
(iv) lim f x
(v) lim f x
(vi) lim f x
(vii) lim f x
(viii) lim f x
x l3
y
x l4
x l0
x l2
x l
1 0
1
x l
(b) State the equations of the horizontal asymptotes. (c) State the equations of the vertical asymptotes. (d) At what numbers is f discontinuous? Explain.
x
y
9–14
■
Use transformations to sketch the graph of the function.
9. y sin 2 x
10. y x 2 2
11. y 1 x 3
12. y 2 sx
1 2
1
1 13. f x x2 14. f x ■
■
0
1x 1 x2
■
■
if x 0 if x 0 ■
■
■
■
■
■
■
15. Determine whether f is even, odd, or neither even nor odd.
(a) (b) (c) (d)
f x 2x 3x 2 f x x 3 x 7 f x cosx 2 f x 1 sin x 5
x
1
2
■
22. Sketch the graph of an example of a function f that satisfies
all of the following conditions: lim f x 2, lim f x 0, x l
lim f x ,
x l3
xl
lim f x 2,
x l3
f is continuous from the right at 3
lim f x ,
x l3
72
■
23–38
CHAPTER 1
FUNCTIONS AND LIMITS
41. If 2x 1 f x x 2 for 0 x 3, find lim x l1 f x.
Find the limit.
■
x 9 x 2 2x 3 2
23. lim cosx sin x
24. lim
x2 9 25. lim 2 x l3 x 2x 3
x2 9 26. lim 2 x l1 x 2x 3
xl0
x l3
h 1 1 h 3
27. lim
h l0
r l9
43– 46
28. lim t l2
vl4
45. lim
xl
4v
4 v
■
1 2x x 2 33. lim x l 1 x 2x 2
1 2x 2 x 4 34. lim x l 5 x 3x 4
v 2 2v 8
xl1
cot 2x csc x
37. lim
xl0
■
■
; 39– 40
1 1 2 x1 x 3x 2
■
■
■
■
■
■
■
■
■
if x 0 if 0 x 3 if x 3
(v) lim f x x l3
■
t3 tan3 2t
(iii) lim f x x l0
(vi) lim f x x l3
■
48. Show that each function is continuous on its domain. State ■
■
the domain. sx 2 9 (a) tx 2 x 2
■
■
■
■
■
4 (b) hx s x x 3 cos x
■ Use the Intermediate Value Theorem to show that there is a root of the equation in the given interval.
49–50
50. 2 sin x 3 2x, ■
■
x l0
49. 2x 3 x 2 2 0,
■
■
x l3
cos2 x x2
■
■
2 sx 4
(b) Where is f discontinuous? (c) Sketch the graph of f .
40. y sx 2 x 1 sx 2 x ■
■
(a) Evaluate each limit, if it exists. (i) lim f x (ii) lim f x
Use graphs to discover the asymptotes of the curve. Then prove what you have discovered.
■
■
sx f x 3 x x 32
■
39. y
x l4
(iv) lim f x
tl0
■
46. lim
x l0
38. lim ■
■
v 4 16
35. lim (sx 2 4x 1 x) 36. lim
■
xl0
1 0 x4
47. Let
32. lim
xl
3 x 0 44. lim s
x l5
4 ss 31. lim s l16 s 16
v l2
Prove the statement using the precise definition of
43. lim 7x 27 8
t 4 t3 8
30. lim
■
a limit.
2
sr r 94
29. lim
42. Prove that lim x l 0 x 2 cos1x 2 0.
■
■
■
■
■
2, 1 0, 1
■
■
■
■
■
■
■
■
DERIVATIVES
2
In this chapter we study a special type of limit, called a derivative, that occurs when we want to find the slope of a tangent line, or a velocity, or any instantaneous rate of change.
2.1
DERIVATIVES AND RATES OF CHANGE The problem of finding the tangent line to a curve and the problem of finding the velocity of an object involve finding the same type of limit, which we call a derivative. THE TANGENT PROBLEM
The word tangent is derived from the Latin word tangens, which means “touching.” Thus a tangent to a curve is a line that touches the curve. In other words, a tangent line should have the same direction as the curve at the point of contact. How can this idea be made precise? For a circle we could simply follow Euclid and say that a tangent is a line that intersects the circle once and only once as in Figure 1(a). For more complicated curves this definition is inadequate. Figure l(b) shows two lines L and T passing through a point P on a curve C. The line L intersects C only once, but it certainly does not look like what we think of as a tangent. The line T, on the other hand, looks like a tangent but it intersects C twice.
T P T
C
L FIGURE 1
(a)
To be specific, let’s look at the problem of trying to find a tangent line T to the parabola y x 2 in the following example.
y
Q { x, ≈}
T V EXAMPLE 1
y=≈
(b)
Find an equation of the tangent line to the parabola y x 2 at the
point P1, 1.
P (1, 1)
SOLUTION We will be able to find an equation of the tangent line T as soon as we 0
FIGURE 2
x
know its slope m. The difficulty is that we know only one point, P, on T, whereas we need two points to compute the slope. But observe that we can compute an approximation to m by choosing a nearby point Qx, x 2 on the parabola (as in Figure 2) and computing the slope mPQ of the secant line PQ. 73
74
■
CHAPTER 2
DERIVATIVES
We choose x 1 so that Q P. Then mPQ
x2 1 x1
What happens as x approaches 1? From Figure 3 we see that Q approaches P along the parabola and the secant lines PQ rotate about P and approach the tangent line T. y
y
y
Q T
T
T
Q Q P
P
0
x
P
0
0
x
x
Q approaches P from the right y
y
y
T
Q
T
P
T
P
P
Q 0
x
Q 0
0
x
x
Q approaches P from the left FIGURE 3
It appears that the slope m of the tangent line is the limit of the slopes of the secant lines as x approaches 1: In Visual 2.1A you can see how the process in Figure 3 works for additional functions.
m lim x l1
x2 1 x 1x 1 lim xl1 x1 x1
lim x 1 1 1 2 x l1
■ Point-slope form for a line through the point x1 , y1 with slope m : y y1 mx x 1
Using the point-slope form of the equation of a line, we find that an equation of the tangent line at 1, 1 is y 1 2x 1
or
y 2x 1
■
We sometimes refer to the slope of the tangent line to a curve at a point as the slope of the curve at the point. The idea is that if we zoom in far enough toward the point, the curve looks almost like a straight line. Figure 4 illustrates this procedure for the curve y x 2 in Example 1. The more we zoom in, the more the parabola looks like a line. In other words, the curve becomes almost indistinguishable from its tangent line.
SECTION 2.1
2
DERIVATIVES AND RATES OF CHANGE
1.5
(1, 1)
2
75
1.1
(1, 1)
0
■
(1, 1)
1.5
0.5
1.1
0.9
FIGURE 4 Zooming in toward the point (1, 1) on the parabola y=≈
Visual 2.1B shows an animation of Figure 4.
In general, if a curve C has equation y f x and we want to find the tangent line to C at the point Pa, f a, then we consider a nearby point Qx, f x, where x a, and compute the slope of the secant line PQ : mPQ
f x f a xa
Then we let Q approach P along the curve C by letting x approach a. If mPQ approaches a number m, then we define the tangent T to be the line through P with slope m. (This amounts to saying that the tangent line is the limiting position of the secant line PQ as Q approaches P. See Figure 5.) y
y
t Q
Q{ x, ƒ }
Q
ƒ-f(a) P
P { a, f(a)}
Q
x-a
0
a
x
x
x
0
FIGURE 5 1 DEFINITION The tangent line to the curve y f x at the point Pa, f a is the line through P with slope f x f a m lim xla xa
provided that this limit exists. Q { a+h, f(a+h)} y
t
There is another expression for the slope of a tangent line that is sometimes easier to use. If h x a, then x a h and so the slope of the secant line PQ is
P { a, f(a)} f(a+h)-f(a)
h 0
FIGURE 6
a
a+h
x
mPQ
f a h f a h
(See Figure 6 where the case h 0 is illustrated and Q is to the right of P. If it happened that h 0, however, Q would be to the left of P.) Notice that as x approaches a, h approaches 0 (because h x a) and so the expression for the slope of the
76
■
CHAPTER 2
DERIVATIVES
tangent line in Definition 1 becomes
2
m lim
hl0
f a h f a h
EXAMPLE 2 Find an equation of the tangent line to the hyperbola y 3x at the
point 3, 1.
SOLUTION Let f x 3x. Then the slope of the tangent at 3, 1 is
m lim
hl0
f 3 h f 3 h
3 3 3 h 1 3h 3h lim lim hl0 hl0 h h y
lim
3 y= x
x+3y-6=0
hl0
h 1 1 lim h l 0 h3 h 3h 3
Therefore, an equation of the tangent at the point 3, 1 is
(3, 1)
y 1 13 x 3
x
0
x 3y 6 0
which simplifies to
The hyperbola and its tangent are shown in Figure 7.
FIGURE 7
■
THE VELOCITY PROBLEM position at time t=a
position at time t=a+h s
0
f(a+h)-f(a)
f(a) f(a+h) FIGURE 8 s
Q { a+h, f(a+h)} P { a, f(a)} h
0
mPQ=
a
a+h
f(a+h)-f(a) average = h velocity
FIGURE 9
t
In Section 1.3 we investigated the motion of a ball dropped from the CN Tower and defined its velocity to be the limiting value of average velocities over shorter and shorter time periods. In general, suppose an object moves along a straight line according to an equation of motion s f t, where s is the displacement (directed distance) of the object from the origin at time t. The function f that describes the motion is called the position function of the object. In the time interval from t a to t a h the change in position is f a h f a. (See Figure 8.) The average velocity over this time interval is displacement f a h f a average velocity time h which is the same as the slope of the secant line PQ in Figure 9. Now suppose we compute the average velocities over shorter and shorter time intervals a, a h . In other words, we let h approach 0. As in the example of the falling ball, we define the velocity (or instantaneous velocity) va at time t a to be the limit of these average velocities:
3
va lim
hl0
f a h f a h
SECTION 2.1
DERIVATIVES AND RATES OF CHANGE
■
77
This means that the velocity at time t a is equal to the slope of the tangent line at P (compare Equations 2 and 3). Now that we know how to compute limits, let’s reconsider the problem of the falling ball. V EXAMPLE 3 Suppose that a ball is dropped from the upper observation deck of the CN Tower, 450 m above the ground. (a) What is the velocity of the ball after 5 seconds? (b) How fast is the ball traveling when it hits the ground?
Recall from Section 1.3: The distance (in meters) fallen after t seconds is 4.9t 2. ■
SOLUTION We first use the equation of motion s f t 4.9t 2 to find the velocity va after a seconds: va lim
hl0
lim
hl0
f a h f a 4.9a h2 4.9a 2 lim hl0 h h 4.9a 2 2ah h 2 a 2 4.92ah h 2 lim hl0 h h
lim 4.92a h 9.8a hl0
(a) The velocity after 5 s is v5 9.85 49 ms. (b) Since the observation deck is 450 m above the ground, the ball will hit the ground at the time t1 when st1 450, that is, 4.9t12 450 This gives t12
450 4.9
t1
and
450 9.6 s 4.9
The velocity of the ball as it hits the ground is therefore
vt1 9.8t1 9.8
450 94 ms 4.9
■
DERIVATIVES
We have seen that the same type of limit arises in finding the slope of a tangent line (Equation 2) or the velocity of an object (Equation 3). In fact, limits of the form lim
h l0
f a h f a h
arise whenever we calculate a rate of change in any of the sciences or engineering, such as a rate of reaction in chemistry or a marginal cost in economics. Since this type of limit occurs so widely, it is given a special name and notation. 4 DEFINITION
The derivative of a function f at a number a, denoted by
f a, is ■
f a lim
f a is read “ f prime of a .”
h l0
if this limit exists.
f a h f a h
78
■
CHAPTER 2
DERIVATIVES
If we write x a h, then h x a and h approaches 0 if and only if x approaches a. Therefore, an equivalent way of stating the definition of the derivative, as we saw in finding tangent lines, is f a lim
5
xla
V EXAMPLE 4
f x f a xa
Find the derivative of the function f x x 2 8x 9 at the
number a. SOLUTION From Definition 4 we have
f a h f a h
f a lim
h l0
lim
a h2 8a h 9 a 2 8a 9 h
lim
a 2 2ah h 2 8a 8h 9 a 2 8a 9 h
lim
2ah h 2 8h lim 2a h 8 h l0 h
h l0
h l0
h l0
2a 8
■
We defined the tangent line to the curve y f x at the point Pa, f a to be the line that passes through P and has slope m given by Equation 1 or 2. Since, by Definition 4, this is the same as the derivative f a, we can now say the following. The tangent line to y f x at a, f a is the line through a, f a whose slope is equal to f a, the derivative of f at a. If we use the point-slope form of the equation of a line, we can write an equation of the tangent line to the curve y f x at the point a, f a:
y
y=≈-8x+9
x
0 (3, _6)
y=_2x FIGURE 10
y f a f ax a V EXAMPLE 5 Find an equation of the tangent line to the parabola y x 2 8x 9 at the point 3, 6.
SOLUTION From Example 4 we know that the derivative of f x x 2 8x 9 at
the number a is f a 2a 8. Therefore, the slope of the tangent line at 3, 6 is f 3 23 8 2. Thus, an equation of the tangent line, shown in Figure 10, is y 6 2x 3
or
y 2x
■
RATES OF CHANGE
Suppose y is a quantity that depends on another quantity x. Thus y is a function of x and we write y f x. If x changes from x 1 to x 2 , then the change in x (also called the increment of x) is x x 2 x 1
SECTION 2.1
Q { ¤, ‡}
y
DERIVATIVES AND RATES OF CHANGE
■
79
and the corresponding change in y is y f x 2 f x 1
P {⁄, fl}
Îy
The difference quotient y f x 2 f x 1 x x2 x1
Îx ⁄
0
¤
x
average rate of change ⫽ mPQ instantaneous rate of change ⫽ slope of tangent at P FIGURE 11
is called the average rate of change of y with respect to x over the interval x 1, x 2 and can be interpreted as the slope of the secant line PQ in Figure 11. By analogy with velocity, we consider the average rate of change over smaller and smaller intervals by letting x 2 approach x 1 and therefore letting x approach 0. The limit of these average rates of change is called the (instantaneous) rate of change of y with respect to x at x x 1 , which is interpreted as the slope of the tangent to the curve y f x at Px 1, f x 1:
6
instantaneous rate of change lim
x l 0
y f x2 f x1 lim x l x x x2 x1 2 1
We recognize this limit as being the derivative f x 1. We know that one interpretation of the derivative f a is as the slope of the tangent line to the curve y f x when x a . We now have a second interpretation: y
The derivative f a is the instantaneous rate of change of y f x with respect to x when x a.
Q
P
x
FIGURE 12
The y-values are changing rapidly at P and slowly at Q.
t
Dt
1980 1985 1990 1995 2000
930.2 1945.9 3233.3 4974.0 5674.2
The connection with the first interpretation is that if we sketch the curve y f x, then the instantaneous rate of change is the slope of the tangent to this curve at the point where x a. This means that when the derivative is large (and therefore the curve is steep, as at the point P in Figure 12), the y-values change rapidly. When the derivative is small, the curve is relatively flat and the y-values change slowly. In particular, if s f t is the position function of a particle that moves along a straight line, then f a is the rate of change of the displacement s with respect to the time t. In other words, f a is the velocity of the particle at time t a. The speed of the particle is the absolute value of the velocity, that is, f a . In the following example we estimate the rate of change of the national debt with respect to time. Here the function is defined not by a formula but by a table of values.
V EXAMPLE 6 Let Dt be the US national debt at time t. The table in the margin gives approximate values of this function by providing end of year estimates, in billions of dollars, from 1980 to 2000. Interpret and estimate the value of D1990.
SOLUTION The derivative D1990 means the rate of change of D with respect to t when t 1990, that is, the rate of increase of the national debt in 1990. According to Equation 5,
D1990 lim
t l1990
Dt D1990 t 1990
80
■
CHAPTER 2
DERIVATIVES
So we compute and tabulate values of the difference quotient (the average rates of change) as follows.
A NOTE ON UNITS The units for the average rate of change Dt are the units for D divided by the units for t , namely, billions of dollars per year. The instantaneous rate of change is the limit of the average rates of change, so it is measured in the same units: billions of dollars per year. ■
t
Dt D1990 t 1990
1980 1985 1995 2000
230.31 257.48 348.14 244.09
From this table we see that D1990 lies somewhere between 257.48 and 348.14 billion dollars per year. [Here we are making the reasonable assumption that the debt didn’t fluctuate wildly between 1980 and 2000.] We estimate that the rate of increase of the national debt of the United States in 1990 was the average of these two numbers, namely D1990 303 billion dollars per year Another method would be to plot the debt function and estimate the slope of the ■ tangent line when t 1990. The rate of change of the debt with respect to time in Example 6 is just one example of a rate of change. Here are a few of the many others: The velocity of a particle is the rate of change of displacement with respect to time. Physicists are interested in other rates of change as well—for instance, the rate of change of work with respect to time (which is called power). Chemists who study a chemical reaction are interested in the rate of change in the concentration of a reactant with respect to time (called the rate of reaction). A steel manufacturer is interested in the rate of change of the cost of producing x tons of steel per day with respect to x (called the marginal cost). A biologist is interested in the rate of change of the population of a colony of bacteria with respect to time. In fact, the computation of rates of change is important in all of the natural sciences, in engineering, and even in the social sciences. All these rates of change can be interpreted as slopes of tangents. This gives added significance to the solution of the tangent problem. Whenever we solve a problem involving tangent lines, we are not just solving a problem in geometry. We are also implicitly solving a great variety of problems involving rates of change in science and engineering.
2.1
EXERCISES 2. (a) Find the slope of the tangent line to the curve y x 3 at
1. (a) Find the slope of the tangent line to the parabola
y x 2x at the point 3, 3 (i) using Definition 1 (ii) using Equation 2 (b) Find an equation of the tangent line in part (a). (c) Graph the parabola and the tangent line. As a check on your work, zoom in toward the point (3, 3) until the parabola and the tangent line are indistinguishable. 2
;
;
the point 1, 1 (i) using Definition 1 (ii) using Equation 2 (b) Find an equation of the tangent line in part (a). (c) Graph the curve and the tangent line in successively smaller viewing rectangles centered at (1, 1) until the curve and the line appear to coincide.
SECTION 2.1
4. y 2x 5x, 3
5. y sx , ■
■
■
3, 2
(1, 1 ■
0, 0 ■
■
■
■
■
■
■
■
7. (a) Find the slope of the tangent to the curve
;
y 3 4x 2 2x 3 at the point where x a. (b) Find equations of the tangent lines at the points 1, 5 and 2, 3. (c) Graph the curve and both tangents on a common screen. 8. (a) Find the slope of the tangent to the curve y 1sx at
;
the point where x a. (b) Find equations of the tangent lines at the points 1, 1 and (4, 12 ). (c) Graph the curve and both tangents on a common screen.
9. The graph shows the position function of a car. Use the
shape of the graph to explain your answers to the following questions. (a) What was the initial velocity of the car? (b) Was the car going faster at B or at C ? (c) Was the car slowing down or speeding up at A, B, and C ? (d) What happened between D and E ?
13. The displacement (in meters) of a particle moving in a
straight line is given by the equation of motion s 1t 2, where t is measured in seconds. Find the velocity of the particle at times t a, t 1, t 2, and t 3. 14. The displacement (in meters) of a particle moving in a
straight line is given by s t 2 8t 18, where t is measured in seconds. (a) Find the average velocity over each time interval: (i) 3, 4 (ii) 3.5, 4 (iii) 4, 5 (iv) 4, 4.5 (b) Find the instantaneous velocity when t 4. (c) Draw the graph of s as a function of t and draw the secant lines whose slopes are the average velocities in part (a) and the tangent line whose slope is the instantaneous velocity in part (b). 15. For the function t whose graph is given, arrange the follow-
ing numbers in increasing order and explain your reasoning: 0
s
81
58 ms, its height (in meters) after t seconds is given by H 58t 0.83t 2. (a) Find the velocity of the arrow after one second. (b) Find the velocity of the arrow when t a. (c) When will the arrow hit the moon? (d) With what velocity will the arrow hit the moon?
1, 3
6. y 2x x 1 2,
■
12. If an arrow is shot upward on the moon with a velocity of
■ Find an equation of the tangent line to the curve at the given point.
3–6
3. y x 1x 2,
DERIVATIVES AND RATES OF CHANGE
t2
t0
t2
t4
E
D
y
C
y=©
B A 0
t
_1
0
1
2
3
4
x
10. Shown are graphs of the position functions of two runners,
A and B, who run a 100-m race and finish in a tie. 16. (a) Find an equation of the tangent line to the graph of
s (meters) 80
y tx at x 5 if t5 3 and t5 4. (b) If the tangent line to y f x at (4, 3) passes through the point (0, 2), find f 4 and f 4.
A
40
17. Sketch the graph of a function f for which f 0 0,
B 0
4
8
f 0 3, f 1 0, and f 2 1.
12
t (seconds)
(a) Describe and compare how the runners run the race. (b) At what time is the distance between the runners the greatest? (c) At what time do they have the same velocity? 11. If a ball is thrown into the air with a velocity of 40 fts, its
height (in feet) after t seconds is given by y 40t 16t 2. Find the velocity when t 2.
18. Sketch the graph of a function t for which t0 t0 0,
t1 1, t1 3, and t2 1.
19. If f x 3x 2 5x, find f 2 and use it to find an equa-
tion of the tangent line to the parabola y 3x 2 5x at the point 2, 2.
20. If tx 1 x 3, find t0 and use it to find an equation of
the tangent line to the curve y 1 x 3 at the point 0, 1.
82
■
CHAPTER 2
DERIVATIVES
21. (a) If Fx 5x1 x 2 , find F2 and use it to find
37. The table shows the estimated percentage P of the popula-
an equation of the tangent line to the curve y 5x1 x 2 at the point 2, 2. (b) Illustrate part (a) by graphing the curve and the tangent line on the same screen.
;
tion of Europe that use cell phones. (Midyear estimates are given.)
22. (a) If Gx 4x x , find Ga and use it to find equa2
3
tions of the tangent lines to the curve y 4x 2 x 3 at the points 2, 8 and 3, 9. (b) Illustrate part (a) by graphing the curve and the tangent lines on the same screen.
;
23–28
■
25. f t
■
26. f x
■
■
x2 1 x2
28. f x s3x 1 ■
■
■
■
■
■
■
29. lim
1 h10 1 h
30. lim
31. lim
2 x 32 x5
32. lim
33. lim
cos h 1 h
34. lim
x l5
h l0
■
■
■
2002
2003
P
28
39
55
68
77
83
h l0
■
t l1
■
■
■
tan x 1 x 4
t4 t 2 t1 ■
■
■
Year
1998
1999
2000
2001
2002
N
1886
2135
3501
4709
5886
(a) Find the average rate of growth (i) from 2000 to 2002 (ii) from 2000 to 2001 (iii) from 1999 to 2000 In each case, include the units. (b) Estimate the instantaneous rate of growth in 2000 by taking the average of two average rates of change. What are its units? (c) Estimate the instantaneous rate of growth in 2000 by measuring the slope of a tangent.
4 16 h 2 s h
x l 4
■
2001
■
■
h l0
2000
is given in the table. (The numbers of locations as of June 30 are given.)
Each limit represents the derivative of some function f at some number a. State such an f and a in each case. 29–34
1999
38. The number N of locations of a popular coffeehouse chain
1 sx 2
27. f x ■
24. f t t 4 5t
2t 1 t3
1998
(a) Find the average rate of cell phone growth (i) from 2000 to 2002 (ii) from 2000 to 2001 (iii) from 1999 to 2000 In each case, include the units. (b) Estimate the instantaneous rate of growth in 2000 by taking the average of two average rates of change. What are its units? (c) Estimate the instantaneous rate of growth in 2000 by measuring the slope of a tangent.
Find f a.
23. f x 3 2x 4x 2
Year
■
39. The cost (in dollars) of producing x units of a certain 35. A warm can of soda is placed in a cold refrigerator. Sketch
the graph of the temperature of the soda as a function of time. Is the initial rate of change of temperature greater or less than the rate of change after an hour? 36. A roast turkey is taken from an oven when its temperature
has reached 185°F and is placed on a table in a room where the temperature is 75°F. The graph shows how the temperature of the turkey decreases and eventually approaches room temperature. By measuring the slope of the tangent, estimate the rate of change of the temperature after an hour.
commodity is Cx 5000 10x 0.05x 2. (a) Find the average rate of change of C with respect to x when the production level is changed (i) from x 100 to x 105 (ii) from x 100 to x 101 (b) Find the instantaneous rate of change of C with respect to x when x 100. (This is called the marginal cost. Its significance will be explained in Section 2.3.) 40. If a cylindrical tank holds 100,000 gallons of water, which
can be drained from the bottom of the tank in an hour, then Torricelli’s Law gives the volume V of water remaining in the tank after t minutes as
T (°F)
Vt 100,000 (1
200
P 100
0
30
60
90
120 150
t (min)
1 60
t)
2
0 t 60
Find the rate at which the water is flowing out of the tank (the instantaneous rate of change of V with respect to t ) as a function of t. What are its units? For times t 0, 10, 20, 30, 40, 50, and 60 min, find the flow rate and the amount of water remaining in the tank. Summarize your findings in a sentence or two. At what time is the flow rate the greatest? The least?
SECTION 2.2
mine is C f x dollars. (a) What is the meaning of the derivative f x? What are its units? (b) What does the statement f 800 17 mean? (c) Do you think the values of f x will increase or decrease in the short term? What about the long term? Explain.
S (mg / L) 16 12 8 4
42. The number of bacteria after t hours in a controlled labora-
0
2
4
6
8
10
12
14
T
73
73
70
69
72
81
88
91
45. The quantity of oxygen that can dissolve in water depends
on the temperature of the water. (So thermal pollution influences the oxygen content of water.) The graph shows how oxygen solubility S varies as a function of the water temperature T. (a) What is the meaning of the derivative ST ? What are its units?
2.2
24
32
40
T (°C)
S (cm/s) 20
0
44. The quantity (in pounds) of a gourmet ground coffee that is
sold by a coffee company at a price of p dollars per pound is Q f p. (a) What is the meaning of the derivative f 8? What are its units? (b) Is f 8 positive or negative? Explain.
16
maximum sustainable swimming speed S of Coho salmon. (a) What is the meaning of the derivative ST ? What are its units? (b) Estimate the values of S15 and S25 and interpret them.
midnight on June 2, 2001. The table shows values of this function recorded every two hours. What is the meaning of T 10? Estimate its value. 0
8
46. The graph shows the influence of the temperature T on the
43. Let Tt be the temperature (in F ) in Dallas t hours after
t
83
(b) Estimate the value of S16 and interpret it.
41. The cost of producing x ounces of gold from a new gold
tory experiment is n f t. (a) What is the meaning of the derivative f 5? What are its units? (b) Suppose there is an unlimited amount of space and nutrients for the bacteria. Which do you think is larger, f 5 or f 10? If the supply of nutrients is limited, would that affect your conclusion? Explain.
■
THE DERIVATIVE AS A FUNCTION
47– 48
■
Determine whether f 0 exists.
47. f x
48. f x ■
■
T (°C)
20
10
x sin
1 x
if x 0 if x 0
0
x 2 sin
1 x
if x 0 if x 0
0
■
■
■
■
■
■
■
■
■
■
THE DERIVATIVE AS A FUNCTION In Section 2.1 we considered the derivative of a function f at a fixed number a: 1
f a lim
hl0
f a h f a h
Here we change our point of view and let the number a vary. If we replace a in Equation 1 by a variable x, we obtain
2
f x lim
hl0
f x h f x h
84
■
CHAPTER 2
DERIVATIVES
Given any number x for which this limit exists, we assign to x the number f x. So we can regard f as a new function, called the derivative of f and defined by Equation 2. We know that the value of f at x, f x, can be interpreted geometrically as the slope of the tangent line to the graph of f at the point x, f x. The function f is called the derivative of f because it has been “derived” from f by the limiting operation in Equation 2. The domain of f is the set x f x exists and may be smaller than the domain of f .
V EXAMPLE 1 The graph of a function f is given in Figure 1. Use it to sketch the graph of the derivative f .
y y=ƒ
SOLUTION We can estimate the value of the derivative at any value of x by drawing
1 0
1
x
FIGURE 1
the tangent at the point x, f x and estimating its slope. For instance, for x 5 we draw the tangent at P in Figure 2(a) and estimate its slope to be about 32 , so f 5 1.5. This allows us to plot the point P5, 1.5 on the graph of f directly beneath P. Repeating this procedure at several points, we get the graph shown in Figure 2(b). Notice that the tangents at A, B, and C are horizontal, so the derivative is 0 there and the graph of f crosses the x-axis at the points A, B, and C, directly y
B m=0 1
m=0
y=ƒ
P mÅ3 2
A
0
1
5
x
m=0
C
Visual 2.2 shows an animation of Figure 2 for several functions.
(a) y
Pª (5, 1.5) y=fª(x)
1
Bª 0
FIGURE 2
Aª
Cª
1
(b)
5
x
SECTION 2.2
THE DERIVATIVE AS A FUNCTION
■
85
beneath A, B, and C. Between A and B the tangents have positive slope, so f x is positive there. But between B and C the tangents have negative slope, so f x is negative there. ■ V EXAMPLE 2
(a) If f x x 3 x, find a formula for f x. (b) Illustrate by comparing the graphs of f and f . SOLUTION
2
(a) When using Equation 2 to compute a derivative, we must remember that the variable is h and that x is temporarily regarded as a constant during the calculation of the limit.
f _2
2
f x lim
hl0
_2
lim
x 3 3x 2h 3xh 2 h 3 x h x 3 x h
lim
3x 2h 3xh 2 h 3 h h
hl0
2
fª
hl0
_2
2
f x h f x x h3 x h x 3 x lim hl0 h h
lim 3x 2 3xh h 2 1 3x 2 1 hl0
_2
FIGURE 3
(b) We use a graphing device to graph f and f in Figure 3. Notice that f x 0 when f has horizontal tangents and f x is positive when the tangents have positive slope. So these graphs serve as a check on our work in part (a). ■ EXAMPLE 3 If f x sx , find the derivative of f . State the domain of f . SOLUTION
f x lim
h l0
lim
h l0
Here we rationalize the numerator.
f x h f x h sx h sx h
lim
lim
x h x 1 lim h l0 sx h sx h (sx h sx )
h l0
h l0
sx h sx sx h sx h sx h sx
1 1 2sx sx sx
We see that f x exists if x 0, so the domain of f is 0, . This is smaller than the domain of f , which is 0, . ■ Let’s check to see that the result of Example 3 is reasonable by looking at the graphs of f and f in Figure 4. When x is close to 0, sx is also close to 0, so f x 1(2sx ) is very large and this corresponds to the steep tangent lines near 0, 0 in Figure 4(a) and the large values of f x just to the right of 0 in Figure 4(b).
86
■
CHAPTER 2
DERIVATIVES
When x is large, f x is very small and this corresponds to the flatter tangent lines at the far right of the graph of f and the horizontal asymptote of the graph of f . y
y
1
1
0
x
1
(a) ƒ=œ„ x
FIGURE 4
EXAMPLE 4 Find f if f x
SOLUTION a c b d ad bc 1 e bd e
0
x
1
(b) f ª (x)=
1 2œ„ x
1x . 2x
1 x h 1x f x h f x 2 x h 2x f x lim lim hl0 hl0 h h lim
1 x h2 x 1 x2 x h h2 x h2 x
lim
2 x 2h x 2 xh 2 x h x 2 xh h2 x h2 x
lim
3h h2 x h2 x
lim
3 3 2 x h2 x 2 x2
hl0
hl0
hl0
hl0
■
OTHER NOTATIONS
If we use the traditional notation y f x to indicate that the independent variable is x and the dependent variable is y, then some common alternative notations for the derivative are as follows: f x y
dy df d f x Df x Dx f x dx dx dx
The symbols D and ddx are called differentiation operators because they indicate the operation of differentiation, which is the process of calculating a derivative. The symbol dydx, which was introduced by Leibniz, should not be regarded as a ratio (for the time being); it is simply a synonym for f x. Nonetheless, it is a very useful and suggestive notation, especially when used in conjunction with increment notation. Referring to Equation 2.1.6, we can rewrite the definition of derivative in Leibniz notation in the form dy y lim x l 0 x dx
SECTION 2.2
Gottfried Wilhelm Leibniz was born in Leipzig in 1646 and studied law, theology, philosophy, and mathematics at the university there, graduating with a bachelor’s degree at age 17. After earning his doctorate in law at age 20, Leibniz entered the diplomatic service and spent most of his life traveling to the capitals of Europe on political missions. In particular, he worked to avert a French military threat against Germany and attempted to reconcile the Catholic and Protestant churches. His serious study of mathematics did not begin until 1672 while he was on a diplomatic mission in Paris. There he built a calculating machine and met scientists, like Huygens, who directed his attention to the latest developments in mathematics and science. Leibniz sought to develop a symbolic logic and system of notation that would simplify logical reasoning. In particular, the version of calculus that he published in 1684 established the notation and the rules for finding derivatives that we use today. Unfortunately, a dreadful priority dispute arose in the 1690s between the followers of Newton and those of Leibniz as to who had invented calculus first. Leibniz was even accused of plagiarism by members of the Royal Society in England. The truth is that each man invented calculus independently. Newton arrived at his version of calculus first but, because of his fear of controversy, did not publish it immediately. So Leibniz’s 1684 account of calculus was the first to be published. ■
THE DERIVATIVE AS A FUNCTION
■
87
If we want to indicate the value of a derivative dydx in Leibniz notation at a specific number a, we use the notation dy dx
dy dx
or xa
xa
which is a synonym for f a. DIFFERENTIABLE FUNCTIONS
A function f is differentiable at a if f a exists. It is differentiable on an open interval a, b [or a, or , a or , ] if it is differentiable at every number in the interval. 3 DEFINITION
Where is the function f x x differentiable?
V EXAMPLE 5
SOLUTION If x 0, then x x and we can choose h small enough that x h 0 and hence x h x h. Therefore, for x 0 we have
f x lim
hl0
lim
hl0
x h x h
x h x h lim lim 1 1 hl0 h hl0 h
and so f is differentiable for any x 0. Similarly, for x 0 we have x x and h can be chosen small enough that x h 0 and so x h x h. Therefore, for x 0,
f x lim
hl0
lim
hl0
x h x h
x h x h lim lim 1 1 hl0 h hl0 h
and so f is differentiable for any x 0. For x 0 we have to investigate f 0 lim
hl0
lim
hl0
f 0 h f 0 h
0 h 0
if it exists
h
Let’s compute the left and right limits separately: lim
h l 0
and
lim
hl0
0 h 0 h
0 h 0 h
lim
h l 0
lim
hl0
h h
h h
lim
h l 0
lim
hl0
h lim 1 1 hl0 h
h lim 1 1 hl0 h
Since these limits are different, f 0 does not exist. Thus f is differentiable at all x except 0.
88
■
CHAPTER 2
DERIVATIVES
A formula for f is given by
y
f x
if x 0 if x 0
1 1
and its graph is shown in Figure 5(b). The fact that f 0 does not exist is reflected geometrically in the fact that the curve y x does not have a tangent line at 0, 0. [See Figure 5(a).] ■
x
0
(a) y=ƒ=| x |
Both continuity and differentiability are desirable properties for a function to have. The following theorem shows how these properties are related.
y 1
4 THEOREM
If f is differentiable at a, then f is continuous at a.
x
0
PROOF To prove that f is continuous at a, we have to show that lim x l a f x f a. We do this by showing that the difference f x f a approaches 0. The given information is that f is differentiable at a, that is,
_1
(b) y=fª(x)
f a lim
FIGURE 5
xla
f x f a xa
exists (see Equation 2.1.5). To connect the given and the unknown, we divide and multiply f x f a by x a (which we can do when x a): f x f a
f x f a x a xa
Thus, using the Product Law and (2.1.5), we can write lim f x f a lim
xla
xla
lim
xla
f x f a x a xa f x f a lim x a xla xa
f a 0 0 To use what we have just proved, we start with f x and add and subtract f a: lim f x lim f a f x f a
xla
xla
lim f a lim f x f a xla
xla
f a 0 f a ■
Therefore, f is continuous at a. |
NOTE The converse of Theorem 4 is false; that is, there are functions that are continuous but not differentiable. For instance, the function f x x is continuous at 0 because lim f x lim x 0 f 0
xl0
xl0
(See Example 6 in Section 1.4.) But in Example 5 we showed that f is not differentiable at 0.
SECTION 2.2
THE DERIVATIVE AS A FUNCTION
■
89
HOW CAN A FUNCTION FAIL TO BE DIFFERENTIABLE?
We saw that the function y x in Example 5 is not differentiable at 0 and Figure 5(a) shows that its graph changes direction abruptly when x 0. In general, if the graph of a function f has a “corner” or “kink” in it, then the graph of f has no tangent at this point and f is not differentiable there. [In trying to compute f a, we find that the left and right limits are different.] Theorem 4 gives another way for a function not to have a derivative. It says that if f is not continuous at a, then f is not differentiable at a. So at any discontinuity (for instance, a jump discontinuity) f fails to be differentiable. A third possibility is that the curve has a vertical tangent line when x a; that is, f is continuous at a and
y
vertical tangent line
lim f x
0
a
xla
x
This means that the tangent lines become steeper and steeper as x l a. Figure 6 shows one way that this can happen; Figure 7(c) shows another. Figure 7 illustrates the three possibilities that we have discussed.
FIGURE 6
y
y
0
0
x
a
y
x
a
0
a
x
FIGURE 7
Three ways for ƒ not to be differentiable at a
(a) A corner
(b) A discontinuity
(c) A vertical tangent
A graphing calculator or computer provides another way of looking at differentiability. If f is differentiable at a, then when we zoom in toward the point a, f a the graph straightens out and appears more and more like a line. (See Figure 8. We saw a specific example of this in Figure 4 in Section 2.1.) But no matter how much we zoom in toward a point like the ones in Figures 6 and 7(a), we can’t eliminate the sharp point or corner (see Figure 9). y
0
y
a
x
0
a
FIGURE 8
FIGURE 9
ƒ is differentiable at a.
ƒ is not differentiable at a.
x
90
■
CHAPTER 2
DERIVATIVES
HIGHER DERIVATIVES
If f is a differentiable function, then its derivative f is also a function, so f may have a derivative of its own, denoted by f f . This new function f is called the second derivative of f because it is the derivative of the derivative of f . Using Leibniz notation, we write the second derivative of y f x as d dx
dy dx
d 2y dx 2
EXAMPLE 6 If f x x 3 x, find and interpret f x. SOLUTION In Example 2 we found that the first derivative is f x 3x 2 1. So
the second derivative is f x lim
2 f·
_1.5
h l0
fª
f
lim
3x h2 1 3x 2 1 h
lim
3x 2 6xh 3h 2 1 3x 2 1 h
h l0
1.5
f x h f x h
h l0
_2
lim 6x 3h 6x
FIGURE 10
In Module 2.2 you can see how changing the coefficients of a polynomial f affects the appearance of the graphs of f , f , and f .
h l0
The graphs of f , f , f are shown in Figure 10. We can interpret f x as the slope of the curve y f x at the point x, f x. In other words, it is the rate of change of the slope of the original curve y f x. Notice from Figure 10 that f x is negative when y f x has negative slope and positive when y f x has positive slope. So the graphs serve as a check on our calculations. ■ In general, we can interpret a second derivative as a rate of change of a rate of change. The most familiar example of this is acceleration, which we define as follows. If s st is the position function of an object that moves in a straight line, we know that its first derivative represents the velocity v t of the object as a function of time: v t st
ds dt
The instantaneous rate of change of velocity with respect to time is called the acceleration at of the object. Thus the acceleration function is the derivative of the velocity function and is therefore the second derivative of the position function: at vt st or, in Leibniz notation, a
dv d 2s 2 dt dt
SECTION 2.2
■
THE DERIVATIVE AS A FUNCTION
91
The third derivative f is the derivative of the second derivative: f f . So f x can be interpreted as the slope of the curve y f x or as the rate of change of f x. If y f x, then alternative notations for the third derivative are y f x
d dx
d2y dx 2
d 3y dx 3
The process can be continued. The fourth derivative f is usually denoted by f 4. In general, the nth derivative of f is denoted by f n and is obtained from f by differentiating n times. If y f x, we write dny dx n
y n f nx
EXAMPLE 7 If f x x 3 x, find f x and f 4x. SOLUTION In Example 6 we found that f x 6x . The graph of the second derivative has equation y 6x and so it is a straight line with slope 6. Since the derivative f x is the slope of f x, we have
f x 6 for all values of x. So f is a constant function and its graph is a horizontal line. Therefore, for all values of x, f 4x 0
■
We have seen that one application of second derivatives occurs in analyzing the motion of objects using acceleration. We will investigate another application of second derivatives in Section 4.3, where we show how knowledge of f gives us information about the shape of the graph of f . In Section 8.7 we will see how second and higher derivatives enable us to represent functions as sums of infinite series.
2.2
EXERCISES 2. (a) f 0
■ Use the given graph to estimate the value of each derivative. Then sketch the graph of f .
1–2
1. (a) f 3
(d) f 0 (g) f 3
(b) f 2 (e) f 1
(b) f 1 (d) f 3 (f ) f 5
(c) f 2 (e) f 4
(c) f 1 (f ) f 2
y y
y=f(x)
y=f(x) 1 1 0
1
x 0 ■
■
■
■
■
x
1 ■
■
■
■
■
■
■
92
■
CHAPTER 2
DERIVATIVES
3. Match the graph of each function in (a)–(d) with the graph
of its derivative in I–IV. Give reasons for your choices. y
(a)
0
x
0
x ■
y
(c)
x
y
I
y
(d)
0
y
11.
y
(b)
0
y
10.
0
■
■
■
■
■
■
■
■
x
■
■
■
12. Shown is the graph of the population function Pt for yeast
cells in a laboratory culture. Use the method of Example 1 to graph the derivative Pt. What does the graph of P tell us about the yeast population?
x
P (yeast cells)
y
II
0
x
500 0
x
y
III
0
0
y
IV
0
x
x
0
5
10
15
t (hours)
13. The graph shows how the average age of first marriage of
x
Japanese men has varied in the last half of the 20th century. Sketch the graph of the derivative function Mt. During which years was the derivative negative?
■ Trace or copy the graph of the given function f . (Assume that the axes have equal scales.) Then use the method of Example 1 to sketch the graph of f below it.
4 –11
4.
y
5.
M
y 27
25 0
0
x
x 1960
6.
y
7.
1970
1980
1990
2000 t
y
14. Make a careful sketch of the graph of the sine function and 0 0
8.
x
y
0
x
below it sketch the graph of its derivative in the same manner as in Exercises 4–11. Can you guess what the derivative of the sine function is from its graph?
9.
x
2 ; 15. Let f x x .
y
0
x
(a) Estimate the values of f 0, f ( 12 ), f 1, and f 2 by using a graphing device to zoom in on the graph of f. (b) Use symmetry to deduce the values of f ( 12 ), f 1, and f 2. (c) Use the results from parts (a) and (b) to guess a formula for f x.
SECTION 2.2
(d) Use the definition of a derivative to prove that your guess in part (c) is correct.
29.
_2
30.
17. f x 2 x 1
18. f x 1.5x 2 x 3.7
1 3
19. f x x 3 3x 5
20. f x x sx
21. tx s1 2x
22. f x
■
■
■
■
■
■
■
3x 1 3x
■
4 x
■
■
0
■
x
2
■
■
■
■
■
■
■
■
; 31. Graph the function f x x s x . Zoom in repeatedly,
first toward the point (1, 0) and then toward the origin. What is different about the behavior of f in the vicinity of these two points? What do you conclude about the differentiability of f ?
; 32. Zoom in toward the points (1, 0), (0, 1), and (1, 0) on
the graph of the function tx x 2 123. What do you notice? Account for what you see in terms of the differentiability of t.
4t 23. Gt t1 ■
■
0
y
_2
17–23 ■ Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative.
93
y
3 ; 16. Let f x x .
(a) Estimate the values of f 0, f ( 12 ), f 1, f 2, and f 3 by using a graphing device to zoom in on the graph of f. (b) Use symmetry to deduce the values of f ( 12 ), f 1, f 2, and f 3. (c) Use the values from parts (a) and (b) to graph f . (d) Guess a formula for f x. (e) Use the definition of a derivative to prove that your guess in part (d) is correct.
■
THE DERIVATIVE AS A FUNCTION
■
■
■
33. The figure shows the graphs of f , f , and f . Identify each
curve, and explain your choices. 24. (a) Sketch the graph of f x s6 x by starting with the
graph of y sx and using the transformations of Section 1.2. (b) Use the graph from part (a) to sketch the graph of f . (c) Use the definition of a derivative to find f x. What are the domains of f and f ? (d) Use a graphing device to graph f and compare with your sketch in part (b).
;
y
a b x
c
25. (a) If f x x 4 2x, find f x.
;
;
(b) Check to see that your answer to part (a) is reasonable by comparing the graphs of f and f . 34. The figure shows graphs of f, f , f , and f . Identify each
26. (a) If f t t 2 st , find f t.
curve, and explain your choices.
(b) Check to see that your answer to part (a) is reasonable by comparing the graphs of f and f .
y
a b c d
27–30 ■ The graph of f is given. State, with reasons, the numbers at which f is not differentiable. 27.
28.
y
y x 0
_2
0
2
x
2
4
x
94
■
CHAPTER 2
DERIVATIVES 3 x has a vertical tangent line at 0, 0. (c) Show that y s (Recall the shape of the graph of f . See Figure 8 in Section 1.2.)
35. The figure shows the graphs of three functions. One is the
position function of a car, one is the velocity of the car, and one is its acceleration. Identify each curve, and explain your choices. y
40. (a) If tx x 23, show that t0 does not exist.
a b
;
c
(b) If a 0, find ta. (c) Show that y x 23 has a vertical tangent line at 0, 0. (d) Illustrate part (c) by graphing y x 23.
41. Show that the function f x x 6 is not differentiable
at 6. Find a formula for f and sketch its graph.
t
0
42. Where is the greatest integer function f x x not differ-
entiable? Find a formula for f and sketch its graph.
43. Recall that a function f is called even if f x f x for
all x in its domain and odd if f x f x for all such x. Prove each of the following. (a) The derivative of an even function is an odd function. (b) The derivative of an odd function is an even function.
Use the definition of a derivative to find f x and f x. Then graph f , f , and f on a common screen and check to see if your answers are reasonable.
; 36 –37
■
36. f x 1x ■
■
■
44. When you turn on a hot-water faucet, the temperature T of
37. f x 1 4x x 2 ■
■
■
■
■
■
■
■
2 3 4 ; 38. If f x 2x x , find f x, f x, f x, and f x.
■
Graph f , f , f , and f on a common screen. Are the graphs consistent with the geometric interpretations of these derivatives?
3 39. Let f x s x.
(a) If a 0, use Equation 2.1.5 to find f a. (b) Show that f 0 does not exist.
2.3
y=c slope=0
0
FIGURE 1
The graph of ƒ=c is the line y=c, so fª(x)=0.
45. Let ᐍ be the tangent line to the parabola y x 2 at the point
1, 1. The angle of inclination of ᐍ is the angle that ᐍ makes with the positive direction of the x-axis. Calculate correct to the nearest degree.
BASIC DIFFERENTIATION FORMULAS
y c
the water depends on how long the water has been running. (a) Sketch a possible graph of T as a function of the time t that has elapsed since the faucet was turned on. (b) Describe how the rate of change of T with respect to t varies as t increases. (c) Sketch a graph of the derivative of T.
x
If it were always necessary to compute derivatives directly from the definition, as we did in the preceding section, such computations would be tedious and the evaluation of some limits would require ingenuity. Fortunately, several rules have been developed for finding derivatives without having to use the definition directly. These formulas greatly simplify the task of differentiation. In this section we learn how to differentiate constant functions, power functions, polynomials, and the sine and cosine functions. Then we use this knowledge to compute rates of change. Let’s start with the simplest of all functions, the constant function f x c. The graph of this function is the horizontal line y c, which has slope 0, so we must have f x 0. (See Figure 1.) A formal proof, from the definition of a derivative, is also easy: f x h f x cc f x lim lim hl0 hl0 h h lim 0 0 hl0
SECTION 2.3
BASIC DIFFERENTIATION FORMULAS
■
95
In Leibniz notation, we write this rule as follows. DERIVATIVE OF A CONSTANT FUNCTION
d c 0 dx
POWER FUNCTIONS
We next look at the functions f x x n, where n is a positive integer. If n 1, the graph of f x x is the line y x, which has slope 1. (See Figure 2.) So
y
y=x slope=1
d x 1 dx
1 0 x
FIGURE 2
The graph of ƒ=x is the line y=x, so fª(x)=1.
(You can also verify Equation 1 from the definition of a derivative.) We have already investigated the cases n 2 and n 3. In fact, in Section 2.2 (Exercises 15 and 16) we found that 2
d x 2 2x dx
d x 3 3x 2 dx
For n 4 we find the derivative of f x x 4 as follows: f x lim
f x h f x x h4 x 4 lim hl0 h h
lim
x 4 4x 3h 6x 2h 2 4xh 3 h 4 x 4 h
lim
4x 3h 6x 2h 2 4xh 3 h 4 h
hl0
hl0
hl0
lim 4x 3 6x 2h 4xh 2 h 3 4x 3 hl0
Thus 3
d x 4 4x 3 dx
Comparing the equations in (1), (2), and (3), we see a pattern emerging. It seems to be a reasonable guess that, when n is a positive integer, ddxx n nx n1. This turns out to be true. THE POWER RULE If n is a positive integer, then
d x n nx n1 dx
96
■
CHAPTER 2
DERIVATIVES
PROOF If f x x n, then
f x lim
hl0
The Binomial Theorem is given on Reference Page 1. ■
f x h f x x hn x n lim hl0 h h
In finding the derivative of x 4 we had to expand x h4. Here we need to expand x hn and we use the Binomial Theorem to do so:
x n nx n1h
f x lim
hl0
nx n1h lim
hl0
lim nx n1 hl0
nn 1 n2 2 x h nxh n1 h n x n 2 h
nn 1 n2 2 x h nxh n1 h n 2 h
nn 1 n2 x h nxh n2 h n1 2
nx n1 because every term except the first has h as a factor and therefore approaches 0.
■
We illustrate the Power Rule using various notations in Example 1. EXAMPLE 1
(a) If f x x 6, then f x 6x 5. (c) If y t 4, then
(b) If y x 1000, then y 1000x 999.
dy 4t 3. dt
(d)
d 3 r 3r 2 dr
■
What about power functions with negative integer exponents? In Exercise 55 we ask you to verify from the definition of a derivative that d dx
1 x
1 x2
We can rewrite this equation as d x 1 1x 2 dx and so the Power Rule is true when n 1. In fact, we will show in the next section [Exercise 55(c)] that it holds for all negative integers. What if the exponent is a fraction? In Example 3 in Section 2.2 we found that d 1 sx dx 2sx which can be written as d 12 x 12 x12 dx
SECTION 2.3
BASIC DIFFERENTIATION FORMULAS
■
97
This shows that the Power Rule is true even when n 12 . In fact, we will show in Section 3.3 that it is true for all real numbers n.
THE POWER RULE (GENERAL VERSION) If n is any real number, then
d x n nx n1 dx
EXAMPLE 2 Differentiate:
(a) f x ■ Figure 3 shows the function y in Example 2(b) and its derivative y. Notice that y is not differentiable at 0 (y is not defined there). Observe that y is positive when y increases and is negative when y decreases.
1 x2
SOLUTION In each case we rewrite the function as a power of x.
(a) Since f x x2, we use the Power Rule with n 2: f x
2
(b)
y yª _3
3
_2
FIGURE 3
y=#œ≈ „
3 (b) y s x2
d 2 x 2 2x 21 2x 3 3 dx x
dy d 3 2 d ( x 23 23 x 231 23 x13 sx ) dx dx dx
■
The Power Rule enables us to find tangent lines without having to resort to the definition of a derivative. It also enables us to find normal lines. The normal line to a curve C at a point P is the line through P that is perpendicular to the tangent line at P. (In the study of optics, one needs to consider the angle between a light ray and the normal line to a lens.) Find equations of the tangent line and normal line to the curve y xsx at the point 1, 1. Illustrate by graphing the curve and these lines. V EXAMPLE 3
SOLUTION The derivative of f x xsx xx 12 x 32 is
f x 32 x 321 32 x 12 32 sx
3
So the slope of the tangent line at (1, 1) is f 1 32 . Therefore, an equation of the tangent line is
tangent
y 1 32 x 1
or
y 32 x 12
normal _1
3
_1
FIGURE 4
The normal line is perpendicular to the tangent line, so its slope is the negative 3 reciprocal of 2, that is, 23. Thus an equation of the normal line is y 1 23 x 1
or
y 23 x 53
We graph the curve and its tangent line and normal line in Figure 4.
■
NEW DERIVATIVES FROM OLD
When new functions are formed from old functions by addition, subtraction, or multiplication by a constant, their derivatives can be calculated in terms of derivatives of
98
■
CHAPTER 2
DERIVATIVES
the old functions. In particular, the following formula says that the derivative of a constant times a function is the constant times the derivative of the function. THE CONSTANT MULTIPLE RULE If c is a constant and f is a differentiable
function, then d d cf x c f x dx dx
GEOMETRIC INTERPRETATION OF THE CONSTANT MULTIPLE RULE ■
PROOF Let tx cf x. Then
tx lim
y
hl0
tx h tx cf x h cf x lim hl0 h h
y=2ƒ
lim c hl0
y=ƒ 0
c lim
hl0
x
Multiplying by c 2 stretches the graph vertically by a factor of 2. All the rises have been doubled but the runs stay the same. So the slopes are doubled, too.
f x h f x h
f x h f x h
(by Law 3 of limits)
cf x
■
EXAMPLE 4
d d 3x 4 3 x 4 34x 3 12x 3 dx dx d d d (b) x 1x 1 x 11 1 dx dx dx (a)
■
The next rule tells us that the derivative of a sum of functions is the sum of the derivatives. THE SUM RULE If f and t are both differentiable, then Using prime notation, we can write the Sum Rule as f t f t ■
d d d f x tx f x tx dx dx dx PROOF Let Fx f x tx. Then
Fx lim
hl0
lim
hl0
lim
hl0
lim
hl0
Fx h Fx h f x h tx h f x tx h
f x h f x tx h tx h h
f x h f x tx h tx lim hl0 h h
f x tx
(by Law 1)
■
SECTION 2.3
BASIC DIFFERENTIATION FORMULAS
■
99
The Sum Rule can be extended to the sum of any number of functions. For instance, using this theorem twice, we get f t h f t h f t h f t h By writing f t as f 1t and applying the Sum Rule and the Constant Multiple Rule, we get the following formula. THE DIFFERENCE RULE If f and t are both differentiable, then
d d d f x tx f x tx dx dx dx
The Constant Multiple Rule, the Sum Rule, and the Difference Rule can be combined with the Power Rule to differentiate any polynomial, as the following examples demonstrate. EXAMPLE 5
d x 8 12x 5 4x 4 10x 3 6x 5 dx d d d d d d x 8 12 x 5 4 x 4 10 x 3 6 x 5 dx dx dx dx dx dx 8x 7 125x 4 44x 3 103x 2 61 0 8x 7 60x 4 16x 3 30x 2 6
■
Find the points on the curve y x 4 6x 2 4 where the tangent line is horizontal. V EXAMPLE 6
SOLUTION Horizontal tangents occur where the derivative is zero. We have
dy d d d x 4 6 x 2 4 dx dx dx dx 4x 3 12x 0 4xx 2 3 Thus dydx 0 if x 0 or x 2 3 0, that is, x s3 . So the given curve has horizontal tangents when x 0, s3 , and s3 . The corresponding points are 0, 4, (s3 , 5), and (s3 , 5). (See Figure 5.) y (0, 4)
0
x
FIGURE 5
The curve y=x$-6x@+4 and its horizontal tangents
{_ œ„ 3, _5}
3, _5} {œ„
■
100
■
CHAPTER 2
DERIVATIVES
THE SINE AND COSINE FUNCTIONS
If we sketch the graph of the function f x sin x and use the interpretation of f x as the slope of the tangent to the sine curve in order to sketch the graph of f (see Exercise 14 in Section 2.2), then it looks as if the graph of f may be the same as the cosine curve (see Figure 6). ƒ=sin x
0
π 2
π
2π
x
π 2
π
2π
x
Visual 2.3 shows an animation of Figure 6. fª(x)
0
FIGURE 6
To prove that this is true we need to use two limits from Section 1.4 (see Equation 5 and Example 11 in that section): lim
l0
sin 1
lim
l0
cos 1 0
d sin x cos x dx
4
PROOF If f x sin x, then
f x lim
hl0
We have used the addition formula for sine. See Appendix A. ■
lim
hl0
lim
hl0
f x h f x sinx h sin x lim hl0 h h sin x cos h cos x sin h sin x h
lim sin x hl0
Note that we regard x as a constant when computing a limit as h l 0 , so sin x and cos x are also constants. ■
cos h 1 h
lim sin x lim hl0
sin x cos h sin x cos x sin h h h
hl0
cos x
sin h h
cos h 1 sin h lim cos x lim hl0 hl0 h h
sin x 0 cos x 1 cos x
■
SECTION 2.3
BASIC DIFFERENTIATION FORMULAS
■
101
Using the same methods as in the proof of Formula 4, one can prove (see Exercise 56) that d cos x sin x dx
5
EXAMPLE 7 Differentiate y 3 sin 4 cos . SOLUTION
dy d d 3 sin 4 cos 3 cos 4 sin d d d
■
EXAMPLE 8 Find the 27th derivative of cos x. SOLUTION The first few derivatives of f x cos x are as follows:
f x sin x f x cos x f x sin x f 4x cos x f 5x sin x Looking for a pattern, we see that the successive derivatives occur in a cycle of length 4 and, in particular, f nx cos x whenever n is a multiple of 4. Therefore f 24x cos x and, differentiating three more times, we have f 27x sin x
■
APPLICATIONS TO RATES OF CHANGE
We discussed velocity and other rates of change in Section 2.1, but now that we know some differentiation formulas we can solve problems involving rates of change more easily. V EXAMPLE 9
The position of a particle is given by the equation s f t t 3 6t 2 9t
where t is measured in seconds and s in meters. (a) Find the velocity at time t. (b) What is the velocity after 2 s? After 4 s? (c) When is the particle at rest? (d) When is the particle moving forward (that is, in the positive direction)? (e) Draw a diagram to represent the motion of the particle. (f ) Find the total distance traveled by the particle during the first five seconds. (g) Find the acceleration at time t and after 4 s.
102
■
CHAPTER 2
DERIVATIVES
(h) Graph the position, velocity, and acceleration functions for 0 t 5. (i) When is the particle speeding up? When is it slowing down? SOLUTION
(a) The velocity function is the derivative of the position function. s f t t 3 6t 2 9t vt
ds 3t 2 12t 9 dt
(b) The velocity after 2 s means the instantaneous velocity when t 2, that is, v2
ds dt
t2
322 122 9 3 ms
The velocity after 4 s is v4 342 124 9 9 ms
(c) The particle is at rest when vt 0, that is, 3t 2 12t 9 3t 2 4t 3 3t 1t 3 0 and this is true when t 1 or t 3. Thus the particle is at rest after 1 s and after 3 s. (d) The particle moves in the positive direction when vt 0, that is, 3t 2 12t 9 3t 1t 3 0
t=3 s=0
t=0 s=0 FIGURE 7
t=1 s=4
s
This inequality is true when both factors are positive t 3 or when both factors are negative t 1. Thus the particle moves in the positive direction in the time intervals t 1 and t 3. It moves backward (in the negative direction) when 1 t 3. (e) Using the information from part (d) we make a schematic sketch in Figure 7 of the motion of the particle back and forth along a line (the s-axis). (f ) Because of what we learned in parts (d) and (e), we need to calculate the distances traveled during the time intervals [0, 1], [1, 3], and [3, 5] separately. The distance traveled in the first second is
f 1 f 0 4 0 4 m From t 1 to t 3 the distance traveled is
f 3 f 1 0 4 4 m From t 3 to t 5 the distance traveled is
f 5 f 3 20 0 20 m The total distance is 4 4 20 28 m. (g) The acceleration is the derivative of the velocity function: at
d 2s dv 6t 12 dt 2 dt
a4 64 12 12 ms 2
SECTION 2.3
25
√
a s
0
5
-12
BASIC DIFFERENTIATION FORMULAS
■
103
(h) Figure 8 shows the graphs of s, v, and a. (i) The particle speeds up when the velocity is positive and increasing (v and a are both positive) and also when the velocity is negative and decreasing (v and a are both negative). In other words, the particle speeds up when the velocity and acceleration have the same sign. (The particle is pushed in the same direction it is moving.) From Figure 8 we see that this happens when 1 t 2 and when t 3. The particle slows down when v and a have opposite signs, that is, when 0 t 1 and when 2 t 3. Figure 9 summarizes the motion of the particle.
FIGURE 8
a
√
In Module 2.3 you can see an animation of Figure 9 with an expression for s that you can choose yourself.
s
5 0 _5
forward
FIGURE 9
t
1
slows down
backward speeds up
slows down
forward speeds up
■
V EXAMPLE 10 Suppose Cx is the total cost that a company incurs in producing x units of a certain commodity. The function C is called a cost function. If the number of items produced is increased from x 1 to x 2 , then the additional cost is C Cx 2 Cx 1 , and the average rate of change of the cost is
C Cx 2 Cx 1 Cx 1 x Cx 1 x x2 x1 x The limit of this quantity as x l 0, that is, the instantaneous rate of change of cost with respect to the number of items produced, is called the marginal cost by economists: marginal cost lim
x l 0
C dC x dx
[Since x often takes on only integer values, it may not make literal sense to let x approach 0, but we can always replace Cx by a smooth approximating function.] Taking x 1 and n large (so that x is small compared to n), we have Cn Cn 1 Cn Thus the marginal cost of producing n units is approximately equal to the cost of producing one more unit [the n 1st unit]. It is often appropriate to represent a total cost function by a polynomial Cx a bx cx 2 dx 3
104
■
CHAPTER 2
DERIVATIVES
where a represents the overhead cost (rent, heat, maintenance) and the other terms represent the cost of raw materials, labor, and so on. (The cost of raw materials may be proportional to x, but labor costs might depend partly on higher powers of x because of overtime costs and inefficiencies involved in large-scale operations.) For instance, suppose a company has estimated that the cost (in dollars) of producing x items is Cx 10,000 5x 0.01x 2 Then the marginal cost function is Cx 5 0.02x The marginal cost at the production level of 500 items is C500 5 0.02500 $15item This gives the rate at which costs are increasing with respect to the production level when x 500 and predicts the cost of the 501st item. The actual cost of producing the 501st item is C501 C500 10,000 5501 0.015012
10,000 5500 0.015002
$15.01 Notice that C500 C501 C500.
2.3 1–24
■
EXERCISES
Differentiate the function. 2. f x s30
3. f x 5x 1
4. Fx 4x 10
5. f x x 3 4x 6
6. f t 2 t 6 3t 4 t
7. f x x 3 sin x
8. y sin t cos t
25
( 12 x) 5
17. y 4
x 4x 3 19. y sx 21. v t 2 23. z ■
1 4 3 t s
22. y
■
■
■
■
■
■
■
■
■
■
■
■
28. y 3x 2 x 3,
1, 2
■
■
■
■
■
■
33. Find
■
■
■
1, 2
■
■
■
■
3 30. G r sr s r
■
■
■
32. ht st 5 sin t ■
■
■
■
■
■
d 99 sin x. dx 99
34. Find the nth derivative of each function by calculating the ■
■
Find the first and second derivatives of the function.
31. tt 2 cos t 3 sin t
sin c 2
■
■
29. f x x 4 3x 3 16x
3 2 24. u s t 2 st 3 ■
■
29–32
x 2 2 sx 20. y x
A B cos y y 10 ■
■
18. tu s2 u s3u
2
■
27. y x sx ,
16. y sx x 1
2
■
1, 9
■ Find an equation of the tangent line to the curve at the given point. Illustrate by graphing the curve and the tangent line on the same screen.
s10 14. Rx x7
13. Vr r 3
■
26. y 1 2x2,
; 27–28
12. Rt 5t 35
4 3
15. F x
■
10. hx x 22x 3
1
3, 3
25. y 6 cos x ,
1
9. f t 4 t 4 8
■
■ Find equations of the tangent line and normal line to the curve at the given point.
25–26
1. f x 186.5
11. y x
■
first few derivatives and observing the pattern that occurs. (a) f x x n (b) f x 1x
■
SECTION 2.3
35. For what values of x does the graph of f x x 2 sin x
BASIC DIFFERENTIATION FORMULAS
■
105
47. If a stone is thrown vertically upward from the surface of
the moon with a velocity of 10 ms, its height (in meters) after t seconds is h 10t 0.83t 2. (a) What is the velocity of the stone after 3 s? (b) What is the velocity of the stone after it has risen 25 m?
have a horizontal tangent? 36. For what values of x does the graph of
f x x 3 3x 2 x 3 have a horizontal tangent? 37. Show that the curve y 6x 3 5x 3 has no tangent line
48. If a ball is thrown vertically upward with a velocity of
with slope 4.
80 fts, then its height after t seconds is s 80t 16t 2. (a) What is the maximum height reached by the ball? (b) What is the velocity of the ball when it is 96 ft above the ground on its way up? On its way down?
38. Find an equation of the tangent line to the curve y x sx
that is parallel to the line y 1 3x.
39. Find an equation of the normal line to the parabola
y x 2 5x 4 that is parallel to the line x 3y 5.
49. Suppose that the cost (in dollars) for a company to produce
x pairs of a new line of jeans is
40. Where does the normal line to the parabola y x x 2
Cx 2000 3x 0.01x 2 0.0002x 3
at the point (1, 0) intersect the parabola a second time? Illustrate with a sketch. 41. The equation of motion of a particle is s t 3 3t, where s
is in meters and t is in seconds. Find (a) the velocity and acceleration as functions of t, (b) the acceleration after 2 s, and (c) the acceleration when the velocity is 0.
(a) Find the marginal cost function. (b) Find C100 and explain its meaning. What does it predict? (c) Compare C100 with the cost of manufacturing the 101st pair of jeans. 50. The cost function for a certain commodity is
42. The equation of motion of a particle is
s 2t 3 7t 2 4t 1, where s is in meters and t is in seconds. (a) Find the velocity and acceleration as functions of t. (b) Find the acceleration after 1 s. (c) Graph the position, velocity, and acceleration functions on the same screen.
;
Cx 84 0.16x 0.0006x 2 0.000003x 3 (a) Find and interpret C100. (b) Compare C100 with the cost of producing the 101st item. 51. A spherical balloon is being inflated. Find the rate of
A particle moves according to a law of motion s f t, t 0, where t is measured in seconds and s in feet. (a) Find the velocity at time t. (b) What is the velocity after 3 s? (c) When is the particle at rest? (d) When is the particle moving in the positive direction? (e) Find the total distance traveled during the first 8 s. (f ) Draw a diagram like Figure 7 to illustrate the motion of the particle. 43– 44
■
43. f t t 3 12t 2 36t 44. f t t 3 9t 2 15t 10 ■
■
■
■
■
■
■
■
■
■
■
■
45. The position function of a particle is given by
s t 3 4.5t 2 7t, t 0. (a) When does the particle reach a velocity of 5 ms? (b) When is the acceleration 0? What is the significance of this value of t ? 46. If a ball is given a push so that it has an initial velocity of
5 ms down a certain inclined plane, then the distance it has rolled after t seconds is s 5t 3t 2. (a) Find the velocity after 2 s. (b) How long does it take for the velocity to reach 35 ms?
increase of the surface area S 4 r 2 with respect to the radius r when r is (a) 1 ft, (b) 2 ft, and (c) 3 ft. What conclusion can you make? 52. If a tank holds 5000 gallons of water, which drains from the
bottom of the tank in 40 minutes, then Torricelli’s Law gives the volume V of water remaining in the tank after t minutes as V 5000 (1 401 t )
2
0 t 40
Find the rate at which water is draining from the tank after (a) 5 min, (b) 10 min, (c) 20 min, and (d) 40 min. At what time is the water flowing out the fastest? The slowest? Summarize your findings. 53. Boyle’s Law states that when a sample of gas is compressed
at a constant temperature, the product of the pressure and the volume remains constant: PV C. (a) Find the rate of change of volume with respect to pressure. (b) A sample of gas is in a container at low pressure and is steadily compressed at constant temperature for 10 minutes. Is the volume decreasing more rapidly at the beginning or the end of the 10 minutes? Explain.
106
■
CHAPTER 2
DERIVATIVES
54. Newton’s Law of Gravitation says that the magnitude F of
the force exerted by a body of mass m on a body of mass M is GmM F r2 where G is the gravitational constant and r is the distance between the bodies. (a) Find dFdr and explain its meaning. What does the minus sign indicate? (b) Suppose it is known that the Earth attracts an object with a force that decreases at the rate of 2 Nkm when r 20,000 km. How fast does this force change when r 10,000 km?
60. (a) Find equations of both lines through the point 2, 3
that are tangent to the parabola y x 2 x. (b) Show that there is no line through the point 2, 7 that is tangent to the parabola. Then draw a diagram to see why.
61. For what values of a and b is the line 2x y b tangent to
the parabola y ax 2 when x 2?
62. Find a parabola with equation y ax 2 bx c that has
slope 4 at x 1, slope 8 at x 1, and passes through the point 2, 15.
63. Find a cubic function y ax 3 bx 2 cx d whose graph
has horizontal tangents at the points 2, 6 and 2, 0.
55. Use the definition of a derivative to show that if f x 1x,
64. A tangent line is drawn to the hyperbola xy c at a point P.
56. Prove, using the definition of derivative, that if f x cos x,
(a) Show that the midpoint of the line segment cut from this tangent line by the coordinate axes is P. (b) Show that the triangle formed by the tangent line and the coordinate axes always has the same area, no matter where P is located on the hyperbola.
then f x 1x 2. (This proves the Power Rule for the case n 1.) then f x sin x.
57. The equation y y 2y sin x is called a differential
equation because it involves an unknown function y and its derivatives y and y. Find constants A and B such that the function y A sin x B cos x satisfies this equation. (Differential equations will be studied in detail in Section 7.6.) 58. Find constants A, B, and C such that the function
y Ax 2 Bx C satisfies the differential equation y y 2y x 2.
xl1
x 1000 1 . x1
66. Draw a diagram showing two perpendicular lines that inter-
sect on the y-axis and are both tangent to the parabola y x 2. Where do these lines intersect? 67. If c 2 , how many lines through the point 0, c are normal 1
59. Draw a diagram to show that there are two tangent lines to
the parabola y x 2 that pass through the point 0, 4. Find the coordinates of the points where these tangent lines intersect the parabola.
2.4
65. Evaluate lim
lines to the parabola y x 2 ? What if c 12 ?
68. Sketch the parabolas y x 2 and y x 2 2x 2. Do you
think there is a line that is tangent to both curves? If so, find its equation. If not, why not?
THE PRODUCT AND QUOTIENT RULES The formulas of this section enable us to differentiate new functions formed from old functions by multiplication or division. THE PRODUCT RULE
| By analogy with the Sum and Difference Rules, one might be tempted to guess, as Leibniz did three centuries ago, that the derivative of a product is the product of the derivatives. We can see, however, that this guess is wrong by looking at a particular example. Let f x x and tx x 2. Then the Power Rule gives f x 1 and tx 2x. But ftx x 3, so ftx 3x 2. Thus ft f t. The correct formula was discovered by Leibniz (soon after his false start) and is called the Product Rule.
We can write the Product Rule in prime notation as ft ft t f ■
THE PRODUCT RULE If f and t are both differentiable, then
d d d f xtx f x tx tx f x dx dx dx
SECTION 2.4
THE PRODUCT AND QUOTIENT RULES
■
107
PROOF Let Fx f xtx. Then
Fx lim
hl0
lim
hl0
Fx h Fx h f x htx h f xtx h
In order to evaluate this limit, we would like to separate the functions f and t as in the proof of the Sum Rule. We can achieve this separation by subtracting and adding the term f x htx in the numerator: Fx lim
hl0
f x htx h f x htx f x htx f xtx h
lim f x h hl0
tx h tx f x h f x tx h h
lim f x h lim hl0
hl0
tx h tx f x h f x lim tx lim hl0 hl0 h h
f xtx txf x Note that lim h l 0 tx tx because tx is a constant with respect to the variable h. Also, since f is differentiable at x, it is continuous at x by Theorem 2.2.4, and so lim h l 0 f x h f x. ■ ■ Figure 1 shows the graphs of the function of Example 1 and its derivative. Notice that y 0 whenever y has a horizontal tangent.
5 yª _4
y
In words, the Product Rule says that the derivative of a product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function. V EXAMPLE 1
4
_5
FIGURE 1
Differentiate y x 2 sin x.
SOLUTION Using the Product Rule, we have
dy d d x2 sin x sin x x 2 dx dx dx x 2 cos x 2x sin x EXAMPLE 2 Differentiate the function f t st a bt.
■ In Example 2, a and b are constants. It is customary in mathematics to use letters near the beginning of the alphabet to represent constants and letters near the end of the alphabet to represent variables.
SOLUTION 1 Using the Product Rule, we have
f t st
d d a bt a bt (st ) dt dt
st b a bt 12 t 12 bst
a bt a 3bt 2st 2st
■
108
■
CHAPTER 2
DERIVATIVES
SOLUTION 2 If we first use the laws of exponents to rewrite f t, then we can proceed directly without using the Product Rule.
f t ast btst at 12 bt 32 f t 12 at12 32 bt 12 which is equivalent to the answer given in Solution 1.
■
Example 2 shows that it is sometimes easier to simplify a product of functions than to use the Product Rule. In Example 1, however, the Product Rule is the only possible method. EXAMPLE 3 If hx xtx and it is known that t3 5 and t3 2, find h3. SOLUTION Applying the Product Rule, we get
hx
d d d xtx x tx tx x dx dx dx
xtx tx h3 3t3 t3 3 2 5 11
Therefore
■
THE QUOTIENT RULE
The following rule enables us to differentiate the quotient of two differentiable functions. THE QUOTIENT RULE If f and t are differentiable, then In prime notation we can write the Quotient Rule as t f ft f t2 t ■
d dx
f x tx
tx
d d f x f x tx dx dx tx 2
PROOF Let Fx f xtx. Then
f x h f x Fx h Fx tx h tx Fx lim lim hl0 hl0 h h lim
hl0
f x htx f xtx h htx htx
We can separate f and t in this expression by subtracting and adding the term f xtx in the numerator: Fx lim
hl0
f x htx f xtx f xtx f xtx h htx htx tx
lim
hl0
f x h f x tx h tx f x h h tx htx
SECTION 2.4
lim tx lim
hl0
hl0
THE PRODUCT AND QUOTIENT RULES
109
f x h f x tx h tx lim f x lim hl0 hl0 h h lim tx h lim tx hl0
■
hl0
txf x f xtx tx 2
Again t is continuous by Theorem 2.2.4, so lim h l 0 tx h tx.
■
In words, the Quotient Rule says that the derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. The Quotient Rule and the other differentiation formulas enable us to compute the derivative of any rational function, as the next example illustrates. We can use a graphing device to check that the answer to Example 4 is plausible. Figure 2 shows the graphs of the function of Example 4 and its derivative. Notice that when y grows rapidly (near 2 ), y is large. And when y grows slowly, y is near 0 . ■
V EXAMPLE 4
Let y
Then x 3 6 y
1.5 yª _4
d d x 2 x 2 x 2 x 2 x 3 6 dx dx x 3 62
x 3 62x 1 x 2 x 23x 2 x 3 62
2x 4 x 3 12x 6 3x 4 3x 3 6x 2 x 3 62
x 4 2x 3 6x 2 12x 6 x 3 62
4 y _1.5
x2 x 2 . x3 6
FIGURE 2
EXAMPLE 5 Find an equation of the tangent line to the curve y sx1 x 2 at
the point (1, 12 ).
SOLUTION According to the Quotient Rule, we have
dy dx
1 x 2
d d (sx ) sx dx 1 x 2 dx 1 x 2 2
1 sx 2x 2sx 1 x 2 2
1 x 2
1 x 2 4x 2 1 3x 2 2sx 1 x 2 2 2sx 1 x 2 2
So the slope of the tangent line at (1, 12 ) is dy dx
x1
1 3 12 1 2 2 2s11 1 4
■
110
■
CHAPTER 2
DERIVATIVES
We use the point-slope form to write an equation of the tangent line at (1, 12 ):
1
y 12 14 x 1
”1, 21 ’ y=
0
FIGURE 3
œ„ x 1+≈
or
y 14 x 34
The curve and its tangent line are graphed in Figure 3. 4
■
NOTE Don’t use the Quotient Rule every time you see a quotient. Sometimes it’s easier to rewrite a quotient first to put it in a form that is simpler for the purpose of differentiation. For instance, although it is possible to differentiate the function
Fx
3x 2 2sx x
using the Quotient Rule, it is much easier to perform the division first and write the function as Fx 3x 2x 12 before differentiating.
TRIGONOMETRIC FUNCTIONS
Knowing the derivatives of the
