Instructor's Solutions Manual for Algebra & Trigonometry and Precalculus, 3 E

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INSTRUCTOR’S SOLUTIONS MANUAL JUDITH A. PENNA Indiana University Purdue University Indianapolis

ALGEBRA AND TRIGONOMETRY THIRD EDITION AND

PRECALCULUS THIRD EDITION

Judith A. Beecher Indiana University Purdue University Indianapolis

Judith A. Penna Indiana University Purdue University Indianapolis

Marvin L. Bittinger Indiana University Purdue University Indianapolis

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This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials.

Reproduced by Pearson Addison-Wesley from QuarkXPress® files. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. ISBN-13: 978-0-321-46481-1 ISBN-10: 0-321-46481-8 1 2 3 4 5 6 OPM 10 09 08 07

Contents Chapter R . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 39 Chapter 2

. . . . . . . . . . . . . . . . . . . . . . . . . 117

Chapter 3

. . . . . . . . . . . . . . . . . . . . . . . . . 185

Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . 273 Chapter 5

. . . . . . . . . . . . . . . . . . . . . . . . . 329

Chapter 6

. . . . . . . . . . . . . . . . . . . . . . . . . 379

Chapter 7

. . . . . . . . . . . . . . . . . . . . . . . . . 419

Chapter 8

. . . . . . . . . . . . . . . . . . . . . . . . . 471

Chapter 9

. . . . . . . . . . . . . . . . . . . . . . . . . 569

Chapter 10

. . . . . . . . . . . . . . . . . . . . . . . . . 653

Chapter R

Basic Concepts of Algebra 16. (−5, ∞)

Exercise Set R.1 √ √ √ √ 1. Whole numbers: 3 8, 0, 9, 25 ( 3 8 = 2, 25 = 5) √ √ √ √ 2. Integers: −12, 3 8, 0, 9, 25 ( 3 8 = 2, 25 = 5) √ √ √ √ 3. Irrational numbers: 7, 5.242242224 . . . , − 14, 5 5, 3 4

(Although there is a pattern in 5.242242224 . . . , there is no repeating block of digits.) √ √ 4. Natural numbers: 3 8, 9, 25 7 √ 2 5. Rational numbers: −12, 5.3, − , 3 8, 0, −1.96, 9, 4 , 3 3 √ 5 25, 7 6. Real numbers: All of them 7 2 5 7. Rational numbers but not integers: 5.3, − , −1.96, 4 , 3 3 7

!5

0

17. This interval is of unlimited extent in the positive direction, and the endpoint 3.8 is not included. Interval notation is (3.8, ∞). 0

√ 18. [ 3, ∞) [ -3

-2

√ 7 10. Real numbers but not integers: 7, 5.3, − , 3 √ √ 2 √ 5 5.242242224 . . ., − 14, 5 5, −1.96, 4 , 3 4, 3 7 11. This is a closed interval, so we use brackets. Interval notation is [−3, 3].

-1

0

1

3 2

3

19. {x|7 < x}, or {x|x > 7}.

This interval is of unlimited extent in the positive direction and the endpoint 7 is not included. Interval notation is (7, ∞).

8. Integers but not whole numbers: −12 9. Integers but not rational numbers: −12, 0

3.8

0

7

20. (−∞, −3) !3

0

21. The endpoints 0 and 5 are not included in the interval, so we use parentheses. Interval notation is (0, 5). 22. [−1, 2]

!5 !4 !3 !2 !1

0 1 2 3 4 5

12. (−4, 4) !4

0

4

13. This is a half-open interval. We use a bracket on the left and a parenthesis on the right. Interval notation is [−4, −1). !4

!1

14. (1, 6] 6

15. This interval is of unlimited extent in the negative direction, and the endpoint −2 is included. Interval notation is (−∞, −2]. !2

0

24. (−9, −5] 25. Both endpoints are included in the interval, so we use brackets. Interval notation is [x, x + h]. 26. (x, x + h]

0

0 1

23. The endpoint −9 is included in the interval, so we use a bracket before the −9. The endpoint −4 is not included, so we use a parenthesis after the −4. Interval notation is [−9, −4).

27. The endpoint p is not included in the interval, so we use a parenthesis before the p. The interval is of unlimited extent in the positive direction, so we use the infinity symbol ∞. Interval notation is (p, ∞). 28. (−∞, q] 29. Since 6 is an element of the set of natural numbers, the statement is true. 30. True

2

Chapter R: Basic Concepts of Algebra

31. Since 3.2 is not an element of the set of integers, the statement is false.

1 57. The sentence 8 · = 1 illustrates the multiplicative inverse 8 property.

32. True

58. Distributive property

11 is an element of the set of rational numbers, 33. Since − 5 the statement is true.

59. The distance of −7.1 from 0 is 7.1, so | − 7.1| = 7.1.

34. False √ 35. Since 11 is an element of the set of real numbers, the statement is false. 36. False 37. Since 24 is an element of the set of whole numbers, the statement is false. 38. True 39. Since 1.089 is not an element of the set of irrational numbers, the statement is true. 40. True 41. Since every whole number is an integer, the statement is true. 42. False 43. Since every rational number is a real number, the statement is true. 44. True 45. Since there are real numbers that are not integers, the statement is false. 46. False 47. The sentence 6 · x = x · 6 illustrates the commutative property of multiplication. 48. Associative property of addition 49. The sentence −3 · 1 = −3 illustrates the multiplicative identity property. 50. Commutative property of addition 51. The sentence 5(ab) = (5a)b illustrates the associative property of multiplication. 52. Distributive property 53. The sentence 2(a+b) = (a+b)2 illustrates the commutative property of multiplication.

60. 86.2 61. The distance of 347 from 0 is 347, so |347| = 347. 62. 54 √ √ √ √ 63. The distance of − 97 from 0 is 97, so | − 97| = 97. 64.

65. The distance of 0 from 0 is 0, so |0| = 0. 66. 15 67. The distance of 68.

56. Additive identity property

√

3

!5! 5 5 5 ! ! from 0 is , so ! ! = . 4 4 4 4

69. | − 5 − 6| = | − 11| = 11, or

|6 − (−5)| = |6 + 5| = |11| = 11

70. |0 − (−2.5)| = |2.5| = 2.5, or | − 2.5 − 0| = | − 2.5| = 2.5

71. | − 2 − (−8)| = | − 2 + 8| = |6| = 6, or

| − 8 − (−2)| = | − 8 + 2| = | − 6| = 6 ! 15 23 ! ! 45 46 ! ! 1 !! 1 ! ! ! ! ! , or 72. ! − ! = ! − ! = ! − ! = 8 12 24 24 24 24 ! 23 15 ! ! 46 45 ! ! 1 ! 1 ! ! ! ! ! ! ! − !=! − !=! != 12 8 24 24 24 24

73. |12.1 − 6.7| = |5.4| = 5.4, or |6.7 − 12.1| = | − 5.4| = 5.4

74. | − 3 − (−14)| = | − 3 + 14| = |11| = 11, or

| − 14 − (−3)| = | − 14 + 3| = | − 11| = 11 ! ! ! ! ! ! ! ! ! ! ! ! 75. ! − 3 − 15 ! = ! − 6 − 15 ! = ! − 21 ! = 21 , or ! 4 8! ! 8 8! ! 8! 8 ! ! ! ! ! ! " #! ! ! ! 15 3 ! ! 15 6 ! ! 21 ! 21 ! 15 3 !=! + !=! + !=! != ! − − !8 4 ! !8 4! ! 8 8! ! 8 ! 8

76.

| − 3.4 − 10.2| = | − 13.6| = 13.6, or

|10.2 − (−3.4)| = |10.2 + 3.4| = |13.6| = 13.6

77.

| − 7 − 0| = | − 7| = 7, or

|0 − (−7)| = |0 + 7| = |7| = 7

54. Additive inverse property 55. The sentence −6(m + n) = −6(n + m) illustrates the commutative property of addition.

12 19

78.

|3 − 19| = | − 16| = 16, or |19 − 3| = |16| = 16

79. Provide an example. For instance, 16÷(8÷2) = 16÷4 = 4, but (16 ÷ 8) ÷ 2 = 2 ÷ 2 = 1.

Exercise Set R.2 80.

3

√

a is a rational number √ when a is the square of a rational number. That is, a is a rational number if there is a rational number c such that a = c2 .

81. Answers may vary. One such number is 0.124124412444 . . . . √ √ 82. Answers may vary. Since − 2.01 ≈ −1.418 and − 2 ≈ −1.414, one such number is −1.415. 1 = 0.0099 and 83. Answers may vary. Since − 101 1 − = −0.01, one such number is −0.00999. 100 √ 84. Answers may vary. One such number is 5.995. with 85. Since 12 + 32 = 10, the hypotenuse of a right triangle √ legs of lengths 1 unit and 3 units has a length of 10 units. ! c!!! 1 ! ! !! 3

1. 3 2.

1 = 7 3

"

−m

a

# 1 = m , a %= 0 a

4. Observe that each exponent is negative. We move each factor to the other side of the fraction bar and change the sign of each exponent. b8 a−2 = b−8 a2 5. Observe that each exponent is negative. We move each factor to the other side of the fraction bar and change the sign of each exponent. t6 t6 m−1 n−12 = 1 12 , or −6 t m n mn12 6. Observe that each exponent is negative. We move each factor to the other side of the fraction bar and change the sign of each exponent. z 11 x−9 y −17 = 9 17 −11 z x y

−

4 3

#0

(For any nonzero real number, a0 = 1.) =1

1 65

13. m−5 · m5 = m−5+5 = m0 = 1 14. n9 · n−9 = n9+(−9) = n0 = 1 15. y 3 · y −7 = y 3+(−7) = y −4 , or

1 y4

16. b−4 · b12 = b−4+12 = b8 17. 73 · 7−5 · 7 = 73+(−5)+1 = 7−1 , or

19. 2x3 · 3x2 = 2 · 3 · x3+2 = 6x5

y4 x−5 = 5 −4 y x

"

12. 62 · 6−7 = 62+(−7) = 6−5 , or

c2 = 10 √ c = 10

3. Observe that each exponent is negative. We move each factor to the other side of the fraction bar and change the sign of each exponent.

8.

11. 58 · 5−6 = 58+(−6) = 52 , or 25

18. 36 · 3−5 · 34 = 36+(−5)+4 = 35

1 = (5.9)4 (5.9)−4

7. 18◦ = 1

10. a0 · a4 = a0+4 = a4

c2 = 12 + 32

Exercise Set R.2 −7

9. x9 · x0 = x9+0 = x9

1 7

20. 3y 4 · 4y 3 = 3 · 4 · y 4+3 = 12y 7 21. (−3a−5 )(5a−7 ) = −3 · 5 · a−5+(−7) = −15a−12 , or 15 − 12 a 22. (−6b−4 )(2b−7 ) = −6 · 2 · b−4+(−7) = −12b−11 , or 12 − 11 b 23. (5a2 b)(3a−3 b4 ) = 5 · 3 · a2+(−3) · b1+4 = 15a−1 b5 , or 15b5 a

24. (4xy 2 )(3x−4 y 5 ) = 4 · 3 · x1+(−4) · y 2+5 = 12x−3 y 7 , or 12y 7 x3

25. (6x−3 y 5 )(−7x2 y −9 ) = 6(−7)x−3+2 y 5+(−9) = 42 −42x−1 y −4 , or − 4 xy 26. (8ab7 )(−7a−5 b2 ) = 8(−7)a1+(−5) b7+2 = −56a−4 b9 , or −

56b9 a4

27. (2x)3 (3x)2 = 23 x3 · 32 x2 = 8 · 9 · x3+2 = 72x5 28. (4y)2 (3y)3 = 16y 2 · 27y 3 = 432y 5 29. (−2n)3 (5n)2 = (−2)3 n3 · 52 n2 = −8 · 25 · n3+2 = −200n5

30. (2x)5 (3x)2 = 25 x5 · 32 x2 = 32 · 9 · x5+2 = 288x7 31.

b40 = b40−37 = b3 b37

32.

a39 = a39−32 = a7 a32

4

33.

Chapter R: Basic Concepts of Algebra x−5 1 = x−5−16 = x−21 , or 21 x16 x

51. Convert 405,000 to scientific notation.

y −24 1 34. −21 = y −24−(−21) = y −24+21 = y −3 , or 3 y y x2 y −2 x3 35. −1 = x2−(−1) y −2−1 = x3 y −3 , or 3 x y y 36.

x3 y −3 x4 = x3−(−1) y −3−2 = x4 y −5 , or 5 −1 2 x y y

37.

32x−4 y 3 32 −4−(−5) 3−8 8x = y = 8xy −5 , or 5 x 4x−5 y 8 4 y

20a5 b−2 20 5−7 −2−(−3) 4b 38. = a b = 4a−2 b, or 2 5a7 b−3 5 a 39. (2ab2 )3 = 23 a3 (b2 )3 = 23 a3 b2·3 = 8a3 b6 40. (4xy 3 )2 = 42 x2 (y 3 )2 = 16x2 y 6 41. (−2x3 )5 = (−2)5 (x3 )5 = (−2)5 x3·5 = −32x15 42. (−3x2 )4 = (−3)4 (x2 )4 = 81x8

c2 d4 c2 d4 = 2 (−5) 25 44. (−4x−5 z −2 )−3 = (−4)−3 (x−5 )−3 (z −2 )−3 = 15 6

x z x z = (−4)3 −64

128n7 " −3 7 #3 2x y (2x−3 y 7 )3 23 x−9 y 21 47. = = = −1 −1 3 z (z ) z −3 8y 21 z 3 8x−9 y 21 , or z −3 x9 " 5 −8 #4 3x y 81x20 y −32 81x20 z 8 48. = , or −2 −8 z z y 32 49.

or 50.

b25 32a20 c10

25p8 q 6 r−15

5

−4 −16 80 −148

p

q r

= (5p q

r )

4 −20 37 −4

q 80 , or 625p16 r148

53. Convert 0.00000039 to scientific notation. We want the decimal point to be positioned between the 3 and the 9, so we move it 7 places to the right. Since 0.00000039 is a number between 0 and 1, the exponent must be negative. 0.00000039 = 3.9 × 10−7 54. Position the decimal point 4 places to the right, between the 9 and the 2. Since 0.00092 is a number between 0 and 1, the exponent must be negative.

55. Convert 234,600,000,000 to scientific notation. We want the decimal point to be positioned between the 2 and the 3, so we move it 11 places to the left. Since 234,600,000,000 is greater than 10, the exponent must be positive. 234, 600, 000, 000 = 2.346 × 1011

57. Convert 0.00104 to scientific notation. We want the decimal point to be positioned between the 1 and the last 0, so we move it 3 places to the right. Since 0.00104 is a number between 0 and 1, the exponent must be negative. 0.00104 = 1.04 × 10−3 58. Position the decimal point 9 places to the right, between the 5 and the 1. Since 0.00000000514 is a number between 0 and 1, the exponent must be negative. 0.00000000514 = 5.14 × 10−9

#−5 = (2a4 b−5 c2 )−5 = 2−5 a−20 b25 c−10 ,

$ 125p12 q −14 r22 %−4

1, 670, 000 = 1.67 × 106

8, 904, 000, 000 = 8.904 × 109

46. (4n−1 )2 (2n3 )3 = 42 n−2 · 23 n9 = 16 · 8 · n−2+9 =

24a10 b−8 c7 12a6 b−3 c5

52. Position the decimal point 6 places to the left, between the 1 and the 6. Since 1,670,000 is greater than 10, the exponent must be positive.

56. Position the decimal point 9 places to the left, between the 8 and the 9. Since 8,904,000,000 is greater than 10, the exponent must be positive.

45. (3m4 )3 (2m−5 )4 = 33 m12 · 24 m−20 = 432 27 · 16m12+(−20) = 432m−8 , or 8 m

"

405, 000 = 4.05 × 105

0.00092 = 9.2 × 10−4

43. (−5c−1 d−2 )−2 = (−5)−2 c−1(−2) d−2(−2) =

15 6

We want the decimal point to be positioned between the 4 and the first 0, so we move it 5 places to the left. Since 405,000 is greater than 10, the exponent must be positive.

=

59. Convert 0.000016 to scientific notation. We want the decimal point to be positioned between the 1 and the 6, so we move it 5 places to the right. Since 0.000016 is a number between 0 and 1, the exponent must be negative. 0.000016 = 1.6 × 10−5 60. Position the decimal point 12 places to the left, between the ones. Since 1,137,000,000,000 is greater than 10, the exponent must be positive. 1, 137, 000, 000, 000 = 1.137 × 1012

Exercise Set R.2

5

61. Convert 8.3 × 10−5 to decimal notation.

73.

The exponent is negative, so the number is between 0 and 1. We move the decimal point 5 places to the left. 8.3 × 10

−5

= (2.6 × 8.5) × (10−18 × 107 )

= 22.1 × 10−11

= 0.000083

62. The exponent is positive, so the number is greater than 10. We move the decimal point 6 places to the right. 4.1 × 106 = 4, 100, 000

(2.6 × 10−18 )(8.5 × 107 )

This is not scientific notation.

= (2.21 × 10) × 10

−11

= 2.21 × 10−10

74.

(6.4 × 1012 )(3.7 × 10−5 ) = 23.68 × 107

= (2.368 × 10) × 107 = 2.368 × 108

63. Convert 2.07 × 10 to decimal notation. 7

The exponent is positive, so the number is greater than 10. We move the decimal point 7 places to the right.

75.

2.07 × 107 = 20, 700, 000 64. The exponent is negative, so the number is between 0 and 1. We move the decimal point 6 places to the left.

= 8 × 10−14

3.15 × 10−6 = 0.00000315 65. Convert 3.496 × 1010 to decimal notation.

The exponent is positive, so the number is greater than 10. We move the decimal point 10 places to the right.

76.

3.496 × 1010 = 34, 960, 000, 000 66. The exponent is positive, so the number is greater than 10. We move the decimal point 11 places to the right.

77.

67. Convert 5.41 × 10−8 to decimal notation.

The exponent is negative, so the number is between 0 and 1. We move the decimal point 8 places to the left.

78.

6.27 × 10−10 = 0.000000000627 69. Convert 2.319 × 108 to decimal notation.

The exponent is positive, so the number is greater than 10. We move the decimal point 8 places to the right. 2.319 × 108 = 231, 900, 000

79. We multiply the number of AUs from Earth to Pluto by the number of miles in 1 AU. 39 × 93 million = 39 × 93, 000, 000

= (3.9 × 10) × (9.3 × 107 )

= (3.9 × 9.3) × (10 × 10 ) = 36.27 × 108

= (3.627 × 10) × 108

)

= 3.627 × 109

= (3.1 × 4.5) × (105 × 10−3 )

= 13.95 × 10

This is not scientific notation.

= (1.395 × 10) × 102 = 1.395 × 103

72.

(9.1 × 10

−17

Writing scientific notation

)(8.2 × 10 ) = 74.62 × 10 3

Writing scientific notation

7

1.67 × 10−24 g = 0.00000000000000000000000167 g

2

1.3 × 104 = 0.25 × 10−6 5.2 × 1010 = (2.5 × 10−1 ) × 10−6 = 2.5 × 10−7

70. The exponent is negative, so the number is between 0 and 1. We move the decimal point 24 places to the left.

(3.1 × 10 )(4.5 × 10

1.8 × 10−3 7.2 × 10−9 1.8 10−3 = × 7.2 10−9 = 0.25 × 106 This is not scientific notation. = 2.5 × 105

68. The exponent is negative, so the number is between 0 and 1. We move the decimal point 10 places to the left.

71.

1.1 × 10−40 = 0.55 × 1031 2.0 × 10−71 = (5.5 × 10−1 ) × 1031

= (2.5 × 10−1 ) × 106

5.41 × 10−8 = 0.0000000541

−3

Writing scientific notation

= 5.5 × 1030

8.409 × 1011 = 840, 900, 000, 000

5

6.4 × 10−7 6.4 10−7 = × 8.0 × 106 8.0 106 = 0.8 × 10−13 This is not scientific notation. = (8 × 10−1 ) × 10−13

−14

= (7.462 × 10) × 10−14 = 7.462 × 10

−13

The distance from Earth to Pluto is about 3.627 × 109 mi. 80.

3.26 × 5.88 × 1012

= 19.1688 × 1012

= (1.91688 × 10) × 1012 = 1.91688 × 1013 mi

6

Chapter R: Basic Concepts of Algebra

81. We multiply the diameter, in nanometers, by the number of meters in 1 nanometer. 360 × 0.000000001

= 3 · 2 − 4 · 22 + 6 · 2 = 3·2−4·4+6·2

= (3.6 × 102 ) × 10−9

= 3.6 × (10 × 10

−9

2

= 3.6 × 10

)

The diameter of the wire is 3.6 × 10 82.

m.

= 3[18 − 6 · 2] = 3[18 − 12] = 3[6] = 18

= 16 ÷ 2 · 256 = 2048

The average cost per mile was about $1.19 × 10 . 7

412 4.12 × 102 = 9, 600, 000 9.6 × 106 ≈ 0.43 × 10−4 ≈ (4.3 × 10

−1

≈ 4.3 × 10

−5

= 23 ÷ 210 ÷ 2−8 = 2−7 ÷ 2−8 =2

) × 10

−4

91.

square miles

=

= = =

= 37, 000, 000, 000 × 3600

Writing scientific notation = (3.7 × 3.6) × (1010 × 103 )

=

= 3.7 × 1010 × 3.6 × 103

Multiplying

= (1.332 × 10) × 1013 = 1.332 × 1014

One gram of radium produces 1.332 × 1014 disintegrations in 1 hour. 2π × 93, 000, 000 7

≈ 5.8 × 10) × 107 ≈ 5.8 × 10 mi

26 · 2−3 ÷ 210 ÷ 2−8

90.

85. First find the number of seconds in 1 hour: # × 60 sec = 3600 sec 60 min 1 hour = 1 " hr × 1" hr 1 min # The number of disintegrations produced in 1 hour is the number of disintegrations per second times the number of seconds in 1 hour. 37 billion × 3600

= 13.32 × 1013

Multiplying and dividing in order from left to right

= 8 · 256

≈ $1.19 × 107

8

16 ÷ 4 · 4 ÷ 2 · 256

= 4 · 4 ÷ 2 · 256

≈ ($1.19 × 102 ) × 105

≈ 58 × 107

3[(2 + 4 · 22 ) − 6(3 − 1)]

89.

$210 × 106 17.6 $210 × 106 = 1.76 × 10 ≈ $119 × 105

= 2π × 9.3 × 10

from left to right

Adding in order

= 3[(2 + 4 · 4) − 6 · 2]

83. The average cost per mile is the total cost divided by the number of miles.

86.

=2

= 3[(2 + 16) − 6 · 2]

On average, about 8.2192 × 104 pieces of luggage were lost each day of the year.

84.

Multiplying

88.

30 million 30 × 106 = 365 3.65 × 102 ≈ 8.2192 × 104

Working inside parentheses Evaluating 22

= 6 − 16 + 12 = −10 + 12

−7

−7

3 · 2 − 4 · 22 + 6(3 − 1)

87.

= 92. = = = = = =

4(8 − 6)2 − 4 · 3 + 2 · 8 31 + 190 4 · 22 − 4 · 3 + 2 · 8 Calculating in the 3+1 numerator and in the denominator 4·4−4·3+2·8 4 16 − 12 + 16 4 4 + 16 4 20 4 5 [4(8 − 6)2 + 4](3 − 2 · 8) 22 (23 + 5) 2 [4 · 2 + 4](3 − 16) 22 (8 + 5) [4 · 4 + 4](−13) 22 · 13 [16 + 4](−13) 4 · 13 20(−13) 52 −260 52 −5

Exercise Set R.2

7

93. Since interest is compounded semiannually, n = 2. Substitute $2125 for P , 6.2% or 0.062 for i, 2 for n, and 5 for t in the compound interest formula. $ i %nt A = P 1+ n $ 0.062 %2·5 = $2125 1 + Substituting 2 2·5 Dividing = $2125(1 + 0.031) = $2125(1.031)2·5

Adding

= $2125(1.031)10

Multiplying 2 and 5

≈ $2125(1.357021264) Evaluating the exponential expression ≈ $2883.670185 Multiplying

≈ $2883.67 Rounding to the nearest cent

$ 0.054 %2·7 ≈ $13, 867.23 94. A = $9550 1 + 2

95. Since interest is compounded quarterly, n = 4. Substitute $6700 for P , 4.5% or 0.045 for i, 4 for n, and 6 for t in the compound interest formula. $ i %nt A = P 1+ n $ 0.045 %4·6 = $6700 1 + Substituting 4 = $6700(1 + 0.01125)4·6 Dividing = $6700(1.01125)4·6 Adding = $6700(1.01125)24 Multiplying 4 and 6 ≈ $6700(1.307991226) Evaluating the exponential expression ≈ $8763.541217 Multiplying

≈ $8763.54 Rounding to the nearest cent

100.

101. Substitute $120,000 for S, 0.06 for r, and 18 for t and solve for P .  $ r %12·t −1 1+  12 S=P r 12  $ 0.06 %12·18 −1 1+ 12  $120, 000 = P  0.06 12 * + (1.005)216 − 1 $120, 000 = P 0.05 $120, 000 ≈ P (387.3532) $309.79 ≈ P

102.

4 · 25 ÷ (10 − 5) = 4 · 25 ÷ 5 = 100 ÷ 5 = 20,

104. (xy · x−y )3 = (x0 )3 = 13 = 1 105. (ta+x · tx−a )4 = (t2x )4 = t2x·4 = t8x 106.

99. Substitute $250 for P , 0.05 for r and 27 for t and perform the resulting computation.  $ r %12·t −1 1+  12 S=P r 12  $ 0.05 %12·27 1+ −1 12  = $250  0.05 12 ≈ $170, 797.30

(mx−b · nx+b )x (mb n−b )x 2

= (mx 2

−bx x2 +bx

n

)(mbx n−bx )

2

= mx n x 107.

but 4 · 25 ÷ 10 − 5 = 100 ÷ 10 − 5 = 10 − 5 = 5.

1 1 98. No; x−2 , or 2 is positive for all x < 0 and x−1 , or x x is negative for all x < 0. Partial confirmation can be obtained by inspecting the graphs of y1 = x−2 and y2 = x−1 for x < 0.

t =70 − 30 = 40  $ 0.045 %12·40 −1 1+ 12  $200, 000 =P  0.045 12 P ≈$149.13

103. (xt · x3t )2 = (x4t )2 = x4t·2 = x8t

$ 0.058 %4·9 ≈ $8185.56 96. A = $4875 1 + 4

97. Yes; find the results with parentheses and without them.

t =65 − 25 = 40 $  0.04 %12·40 1+ −1 12  ≈ $118, 196.13 S =$100  0.04 12

, (3xa y b )3 -2 , 27x3a y 3b -2 = (−3xa y b )2 9x2a y 2b . a b /2 = 3x y = 9x2a y 2b

,$ xr %2 $ x2r %−2 -−3

108. = =

yt y 4t ,$ x2r %$ x−4r %-−3 y 2t y −8t $ x−2r %−3

y −6t x6r = 18t , or x6r y −18t y

8

Chapter R: Basic Concepts of Algebra 13.

Exercise Set R.3

(a − b)(2a3 − ab + 3b2 )

= (a − b)(2a3 ) + (a − b)(−ab) + (a − b)(3b2 )

Using the distributive property

1. −5y 4 + 3y 3 + 7y 2 − y − 4 = −5y 4 + 3y 3 + 7y 2 + (−y) + (−4)

= 2a − 2a b − a2 b + ab2 + 3ab2 − 3b3 4

3

Using the distributive property three more times

Terms: −5y 4 , 3y 3 , 7y 2 , −y, −4

The degree of the term of highest degree, −5y 4 , is 4. Thus, the degree of the polynomial is 4. 2. 2m3 − m2 − 4m + 11 = 2m3 + (−m2 ) + (−4m) + 11

= 2a4 − 2a3 b − a2 b + 4ab2 − 3b3 Collecting like terms 14.

Terms: 2m , −m , −4m, 11 3

2

= (n + 1)(n2 ) + (n + 1)(−6n) + (n + 1)(−4)

The degree of the term of highest degree, 2m3 , is 3. Thus, the degree of the polynomial is 3. 3. 3a4 b − 7a3 b3 + 5ab − 2 = 3a4 b + (−7a3 b3 ) + 5ab + (−2)

= n3 + n2 − 6n2 − 6n − 4n − 4

= n3 − 5n2 − 10n − 4 15.

Terms: 3a4 b, −7a3 b3 , 5ab, −2

The degrees of the terms are 5, 6, 2, and, 0, respectively, so the degree of the polynomial is 6.

4. 6p3 q 2 − p2 q 4 − 3pq 2 + 5 = 6p3 q 2 + (−p2 q 4 ) + (−3pq 2 ) + 5 Terms: 6p3 q 2 , −p2 q 4 , −3pq 2 , 5

(5x2 y − 2xy 2 + 3xy − 5)+

(−2x2 y − 3xy 2 + 4xy + 7)

= x + 2x − 15 2

17.

2

19.

(2x + 3y + z − 7) + (4x − 2y − z + 8)+

(2a + 3)(a + 5) = 2a2 + 10a + 3a + 15

Using FOIL

= 2a + 13a + 15

Collecting like terms

(2x + 3y)(2x + y) = 4x + 8xy + 3y

22.

= 3x + 2y − 2z − 3

23.

Using FOIL

2

(2a − 3b)(2a − b) = 4a2 − 2ab − 6ab + 3b2 =

4a2 − 8ab + 3b2 (y + 5)2

= y 2 + 2 · y · 5 + 52

8. 7x2 + 12xy − 2x − y − 9

[(A + B)2 = A2 + 2AB + B 2 ]

(3x2 − 2x − x3 + 2) − (5x2 − 8x − x3 + 4)

= (3x2 − 2x − x3 + 2) + (−5x2 + 8x + x3 − 4)

= (3 − 5)x2 + (−2 + 8)x + (−1 + 1)x3 + (2 − 4) = −2x2 + 6x − 2

= y 2 + 10y + 25 24. (y + 7)2 = y 2 + 2 · y · 7 + 72 = y 2 + 14y + 49 25.

(x − 4)2

= x2 − 2 · x · 4 + 42

10. −4x2 + 8xy − 5y 2 + 3

[(A − B)2 = A2 − 2AB + B 2 ]

(x − 3x + 4x) − (3x + x − 5x + 3) 2

Collecting like terms

2

= (2 + 4 − 3)x + (3 − 2 + 1)y + (1 − 1 − 2)z+

4

Using FOIL

= x2 + 10x + 24

= 4x2 + 2xy + 6xy + 3y 2

(−7 + 8 − 4)

11.

= x2 + 4x + 6x + 24

2

21.

(−3x + y − 2z − 4)

9.

(x + 6)(x + 4)

20. (3b + 1)(b − 2) = 3b2 − 6b + b − 2 = 3b2 − 5b − 2

6. 2x y − 7xy + 8xy + 5 7.

Collecting like terms

18. (n − 5)(n − 8) = n2 − 8n − 5n + 40 = n2 − 13n + 40

(−5 + 7)

2

Using FOIL

16. (y − 4)(y + 1) = y 2 + y − 4y − 4 = y 2 − 3y − 4

= (5 − 2)x2 y + (−2 − 3)xy 2 + (3 + 4)xy+ = 3x2 y − 5xy 2 + 7xy + 2

(x + 5)(x − 3)

= x2 − 3x + 5x − 15

The degrees of the terms are 5, 6, 3, and 0, respectively, so the degree of the polynomial is 6.

5.

(n + 1)(n2 − 6n − 4)

3

2

= (x4 − 3x2 + 4x) + (−3x3 − x2 + 5x − 3)

= x4 − 3x3 + (−3 − 1)x2 + (4 + 5)x − 3 = x4 − 3x3 − 4x2 + 9x − 3

12. 2x4 − 5x3 − 5x2 + 10x − 5

= x2 − 8x + 16

26. (a − 6)2 = a2 − 2 · a · 6 + 62 = a2 − 12a + 36 27.

(5x − 3)2

= (5x)2 − 2 · 5x · 3 + 32

[(A − B)2 = A2 − 2AB + B 2 ]

= 25x2 − 30x + 9

Exercise Set R.3

9

28. (3x − 2)2 = (3x)2 − 2 · 3x · 2 + 22 = 9x2 − 12x + 4 29.

(2x + 3y)2

44. Algebraically: Choose specific values for A and B, A %= 0, B %= 0, and evaluate (A + B)2 and A2 + B 2 . For example, (2 + 3)2 = 52 = 25, but 22 + 32 = 4 + 9 = 13.

= (2x)2 + 2(2x)(3y) + (3y)2

Geometrically: Show that the area of a square with side A + B is not equal to A2 + B 2 . See the figure below.

[(A+B)2 = A2 +2AB +B 2 ] = 4x2 + 12xy + 9y 2 30.

(5x + 2y)2 = (5x)2 + 2 · 5x · 2y + (2y)2 =

25x2 + 20xy + 4y 2 31.

(2x2 − 3y)2

= (2x2 )2 − 2(2x2 )(3y) + (3y)2

[(A − B)2 = A2 − 2AB + B 2 ] = 4x − 12x y + 9y 2 4

32.

(4x2 − 5y)2 = (4x2 )2 − 2 · 4x2 · 5y + (5y)2 = (a + 3)(a − 3)

= a2 − 32

= a2 − 9

[(A + B)(A − B) = A2 − B 2 ]

34. (b + 4)(b − 4) = b2 − 42 = b2 − 16 35.

(2x − 5)(2x + 5)

= (2x)2 − 52

= 4x2 − 25

[(A + B)(A − B) = A2 − B 2 ]

45.

46.

47.

38. (3x + 5y)(3x − 5y) = (3x)2 − (5y)2 = 9x2 − 25y 2

48.

49.

A

B

= 4x + 12xy + 9y − 16 2

50.

43. No; if the leading coefficients of the polynomials are additive inverses, the degree of the sum is less than n. For example, the sum of the second degree polynomials x2 +x−1 and −x2 + 4 is x + 3, a first degree polynomial.

[(2x − 1)2 − 1]2

= [4x2 − 4x + 1 − 1]2 = [4x2 − 4x]2

= (4x2 )2 − 2(4x2 )(4x) + (4x)2 = 16x4 − 32x3 + 16x2 51.

(xa−b )a+b = x(a−b)(a+b) 2

= xa 52.

−b2

(tm+n )m+n · (tm−n )m−n 2

= tm

= x −1

= y 4 − 16

(x − 1)(x2 + x + 1)(x3 + 1)

= x6 − 1

4

= (y 2 − 4)(y 2 + 4)

(x3m − t5n )2 = (x3m )2 − 2 · x3m · t5n + (t5n )2 =

= (x3 )2 − 12

25x2 + 20xy + 4y 2 − 9

(y − 2)(y + 2)(y 2 + 4)

(an + bn )2 = (an )2 + 2 · an · bn + (bn )2

= (x3 − 1)(x3 + 1)

(5x + 2y + 3)(5x + 2y − 3) = (5x + 2y)2 − 32 =

= (x2 − 1)(x2 + 1)

(ta + 4)(ta − 7) = (ta )2 − 7ta + 4ta − 28 =

= (x3 − x2 + x2 − x + x − 1)(x3 + 1)

2

(x + 1)(x − 1)(x2 + 1)

(an + bn )(an − bn ) = (an )2 − (bn )2

= [(x − 1)x2 + (x − 1)x + (x − 1) · 1](x3 + 1)

(2x + 3y + 4)(2x + 3y − 4)

= (2x + 3y)2 − 42

42.

B2

x6m − 2x3m t5n + t10n

= [(2x + 3y) + 4][(2x + 3y) − 4]

41.

AB

= a2n + 2an bn + b2n

[(A−B)(A+B) = A2 −B 2 ]

= 9x2 − 4y 2

40.

B

t2a − 3ta − 28

(3x − 2y)(3x + 2y)

= (3x)2 − (2y)2

39.

AB

= a2n − b2n

36. (4y − 1)(4y + 1) = (4y)2 − 12 = 16y 2 − 1 37.

A2

2

16x4 − 40x2 y + 25y 2 33.

A

+2mn+n2 2

= t2m 53.

+2n2

2

· tm

−2mn+n2

(a + b + c)2 = (a + b + c)(a + b + c) = (a + b + c)(a) + (a + b + c)(b) + (a + b + c)(c) = a2 + ab + ac + ab + b2 + bc + ac + bc + c2 = a2 + b2 + c2 + 2ab + 2ac + 2bc

10

Chapter R: Basic Concepts of Algebra 18.

Exercise Set R.4

= x2 (x − 1) − 6(x − 1) = (x − 1)(x2 − 6)

1. 2x − 10 = 2 · x − 2 · 5 = 2(x − 5)

19. p2 + 6p + 8

2. 7y + 42 = 7 · y + 7 · 6 = 7(y + 6)

We look for two numbers with a product of 8 and a sum of 6. By trial, we determine that they are 2 and 4.

3. 3x4 − 9x2 = 3x2 · x2 − 3x2 · 3 = 3x2 (x2 − 3)

p2 + 6p + 8 = (p + 2)(p + 4)

4. 20y 2 − 5y 5 = 5y 2 · 4 − 5y 2 · y 3 = 5y 2 (4 − y 3 ) 5. 4a2 − 12a + 16 = 4 · a2 − 4 · 3a + 4 · 4 = 4(a2 − 3a + 4) 6. 6n + 24n − 18 = 6 · n + 6 · 4n − 6 · 3 = 6(n + 4n − 3) 2

2

7. a(b − 2) + c(b − 2) = (b − 2)(a + c) 8. a(x2 − 3) − 2(x2 − 3) = (x2 − 3)(a − 2) 9.

x + 3x + 6x + 18 3

2

= x2 (x + 3) + 6(x + 3) = (x + 3)(x2 + 6) 10.

3x3 − x2 + 18x − 6

= x2 (3x − 1) + 6(3x − 1) = (3x − 1)(x2 + 6)

11.

y 3 − y 2 + 3y − 3

= y 2 (y − 1) + 3(y − 1) = (y − 1)(y + 3) 2

12.

y − y + 2y − 2 3

2

= y 2 (y − 1) + 2(y − 1) = (y − 1)(y 2 + 2)

13.

24x3 − 36x2 + 72x − 108

= 12(2x3 − 3x2 + 6x − 9)

= 12[x2 (2x − 3) + 3(2x − 3)] = 12(2x − 3)(x2 + 3)

14.

5a − 10a + 25a − 50 3

2

= 5(a3 − 2a2 + 5a − 10)

= 5[a2 (a − 2) + 5(a − 2)] = 5(a − 2)(a2 + 5)

15.

a − 3a − 2a + 6 3

2

= a2 (a − 3) − 2(a − 3) = (a − 3)(a2 − 2)

16.

t + 6t − 2t − 12 3

2

= t2 (t + 6) − 2(t + 6) = (t + 6)(t2 − 2)

17.

x3 − x2 − 5x + 5

= x (x − 1) − 5(x − 1) 2

= (x − 1)(x2 − 5)

x3 − x2 − 6x + 6

2

20. Note that (−5)(−2) = 10 and −5 + (−2) = −7. Then w2 − 7w + 10 = (w − 5)(w − 2).

21. x2 − 8x + 12

We look for two numbers with a product of 12 and a sum of −8. By trial, we determine that they are −2 and −6. x2 − 8x + 12 = (x − 2)(x − 6)

22. Note that 1 · 5 = 5 and 1 + 5 = 6. Then x2 + 6x + 5 = (x + 1)(x + 5).

23. t2 + 8t + 15 We look for two numbers with a product of 15 and a sum of 8. By trial, we determine that they are 3 and 5. t2 + 8t + 15 = (t + 3)(t + 5) 24. Note that 3 · 9 = 27 and 3 + 9 = 12. Then y 2 + 12y + 27 = (y + 3)(y + 9).

25. x2 − 6xy − 27y 2

We look for two numbers with a product of −27 and a sum of −6. By trial, we determine that they are 3 and −9. x2 − 6xy − 27y 2 = (x + 3y)(x − 9y)

26. Note that 3(−5) = −15 and 3 + (−5) = −2. Then t2 − 2t − 15 = (t + 3)(t − 5).

27. 2n2 − 20n − 48 = 2(n2 − 10n − 24)

Now factor n2 − 10n − 24. We look for two numbers with a product of −24 and a sum of −10. By trial, we determine that they are 2 and −12. Then n2 − 10n − 24 = (n + 2)(n − 12). We must include the common factor, 2, to have a factorization of the original trinomial. 2n2 − 20n − 48 = 2(n + 2)(n − 12)

28. 2a2 − 2ab − 24b2 = 2(a2 − ab − 12b2 )

Note that −4 · 3 = −12 and −4 + 3 = −1. Then 2a2 − 2ab − 24b2 = 2(a − 4b)(a + 3b).

29. y 4 − 4y 2 − 21 = (y 2 )2 − 4y 2 − 21

We look for two numbers with a product of −21 and a sum of −4. By trial, we determine that they are 3 and −7. y 4 − 4y 2 − 21 = (y 2 + 3)(y 2 − 7)

30. Note that 9(−10) = −90 and 9 + (−10) = −1. Then m4 − m2 − 90 = (m2 + 9)(m2 − 10).

Exercise Set R.4

11

31. y 4 + 9y 3 + 14y 2 = y 2 (y 2 + 9y + 14) Now factor y 2 + 9y + 14. Look for two numbers with a product of 14 and a sum of 9. The numbers are 2 and 7. Then y 2 + 9y + 14 = (y + 2)(y + 7). We must include the common factor, y 2 , in order to have a factorization of the original trinomial. y 4 + 9y 3 + 14y 2 = y 2 (y + 2)(y + 7) 32. 3z 3 − 21z 2 + 18z = 3z(z 2 − 7z + 6) = 3z(z − 1)(z − 6) 33. 2x3 − 2x2 y − 24xy 2 = 2x(x2 − xy − 12y 2 )

Now factor x2 − xy − 12y 2 . Look for two numbers with a product of −12 and a sum of −1. The numbers are −4 and 3. Then x2 −xy−12y 2 = (x−4y)(x+3y). We must include the common factor, 2x, in order to have a factorization of the original trinomial. 2x3 − 2x2 y − 24xy 2 = 2x(x − 4y)(x + 3y)

34. a3 b−9a2 b2 +20ab3 = ab(a2 −9ab+20b2 ) = ab(a−4b)(a−5b)

38. 6x2 − 7x − 20 = (3x + 4)(2x − 5) 39. 4x2 + 15x + 9 We use the FOIL method. 1. There is no common factor other than 1 or −1. 2. The factorization must be of the form (4x+ )(x+ ) or (2x+ )(2x+ ).

3. Factor the constant term, 9. The possibilities are 1 · 9, −1(−9), 3 · 3, and −3(−3). The first two pairs of factors can be written in the opposite order as well: 9 · 1, −9(−1).

4. Find a pair of factors for which the sum of the outside and the inside products is the middle term, 15x. By trial, we determine that the factorization is (4x + 3)(x + 3). 40. 2y 2 + 7y + 6 = (2y + 3)(y + 2) 41. 2y 2 + y − 6

We use the grouping method.

35. 2n2 + 9n − 56

We use the FOIL method. 1. There is no common factor other than 1 or −1. 2. The factorization must be of the form (2n+ )(n+ ).

3. Factor the constant term, −56. The possibilities are −1 · 56, 1(−56), −2 · 28, 2(−28), −4 · 16, 4(−16), −7 · 8, and 7(−8). The factors can be written in the opposite order as well: 56(−1), −56 · 1, 28(−2), −28 · 2, 16(−4), −16 · 4, 8(−7), and −8 · 7.

4. Find a pair of factors for which the sum of the outside and the inside products is the middle term, 9n. By trial, we determine that the factorization is (2n − 7)(n + 8). 36. 3y 2 + 7y − 20 = (3y − 5)(y + 4) 37. 12x2 + 11x + 2 We use the grouping method. 1. There is no common factor other than 1 or −1.

2. Multiply the leading coefficient and the constant: 12 · 2 = 24.

3. Try to factor 24 so that the sum of the factors is the coefficient of the middle term, 11. The factors we want are 3 and 8. 4. Split the middle term using the numbers found in step (3): 11x = 3x + 8x 5. Factor by grouping. 12x2 + 11x + 2 = 12x2 + 3x + 8x + 2 = 3x(4x + 1) + 2(4x + 1) = (4x + 1)(3x + 2)

1. There is no common factor other than 1 or −1.

2. Multiply the leading coefficient and the constant: 2(−6) = −12. 3. Try to factor −12 so that the sum of the factors is the coefficient of the middle term, 1. The factors we want are 4 and −3.

4. Split the middle term using the numbers found in step (3): y = 4y − 3y

5. Factor by grouping.

2y 2 + y − 6 = 2y 2 + 4y − 3y − 6

= 2y(y + 2) − 3(y + 2) = (y + 2)(2y − 3)

42. 20p2 − 23p + 6 = (4p − 3)(5p − 2) 43. 6a2 − 29ab + 28b2

We use the FOIL method. 1. There is no common factor other than 1 or −1.

2. The factorization must be of the form (6x+ )(x+ ) or (3x+ )(2x+ ).

3. Factor the coefficient of the last term, 28. The possibilities are 1 · 28, −1(−28), 2 · 14, −2(−14), 4 · 7, and −4(−7). The factors can be written in the opposite order as well: 28·1, −28(−1), 14·2, −14(−2), 7 · 4, and −7(−4).

4. Find a pair of factors for which the sum of the outside and the inside products is the middle term, −29. Observe that the second term of each binomial factor will contain a factor of b. By trial, we determine that the factorization is (3a−4b)(2a−7b). 44. 10m2 + 7mn − 12n2 = (5m − 4n)(2m + 3n)

12

Chapter R: Basic Concepts of Algebra

45. 12a2 − 4a − 16

59.

We will use the grouping method. 1. Factor out the common factor, 4. 12a2 − 4a − 16 = 4(3a2 − a − 4)

2. Now consider 3a2 − a − 4. Multiply the leading coefficient and the constant: 3(−4) = −12.

3. Try to factor −12 so that the sum of the factors is the coefficient of the middle term, −1. The factors we want are −4 and 3.

4. Split the middle term using the numbers found in step (3):

= (2z + 3)2

60. 9z 2 − 12z + 4 = (3z − 2)2 61.

62. 1 + 10x + 25x2 = (1 + 5x)2 63.

= a(a2 + 2 · a · 12 + 122 )

= a(a + 12)2

3a2 − a − 4 = 3a2 − 4a + 3a − 4

= a(3a − 4) + (3a − 4) = (3a − 4)(a + 1)

64. y 3 − 18y 2 + 81y = y(y 2 − 18y + 81) = y(y − 9)2 65.

12a2 − 4a − 16 = 4(3a − 4)(a + 1) 46. 12a2 − 14a − 20 = 2(6a2 − 7a − 10) = 2(6a + 5)(a − 2) m2 − 4 = m2 − 22

= (m + 2)(m − 2)

= 4(p − q)2

66. 5a2 − 10ab + 5b2 = 5(a2 − 2ab + b2 ) = 5(a − b)2 67.

= (x + 2)(x2 − 2x + 4)

69.

= (2z + 9)(2z − 9)

70. n3 + 216 = (n + 6)(n2 − 6n + 36)

50. 16x2 − 9 = (4x − 3)(4x + 3)

71.

52. 8a2 − 8b2 = 8(a2 − b2 ) = 8(a + b)(a − b) 4xy 4 − 4xz 2 = 4x(y 4 − z 2 )

= 2(y − 4)(y 2 + 4y + 16) 72. 8t3 − 8 = 8(t3 − 1) = 8(t − 1)(t2 + t + 1) 73.

= 4x[(y 2 )2 − z 2 ]

55.

7pq − 7py = 7p(q − y )

= 7p[(q ) − (y ) ]

74.

= 7p(q 2 + y 2 )(q + y)(q − y) 25ab4 − 25az 4 = 25a(b4 − z 4 )

= 25a(b2 + z 2 )(b2 − z 2 )

75.

57.

y − 6y + 9 = y − 2 · y · 3 + 3 2

= (y − 3)

2

t6 + 1 = (t2 )3 + 13 = (t2 + 1)(t4 − t2 + 1)

76. 27x6 − 8 = (3x2 − 2)(9x4 + 6x2 + 4) 77.

= 25a(b2 + z 2 )(b + z)(b − z) 2

250z 4 − 2z = 2z(125z 3 − 1)

= 2z(5z − 1)(25z 2 + 5z + 1)

2 2

= 7p(q 2 + y 2 )(q 2 − y 2 )

56.

= 3a2 (a − 2)(a2 + 2a + 4)

4

2 2

3a5 − 24a2 = 3a2 (a3 − 8)

= 3a2 (a3 − 23 )

= 4x(y 2 + z)(y 2 − z) 54. 5x2 y − 5yz 4 = 5y(x2 − z 4 ) = 5y(x + z 2 )(x − z 2 )

2y 3 − 128 = 2(y 3 − 64) = 2(y 3 − 43 )

51. 6x2 − 6y 2 = 6(x2 − y 2 ) = 6(x + y)(x − y)

4

m3 − 1 = m3 − 13

= (m − 1)(m2 + m + 1)

4z 2 − 81 = (2z)2 − 92

4

x3 + 8 = x3 + 23

68. y 3 − 64 = (y − 4)(y 2 + 4y + 16)

48. z 2 − 81 = (z + 9)(z − 9)

4

4p2 − 8pq + 4q 2

= 4(p2 − 2pq + q 2 )

We must include the common factor to get a factorization of the original trinomial.

53.

a3 + 24a2 + 144a = a(a2 + 24a + 144)

5. Factor by grouping.

49.

1 − 8x + 16x2 = 12 − 2 · 1 · 4x + (4x)2 = (1 − 4x)2

−a = −4a + 3a

47.

4z 2 + 12z + 9 = (2z)2 + 2 · 2z · 3 + 32

18a2 b − 15ab2 = 3ab · 6a − 3ab · 5b = 3ab(6a − 5b)

78. 4x2 y + 12xy 2 = 4xy(x + 3y)

2

79.

58. x + 8x + 16 = (x + 4) 2

2

x3 − 4x2 + 5x − 20 = x2 (x − 4) + 5(x − 4) = (x − 4)(x2 + 5)

Exercise Set R.4 80.

13

z 3 + 3z 2 − 3z − 9 = z 2 (z + 3) − 3(z + 3) = (z + 3)(z 2 − 3)

81.

93. 6x2 + 7x − 3

We use the grouping method. 1. There is no common factor other than 1 or −1.

8x2 − 32 = 8(x2 − 4)

2. Multiply the leading coefficient and the constant: 6(−3) = −18.

= 8(x + 2)(x − 2)

82. 6y 2 − 6 = 6(y 2 − 1) = 6(y + 1)(y − 1)

3. Try to factor −18 so that the sum of the factors is the coefficient of the middle term, 7. The factors we want are 9 and −2.

83. 4y 2 − 5

There are no common factors. We might try to factor this polynomial as a difference of squares, but there is no integer which yields 5 when squared. Thus, the polynomial is prime.

4. Split the middle term using the numbers found in step (3): 7x = 9x − 2x

5. Factor by grouping.

84. There are no common factors and there is no integer which yields 7 when squared, so 16x2 − 7 is prime. 85.

6x2 + 7x − 3 = 6x2 + 9x − 2x − 3

= 3x(2x + 3) − (2x + 3)

m2 − 9n2 = m2 − (3n)2

= (m + 3n)(m − 3n)

86. 25t − 16 = (5t + 4)(5t − 4) 2

= (2x + 3)(3x − 1) 94. 8x2 + 2x − 15 = (4x − 5)(2x + 3) 95.

87. x + 9x + 20 2

We look for two numbers with a product of 20 and a sum of 9. They are 4 and 5. x2 + 9x + 20 = (x + 4)(x + 5)

= (y − 9)2

96. n2 + 2n + 1 = (n + 1)2 97.

89. y 2 − 6y + 5

We look for two numbers with a product of 5 and a sum of −6. They are −5 and −1. y 2 − 6y + 5 = (y − 5)(y − 1)

98. 4z 2 + 20z + 25 = (2z + 5)2 99.

100. x2 y 2 − 16xy + 64 = (xy − 8)2 101.

91. 2a2 + 9a + 4

102.

103.

2. The factorization must be of the form (2a+ )(a+ ).

92. 3b2 − b − 2 = (3b + 2)(b − 1)

21x2 y + 2xy − 8y = y(21x2 + 2x − 8)

= y(7x − 4)(3x + 2)

We use the FOIL method.

4. Find a pair of factors for which the sum of the outside and the inside products is the middle term, 9a. By trial, we determine that the factorization is (2a + 1)(a + 4).

4ax2 + 20ax − 56a = 4a(x2 + 5x − 14) = 4a(x + 7)(x − 2)

x2 − 4x − 21 = (x − 7)(x + 3)

3. Factor the constant term, 4. The possibilities are 1 · 4, −1(−4), and 2 · 2. The first two pairs of factors can be written in the opposite order as well: 4 · 1, −4(−1).

x2 y 2 − 14xy + 49 = (xy)2 − 2 · xy · 7 + 72 = (xy − 7)2

90. Note that −7(3) = −21 and −7 + 3 = −4.

1. There is no common factor other than 1 or −1.

9z 2 − 24z + 16 = (3z)2 − 2 · 3z · 4 + 42 = (3z − 4)2

88. Note that 3(−2) = −6 and 3 + (−2) = 1. Then y 2 + y − 6 = (y + 3)(y − 2).

y 2 − 18y + 81 = y 2 − 2 · y · 9 + 92

3z 3 − 24 = 3(z 3 − 8)

= 3(z 3 − 23 )

= 3(z − 2)(z 2 + 2z + 4) 104.

4t3 + 108 = 4(t3 + 27) = 4(t + 3)(t2 − 3t + 9)

105.

16a7 b + 54ab7 = 2ab(8a6 + 27b6 ) = 2ab[(2a2 )3 + (3b2 )3 ] = 2ab(2a2 + 3b2 )(4a4 − 6a2 b2 + 9b4 )

106.

24a2 x4 − 375a8 x

= 3a2 x(8x3 − 125a6 )

= 3a2 x(2x − 5a2 )(4x2 + 10a2 x + 25a4 )

14

Chapter R: Basic Concepts of Algebra

107.

121. A2 + B 2 can be factored when A and B have a common factor. For example, let A = 2x and B = 10. Then A2 + B 2 = 4x2 + 100 = 4(x2 + 25).

y 3 − 3y 2 − 4y + 12

= y 2 (y − 3) − 4(y − 3) = (y − 3)(y 2 − 4)

122.

= (y − 3)(y + 2)(y − 2) 108.

= (A + (−B))(A2 − A(−B) + (−B)2 )

p3 − 2p2 − 9p + 18

= (A − B)(A2 + AB + B 2 )

= p2 (p − 2) − 9(p − 2)

123.

= (p − 2)(p2 − 9)

109.

x −x +x−1 3

2

= (x − 1)(x + 1) 2

124.

Substituting y 2 for u

11x2 + x4 − 80

= (x − 1)(x − 1)

= (u + 16)(u − 5)

2

= (x − 1)(x + 1)(x − 1), or (x − 1) (x + 1) 2

125.

5m4 − 20 = 5(m4 − 4)

= 5(m2 + 2)(m2 − 2)

112. 2x2 − 288 = 2(x2 − 144) = 2(x + 12)(x − 12) 113.

2x3 + 6x2 − 8x − 24

= 2(x3 + 3x2 − 4x − 12)

126. 127.

= 2[x (x + 3) − 4(x + 3)] 2

= 2(x + 3)(x2 − 4)

= 2(x + 3)(x + 2)(x − 2) 114.

128.

3x3 + 6x2 − 27x − 54

= 3(x + 2x − 9x − 18) 3

2

129.

= 3[x2 (x + 2) − 9(x + 2)] = 3(x + 2)(x − 9) 2

= 3(x + 2)(x + 3)(x − 3) 4c2 − 4cd − d2 = (2c)2 − 2 · 2c · d − d2 = (2c − d)2

130. 131.

116. 9a2 − 6ab + b2 = (3a − b)2 3

3 2

3

We look for two numbers with a product of −20 and a sum of 8. They are 10 and −2. m6 + 8m3 − 20 = (m3 + 10)(m3 − 2)

p − 64p4 = p(1 − 64p3 )

= p[1 − (4p) ] 3

3

= p(1 − 4p)(1 + 4p + 16p2 ) 120. 125a − 8a4 = a(125 − 8a3 ) = a(5 − 2a)(25 + 10a + 4a2 )

Substituting u for x2

= (x2 + 16)(x2 − 5)

Substituting x2 for u

8 2 2 8 + y = y2 + y − 49 7 7 49 $ 2% 4 %$ y− = y+ 7 7 " #" # 27 3 3 27 9 3 t2 − + t = t2 + t − = t+ t− 100 5 5 100 10 10 " #2 9 3 3 x2 + 3x + = x2 + 2 · x · + 4 2 2 " #2 3 = x+ 2 " #2 25 5 x2 − 5x + = x− 4 2 " #2 1 1 1 x2 − x + = x2 − 2 · x · + 4 2 2 #2 " 1 = x− 2 " #2 2 1 1 x2 − x + = x − 3 9 3 y2 −

(x + h)3 − x3

= (x + h − x)(x2 + 2xh + h2 + x2 + xh + x2 ) = h(3x2 + 3xh + h2 ) 132.

(x + 0.01)2 − x2

= (x + 0.01 + x)(x + 0.01 − x)

x4 − 37x2 + 36 = (x2 − 1)(x2 − 36)

= (x + 1)(x − 1)(x + 6)(x − 6)

= u2 + 11u − 80

= [(x + h) − x][(x + h)2 + x(x + h) + x2 ]

117. m + 8m − 20 = (m ) + 8m − 20 6

119.

= (y 2 + 12)(y 2 − 7) = x4 + 11x2 − 80

2

= x (x − 1) − (x − 1)

118.

Substituting u for y 2

x −x −x+1 3 2

115.

= u2 + 5u − 84

= (u + 12)(u − 7)

= x2 (x − 1) + (x − 1) 110.

y 4 − 84 + 5y 2

= y 4 + 5y 2 − 84

= (p − 2)(p + 3)(p − 3)

111.

A3 − B 3 = A3 + (−B)3

= 0.01(2x + 0.01), or 0.02(x + 0.005) 133.

(y − 4)2 + 5(y − 4) − 24

= u2 + 5u − 24

= (u + 8)(u − 3)

Substituting u for y − 4

= (y − 4 + 8)(y − 4 − 3) Substituting y − 4 for u = (y + 4)(y − 7)

Exercise Set R.5 134.

15

6(2p + q)2 − 5(2p + q) − 25

= 6u2 − 5u − 25

Substituting u for 2p + q

= (3u + 5)(2u − 5)

= [3(2p + q) + 5][2(2p + q) − 5] = (6p + 3q + 5)(4p + 2q − 5) 135.

4.

Substituting 2p + q for u

x2n + 5xn − 24 = (xn )2 + 5xn − 24 = (xn + 8)(xn − 3)

5.

136. 4x2n − 4xn − 3 = (2xn − 3)(2xn + 1) 137.

x2 + ax + bx + ab = x(x + a) + b(x + a) = (x + a)(x + b)

138.

6.

bdy 2 + ady + bcy + ac = dy(by + a) + c(by + a) = (by + a)(dy + c)

139.

25y 2m − (x2n − 2xn + 1) n

= (5y m + xn − 1)(5y m − xn + 1) x6a − t3b = (x2a )3 − (tb )3 =

(x2a − tb )(x4a + x2a tb + t2b )

141.

(y − 1)4 − (y − 1)2

8.

= (y − 1)2 [(y − 1)2 − 1]

= (y − 1)2 [y 2 − 2y + 1 − 1] = (y − 1)2 (y 2 − 2y) = y(y − 1)2 (y − 2)

142.

x6 − 2x5 + x4 − x2 + 2x − 1

= x4 (x2 − 2x + 1) − (x2 − 2x + 1) = (x2 − 2x + 1)(x4 − 1)

9.

= (x − 1)2 (x2 + 1)(x2 − 1)

10.

= (x2 + 1)(x + 1)(x − 1)3

11.

= (x − 1)2 (x2 + 1)(x + 1)(x − 1)

3.

3x − 3 x(x − 1) The denominator is 0 when the factor x = 0 and also when x − 1 = 0, or x = 1. The domain is {x|x is a real number and x %= 0 and x %= 1}.

! ! (x − 1)(x+3) ! x−1 x2 + 2x − 3 = x2 − 9 (x+3) ! (x − 3) = x − 3 x2

(x + 2)(x−2) x + 2 x2 − 4 = = − 4x + 4 (x − 2)(x−2) x − 2

x(x2 − 6x + 9) x3 − 6x2 + 9x = x3 − 3x2 x2 (x − 3)

!

x /(x−3)(x − 3) x / · x(x−3) x−3 = x

1. Since −

2. Since 8 − x = 0 when x = 8, the domain is {x|x is a real number and x %= 8}.

7x2 + 11x − 6 7x2 + 11x − 6 = 2 x(x − x − 6) x(x − 3)(x + 2) The denominator is 0 when x = 0 or when x − 3 = 0 or when x + 2 = 0. Now x − 3 = 0 when x = 3 and x + 2 = 0 when x = −2. Thus, the domain is {x|x is a real number and x %= 0 and x %= 3 and x %= −2}.

=

Exercise Set R.5 3 is defined for all real numbers, the domain is 4 {x|x is a real number}.

(x2 − 4)(x + 1) (x2 − 4)(x + 1) = (x + 2)(x2 − 1) (x + 2)(x + 1)(x − 1) x + 2 = 0 when x = −2; x + 1 = 0 when x = −1; x − 1 = 0 when x = 1. The domain is {x|x is a real number and x %= −2 and x %= −1 and x %= 1}. 7x2 − 28x + 28 7x2 − 28x + 28 = (x2 −4)(x2 +3x−10) (x+2)(x−2)(x+5)(x−2) We see that x + 2 = 0 when x = −2, x − 2 = 0 when x = 2, and x + 5 = 0 when x = −5. Thus, the domain is {x|x is a real number and x %= −2 and x %= 2 and x %= −5}.

2

= [5y m + (xn − 1)][5y m − (xn − 1)]

140.

x+5 x+5 = x2 + 4x − 5 (x + 5)(x − 1) We see that x + 5 = 0 when x = −5 and x − 1 = 0 when x = 1. Thus, the domain is {x|x is a real number and x %= −5 and x %= 1}.

7. We first factor the denominator completely.

= (5y ) − (x − 1) m 2

15x − 10 2x(3x − 2) Since 2x = 0 when x = 0 and 3x − 2 = 0 when 2 x = , the domain is 3 1 0 ! ! 2 . x!!x is a real number and x %= 0 and x %= 3

12.

!

y 5 − 5y 4 + 4y 3 y 3 (y 2 − 5y + 4) = y 3 − 6y 2 + 8y y(y 2 − 6y + 8) = =

! !

y/ · y · y · (y − 1)(y−4) y/(y − 2)(y−4) y 2 (y − 1) y−2

16 13.

Chapter R: Basic Concepts of Algebra 6y 2 + 12y − 48 6(y 2 + 2y − 8) = 2 3y − 9y + 6 3(y 2 − 3y + 2)

! !

2·3 / · (y + 4)(y−2) /(y − 1)(y−2) 3

=

14.

=

2(y + 4) y−1

=

=

2(x2 − 10x + 25) 2x2 − 20x + 50 = 2 10x − 30x − 100 10(x2 − 3x − 10)

! !

15.

16.

17.

!

=

(x2 − y 2 ) · 1 = (x − y)2 (x + y)

=

=

! ! ! ! ! !

(r−s)(r−s)(r+s) · 1 (r+s)(r−s)(r−s) =1 =

x2 − 2x − 35 4x3 − 9x · 2x3 − 3x2 7x − 49

19. = =

!

$

20. = =

!

$

$

!

(x − 5)(3x + 2) 7x

a2 − 2a − 8 a2 − a − 6 · 2 2 a − 7a + 12 a − 3a − 10

21.

! ! ! ! ! !

(a−3)(a + 2)(a−4)(a+2) (a−4)(a−3)(a − 5)(a+2) a+2 = a−5

=

! ! ! !

=

3(x − 4) 2(x + 4) a2 − a − 2 a2 − 2a ÷ a2 − a − 6 2a + a2

=

a2 − a − 2 2a + a2 · a2 − a − 6 a2 − 2a

!

!

! !

x2 − y 2 x2 + xy + y 2 · x3 − y 3 x2 + 2xy + y 2

27.

(x+7)(x − 5)(x) " (3x + 2)(3x−2) x / · x(3x−2)(7)(x+7)

3x + 12 (x − 4)2 · 2x − 8 (x + 4)2

(a−2)(a + 1)(a) " (2+a) (a − 3)(a+2)(a) " (a−2) a+1 = a−3

$

3

! ! ! !

3(x+4)(x−4)(x − 4) 2(x−4)(x+4)(x + 4)

=

!

x + 2x − 35 9x − 4x · 3x3 − 2x2 7x + 49 2

a2 − b2 x − y · x−y a+b

=

26.

(x−7)(x + 5)(x) " (2x + 3)(2x−3) x / · x(2x−3)(7)(x−7) (x + 5)(2x + 3) 7x

! !

!

3x + 12 (x + 4)2 ÷ 2x − 8 (x − 4)2

25.

! ! ! !

(r − s)(r2 − s2 ) r − s r2 − s2 · = r + s (r − s)2 (r + s)(r − s)2

m2 − n 2 r + s · r+s m−n

(a+b)(a − b)(x−y) (x−y)(a+b) · 1 = a−b

=

(x+y)(x−y) · 1 = (x−y)(x − y)(x+y) 1 = x−y

18.

(a + 3)2 (a − 6)(a + 4)

a+b a2 − b2 ÷ x−y x−y

24.

! !

−1(x−6) 6−x = x2 − 36 (x + 6)(x−6) −1 1 = , or − x+6 x+6 1 x2 − y 2 · (x − y)2 x + y

!

(m + n)(m−n)(r+s) (r+s)(m − n) = m+n =

!

−1(x−4) 4−x = x2 + 4x − 32 (x−4)(x + 8) −1 1 = , or − x+8 x+8

! ! !

(a−4)(a + 3)(a+3)(a−2) (a−2)(a−4)(a − 6)(a+4)

m2 − n 2 m−n ÷ r+s r+s

23.

/(x−5)(x − 5) 2 / · 5 · (x−5)(x + 2) 2 x−5 = 5(x + 2) =

a2 − a − 12 a2 + a − 6 · a2 − 6a + 8 a2 − 2a − 24

22.

=

(x + y)(x − y)(x2 + xy + y 2 ) (x − y)(x2 + xy + y 2 )(x + y)(x + y)

1 x+y 1 = x+y 1 = x+y =

·

(x + y)(x − y)(x2 + xy + y 2 ) (x + y)(x − y)(x2 + xy + y 2 )

·1

Removing a factor of 1

Exercise Set R.5 c3 + 8 c2 − 2c + 4 ÷ c2 − 4 c2 − 4c + 4

28. = =

34.

(c + 2)(c2 − 2c + 4)(c − 2)(c − 2) (c + 2)(c − 2)(c2 − 2c + 4) 2

(x − y)2 − z 2 x + y − z · = (x + y)2 − z 2 x − y + z

35. = = = 36.

(x − y + z)(x − y − z)(x + y − z) = (x + y + z)(x + y − z)(x − y + z)

=

=

=

(x − y + z)(x + y − z) x − y − z · (x − y + z)(x + y − z) x + y + z x−y−z = 1· Removing a factor of 1 x+y+z x−y−z = x+y+z

37. =

(a + b)2 − 9 a − b − 3 · (a − b)2 − 9 a + b + 3

30. =

=

(a + b + 3)(a + b − 3)(a − b − 3) (a − b + 3)(a − b − 3)(a + b + 3)

(a + b + 3)(a − b − 3) a + b − 3 · (a + b + 3)(a − b − 3) a − b + 3 a+b−3 = a−b+3

= =

=

31.

32.

33.

38.

1 5+1 5 + = 2x 2x 2x 6 = 2x 2/ · 3 = 2/ · x 3 = x

4 6 10 − = 9y 9y 9y 3/ · 2 = 3/ · 3y 2 = 3y 3 2a + 2a + 3 2a + 3 3 + 2a = 2a + 3 =1

! !

2(a+b) 1 · (a+b) =2

=

x−y+z (x − y)2 − z 2 ÷ (x + y)2 − z 2 x+y−z

29.

a − 3b a + 5b 2a + 2b + = a+b a+b a+b

c3 + 8 c2 − 4c + 4 · c2 − 4 c2 − 2c + 4

(c + 2)(c − 2c + 4)(c − 2) c − 2 · (c + 2)(c2 − 2c + 4)(c − 2) 1 = c−2 =

17

3 5 − , LCD is 8z 4z 8z 5 2 3 · − 4z 2 8z 3 10 − 8z 8z 7 8z 5 12 + , LCD is x2 y 2 x2 y xy 2 12y 5x + 2 2 x2 y 2 x y 12y + 5x x2 y 2 3 2 + x + 2 x2 − 4 2 3 + , LCD is (x+2)(x−2) x + 2 (x + 2)(x − 2) x−2 2 3 · + x + 2 x − 2 (x + 2)(x − 2) 3x − 6 2 + (x + 2)(x − 2) (x + 2)(x − 2) 3x − 4 (x + 2)(x − 2)

2 5 − a − 3 a2 − 9 2 5 = − , LCD is (a + 3)(a − 3) a − 3 (a + 3)(a − 3) 5(a + 3) − 2 (a + 3)(a − 3) 5a + 15 − 2 = (a + 3)(a − 3) 5a + 13 = (a + 3)(a − 3) =

39.

y − y 2 − y − 20 y = (y + 4)(y − 5) y = (y + 4)(y − 5) y = (y + 4)(y − 5) y − (2y − 10) (y + 4)(y − 5) y − 2y + 10 = (y + 4)(y − 5) −y + 10 = (y + 4)(y − 5)

=

2 y+4 2 − , LCD is (y + 4)(y − 5) y+4 2 y−5 − · y+4 y−5 2y − 10 − (y + 4)(y − 5)

18 40.

Chapter R: Basic Concepts of Algebra 6 5 − y 2 + 6y + 9 y + 3 5 6 − , LCD is (y + 3)2 = (y + 3)2 y+3 6 − 5(y + 3) (y + 3)2 6 − 5y − 15 = (y + 3)2 −5y − 9 = (y + 3)2 =

41. = = = =

x − 5y 3 + x + y x2 − y 2 3 x − 5y + , LCD is (x + y)(x − y) x + y (x + y)(x − y) x−y x − 5y 3 · + x + y x − y (x + y)(x − y) 3x − 3y x − 5y + (x + y)(x − y) (x + y)(x − y) 4x − 8y (x + y)(x − y) a2 + 1 a − 1 − a2 − 1 a + 1

42. = =

46.

47.

48.

a2 + 1 − (a − 1)(a − 1) (a + 1)(a − 1)

a2 + 1 − a2 + 2a − 1 (a + 1)(a − 1) 2a = (a + 1)(a − 1)

44.

45.

b a −b a + = + a−b b−a a−b a−b a−b = a−b =1 y x − 2x − 3y 3y − 2x −1 y x − · = 2x − 3y −1 3y − 2x x −y = − 2x − 3y 2x − 3y x+y [x − (−y) = x + y] = 2x − 3y

=

9x2 − 7x − 2 7x − 14 + (3x+4)(x−2)(x−1) (3x+4)(x−1)(x−2)

=

9x2 − 16 (3x + 4)(x − 2)(x − 1)

$ $

(3x+4)(3x − 4) (3x+4)(x − 2)(x − 1) 3x − 4 = (x − 2)(x − 1)

a−1 a2 + 1 − , LCD is (a + 1)(a − 1) (a + 1)(a − 1) a + 1

y 2 + y−1 1−y −1 2 y + · = y − 1 −1 1 − y y −2 = + y−1 y−1 y−2 = y−1

9x + 2 7 + 3x2 − 2x − 8 3x2 + x − 4 7 9x + 2 + , = (3x + 4)(x − 2) (3x + 4)(x − 1) LCD is (3x + 4)(x − 2)(x − 1) x−1 7 x−2 9x + 2 · + · = (3x+4)(x−2) x − 1 (3x+4)(x−1) x − 2

=

=

43.

3a 2a 3a −2a − = − 3a − 2b 2b − 3a 3a − 2b 3a − 2b 5a = 3a − 2b

2y 3y − y 2 − 7y + 10 y 2 − 8y + 15 2y 3y = − , (y − 2)(y − 5) (y − 5)(y − 3) LCD is (y − 2)(y − 5)(y − 3) 3y(y − 3) − 2y(y − 2) = (y − 2)(y − 5)(y − 3) = =

3y 2 − 9y − 2y 2 + 4y (y − 2)(y − 5)(y − 3) y 2 − 5y (y − 2)(y − 5)(y − 3)

! !

y(y−5) (y − 2)(y−5)(y − 3) y = (y − 2)(y − 3) =

49.

4b ab 5a + 2 + 2 a−b a −b a+b ab 4b 5a + + , = a − b (a + b)(a − b) a + b LCD is (a + b)(a − b) 5a a+b ab 4b a−b = · + + · a − b a + b (a + b)(a − b) a + b a − b =

ab 4ab − 4b2 5a2 + 5ab + + (a+b)(a−b) (a+b)(a−b) (a+b)(a−b)

=

5a2 + 10ab − 4b2 (a + b)(a − b)

Exercise Set R.5 50.

51.

6a 3b 5 + + a − b b − a a2 − b2 6a −3b 5 = + + , a − b a − b (a + b)(a − b) LCD is (a + b)(a − b) 6a(a + b) + 3b(a + b) + 5 = (a + b)(a − b)

1 x x2 + 2 + + 2 x+1 2−x x −x−2

53. = =

6a2 + 6ab + 3ab + 3b2 + 5 = (a + b)(a − b)

=

=

=

6a2 + 9ab + 3b2 + 5 (a + b)(a − b)

x+8 3x − 2 7 − + x + 2 4 − x2 4 − 4x + x2 7 x+8 3x − 2 = , − + x + 2 (2 + x)(2 − x) (2 − x)2 LCD is (2 + x)(2 − x)2 =

= = = 52.

19

7 (2 − x)2 x+8 · − 2 + x (2 − x)2 (2 + x)(2 − x) 3x − 2 (2 − x)2

2−x + 2−x 2+x · 2+x

=

= =

1 −x x2 + 2 + + , x + 1 x − 2 (x + 1)(x − 2) LCD is (x + 1)(x − 2)

x−2 −x x + 1 x2 + 2 1 · + · + x + 1 x − 2 x − 2 x + 1 (x + 1)(x − 2) −x2 − x x2 + 2 x−2 + + (x+1)(x−2) (x+1)(x−2) (x+1)(x−2)

x − 2 − x2 − x + x2 + 2 (x + 1)(x − 2) 0 = (x + 1)(x − 2) =0

·

54.

28−28x+7x2 −(16−6x−x2 )+3x2 +4x−4 (2 + x)(2 − x)2

28 − 28x + 7x2 − 16 + 6x + x2 + 3x2 + 4x − 4 (2 + x)(2 − x)2

11x2 − 18x + 8 11x2 − 18x + 8 , or 2 (2 + x)(2 − x) (x + 2)(x − 2)2

x−6 x−1 x+1 − − x − 2 x + 2 4 − x2 x−1 x+1 x−6 = − − x − 2 x + 2 (2 + x)(2 − x) x−6 1−x x+1 = − − , 2 − x x + 2 (2 + x)(2 − x) LCD is (2 + x)(2 − x) (1 − x)(2 + x) − (x + 1)(2 − x) − (x − 6) = (2 + x)(2 − x) 2 − x − x2 + x2 − x − 2 − x + 6 (2 + x)(2 − x) 6 − 3x = (2 + x)(2 − x) =

! !

3(2−x) (2 + x)(2−x) 3 = 2+x

6(3 − x)2 − (x + 4)(3 − x) + (2x − 3)(3 + x) (3 + x)(3 − x)2

=

54 − 36x + 6x2 + x2 + x − 12 + 2x2 + 3x − 9 (3 + x)(3 − x)2

33 − 32x + 9x2 9x2 − 32x + 33 , or 2 (3 + x)(3 − x) (x + 3)(x − 3)2

−1 x x2 + 2 1 + · + x + 1 −1 2 − x (x + 1)(x − 2)

=

6 x+4 2x − 3 − + x + 3 9 − x2 9 − 6x + x2 6 x+4 2x − 3 = − + , x + 3 (3 + x)(3 − x) (3 − x)2 LCD is (3 + x)(3 − x)2 =

x x2 + 2 1 + + x + 1 2 − x (x + 1)(x − 2)

55.

x2 − y 2 x2 − y 2 y xy = · x−y xy x−y y (x + y)(x−y) " y = x" y (x−y) x+y = x

! !

20 56.

Chapter R: Basic Concepts of Algebra a−b a−b ab b = · b (a + b)(a − b) a2 − b2 ab a "(a−b) b = /b (a + b)(a−b) a = a+b

61.

! !

57.

! !

(x+y)(x − y) (x+y) · 1 = x−y

!

!

c2 − 2c + 4 = c

ab − a a b = b b ab − b b− a a a a(b − 1) = · b b(a − 1)

a−

a2 (b − 1) = 2 b (a − 1)

(x2 + xy + y 2 )(xy) (x − y)(x2 + xy + y 2 )

xy x2 + xy + y 2 · x2 + xy + y 2 x − y xy = 1· x−y xy = x−y 62.

b2 a3 + b3 a2 + b a = 2 ab a2 − ab + b2 a − ab + b2 =

(a+b)(a2 −ab+b2 ) 1 · 2 ab a − ab + b2

a + b a2 − ab + b2 · 2 ab a − ab + b2 a+b = ab =

63.

c2 8 8 c · + 2 c2 = c2 c 2 c 2 1+ 1· + c c c 3 c +8 2 = c c+2 c c3 + 8 c = · c2 c+2 (c+2)(c2 − 2c + 4)c / /c · c(c+2)

xy x3 − y 3

=

c+

=

60.

=

! !

(a + b)(a−b) −1 · (a−b) = −a − b

x2 + xy + y 2 x3 − y 3 xy

= (x2 + xy + y 2 ) ·

b a − 2 2 ab b a = a −b Multiplying by 1 1 b−a ab − a b (a + b)(a − b) = b−a =

59.

=

x y x y − − y x y x xy = · , LCM is xy 1 1 1 1 xy + + y x y x $x y % − (xy) y x = $ % 1 1 + (xy) y x x2 − y 2 = x+y =

58.

x2 + xy + y 2 x2 + xy + y 2 = 2 2 2 x y x x y2 y − · − · y x y x x y

a 1 a· − a− a − a−1 a a = = 1 a a + a−1 a+ a· + a a 2 a −1 = 2a a +1 a a a2 − 1 = · 2 a a +1 =

64.

1 a 1 a

a2 − 1 a2 + 1

1 1 + x−1 + y −1 x y = 1 1 x−3 + y −3 + x3 y3 $1 1% 3 3 + (x y ) x y = $ % 1 1 + 3 (x3 y 3 ) x3 y x2 y 3 + x3 y 2 = y 3 + x3 = =

!

x2 y 2 (y+x) (y+x)(y 2 − yx + x2 )

!

x2 y 2 y 2 − yx + x2

Exercise Set R.5 65.

21

1 1 2 + x−3 x+3 = x−3 3 4 3 − x−1 x+2 x−1

x+3 2 · + x+3 x+3 x+2 4 · − x+2 x+2

x + 3 + 2(x − 3) (x − 3)(x + 3) = 3(x + 2) − 4(x − 1) (x − 1)(x + 2)

x−3 · x−3 x−1 · x−1

x + 3 + 2x − 6 (x − 3)(x + 3) = 3x + 6 − 4x + 4 (x − 1)(x + 2) 3x − 3 (x − 3)(x + 3) = −x + 10 (x − 1)(x + 2) =

=

66.

5x − 10 − 3x − 3 (x + 1)(x − 2) = x + 2 + 2x − 10 (x − 5)(x + 2) 2x − 13 (x + 1)(x − 2) = 3x − 8 (x − 5)(x + 2)

=

67.

2x − 13 (x − 5)(x + 2) · (x + 1)(x − 2) 3x − 8

(2x − 13)(x − 5)(x + 2) (x + 1)(x − 2)(3x − 8)

a a 1+a + 1−a a = 1−a a 1−a 1−a + a 1+a a

a 1+a 1−a · + · a a 1−a 1+a a a · + · 1+a 1+a a

a2 + (1 − a2 ) a(1 − a) = (1 − a2 ) + a2 a(1 + a) 1 a /(1 + a) = · a /(1 − a) 1 1+a = 1−a

2

x /(1 − x) x /(1 + x) 1−x = 1+x

=

69.

2 1 2 1 1 1 + + + + 2 2 a2 ab b2 = a2 ab b2 · a b , 1 1 1 1 a2 b2 − 2 − 2 a2 b a2 b LCM is a2 b2 2 2 b + 2ab + a = b2 − a2

! !

(b+a)(b + a) (b+a)(b − a) b+a = b−a =

(3x − 3)(x − 1)(x + 2) , or (x − 3)(x + 3)(−x + 10)

5(x − 2) − 3(x + 1) 3 5 − (x + 1)(x − 2) x+1 x−2 = 1 2 x + 2 + 2(x − 5) + x−5 x+2 (x − 5)(x + 2)

=

68.

(x − 1)(x + 2) 3x − 3 · (x − 3)(x + 3) −x + 10

3(x − 1)2 (x + 2) (x − 3)(x + 3)(−x + 10)

1−x +x 1−x x + x 1 + x = x(1 + x) 1+x x 1 − x2 + x2 + x 1−x x(1 − x) 1 x(1 − x) = · x(1 + x) 1 2

70.

1 1 − 2 y 2 − x2 x2 y = 2 2 1 1 y − 2xy + x2 − + x2 xy y 2 x2 y 2 Multiplying by 2 2 x y

! !

(y + x)(y−x) (y − x)(y−x) y+x = y−x =

71. When the least common denominator is used, the multiplication in the numerators is often simpler and there is usually less simplification required after the addition or subtraction is performed. 72. When there are three or more different binomial denominators, as in Exercise 65, Method 2 is usually preferable. Otherwise, some might prefer Method 1 while others will prefer Method 2. 73.

x2 + 2xh + h2 − x2 (x + h)2 − x2 = h h =

2xh + h2 h

h/(2x + h) h/ · 1 = 2x + h

=

22 74.

Chapter R: Basic Concepts of Algebra x−x−h 1 1 − x + h x = x(x + h) h h −h 1 = · x(x + h) h

78.

1+

−1 · h/ xh /(x + h) −1 = x(x + h)

=

75.

=

3x h + 3xh + h h 2

1

= 1+

= 1+

x2 − x2 − 2xh − h2 1 1 − 2 2 (x + h) x x2 (x + h)2 = h h −2xh − h2 1 = 2 · x (x + h)2 h

= 1+

h/(−2x − h) x2 h/(x + h)2 −2x − h = 2 x (x + h)2

=

=

 5  x+1 (x + 1) + (x − 1) 5 +1  x−1    x−1  x+1  =  (x + 1) − (x − 1)  −1 x−1 x−1 * +5 2x x−1 = · x−1 2 * + /x(x−1) 2 # 5 = 1·2 /(x−1) # = x5

1+

1+

1 1+

1+

1

1 x+1 x 1

x x+1

1

1+

1 2x + 1 x+1

1 x+1 1+ 2x + 1 1 3x + 2 2x + 1 2x + 1 3x + 2

5x + 3 3x + 2

n(n + 1)(n + 2) (n + 1)(n + 2) + 2·3 2

79. =



1

1 1+ x

3

h/(3x2 + 3xh + h2 ) h/ · 1 = 3x2 + 3xh + h2

77.

1+

= 1+

= 1+

=

76.

1+

1

= 1+

x3 + 3x2 h + 3xh2 + h3 − x3 (x + h)3 − x3 = h h 2

1

= =

n(n + 1)(n + 2) (n + 1)(n + 2) 3 + · , 2·3 2 3 LCD is 2 · 3 n(n + 1)(n + 2) + 3(n + 1)(n + 2) 2·3 (n + 1)(n + 2)(n + 3) 2·3

Factoring the numerator by grouping

n(n+1)(n+2)(n+3) (n+1)(n+2)(n+3) + 2·3·4 2·3

80.

n(n+1)(n+2)(n+3) + 4(n+1)(n+2)(n+3) , 2·3·4 LCD is 2 · 3 · 4 (n + 1)(n + 2)(n + 3)(n + 4) = 2·3·4

=

Exercise Set R.6

23

x2 − 9 5x2 − 15x + 45 x2 + x · + x3 + 27 x2 − 2x − 3 4 + 2x

81. =

x2 + x (x + 3)(x − 3)(5)(x2 − 3x + 9) + 2 (x + 3)(x − 3x + 9)(x − 3)(x + 1) 4 + 2x

5 x2 + x (x + 3)(x − 3)(x2 − 3x + 9) = · + (x + 3)(x − 3)(x2 − 3x + 9) x + 1 4 + 2x = 1· =

5 x +x + x + 1 4 + 2x 2

12. 13. 14. 15. 16. 17.

x2 + x 5 + x + 1 2(2 + x)

5 · 2(2 + x) + (x2 + x)(x + 1) = 2(x + 1)(2 + x)

18. 19.

=

20 + 10x + x + 2x + x 2(x + 1)(2 + x)

=

x3 + 2x2 + 11x + 20 2(x + 1)(2 + x)

21.

x2 + 2x − 3 x2 − 1 2x + 1 ÷ − x2 − x − 12 x2 − 16 x2 + 2x + 1

22.

82. =

3

2

x + 2x − 3 x − 16 2x + 1 · − 2 x2 − x − 12 x2 − 1 x + 2x + 1 2

2

(x+3) # (x−1) # (x + 4)(x−4) # − 2x + 1 = (x−4) # (x+3) # (x + 1)(x−1) # x2 + 2x + 1 2x + 1 x+4 = − x + 1 (x + 1)(x + 1) =

(x + 4)(x + 1) − (2x + 1) (x + 1)(x + 1)

x2 + 5x + 4 − 2x − 1 = (x + 1)(x + 1) =

x2 + 3x + 3 (x + 1)2

Exercise Set R.6

20.

23. 24. 25.

26. 27.

28. 29. 30.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

4 (−11)2 = | − 11| = 11 4 (−1)2 = | − 1| = 1 4 4 16y 2 = (4y)2 = |4y| = 4|y| √ 36t2 = |6t| = 6|t| 4 (b + 1)2 = |b + 1| 4 (2c − 3)2 = |2c − 3| 4 √ 3 −27x3 = 3 (−3x)3 = −3x 4 3 −8y 3 = −2y √ 4 4 81x8 = 4 (3x2 )4 = |3x2 | = 3x2 √ 4 16z 12 = |2z 3 | = 2|z 3 | = 2z 2 |z| √ √ 5 5 32 = 25 = 2

31. 32. 33.

34.

√ 5

−32 = −2 √ √ √ √ 180 = 36 · 5 = 36 · 5 = 6 5 √ √ √ 48 = 16 · 3 = 4 3 √ √ √ √ √ 72 = 36 · 2 = 36 · 2 = 6 2 √ √ √ 250 = 25 · 10 = 5 10 √ √ √ √ √ 3 54 = 3 27 · 2 = 3 27 · 3 2 = 3 3 2 √ √ √ 3 135 = 3 27 · 5 = 3 3 5 √ √ √ √ 128c2 d4 = 64c2 d4 · 2 = |8cd2 | 2 = 8 2 |c|d2 √ √ √ 162c4 d6 = 81c4 · d6 · 2 = 9c2 |d3 | 2 = √ 9 2c2 d2 |d| 4 4 √ 4 4 48x6 y 4 = 4 16x4 y 4 · 3x2 = |2xy| 3x2 = √ 4 2|x||y| 3x2 √

√ 4

√ 4 81m4 n8 · 3mn2 = 3|m|n2 3mn2 4 √ x2 − 4x + 4 = (x − 2)2 = |x − 2| 4 √ x2 + 16x + 64 = (x + 8)2 = |x + 8| √ √ √ √ √ 10 30 = 10 · 30 = 300 = 100 · 3 = √ √ √ 100 · 3 = 10 3 √ √ √ √ √ 28 14 = 28 · 14 = 14 · 2 · 14 = 14 2 √ √ √ √ √ 12 33 = 12 · 33 = 3 · 4 · 3 · 11 = 32 · 4 · 11 = √ √ 3 · 2 11 = 6 11 √ √ √ √ √ 15 35 = 15 · 35 = 3 · 5 · 5 · 7 = 5 21 4 4 4 √ √ 2x3 y 12xy = 24x4 y 2 = 4x4 y 2 · 6 = 2x2 y 6 4 4 4 √ √ 3y 4 z 20z = 60y 4 z 2 = 4y 4 z 2 · 15 = 2y 2 z 15 4 4 4 √ √ 3 3x2 y 3 36x = 3 108x3 y = 3 27x3 · 4y = 3x 3 4y 4 4 4 √ 5 5 8x3 y 4 5 4x4 y = 5 32x7 y 5 = 2xy x2 4 4 4 3 2(x + 4) 3 4(x + 4)4 = 3 8(x + 4)5 4 = 3 8(x + 4)3 · (x + 4)2 4 = 2(x + 4) 3 (x + 4)2 4 4 4 3 4(x + 1)2 3 18(x + 1)2 = 3 72(x + 1)4 4 = 3 8(x + 1)3 · 9(x + 1) 4 = 2(x + 1) 3 9(x + 1) 243m5 n10 =

√ 4

35.

5

m12 n24 = 64

36.

5

m2 n 3 m16 n24 = 28 2

6

8

5 6

$ m2 n 4 % 6 2

=

m2 n4 2

24

37. 38.

39.

40.

41.

42. 43.

44. 45.

Chapter R: Basic Concepts of Algebra 5 √ 3 √ 40m 3 3 40m √ = = 8=2 3 5m 5m 5 √ 40xy 40xy 4 √ = = 5y 8x 8x 5 5 √ 3 3x2 3x2 1 1 3 3 √ = = = 3 24x5 8x3 2x 24x5 5 √ √ 128a2 b4 128a2 b4 √ = = 8ab3 = 16ab 16ab √ √ 2 4 · 2 · a · b · b = 2b 2ab 5 5 4 3 3 64a 3 64 · a · a = 27b3 27 · b3 √ √ 3 64a3 3 a √ = 3 27b3 √ 3 4a a = 3b 6 √ √ 9x7 9 · x6 · x 3x3 x 4 = = 16y 8 4y 4 16 · y 8 6 6 7x3 7 · x2 · x = 6 36y 36 · y 6 √ √ x2 7x = 4 36y 6 √ x 7x = 6y 3 5 5 √ √ 3 y 3 y 2yz y 3 3 √ = = = 3 4 3 3 250z 125z 5z 125z √ √ √ √ 9 50 + 6 2 = 9 25 · 2 + 6 2 √ √ = 9·5 2+6 2 √ √ = 45 2 + 6 2 √ = (45 + 6) 2 √ = 51 2

√ √ √ √ √ √ 46. 11 27 − 4 3 = 11 · 3 3 − 4 3 = 33 3 − 4 3 = √ 29 3 √ √ √ √ √ √ 47. 6 20 − 4 45 + 80 = 6 4 · 5 − 4 9 · 5 + 16 · 5 √ √ √ = 6·2 5−4·3 5+4 5 √ √ √ = 12 5 − 12 5 + 4 5 √ = (12 − 12 + 4) 5 √ =4 5 √ √ √ √ √ √ 48. 2 32 + 3 8 − 4 18 = 2 · 4 2 + 3 · 2 2 − 4 · 3 2 = √ √ √ √ 8 2 + 6 2 − 12 2 = 2 2

49. = = = = 50. = = 51. =

√ √ √ 8 2x2 − 6 20x − 5 8x2 √ √ √ 8x 2 − 6 4 · 5x − 5 4x2 · 2 √ √ √ 8x 2 − 6 · 2 5x − 5 · 2x 2 √ √ √ 8x 2 − 12 5x − 10x 2 √ √ −2x 2 − 12 5x √ √ √ 3 3 2 8x2 + 5 27x2 − 3 x3 √ √ √ 3 3 4 x2 + 15 x2 − 3x x √ √ 3 19 x2 − 3x x √ 87√ √ 8 7√ 3− 2 3+ 2 7√ 82 7√ 82 3 − 2

= 3−2 =1 52. 53.

54.

55.

√ √ √ √ ( 8 + 2 5)( 8 − 2 5) = 8 − 4 · 5 = −12 √ √ √ √ (2 3 + 5)( 3 − 3 5) √ √ √ √ √ √ √ √ = 2 3· 3−2 3·3 5+ 5· 3− 5·3 5 √ √ = 2 · 3 − 6 15 + 15 − 3 · 5 √ √ = 6 − 6 15 + 15 − 15 √ = −9 − 5 15 √ √ √ √ ( 6 − 4 7)(3 6 + 2 7) √ √ = 3 · 6 + 2 42 − 12 42 − 8 · 7 √ √ = 18 + 2 42 − 12 42 − 56 √ = −38 − 10 42 √ √ √ (1 + 3)2 = 12 + 2 · 1 · 3 + ( 3)2 √ = 1+2 3+3 √ = 4+2 3

√ √ √ 56. ( 2 − 5)2 = 2 − 10 2 + 25 = 27 − 10 2 √ √ √ √ √ √ 57. ( 5 − 6)2 = ( 5)2 − 2 5 · 6 + ( 6)2 √ = 5 − 2 30 + 6 √ = 11 − 2 30 √ √ √ √ 58. ( 3 + 2)2 = 3 + 2 6 + 2 = 5 + 2 6

59. We use the Pythagorean theorem to find b, the airplane’s horizontal distance from the airport. We have a = 3700 and c = 14, 200. c2 = a2 + b2 14, 2002 = 37002 + b2 201, 640, 000 = 13, 690, 000 + b2 187, 950, 000 = b2 13, 709.5 ≈ b

The airplane is about 13, 709.5 ft horizontally from the airport.

Exercise Set R.6 60.

25

%& &&&& c %%%% &&& a %%% &&&& %%%% ( ' b

64.

x

2 mi = 2 · 5280 ft = 10, 560 ft

S

a2 = (5281)2 − (5280)2 a2 = 10, 561

A= A= A= 62.

x2 = 25

$ a %2

= a2

2

Pythagorean theorem

x=5

a2 = a2 4

3a2 4 5 3a2 h= 4 a√ 3 h= 2 the result of part (a) we have 1 · base · height 2 $a a % 1 a√ a· 3 + =a 2 2 2 2 2√ a 3 4

65. 66. 67. 68. 69.

c2 = s2 + s2 c2 = 2s2 √ c=s 2

70. Length of third side

63.

71.

x ) )

√ )8 2 ) ) )

x

x √ x2 + x2 = (8 2)2 2x = 128 2

x2 = 64 x=8

)C * * * x *y * ) )* x D x R

4x2 = 100

h2 =

b) Using

x )

(2x)2 = 100

a ≈ 102.8 ft

h2 +

B x *) ) * y* )y ) *

Q

x

* A* ) ) ) x y)

2 mi + 2 ft = 10, 562 ft 1 Then b = · 10, 560 ft = 5280 ft and 2 1 c = · 10, 562 ft = 5281 ft. 2 a2 = c2 − b2

61. a) h2 +

P

x ) Pythagorean theorem 72.

A = y 2 = x2 + x2 = 52 + 52 = 25 + 25 = 50 ft2 5 5 5 √ √ 2 2 3 6 6 6 = · = =√ = 3 3 3 9 3 9 5 5 5 √ 3 3 7 21 21 = · = = 7 7 7 49 7 √ √ √ √ √ 3 3 3 3 5 5 32 10 10 √ = √ · √ = √ = 3 3 3 3 2 4 4 2 8 √ √ √ √ √ 3 3 3 3 7 7 35 35 35 √ = √ · √ = √ = 3 3 3 3 5 25 25 5 125 5 5 5 √ 3 3 48 3 16 3 16 3 48 = · = = √ = 3 9 9 3 27 27 √ √ 3 8·6 236 = 3 3 5 5 5 √ 3 25 75 75 3 3 3 3 3 = · = = 5 5 25 125 5 √ 6 6 3− 5 √ = √ · √ 3+ 5 3+ 5 3− 5 √ 8 7 6 3− 5 = 9−5 √ 8 7 6 3− 5 = 4 √ 8 7 √ 3 3− 5 9−3 5 = = 2 2 √ 2 3+1 2 √ = √ ·√ 3−1 3−1 3+1 √ /( 3 + 1) 2 = /·1 2 √ = 3+1

26 73.

74.

75.

76. 77. 78. 79. 80. 81. 82. 83.

84.

Chapter R: Basic Concepts of Algebra √ √ √ √ 1− 2 2 3+ 6 1− 2 √ √ = √ √ · √ √ 2 3− 6 2 3− 6 2 3+ 6 √ √ √ √ 2 3 + 6 − 2 6 − 12 = 4·3−6 √ √ √ √ 2 3+ 6−2 6−2 3 = 12 − 6 √ √ − 6 6 = , or − 6 6 √ √ √ √ 5+4 5+4 2−3 7 √ √ = √ √ ·√ √ 2+3 7 2+3 7 2−3 7 √ √ √ √ 10 − 3 35 + 4 2 − 12 7 = 2−9·7 √ √ √ √ 10 − 3 35 + 4 2 − 12 7 = −61 √ √ 6 6 m+ n √ = √ √ ·√ √ √ m− n m− n m+ n √ √ 6( m + n) √ = √ 2 ( m) − ( n)2 √ √ 6 m+6 n = m−n √ √ √ √ 3 3 v− w 3 v−3 w √ √ =√ √ ·√ √ = v−w v+ w v+ w v− w √ √ √ √ 12 12 3 36 6 = ·√ = √ = √ 5 5 3 5 3 5 3 √ √ √ √ 10 50 50 2 100 = ·√ = √ = √ 3 3 2 3 2 3 2 5 5 5 √ 3 49 7 343 3 7 3 7 3 343 = · = = √ = √ 3 3 2 2 49 98 98 98 5 5 5 4 8 2 3 2 3 2 3 = · = = √ 3 5 5 4 20 20 √ √ √ √ 11 11 11 121 11 √ = √ ·√ = √ =√ 3 3 11 33 33 √ √ √ √ 5 5 5 25 5 √ =√ ·√ =√ =√ 2 2 5 10 10 √ √ √ 9− 5 9+ 5 9− 5 √ = √ · √ 3− 3 3− 3 9+ 5 √ 92 − ( 5)2 √ √ √ = 27 + 3 5 − 9 3 − 15 81 − 5 √ √ √ = 27 + 3 5 − 9 3 − 15 76 √ √ √ = 27 + 3 5 − 9 3 − 15 √ √ √ 8− 6 8+ 6 8− 6 √ = √ · √ 5− 2 5− 2 8+ 6 64 − 6 √ √ √ = 40 + 5 6 − 8 2 − 12 58 √ √ √ = 40 + 5 6 − 8 2 − 2 3

85.

86.

√ √ √ √ √ a+ b a+ b a− b √ = ·√ 3a 3a a− b √ √ ( a)2 − ( b)2 √ = √ 3a( a − b) a−b √ = √ 3a a − 3a b √ √ √ √ √ √ p− q p− q p+ q = · √ √ √ √ 1+ q 1+ q p+ q p−q = √ √ √ p + q + pq + q √

87. x3/4 =

√ 4

x3 4 = 5 y2

88. y 2/5

89. 163/4 = (161/4 )3 =

7√ 4

83

16

= 23 = 8

√ 90. 47/2 = ( 4)7 = 27 = 128 91. 125−1/3 =

1 1 1 = √ = 3 5 1251/3 125

7√ 5

8−4 1 32 = 2−4 = 16 √ 5 √ 4 a5/4 a4a a5 a 4 √ 93. a5/4 b−3/4 = 3/4 = √ = , or a 4 3 4 3 b3 b b b 6 x2 2/5 −1/5 94. x y = 5 y 92. 32−4/5 =

√ √ √ 3 3 3 m5 n7 = m5 n7 = mn2 m2 n 4 4 4 4 = 6 p7 6 q 11 = 6 p7 q 11 = pq 6 pq 5

95. m5/3 n7/3 = 96. p7/6 q 11/6 97. 98.

7√ 4 √ 5

85

13

=

√ 3

√ 4

135 = 135/4

173 = 173/5

√ 3

202 = 202/3 7 √ 84 100. 5 12 = 124/5 99.

101. 102. 103. 104. 105. 106.

4 √ 3

11 =

4 √ 3

7√

11

81/3

= (111/2 )1/3 = 111/6

7 = (71/4 )1/3 = 71/12 √ √ 5 3 5 = 51/2 · 51/3 = 51/2+1/3 = 55/6 √ √ 3 2 2 = 21/3 · 21/2 = 25/6 √ 5 322 = 322/5 = (321/5 )2 = 22 = 4 √ 3 642 = 642/3 = (641/3 )2 = 42 = 16 4

107. (2a3/2 )(4a1/2 ) = 8a3/2+1/2 = 8a2 √ 108. (3a5/6 )(8a2/3 ) = 24a3/2 = 24a a 109.

$ x6 %1/2 $ x6 %1/2 x3 x3 b2 = 2 −4 = −2 , or −4 9b 3 b 3b 3

Exercise Set R.7

27

√ √ 3 x1/3 x y3x = 1/2 −1 = −1 , or 2y 2 4 y

110.

$ x2/3 %1/2

111.

x2/3 y 5/6 √ = x2/3−(−1/3) y 5/6−1/2 = xy 1/3 = x 3 y x−1/3 y 1/2

112.

√ a1/2 b5/8 4 = a1/4 b1/4 = ab 1/4 3/8 a b

4y −2

1 2

5 2

113. (m1/2 n5/2 )2/3 = m 2 · 3 n 2 · 3 = m1/3 n5/3 = √ √ √ √ 3 3 3 3 m n5 = mn5 = n mn2 4 114. (x5/3 y 1/3 z 2/3 )3/5 = xy 1/5 z 2/5 = x 5 yz 2 115.

116.

117.

119.

120.

121.

127.

4

1 + x2 + √

1 1 + x2

√ 4 1 + x2 1 1 + x2 √ √ 1 + x2 · + · 2 2 1+x 1+x 1 + x2 √ √ (1 + x2 ) 1 + x2 1 + x2 = + 1 + x2 1 + x2 √ (2 + x2 ) 1 + x2 = 1 + x2 =

m2/3 (m7/4 − m5/4 ) = m29/12 − m23/12 = √ √ √ √ 12 12 12 12 m29 − m23 = m2 m5 − m m11 √ √ 3 6 2 = 61/3 21/2 = 62/6 23/6

√ √ 2 4 8 = 21/2 (23 )1/4 = 21/2 23/4 = 25/4 = √ √ 4 25 = 2 4 2 4 √ 4 xy 3 x2 y = (xy)1/4 (x2 y)1/3 = (xy)3/12 (x2 y)4/12 . /1/12 = (xy)3 (x2 y)4 . /1/12 = x3 y 3 x8 y 4 4 = 12 x11 y 7 √ √ 3 ab2 ab = (ab2 )1/3 (ab)1/2 = (ab2 )2/6 (ab)3/6 = √ √ 4 6 6 6 (ab2 )2 (ab)3 = a5 b7 = b a5 b 4 √ 7 √ 81/3 3 a4 a3 = a4 a3 = (a4 a3/2 )1/3

√ (x + y)2 3 x + y (x + y)2/4 (x + y)1/3 4 = = 3 (x + y)3/2 (x + y) 1 (x + y)−2/3 = 4 3 (x + y)2 4 4

125. Choose √ for a and b (a %= 0, b %= 0) and show √ specific√values that √ a + b %= a+ let a = 9 and b = 16. √ b. For example √ √ Then 9 + 16 = 25 = 5, but 9 + 16 = 3 + 4 = 7. √ √ √ 126. Observe that 26 > 25, so 26 > √ √ 25, or 26 > 5. Then 10 26 − 50 > 10 · 5 − 50, or 10 26 − 50 > 0.

a3/4 (a2/3 + a4/3 ) = a3/4+2/3 + a3/4+4/3 = √ √ 12 12 a17 + a25 = a17/12 + a25/12 = √ √ 12 a a5 + a2 12 a

= (62 23 )1/6 √ = 6 36 · 8 √ = 6 288 118.

124.

128.

4 x2 1 − x2 − √ 2 1 − x2 √ 4 x2 1 − x2 = 1 − x2 − 2(1 − x2 )

√ √ 2(1 − x2 ) 1 − x2 − x2 1 − x2 2(1 − x2 ) √ 2 (2 − 3x ) 1 − x2 = 2(1 − x2 )

=

129. 130.

9 4

√

a

a

:√a

$ √ %√ a = a a/2 = aa/2

(2a3 b5/4 c1/7 )4 16a12 b5 c4/7 = (54a−2 b2/3 c6/5 )−1/3 54−1/3 a2/3 b−2/9 c−2/5 √ = 16 3 54a34/3 b47/9 c34/35 = 24 · 3 · 21/3 a34/3 b47/9 c34/35

= (a11/2 )1/3 = a11/6 √ 6 = a11 √ 6 = a a5

122.

123.

4 √ 3 a3 a2 = (a3 · a2/3 )1/2 = (a11/3 )1/2 = a11/6 = √ √ 6 6 a11 = a a5

= 3 · 213/3 a34/3 b47/9 c34/35 , or 48 · 21/3 a34/3 b47/9 c34/35

Exercise Set R.7 1.

(a + x)26/12 (a + x)3/12

= (a + x)23/12 4 = 12 (a + x)23 4 = (a + x) 12 (a + x)11

6x − 15 = 45

6x = 60

4 4 (a + x)3 3 (a + x)2 (a + x)3/2 (a + x)2/3 √ = 4 (a + x)1/4 a+x =

Rationalizing the denominator

x = 10

Adding 15 Dividing by 6

The solution is 10. 2.

4x − 7 = 81

4x = 88 x = 22

28 3.

Chapter R: Basic Concepts of Algebra 5x − 10 = 45

14.

5x = 55

Adding 10

x = 11

−8x = −20 5 x= 2

Dividing by 5

The solution is 11. 4.

15.

6x − 7 = 11

9t = −9 t = −1

Subtracting 4 Dividing by 9

The solution is −3. 16.

17.

8x + 48 = 3x − 12 5x = −60

−8x = −8

Subtracting 3x Dividing by 5

18.

y = −5

15x + 40 = 8x − 9

19.

−3x + 14 = 14

7y − 1 = 23 − 5y

12y − 1 = 23

The solution is 0.

Dividing by 12

The solution is 2.

20.

12 = 5y 12 =y 5

6x = 30 x=5

The solution is −1. 12.

Subtracting 12x

24 = 10t + 25

9t − 4 = 14 + 15t

The solution is − 22.

t = −3

5 − 4a = a − 13 5 − 5a = −13 −5a = −18 18 a= 5

The solution is

9 = 4(3y − 2) 9 = 12y − 8

Subtracting a Subtracting 5

18 . 5

24 = 5(2t + 5) −1 = 10t 1 =t − 10

Dividing by −9

−6t = 18

13.

21.

Adding 4

x = −1

3(y + 4) = 8y 3y + 12 = 8y

3x − 15 = 15 − 3x

3x − 4 = 5 + 12x

Subtracting 14

x=0

Adding 1

y=2

Subtracting 5x

−3x = 0

Adding 5y

12y = 24

−9x = 9

2(x + 7) = 5x + 14 2x + 14 = 5x + 14

x = −7

−9x − 4 = 5

20 − 4y = 10 − 6y 2y = −10

7x = −49

11.

Subtracting 11

The solution is 1.

Subtracting 48

x = −12

Subtracting 5x

x=1

The solution is −12.

10.

11 − 3x = 5x + 3 11 − 8x = 3

5x + 48 = −12

9.

5x − 8 = 2x − 8 3x = 0

x = −4

8.

Dividing by 2

x=0

5x + 7 = −13 5x = −20

7.

Adding 7

m = −3

The solution is −1. 6.

Subtracting m

2m = −6

x=3 9t + 4 = −5

3m − 7 = −13 + m 2m − 7 = −13

6x = 18

5.

6 − 7x = x − 14

Dividing by −5

17 = 12y 17 =y 12

Subtracting 25 Dividing by 10 1 . 10

Exercise Set R.7 23.

29 30.

5y − (2y − 10) = 25 5y − 2y + 10 = 25 3y + 10 = 25

Subtracting 10

y=5

Dividing by 3

The solution is 5. 24.

(y − 9)(y + 5) = 0

Collecting like terms

3y = 15

y − 9 = 0 or y + 5 = 0 y = 9 or

31.

8x − (3x − 5) = 40

x = 0 or

32.

x=7

Collecting like terms

22x + 42 = 9

Adding x

22x = −33 3 x=− 2 3 The solution is − . 2

33.

or y + 3 = 0

y = −3 or

The solution is −3. 34.

n2 + 4n + 4 = 0 n+2 = 0

or n + 2 = 0

n = −2 or

x = −3

35.

12y − 10 = 5y + 10 7y − 10 = 10

7y = 20 20 y= 7 20 . The solution is 7

n = −2

x2 + 100 = 20x x − 20x + 100 = 0

Subtracting 20x

2

4(3y − 1) − 6 = 5(y + 2) 12y − 4 − 6 = 5y + 10

y = −3

(n + 2)(n + 2) = 0

19x = −57

(x − 10)(x − 10) = 0

x − 10 = 0 or x − 10 = 0

Collecting like terms

x = 10 or

Subtracting 5y Adding 10 Dividing by 7

x = 10

The solution is 10. 36.

y 2 + 25 = 10y y − 10y + 25 = 0 2

(y − 5)(y − 5) = 0

y − 5 = 0 or y − 5 = 0

3(2n − 5) − 7 = 4(n − 9) 6n − 15 − 7 = 4n − 36 6n − 22 = 4n − 36

y = 5 or

37.

2n = −14

y=5

x2 − 4x − 32 = 0

(x − 8)(x + 4) = 0

n = −7

x − 8 = 0 or x + 4 = 0 x = 8 or

2

x=4

The solutions are −7 and 4.

x = −4

The solutions are 8 and −4.

Factoring

or x − 4 = 0 Principle of zero products

x = −7 or

y 2 + 6y + 9 = 0 y+3 = 0

Dividing by 22

18x + 72 = 15 − x

x+7 = 0

t=9

(y + 3)(y + 3) = 0

18x + 72 = 20 − x − 5

(x + 7)(x − 4) = 0

t2 − 9t = 0

t = 0 or

Subtracting 42

9(2x + 8) = 20 − (x + 5)

x + 3x − 28 = 0

x=8

t = 0 or t − 9 = 0

21x + 42 = 9 − x

29.

Principle of zero products

t(t − 9) = 0

7(3x + 6) = 11 − (x + 2) 21x + 42 = 11 − x − 2

28.

Factoring

The solutions are 0 and 8.

5x = 35

27.

x2 − 8x = 0

x = 0 or x − 8 = 0

5x + 5 = 40

26.

y = −5

x(x − 8) = 0

8x − 3x + 5 = 40

25.

y 2 − 4y − 45 = 0

38.

t2 + 12t + 27 = 0 (t + 9)(t + 3) = 0 t+9 = 0

or t + 3 = 0

t = −9 or

t = −3

30 39.

Chapter R: Basic Concepts of Algebra 45.

3y 2 + 8y + 4 = 0

14 = x2 − 5x

(3y + 2)(y + 2) = 0 3y + 2 = 0

The solutions are − 40.

0 = x2 − 5x − 14

or y + 2 = 0

3y = −2 or 2 y = − or 3

0 = (x − 7)(x + 2)

y = −2

x − 7 = 0 or x + 2 = 0

y = −2

x = 7 or

2 and −2. 3

41.

0 = x2 − 2x − 24

or 3y + 1 = 0

3y = −4 or 4 y = − or 3

0 = (x − 6)(x + 4)

3y = −1 1 y=− 3

12z + z = 6 2

x − 6 = 0 or x + 4 = 0 x = 6 or

47.

12z 2 + z − 6 = 0

3z = 2 2 z= 3 3 2 The solutions are − and . 4 3

x = −6 or

48.

(y + 9)(y − 9) = 0 y+9 = 0

12a − 5a − 28 = 0

(t + 5)(t − 5) = 0 t+5 = 0 51.

2x2 − 20 = 0

x2 = 10

√ 10 or x = − 10 Principle of square roots √ √ √ The solutions are 10 and − 10, or ± 10. x=

(3n + 2)(7n − 5) = 0

7n = 5 5 n= 7

t=5

2x2 = 0

21n − n − 10 = 0

or 7n − 5 = 0

or t − 5 = 0

t = −5 or

2

3n = −2 or 2 n = − or 3

t2 = 25 t2 − 25 = 0

or 4a − 7 = 0

21n2 − 10 = n

3n + 2 = 0

z = 12

The solutions are −12 and 12. 50.

4a = 7 7 a= 4 7 4 The solutions are − and . 3 4

or z − 12 = 0

z = −12 or

3a = −4 or 4 a = − or 3

44.

z 2 = 144

z + 12 = 0

x=2

(3a + 4)(4a − 7) = 0

y=9

(z + 12)(z − 12) = 0

x=2

12a − 28 = 5a

3a + 4 = 0

or y − 9 = 0

z 2 − 144 = 0

or x − 2 = 0

2

2

y 2 − 81 = 0

49.

(6x + 5)(x − 2) = 0

43.

x=6

y = −9 or

6x2 − 7x − 10 = 0

6x = −5 or 5 x = − or 6

or x − 6 = 0

The solutions are −6 and 6.

6x2 − 7x = 10

6x + 5 = 0

x2 − 36 = 0

x+6 = 0

or 3z − 2 = 0

4z = −3 or 3 z = − or 4

42.

x = −4

(x + 6)(x − 6) = 0

(4z + 3)(3z − 2) = 0 4z + 3 = 0

24 = x(x − 2) 24 = x2 − 2x

(3y + 4)(3y + 1) = 0 3y + 4 = 0

x = −2

The solutions are 7 and −2. 46.

9y 2 + 15y + 4 = 0

14 = x(x − 5)

52.

√

3y 2 − 15 = 0

3y 2 = 15

y=

√

y2 = 5

√ 5 or y = − 5

Exercise Set R.7 53.

31 59.

6z 2 − 18 = 0

x − {3x − [2x − (−2x + 1)]} = x + 7

z2 = 3

√ z = 3 or z = − 3 √ √ √ The solutions are 3 and − 3, or ± 3. √

54.

x − {3x − [2x − (5x − (7x − 1))]} = x + 7 x − {3x − [2x − (5x − 7x + 1)]} = x + 7

6z = 18 2

x − {3x − [2x + 2x − 1]} = x + 7 x − {3x − [4x − 1]} = x + 7

x − {3x − 4x + 1} = x + 7

5x2 − 75 = 0

x − {−x + 1} = x + 7

5x2 = 75

x=

√

x+x−1 = x+7

x2 = 15

2x − 1 = x + 7

√ 15 or x = − 15

55. To eliminate a term from one side of an equation, add the opposite of the term. To eliminate a coefficient, multiply by its reciprocal. 56. If the solutions of a quadratic equation are −3 and 4, then we have: x = −3 or x=4 x+3 = 0

x−1 = 7

x=8

The solution is 8. 60.

23−2[4+3(x−1)]+5[x−2(x+3)]=7{x−2[5−(2x+3)]} 23 − 2[4 + 3x − 3] + 5[x − 2x − 6]=7{x − 2[5 − 2x − 3]} 23 − 2[3x + 1] + 5[−x − 6]=7{x − 2[−2x + 2]} 23 − 6x − 2 − 5x − 30=7{x + 4x − 4}

or x − 4 = 0.

−11x − 9=7{5x − 4}

Then the equation can be written (x + 3)(x − 4) = 0, or x2 − x − 12 = 0. 57.

−11x − 9=35x − 28 −46x=−19 19 x= 46

3[5 − 3(4 − t)] − 2 = 5[3(5t − 4) + 8] − 26 3[5 − 12 + 3t] − 2 = 5[15t − 12 + 8] − 26 3[−7 + 3t] − 2 = 5[15t − 4] − 26 −21 + 9t − 2 = 75t − 20 − 26

61.

x(5x + 6)(4x + 1)(3x − 2) = 0

9t − 23 = 75t − 46

x = 0 or 5x+6 = 0

−66t − 23 = −46

−66t = −23 23 t= 66 23 The solution is . 66 58.

6[4(8 − y) − 5(9 + 3y)] − 21 = −7[3(7 + 4y) − 4]

or 3x−2 = 0

5x = −6 or 4x = −1 or 1 6 x = − or x = 0 or x = − or 5 4 1 2 6 The solutions are 0, − , − , and . 5 4 3 62.

3x = 2 2 x= 3

(3x2 + 7x − 20)(x2 − 4x) = 0

(3x − 5)(x + 4)(x)(x − 4) = 0

3x − 5 = 0 or x + 4 = 0 or x = 0 or x − 4 = 0 5 x = −4 or x = 0 or x=4 x = or 3

6[−13 − 19y] − 21 = −7[17 + 12y] −78 − 114y − 21 = −119 − 84y −30y = −20 2 y= 3

or 4x+1 = 0

x = 0 or

6[32 − 4y − 45 − 15y] − 21 = −7[21 + 12y − 4]

−114y − 99 = −119 − 84y

(5x2 + 6x)(12x2 − 5x − 2) = 0

63.

3x3 + 6x2 − 27x − 54 = 0

3(x3 + 2x2 − 9x − 18) = 0

3[x2 (x + 2) − 9(x + 2)] = 0

Factoring by grouping

3(x + 2)(x2 − 9) = 0

3(x + 2)(x + 3)(x − 3) = 0

x+2 = 0

or x + 3 = 0

x = −2 or

or x − 3 = 0

x = −3 or

The solutions are −2, −3, and 3.

x=3

32

Chapter R: Basic Concepts of Algebra

64.

2x3 + 6x2 = 8x + 24

16.

2x3 + 6x2 − 8x − 24 = 0

17. The exponent is positive, so the number is greater than 10. We move the decimal point 6 places to the right.

2(x3 + 3x2 − 4x − 12) = 0

2[x (x + 3) − 4(x + 3)] = 0 2

3.261 × 106 = 3, 261, 000

2(x + 3)(x2 − 4) = 0

2(x + 3)(x + 2)(x − 2) = 0

x+3 = 0

or x + 2 = 0

x = −3 or

or x − 2 = 0

x = −2 or

18. The exponent is negative, so the number is between 0 and 1. We move the decimal point 4 places to the left. 4.1 × 10−4 = 0.00041

x=2

Chapter R Review Exercises 1. True 2. For any real number a, a %= 0, and any integers m and n, am · an = am+n . Thus the given statement is false.

19. Position the decimal point 2 places to the right, between the 1 and the 4. Since 0.01432 is a number between 0 and 1, the exponent must be negative. 0.01432 = 1.432 × 10−2 20. Position the decimal point 4 places to the left, between the 4 and the 3. Since 43,210 is greater than 10, the exponent must be positive. 43, 210 = 4.321 × 104

3. True 4. True

21.

5. Integers: 12, −3, −1, −19, 31, 0 6. Natural numbers: 12, 31 4 2 1 −43.89, 12, −3, − , −1, − , 7 , −19, 31, 0 5 3 3

8. Real numbers: All of them √ √ 9. Irrational numbers: 7, 3 10

= 7.8125 × 10−22 22.

= (5.46 × 10) × 10−33 = 5.46 × 10−32

23.

|3 − (−7)| = |3 + 7| = |10| = 10

15.

53 − [2(42 − 32 − 6)]3

= 53 − [2(16 − 9 − 6)]3 = 53 − [2(1)]3 = 53 − [2]3 = 125 − 8 = 117

(7a2 b4 )(−2a−4 b3 ) = 7(−2)a2+(−4) b4+3

12. | − 3.5| = 3.5 13. |16| = 16

(8.4 × 10−17 )(6.5 × 10−16 )

= 54.6 × 10−33

10. Whole numbers: 12, 31, 0 11. [−3, 5)

2.5 × 10−8 3.2 × 1013 2.5 10−8 = × 13 3.2 10 = 0.78125 × 10−21

= (7.8125 × 10−1 ) × 10−21

7. Rational numbers:

14. | − 7 − 3| = | − 10| = 10, or

34 − (6 − 7)4 34 − (−1)4 81 − 1 80 = = = = −10 3 4 2 −2 23 − 24 8 − 16 −8

= −14a−2 b7 , or −

14b7 a2

54x6 y −4 z 2 9x−3 y 2 z −4 54 6−(−3) −4−2 2−(−4) = y z x 9 6x9 z 6 = 6x9 y −6 z 6 , or y6 √ √ 4 25. 4 81 = 34 = 3 √ 26. 5 −32 = −2 24.

Chapter R Review Exercises 27.

1 b− b − a−1 a = 1 a − b−1 a− b a 1 b· − a a = b 1 a· − b b ab 1 − a = a ab 1 − b b ab − 1 a = ab − 1 b ab − 1 b = · a ab − 1 (ab−1)b = a(ab−1) b = a x2 y2 + y x y 2 − xy + x2

! !

28.

x3 + y 3 xy = 2 y − xy + x2

29.

= 3−7

31.

32.

√

(5a + 4b)(2a − 3b)

= 10a2 − 15ab + 8ab − 12b2

= 10a2 − 7ab − 12b2 34.

(5xy 4 − 7xy 2 + 4x2 −3)−(−3xy 4 + 2xy 2 −2y+ 4)

= (5xy 4 −7xy 2 + 4x2 − 3)+(3xy 4 −2xy 2 + 2y − 4)

= (5 + 3)xy 4 + (−7 − 2)xy 2 + 4x2 + 2y + (−3 − 4) = 8xy 4 − 9xy 2 + 4x2 + 2y − 7

35.

x3 + 2x2 − 3x − 6

= x2 (x + 2) − 3(x + 2) = (x + 2)(x2 − 3)

36.

12a3 − 27ab4

= 3a(4a2 − 9b4 )

= 3a(2a + 3b2 )(2a − 3b2 ) 37.

24x + 144 + x2 = x2 + 24x + 144 = (x + 12)2

38.

9x3 + 35x2 − 4x

= x(9x2 + 35x − 4) 39.

(x + t)(x2 − xt + t2 )

= (x + t)(x2 ) + (x + t)(−xt) + (x + t)(t2 ) = x3 + x2 t − x2 t − xt2 + xt2 + t3

8x3 − 1

= (2x)3 − 13

= (2x − 1)(4x2 + 2x + 1) 40.

27x6 + 125y 6 = (3x2 + 5y 2 )(9x4 − 15x2 y 2 + 25y 4 )

41.

6x3 + 48 = 6(x3 + 8) = 6(x + 2)(x2 − 2x + 4)

42.

= −4

√ 2)2 = 25x4 − 10 2x2 + 2 √ √ √ 25 25 5 8 5+ √ = 8 5+ √ · √ 5 5 5 √ √ 25 5 = 8 5+ 5 √ √ = 8 5+5 5 √ = 13 5

= x3 + t3

33.

= x(9x − 1)(x + 4)

x3 + y 3 1 = · 2 xy y − xy + x2 (x + y)(x2 − xy + y 2 ) = xy(y 2 − xy + x2 ) x + y x2 − xy + y 2 = · 2 xy x − xy + y 2 x+y = xy √ √ √ √ √ √ ( 3 − 7)( 3 + 7) = ( 3)2 − ( 7)2

30. (5x2 −

33

4x3 − 4x2 − 9x + 9

= 4x2 (x − 1) − 9(x − 1) = (x − 1)(4x2 − 9)

= (x − 1)(2x + 3)(2x − 3) 43. 9x2 − 30x + 25 = (3x − 5)2 44. 18x2 − 3x + 6 = 3(6x2 − x + 2) 45. a2 b2 − ab − 6 = (ab − 3)(ab + 2) 46.

3x2 − 12 x−2 ÷ x2 + 4x + 4 x + 2 x+2 3x2 − 12 = 2 · x + 4x + 4 x − 2 3(x+2)(x−2)(x+2) = (x+2)(x+2)(x−2) =3

! ! ! ! ! !

34 47.

Chapter R: Basic Concepts of Algebra x 4 − x2 + 9x + 20 x2 + 7x + 12 4 x − = (x + 5)(x + 4) (x + 4)(x + 3)

LCD is (x + 5)(x + 4)(x + 3) x x+3 x x+5 = · − · (x + 5)(x + 4) x + 3 (x + 4)(x + 3) x + 5 x(x + 3) − 4(x + 5) = (x + 5)(x + 4)(x + 3) x2 + 3x − 4x − 20 = (x + 5)(x + 4)(x + 3) x2 − x − 20 = (x + 5)(x + 4)(x + 3) (x − 5)(x+4) = (x + 5)(x+4)(x + 3) x−5 = (x + 5)(x + 3) 4 4 y 5 · 3 y 2 = (y 5 )1/2 (y 2 )1/3 = y 5/2 · y 2/3 = y 19/6 48. 4 √ = 6 y 19 = y 3 6 y

55.

2x − 7 = −9 2x = −2 x = −1

The solution is −1. 56.

49.

√ (a + b)3 3 a + b 4 6 (a + b)7

(a + b)3/2 (a + b)1/3 (a + b)7/6 = (a + b)3/2+1/3−7/6 =

x=3

57.

√ 5

12x − 6 = −x − 7 13x − 6 = −7

13x = −1 1 x=− 13

The solution is − 58.

y+8 = 0 59.

x2 − x = 20

x − 5 = 0 or x + 4 = 0

√ 5

x = 5 or

The guy wire is about 18.8 ft long.

x = −4

The solutions are 5 and −4. 60.

2x2 + 11x − 6 = 0

(2x − 1)(x + 6) = 0

2x − 1 = 0 or x + 6 = 0 1 or x = −6 x= 2 61.

x(x − 2) = 3

x2 − 2x = 3

x2 − 2x − 3 = 0

(x + 1)(x − 3) = 0 x+1 = 0

or x − 3 = 0

x = −1 or

x=3

The solutions are −1 and 3.

c2 = 353

x=7

y = −8

x − x − 20 = 0 2

c2 = 64 + 289

2x = 14

or y + 8 = 0

y = −8 or

c2 = 82 + 172

2x − 7 = 7

y 2 + 16y + 64 = 0 (y + 8)(y + 8) = 0

53. a = 8 and b = 17. Find c. c2 = a2 + b2

54.

1 . 13

(x − 5)(x + 4) = 0

50. b7/5 = b7 = b b2 5 " 32 16 #1/8 32 n16 m n m4 n 2 8 m 51. = = 8 8 3 3 3 √ √ √ 4− 3 5− 3 4− 3 52. √ = √ · √ 5+ 3 5+ 3 5− 3 √ √ 20 − 4 3 − 5 3 + 3 = 25 − 3 √ 23 − 9 3 = 22

c ≈ 18.8

6(2x − 1) = 3 − (x + 10) 12x − 6 = 3 − x − 10

= (a + b)9/6+2/6−7/6 = (a + b)2/3 4 = 3 (a + b)2

8 − 3x = −7 + 2x −5x = −15

! !

4

5x − 7 = 3x − 9

62.

y 2 − 16 = 0

(y + 4)(y − 4) = 0 y+4 = 0

or y − 4 = 0

y = −4 or

y=4

Chapter R Test 63.

35 " #240 0.075 0.075 1 +  12 12 M = $117, 000  #240  " 0.075 −1 1+ 12 ≈ $942.54

n2 − 7 = 0



n2 = 7 √ √ n = 7 or n = − 7 √ √ √ The solutions are 7 and − 7, or ± 7. 64.

128 ÷ (−2)3 ÷ (−2) · 3 = 128 ÷ (−8) ÷ (−2) · 3 = −16 ÷ (−2) · 3 = 8·3 = 24 Answer B is correct.

65.

9x2 − 36y 2 = 9(x2 − 4y 2 )

= 9[x2 − (2y)2 ]

66. Anya is probably not following the rules for order of operations. She is subtracting 6 from 15 first, then dividing the difference by 3, and finally multiplying the quotient by 4. The correct answer is 7. 67. When the number 4 is raised to a positive integer power, the last digit of the result is 4 or 6. Since the calculator returns 4.398046511 × 1012 , or 4,398,046,511,000, we can conclude that this result is an approximation. 68. Substitute $98, 000 − $16, 000, or $82,000, for P , 0.065 for r, and 12 · 25, or 300, for n and perform the resulting computation. " #n   r r 1 +   12  #12 M =P n  " r 1+ −1 12 " #300   0.065 0.065 1 +   12 12  = $82, 000  #300   " 0.065 1+ −1 12 ≈ $553.67 69. Substitute $124, 000 − $20, 000, or $104,000 for P , 0.0575 for r, and 12 · 30, or 360, for n and perform the resulting computation. " #n   r r 1 +   12  #12 M =P n  " r −1 1+ 12 " #360   0.0575 0.0575 1 +  12  12  = $104, 000  #360  "  0.0575 1+ −1 12 ≈ $606.92 70. P = $135, 000 − $18, 000 = $117, 000, r = 0.075, n = 12 · 20 = 240.

  

71. P = $151, 000 − $21, 000 = $130, 000, r = 0.0625, n = 12 · 25 = 300. " #300   0.0625 0.0625 1+   12 12  M = $130, 000  #300   " 0.0625 −1 1+ 12 ≈ $857.57 72.

(xn + 10)(xn − 4) = (xn )2 − 4xn + 10xn − 40 = x2n + 6xn − 40

= 9(x + 2y)(x − 2y)

Answer C is correct.



73.

(ta + t−a )2 = (ta )2 + 2 · ta · t−a + (t−a )2 = t2a + 2 + t−2a

74.

(y b − z c )(y b + z c ) = (y b )2 − (z c )2 = y 2b − z 2c

75.

(an −bn )3 = (an −bn )(an −bn )2

= (an − bn )(a2n − 2an bn + b2n )

= a3n−2a2n bn +an b2n−a2n bn +2an b2n−b3n = a3n − 3a2n bn + 3an b2n − b3n

76.

y 2n + 16y n + 64 = (y n )2 + 16y n + 64 = (y n + 8)2

77.

x2t − 3xt − 28 = (xt )2 − 3xt − 28 = (xt − 7)(xt + 4)

78.

m6n − m3n = m3n (m3n − 1)

= m3n [(mn )3 − 13 ]

= m3n (mn − 1)(m2n + mn + 1)

Chapter R Test 1. a) Integers: −8, 0, 36

1 11 , 0, −5.49, 36, 10 3 6 1 11 , −5.49, 10 c) Rational numbers but not integers: 3 6 d) Integers but not natural numbers: −8, 0 ! ! ! 14 ! 14 ! 2. ! − !! = 5 5 b) Rational numbers: −8,

3. |19.4| = 19.4

4. | − 1.2xy| = | − 1.2||x||y| = 1.2|x||y| 5. (−3, 6]

36

Chapter R: Basic Concepts of Algebra

6. | − 7 − 5| = | − 12| = 12, or

18.

|5 − (−7)| = |5 + 7| = |12| = 12

7.

32 ÷ 2 − 12 ÷ 4 · 3 3

19. 20.

= 32 ÷ 8 − 12 ÷ 4 · 3

= 4 − 12 ÷ 4 · 3 = 4−3·3 = 4−9

21.

= −5

8. Position the decimal point 5 places to the right, between the 3 and the 6. Since 0.0000367 is a number between 0 and 1, the exponent must be negative.

22.

0.0000367 = 3.67 × 10−5 9. The exponent is positive, so the number is greater than 10. We move the decimal point 6 places to the right. 4.51 × 106 = 4, 510, 000 10.

2.7 × 10 = 0.75 × 107 3.6 × 10−3 = (7.5 × 10−1 ) × 107 = 7.5 × 10

6

23. y 2 − 3y − 18 = (y + 3)(y − 6) 25. 2n2 + 5n − 12 = (2n − 3)(n + 4) 26. 8x2 − 18 = 2(4x2 − 9) = 2(2x + 3)(2x − 3) 27. m3 − 8 = (m − 2)(m2 + 2m + 4)

1 x3

28.

12. (2y 2 )3 (3y 4 )2 = 23 y 6 · 32 y 8 = 8 · 9 · y 6+8 = 72y 14 13.

= −3 · 5 · a5+(−1) · b−4+3

14.

15a4 b

(3x4 − 2x2 + 6x) − (5x3 − 3x2 + x)

= 3x − 2x + 6x − 5x + 3x − x 4

2

3

2

= 3x4 − 5x3 + x2 + 5x

15. (x + 3)(2x−5) = 2x2 −5x + 6x−15 = 2x2 + x−15 16. (2y − 1)2 = (2y)2 − 2 · 2y · 1 + (1)2 = 4y 2 − 4y + 1 17.

x x y y x y − · − · y x y x x y = x+y x+y x2 y2 − xy xy = x+y x2 − y 2 xy = x+y =

x2 − y 2 1 · xy x+y

!

(x+y)(x − y) xy(x+y) x−y = xy

=

!

x2 − 25 x2 + x − 6 · x2 + 8x + 15 x2 − 4x + 4 (x2 + x − 6)(x2 − 25) = 2 (x + 8x + 15)(x2 − 4x + 4) (x+3)(x−2)(x+5)(x − 5) = (x+3)(x+5)(x−2)(x − 2) x−5 = x−2

! ! ! ! ! !

(−3a5 b−4 )(5a−1 b3 )

= −15a4 b−1 , or −

√ √ √ √ 54 = 9 · 6 = 9 6 = 3 6 √ √ √ √ √ 3 40 = 3 8 · 5 = 3 8 3 5 = 2 3 5 √ √ √ √ 3 75 + 2 27 = 3 25 · 3 + 2 9 · 3 √ √ = 3·5 3+2·3 3 √ √ = 15 3 + 6 3 √ = 21 3 √ √ √ √ 18√ 10 = √ 18 · 10 = 2 · 3 · 3 · 2 · 5 = 2·3 5=6 5 √ √ (2 + 3)(5 − 2 3) √ √ = 2·5−4 3+5 3−2·3 √ √ = 10 − 4 3 + 5 3 − 6 √ = 4+ 3

24. x3 + 10x2 + 25x = x(x2 + 10x + 25) = x(x + 5)2

4

11. x−8 · x5 = x−8+5 = x−3 , or

√

29.

x 3 − x2 − 1 x2 + 4x − 5 3 x = − (x + 1)(x − 1) (x − 1)(x + 5)

LCD is (x + 1)(x − 1)(x + 5) x+ 5 3 x+ 1 x · − · = (x+ 1)(x−1) x + 5 (x−1)(x + 5) x+ 1 = = =

x(x + 5) − 3(x + 1) (x + 1)(x − 1)(x + 5) x2 + 5x − 3x − 3 (x + 1)(x − 1)(x + 5) x2 + 2x − 3 (x + 1)(x − 1)(x + 5)

! !

(x + 3)(x−1) (x + 1)(x−1)(x + 5) x+3 = (x + 1)(x + 5) √ √ 5 5 7+ 3 35 + 5 3 √ = √ · √ = = 30. 49 − 3 7− 3 7− 3 7+ 3 √ 35 + 5 3 46 =

Chapter R Test

37

√ 7 31. t5/7 = t5 √ 32. ( 5 7)3 = (71/5 )3 = 73/5 33. a = 5 and b = 12. Find c. c2 = a2 + b2 c2 = 52 + 122 c2 = 25 + 144 c2 = 169 c = 13 The guy wire is 13 ft long. 34.

7x − 4 = 24

7x = 28 x=4

The solution is 4. 35.

3(y − 5) + 6 = 8 − (y + 2) 3y − 15 + 6 = 8 − y − 2 3y − 9 = −y + 6

4y − 9 = 6

4y = 15 15 y= 4 15 The solution is . 4 36.

2x2 + 5x + 3 = 0 (2x + 3)(x + 1) = 0 2x + 3 = 0

or x + 1 = 0

2x = −3 or 3 x = − or 2 The solutions are − 37.

x = −1 x = −1

3 and −1. 2

z 2 − 11 = 0

z 2 = 11

√ 11 or z = − 11 √ √ √ The solutions are 11 and − 11, or ± 11. z=

38.

√

(x − y − 1)2

= [(x − y) − 1]2

= (x − y)2 − 2(x − y)(1) + 12

= x2 − 2xy + y 2 − 2x + 2y + 1

Chapter 1

Graphs, Functions, and Models y

Exercise Set 1.1 4

1. To graph (4, 0) we move from the origin 4 units to the right of the y-axis. Since the second coordinate is 0, we do not move up or down from the x-axis. To graph (−3, −5) we move from the origin 3 units to the left of the y-axis. Then we move 5 units down from the x-axis. To graph (−1, 4) we move from the origin 1 unit to the left of the y-axis. Then we move 4 units up from the x-axis. To graph (0, 2) we do not move to the right or the left of the y-axis since the first coordinate is 0. From the origin we move 2 units up. To graph (2, −2) we move from the origin 2 units to the right of the y-axis. Then we move 2 units down from the x-axis.

(!1, 4) 4 (4, 0)

2

!4 !2

4

x

(!3, !5) !4

2.

(!5, 0) (!4, !2)

(4, 0) 2

4

x

!2 !4

x

!2 (2, !1) !4

4.

y 4

(!5, 2)

2

(!5, 0)

(4, 0) 2

!4 !2

4

x

!2 !4

(!1, !5)

(4, !3)

y = 2x − 3

(1, 4)

2

!4 !2

(5, 1) 4

!4 !2

7. To determine whether (1, −1) is a solution, substitute 1 for x and −1 for y.

y 4

(2, 3) (0, 1)

6. (1995, 21.6), (2001, 19.0), (2002, 16.9), (2003, 15.8), (2004, 15.6)

(2, !2)

!2

2

5. The first coordinate represents the year and the corresponding second coordinate represents total advertisement spending, in millions of dollars. The ordered pairs are (2000, 310.6), (2001, 311.2), (2002, 348.2), (2003, 361.6), (2004, 435.8), and (2005, 467.7).

y

2 (0, 2)

(!5, 1)

(2, !4)

3. To graph (−5, 1) we move from the origin 5 units to the left of the y-axis. Then we move 1 unit up from the x-axis. To graph (5, 1) we move from the origin 5 units to the right of the y-axis. Then we move 1 unit up from the x-axis. To graph (2, 3) we move from the origin 2 units to the right of the y-axis. Then we move 3 units up from the x-axis. To graph (2, −1) we move from the origin 2 units to the right of the y-axis. Then we move 1 unit down from the x-axis. To graph (0, 1) we do not move to the right or the left of the y-axis since the first coordinate is 0. From the origin we move 1 unit up.

−1 ?! 2 · 1 − 3 ! ! 2−3 ! −1 ! −1 TRUE

The equation −1 = −1 is true, so (1, −1) is a solution.

To determine whether (0, 3) is a solution, substitute 0 for x and −3 for y. y = 2x − 3

3 ?! 2 · 0 − 3 ! ! 0−3 ! 3 ! −3 FALSE

The equation 3 = −3 is false, so (0, 3) is not a solution. 8. For (2, 5):

y = 3x − 1

5 ?! 3 · 2 − 1 ! ! 6−1 ! 5 ! 5 TRUE

(2, 5) is a solution.

40

Chapter 1: Graphs, Functions, and Models For (−2, −5):

y = 3x − 1

−5 ?! 3(−2) − 1 ! ! −6 − 1 ! FALSE −5 ! −7

(−2, −5) is not a solution. "2 3# 2 is a solution, substitute 9. To determine whether , 3 4 3 3 for x and for y. 4 6x − 4y = 1 6·

2 3 −4· ? 1 3 4 !! ! 4−3 ! ! 1 ! 1 TRUE

"2 3# , is a solution. The equation 1 = 1 is true, so 3 4 " 3# is a solution, substitute 1 for To determine whether 1, 2 3 x and for y. 2 6x − 4y = 1 3 ? 1 2 !! ! 6−6 ! ! 0 ! 1 FALSE

" 3# To determine whether 0, is a solution, substitute 0 for 5 3 a and for b. 5 2a + 5b = 3 3 ? 3 5 !! ! 0+3 ! ! 3 ! 3 TRUE " 3# is a solution. The equation 3 = 3 is true, so 0, 5 " 3# : 12. For 0, 3m + 4n = 6 2 3 3·0+4· ? 6 2 !! ! 0+6 ! ! 6 ! 6 TRUE " 3# is a solution. 0, 2 2·0+5·

For

"2 3

# ,1 :

6·1−4·

3# is not a solution. The equation 0 = 1 is false, so 1, 2 "

x2 + y 2 = 9

10. For (1.5, 2.6):

(1.5)2 + (2.6)2 ?! 9 ! 2.25 + 6.76 ! ! 9.01 ! 9 FALSE

(1.5, 2.6) is not a solution. x2 + y 2 = 9 For (−3, 0):

(−3)2 + 02 ?! 9 ! 9+0 ! ! 9 ! 9 TRUE

(−3, 0) is a solution.

" 1 4# is a solution, substitute 11. To determine whether − , − 2 5 1 4 − for a and − for b. 2 5 2a + 5b = 3 " 4# " 1# +5 − ? 3 2 − 2 5 !! ! −1 − 4 ! ! −5 ! 3 FALSE " 1 4# is not a soluThe equation −5 = 3 is false, so − , − 2 5 tion.

"2

3

3m + 4n = 6 3·

# , 1 is a solution.

2 +4·1 ? 6 3 !! ! 2+4 ! ! 6 ! 6 TRUE

13. To determine whether (−0.75, 2.75) is a solution, substitute −0.75 for x and 2.75 for y. x2 − y 2 = 3

(−0.75)2 − (2.75)2 ?! 3 ! 0.5625 − 7.5625 ! ! −7 ! 3 FALSE

The equation −7 = 3 is false, so (−0.75, 2.75) is not a solution. To determine whether (2, −1) is a solution, substitute 2 for x and −1 for y. x2 − y 2 = 3

22 − (−1)2 ?! 3 ! 4−1 ! ! 3 ! 3 TRUE

The equation 3 = 3 is true, so (2, −1) is a solution. 14. For (2, −4):

5x + 2y 2 = 70

5 · 2 + 2(−4)2 ?! 70 ! 10 + 2 · 16 ! ! 10 + 32 !! 42 ! 70 FALSE

(2, −4) is not a solution.

Exercise Set 1.1

41 5x + 2y 2 = 70

For (4, −5):

We plot the intercepts and draw the line that contains them. We could find a third point as a check that the intercepts were found correctly.

5 · 4 + 2(−5) ?! 70 ! 20 + 2 · 25 ! ! 20 + 50 !! 70 ! 70 TRUE 2

y 4

(4, −5) is a solution.

5x = −15

2x # y " 4

18.

y

2

To find the y-intercept we replace x with 0 and solve for y.

(2, 0) 2

!4 !2

19. Graph 4y − 3x = 12.

To find the x-intercept we replace y with 0 and solve for x. 4 · 0 − 3x = 12 −3x = 12

x = −4

y

The x-intercept is (−4, 0).

(0, 5)

4

To find the y-intercept we replace x with 0 and solve for y.

2 2

!4 !2

4

x

4y − 3 · 0 = 12

!2

4y = 12

!4

y=3

5x ! 3y " !15

The y-intercept is (0, 3).

y

We plot the intercepts and draw the line that contains them. We could find a third point as a check that the intercepts were found correctly.

4 2

(4, 0) 2

!4 !2 !2

x

3x # y " 6

We plot the intercepts and draw the line that contains them. We could find a third point as a check that the intercepts were found correctly.

16.

4

!2

5 · 0 − 3y = −15

(!3, 0)

(0, 6)

4

The x-intercept is (−3, 0).

The y-intercept is (0, 5).

x

!4

6

y=5

4

!2

x = −3

−3y = −15

(2, 0) 2

!4 !2

15. Graph 5x − 3y = −15.

To find the x-intercept we replace y with 0 and solve for x. 5x − 3 · 0 = −15

(0, 4)

2

4

x y

(0, !2)

!4

4

2x ! 4y " 8

2x = 4 x=2 The x-intercept is (2, 0). To find the y-intercept we replace x with 0 and solve for y. 2·0+y = 4 y=4

The y-intercept is (0, 4).

(0, 3) 2

!4 !2

17. Graph 2x + y = 4. To find the x-intercept we replace y with 0 and solve for x. 2x + 0 = 4

2

(!4, 0)

4

x

!2 !4

4y ! 3x " 12

20.

y

(!3, 0)

4 2 2

!4 !2 !2

4

(0, !2)

!4

3y # 2x " !6

x

42

Chapter 1: Graphs, Functions, and Models

21. Graph y = 3x + 5.

24.

y

We choose some values for x and find the corresponding y-values.

6

2

When x = −1, y = 3x + 5 = 3(−1) + 5 = −3 + 5 = 2. When x = 0, y = 3x + 5 = 3 · 0 + 5 = 0 + 5 = 5

We list these points in a table, plot them, and draw the graph. x

y

(x, y)

−3 −4 (−3, −4) −1

2

(−1, 2)

0

5

(0, 5)

x#y"4

4

When x = −3, y = 3x + 5 = 3(−3) + 5 = −9 + 5 = −4.

2

!4 !2

4

x

!4

3 25. Graph y = − x + 3. 4 By choosing multiples of 4 for x, we can avoid fraction values for y. Make a table of values, plot the points in the table, and draw the graph. x

y

(x, y)

−4 6 (−4, 6)

y 6

y " 3x # 5

0

3

(0, 3)

4

0

(4, 0)

2 2

!4

4

y x

!2

4 2

22.

y 4

!2 !4

2 2

!4 !2

2

!4 !2

y " !2x ! 1

4

x

!2

26.

4

x

3 y " !!x 4 #3

y

!4 4 2

23. Graph x − y = 3.

x

y

2

!4 !2

Make a table of values, plot the points in the table, and draw the graph.

!2

4

x

3y ! 2x " 3

!4

(x, y)

−2 −5 (−2, −5) 0

−3

(0, −3)

3

0

(3, 0)

27. Graph 5x − 2y = 8.

We could solve for y first. 5x − 2y = 8

−2y = −5x + 8 5 y = x−4 2

y 4

x!y"3

2 2

!4 !2 !2 !4

4

x

Subtracting 5x on both sides 1 Multiplying by − on both 2 sides

By choosing multiples of 2 for x we can avoid fraction values for y. Make a table of values, plot the points in the table, and draw the graph. x

y

(x, y)

0 −4 (0, −4) 2

1

(2, 1)

4

6

(4, 6)

Exercise Set 1.1

43

y 4 2 2

!4 !2

4

x

!2 !4

x

y

(x, y)

−5

0

(−5, 0)

0

−2 (0, −2)

5

−4 (5, −4)

5x ! 2y " 8 y

28.

4

y 4

4 y " 2 ! !x 3

!6

2 2

!4 !2

4

2x # 5y " !10

2 !2

2

x

4

x

!4

x

!2 !4

32.

y 4

29. Graph x − 4y = 5.

x

!2 !4

(x, y)

y

!6

−3 −2 (−3, −2) 1

−1

(1, −1)

5

0

(5, 0)

33. Graph y = −x2 .

Make a table of values, plot the points in the table, and draw the graph. x

x ! 4y " 5

4

6

!4

y

0

0

(0, 0)

1

−1

(1, −1)

2

−4

(2, −4) y

4 2 4

4

x

!2

x

!4

!2 !4

2

!4 !2 2

!4 !2

(x, y)

−1 −1 (−1, −1)

x

!2

30.

y

−2 −4 (−2, −4)

2 !2

4x ! 3y " 12

!8

y 4

2

!4 !2

Make a table of values, plot the points in the table, and draw the graph.

6x ! y " 4

y " !x 2

!6 !8

31. Graph 2x + 5y = −10.

In this case, it is convenient to find the intercepts along with a third point on the graph. Make a table of values, plot the points in the table, and draw the graph.

34.

y 8 6 4

y " x2

2 !4 !2

2

4

x

44

Chapter 1: Graphs, Functions, and Models

35. Graph y = x2 − 3.

38.

y

Make a table of values, plot the points in the table, and draw the graph. x

y

(x, y)

−3

6

(−3, 6)

4 2

−3

(0, −3)

1

−2

(1, −2)

3

6

(3, 6)

!4

6 4

y " x2 ! 3 2

!4 !2

4

x

!2

36.

y y " 4 ! x2 2 4

!4

x

!2 !4

37. Graph y = −x2 + 2x + 3.

Make a table of values, plot the points in the table, and draw the graph. x

(x, y)

y

0

(−1, 0)

0

3

(0, 3)

1

4

(1, 4)

2

3

(2, 3)

3

0

(3, 0)

4

−5

(4, −5) y

y " !x 2 # 2x # 3 4

!4 !8 !12

43. Either point can be considered as (x1 , y1 ). $ d = (−4.2 − 2.1)2 + [3 − (−6.4)]2 $ √ = (−6.3)2 + (9.4)2 = 128.05 ≈ 11.316 %& ' ()2 ' (2 3 3 2 44. d = − − − + −4− = 5 5 3 %' (2 14 14 = − 3 3 45. Either point can be considered as (x1 , y1 ). %' (2 1 5 − − + (4 − 4)2 d= 2 2 $ √ = (−3)2 + 02 = 9 = 3

$ d = [0.6 − (−8.1)]2 + [−1.5 − (−1.5)]2 = $ (8.7)2 = 8.7

47. Either point can be considered as (x1 , y1 ). * √ √ √ d = (− 6 − 3)2 + (0 − (− 5))2 $ $ √ √ = 6 + 2 18 + 3 + 5 = 14 + 2 9 · 2 $ $ √ √ = 14 + 2 · 3 2 = 14 + 6 2 ≈ 4.742 * √ $ √ √ 48. d = (− 2−0)2 +(1− 7)2 = 2+1−2 7+7 = $ √ 10 − 2 7 ≈ 2.170

4 !8 !4

y " x # 2x ! 1

41. Either point can be considered as (x1 , y1 ). $ d = (6 − 9)2 + (−1 − 5)2 $ √ = (−3)2 + (−6)2 = 45 ≈ 6.708 $ √ 42. d = (−4 − (−1))2 + (−7 − 3)2 = 109 ≈ 10.440

46.

−2 −5 (−2, −5) −1

x

39. Either point can be considered as (x1 , y1 ). $ d = (4 − 5)2 + (6 − 9)2 $ √ = (−1)2 + (−3)2 = 10 ≈ 3.162 $ √ 40. d = (−3 − 2)2 + (7 − 11)2 = 41 ≈ 6.403

y

2

4 2

−1 −2 (−1, −2) 0

2

!4

8

x

49. Either point can be considered as (x1 , y1 ). $ √ d = (0 − a)2 + (0 − b)2 = a2 + b2 $ √ 50. d = [r − (−r)]2 + [s − (−s)]2 = 4r2 + 4s2 = √ 2 r2 + s2

Exercise Set 1.1 51. First we find the length of the diameter: $ d = (−3 − 9)2 + (−1 − 4)2 $ √ = (−12)2 + (−5)2 = 169 = 13

The length of the radius is one-half the length of the di1 ameter, or (13), or 6.5. 2 $ √ 52. Radius = (−3 − 0)2 + (5 − 1)2 = 25 = 5 Diameter = 2 · 5 = 10

53. First we find the distance between each pair of points. For (−4, 5) and (6, 1): $ d = (−4 − 6)2 + (5 − 1)2 $ √ = (−10)2 + 42 = 116

For (−4, 5) and (−8, −5): $ d = (−4 − (−8))2 + (5 − (−5))2 √ √ = 42 + 102 = 116

For (6, 1) and (−8, −5): $ d = (6 − (−8))2 + (1 − (−5))2 √ √ = 142 + 62 = 232 √ √ √ Since ( 116)2 + ( 116)2 = ( 232)2 , the points could be the vertices of a right triangle. 54. For (−3, 1) and (2, −1): $ √ d = (−3 − 2)2 + (1 − (−1))2 = 29 For (−3, 1) and (6, 9): $ √ d = (−3 − 6)2 + (1 − 9)2 = 145

For (2, −1) and (6, 9): $ √ d = (2 − 6)2 + (−1 − 9)2 = 116 √ √ √ Since ( 29)2 + ( 116)2 = ( 145)2 , the points could be the vertices of a right triangle. 55. First we find the distance between each pair of points. For (−4, 3) and (0, 5): $ d = (−4 − 0)2 + (3 − 5)2 $ √ = (−4)2 + (−2)2 = 20

For (−4, 3) and (3, −4): $ d = (−4 − 3)2 + [3 − (−4)]2 $ √ = (−7)2 + 72 = 98

For (0, 5) and (3, −4): $ d = (0 − 3)2 + [5 − (−4)]2 $ √ = (−3)2 + 92 = 90 √ The greatest distance is 98, so if the points are the vertices triangle, √ of a right √ √ then it is the hypotenuse. But ( 20)2 + ( 90)2 $= ( 98)2 , so the points are not the vertices of a right triangle. 56. See the graph of this rectangle in Exercise 67. The segments with endpoints (−3, 4), (2, −1) and (5, 2), (0, 7) are one pair of opposite sides. We find the length of each of these sides.

45 For (−3, 4), (2, −1): $ √ d = (−3 − 2)2 + (4 − (−1))2 = 50

For (5, 2), (0, 7): $ √ d = (5 − 0)2 + (2 − 7)2 = 50

The segments with endpoints (2, −1), (5, 2) and (0, 7), (−3, 4) are the second pair of opposite sides. We find their lengths. For (2, −1), (5, 2): $ √ d = (2 − 5)2 + (−1 − 2)2 = 18

For (0, 7), (−3, 4): $ √ d = (0 − (−3))2 + (7 − 4)2 = 18

The endpoints of the diagonals are (−3, 4), (5, 2) and (2, −1), (0, 7). We find the length of each. For (−3, 4), (5, 2): $ √ d = (−3 − 5)2 + (4 − 2)2 = 68

For (2, −1), (0, 7): $ √ d = (2 − 0)2 + (−1 − 7)2 = 68

The opposite sides of the quadrilateral are the same length and the diagonals are the same length, so the quadrilateral is a rectangle. 57. We use the midpoint formula. ( ' ( ' 8 12 4 + (−12) −9 + (−3) , = − ,− = (−4, −6) 2 2 2 2 ( ' ( ' 7 + 9 −2 + 5 3 58. , = 8, 2 2 2 59. We use the midpoint formula. ( ' ( ' 9.9 9.9 6.1 + 3.8 −3.8 + (−6.1) , ,− = = 2 2 2 2

(4.95, −4.95) ( ' −0.5 + 4.8 −2.7 + (−0.3) , = (2.15, −1.5) 60. 2 2

61. We use the midpoint formula. ( ' ( ' ( ' 12 13 13 −6 + (−6) 5 + 8 , = − , = − 6, 2 2 2 2 2 ( ' 1 + (−1) −2 + 2 , = (0, 0) 62. 2 2 63. We use the midpoint formula. 3 5 , + 5 13 , + 1 " 2# − + − − + − 6 3 , 5 4 = 6 , 20 = 2 2 2 2 ' ( 5 13 − , 12 40 + 2 " 2# 1 4 , ' ( + − + 9 5 , 3 5 = − 4 , 17 64. 2 2 45 30 65. We use the midpoint formula. √ ( ' √ ( ' ( '√ √ 3 4 3 3 3 + 3 3 −1 + 4 , = , = 2 3, 2 2 2 2 2

46

66. 67.

Chapter 1: Graphs, Functions, and Models √ √ ( ' √ ( ' √ − 5+ 5 2+ 7 2+ 7 , = 0, 2 2 2

For the side with vertices (7, −6) and (12, 6): ' ( ' ( 7 + 12 −6 + 6 19 = , ,0 2 2 2 For the side with vertices (12, 6) and (0, 11): ( ' ( ' 17 12 + 0 6 + 11 , = 6, 2 2 2 For the side with vertices (0, 11) and (−5, −1): ( ' ( ' 5 0 + (−5) 11 + (−1) , = − ,5 2 2 2 For the quadrilateral whose vertices are the points ( ' found 7 above, one pair of opposite sides has endpoints 1, − , 2 ' ( ' ( ' ( 19 17 5 , 0 and 6, , − , 5 . The length of each of 2 2 √ 2 338 . The other pair of opposite sides has these sides is 2( ' ( ' ( ' ( ' 17 5 7 19 , 0 , 6, and − , 5 , 1, − . endpoints 2 2 2 2 √ 338 The length of each of these sides is also . The end2 ( ' 7 points of the diagonals of the quadrilateral are 1, − , 2 ' ( ' ( ' ( 17 19 5 and , 0 , − , 5 . The length of each di6, 2 2 2 agonal is 13. Since the four sides of the quadrilateral are the same length and the diagonals are the same length, the midpoints are vertices of a square.

y 8 6 4 2 4

!4 !2

6 x

!2

For the side with vertices (−3, 4) and (2, −1): ( ' ( ' 1 3 −3 + 2 4 + (−1) , = − , 2 2 2 2

For the side with vertices (2, −1) and (5, 2): ' ( ' ( 7 1 2 + 5 −1 + 2 , = , 2 2 2 2 For the side with vertices (5, 2) and (0, 7): ( ' ( ' 5 9 5+0 2+7 , = , 2 2 2 2

For the side with vertices (0, 7) and (−3, 4): ( ' ( ' 3 11 0 + (−3) 7 + 4 , = − , 2 2 2 2 For the quadrilateral whose vertices are the points found above, the diagonals have endpoints ( ' ( ' ( ' ( ' 5 9 7 1 3 11 1 3 − , , , and , , − , . 2 2 2 2 2 2 2 2 We find the length of each of these diagonals. ( ' ( ' 5 9 1 3 , , : For − , 2 2 2 2 %' (2 ' (2 3 9 1 5 d= + − − − 2 2 2 2 $ √ = (−3)2 + (−3)2 = 18 ( ' ( ' 7 1 3 11 , , − , : For 2 2 2 2 %' ' ((2 ' (2 3 1 11 7 − − + − d= 2 2 2 2 $ √ 2 2 = 5 + (−5) = 50 Since the diagonals do not have the same lengths, the midpoints are not vertices of a rectangle.

68.

y 12

69. We use the midpoint formula. √ √ ( '√ ( '√ 7 + 2 −4 + 3 7+ 2 1 , = ,− 2 2 2 2 √ √ ( ' √ √ ( ' −3 + 1 5+ 2 5+ 2 , 70. = − 1, 2 2 2 71.

72.

(x − h)2 + (y − k)2 = r2 ' (2 5 (x − 2)2 + (y − 3)2 = 3 25 (x − 2)2 + (y − 3)2 = 9 2 2 (x − 4) + (y − 5) = (4.1)2

Substituting

(x − 4)2 + (y − 5)2 = 16.81

73. The length of a radius is the distance between (−1, 4) and (3, 7): $ r = (−1 − 3)2 + (4 − 7)2 $ √ = (−4)2 + (−3)2 = 25 = 5 (x − h)2 + (y − k)2 = r2

8

[x − (−1)]2 + (y − 4)2 = 52

4 4

8

12

x

!4

For the side with vertices (−5, −1) and (7, −6): ( ' ( ' 7 −5 + 7 −1 + (−6) , = 1, − 2 2 2

(x + 1)2 + (y − 4)2 = 25

74. Find the length of a radius: $ √ r = (6 − 1)2 + (−5 − 7)2 = 169 = 13 (x − 6)2 + [y − (−5)]2 = 132

(x − 6)2 + (y + 5)2 = 169

Exercise Set 1.1

47

75. The center is the midpoint of the diameter: ' ( 7 + (−3) 13 + (−11) , = (2, 1) 2 2 Use the center and either endpoint of the diameter to find the length of a radius. We use the point (7, 13): $ r = (7 − 2)2 + (13 − 1)2 √ √ = 52 + 122 = 169 = 13 (x − h) + (y − k) = r 2

2

y 8 4 !8

4

81.

x2 + (y − 3)2 = 16

(x − 0)2 + (y − 3)2 = 42

Center: (0, 3); radius: 4

76. The points (−9, 4) and (−1, −2) are opposite vertices of the square and hence endpoints of a diameter of the circle. We use these points to find the center and radius. ( ' −9 + (−1) 4 + (−2) = (−5, 1) , Center: 2 2 1$ 1 Radius: (−9−(−1))2 +(4−(−2))2 = ·10 = 5 2 2 [x − (−5)]2 + (y − 1)2 = 52

y 8 6 4 2

(x + 5)2 + (y − 1)2 = 25

!4

(x − h) + (y − k) = r 2

!2

2

x 2 # (y ! 3)2 " 16 82.

(x + 2)2 + y 2 = 100 [x − (−2)]2 + (y − 0)2 = 102

(x + 2)2 + (y − 3)2 = 4

Center: (−2, 0); radius: 10

78. Since the center is 5 units below the x-axis and the circle is tangent to the x-axis, the length of a radius is 5.

y

(x − 4)2 + [y − (−5)]2 = 52

8

(x − 4)2 + (y + 5)2 = 25

4

x2 + y 2 = 4 (x − 0)2 + (y − 0)2 = 22

!12

!8

!4

4

Center: (0, 0); radius: 2

!8

(x # 2)2 # y 2 " 100

x2 # y 2 " 4

2

83. 2

4

!2

x

(x − 1)2 + (y − 5)2 = 36

(x − 1)2 + (y − 5)2 = 62

Center: (1, 5); radius: 6

!4

80.

8 x

!4

y

!2

x

2

[x − (−2)]2 + (y − 3)2 = 22

!4

4

!2

77. Since the center is 2 units to the left of the y-axis and the circle is tangent to the y-axis, the length of a radius is 2.

4

x

!8

(x − 2) + (y − 1) = 169

79.

8

!4

2

2

!4

2

(x − 2)2 + (y − 1)2 = 132 2

x 2 # y 2 " 81

y 12

x2 + y 2 = 81

8

(x − 0)2 + (y − 0)2 = 92

Center: (0, 0); radius: 9

4 !8

!4

4

8

!4

(x ! 1)2 # (y ! 5)2 " 36

x

48 84.

Chapter 1: Graphs, Functions, and Models 89. From the graph we see that the center of the circle is (5, −5) and the radius is 15. The equation of the circle is (x − 5)2 + [y − (−5)]2 = 152 , or (x − 5)2 + (y + 5)2 = 152 .

(x − 7)2 + (y + 2)2 = 25

(x − 7) + [y − (−2)] = 5 2

2

2

Center: (7, −2); radius: 5

90. Center: (−8, 2), radius: 4

y

Equation: [x − (−8)]2 + (y − 2)2 = 42 , or (x + 8)2 + (y − 2)2 = 42

8

91. Graph (b) is the graph of y = 3 − x.

4 !4

4

8

12

x

!4

93. Graph (a) is the graph of y = x2 + 2x + 1.

!8

94. Graph (c) is the graph of y = 8 − x2 .

(x ! 7)2 # (y # 2)2 " 25 85.

92. Graph (d) is the graph of 2x − y = 6.

(x + 4)2 + (y + 5)2 = 9

95. First solve the equation for y: y = −4x + 7. Enter the equation in this form, select the standard window, and graph the equation.

[x − (−4)]2 + [y − (−5)]2 = 32

4x # y " 7

Center: (−4, −5); radius: 3

10

y 2 !6

!4

10

!10

!2

2

x

!2

!10

!4

5x # y " !8

96.

10

!6 !8

(x # 4)2 # (y # 5)2 " 9 86.

10

!10

(x + 1)2 + (y − 2)2 = 64

!10

[x − (−1)]2 + (y − 2)2 = 82

Center: (−1, 2); radius: 8

97. Enter the equation, select the standard window, and graph the equation.

y 12

y " ax # 2

8

10

4 !8

!4

4

8

10

!10

x

!4 !8

(x #

1)2 #

(y !

!10

2)2 "

64

87. From the graph we see that the center of the circle is (−2, 1) and the radius is 3. The equation of the circle is [x − (−2)]2 + (y − 1)2 = 32 , or (x + 2)2 + (y − 1)2 = 32 .

98.

y " wx ! 4 10

10

!10

88. Center: (3, −5), radius: 4

Equation: (x − 3)2 + [y − (−5)]2 = 42 , or (x − 3)2 + (y + 5)2 = 42

!10

Exercise Set 1.1

49

99. Enter the equation, select the standard window, and graph the equation.

[−4, 4, −4, 4]

y " x2 # 6 10

10

!10

We see that the standard window is a better choice for this graph.

!10

100.

y " 5 ! x2

104. Standard window:

10

10

!10

!10

101. Enter the equation, select the standard window, and graph the equation.

[−15, 15, −10, 30], Xscl = 3, Yscl = 5

y " 2 ! x2 10

10

!10

We see that [−15, 15, −10, 30] is a better choice for this graph.

!10

102.

y " x 2 ! 5x # 3

105. Standard window:

10

10

!10

!10

103. Standard window: [−1, 1, −0.3, 0.3], Xscl = 0.1, Yscl = 0.1

We see that [−1, 1, −0.3, 0.3] is a better choice for this graph.

50

Chapter 1: Graphs, Functions, and Models

106. Standard window:

112.

(x # 1)2 # (y ! 2)2 " 64 10

15

!15

!10

113. The Pythagorean theorem is used to derive the distance formula, and the distance formula is used to derive the equation of a circle in standard form.

[−3, 3, −3, 3]

We see that the standard window is a better choice for this graph. 107. Square the viewing window. For the graph shown, one possibility is [−12, 9, −4, 10]. 108. Square the viewing window. For the graph shown, one possibility is [−10, 20, −15, 5]. 109.

x2 # y2 " 4 y1 " !4 ! x 2, y2 " !!4 ! x 2 6

9

!9

!6

110.

(x # 2)2 # y 2 " 100 10

15

!15

!10

111.

(x ! 1)2 # (y ! 5)2 " 36 y1 " 5 #!36 ! (x ! 1)2, y2 " 5 !!36 ! (x ! 1)2 11

9

!9 !1

114. Let A = (a, b) and B = (c, d). The coordinates of a point ( ' a+c b+d , . C one-half of the way from A to B are 2 2 A point D that is one-half of the way from C to B 1 1 1 3 is + · , or of the way from A to B. Its co2 2 2 ' 4 ( ' ( a+c b+d +d a + 3c b + 3d 2 +c , 2 , or , . ordinates are 2 2 4 4 Then a point E that is one-half of the way from D to B 3 1 1 7 is + · , or of the way from A to B. Its coordinates 4' 2 4 8 ( ' ( a+3c b+3d +d a + 7c b + 7d 4 +c , 4 , are , or . 2 2 8 8 115. If the point (p, q) is in the fourth quadrant, then p > 0 and q < 0. If p > 0, then −p < 0 so both coordinates of the point (q, −p) are negative and (q, −p) is in the third quadrant. 116. Use the distance formula: % (2 ' 1 1 − = d = (a + h − a)2 + a+h a % % (2 ' −h h2 h2 + = h2 + 2 = a(a + h) a (a + h)2 % % h2 a2 (a + h)2 + h2 h2 (a2 (a + h)2 + 1) = = a2 (a + h)2 a2 (a + h)2 ! ! !$ ! h ! ! ! a2 (a + h)2 + 1 ! ! a(a + h) ! Find the midpoint: ' ( ' 1 1 ( 2a + h 2a + h a + a + h a + a+h , = , 2 2 2 2a(a + h)

117. Use the distance formula. Either point can be considered as (x1 , y1 ). * √ √ d = (a + h − a)2 + ( a + h − a)2 $ √ = h2 + a + h − 2 a2 + ah + a $ √ = h2 + 2a + h − 2 a2 + ah Next we use the midpoint formula. ( ' ( ' √ √ √ √ a+ a+h a+ a+h a+a+h 2a+h , = , 2 2 2 2

Exercise Set 1.1 118.

51

C = 2πr

Since d1 + d2 = d3 , the points are collinear.

10π = 2πr

123. Label the drawing with additional information and lettering.

5=r Then [x−(−5)]2 +(y −8)2 = 52 , or (x+5)2 +(y −8)2 = 25.

30 ft

119. First use the formula for the area of a circle to find r2 : A = πr2

a1 { 20

36π = πr2

20

b

36 = r2 Then we have: (x − h)2 + (y − k)2 = r2

a2

20 ft 10

(x − 2)2 + [y − (−7)]2 = 36 (x − 2)2 + (y + 7)2 = 36

Find b using the Pythagorean theorem.

120. Let the point be (x, 0). We set the distance from (−4, −3) to (x, 0) equal to the distance from (−1, 5) to (x, 0) and solve for x. $ $ (−4 − x)2 + (−3 − 0)2 = (−1 − x)2 + (5 − 0)2 √ √ 16 + 8x + x2 + 9 = 1 + 2x + x2 + 25 √ √ x2 + 8x + 25 = x2 + 2x + 26

b2 + 102 = 202 b2 + 100 = 400 b2 = 300 √ b = 10 3

x2 + 8x + 25 = x2 + 2x + 26 Squaring both sides 8x + 25 = 2x + 26

The point is

'

( 1 ,0 . 6

6x = 1 1 x= 6

Find a1 :

a1 = 20 − b ≈ 20 − 17.3 ≈ 2.7 ft Find a2 :

a2 = 2b + a1 ≈ 2(17.3) + 2.7 ≈ 37.3 ft 124. a) When the circle is positioned on a coordinate system as shown in the text, the center lies on the y-axis and is equidistant from (−4, 0) and (0, 2). Let (0, y) be the coordinates of the center. $ $ (−4−0)2 +(0−y)2 = (0−0)2 +(2−y)2

121. Let (0, y) be the required point. We set the distance from (−2, 0) to (0, y) equal to the distance from (4, 6) to (0, y) and solve for y. $ $ [0 − (−2)]2 + (y − 0)2 = (0 − 4)2 + (y − 6)2 $ $ 4 + y 2 = 16 + y 2 − 12y + 36

42 + y 2 = (2 − y)2

16 + y 2 = 4 − 4y + y 2 12 = −4y

4 + y 2 = 16 + y 2 − 12y + 36 Squaring both sides −48 = −12y

−3 = y

The center of the circle is (0, −3).

b) Use the point (−4, 0) and the center (0, −3) to find the radius. (−4 − 0)2 + [0 − (−3)]2 = r2

4=y

The point is (0, 4). 122. We first find the distance between each pair of points. For (−1, −3) and (−4, −9): $ d1 = [−1 − (−4)]2 + [−3 − (−9)]2 √ √ = 32 + 62 = 9 + 36 √ √ = 45 = 3 5 For (−1, −3) and (2, 3): $ d2 = (−1 − 2)2 + (−3 − 3)2 $ √ = (−3)2 + (−6)2 = 9 + 36 √ √ = 45 = 3 5 For (−4, −9) and (2, 3): $ d3 = (−4 − 2)2 + (−9 − 3)2 $ √ = (−6)2 + (−12)2 = 36 + 144 √ √ = 180 = 6 5

b ≈ 17.3

25 = r2 5=r

The radius is 5 ft. 125.

x2 + y 2 = 1 ' √ (2 ' (2 3 1 + − ? 1 2 2 ! ! 3 1 !! + ! 4 4 ! ! 1 ! 1 TRUE ( '√ 3 1 ,− lies on the unit circle. 2 2

52 126.

Chapter 1: Graphs, Functions, and Models x2 + y 2 = 1 02 + (−1)2 ?! 1 1 ! 1 TRUE

(0, −1) lies on the unit circle. 127.

128.

x2 + y 2 = 1 ' √ (2 ' √ (2 2 2 + ? 1 2 2 ! ! 2 2 !! + ! 4 4 ! ! 1 ! 1 TRUE '√ √ ( 2 2 , lies on the unit circle. 2 2 x2 + y 2 = 1 ' (2 ' √ (2 1 3 + − ? 1 2 2 ! ! 1 3 !! + ! 4 4 ! ! 1 ! 1 TRUE √ ( ' 3 1 ,− lies on the unit circle. 2 2

129. a), b) See the answer section in the text. ( ' b h , by the midpoint formula. 130. The coordinates of P are 2 2 By the distance formula, each of the distances √ from P to b2 + h 2 . (0, h), from P to (0, 0), and from P to (b, 0) is 2

Exercise Set 1.2 1. This correspondence is a function, because each member of the domain corresponds to exactly one member of the range. 2. This correspondence is a function, because each member of the domain corresponds to exactly one member of the range. 3. This correspondence is a function, because each member of the domain corresponds to exactly one member of the range. 4. This correspondence is not a function, because there is a member of the domain (1) that corresponds to more than one member of the range (4 and 6). 5. This correspondence is not a function, because there is a member of the domain (m) that corresponds to more than one member of the range (A and B). 6. This correspondence is a function, because each member of the domain corresponds to exactly one member of the range.

7. This correspondence is a function, because each member of the domain corresponds to exactly one member of the range. 8. This correspondence is a function, because each member of the domain corresponds to exactly one member of the range. 9. This correspondence is a function, because each car has exactly one license number. 10. This correspondence is not a function, because we can safely assume that at least one person uses more than one doctor. 11. This correspondence is a function, because each member of the family has exactly one eye color. 12. This correspondence is not a function, because we can safely assume that at least one band member plays more than one instrument. 13. This correspondence is not a function, because at least one student will have more than one neighboring seat occupied by another student. 14. This correspondence is a function, because each bag has exactly one weight. 15. The relation is a function, because no two ordered pairs have the same first coordinate and different second coordinates. The domain is the set of all first coordinates: {2, 3, 4}.

The range is the set of all second coordinates: {10, 15, 20}.

16. The relation is a function, because no two ordered pairs have the same first coordinate and different second coordinates. Domain: {3, 5, 7} Range: {1}

17. The relation is not a function, because the ordered pairs (−2, 1) and (−2, 4) have the same first coordinate and different second coordinates. The domain is the set of all first coordinates: {−7, −2, 0}.

The range is the set of all second coordinates: {3, 1, 4, 7}. 18. The relation is not a function, because of each of the ordered pairs has the same first coordinate and different second coordinates. Domain: {1}

Range: {3, 5, 7, 9} 19. The relation is a function, because no two ordered pairs have the same first coordinate and different second coordinates. The domain is the set of all first coordinates: {−2, 0, 2, 4, −3}.

The range is the set of all second coordinates: {1}.

Exercise Set 1.2 20. The relation is not a function, because the ordered pairs (5, 0) and (5, −1) have the same first coordinates and different second coordinates. This is also true of the pairs (3, −1) and (3, −2). Domain: {5, 3, 0}

Range: {0, −1, −2}

21. From the graph we see that, when the input is 1, the output is −2, so h(1) = −2. When the input is 3, the output is 2, so h(3) = 2. When the input is 4, the output is 1, so h(4) = 1. 22. t(−4) = 3; t(0) = 3; t(3) = 3 23. From the graph we see that, when the input is −4, the output is 3, so s(−4) = 3. When the input is −2, the output is 0, so s(−2) = 0. When the input is 0, the output is −3, so s(0) = −3.

5 3 ; g(−1) = −3; g(0) = − 2 2 25. From the graph we see that, when the input is −1, the output is 2, so f (−1) = 2. When the input is 0, the output is 0, so f (0) = 0. When the input is 1, the output is −2, so f (1) = −2. 24. g(−4) =

26. g(−2) = 4; g(0) = −4; g(2.4) = −2.6176

27. g(x) = 3x2 − 2x + 1 a) g(0) = 3 · 02 − 2 · 0 + 1 = 1 b) g(−1) = 3(−1)2 − 2(−1) + 1 = 6 c) g(3) = 3 · 32 − 2 · 3 + 1 = 22 d) g(−x) = 3(−x)2 − 2(−x) + 1 = 3x2 + 2x + 1 e) g(1 − t) = 3(1 − t)2 − 2(1 − t) + 1 =

3(1−2t+t2 )−2(1−t)+1 = 3−6t+3t2 −2+2t+1 =

3t2 − 4t + 2

28. f (x) = 5x2 + 4x a) f (0) = 5 · 02 + 4 · 0 = 0 + 0 = 0

b) f (−1) = 5(−1)2 + 4(−1) = 5 − 4 = 1

c) f (3) = 5 · 32 + 4 · 3 = 45 + 12 = 57 d) f (t) = 5t2 + 4t

e) f (t − 1) = 5(t − 1)2 + 4(t − 1) = 5t2 − 6t + 1

29. g(x) = x3 a) g(2) = 23 = 8 b) g(−2) = (−2)3 = −8 c) g(−x) = (−x)3 = −x3 d) g(3y) = (3y)3 = 27y 3 e) g(2 + h) = (2 + h)3 = 8 + 12h + 6h2 + h3 30. f (x) = 2|x| + 3x a) f (1) = 2|1| + 3 · 1 = 2 + 3 = 5

b) f (−2) = 2| − 2| + 3(−2) = 4 − 6 = −2

c) f (−x) = 2| − x| + 3(−x) = 2|x| − 3x

d) f (2y) = 2|2y| + 3 · 2y = 4|y| + 6y e) f (2 − h) = 2|2 − h| + 3(2 − h) = 2|2 − h| + 6 − 3h

53 x−4 x+3 1 5−4 a) g(5) = = 5+3 8 4−4 b) g(4) = =0 4+7 −7 −3 − 4 c) g(−3) = = −3 + 3 0 Since division by 0 is not defined, g(−3) does not exist. −16.25 − 4 −20.25 81 d) g(−16.25) = = = −16.25 + 3 −13.25 53 x+h−4 e) g(x + h) = x+h+3 x 32. f (x) = 2−x 2 2 a) f (2) = = 2−2 0 Since division by 0 is not defined, f (2) does not exist. 31. g(x) =

1 =1 2−1 −16 −16 8 c) f (−16) = = =− 2 − (−16) 18 9 −x −x = d) f (−x) = 2 − (−x) 2+x 2 2 ( ' − − 1 2 3 3 e) f − (= ' =− = 8 2 3 4 2− − 3 3 x g(x) = √ 33. 1 − x2 0 0 0 g(0) = √ = √ = =0 1 1 1 − 02 −1 −1 −1 −1 g(−1) = $ =√ =√ = 0 1−1 0 1 − (−1)2 b) f (1) =

Since division by 0 is not defined, g(−1) does not exist. 5 5 5 =√ =√ g(5) = √ 1 − 25 −24 1 − 52 √ Since −24 is not defined as a real number, g(5) does not exist as a real number. 1 1 1 ' ( 1 = % 2' ( = - 2 = -2 = g 2 2 1 3 1 1− 1− 4 4 2 1 √ 1·2 1 3 1 2 √2 = · √ = √ = √ , or 2 3 3 3 2 3 3 2 √ 34. h(x) = x + x2 − 1 √ √ h(0) = 0 + 02 − 1 = 0 + −1 √ Since −1 is not defined as a real number, h(0) does not exist as a real number. √ √ h(2) = 2 + 22 − 1 = 2 + 3 $ √ h(−x) = −x + (−x)2 − 1 = −x + x2 − 1

54

Chapter 1: Graphs, Functions, and Models

35. This is not the graph of a function, because we can find a vertical line that crosses the graph more than once.

42. This is not the graph of a function, because we can find a vertical line that crosses the graph more than once.

y

x

36. This is not the graph of a function, because we can find a vertical line that crosses the graph more than once. y

43. We can substitute any real number for x. Thus, the domain is the set of all real numbers, or (−∞, ∞). 44. We can substitute any real number for x. Thus, the domain is the set of all real numbers, or (−∞, ∞). 45. The input 0 results in a denominator of 0. Thus, the domain is {x|x $= 0}, or (−∞, 0) ∪ (0, ∞).

x

37. This is the graph of a function, because there is no vertical line that crosses the graph more than once. 38. This is the graph of a function, because there is no vertical line that crosses the graph more than once. 39. This is the graph of a function, because there is no vertical line that crosses the graph more than once. 40. This is the graph of a function, because there is no vertical line that crosses the graph more than once. 41. This is not the graph of a function, because we can find a vertical line that crosses the graph more than once.

46. The input 0 results in a denominator of 0. Thus, the domain is {x|x $= 0}, or (−∞, 0) ∪ (0, ∞). 47. We can substitute any real number in the numerator, but we must avoid inputs that make the denominator 0. We find these inputs. 2−x = 0

2=x

The domain is {x|x $= 2}, or (−∞, 2) ∪ (2, ∞). 48. We find the inputs that make the denominator 0: x+4 = 0 x = −4

The domain is {x|x $= −4}, or (−∞, −4) ∪ (−4, ∞). 49. We find the inputs that make the denominator 0: x2 − 4x − 5 = 0

(x − 5)(x + 1) = 0

x − 5 = 0 or x + 1 = 0 x = 5 or

x = −1

The domain is {x|x $= 5 and x $= −1}, or (−∞, −1) ∪ (−1, 5) ∪ (5, ∞). 50. We can substitute any real number in the numerator. Find the inputs that make the denominator 0: 3x2 − 10x − 8 = 0

(3x + 2)(x − 4) = 0

2 x = − or x = 4 3 . ! / ! 2 Domain: x!!x $= − and x $= 4 , or 3 ( ' ( ' 2 2 ∪ , 4 ∪ (4, ∞) − ∞, − 3 3

Exercise Set 1.2 51. We can substitute any real number for which the radicand is nonnegative. We see that 8 − x ≥ 0 for x ≤ 8, so the domain is {x|x ≤ 8}, or (−∞, 8].

55 63.

y 5 4

52. We can substitute any real number for x. Thus, the domain is the set of all real numbers, or (−∞, ∞).

3 2 1

53. We can substitute any real number for x. Thus, the domain is the set of all real numbers, or (−∞, ∞).

-5 -4 -3 -2 -1 0 -1

-5

To find the domain we look for the inputs on the x-axis that correspond to a point on the graph. We see that each point on the x-axis corresponds to a point on the graph so the domain is the set of all real numbers, or (−∞, ∞).

55. The inputs on the x-axis that correspond to points on the graph extend from 0 to 5, inclusive. Thus, the domain is {x|0 ≤ x ≤ 5}, or [0, 5].

To find the range we look for outputs on the y-axis. The number 0 is the smallest output, and every number greater than 0 is also an output. Thus, the range is [0, ∞).

The outputs on the y-axis extend from 0 to 3, inclusive. Thus, the range is {y|0 ≤ y ≤ 3}, or [0, 3].

64.

The outputs on the y-axis start at −3 and increase without bound. Thus, the range is [−3, ∞).

61. The inputs on the x-axis extend from −5 to 3, inclusive. Thus, the domain is [−5, 3]. The outputs on the y-axis extend from −2 to 2, inclusive. Thus, the range is [−2, 2]. 62. The inputs on the x-axis extend from −2 to 4, inclusive. Thus, the domain is [−2, 4]. The only output is 4. Thus, the range is {4}.

1 2 3 4 5

-2 -3 -4 -5

58. The inputs on the x-axis that correspond to points on the graph extend from −2 to 1, inclusive. Thus, the domain is {x| − 2 ≤ x ≤ 1}, or [−2, 1].

60. The graph extends to the left and to the right without bound. Thus, the domain is the set of all real numbers, or (−∞, ∞).

x

-5 -4 -3 -2 -1 0 -1

The outputs on the y-axis extend from −1 to 1, inclusive. Thus, the range is {y| − 1 ≤ y ≤ 1}, or [−1, 1].

The only output is −3, so the range is {−3}.

f(x) = |x | – 2

3 2 1

57. The inputs on the x-axis that correspond to points on the graph extend from −2π to 2π inclusive. Thus, the domain is {x| − 2π ≤ x ≤ 2π}, or [−2π, 2π].

59. The graph extends to the left and to the right without bound. Thus, the domain is the set of all real numbers, or (−∞, ∞).

y 5 4

The outputs on the y-axis extend from −4 up to but not including 1. Thus, the range is {y|−4 ≤ y < 1}, or [−4, 1).

The outputs on the y-axis extend from −1 to 4, inclusive. Thus, the range is {y| − 1 ≤ y ≤ 4}, or [−1, 4].

x 1 2 3 4 5

-2 -3 -4

54. In the numerator we can substitute any real number for which the radicand is nonnegative. We see that x + 1 ≥ 0 for x ≥ −1. The denominator is 0 when x = 0, so 0 cannot be an input. Thus the domain is {x|x ≥ −1 and x $= 0}, or [−1, 0) ∪ (0, ∞).

56. The inputs on the x-axis that correspond to points on the graph extend from −3 up to but not including 5. Thus, the domain is {x| − 3 ≤ x < 5}, or [−3, 5).

f(x) = |x |

Domain: all real numbers, or (−∞, ∞) Range: [−2, ∞) 65.

y 5 4 3 2 1 -5 -4 -3 -2 -1 0 -1

f(x) = 9 – x 2

x 1 2 3 4 5

-2 -3 -4 -5

The inputs on the x-axis extend from −3 to 3, so the domain is [−3, 3]. The outputs on the y-axis extend from 0 to 3, so the range is [0, 3].

56

Chapter 1: Graphs, Functions, and Models y

66.

69.

6

y

5 4 3

10 8

2 1 -6 -5 -4 -3 -2 -1 0 -1 -2

6 4 2

x 1 2 3 4 5 6

f(x) = – 25 – x 2

-10

The largest input on the x-axis is 7 and every number less than 7 is also an input. Thus, the domain is (−∞, 7].

Range: [−5, 0]

The number 0 is the smallest output, and every number greater than 0 is also an output. Thus, the range is [0, ∞).

y 5 4

70.

3 2 1

y 10 8

f(x) = (x – 1)3 + 2 x

-5 -4 -3 -2 -1 0 -1

2 4 6 8 10

-4 -6 -8 -10

Each point on the x-axis corresponds to a point on the graph, so the domain is the set of all real numbers, or (−∞, ∞). Each point on the y-axis also corresponds to a point on the graph, so the range is the set of all real numbers, (−∞, ∞). y

Domain: [−8, ∞) Range: [0, ∞) 71.

y 5 4 f(x) = – x 2 + 4 x – 1

5 4

3 2 1

3 2 1

-2 -3 -4

x+8 x

-10 -8 -6 -4 -2 0 -2

-5

-5 -4 -3 -2 -1 0 -1

f(x) =

6 4 2

1 2 3 4 5

-2 -3 -4

68.

2 4 6 8 10

-4 -6 -8

Domain: [−5, 5] 67.

7–x x

-10 -8 -6 -4 -2 0 -2

-3 -4 -5 -6

f(x) =

x 1 2 3 4 5 f(x) = (x – 2)4 + 1

-5 -4 -3 -2 -1 0 -1

x 1 2 3 4 5

-2 -3 -4 -5

-5

Domain: all real numbers, or (−∞, ∞) Range: [1, ∞)

Each point on the x-axis corresponds to a point on the graph, so the domain is the set of all real numbers, or (−∞, ∞).

The largest output is 3 and every number less than 3 is also an output. Thus, the range is (−∞, 3].

Exercise Set 1.2

57

72.

−21.8, g(5.08) ≈ −130.4, and g(10.003) ≈ −468.3.

y 6

f(x) = 2 x 2 – x 4 + 5

78. a)

5 4 3 2 1 -6 -5 -4 -3 -2 -1 0 -1 -2

x 1 2 3 4 5 6

We see that f (−4) = −0.25. The ERROR message in the table indicates that f (−6) does not exist.

-3 -4 -5 -6

b)

Domain: all real numbers, or (−∞, ∞) Range: (−∞, 6]

73. E(t) = 1000(100 − t) + 580(100 − t)2

a) E(99.5) = 1000(100−99.5)+580(100−99.5)2 = 1000(0.5) + 580(0.5)2 = 500 + 580(0.25) = 500 + 145 = 645 m above sea level

b) E(100) = 1000(100 − 100) + 580(100 − 100)2 = 1000 · 0 + 580(0)2 = 0 + 0

= 0 m above sea level, or at sea level 74. a) V (18) = 0.4123(18) + 13.2617 ≈ $20.68 V (25) = 0.4123(25) + 13.2617 ≈ $23.57 b) Solve: 30 = 0.4123x + 13.2617 x ≈ 41, so it will take about $30 to equal the value of $1 in 1913 approximately 41 yr after 1990, or in 2031. 75.

T (0.5) = 0.51.31 ≈ 0.4 acre

T (10) = 101.31 ≈ 20.4 acres T (20) = 201.31 ≈ 50.6 acres

T (100) = 1001.31 ≈ 416.9 acres 76.

T (200) = 2001.31 ≈ 1033.6 acres

The ERROR message in the table indicates that g(−5) does not exist. We see that g(1) = 3. 79. A function is a correspondence between two sets in which each member of the first set corresponds to exactly one member of the second set. 80. The domain of a function is the set of all inputs of the function. The range is the set of all outputs. The range depends on the domain. 81. To determine whether (0, −7) is a solution, substitute 0 for x and −7 for y. y = 0.5x + 7

−7 ?! 0.5(0) + 7 ! ! 0+7 ! −7 ! 7 FALSE

The equation −7 = 7 is false, so (0, −7) is not a solution.

To determine whether (8, 11) is a solution, substitute 8 for x and 11 for y. y = 0.5x + 7 11 ?! 0.5(8) + 7 ! ! 4+7 ! 11 ! 11 TRUE

We see that h(−11) = 57, 885, h(7) = 4017, and h(15) = 119, 241. 77.

Rounding to the nearest tenth, we see that g(−2.1) ≈

The equation 11 = 11 is true, so (8, 11) is a solution. # "4 , −2 : 15x − 10y = 32 82. For: 5 4 15 · − 10(−2) ? 32 5 ! ! 12 + 20 !! ! 32 ! 32 TRUE ( ' 4 , −2 is a solution. 5

58

Chapter 1: Graphs, Functions, and Models

For:

'

" 11 1 # , : 5 10

11 1 , 5 10

(

88.

15x − 10y = 32

15 ·

4

11 1 − 10 · 5 10 33 − 1

? 32 ! ! ! ! ! 32 ! 32

is a solution.

y

!4 !2

TRUE

y

x

2

4

x

y 4

Make a table of values, plot the points in the table, and draw the graph. x

4

!4

89.

83. Graph y = (x − 1)2 .

2 !2

2 !4 !2

(x, y)

!2

−1 4 (−1, 4)

y

!4

0

1

(0, 1)

4

1

0

(1, 0)

2

2

1

(2, 1)

3

4

(3, 4)

!4

!2

90. 2

!2

4

x

y " (x !1)2

!4

f (x − 1) = 5x

f (6) = f (7 − 1) = 5 · 7 = 35 91. First find the value of x for which x + 3 = −1. x + 3 = −1 x = −4

84.

Then we have:

y

g(x + 3) = 2x + 1

2 !4

!2

2 !2 !4

4

g(−1) = g(−4 + 3) = 2(−4) + 1 = −8 + 1 = −7

x

92. f (x) = |x + 3| −| x − 4|

1

y"3x!6

a) If x is in the interval (−∞, −3), then x + 3 < 0 and x − 4 < 0. We have: f (x) = |x + 3| −| x − 4| = −(x + 3) − [−(x − 4)]

85. Graph −2x − 5y = 10.

Make a table of values, plot the points in the table, and draw the graph. x

y

(x, y)

−5

0

(−5, 0)

0 5

4 2 !4

!2

2

4

x

!2

−4 (5, −4)

!4

!2x ! 5y " 10

86.

= x + 3 − [−(x − 4)] = x + 3 − (−x + 4) = x+3+x−4

c) If x is in the interval [4, ∞), then x + 3 > 0 and x − 4 ≥ 0. We have: f (x) = |x + 3| −| x − 4|

4 2 2

4

x

! 3)2

# y2

= x + 3 − (x − 4) = x+3−x+4

!2 !4

b) If x is in the interval [−3, 4), then x + 3 ≥ 0 and x − 4 < 0. We have: f (x) = |x + 3| −| x − 4|

= 2x − 1

y

!4 !2

= −x − 3 + x − 4 = −7

y

−2 (0, −2)

= −(x + 3) − (−x + 4)

(x

"4

87. Answers may vary. Two possibilities are f (x) = x, g(x) = x + 1 and f (x) = x2 , g(x) = x2 − 4.

=7

93. f (x) = |x| + |x − 1|

a) If x is in the interval (−∞, 0), then x < 0 and x − 1 < 0. We have:

Exercise Set 1.3 f (x) = |x| + |x − 1|

= −x − (x − 1) = −x − x + 1 = −2x + 1

b) If x is in the interval [0, 1), then x ≥ 0 and x−1 < 0. We have: f (x) = |x| + |x − 1|

59 8. m = 9. m = 10. m =

= x − (x − 1)

11. m =

=1

12. m =

= x−x+1

c) If x is in the interval [1, ∞), then x > 0 and x − 1 ≥ 0. We have: f (x) = |x| + |x − 1| = x+x−1 = 2x − 1

Exercise Set 1.3

b) Yes. Each output is 3 more than the one that precedes it.

16. m =

18. m =

c) No. Constant changes in inputs do not result in constant changes in outputs.

19. m =

3. a) Yes. Each input is 15 more than the one that precedes it.

20. m =

c) No. Constant changes in inputs do not result in constant changes in outputs. 4. a) Yes. Each input is 2 more than the one that precedes it.

y2 − y1 6 − (−9) 15 5 = = =− x2 − x1 −5 − 4 −9 3

−13 − (−1) −12 3 = =− 2 − (−6) 8 2

−2 − (−2) 0 y2 − y1 = = =0 x2 − x1 4−2 2

−14 7 −6 − 8 = =− 7 − (−9) 16 8 " ! 6 3 3 − − 6 y2 − y1 5 5 5 = 17. m = = =− 1 1 x2 − x1 −1 5 − − 2 2

b) No. The change in the outputs varies.

b) No. The change in the outputs varies.

−1 − 7 −8 = = −1 5 − (−3) 8

−0.4 − (−0.1) −0.3 y2 − y1 = = = 0.3 x2 − x1 −0.3 − 0.7 −1 20 13 7 5 ! 1" − + − − − − 7 4 28 28 28 = = = 14. m = 2 ! 3" 8 21 29 − − + 7 4 28 28 28 13 13 28 − · =− 28 29 29 15. m =

2. a) Yes. Each input is 10 more than the one that precedes it.

2−4 −2 1 y2 − y1 = = = x2 − x1 −1 − 9 −10 5

13. m =

1. a) Yes. Each input is 1 more than the one that precedes it.

c) Yes. Constant changes in inputs result in constant changes in outputs.

1 − (−4) 5 = 5 − (−3) 8

−2.16 − 4.04 −6.2 62 31 = =− =− 3.14 − (−8.26) 11.4 114 57 −5 − (−13) 8 1 y2 − y1 = = =− x2 − x1 −8 − 16 −24 3

7 − (−7) 14 = 10 − (−10) 0 The slope is not defined.

y2 − y1 2 − (−3) 5 = = x2 − x1 π−π 0 Since division by 0 is not defined, the slope is not defined.

21. m =

0 −4 − (−4) √ = √ =0 0.56 − 2 0.56 − 2

b) Yes. Each output is 4 less than the one that precedes it.

22. m =

c) Yes. Constant changes in inputs result in constant changes in outputs.

23. We have the points (4, 3) and (−2, 15). 15 − 3 12 y2 − y1 = = −2 = m= x2 − x1 −2 − 4 −6

5. Two points on the line are (0, 3) and (5, 0). 0−3 −3 3 y2 − y1 = = , or − m= x2 − x1 5−0 5 5 3 0 − (−3) = −2 − (−2) 0 The slope is not defined.

6. m =

7. m =

0 3−3 y2 − y1 = =0 = x2 − x1 3−0 3

−6 3 −5 − 1 = = −4 − 4 −8 4 $ # $ # 11 1 1 , and − 1, − . 25. We have the points 5 2 2 11 1 # $ − − −6 5 y2 − y1 2 2 = = −6 · − = =5 m= 1 6 x2 − x1 6 −1 − − 5 5

24. m =

60

Chapter 1: Graphs, Functions, and Models

13 10 # $ − (−1) 13 3 1 3 26. m = = 3 = · − =− 2 26 3 26 2 − −8 − 3 3

y 4 2

1 27. y = − x + 3 2 We find and plot two ordered pairs on the line and connect the points. 1 When x = 0, y = − · 0 + 3 = 3. 2 1 When x = 2, y = − · 2 + 3 = −1 + 3 = 2. 2

2

!4

4

x

2

4

x

!2 !4

Two points on the graph are (−3, 2) and (1, 0). −2 1 0−2 = =− m= 1 − (−3) 4 2

31. 5x + 2y = 10

y 4

We find and plot two ordered pairs on the line and connect the points.

2

First let y = 0 and solve for x. 2

!4 !2

4

x

!4

5x + 2 · 0 = 10

5x = 10

!2

m=

x # 2y " 1

!4 !2

1 y " !!x 2 #3

1 Since the equation is in the form y = mx+b with m = − , 2 1 we know that the slope is − . 2

2y ! 3x " !6

y

2 2

x

Now find the slope. y2 − y 1 −3 − 0 −3 3 m= = = = x2 − x1 0−2 −2 2

4

!2

28.

!4

4

!4 !2

4

!2

30.

y

2

!4 !2

x=2

3 y " !x 2 !4

The point (2, 0) is on the graph.

3 2

Next let x = 0 and solve for y.

29. 2y − 3x = −6

We find and plot two ordered pairs on the line and connect the points. First let y = 0 and solve for x. 2 · 0 − 3x = −6

5 · 0 + 2y = 10

2y = 10 y=5

The point (0, 5) is on the graph. Plot the points (2, 0) and (0, 5) and connect them.

−3x = −6

y

x=2

The point (2, 0) is on the graph.

4

Next let x = 0 and solve for y.

2

2y − 3 · 0 = −6 2y = −6 y = −3

The point (0, −3) is on the graph.

Plot the points (2, 0) and (0, −3) and connect them.

5x # 2y " 10

2

!4 !2

4

!2 !4

Now find the slope 5 5 5−0 = =− m= 0−2 −2 2

x

Exercise Set 1.3 32.

61 3 is a horizontal line, so the slope is 4 0. (We also see this if we write the equation in the form 3 y = 0x + .) 4

40. The graph of y =

y 4 2 2

!4 !2

4

x

!2 !4

2y ! x " 8

Two points on the graph are (0, 4) and (2, 5). 1 5−4 = m= 2−0 2

2 3 This is the equation of a horizontal line whose graph is 2 units below the x-axis. 3

33. y = −

y 4 2 2

!4 !2 !2 !4

4

x

2 y " !! 3

The slope of a horizontal line is 0. 34.

y x "3

4 2 2

!4 !2

4

x

!2 !4

The slope of a vertical line is not defined. 35. y = 1.3x − 5 is in the form y = mx + b with m = 1.3, so the slope is 1.3. 36. −

2 5

37. The graph of x = −2 is a vertical line, so the slope is not defined. 11 3 38. Solving 3x − 4y = −11 for y, we get y = x + , so the 4 4 3 slope is . 4 39. First solve the equation for y. 10y + x = 9 10y = −x + 9 9 1 y = − x+ 10 10 9 1 is in the form y = mx + b The equation y = − x + 10 10 1 1 with m = − , so the slope is − . 10 10

41. We have the points (1998, 1.1) and (2003, 1.3). We find the average rate of change, or slope. 0.2 1.3 − 1.1 = = 0.04 m= 2003 − 1998 5 The average rate of change over the 5-yr period was 0.04 billion, or 40 million, visits per year. 38.3 40.3 − 2 = = 1.915 2005 − 1985 20 The average rate of change over the 20-year period was 1.915 million cases per year.

42. m =

43. The change in attendance was 107, 097 − 59, 368 = 47, 729. We divide this number by the number of races, 29. 47, 729 ≈ 1646 29 The average rate of change in attendance for the 29 races was about 1646 per race. −8.0 8.3 − 16.3 = ≈ −0.89 2004 − 1995 9 The average rate of change over the 9-yr period was about −0.89% per year.

44. m =

45. We have the points (1994, 43) and (2004, 18.1). We find the average rate of change, or slope. −24.9 18.1 − 43 = = −2.49 m= 2004 − 1994 10 The average rate of change over the 10-yr period was −2.49 per 1000 women per year.

5625 8900 − 3275 = ≈ 401.79 2006 − 1992 14 The average rate of change over the 14-yr period was about $401.79 per year.

46. m =

47. First we convert the times to minutes. 1 min 10 sec 1 10 sec = 10 sec · = · · min = min 60 sec 60 sec 6 1 1 Thus, 31 min 10 sec = 31 min + min = 31 min, or 6 6 187 min. 6 1 hr = 60 min 38 sec 19 1 min = · · min = min 38 sec = 38 sec · 60 sec 60 sec 30 19 Thus, 1 hr 1 min 38 sec = 60 min + 1 min + min = 30 1849 19 min, or min. 61 30 30 Now we find the speed. 5 5 30 10 − 5 = = =5· = 1849 187 1849 935 914 914 − − 30 6 30 30 30 / · 75 2 75 1 150 = = ≈ 914 / · 457 2 457 6 1 75 mi per minute, or about mi per Lucie’s speed was 457 6 minute.

62 48.

Chapter 1: Graphs, Functions, and Models 3 1 − 4 6 Typing rate = 6 7 12 = 6 7 of the paper per hour = 72

49. a) W (h) = 4h − 130

b) W (62) = 4 · 62 − 130 = 248 − 130 = 118 lb

c) Both the height and weight must be positive. Solving h > 0 and 4h−130 > 0, we find that the domain of the function is {h|h > 32.5}, or (32.5, ∞).

1 d+1 33 1 a) P (0) = · 0 + 1 = 1 atm 33 1 5 P (5) = ·5+1=1 atm 33 33 10 1 · 10 + 1 = 1 atm P (10) = 33 33 1 P (33) = · 33 + 1 = 2 atm 33 1 233 2 P (200) = · 200 + 1 = atm, or 7 atm 33 33 33 b) The depth must be nonnegative, so the domain is {d|d ≥ 0}, or [0, ∞).

50. P (d) =

51. D(F ) = 2F + 115 a) D(0) = 2 · 0 + 115 = 115 ft

D(−20) = 2(−20) + 115 = −40 + 115 = 75 ft D(10) = 2 · 10 + 115 = 20 + 115 = 135 ft D(32) = 2 · 32 + 115 = 64 + 115 = 179 ft

b) Below −57.5◦ , stopping distance is negative; above 32◦ , ice doesn’t form. 52. a)

M (x) = 2.89x + 70.64 M (26) = 2.89(26) + 70.64 = 145.78 cm

b) The length of the humerus must be positive, so the domain is {x|x > 0}, or (0, ∞). Realistically, however, we might expect the length of the humerus to be between 20 cm and 60 cm, so the domain could be {x|20 ≤ x ≤ 60}, or [20, 60]. Answers may vary.

11 5 11r + 5 = r+ 53. a) D(r) = 10 10 10 11 . The slope is 10

11 For each mph faster the car travels, it takes ft 10 longer to stop.

b)

60 11 · 5 + 5 = = 6 ft 10 10 11 · 10 + 5 115 D(10) = = = 11.5 ft 10 10 11 · 20 + 5 225 = = 22.5 ft D(20) = 10 10 D(5) =

11 · 50 + 5 555 = = 55.5 ft 10 10 720 11 · 65 + 5 = = 72 ft D(65) = 10 10 1 c) The speed cannot be negative. D(0) = which 2 1 ft before stopsays that a stopped car travels 2 ping. Thus, 0 is not in the domain. The speed can be positive, so the domain is {r|r > 0}, or (0, ∞). D(50) =

54. V (t) = $5200 − $512.50t

a) V (0) = $5200 − $512.50(0) = $5200 − $0 = $5200

V (1) = $5200 − $512.50(1) =

$5200 − $512.50 = $4687.50

V (2) = $5200 − $512.50(2) =

$5200 − $1025 = $4175

V (3) = $5200 − $512.50(3) =

$5200 − $1537.50 = $3662.50

V (8) = $5200 − $512.50(8) =

$5200 − $4100 = $1100

b) Since the time must be nonnegative and not more than 8 years, the domain is [0, 8]. The value starts at $5200 and declines to $1100, so the range is [1100, 5200]. 55. C(t) = 60 + 29t C(6) = 60 + 29 · 6 = $234 56. C(t) = 65 + 80t C(8) = 65 + 80 · 8 = $705 57. Let x = the number of shirts produced. C(x) = 800 + 3x C(75) = 800 + 3 · 75 = $1025 58. Let x = the number of rackets restrung. C(x) = 950 + 18x C(150) = 950 + 18 · 150 = $3650 59. Left to the student. 60. The sign of the slope indicates the slant of a line. A line that slants up from left to right has positive slope because corresponding changes in x and y have the same sign. A line that slants down from left to right has negative slope, because corresponding changes in x and y have opposite signs. A horizontal line has zero slope, because there is no change in y for a given change in x. A vertical line has undefined slope, because there is no change in x for a given change in y and division by 0 is undefined. The larger the absolute value of slope, the steeper the line. This is because a larger absolute value corresponds to a greater change in y, compared to the change in x, than a smaller absolute value. 61. A vertical line (x = a) crosses the graph more than once.

Exercise Set 1.4

63

62. f (5) = 52 − 3 · 5 = 10 63. 64.

72.

f (x) = x2 − 3x

f (−5) = (−5)2 − 3(−5) = 25 + 15 = 40 f (x) = x2 − 3x

f (−a) = (−a) − 3(−a) = a + 3a 2

2

65. f (x) = x2 − 3x

f (a + h) = (a + h)2 − 3(a + h) = a2 + 2ah + h2 − 3a − 3h

66. We make a drawing and label it. Let h = the height of the triangle, in feet.

5 ft h x

73. False. For example, let f (x) = x + 1. Then f (cd) = cd + 1, but f (c)f (d) = (c + 1)(d + 1) = cd + c + d + 1 '= cd + 1 for c '= −d. 74. False. For example, let f (x) = x + 1. Then f (c + d) = c + d + 1, but f (c) + f (d) = c + 1 + d + 1 = c + d + 2. 75. False. For example, let f (x) = x + 1. Then f (c − d) = c − d + 1, but f (c) − f (d) = c + 1 − (d + 1) = c − d. 76. False. For example, let f (x) = x+1. Then f (kx) = kx+1, but kf (x) = k(x + 1) = kx + k '= kx + 1 for k '= 1. 77.

m(x + 2) + b = mx + b + 2 mx + 2m + b = mx + b + 2

x + h = 25 2

2m = 2

x2 = 25 − h2 √ x = 25 − h2

We know that the grade of the treadmill is 8%, or 0.08. Then we have h = 0.08 x √ h √ = 0.08 Substituting 25 − h2 for x 2 25 − h h2 = 0.0064 Squaring both sides 25 − h2 h2 = 0.16 − 0.0064h2 1.0064h2 = 0.16

0.16 h = 1.0064 h ≈ 0.4 ft

f (x) = mx + b f (x + 2) = f (x) + 2

Using the Pythagorean theorem we have: 2

3(a + h) + 1 − (3a + 1) = a+h−a 3a + 3h + 1 − 3a − 1 3h = =3 h h

m=

m=1 Thus, f (x) = 1 · x + b, or f (x) = x + b. 78.

3mx + b = 3(mx + b) 3mx + b = 3mx + 3b b = 3b 0 = 2b 0=b Thus, f (x) = mx + 0, or f (x) = mx.

Exercise Set 1.4

2

67. m =

−2d − (−d) −2d + d −d d y2 − y1 = = = =− x2 − x1 9c − (−c) 9c + c 10c 10c

s−s−t s − (s + t) = r−r 0 The slope is not defined.

68. m =

69. m =

y2 − y1 z−z 0 = = = x2 − x1 z − q − (z + q) z−q−z−q

0 =0 −2q 70. m =

p − q − (p + q) p−q−p−q −2q = = = a + b − (−a − b) a+b+a+b 2a + 2b

q a+b (a+h)2 −a2 y2 − y1 a2 +2ah+h2 −a2 = = = m= x2 − x1 a+h−a h

− 71.

2ah + h2 h(2a + h) = = 2a + h h h

1. y =

3 x−7 5

3 The equation is in the form y = mx + b where m = 5 3 and b = −7. Thus, the slope is , and the y-intercept is 5 (0, −7). 2. f (x) = −2x + 3

Slope: −2; y-intercept: (0, 3)

3. x = −

2 5

2 This is the equation of a vertical line unit to the left 5 of the y-axis. The slope is not defined, and there is no y-intercept.

4. y =

4 4 =0·x+ 7 7

Slope: 0; y-intercept:

#

0,

4 7

$

64

Chapter 1: Graphs, Functions, and Models

1 1 5. f (x) = 5 − x, or f (x) = − x + 5 2 2 The second equation is in the form y = mx + b where 1 1 m = − and b = 5. Thus, the slope is − and the y2 2 intercept is (0, 5). 3 6. y = 2 + x 7 3 Slope: ; y-intercept: (0, 2) 7 7. Solve the equation for y. 3x + 2y = 10 2y = −3x + 10 3 y = − x+5 2 3 Slope: − ; y-intercept: (0, 5) 2 8. 2x − 3y = 12

−3y = −2x + 12 2 y = x−4 3 2 Slope: ; y-intercept: (0, −4) 3 9. y = −6 = 0 · x − 6 Slope: 0; y-intercept: (0, −6)

10. x = 10 This is the equation of a vertical line 10 units to the right of the y-axis. The slope is not defined, and there is no y-intercept. 11. Solve the equation for y. 5y − 4x = 8

12.

4 ; y-intercept: 5

5x − 2y + 9 = 0

16. We see that the y-intercept is (0, −1). Another point on the graph is (3, 1). 2 1 − (−1) = m= 3−0 3 2 The equation is y = x − 1. 3 17. We see that the y-intercept is (0, −3). This is a horizontal line, so the slope is 0. We have m = 0 and b = −3, so the equation is y = 0 · x − 3, or y = −3. 18. We see that the y-intercept is (0, 0). Another point on the graph is (3, 3). 3 3−0 = =1 m= 3−0 3 The equation is y = 1 · x + 0, or y = x. 2 for m and 4 for b in the slope-intercept 19. We substitute 9 equation. y = mx + b 2 y = x+4 9 3 20. y = − x + 5 8

21. We substitute −4 for m and −7 for b in the slope-intercept equation. y = mx + b y = −4x − 7

5y = 4x + 8 8 4 y = x+ 5 5

Slope:

15. We see that the y-intercept is (0, 0). Another point on the graph is (3, −3). Use these points to find the slope. y2 − y1 −3 − 0 −3 m= = = = −1 x2 − x1 3−0 3 We have m = −1 and b = 0, so the equation is y = −1 · x + 0, or y = −x.

22. y = #

0,

8 5

$

−2y = −5x − 9 9 5 y = x+ 2 2 # $ 5 9 Slope: ; y-intercept: 0, 2 2 13. We see that the y-intercept is (0, −2). Another point on the graph is (1, 2). Use these points to find the slope. 2 − (−2) 4 y2 − y1 = = =4 m= x2 − x1 1−0 1 We have m = 4 and b = −2, so the equation is y = 4x − 2. 14. We see that the y-intercept is (0, 2). Another point on the graph is (4, −1). 3 −1 − 2 =− m= 4−0 4 3 The equation is y = − x + 2. 4

2 x−6 7

3 23. We substitute −4.2 for m and for b in the slope-intercept 4 equation. y = mx + b 3 y = −4.2x + 4 3 24. y = −4x − 2 25. Using the point-slope equation: y − y1 = m(x − x1 ) 2 Substituting y − 7 = (x − 3) 9 2 2 y−7 = x− 9 3 2 19 y = x+ Slope-intercept equation 9 3 Using the slope-intercept equation: 2 for m, 3 for x, and 7 for y in the slopeSubstitute 9 intercept equation and solve for b.

Exercise Set 1.4

65

y = mx + b 2 7 = ·3+b 9 2 7 = +b 3 19 =b 3 2 19 Now substitute for m and for b in y = mx + b. 9 3 19 2 y = x+ 9 3 26. Using the point-slope equation: 3 y − 6 = − (x − 5) 8 63 3 y = − x+ 8 8 Using the slope-intercept equation: 3 6 = − ·5+b 8 63 =b 8 63 3 We have y = − x + . 8 8 27. Using the point-slope equation: y − y1 = m(x − x1 )

y − (−2) = 3(x − 1) y + 2 = 3x − 3 y = 3x − 5

Using the slope-intercept equation: y = mx + b 3 −1 = − (−4) + b 5 12 −1 = +b 5 17 =b − 5 17 3 Then we have y = − x − . 5 5 30. Using the point-slope equation: 2 y − (−5) = (x − (−4)) 3 7 2 y = x− 3 3 Using the slope-intercept equation: 2 −5 = (−4) + b 3 7 − =b 3 2 7 We have y = x − . 3 3 31. First we find the slope. −9 −4 − 5 = = −3 m= 2 − (−1) 3 Using the point-slope equation: Using the point (−1, 5), we get

Slope-intercept equation

Using the slope-intercept equation: y = mx + b −2 = 3 · 1 + b −2 = 3 + b −5 = b

Then we have y = 3x − 5. 28. Using the point-slope equation: y − 1 = −2(x − (−5)) y = −2x − 9

Using the slope-intercept equation: 1 = −2(−5) + b

−9 = b

We have y = −2x − 9. 29. Using the point-slope equation: y − y1 = m(x − x1 ) 3 y − (−1) = − (x − (−4)) 5 3 y + 1 = − (x + 4) 5 3 12 y+1 = − x− 5 5 17 3 Slope-intercept y = − x− 5 5 equation

y − 5 = −3(x − (−1)), or y − 5 = −3(x + 1).

Using the point (2, −4), we get

y − (−4) = −3(x − 2), or y + 4 = −3(x − 2).

In either case, the slope-intercept equation is y = −3x + 2.

Using the slope-intercept equation and the point (−1, 5): y = mx + b 5 = −3(−1) + b 5 = 3+b 2=b Then we have y = −3x + 2. 32. First we find the slope. −10 −11 − (−1) = = −2 m= 7−2 5 Using the point-slope equation: Using (2, −1): y − (−1) = −2(x − 2), or y + 1 = −2(x − 2)

Using (7, −11): y − (−11) = −2(x − 7), or y + 11 = −2(x − 7)

In either case, we have y = −2x + 3.

Using the slope-intercept equation and the point (2, −1): −1 = −2 · 2 + b 3=b

We have y = −2x + 3.

66

Chapter 1: Graphs, Functions, and Models

33. First we find the slope. 4−0 4 1 m= = =− −1 − 7 −8 2 Using the point-slope equation: Using the point (7, 0), we get 1 y − 0 = − (x − 7). 2 Using the point (−1, 4), we get 1 y − 4 = − (x − (−1)), or 2 1 y − 4 = − (x + 1). 2 In either case, the slope-intercept equation is 7 1 y =− x+ . 2 2 Using the slope-intercept equation and the point (7, 0): 1 0 = − ·7+b 2 7 =b 2 7 1 Then we have y = − x + . 2 2

1 37. Graph y = − x − 3. 2 Plot the y-intercept, (0, −3). We can think of the slope −1 as . Start at (0, −3) and find another point by moving 2 down 1 unit and right 2 units. We have the point (2, −4). 1 . Then we can start We could also think of the slope as −2 at (0, −3) and get another point by moving up 1 unit and left 2 units. We have the point (−2, −2). Connect the three points to draw the graph. y 4 2 !4 !2

y + 5 = −6(x + 1)

38.

x

2

4

x

y 4 2 !4 !2 !2 !4

3 y " !x 2 #1

39. Graph f (x) = 3x − 1.

Plot the y-intercept, (0, −1). We can think of the slope 3 as . Start at (0, −1) and find another point by moving 1 up 3 units and right 1 unit. We have the point (1, 2). We can move from the point (1, 2) in a similar manner to get a third point, (2, 5). Connect the three points to draw the graph.

In either case, we have y = −6x − 11.

Using the slope-intercept equation and the point (−1, −5): −5 = −6(−1) + b

−11 = b

We have y = −6x − 11.

y

35. First we find the slope. 2 −4 − (−6) = m= 3−0 3 We know the y-intercept is (0, −6), so we substitute in the slope-intercept equation.

4 2 2

!4 !2

4

x

!2

y = mx + b 2 y = x−6 3 36. First we find the slope. 4 4 −0 4 5 5 = = m= 0 − (−5) 5 25 $ # 4 , so we substitute in the We know the y-intercept is 0, 5 slope-intercept equation. 4 4 y= x+ 25 5

4

!4

Using (−3, 7): y − 7 = −6(x − (−3)), or y − 7 = −6(x + 3)

2 !2

34. First we find the slope. −12 −5 − 7 = = −6 m= −1 − (−3) 2 Using the point-slope equation:

Using (−1, −5): y − (−5) = −6(x − (−1)), or

1 y " !!x 2 !3

!4

40.

f(x) " 3x ! 1

y 4

f(x) " !2x # 5

2 2

!4 !2 !2 !4

4

x

Exercise Set 1.4

67

41. First solve the equation for y.

44.

y

3x − 4y = 20

4

−4y = −3x + 20 3 y = x−5 4

2

y

3x ! 4y " 20 2

!4 !2

4

4

x

!2

Plot the y-intercept, (0, −5). Then using the slope,

2

2

!4 !2

3 , 4 start at (0, −5) and find another point by moving up 3 units and right 4 units. We have the point (4, −2). We can move from the point (4, −2) in a similar manner to get a third point, (8, 1). Connect the three points to draw the graph.

5y ! 2x " !20

!4 !6

45. We have the points (1, 4) and (−2, 13). First we find the slope. 13 − 4 9 m= = = −3 −2 − 1 −3 We will use the point-slope equation, choosing (1, 4) for the given point. y − 4 = −3(x − 1)

x

y − 4 = −3x + 3

!2

y = −3x + 7, or

!4

h(x) = −3x + 7

!6

9 4 3 − (−6) # $= =9· =4 9 1 9 2− − 4 4 Using the point-slope equation and the point (2, 3):

46. m =

42.

y 6

2x # 3y " 15

y − 3 = 4(x − 2)

4

y − 3 = 4x − 8

2 2

!4 !2

4

x

43. First solve the equation for y. x + 3y = 18 3y = −x + 18 1 y = − x+6 3 Plot the y-intercept, (0, 6). We can think of the slope as −1 . Start at (0, 6) and find another point by moving down 3 1 unit and right 3 units. We have the point (3, 5). We can move from the point (3, 5) in a similar manner to get a third point, (6, 4). Connect the three points and draw the graph. y 6 4 2

x # 3y " 18 2

!4 !2 !2

y = 4x − 5, or

g(x) = 4x − 5

!2

4

x

47. We have the points (5, 1) and (−5, −3). First we find the slope. −4 2 −3 − 1 = = m= −5 − 5 −10 5 We will use the slope-intercept equation, choosing (5, 1) for the given point. y = mx + b 2 1 = ·5+b 5 1 = 2+b −1 = b

2 x − 1. 5 2−3 −1 1 48. m = = =− 0 − (−3) 3 3 Using the slope-intercept equation and the point (0, 2), 1 which is the y-intercept, we have h(x) = − x + 2. 3 Then we have f (x) =

49. The equation of the horizontal line through (0, −3) is of the form y = b where b is −3. We have y = −3. The equation of the vertical line through (0, −3) is of the form x = a where a is 0. We have x = 0.

50. Horizontal line: y = 7 1 Vertical line: x = − 4

68

Chapter 1: Graphs, Functions, and Models $ 2 , −1 is 11 of the form y = b where b is −1. We have y = −1. $ # 2 The equation of the vertical line through , −1 is of 11 2 2 the form x = a where a is . We have x = . 11 11

51. The equation of the horizontal line through

#

52. Horizontal line: y = 0 Vertical line: x = 0.03 3 26 and − . Their product is −1, so the 53. The slopes are 3 26 lines are perpendicular. 1 54. The slopes are −3 and − . The slopes are not the same 3 and their product is not −1, so the lines are neither parallel nor perpendicular. 2 2 55. The slopes are and − . The slopes are not the same and 5 5 their product is not −1, so the lines are neither parallel nor perpendicular. $ # 3 = 1.5 and the y-intercepts, 56. The slopes are the same 2 −8 and 8, are different, so the lines are parallel. 57. We solve each equation for y. x + 2y = 5

2x + 4y = 8

5 1 y = − x+ 2 2

1 y = − x+2 2 1 1 We see that m1 = − and m2 = − . Since the slopes are 2 2 5 the same and the y-intercepts, and 2, are different, the 2 lines are parallel. 58.

2x − 5y = −3 2x + 5y = 4 3 2 4 2 y = − x+ y = x+ 5 5 5 5 2 2 4 m1 = , m2 = − ; m1 '= m2 ; m1 m2 = − '= −1 5 5 25 The lines are neither parallel nor perpendicular.

59. We solve each equation for y. y = 4x − 5

60.

4y = 8 − x 1 y = − x+2 4

1 We see that m1 = 4 and m2 = − . Since 4 $ # 1 = −1, the lines are perpendicular. m 1 m2 = 4 − 4 y = 7 − x, y = x+3

m1 = −1, m2 = 1; m1 m2 = −1 · 1 = −1

The lines are perpendicular.

61. y =

2 2 x + 1; m = 7 7

2 . We 7 2 use the point-slope equation for a line with slope and 7 containing the point (3, 5): The line parallel to the given line will have slope

y − y1 = m(x − x1 ) 2 y − 5 = (x − 3) 7 2 6 y−5 = x− 7 7 2 29 Slope-intercept form y = x+ 7 7 The slope of the line perpendicular to the given line is the 2 7 opposite of the reciprocal of , or − . We use the point7 2 7 slope equation for a line with slope − and containing the 2 point (3, 5): y − y1 = m(x − x1 ) 7 y − 5 = − (x − 3) 2 21 7 y−5 = − x+ 2 2 31 7 Slope-intercept form y = − x+ 2 2 62. f (x) = 2x + 9 1 1 m = 2, − = − m 2 Parallel line: y − 6 = 2(x − (−1)) y = 2x + 8

1 Perpendicular line: y − 6 = − (x − (−1)) 2 11 1 y =− x+ 2 2 63. y = −0.3x + 4.3; m = −0.3

The line parallel to the given line will have slope −0.3. We use the point-slope equation for a line with slope −0.3 and containing the point (−7, 0): y − y1 = m(x − x1 )

y − 0 = −0.3(x − (−7)) y = −0.3x − 2.1

Slope-intercept form

The slope of the line perpendicular to the given line is the 10 1 = . opposite of the reciprocal of −0.3, or 0.3 3 10 We use the point-slope equation for a line with slope 3 and containing the point (−7, 0): y − y1 = m(x − x1 ) 10 y−0 = (x − (−7)) 3 70 10 x+ Slope-intercept form y= 3 3

Exercise Set 1.4 64.

69

2x + y = −4

y = −2x − 4

1 1 = m 2 Parallel line: y − (−5) = −2(x − (−4))

m = −2, −

y = −2x − 13

Perpendicular line: y − (−5) =

1 (x − (−4)) 2

1 y = x−3 2 65.

3x + 4y = 5 4y = −3x + 5 5 3 3 y = − x+ ; m=− 4 4 4

3 The line parallel to the given line will have slope − . We 4 3 use the point-slope equation for a line with slope − and 4 containing the point (3, −2):

y − y1 = m(x − x1 ) 3 y − (−2) = − (x − 3) 4 9 3 y+2 = − x+ 4 4 3 1 y = − x+ Slope-intercept form 4 4 The slope of the line perpendicular to the given line is the 4 3 opposite of the reciprocal of − , or . We use the point4 3 4 slope equation for a line with slope and containing the 3 point (3, −2): y − y1 = m(x − x1 ) 4 y − (−2) = (x − 3) 3 4 y+2 = x−4 3 4 y = x − 6 Slope-intercept form 3

66.

67. x = −1 is the equation of a vertical line. The line parallel to the given line is a vertical line containing the point (3, −3), or x = 3.

The line perpendicular to the given line is a horizontal line containing the point (3, −3), or y = −3.

68. y = −1 is a horizontal line. Parallel line: y = −5

Perpendicular line: x = 4 69. x = −3 is a vertical line and y = 5 is a horizontal line, so it is true that the lines are perpendicular. 70. The slope of y = 2x − 3 is 2, and the slope of y = −2x − 3 is −2. Since 2(−2) = −4 '= −1, it is false that the lines are perpendicular. 2 , and different y5 intercepts, (0, 4) and (0, −4), so it is true that the lines are parallel.

71. The lines have the same slope,

3 72. y = 2 is a horizontal line 2 units above the x-axis; x = − 4 3 is a vertical line unit to the left of the y-axis. Thus it is 4 3 true that their intersection is the point unit to the left 4 $ # 3 of the y-axis and 2 units above the x-axis, or − , 2 . 4 73. x = −1 and x = 1 are both vertical lines, so it is false that they are perpendicular. 4 2 2 74. The slope of 2x+3y = 4, or y = − x + is − ; the slope 3 3 3 3 3 2 3 of 3x − 2y = 4, or y = x − 2, is . Since − · = −1, it 2 2 3 2 is true that the lines are perpendicular. 75. a) Answers will vary depending on the data points used. We will use (0, 96.7) and (6, 121.2). 24.5 121.2 − 96.7 = ≈ 4.08 m= 6−0 6 We know the y-intercept is (0, 96.7), so we use the slope-intercept equation. y = mx + b

y = 4.2(x − 3) + 1

y = 4.2x − 11.6 1 1 5 m = 4.2; − = − =− m 4.2 21 Parallel line: y − (−2) = 4.2(x − 8) y = 4.2x − 35.6

5 (x − 8) 21 5 2 y =− x− 21 21

Perpendicular line: y − (−2) = −

y = 4.08x − 96.7

Since we have 245 5 · 49 / 49 967 24.5 = = = and 96.7 = , 6 60 / · 12 5 12 10 49 967 x+ . we can also write the equation as y = 12 10 b) In 2007, x = 2007 − 1995 = 12. y = 4.08(12) + 96.7 = 145.66

The number of twin births in 2007 will be about 145.66 thousand, or 145,660. In 2010, x = 2010 − 1995 = 15. y = 4.08(15) + 96.7 = 157.9

The number of twin births in 2010 will be about 157.9 thousand, or 157,900.

70

Chapter 1: Graphs, Functions, and Models

76. a) Answers will vary depending on the data points used. We will use (2, 4551) and (8, 6885). 2334 6885 − 4551 = = 389 m= 8−2 6 Using the point-slope equation: y − 4551 = 389(x − 2)

y − 4551 = 389x − 778

y = 389x + 3773

b) 2008: y = 389(15) + 3773 = 9608 triplet births 2012: y = 389(19) + 3773 = 11, 164 triplet births 77. Answers will vary depending on the data points used. We will use (1, 47.37) and (4, 50.64). 3.27 50.64 − 47.37 = = 1.09 m= 4−1 3 Using the point-slope equation and the point (1, 47.37): y − 47.37 = 1.09(x − 1)

y − 47.37 = 1.09x − 1.09

y = 1.09x + 46.28

In 2007, x = 2007 − 2000 = 7. y = 1.09(7) + 46.28 = $53.91

In 2009, x = 2009 − 2000 = 9 y = 1.09(9) + 46.28 = $56.09

78. Answers will vary depending on the data points used. We will use (0, 146) and (3, 203). 57 203 − 146 = = 19 m= 3−0 3 Substituting in the slope-intercept equation, we have y = 19x + 146. 2006: y = 19 · 6 + 146 = $260 billion

2011: y = 19 · 11 + 146 = $355 billion

79. Answers will vary depending on the data points used. We will use (10, 321.10) and (30, 844.60). 523.5 844.60 − 321.10 = = 26.175 m= 30 − 10 20 Using the slope-intercept equation and the point (10, 321.10): y = mx + b 321.10 = 26.175(10) + b 321.10 = 261.75 + b 59.35 = b The equation is y = 26.175x + 59.35. In 2005, x = 2005 − 1970 = 35.

y = 26.175(35) + 59.35 ≈ $975.48 In 2010, x = 2010 − 1970 = 40.

y = 26.175(40) + 59.35 = $1106.35 In 2020, x = 2020 − 1970 = 50.

y = 26.175(50) + 59.35 = $1368.10

80. Answers will vary depending on the data points used. We will use (0, 20, 423) and (20, 11, 358). 11, 358 − 20, 423 −9065 m= = = −453.25 20 − 0 20 Substituting in the slope-intercept equation we have y = −453.25x + 20, 423. 2008: y = −453.25(38) + 20, 423 = 3199.5

We estimate the number of sheep and lambs on farms in 2008 to be 3199.5 thousand, or 3,199,500. 2013: y = −453.25(43) + 20, 423 = 933.25

There will be about 933.25 thousand, or 933,250 sheep and lambs on farms in 2013.

81. a) Using the linear regression feature on a graphing calculator, we get M = 0.2H + 156. b) For H = 40: M = 0.2(40) + 156 = 164 beats per minute For H = 65: M = 0.2(65) + 156 = 169 beats per minute For H = 76: M = 0.2(76) + 156 ≈ 171 beats per minute For H = 84: M = 0.2(84) + 156 ≈ 173 beats per minute c) r = 1; all the data points are on the regression line so it should be a good predictor. 82. a) y = 0.072050673x + 81.99920823 b) For x = 24: y = 0.072050673(24) + 81.99920823 ≈ 84% For x = 6: y = 0.072050673(6) + 81.99920823 ≈ 82%

For x = 18: y = 0.072050673(18) + 81.99920823 ≈ 83%

c) r = 0.0636; since there is a very low correlation, the regression line is not a good predictor. 83. a) Using the linear regression feature on a graphing calculator, we get y = 0.02x + 1.77, where x is the number of years after 1999 and y is in billions of dollars. b) In 2005, x = 2005 − 1999 = 6, and y = 0.02(6) + 1.77 = $1.89 billion. In 2010, x = 2010 − 1999 = 11, and y = 0.02(11) + 1.77 = $1.99 billion. c) r ≈ 0.3162; since there is a low correlation, the regression line does not fit the data closely. 84. a) y = 342.5142857x + 4165.761905, where x is the number of years after 1993 b) In 2008, y = 342.5142857(15) + 4165.761905 ≈ 9303 triplet births. This value is only 305 less than the value found in Exercise 76. c) r ≈ 0.9402; the line fits the data well.

Exercise Set 1.5

71

85. a) Using the linear regression feature on a graphing calculator, we get y = 4.055x + 96.78, where x is the number of years after 1995 and y is in thousands. b) In 2010, x = 2010 − 1995 = 15.

y = 4.055(15) + 96.78 = 157.605 thousand, or 157,605 This value is only 295 less than the value found in Exercise 75.

c) r ≈ 0.9980; the line fits the data well. 86. A linear function cannot be used to express the yearly salary as a function of the number of years the employee has worked because the increases in salary are not a constant amount. Each increase is larger than the preceding one because it is calculated on a larger base amount than the preceding one. % % % % % 3% %2% 3 87. %% − %% > %% %%, so the line with slope − is steeper. 4 5 4 −14 −7 − 7 = 5−5 0 The slope is not defined.

88. m =

89.

y2 − y1 x2 − x1 −1 − (−8) −1 + 8 = = −5 − 2 −7 7 = = −1 −7

m=

d 5 90. r = = 2 2

# $2 5 2 25 x2 + (y − 3)2 = , or 4 2 2 x + (y − 3) = 6.25

(x − 0)2 + (y − 3)2 =

91.

(x − h)2 + (y − k)2 = r2 # $2 9 2 2 [x − (−7)] + [y − (−1)] = 5 81 (x + 7)2 + (y + 1)2 = 25 920.58 = 0.067 13, 740 The road grade is 6.7%.

92. m =

We find an equation of the line with slope 0.067 and containing the point (13, 740, 920.58): y − 920.58 = 0.067(x − 13, 740) y − 920.58 = 0.067x − 920.58 y = 0.067x

93. The slope of the line containing (−3, k) and (4, 8) is 8−k 8−k = . 4 − (−3) 7

The slope of the line containing (5, 3) and (1, −6) is −6 − 3 −9 9 = = . 1−5 −4 4 The slopes must be equal in order for the lines to be parallel: 9 8−k = 7 4 32 − 4k = 63 Multiplying by 28 −4k = 31 31 k = − , or − 7.75 4

94. The slope of the line containing (−1, 3) and (2, 9) is 6 9−3 = = 2. 2 − (−1) 3 1 Then the slope of the desired line is − . We find the 2 equation of that line: 1 y − 5 = − (x − 4) 2 1 y−5 = − x+2 2 1 y = − x+7 2

Exercise Set 1.5 1. a) For x-values from −5 to 1, the y-values increase from −3 to 3. Thus the function is increasing on the interval (−5, 1). b) For x-values from 3 to 5, the y-values decrease from 3 to 1. Thus the function is decreasing on the interval (3, 5). c) For x-values from 1 to 3, y is 3. Thus the function is constant on (1, 3). 2. a) For x-values from 1 to 3, the y-values increase from 1 to 2. Thus, the function is increasing on the interval (1, 3). b) For x-values from −5 to 1, the y-values decrease from 4 to 1. Thus the function is decreasing on the interval (−5, 1). c) For x-values from 3 to 5, y is 2. Thus the function is constant on (3, 5). 3. a) For x-values from −3 to −1, the y-values increase from −4 to 4. Also, for x-values from 3 to 5, the y-values increase from 2 to 6. Thus the function is increasing on (−3, −1) and on (3, 5).

b) For x-values from 1 to 3, the y-values decrease from 3 to 2. Thus the function is decreasing on the interval (1, 3). c) For x-values from −5 to −3, y is 1. Thus the function is constant on (−5, −3).

72

Chapter 1: Graphs, Functions, and Models

4. a) For x-values from 1 to 2, the y-values increase from 1 to 2. Thus the function is increasing on the interval (1, 2). b) For x-values from −5 to −2, the y-values decrease from 3 to 1. For x-values from −2 to 1, the y-values decrease from 3 to 1. And for x-values from 3 to 5, the y-values decrease from 2 to 1. Thus the function is decreasing on (−5, −2), on (−2, 1), and on (3, 5).

c) For x-values from 2 to 3, y is 2. Thus the function is constant on (2, 3).

14. From the graph we see that a relative minimum value of 2 occurs at x = 1. There is no relative maximum value. The graph starts falling, or decreasing, from the left and stops decreasing at the relative minimum. From this point, the graph increases. Thus the function is increasing on (1, ∞) and is decreasing on (−∞, 1). 15. From the graph we see that a relative maximum value of the function is 2.370. It occurs at x = −0.667. We also see that a relative minimum value of 0 occurs at x = 2. The graph starts rising, or increasing, from the left and stops increasing at the relative maximum. From this point it decreases to the relative minimum and then increases again. Thus the function is increasing on (−∞, −0.667) and on (2, ∞). It is decreasing on (−0.667, 2).

5. a) For x-values from −∞ to −8, the y-values increase from −∞ to 2. Also, for x-values from −3 to −2, the y-values increase from −2 to 3. Thus the function is increasing on (−∞, −8) and on (−3, −2). b) For x-values from −8 to −6, the y-values decrease from 2 to −2. Thus the function is decreasing on the interval (−8, −6). c) For x-values from −6 to −3, y is −2. Also, for xvalues from −2 to ∞, y is 3. Thus the function is constant on (−6, −3) and on (−2, ∞).

6. a) For x-values from 1 to 4, the y-values increase from 2 to 11. Thus the function is increasing on the interval (1, 4). b) For x-values from −1 to 1, the y-values decrease from 6 to 2. Also, for x-values from 4 to ∞, the yvalues decrease from 11 to −∞. Thus the function is decreasing on (−1, 1) and on (4, ∞).

16. From the graph we see that a relative maximum value of 2.921 occurs at x = 3.601. A relative minimum value of 0.995 occurs at x = 0.103. The graph starts decreasing from the left and stops decreasing at the relative minimum. From this point it increases to the relative maximum and then decreases again. Thus the function is increasing on (0.103, 3.601) and is decreasing on (−∞, 0.103) and on (3.601, ∞). 17.

y 5 4 3 2 1

c) For x-values from −∞ to −1, y is 3. Thus the function is constant on (−∞, −1).

f(x) =x 2 x

-5 -4 -3 -2 -1 0 -1

1 2 3 4 5

-2 -3 -4

7. The x-values extend from −5 to 5, so the domain is [−5, 5]. The y-values extend from −3 to 3, so the range is [−3, 3].

-5

8. Domain: [−5, 5]; range: [1, 4] 9. The x-values extend from −5 to −1 and from 1 to 5, so the domain is [−5, −1] ∪ [1, 5]. The y-values extend from −4 to 6, so the range is [−4, 6].

10. Domain: [−5, 5]; range: [1, 3]

The function is increasing on (0, ∞) and decreasing on (−∞, 0). We estimate that the minimum is 0 at x = 0. There are no maxima. 18.

y 5 4

11. The x-values extend from −∞ to ∞, so the domain is (−∞, ∞).

The y-values extend from −∞ to 3, so the range is (−∞, 3].

12. Domain: (−∞, ∞); range: (−∞, 11]

-5 -4 -3 -2 -1 0 -1

13. From the graph we see that a relative maximum value of the function is 3.25. It occurs at x = 2.5. There is no relative minimum value. The graph starts rising, or increasing, from the left and stops increasing at the relative maximum. From this point, the graph decreases. Thus the function is increasing on (−∞, 2.5) and is decreasing on (2.5, ∞).

f(x) = 4 –x 2

3 2 1

x 1 2 3 4 5

-2 -3 -4 -5

Increasing: (−∞, 0) Decreasing: (0, ∞)

Maximum: 4 at x = 0 Minima: none

Exercise Set 1.5

73

19.

22.

y 5 4

y 10 8

f(x) = 5 – |x |

3 2 1

6 f(x) = –x 2 – 8x – 9 4 2 x

x

-5 -4 -3 -2 -1 0 -1

1 2 3 4 5

-10 -8 -6 -4 -2 0 -2

-2 -3 -4

-4 -6 -8

-5

-10

The function is increasing on (−∞, 0) and decreasing on (0, ∞). We estimate that the maximum is 5 at x = 0. There are no minima. 20.

3 2 1

x 1 2 3 4 5

-2 -3 -4

Decreasing: (−∞, −3)

Maxima: none

A(x) = x(30 − x)

25. After t minutes, the balloon has risen 120t ft. We use the Pythagorean theorem.

Minimum: −5 at x = −3

[d(t)]2 = (120t)2 + (400)2 & d(t) = (120t)2 + (400)2

y 5 4

We only considered the positive square root since distance must be nonnegative.

3 2 1

-2 -3 -4

23. If x = the length of the rectangle, in meters, then the 60 − 2x , or 30 − x. We use the formula Area = width is 2 length × width:

24. Let h = the height of the scarf, in inches. Then the length of the base = 2h − 7. 1 A(h) = (2h − 7)(h) 2 7 A(h) = h2 − h 2

Increasing: (−3, ∞)

-5 -4 -3 -2 -1 0 -1

Decreasing: (−4, ∞)

Maximum: 7 at x = −4

A(x) = 30x − x2

f(x) =|x +3| – 5

-5

21.

Increasing: (−∞, −4)

Minima: none

y 5 4

-5 -4 -3 -2 -1 0 -1

2 4 6 8 10

x 1 2 3 4 5 f(x) =x 2 – 6x + 10

26. Use the Pythagorean theorem. [h(d)]2 + (3700)2 = d2 [h(d)]2 = d2 − (3700)2 & h(d) = d2 − (3700)2 Taking the positive square root

-5

The function is decreasing on (−∞, 3) and increasing on (3, ∞). We estimate that the minimum is 1 at x = 3. There are no maxima.

27. Let w = the width of the rectangle. Then the 40 − 2w , or 20 − w. Divide the rectangle into length = 2 quadrants as shown below.

20 – w

w

74

Chapter 1: Graphs, Functions, and Models In each quadrant there are two congruent triangles. One triangle is part of the rhombus and both are part of the rectangle. Thus, in each quadrant the area of the rhombus is one-half the area of the rectangle. Then, in total, the area of the rhombus is one-half the area of the rectangle. 1 A(w) = (20 − w)(w) 2 A(w) = 10w −

33. a) When a square with sides of length x are cut from each corner, the length of each of the remaining sides of the piece of cardboard is 12 − 2x. Then the dimensions of the box are x by 12− 2x by 12− 2x. We use the formula Volume = length × width × height to find the volume of the box: V (x) = (12 − 2x)(12 − 2x)(x) V (x) = (144 − 48x + 4x2 )(x)

w2 2

28. Let w = the width, in feet. Then the length = or 23 − w. A(w) = (23 − w)w

46 − 2w , 2

A(w) = 23w − w2

29. We will $ in # use similar triangles, expressing all distances 1 s ft, and d yd = 3d ft We feet. 6 in. = ft, s in. = 2 12 have 1 3d 2 = s 7 12 1 s · 3d = 7 · 12 2 sd 7 = 4 2 4 7 d = · , so s 2 14 d(s) = . s 30. The volume of the tank is the sum of the volume of a sphere with radius r and a right circular cylinder with radius r and height 6 ft. 4 V (r) = πr3 + 6πr2 3

V (x) = 144x − 48x2 + 4x3

This can also be expressed as V (x) = 4x(x − 6)2 , or V (x) = 4x(6 − x)2 .

b) The length of the sides of the square corners that are cut out must be positive and less than half the length of a side of the piece of cardboard. Thus, the domain of the function is {x|0 < x < 6}, or (0, 6).

c) We see from the graph that the maximum value of the area function on the interval (0, 6) appears to be 128 when x = 2. When x = 2, then 12 − 2x = 12 − 2 · 2 = 8, so the dimensions that yield the maximum volume are 8 cm by 8 cm by 2 cm. 34. a) V (x) = 8x(14 − 2x), or 112x − 16x2 ' % ( % 14 b) The domain is x%%0 < x < , or 2 {x|0 < x < 7}, or (0, 7).

c) The maximum occurs when x = 3.5, so the file should be 3.5 in. tall. ' x + 4, for x ≤ 1, 35. g(x) = 8 − x, for x > 1 Since −4 ≤ 1, g(−4) = −4 + 4 = 0. Since 0 ≤ 1, g(0) = 0 + 4 = 4. Since 1 ≤ 1, g(1) = 1 + 4 = 5.

Since 3 > 1, g(3) = 8 − 3 = 5.

31. a) If the length = x feet, then the width = 30 − x feet. A(x) = x(30 − x)

36. f (x) =

A(x) = 30x − x2

b) The length of the rectangle must be positive and less than 30 ft, so the domain of the function is {x|0 < x < 30}, or (0, 30).

c) We see from the graph that the maximum value of the area function on the interval (0, 30) appears to be 225 when x = 15. Then the dimensions that yield the maximum area are length = 15 ft and width = 30 − 15, or 15 ft. 32. a) A(x) = x(360 − 3x), or 360x − 3x ( ' % % 360 b) The domain is x%%0 < x < , or 3 {x|0 < x < 120}, or (0, 120). 2

c) The maximum value occurs when x = 60 so the width of each corral should be 60 yd and the total length of the two corrals should be 360 − 3 · 60, or 180 yd.

   3,

for x ≤ −2,

1   x + 6, for x > −2 2

f (−5) = 3

f (−2) = 3 1 f (0) = · 0 + 6 = 6 2 1 f (2) = · 2 + 6 = 7 2 37. h(x) =

-

−3x − 18, for x < −5, 1, for −5 ≤ x < 1, x + 2, for x ≥ 1

Since −5 is in the interval [−5, 1), h(−5) = 1. Since 0 is in the interval [−5, 1), h(0) = 1. Since 1 ≥ 1, h(1) = 1 + 2 = 3. Since 4 ≥ 1, h(4) = 4 + 2 = 6.

Exercise Set 1.5

75

   −5x − 8, for x < −2,    1 38. f (x) = x + 5, for −2 ≤ x ≤ 4, 2     10 − 2x, for x > 4

Since −4 < −2, f (−4) = −5(−4) − 8 = 12. 1 Since −2 is in the interval [−2, 4], f (−2) = (−2) + 5 = 4. 2 1 Since 4 is in the interval [−2, 4], f (4) = · 4 + 5 = 7. 2 Since 6 > 4, f (6) = 10 − 2 · 6 = −2.

1  x, for x < 0, 39. f (x) = 2  x + 3, for x ≥ 0

1 x for 2 inputs x less than 0. Then graph f (x) = x + 3 for inputs x greater than or equal to 0. We create the graph in two parts. Graph f (x) =

y

42. h(x) =

2x − 1, for x < 2 2 − x, for x ≥ 2

'

y 4 2 2

!4 !2

4

x

!2 !4

  x + 1, for x ≤ −3,       for −3 < x < 4 43. f (x) = −1,     1   for x ≥ 4  2 x,

We create the graph in three parts. Graph f (x) = x + 1 for inputs x less than or equal to −3. Graph f (x) = −1 for inputs greater than −3 and less than 4. Then graph 1 f (x) = x for inputs greater than or equal to 4. 2 y

4

4

2 2

!4 !2

4

2

x

!2

2

!4 !2

4

x

!2

!4

!4

 1  − x + 2, for x ≤ 0, 3 40. f (x) =  x − 5, for x > 0

44. f (x) =

y

4 2

!4 !2

4

x

!2

4

x

!4

 3  − x + 2, for x < 4, 4 41. f (x) =  −1 + x, for x ≥ 4 We create the graph in two parts. Graph 3 f (x) = − x + 2 for inputs x less than 4. Then graph 4 f (x) = −1 + x for inputs x greater than or equal to 4. y 4 2 2 !4

2

!4 !2

!4

!2

4, for x ≤ −2, x + 1, for −2 < x < 3 −x, for x ≥ 3 y

4

!4 !2

-

4

x

1   x − 1, for x < 0,   2 45. g(x) = 3, for 0 ≤ x ≤ 1      −2x, for x > 1

1 x−1 2 for inputs less than 0. Graph g(x) = 3 for inputs greater than or equal to 0 and less than or equal to 1. Then graph g(x) = −2x for inputs greater than 1. We create the graph in three parts. Graph g(x) =

76

Chapter 1: Graphs, Functions, and Models 49. f (x) = int(x)

y 4

See Example 8.

2

50. f (x) = 2 int(x) 2

!4 !2

4

x

!2 !4

 2   x − 9 , for x '= −3, x+3 46. f (x) =   5, for x = −3 y

4 2 2

!4 !2

4

x

This function can be defined by a piecewise function with an infinite  number of statements: .      .    .     −4, for −2 ≤ x < −1,    −2, for −1 ≤ x < −0, f (x) = 0, for 0 ≤ x < 1,     2, for 1 ≤ x < 2,     .    .     . y

!2

6

!4

4

!6

47. f (x) =

2

 2,  

for x = 5,

x − 25   , for x '= 5 x−5 When x '= 5, the denominator of (x2 − 25)/(x − 5) is nonzero so we can simplify: x2 − 25 (x + 5)(x − 5) = = x + 5. x−5 x−5 Thus, f (x) = x + 5, for x '= 5.

2

!4

y 8 2 4

!8 !4

8

x

x

f(x) " 2!x"

2

The graph of this part of the function consists of a line with a “hole” at the point (5, 10), indicated by an open dot. At x = 5, we have f (5) = 2, so the point (5, 2) is plotted below the open dot.

4

51. f (x) = 1 + int(x) This function can be defined by a piecewise function with an infinite  number of statements:  .     .    .     −1, for −2 ≤ x < −1,     0, for −1 ≤ x < 0, f (x) = 1, for 0 ≤ x < 1,    2, for 1 ≤ x < 2,     .    .      . y

!4

4

!8

2

 2   x + 3x + 2 , for x '= −1, x+1 48. f (x) =   7, for x = −1 y

8 6 4 2 !4 !2

2

4

x

2

!4

4

x

!2 !4

g(x) " 1 # #x$

1 int(x) − 2 2 This function can be defined by a piecewise function with an infinite number of statements:

52. f (x) =

Exercise Set 1.5    .     .     .     −2 12 ,    −2, f (x) = −1 1 ,  2    −1,    .     .     . 

77 h(x) = for for for for

4

h(x) " q !x" ! 2

2 2

!4 !2

x

!2 !4

53. From the graph we see that the domain is (−∞, ∞) and the range is (−∞, 0) ∪ [3, ∞). 54. Domain: (−∞, ∞); range: (−5, ∞) 55. From the graph we see that the domain is (−∞, ∞) and the range is [−1, ∞). 56. Domain: (∞, ∞); range: (−∞, 3) 57. From the graph we see that the domain is (−∞, ∞) and the range is (−∞, −2] ∪ {−1} ∪ [2, ∞), or {y|y ≤ −2 or y = −1 or y ≥ 2}.

|x|, for x < 3, −2, for x ≥ 3

This can also be expressed as follows: −x, for x ≤ 0, h(x) = x, for 0 < x < 3, −2, for x ≥ 3 It can also be expressed as follows: −x, for x < 0, h(x) = x, for 0 ≤ x < 3, −2, for x ≥ 3

−1 ≤ x < 0, 0 ≤ x < 1, 1 ≤ x < 2, 2 ≤ x < 3,

y

'

63. From the graph we see that the domain is [−5, 3] and the range is (−3, 5). Finding the slope of each segment and using the slope-intercept or point-slope formula, we find that an equation for the function is: x + 8, for −5 ≤ x < −3, h(x) = 3, for −3 ≤ x ≤ 1, 3x − 6, for 1 < x ≤ 3 64. Domain: [−4, ∞); range: [−2, 4] −2x − 4, for −4 ≤ x ≤ −1, f (x) = x − 1, for −1 < x < 2, 2, for x ≥ 2 This can also be expressed as: −2x − 4, for −4 ≤ x < −1, f (x) = x − 1, for −1 ≤ x < 2, 2, for x ≥ 2

65.

58. Domain: (−∞, ∞); range: (−∞, −3] ∪ (−1, 4] 59. From the graph we see that the domain is (−∞, ∞) and the range is {−5, −2, 4}. An equation for the function is: −2, for x < 2, f (x) = −5, for x = 2, 4, for x > 2

Beginning at the left side of the window, the graph first drops as we move to the right. We see that the function is decreasing on (−∞, 1). We then find that the function is increasing on (1, 3) and decreasing again on (3, ∞). The MAXIMUM and MINIMUM features also show that the relative maximum is −4 at x = 3 and the relative minimum is −8 at x = 1.

60. Domain: (−∞, ∞); range: {y|y = −3 or y ≥ 0} ' −3, for x < 0, g(x) = x, for x ≥ 0 61. From the graph we see that the domain is (−∞, ∞) and the range is (−∞, −1] ∪ [2, ∞). Finding the slope of each segment and using the slope-intercept or point-slope formula, we find that an equation for the function is: x, for x ≤ −1, g(x) = 2, for −1 < x ≤ 2, x, for x > 2 This can also be expressed as follows: x, for x ≤ −1, g(x) = 2, for −1 < x < 2, x, for x ≥ 2

62. Domain: (−∞, ∞); range: {−2} ∪ [0, ∞), or {y|y = −2 or y ≥ 0}. An equation for the function is:

66.

Increasing: (−∞, −2.573), (3.239, ∞) Decreasing: (−2.573, 3.239)

Relative maximum: 4.134 at x = −2.573

Relative minimum: −15.497 at x = 3.239

78

Chapter 1: Graphs, Functions, and Models −4 . x2 + 1 Increasing: (0, ∞)

67.

72. Graph y =

Decreasing: (−∞, 0) √ 73. Graph y = x 4 − x2 , for −2 ≤ x ≤ 2. Increasing: (−1.414, 1.414)

We find that the function is increasing on (−1.552, 0) and on (1.552, ∞) and decreasing on (−∞, −1.552) and on (0, 1.552). The relative maximum is 4.07 at x = 0 and the relative minima are −2.314 at x = −1.552 and −2.314 at x = 1.552. 68.

Decreasing: (−2, −1.414), (1.414, 2) √ 74. Graph y = −0.8x 9 − x2 , for −3 ≤ x ≤ 3. Increasing: (−3, −2.121), (2.121, 3)

Decreasing: (−2.121, 2.121)

75. a) The length of a diameter of the circle (and a diagonal of the rectangle) is 2 · 8, or 16 ft. Let l = the length of the rectangle. Use the Pythagorean theorem to write l as a function of x. x2 + l2 = 162 x2 + l2 = 256 l2 = 256 − x2 & l = 256 − x2

Increasing: (−3, ∞)

Since the length must be positive, we considered only the positive square root.

Relative maxima: none

Use the formula Area = length × width to find the area of the rectangle: √ A(x) = x 256 − x2

Decreasing: (−∞, −3)

Relative minimum: 9.78 at x = −3 69. a)

y " !0.1x 2 # 1.2x # 98.6 110

b) The width of the rectangle must be positive and less than the diameter of the circle. Thus, the domain of the function is {x|0 < x < 16}, or (0, 16).

c) 0 90

12

b) Using the MAXIMUM feature we find that the relative maximum is 102.2 at t = 6. Thus, we know that the patient’s temperature was the highest at t = 6, or 6 days after the onset of the illness and that the highest temperature was 102.2◦ F. 70. a)

y " x%256 ! x 2 150

y " !x 2 # 300x # 6 50,000

0

0

16

d) Using the MAXIMUM feature, we find that the maximum area x is about 11.314. When x ≈ √ occurs when & 256 − (11.314)2 ≈ 11.313. 11.314, 256 − x2 ≈ Thus, the dimensions that maximize the area are about 11.314 ft by 11.313 ft. (Answers may vary slightly due to rounding differences.) 76. a) Let h(x) = the height of the box.

0

0

300

b) 22, 506 at a = 150 c) The greatest number of games will be sold when $150 thousand is spent on advertising. For that amount, 22,506 games will be sold. 8x . x2 + 1 Increasing: (−1, 1)

71. Graph y =

Decreasing: (−∞, −1), (1, ∞)

320 = x · x · h(x) 320 = h(x) x2 Area of the bottom: x2 # $ 320 320 Area of each side: x , or x2 x

Area of the top: x2

$ # 320 + 1 · x2 C(x) = 1.5x2 + +4(2.5) x 3200 C(x) = 2.5x2 + x

Exercise Set 1.6

79 h 10 = 6−r 6 5 10 (6 − r) = (6 − r) h= 6 3 30 − 5r h= 3 30 − 5r Thus, h(r) = . 3 b) V = πr2 h " ! 30 − 5r Substituting for h V (r) = πr2 3

b) The length of the base must be positive, so the domain of the function is {x|x > 0}, or (0, ∞).

c)

y ! 2.5x2 "

3200 x

1000

0

20

0

d) Using the MIMIMUM feature, we find that the minimum cost occurs when x ≈ 8.618. Thus, the dimensions that minimize the cost are about 320 , or about 4.309 ft. 8.618 ft by 8.618 ft by (8.618)2

c) We first express r in terms of h. 30 − 5r h= 3 3h = 30 − 5r 5r = 30 − 3h 30 − 3h r= 5 V = πr2 h ! "2 30 − 3h h V (h) = π 5

77. Some possibilities are outdoor temperature during a 24 hour period, sales of a new product, and temperature during an illness. 78. For continuous functions, relative extrema occur at points for which the function changes from increasing to decreasing or vice versa.

Substituting for r "2 ! 30 − 3h . We can also write V (h) = πh 5

79. Function; domain; range; domain; exactly one; range 80. Midpoint formula 81. x-intercept

86. a) The distance from A to S is 4 − x.

82. Constant; identity

Using the Pythagorean √ theorem, we find that the distance from S to C is 1 + x2 . √ Then C(x) = √ 3000(4−x)+5000 1 + x2 , or 12, 000− 3000x + 5000 1 + x2 . √ b) Graph y = 12, 000 − 3000x + 5000 1 + x2 in a window such as [0, 5, 10, 000, 20, 000], Xscl = 1, Yscl = 1000. Using the MINIMUM feature, we find that cost is minimized when x = 0.75, so the line should come to shore 0.75 mi from B.

83. If [int(x)]2 = 25, then int(x) = −5 or int(x) = 5. For −5 ≤ x < −4, int(x) = −5. For 5 ≤ x < 6, int(x) = 5. Thus, the possible inputs for x are {x| − 5 ≤ x < −4 or 5 ≤ x < 6}. 84. If int(x + 2) = −3, then −3 ≤ x + 2 < −2, or −5 ≤ x < −4. The possible inputs for x are {x| − 5 ≤ x < −4}. 85. a) We add labels to the drawing in the text.

Exercise Set 1.6

E

1.

= (52 − 3) + (2 · 5 + 1)

D

10

= 25 − 3 + 10 + 1 = 33

h

2. A

r

(f + g)(5) = f (5) + g(5)

B

6–r

6

We write a proportion involving the lengths of the sides of the similar triangles BCD and ACE. Then we solve it for h.

(f g)(0) = f (0) · g(0)

= (02 − 3)(2 · 0 + 1)

C

3.

= −3(1) = −3

(f − g)(−1) = f (−1) − g(−1)

= ((−1)2 − 3) − (2(−1) + 1)

= −2 − (−1) = −2 + 1 = −1

80 4.

Chapter 1: Graphs, Functions, and Models (f g)(2) = f (2) · g(2)

11.

= (2 − 3)(2 · 2 + 1) 2

5.

= 1·5=5 # 1$ " ! f − 1 = # 2$ (f /g) − 1 2 g − 2 # 1 $2 −3 − = # 2 $ 1 +1 2 − 2 1 −3 = 4 −1 + 1 11 − 4 = 0 " ! 1 does not Since division by 0 is not defined, (f /g) − 2 exist.

6.

(f − g)(0) = f (0) − g(0)

= (02 − 3) − (2 · 0 + 1)

7.

8.

9.

= −3 − 1 = −4 " ! " ! " ! 1 1 1 (f g) − =f − ·g − 2 2 2 "2 " & %! &% ! 1 1 −3 2 − +1 = − 2 2 11 = − ·0=0 4√ √ f (− 3) √ (f /g)(− 3) = g(− 3) √ (− 3)2 − 3 √ = 2(− 3) + 1 0 √ = =0 −2 3 + 1 (g − f )(−1) = g(−1) − f (−1)

= [2(−1) + 1] − [(−1)2 − 3]

= (−2 + 1) − (1 − 3) = −1 − (−2) = −1 + 2 =1 10.

# 1$ " ! g − 1 = # 2$ (g/f ) − 1 2 f − 2 # 1$ 2 − +1 2 = # 1 $2 − −3 2 0 = 11 − 4 =0

12.

(h − g)(−4) = h(−4) − g(−4) √ = (−4 + 4) − −4 − 1 √ = 0 − −5 √ Since −5 is not a real number, (h−g)(−4) does not exist. (gh)(10) = g(10) · h(10) √ = 10 − 1(10 + 4) √ = 9(14) = 3 · 14 = 42

g(1) h(1) √ 1−1 = 1+4 √ 0 = 5 0 = =0 5 h(1) 14. (h/g)(1) = g(1) 1+4 = √ 1−1 5 = 0 Since division by 0 is not defined, (h/g)(1) does not exist.

13.

(g/h)(1) =

15.

(g + h)(1) = g(1) + h(1) √ = 1 − 1 + (1 + 4) √ = 0+5 = 0+5=5

16.

(hg)(3) = h(3) · g(3) √ = (3 + 4) 3 − 1 √ =7 2

17. f (x) = 2x + 3, g(x) = 3 − 5x a) The domain of f and of g is the set of all real numbers, or (−∞, ∞). Then the domain of f + g, f − g, f f , 3 and f g is also (−∞, ∞). For f /g we must exclude 5 #3$ since g = 0. Then the domain of f /g is 5 # $ 3$ #3 ∪ , ∞ . For g/f we must exclude − ∞, 5 5 # 3$ 3 = 0. The domain of g/f is − since f − 2 2 $ # $ # 3 3 ∪ − ,∞ . − ∞, − 2 2 b) (f + g)(x) = f (x) + g(x) = (2x + 3) + (3 − 5x) = −3x + 6 (f − g)(x) = f (x) − g(x) = (2x + 3) − (3 − 5x) = 2x + 3 − 3 + 5x = 7x

(f g)(x) = f (x) · g(x) = (2x + 3)(3 − 5x) = 6x − 10x2 + 9 − 15x = −10x2 − 9x + 9

(f f )(x) = f (x) · f (x) = (2x + 3)(2x + 3) = 4x2 + 12x + 9

Exercise Set 1.6 (f /g)(x) = (g/f )(x) =

81 f (x) 2x + 3 = g(x) 3 − 5x

g(x) 3 − 5x = f (x) 2x + 3

18. f (x) = −x + 1, g(x) = 4x − 2

g, " f + g, f − g, f g, and f f is a) The domain of f , ! 1 = 0, the domain of f /g is (−∞, ∞). Since g 2 $ # # $ 1 1 ∪ − ∞, , ∞ . Since f (1) = 0, the domain of 2 2 g/f is (−∞, 1) ∪ (1, ∞). b) (f + g)(x) = (−x + 1) + (4x − 2) = 3x − 1

(f − g)(x) = (−x + 1) − (4x − 2) = −x + 1 − 4x + 2 = −5x + 3 (f g)(x) = (−x + 1)(4x − 2) = −4x2 + 6x − 2

(f f )(x) = (−x + 1)(−x + 1) = x2 − 2x + 1

−x + 1 4x − 2 4x − 2 (g/f )(x) = −x + 1 √ 19. f (x) = x − 3, g(x) = x + 4 (f /g)(x) =

a) Any number can be an input in f , so the domain of f is the set of all real numbers, or (−∞, ∞).

The inputs of g must be nonnegative, so we have x + 4 ≥ 0, or x ≥ −4. Thus, the domain of g is [−4, ∞). The domain of f + g, f − g, and f g is the set of all numbers in the domains of both f and g. This is [−4, ∞).

The domain of f f is the domain of f , or (−∞, ∞). The domain of f /g is the set of all numbers in the domains of f and g, excluding those for which g(x) = 0. Since g(−4) = 0, the domain of f /g is (−4, ∞).

The domain of g/f is the set of all numbers in the domains of g and f , excluding those for which f (x) = 0. Since f (3) = 0, the domain of g/f is [−4, 3) ∪ (3, ∞). √ b) (f + g)(x) = f (x) + g(x) = x − 3 + x + 4 √ (f − g)(x) = f (x) − g(x) = x − 3 − x + 4 √ (f g)(x) = f (x) · g(x) = (x − 3) x + 4 ' (2 (f f )(x) = f (x) = (x − 3)2 = x2 − 6x + 9 f (x) x−3 =√ g(x) x+4 √ g(x) x+4 = (g/f )(x) = f (x) x−3

(f /g)(x) =

20. f (x) = x + 2, g(x) =

√

x−1

a) The domain of f is (−∞, ∞). The domain of g consists of all the values of x for which x − 1 is nonnegative, or [1, ∞). Then the domain of f + g, f − g, and f g is [1, ∞). The domain of f f is (−∞, ∞). Since g(1) = 0, the domain of f /g is (1, ∞). Since f (−2) = 0 and −2 is not in the domain of g, the domain of g/f is [1, ∞). √ b) (f + g)(x) = x + 2 + x − 1 √ (f − g)(x) = x + 2 − x + 1 √ (f g)(x) = (x + 2) x − 1 (f f )(x) = (x + 2)(x + 2) = x2 + 4x + 4 x+2 (f /g)(x) = √ x−1 √ x−1 (g/f )(x) = x+2

21. f (x) = 2x − 1, g(x) = −2x2

a) The domain of f and of g is (−∞, ∞). Then the domain of f + g, f − g, f g, and f f is (−∞, ∞). For f /g, we must exclude 0 since g(0) = 0. The domain of f /g is (−∞, 0) ∪ (0, ∞). For g/f , we #1$ 1 must exclude since f = 0. The domain of 2 # $2 1$ #1 g/f is − ∞, ∪ ,∞ . 2 2 b) (f + g)(x) = f (x) + g(x) = (2x − 1) + (−2x2 ) = −2x2 + 2x − 1 (f − g)(x) = f (x) − g(x) = (2x − 1) − (−2x2 ) = 2x2 + 2x − 1 (f g)(x) = f (x) · g(x) = (2x − 1)(−2x2 ) = −4x3 + 2x2 (f f )(x) = f (x) · f (x) = (2x − 1)(2x − 1) = 4x2 − 4x + 1 (f /g)(x) = (g/f )(x) =

2x − 1 f (x) = g(x) −2x2

g(x) −2x2 = f (x) 2x − 1

22. f (x) = x2 − 1, g(x) = 2x + 5

a) The domain of f and of g is the set of all real numbers, or (−∞, ∞). Then the domain ! of f"+ g, f − g, 5 = 0, the f g and f f is (−∞, ∞). Since g − 2 $ # $ # 5 5 ∪ − , ∞ . Since domain of f /g is − ∞, − 2 2 f (1) = 0 and f (−1) = 0, the domain of g/f is (−∞, −1) ∪ (−1, 1) ∪ (1, ∞).

b) (f + g)(x) = x2 − 1 + 2x + 5 = x2 + 2x + 4

(f − g)(x) = x2 − 1 − (2x + 5) = x2 − 2x − 6 (f g)(x) = (x2 −1)(2x+5) = 2x3 +5x2 −2x−5

(f f )(x) = (x2 − 1)2 = x4 − 2x2 + 1 x2 − 1 2x + 5 2x + 5 (g/f )(x) = 2 x −1

(f /g)(x) =

82

Chapter 1: Graphs, Functions, and Models

23. f (x) =

√

x − 3, g(x) =

√

x+3

a) Since f (x) is nonnegative for values of x in [3, ∞), this is the domain of f . Since g(x) is nonnegative for values of x in [−3, ∞), this is the domain of g. The domain of f +g, f −g, and f g is the intersection of the domains of f and g, or [3, ∞). The domain of f f is the same as the domain of f , or [3, ∞). For f /g, we must exclude −3 since g(−3) = 0. This is not in [3, ∞), so the domain of f /g is [3, ∞). For g/f , we must exclude 3 since f (3) = 0. The domain of g/f is (3, ∞). √ √ b) (f + g)(x) = f (x) + g(x) = x − 3 + x + 3 √ √ (f − g)(x) = f (x) − g(x) = x − 3 − x + 3 √ √ √ (f g)(x) = f (x) · g(x) = x−3 · x + 3 = x2 −9 √ √ (f f )(x) = f (x) · f (x) = x − 3 · x − 3 = |x − 3| √ x−3 (f /g)(x) = √ x+3 √ x+3 (g/f )(x) = √ x−3 √ √ 24. f (x) = x, g(x) = 2 − x

a) The domain of f is [0, ∞). The domain of g is (−∞, 2]. Then the domain of f + g, f − g, and f g is [0, 2]. The domain of f f is the same as the domain of f , [0, ∞). Since g(2) = 0, the domain of f /g is [0, 2). Since f (0) = 0, the domain of g/f is (0, 2]. √ √ b) (f + g)(x) = x + 2 − x √ √ (f − g)(x) = x − 2 − x √ √ √ (f g)(x) = x · 2 − x = 2x − x2 √ √ √ (f f )(x) = x · x = x2 = |x| √ x (f /g)(x) = √ 2−x √ 2−x (g/f )(x) = √ x

25. f (x) = x + 1, g(x) = |x|

a) The domain of f and of g is (−∞, ∞). Then the domain of f + g, f − g, f g, and f f is (−∞, ∞). For f /g, we must exclude 0 since g(0) = 0. The domain of f /g is (−∞, 0) ∪ (0, ∞). For g/f , we must exclude −1 since f (−1) = 0. The domain of g/f is (−∞, −1) ∪ (−1, ∞).

b) (f + g)(x) = f (x) + g(x) = x + 1 + |x| (f − g)(x) = f (x) − g(x) = x + 1 − |x| (f g)(x) = f (x) · g(x) = (x + 1)|x|

(f f )(x) = f (x)·f (x) = (x+1)(x+1) = x2 + 2x + 1

(f /g)(x) = (g/f )(x) =

x+1 |x|

|x| x+1

26. f (x) = 4|x|, g(x) = 1 − x

a) The domain of f and of g is (−∞, ∞). Then the domain of f +g, f −g, f g, and f f is (−∞, ∞). Since g(1) = 0, the domain of f /g is (−∞, 1) ∪ (1, ∞). Since f (0) = 0, the domain of g/f is (−∞, 0) ∪ (0, ∞).

b) (f + g)(x) = 4|x| + 1 − x

(f − g)(x) = 4|x| − (1 − x) = 4|x| − 1 + x

(f g)(x) = 4|x|(1 − x) = 4|x| − 4x|x| (f f )(x) = 4|x| · 4|x| = 16x2

4|x| 1−x 1−x (g/f )(x) = 4|x| (f /g)(x) =

27. f (x) = x3 , g(x) = 2x2 + 5x − 3

a) Since any number can be an input for either f or g, the domain of f , g, f + g, f − g, f g, and f f is the set of all real numbers, or (−∞, ∞). ! " 1 = 0, the domain of f /g Since g(−3) = 0 and g 2 $ # 1$ #1 ∪ ,∞ . is (−∞, −3) ∪ − 3, 2 2 Since f (0) = 0, the domain of g/f is (−∞, 0) ∪ (0, ∞).

b) (f + g)(x) = f (x) + g(x) = x3 + 2x2 + 5x − 3

(f − g)(x) = f (x)−g(x) = x3 −(2x2 +5x−3) = x3 − 2x2 − 5x + 3

(f g)(x) = f (x) · g(x) = x3 (2x2 + 5x − 3) = 2x5 + 5x4 − 3x3

(f f )(x) = f (x) · f (x) = x3 · x3 = x6 (f /g)(x) = (g/f )(x) =

f (x) x3 = 2 g(x) 2x + 5x − 3

2x2 + 5x − 3 g(x) = f (x) x3

28. f (x) = x2 − 4, g(x) = x3

a) The domain of f and of g is (−∞, ∞). Then the domain of f +g, f −g, f g, and f f is (−∞, ∞). Since g(0) = 0, the domain of f /g is (−∞, 0) ∪ (0, ∞). Since f (−2) = 0 and f (2) = 0, the domain of g/f is (−∞, −2) ∪ (−2, 2) ∪ (2, ∞).

b) (f + g)(x) = x2 − 4 + x3 , or x3 + x2 − 4

(f − g)(x) = x2 − 4 − x3 , or − x3 + x2 − 4

(f g)(x) = (x2 − 4)(x3 ) = x5 − 4x3

(f f )(x) = (x2 − 4)(x2 − 4) = x4 − 8x2 + 16

(f /g)(x) = (g/f )(x) =

x2 − 4 x3 x3 −4

x2

Exercise Set 1.6

83

4 1 , g(x) = x+1 6−x a) Since x + 1 = 0 when x = −1, we must exclude −1 from the domain of f . It is (−∞, −1) ∪ (−1, ∞). Since 6 − x = 0 when x = 6, we must exclude 6 from the domain of g. It is (−∞, 6)∪(6, ∞). The domain of f + g, f − g, and f g is the intersection of the domains of f and g, or (−∞, −1) ∪ (−1, 6) ∪ (6, ∞). The domain of f f is the same as the domain of f , or (−∞, −1) ∪ (−1, ∞). Since there are no values of x for which g(x) = 0 or f (x) = 0, the domain of f /g and g/f is (−∞, −1) ∪ (−1, 6) ∪ (6, ∞). 1 4 + b) (f + g)(x) = f (x) + g(x) = x+1 6−x 4 1 (f − g)(x) = f (x) − g(x) = − x+1 6−x 1 4 4 · = (f g)(x) = f (x)·g(x) = x+1 6−x (x+1)(6−x) 4 4 16 (f f )(x) = f (x)·f (x) = , or · = x + 1 x + 1 (x + 1)2 16 x2 + 2x + 1

29. f (x) =

4 x + 1 = 4 (f /g)(x) = 1 x+1 6−x 1 1 (g/f )(x) = 6 − x = 4 6−x x+1

·

6−x 4(6 − x) = 1 x+1

·

x+1 x+1 = 4 4(6 − x)

2 x−5 a) The domain of f is (−∞, ∞). Since x − 5 = 0 when x = 5, the domain of g is (−∞, 5)∪(5, ∞). Then the domain of f + g, f − g, and f g is (−∞, 5) ∪ (5, ∞). The domain of f f is (−∞, ∞). Since there are no values of x for which g(x) = 0, the domain of f /g is (−∞, 5) ∪ (5, ∞). Since f (0) = 0, the domain of g/f is (−∞, 0) ∪ (0, 5) ∪ (5, ∞). 2 2 b) (f + g)(x) = 2x + x−5 2 (f − g)(x) = 2x2 − x−5 2 4x2 (f g)(x) = 2x2 · = x−5 x−5 (f f )(x) = 2x2 · 2x2 = 4x4

30. f (x) = 2x2 , g(x) =

2x2 x−5 = x2 (x−5) = x3 −5x2 = 2x2 · 2 2 x−5 2 2 1 1 1 x−5 · = (g/f )(x) = = = 2x2 x−5 2x2 x2 (x−5) x3 −5x2 (f /g)(x) =

1 , g(x) = x − 3 x a) Since f (0) is not defined, the domain of f is (−∞, 0) ∪ (0, ∞). The domain of g is (−∞, ∞). Then the domain of f + g, f − g, f g, and f f is (−∞, 0) ∪ (0, ∞). Since g(3) = 0, the domain of f /g is (−∞, 0) ∪ (0, 3) ∪ (3, ∞). There are no values of x for which f (x) = 0, so the domain of g/f is (−∞, 0) ∪ (0, ∞). 1 b) (f + g)(x) = f (x) + g(x) = + x − 3 x 1 1 (f −g)(x) = f (x)−g(x) = −(x−3) = −x + 3 x x x−3 3 1 , or 1 − (f g)(x) = f (x)·g(x) = ·(x−3) = x x x 1 1 1 (f f )(x) = f (x) · f (x) = · = 2 x x x 1 f (x) 1 1 1 x (f /g)(x) = = = · = g(x) x−3 x x−3 x(x − 3)

31. f (x) =

x g(x) x−3 = (x−3) · = x(x−3), or = 1 f (x) 1 x 2 x − 3x

(g/f )(x) =

1 x a) The domain of f (x) is [−6, ∞). The domain of g(x) is (−∞, 0) ∪ (0, ∞). Then the domain of f + g, f − g, and f g is [−6, 0) ∪ (0, ∞). The domain of f f is [−6, ∞). Since there are no values of x for which g(x) = 0, the domain of f /g is [−6, 0)∪(0, ∞). Since f (−6) = 0, the domain of g/f is (−6, 0) ∪ (0, ∞). √ 1 b) (f + g)(x) = x + 6 + x √ 1 (f − g)(x) = x + 6 − x √ √ 1 x+6 (f g)(x) = x + 6 · = x x √ √ (f f )(x) = x + 6 · x + 6 = |x + 6|

32. f (x) =

√

x + 6, g(x) =

√

√ x+6 √ x = x+6· =x x+6 1 1 x 1 1 1 1 x (g/f )(x) = √ = ·√ = √ x x+6 x+6 x x+6 (f /g)(x) =

33. From the graph we see that the domain of F is [2, 11] and the domain of G is [1, 9]. The domain of F + G is the set of numbers in the domains of both F and G. This is [2, 9]. 34. The domain of F − G and F G is the set of numbers in the domains of both F and G. (See Exercise 33.) This is [2, 9]. The domain of F/G is the set of numbers in the domains of both F and G, excluding those for which G = 0. Since G > 0 for all values of x in its domain, the domain of F/G is [2, 9].

84

Chapter 1: Graphs, Functions, and Models

35. The domain of G/F is the set of numbers in the domains of both F and G (See Exercise 33.), excluding those for which F = 0. Since F (3) = 0, the domain of G/F is [2, 3) ∪ (3, 9]. 36.

43.

y 6

G#F

4 2

y

2

8

4

6

8

10

x

#2

6 4

44.

F "G

2

y 4

2

4

6

8

10

x

#2

37.

2

6

8

10

x

#4

4

G #F

45. a) P (x) = R(x) − C(x) = 60x − 0.4x2 − (3x + 13) =

2 2

4

6

8

10

x

#2

60x − 0.4x2 − 3x − 13 = −0.4x2 + 57x − 13

b) R(100) = 60·100−0.4(100)2 = 6000−0.4(10, 000) =

#4

6000 − 4000 = 2000

C(100) = 3 · 100 + 13 = 300 + 13 = 313

P (100) = R(100) − C(100) = 2000 − 313 = 1687

y 6

46. a) P (x) = 200x − x2 − (5000 + 8x) = 200x − x2 − 5000 − 8x = −x2 + 192x − 5000

F #G

4 2 2

4

6

8

10

x

#2

b) R(175) = 200(175) − 1752 = 4375 C(175) = 5000 + 8 · 175 = 6400

P (175) = R(175) − C(175) = 4375 − 6400 = −2025

#4

39. From the graph, we see that the domain of F is [0, 9] and the domain of G is [3, 10]. The domain of F + G is the set of numbers in the domains of both F and G. This is [3, 9]. 40. The domain of F − G and F G is the set of numbers in the domains of both F and G. (See Exercise 39.) This is [3, 9]. The domain of F/G is the set of numbers in the domains of both F and G, excluding those for which G = 0. Since G > 0 for all values of x in its domain, the domain of F/G is [3, 9]. 41. The domain of G/F is the set of numbers in the domains of both F and G (See Exercise 39.), excluding those for which F = 0. Since F (6) = 0 and F (8) = 0, the domain of G/F is [3, 6) ∪ (6, 8) ∪ (8, 9]. 42.

4

#2

y 6

38.

F#G

2

y 10 8

F"G

6 4 2 2

4

6

8

10

x

(We could also use the function found in part (a) to find P (175).) 3 47. f (x) = − x + 10 5 3 3 3 f (x + h) = − (x + h) + 10 = − x − h + 10 5 5 5 ! " 3 3 3 − x − h + 10 − − x + 10 f (x + h) − f (x) 5 5 5 = h h 3 3 3 − x − h + 10 + x − 10 5 5 5 = h 3 − h 3 = 5 =− h 5 48. f (x) = x − 4 x + h − 4 − (x − 4) f (x + h) − f (x) = h h x+h−4−x+4 = h h = =1 h

Exercise Set 1.6

85 54. f (x) = 5x2 + 4x

49. f (x) = x2 + 1

f (x+h)−f (x) (5x2+10xh+5h2+4x+4h)−(5x2+4x) = = h h 10xh + 5h2 + 4h = 10x + 5h + 4 h

f (x + h) = (x + h)2 + 1 = x2 + 2xh + h2 + 1 f (x + h) − f (x) x + 2xh + h + 1 − (x + 1) = h h 2

2

2

=

x2 + 2xh + h2 + 1 − x2 − 1 h

=

2xh + h2 h

55. f (x) = 4 + 5|x| f (x + h) = 4 + 5|x + h| 4 + 5|x + h| − (4 + 5|x|) f (x + h) − f (x) = h h

h(2x + h) h h 2x + h = · h 1 = 2x + h

=

50. f (x) = 2 − x2

f (x + h) − f (x) 2 − (x + h) − (2 − x ) = = h h 2

2

2 − x2 − 2xh − h2 − 2 + x2 −2xh − h2 = = h h h(−2x − h) = −2x − h h

1 52. f (x) = − x + 7 2

# 1 $ 1 − (x + h) + 7 − − + 7 f (x + h) − f (x) 2 = 2 = h h 1 1 1 1 − h − x− h+7+ −7 1 2 2 2 2 = =− h h 2

53. f (x) = 3x2 − 2x + 1

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

3(x2 + 2xh + h2 ) − 2(x + h) + 1 = 3x2 + 6xh + 3h2 − 2x − 2h + 1

f (x) = 3x2 − 2x + 1 f (x + h) − f (x) = h 2 2 (3x + 6xh + 3h −2x−2h + 1)−(3x2 − 2x + 1) = h 2 2 2 3x + 6xh + 3h − 2x − 2h + 1 − 3x + 2x − 1 = h 2 h(6x + 3h − 2) 6xh + 3h − 2h = = h h·1 h 6x + 3h − 2 · = 6x + 3h − 2 h 1

4 + 5|x + h| − 4 − 5|x| h

=

5|x + h| − 5|x| h

56. f (x) = 2|x| + 3x f (x+h)−f (x) (2|x+h|+3x+3h)−(2|x|+3x) = = h h 2|x + h| − 2|x| + 3h h 57. f (x) = x3 f (x + h) = (x + h)3 = x3 + 3x2 h + 3xh2 + h3 f (x) = x3

51. f (x) = 3x − 5

f (x + h) = 3(x + h) − 5 = 3x + 3h − 5 3x + 3h − 5 − (3x − 5) f (x + h) − f (x) = h h 3x + 3h − 5 − 3x + 5 = h 3h =3 = h

=

f (x + h) − f (x) x3 + 3x2 h + 3xh2 + h3 − x3 = = h h h(3x2 + 3xh + h2 ) 3x2 h + 3xh2 + h3 = = h h·1 h 3x2 + 3xh + h2 · = 3x2 + 3xh + h2 h 1 58.

f (x) = x3 − 2x

f (x+h)−f (x) (x+h)3 −2(x+h)−(x3 −2x) = = h h

x3 + 3x2 h + 3xh2 + h3 −2x − 2h−x3 +2x = h 3x2 h+3xh2 + h3 −2h h(3x2 +3xh + h2 −2) = = h h 3x2 + 3xh + h2 − 2 59. f (x) =

x−4 x+3

x+h−4 x−4 − f (x + h) − f (x) x = +h+3 x+3 = h h x+h−4 x−4 − x + h + 3 x + 3 · (x + h + 3)(x + 3) = h (x + h + 3)(x + 3) (x + h − 4)(x + 3) − (x − 4)(x + h + 3) = h(x + h + 3)(x + 3) x2+hx−4x+3x+3h−12−(x2+hx+3x−4x−4h−12) = h(x+h+3)(x+3) x2 + hx − x + 3h − 12 − x2 − hx + x + 4h + 12 = h(x + h + 3)(x + 3) h 7 7h = · = h(x + h + 3)(x + 3) h (x + h + 3)(x + 3)

86

Chapter 1: Graphs, Functions, and Models 7 (x + h + 3)(x + 3)

60. f (x) =

71.

x 2−x

x+h x − f (x + h) − f (x) 2 − (x + h) 2 − x = = h h (x + h)(2 − x) − x(2 − x − h) (2 − x − h)(2 − x) = h

The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞). 72.

2x − x2 + 2h − hx − 2x + x2 + hx (2 − x − h)(2 − x) = h 2h (2 − x − h)(2 − x) = h

61. 62. 63.

f (1 + 2 − 6) = f (−3) = 3(−3) + 1 = −9 + 1 = −8

73.

74.

(g ◦ f )(−2) = g(f (−2)) = g(3(−2) + 1) = g(−5) =

(−5)2 − 2(−5) − 6 = 25 + 10 − 6 = 29

(h ◦ f )(1) = h(f (1)) = h(3 · 1 + 1) = h(3 + 1) =

65.

66.

67. 68. 69.

! # $" !# $ " #1$ #1$ 1 1 3 =g h =g = (g ◦ h) =g 2 2 2 8 # 1 $2 # 1 $ 1 1 399 15 −2 −6 = − −6=− , or − 6 8 8 64 4 64 64 (g ◦ f )(5) = g(f (5)) = g(3 · 5 + 1) = g(15 + 1) =

g(16) = 162 − 2 · 16 − 6 = 218 ! # $" !# $ " #1$ #1$ 1 1 2 =f g −2 −6 = =f (f ◦ g) 3 3 3 3 # 59 $ #1 2 $ # 59 $ 56 =3 − + 1=− f − − 6 =f − 9 3 9 9 3 (f ◦ h)(−3) = f (h(−3)) = f ((−3)3 ) = f (−27) = 3(−27) + 1 = −81 + 1 = −80

(h ◦ g)(3) = h(g(3)) = h(32 − 2 · 3 − 6) = h(9 − 6 − 6) = h(−3) = (−3)3 = −27

(f ◦ g)(x) = f (g(x)) = f (x − 3) = x − 3 + 3 = x (g ◦ f )(x) = g(f (x)) = g(x + 3) = x + 3 − 3 = x

The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞). ! " 4 5 5 70. (f ◦ g)(x) = f x = · x=x 4 5 4 ! " 4 5 4 x = · x=x (g ◦ f )(x) = g 5 4 5 The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞).

(f ◦ g)(x) = f (g(x)) = f (4x−3) = (4x−3)2 −3 = 16x2 − 24x + 9 − 3 = 16x2 − 24x + 6 (g ◦ f )(x) = g(f (x)) = g(x2 −3) = 4(x2 −3)−3 = 4x2 − 12 − 3 = 4x2 − 15

The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞). (f ◦ g)(x) = f (2x − 7) = 4(2x − 7)2 − (2x − 7) + 10 = 4(4x2 − 28x + 49) − (2x − 7) + 10 = 16x2 − 112x + 196 − 2x + 7 + 10 = 16x2 − 114x + 213 (g ◦ f )(x) = g(4x2 − x + 10) = 2(4x2 − x + 10) − 7 = 8x2 − 2x + 20 − 7 = 8x2 − 2x + 13

The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞).

h(4) = 43 = 64 64.

(f ◦ g)(x) = f (x2 + 5) = 3(x2 + 5) − 2 = 3x2 + 15 − 2 = 3x2 + 13 (g ◦ f )(x) = g(3x−2) = (3x−2)2 +5 = 9x2 −12x+4+5 = 9x2 − 12x + 9

The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞).

1 2 2h · = (2 − x − h)(2 − x) h (2 − x − h)(2 − x)

(f ◦ g)(−1) = f (g(−1)) = f ((−1)2 − 2(−1) − 6) =

(f ◦ g)(x) = f (g(x)) = f (3x2 −2x−1) = 3x2 −2x−1+1 = 3x2 − 2x (g ◦ f )(x) = g(f (x)) = g(x+1) = 3(x+1)2 −2(x+1)−1 = 3(x2 +2x+1)−2(x+1)−1 = 3x2 +6x+3−2x−2−1 = 3x2 + 4x

75.

4 = = 1 5 1−5· 1− x x x 4x 4 =4· = x−5 x−5 x−5 x # 4 $ 1 (g ◦ f )(x) = g(f (x)) = g = = 4 1 − 5x 1 − 5x 1 − 5x 1 − 5x = 1· 4 4 ) * + * 1 The domain of f is x**x )= and the domain of g is 5 {x|x )= 0}. Consider the domain of f ◦ g. Since 0 is not in 1 the domain of g, 0 is not in the domain of f ◦ g. Since 5 1 is not in the domain of f , we know that g(x) cannot be . 5 1 We find the value(s) of x for which g(x) = . 5 1 1 = x 5 5=x Multiplying by 5x (f ◦ g)(x) = f (g(x)) = f

#1$ x

=

4

Thus 5 is also not in the domain of f ◦ g. Then the domain of f ◦g is {x|x )= 0 and x )= 5}, or (−∞, 0)∪(0, 5)∪(5, ∞). 1 Now consider the domain of g ◦ f . Recall that is not in 5 the domain of f , so it is not in the domain of g ◦ f . Now 0 is not in the domain of g but f (x) is never 0, so the domain # $ , * 1$ #1 1* , or − ∞, ∪ ,∞ . of g ◦ f is x*x )= 5 5 5

Exercise Set 1.6 76.

87

2x + 1 6 =6· = 1 1 2x + 1 6(2x + 1), or 12x + 6 #6$ 1 1 1 = = = = (g ◦ f )(x) = g 6 12 12 +x x 2· +1 +1 x x x x x 1· = 12 + x 12 + x The domain of f is {x|x )= 0} and the domain of g , * 1* is x*x )= − . Consider the domain of f ◦ g. Since 2 1 1 − is not in the domain of g, − is not in the domain 2 2 of f ◦ g. Now 0 is not in the domain of f but g(x) , * 1* is never 0, so the domain of f ◦ g is x*x )= − , or 2 # $ 1$ # 1 − ∞, − ∪ − ,∞ . 2 2 Now consider the domain of g ◦ f . Since 0 is not in the domain of f , then 0 is not in the domain of g ◦ f . Also, 1 since − is not in the domain of g, we find the value(s) of 2 1 x for which f (x) = − . 2 1 6 =− x 2 −12 = x , * * Then the domain of g ◦ f is x*x )= −12 and x )= 0 , or (f ◦ g)(x) = f

#

1 $ = 2x + 1

(−∞, −12) ∪ (−12, 0) ∪ (0, ∞). " ! x+7 77. (f ◦ g)(x) = f (g(x)) = f = 3 ! " x+7 3 −7=x+7−7=x 3 (3x − 7) + 7 = (g ◦ f )(x) = g(f (x)) = g(3x − 7) = 3 3x =x 3 The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞). 78.

(f ◦ g)(x) = f (1.5x + 1.2) =

2 4 (1.5x + 1.2) − 3 5

4 x + 0.8 − = x 5 " ! " ! 4 2 4 2 x− = 1.5 x − + 1.2 = (g ◦ f )(x) = g 3 5 3 5 x − 1.2 + 1.2 = x

The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞). √ √ 79. (f ◦ g)(x) = f (g(x)) = f ( x) = 2 x + 1 √ (g ◦ f )(x) = g(f (x)) = g(2x + 1) = 2x + 1 The domain of f is (−∞, ∞) and the domain of g is {x|x ≥ 0}. Thus the domain of f ◦ g is {x|x ≥ 0}, or [0, ∞).

Now consider the domain of g ◦f . There are no restrictions on the domain of f , but the domain of g is {x|x ≥ 0}. Since

, * 1 1* f (x) ≥ 0 for x ≥ − , the domain of g ◦ f is x*x ≥ − , 2 2 $ . 1 or − , ∞ . 2 √ 80. (f ◦ g)(x) = f (2 − 3x) = 2 − 3x √ √ (g ◦ f )(x) = g( x) = 2 − 3 x

The domain of f is {x|x ≥ 0} and the domain of g is 2 (−∞, ∞). Since g(x) ≥ 0 when x ≤ , the domain of f ◦ g 3 ! & 2 is − ∞, . 3 Now consider the domain of g ◦ f . Since the domain of f is {x|x ≥ 0} and the domain of g is (−∞, ∞), the domain of g ◦ f is {x|x ≥ 0}, or [0, ∞).

81.

(f ◦ g)(x) = f (g(x)) = f (0.05) = 20 (g ◦ f )(x) = g(f (x)) = g(20) = 0.05

The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞). √ 82. (f ◦ g)(x) = ( 4 x)4 = x √ 4 (g ◦ f )(x) = x4 = |x| The domain of f is (−∞, ∞) and the domain of g is {x|x ≥ 0}, so the domain of f ◦ g is {x|x ≥ 0}, or [0, ∞).

Now consider the domain of g ◦f . There are no restrictions on the domain of f and f (x) ≥ 0 for all values of x, so the domain is (−∞, ∞). 83.

(f ◦ g)(x) = f (g(x)) = f (x2 − 5) = √ √ x2 − 5 + 5 = x2 = |x| √ (g ◦ f )(x) = g(f (x)) = g( x + 5) = √ ( x + 5)2 − 5 = x + 5 − 5 = x

The domain of f is {x|x ≥ −5} and the domain of g is (−∞, ∞). Since x2 ≥ 0 for all values of x, then x2 −5 ≥ −5 for all values of x and the domain of g ◦ f is (−∞, ∞).

Now consider the domain of f ◦g. There are no restrictions on the domain of g, so the domain of f ◦ g is the same as the domain of f , {x|x ≥ −5}, or [−5, ∞). √ 84. (f ◦ g)(x) = ( 5 x + 2)5 − 2 = x + 2 − 2 = x √ √ 5 (g ◦ f )(x) = 5 x5 − 2 + 2 = x5 = x

The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞). √ √ 85. (f ◦ g)(x) = f (g(x)) = f ( 3 − x) = ( 3 − x)2 + 2 = 3−x+2=5−x

(g ◦ f )(x) = g(f (x)) = g(x2 + 2) = √ √ 3 − x2 − 2 = 1 − x2

3 − (x2 + 2) =

/

The domain of f is (−∞, ∞) and the domain of g is {x|x ≤ 3}, so the domain of f ◦ g is {x|x ≤ 3}, or (−∞, 3].

Now consider the domain of g ◦f . There are no restrictions on the domain of f and the domain of g is {x|x ≤ 3}, so we find the values of x for which f (x) ≤ 3. We see that x2 + 2 ≤ 3 for −1 ≤ x ≤ 1, so the domain of g ◦ f is {x| − 1 ≤ x ≤ 1}, or [−1, 1].

88 86.

Chapter 1: Graphs, Functions, and Models √ √ (f ◦ g)(x) = f ( x2 − 25) = 1 − ( x2 − 25)2 =

1 +2 x − 2 = 1 x−2 1 + 2x − 4 2x − 3 x − 2 = = x−2 1 1 x−2 x−2 2x − 3 x − 2 · = 2x − 3 = x−2 1 The domain of f is {x|x )= 2} and the domain of g is {x|x )= 0}, so 0 is not in the domain of f ◦ g. We find the values of x for which g(x) = 2. x+2 =2 x x + 2 = 2x 1 (g ◦ f )(x) = g x−2 !

1 − (x2 − 25) = 1 − x2 + 25 = 26 − x2 / (g ◦ f )(x) = g(1 − x2 ) = (1 − x2 )2 − 25 = √ √ 1 − 2x2 + x4 − 25 = x4 − 2x2 − 24

The domain of f is (−∞, ∞) and the domain of g is {x|x ≤ −5 or x ≥ 5}, so the domain of f ◦ g is {x|x ≤ −5 or x ≥ 5}, or (−∞, −5] ∪ [5, ∞).

Now consider the domain of g ◦ f . There are no restrictions on the domain of f and the domain of g is {x|x ≤ −5 or x ≥ 5}, so we find the values of x for 2 which f (x) ≤ √ ≥ 5. We 2see that 1 − x ≤ −5 √ −5 or f (x) when x ≤ − 6 or x ≥ 6 and 1 − x√≥ 5 has no solution, √ so the domain of g ◦ f is {x|x ≤ − 6 or x ≥ 6}, or √ √ (−∞, − 6] ∪ [ 6, ∞). ! " 1 87. (f ◦ g)(x) = f (g(x)) = f = 1+x 1 1− 1+x 1 1+x !

"

2=x Then the domain of f ◦ g is (−∞, 0) ∪ (0, 2) ∪ (2, ∞).

1+x−1 = 1+x = 1 1+x

1+x x · =x 1+x 1 ! " 1−x (g ◦ f )(x) = g(f (x)) = g = x

1+

!

1 1 "= = x+1−x 1−x x x

x 1 =1· =x 1 1 x The domain of f is {x|x )= 0} and the domain of g is {x|x )= −1}, so we know that −1 is not in the domain of f ◦ g. Since 0 is not in the domain of f , values of x for which g(x) = 0 are not in the domain of f ◦ g. But g(x) is never 0, so the domain of f ◦ g is {x|x )= −1}, or (−∞, −1) ∪ (−1, ∞). Now consider the domain of g ◦ f . Recall that 0 is not in the domain of f . Since −1 is not in the domain of g, we know that g(x) cannot be −1. We find the value(s) of x for which f (x) = −1. 1−x = −1 x 1 − x = −x Multiplying by x 1=0

False equation

We see that there are no values of x for which f (x) = −1, so the domain of g ◦ f is {x|x )= 0}, or (−∞, 0) ∪ (0, ∞). " ! 1 x+2 88. (f ◦ g)(x) = f = x+2 x −2 x 1 1 = = x + 2 − 2x −x + 2 x x x x x = , or = 1· −x + 2 −x + 2 2−x

"

89.

Now consider the domain of g ◦ f . Since the domain of f is {x|x )= 2}, we know that 2 is not in the domain of g ◦ f . Since the domain of g is {x|x )= 0}, we find the values of x for which f (x) = 0. 1 =0 x−2 1=0 We get a false equation, so there are no such values. Then the domain of g ◦ f is (−∞, 2) ∪ (2, ∞). (f ◦ g)(x) = f (g(x)) = f (x + 1) =

(x + 1)3 − 5(x + 1)2 + 3(x + 1) + 7 =

x3 + 3x2 + 3x + 1 − 5x2 − 10x − 5 + 3x + 3 + 7 = x3 − 2x2 − 4x + 6

(g ◦ f )(x) = g(f (x)) = g(x3 − 5x2 + 3x + 7) =

x3 − 5x2 + 3x + 7 + 1 = x3 − 5x2 + 3x + 8

90.

The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞). (g ◦ f )(x) = x3 + 2x2 − 3x − 9 − 1 = x3 + 2x2 − 3x − 10

(g ◦ f )(x) = (x − 1)3 + 2(x − 1)2 − 3(x − 1) − 9 =

x3 − 3x2 + 3x − 1 + 2x2 − 4x + 2 − 3x + 3 − 9 = x3 − x2 − 4x − 5

The domain of f and of g is (−∞, ∞), so the domain of f ◦ g and of g ◦ f is (−∞, ∞).

91. h(x) = (4 + 3x)5

This is 4 + 3x to the 5th power. The most obvious answer is f (x) = x5 and g(x) = 4 + 3x. √ 92. f (x) = 3 x, g(x) = x2 − 8 1 (x − 2)4 This is 1 divided by (x − 2) to the 4th power. One obvious 1 answer is f (x) = 4 and g(x) = x − 2. Another possibility x 1 is f (x) = and g(x) = (x − 2)4 . x

93. h(x) =

Exercise Set 1.7 1 94. f (x) = √ , g(x) = 3x + 7 x 95. f (x) =

x−1 , g(x) = x3 x+1

96. f (x) = |x|, g(x) = 9x2 − 4 2 + x3 2 − x3 √ 4 98. f (x) = x , g(x) = x − 3

97. f (x) = x6 , g(x) =

x−5 x+2 √ √ 100. f (x) = 1 + x, g(x) = 1 + x 99. f (x) =

√

x, g(x) =

101. f (x) = x3 − 5x2 + 3x − 1, g(x) = x + 2

89 109. Equations (a) − (f ) are in the form y = mx + b, so we can read the y-intercepts directly from the equations. Equa2 tions (g) and (h) can be written in this form as y = x − 2 3 and y = −2x + 3, respectively. We see that only equation (c) has y-intercept (0, 1). 110. None (See Exercise 109.) 111. If a line slopes down from left to right, its slope is negative. The equations y = mx + b for which m is negative are (b), (d), (f), and (h). (See Exercise 109.) 112. The equation for which |m| is greatest is the equation with the steepest slant. This is equation (b). (See Exercise 109.) 113. The only equation that has (0, 0) as a solution is (a).

102. f (x) = 2x5/3 + 5x2/3 , g(x) = x − 1, or

114. Equations (c) and (g) have the same slope. (See Exercise 109.)

103. f (x) = (t ◦ s)(x) = t(s(x)) = t(x − 3) = x − 3 + 4 = x + 1

115. Only equations (c) and (g) have the same slope and different y-intercepts. They represent parallel lines.

f (x) = 2x5 + 5x2 , g(x) = (x − 1)1/3 We have f (x) = x + 1.

104. a) Use the distance formula, distance = rate × time. Substitute 3 for the rate and t for time. r(t) = 3t b) Use the formula for the area of a circle. A(r) = πr2 c) (A ◦ r)(t) = A(r(t)) = A(3t) = π(3t)2 = 9πt2

This function gives the area of the ripple in terms of time t.

105.

116. The only equations for which the product of the slopes is −1 are (a) and (f). 117. Answers may vary. One example is f (x) = 2x + 5 and x−5 . Other examples are found in Exercises 69, g(x) = 2 70, 77, 78, 84, and 87. 1 1 , g(x) = x+7 x−3 , * 7* , and the domain of g(x) 119. The domain of h(x) is x*x )= 3 7 is {x|x )= 3}, so and 3 are not in the domain of (h/g)(x). 3 We must also exclude the value of x for which g(x) = 0. 118. Answers may vary; f (x) =

x4 − 1 =0 5x − 15 x4 − 1 = 0 x =1 4

Multiplying by 5x − 15

x = ±1

Then the domain of (h/g)(x) is , * 7 * x*x )= and x )= 3 and x )= −1 and x )= 1 , or 3 # 7$ #7 $ (−∞, −1) ∪ (−1, 1) ∪ 1, ∪ , 3 ∪ (3, ∞). 3 3

106.

120. The domain of h + f , h − f , and hf consists of all numbers that are in the domain of both h and f , or {−4, 0, 3}. 107. The graph of y = (h − g)(x) will be the same as the graph of y = h(x) moved down b units. 108. The values of x that must be excluded from the domain of (f ◦ g)(x) are the value of x that are not in the domain of g as well as the values of x for which g(x) takes values that are not in the domain of f . The values of x that must be excluded from the domain of (g ◦ f )(x) are the values of x that are not in the domain of f as well as the values of x for which f (x) takes values that are not in the domain of g.

The domain of h/f consists of all numbers that are in the domain of both h and f , excluding any for which the value of f is 0, or {−4, 0}.

Exercise Set 1.7 1. If the graph were folded on the x-axis, the parts above and below the x-axis would not coincide, so the graph is not symmetric with respect to the x-axis. If the graph were folded on the y-axis, the parts to the left and right of the y-axis would coincide, so the graph is symmetric with respect to the y-axis.

90

Chapter 1: Graphs, Functions, and Models If the graph were rotated 180◦ , the resulting graph would not coincide with the original graph, so it is not symmetric with respect to the origin.

7.

5 4 3 2 1

2. If the graph were folded on the x-axis, the parts above and below the x-axis would not coincide, so the graph is not symmetric with respect to the x-axis.

#5 #4 #3 #2 #1 #1 #2 #3 #4 #5

If the graph were folded on the y-axis, the parts to the left and right of the y-axis would coincide, so the graph is symmetric with respect to the y-axis.

Simplifying

y = |x| − 2

Original equation

y = |x| − 2

Simplifying

y = | − x| − 2

4. If the graph were folded on the x-axis, the parts above and below the x-axis would not coincide, so the graph is not symmetric with respect to the x-axis.

Replacing x by −x

The last equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Test algebraically for symmetry with respect to the origin:

If the graph were folded on the y-axis, the parts to the left and right of the y-axis would not coincide, so the graph is not symmetric with respect to the y-axis.

y = |x| − 2

Original equation

−y = | − x| − 2

If the graph were rotated 180◦ , the resulting graph would coincide with the original graph, so it is symmetric with respect to the origin.

If the graph were rotated 180◦ , the resulting graph would coincide with the original graph, so it is symmetric with respect to the origin.

y = −|x| + 2

Replacing y by −y

Test algebraically for symmetry with respect to the y-axis:

If the graph were rotated 180 , the resulting graph would not coincide with the original graph, so it is not symmetric with respect to the origin.

If the graph were folded on the y-axis, the parts to the left and right of the y-axis would coincide, so the graph is symmetric with respect to the y-axis.

Original equation

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis.

◦

6. If the graph were folded on the x-axis, the parts above and below the x-axis would coincide, so the graph is symmetric with respect to the x-axis.

y = |x| − 2

−y = |x| − 2

If the graph were folded on the y-axis, the parts to the left and right of the y-axis would not coincide, so the graph is not symmetric with respect to the y-axis.

If the graph were rotated 180◦ , the resulting graph would coincide with the original graph, so it is symmetric with respect to the origin.

y ! | x| " 2

Test algebraically for symmetry with respect to the x-axis:

3. If the graph were folded on the x-axis, the parts above and below the x-axis would coincide, so the graph is symmetric with respect to the x-axis.

If the graph were folded on the y-axis, the parts to the left and right of the y-axis would not coincide, so the graph is not symmetric with respect to the y-axis.

x

1 2 3 4 5

The graph is symmetric with respect to the y-axis. It is not symmetric with respect to the x-axis or the origin.

If the graph were rotated 180◦ , the resulting graph would not coincide with the original graph, so it is not symmetric with respect to the origin.

5. If the graph were folded on the x-axis, the parts above and below the x-axis would not coincide, so the graph is not symmetric with respect to the x-axis.

y

Replacing x by −x and y by −y

−y = |x| − 2

Simplifying

y = −|x| + 2

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. 8.

y 10 8

y = |x + 5|

6 4 2 -10 -8 -6 -4 -2

-2

x 2

4

6

8 10

-4 -6 -8 -10

The graph is not symmetric with respect to the x-axis, the y-axis, or the origin. Test algebraically for symmetry with respect to the x-axis: y = |x + 5|

−y = |x + 5|

Original equation Replacing y by −y

y = −|x + 5| Simplifying

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis.

Exercise Set 1.7

91

Test algebraically for symmetry with respect to the y-axis: y = |x + 5|

10.

y

Original equation

5

y = | − x + 5| Replacing x by −x

4 3

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis.

1

Test algebraically for symmetry with respect to the origin: y = |x + 5|

-5 -4 -3 -2 -1

Original equation

4

5

-5

The graph is not symmetric with respect to the x-axis, the y-axis, or the origin. Test algebraically for symmetry with respect to the x-axis:

y

2x − 5 = 3y

5 4

Original equation

2x − 5 = 3(−y)

3 2

5y = 4x + 5

1 1

2

3

4

Test algebraically for symmetry with respect to the y-axis: 2x − 5 = 3y

-4 -5

5y = 4x + 5

Test algebraically for symmetry with respect to the origin: 2x − 5 = 3y

Replacing y by −y

5y = −4x − 5

Original equation

2x + 5 = 3y

11.

y

Simplifying

5 4

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis.

3 2

Test algebraically for symmetry with respect to the origin: 5y = 4x + 5

−5y = −4x + 5

Original equation Replacing x by −x and y by −y Simplifying

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin.

Simplifying

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin.

Test algebraically for symmetry with respect to the y-axis: Replacing x by −x

Replacing x by −x and y by −y

−2x − 5 = −3y

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis.

5y = 4(−x) + 5

Original equation

2(−x) − 5 = 3(−y)

Simplifying

5y = 4x + 5

Simplifying

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis.

Original equation

5(−y) = 4x + 5

Replacing x by −x

−2x − 5 = 3y

Test algebraically for symmetry with respect to the x-axis:

5(−y) = 4(−x) + 5

Original equation

2(−x) − 5 = 3y

The graph is not symmetric with respect to the x-axis, the y-axis, or the origin.

5y = −4x + 5

Simplifying

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis.

x

5

-3

−5y = 4x + 5

Replacing y by −y

−2x + 5 = 3y

-2

5y = 4x − 5

3

-4

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin.

-1

2

-3

y = −| − x + 5| Simplifying

-5 -4 -3 -2 -1

x 1

-1 -2

−y = | − x + 5| Replacing x by −x and y by −y

9.

2x — 5 = 3y

2

1 -5 -4 -3 -2 -1

-1

x 1

2

3

4

5

5y = 2 x 2 – 3

-2 -3 -4 -5

The graph is symmetric with respect to the y-axis. It is not symmetric with respect to the x-axis or the origin. Test algebraically for symmetry with respect to the x-axis: 5y = 2x2 − 3

5(−y) = 2x − 3 2

−5y = 2x2 − 3

5y = −2x + 3 2

Original equation Replacing y by −y Simplifying

92

Chapter 1: Graphs, Functions, and Models The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis.

13.

y 5 4 3 2 1

Test algebraically for symmetry with respect to the y-axis: 5y = 2x2 − 3

Original equation

5y = 2(−x) − 3

Replacing x by −x

2

5y = 2x − 3 2

!5

The graph is not symmetric with respect to the x-axis or the y-axis. It is symmetric with respect to the origin.

Test algebraically for symmetry with respect to the origin: 5y = 2x − 3

Original equation

2

5(−y) = 2(−x)2 − 3

Test algebraically for symmetry with respect to the x-axis: 1 y= Original equation x 1 Replacing y by −y −y = x 1 y=− Simplifying x The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis.

Replacing x by −x and y by −y

−5y = 2x2 − 3

Simplifying

5y = −2x2 + 3

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. 12.

y 5

Test algebraically for symmetry with respect to the y-axis: 1 Original equation y= x 1 y= Replacing x by −x −x 1 Simplifying y=− x The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis.

4 3

x 2 + 4 = 3y

2 1 -5 -4 -3 -2 -1

-1

x 1

2

3

4

5

-2 -3 -4 -5

Test algebraically for symmetry with respect to the origin: 1 Original equation y= x 1 −y = Replacing x by −x and y by −y −x 1 y= Simplifying x The last equation is equivalent to the original equation, so the graph is symmetric with respect to the origin.

The graph is symmetric with respect to the y-axis. It is not symmetric with respect to the x-axis or the origin. Test algebraically for symmetry with respect to the x-axis: x2 + 4 = 3y

Original equation

x2 + 4 = 3(−y)

Replacing y by −y

−x − 4 = 3y 2

Simplifying

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Test algebraically for symmetry with respect to the y-axis: x2 + 4 = 3y (−x)2 + 4 = 3y x + 4 = 3y 2

Original equation Replacing x by −x

Test algebraically for symmetry with respect to the origin: x2 + 4 = 3y (−x) + 4 = 3(−y) 2

x2 + 4 = −3y

−x2 − 4 = 3y

14.

y 5 4

4 y=–— x

3

The last equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Original equation Replacing x by −x and y by −y Simplifying

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin.

x

1 2 3 4 5

!5

The last equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis.

1 y!" x

2 1 –5 –4 –3 –2 –1 0 –1

x 1

2

3

4

5

–2 –3 –4 –5

The graph is not symmetric with respect to the x-axis or the y-axis. It is symmetric with respect to the origin.

Exercise Set 1.7

93

Test algebraically for symmetry with respect to the x-axis: 4 y=− Original equation x 4 Replacing y by −y −y = − x 4 y= Simplifying x The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Test algebraically for symmetry with respect to the y-axis: 4 Original equation y=− x 4 Replacing x by −x y=− −x 4 Simplifying y= x The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Test algebraically for symmetry with respect to the origin: 4 Original equation y=− x 4 −y = − Replacing x by −x and y by −y −x 4 Simplifying y=− x The last equation is equivalent to the original equation, so the graph is symmetric with respect to the origin. 15. Test for symmetry with respect to the x-axis: 5x − 5y = 0 Original equation 5x − 5(−y) = 0

5x + 5y = 0

Replacing y by −y Simplifying

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis: 5x − 5y = 0 Original equation 5(−x) − 5y = 0 −5x − 5y = 0

Replacing x by −x

Simplifying

Test for symmetry with respect to the origin: 5x − 5y = 0 Original equation −5x + 5y = 0

Replacing x by −x and y by −y Simplifying

5x − 5y = 0

16. Test for symmetry with respect to the x-axis: 6x + 7y = 0 Original equation 6x − 7y = 0

6x + 7y = 0

Original equation

6(−x) + 7y = 0

Replacing x by −x

6x − 7y = 0

Simplifying

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Test for symmetry with respect to the origin: 6x + 7y = 0 6(−x) + 7(−y) = 0 6x + 7y = 0

Original equation Replacing x by −x and y by −y Simplifying

The last equation is equivalent to the original equation, so the graph is symmetric with respect to the origin. 17. Test for symmetry with respect to the x-axis: 3x2 − 2y 2 = 3

Original equation

3x − 2y = 3

Simplifying

3x − 2(−y)2 = 3 2

2

2

Replacing y by −y

The last equation is equivalent to the original equation, so the graph is symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis: 3x2 − 2y 2 = 3

Original equation

3x − 2y = 3

Simplifying

3(−x)2 − 2y 2 = 3 2

2

Replacing x by −x

The last equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Test for symmetry with respect to the origin: 3x2 − 2y 2 = 3

Original equation

3x2 − 2y 2 = 3

Simplifying

3(−x) − 2(−y) = 3 2

2

Replacing x by −x and y by −y

The last equation is equivalent to the original equation, so the graph is symmetric with respect to the origin. 5y = 7x2 − 2x

5(−y) = 7x2 − 2x

Replacing y by −y Simplifying

Original equation Replacing y by −y

5y = −7x + 2x Simplifying 2

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis: 5y = 7x2 − 2x

Original equation

5y = 7x2 + 2x

Simplifying

5y = 7(−x) − 2(−x) 2

The last equation is equivalent to the original equation, so the graph is symmetric with respect to the origin.

6x + 7(−y) = 0

Test for symmetry with respect to the y-axis:

18. Test for symmetry with respect to the x-axis:

5x + 5y = 0

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis.

5(−x) − 5(−y) = 0

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis.

Replacing x by −x

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis.

94

Chapter 1: Graphs, Functions, and Models Test for symmetry with respect to the origin: 5y = 7x − 2x

Original equation

2

5(−y) = 7(−x)2 − 2(−x) −5y = 7x2 + 2x

Replacing x by −x and y by −y Simplifying

5y = −7x2 − 2x

21. Test for symmetry with respect to the x-axis: 2x4 + 3 = y 2

Original equation

2x4 + 3 = (−y)2

Replacing y by −y

2x + 3 = y 4

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. 19. Test for symmetry with respect to the x-axis: y = |2x|

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin.

The last equation is equivalent to the original equation, so the graph is symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis:

Original equation

−y = |2x|

Replacing y by −y

y = −|2x| Simplifying

Simplifying

2

2x4 + 3 = y 2

Original equation

2(−x)4 + 3 = y 2 2x + 3 = y 4

Replacing x by −x Simplifying

2

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis.

The last equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis.

Test for symmetry with respect to the y-axis:

Test for symmetry with respect to the origin:

y = |2x|

Original equation

y = | − 2x|

Simplifying

y = |2(−x)| Replacing x by −x

The last equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Test for symmetry with respect to the origin: y = |2x|

2(−x) + 3 = (−y)

2

−y = |2(−x)| Replacing x by −x and y by −y −y = | − 2x| Simplifying −y = |2x|

y = −|2x|

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. 20. Test for symmetry with respect to the x-axis: y 3 = 2x2

Original equation

(−y) = 2x

Replacing y by −y

2

−y 3 = 2x2

Simplifying

y 3 = −2x2

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis: y 3 = 2x2

Original equation

y 3 = 2(−x)2

Replacing x by −x

y = 2x

2

Simplifying

The last equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Test for symmetry with respect to the origin: y 3 = 2x2

Original equation

(−y) = 2(−x) 3

−y 3 = 2x2

y 3 = −2x2

2

Replacing x by −x and y by −y Simplifying

Replacing x by −x and y by −y Simplifying

The last equation is equivalent to the original equation, so the graph is symmetric with respect to the origin. 22. Test for symmetry with respect to the x-axis:

Original equation

3

Original equation

4

2x4 + 3 = y 2

y = |2x|

3

2x4 + 3 = y 2

2y 2 = 5x2 + 12 2(−y)2 = 5x2 + 12 2y = 5x + 12 2

2

Original equation Replacing y by −y Simplifying

The last equation is equivalent to the original equation, so the graph is symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis: 2y 2 = 5x2 + 12

Original equation

2y = 5(−x) + 12

Replacing x by −x

2

2

2y 2 = 5x2 + 12

Simplifying

The last equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Test for symmetry with respect to the origin: 2y 2 = 5x2 + 12 2(−y) = 5(−x) + 12 2

2

2y 2 = 5x2 + 12

Original equation Replacing x by −x and y by −y Simplifying

The last equation is equivalent to the original equation, so the graph is symmetric with respect to the origin. 23. Test for symmetry with respect to the x-axis: 3y 3 = 4x3 + 2 3(−y) = 4x + 2 3

3

−3y 3 = 4x3 + 2

3y = −4x − 2 3

Original equation Replacing y by −y Simplifying

3

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis.

Exercise Set 1.7

95

Test for symmetry with respect to the y-axis: 3y 3 = 4x3 + 2 Original equation 3y 3 = 4(−x)3 + 2 3y 3 = −4x3 + 2

Replacing x by −x Simplifying

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Test for symmetry with respect to the origin: Original equation 3y 3 = 4x3 + 2 3(−y)3 = 4(−x)3 + 2 −3y 3 = −4x3 + 2

Replacing x by −x and y by −y Simplifying

3y 3 = 4x3 − 2

24. Test for symmetry with respect to the x-axis: 3x = |y| Original equation 3x = | − y| Replacing y by −y Simplifying

The last equation is equivalent to the original equation, so the graph is symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis: 3x = |y| Original equation 3(−x) = |y| Replacing x by −x −3x = |y| Simplifying

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Test for symmetry with respect to the origin: 3x = |y| Original equation

3(−x) = | − y| Replacing x by −x and y by −y −3x = |y|

Simplifying

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. 25. Test for symmetry with respect to the x-axis: xy = 12 Original equation x(−y) = 12 −xy = 12

Replacing y by −y Simplifying

xy = −12

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis: xy = 12 Original equation −xy = 12

xy = −12

Replacing x by −x

Simplifying

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Test for symmetry with respect to the origin: xy = 12 Original equation −x(−y) = 12

xy = 12

26. Test for symmetry with respect to the x-axis: xy − x2 = 3

Original equation

xy + x = −3

Simplifying

x(−y) − x2 = 3

Replacing x by −x and y by −y Simplifying

Replacing y by −y

2

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis: xy − x2 = 3

Original equation

xy + x = −3

Simplifying

−xy − (−x)2 = 3 2

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin.

3x = |y|

The last equation is equivalent to the original equation, so the graph is symmetric with respect to the origin.

Replacing x by −x

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Test for symmetry with respect to the origin: xy − x2 = 3

Original equation

xy − x2 = 3

Simplifying

−x(−y) − (−x) = 3 2

Replacing x by −x and y by −y

The last equation is equivalent to the original equation, so the graph is symmetric with respect to the origin. 27. x-axis: Replace y with −y; (−5, −6) y-axis: Replace x with −x; (5, 6)

Origin: Replace x with −x and y with −y; (5, −6) " ! 7 ,0 28. x-axis: Replace y with −y; 2 " ! 7 y-axis: Replace x with −x; − , 0 2 " ! 7 Origin: Replace x with −x and y with −y; − , 0 2 29. x-axis: Replace y with −y; (−10, 7)

y-axis: Replace x with −x; (10, −7)

Origin: Replace x with −x and y with −y; (10, 7) " ! 3 30. x-axis: Replace y with −y; 1, − 8 " ! 3 y-axis: Replace x with −x; − 1, 8 ! " 3 Origin: Replace x with −x and y with −y; − 1, − 8 31. x-axis: Replace y with −y; (0, 4)

y-axis: Replace x with −x; (0, −4)

Origin: Replace x with −x and y with −y; (0, 4) 32. x-axis: Replace y with −y; (8, 3)

y-axis: Replace x with −x; (−8, −3)

Origin: Replace x with −x and y with −y; (−8, 3) 33. The graph is symmetric with respect to the y-axis, so the function is even.

96

Chapter 1: Graphs, Functions, and Models

34. The graph is symmetric with respect to the y-axis, so the function is even.

45.

1 1 = 2 (−x)2 x f (x) = f (−x), so f is even. 46.

−f (x) = −(x − |x|) = −x + |x|

f (x) )= f (−x), so f is not even.

38. The graph is not symmetric with respect to either the yaxis or the origin, so the function is neither even nor odd. f (x) = −3x3 + 2x

f (−x) = −3(−x)3 + 2(−x) = 3x3 − 2x

−f (x) = −(−3x3 + 2x) = 3x3 − 2x

f (−x) = −f (x), so f is odd. 40.

f (x) = 7x3 + 4x − 2

f (−x) = 7(−x)3 + 4(−x) − 2 = −7x3 − 4x − 2

−f (x) = −(7x3 + 4x − 2) = −7x3 − 4x + 2

f (x) )= f (−x), so f is not even.

f (x) = x − |x|

f (−x) = (−x) − |(−x)| = −x − |x|

37. The graph is not symmetric with respect to either the yaxis or the origin, so the function is neither even nor odd.

39.

f (−x) )= −f (x), so f is not odd.

Thus, f (x) = x − |x| is neither even nor odd. 47.

f (x) = 8 f (−x) = 8

f (x) = f (−x), so f is even. √ 48. f (x) = x2 + 1 / √ f (−x) = (−x)2 + 1 = x2 + 1 f (x) = f (−x), so f is even.

49. Shift the graph of f (x) = x2 right 3 units.

f (−x) )= −f (x), so f is not odd.

y

Thus, f (x) = 7x3 + 4x − 2 is neither even nor odd. 41.

4

f (x) = 5x2 + 2x4 − 1

f (−x) = 5(−x)2 + 2(−x)4 − 1 = 5x2 + 2x4 − 1

f (x) = f (−x), so f is even.

1 x2

f (−x) =

35. The graph is symmetric with respect to the origin, so the function is odd. 36. The graph is not symmetric with respect to either the yaxis or the origin, so the function is neither even nor odd.

f (x) =

2 2

#4 #2

4

x

#2

42.

f (x) = x +

1 x

1 1 = −x − f (−x) = −x + −x x " ! 1 1 −f (x) = − x + = −x − x x

#4

50. Shift the graph of g(x) = x2 up

4

f (x) = x

17

f (−x) = (−x)17 = −x17

2 2

#4 #2

−f (x) = −x17

f (−x) = −f (x), so f is odd. √ 44. f (x) = 3 x √ √ f (−x) = 3 −x = − 3 x √ −f (x) = − 3 x

1 unit. 2

y

f (−x) = −f (x), so f is odd. 43.

f(x) ! (x # 3)2

#2 #4

4

x

1 g(x) ! x 2 " # 2

51. Shift the graph of g(x) = x down 3 units. y

f (−x) = −f (x), so f is odd.

4 2 2

#4 #2

4

#2 #4

g(x) ! x # 3

x

Exercise Set 1.7

97

52. Reflect the graph of g(x) = x across the x-axis and then shift it down 2 units. y

57. First stretch the graph of h(x) = x vertically by multiplying each y-coordinate by 3. Then reflect it across the x-axis and shift it up 3 units. y

4 2

g (x) ! #x # 2 2

#4 #2

4

4

h(x) ! #3x " 3

2

x

#2

2

#4 #2

4

x

#2

#4

#4

53. Reflect the graph of h(x) =

√

x across the x-axis. 58. First stretch the graph of f (x) = x vertically by multiplying each y-coordinate by 2. Then shift it up 1 unit.

y 4

y

h(x) ! #√x

2 4 2

#4 #2

4

x

2

#2

2

#4 #2

#4

4

x

#2

54. Shift the graph of g(x) =

√

y 4

#4

x right 1 unit.

g(x) ! √  x #1

59. First shrink the graph of h(x) = |x| vertically by multiply1 ing each y-coordinate by . Then shift it down 2 units. 2

2

y 2

#4 #2

f(x) ! 2x " 1

4

x

#2

4

#4

2

1 up 4 units. x

55. Shift the graph of h(x) =

1 h(x) ! #|x| #2 2

2

#4 #2

4

x

#2 #4

y

60. Reflect the graph of g(x) = |x| across the x-axis and shift it up 2 units.

8 6 4 2

y 1 h(x) ! # x "4 2

#4 #2

4

4

x

56. Shift the graph of g(x) =

1 right 2 units. x

2

#4 #2 #2 #4

y 4 2 2

#2 #2 #4

g(x) ! #|x| " 2

2

#2

4

6

x

1 g(x) ! # ## x# # 2

4

x

98

Chapter 1: Graphs, Functions, and Models

61. Shift the graph of g(x) = x3 right 2 units and reflect it across the x-axis.

66. Reflect the graph of h(x) = x3 across the y-axis. y

y 4 4

g (x) ! #(x #

2)3

h(x) ! (#x)3

2

2 2

#4 #2 2

#4 #2

4

x

4

x

#2

#2

#4

#4

62. Shift the graph of f (x) = x left 1 unit. 3

√

67. Shift the graph of f (x) =

x left 2 units.

y

y

4 4

2

2 2

#4 #2 2

#4 #2

4

x

#2 #4

#4

f(x) ! (x " 1)3

63. Shift the graph of g(x) = x2 left 1 unit and down 1 unit. y

x

f(x) ! √  x "2

√

x right 1 unit. Shrink it 1 vertically by multiplying each y-coordinate by and then 2 reflect it across the x-axis.

68. First shift the graph of f (x) =

y

4 2

4 2

#4 #2

4

#2

4

x

2

1

f(x) ! ##√ x #1 2 

#2 #4

g(x) ! (x "

1)2

#1

2

#2

4

6

x

#2

64. Reflect the graph of h(x) = x2 across the x-axis and down 4 units. y

#4

√ 3

69. Shift the graph of f (x) =

x down 2 units.

y 2

#4 #2 #2

4

x 4

h (x) ! #x 2 # 4

3

f(x) ! √x # 2

2

#4 #6

2

#4 #2

4

x

#2

#8

#4

65. First shrink the graph of g(x) = x3 vertically by multiply1 ing each y-coordinate by . Then shift it up 2 units. 3

70. Shift the graph of h(x) = y

y

4

4

2

2 2

#4 #2 #2 #4

√ 3

4

x

1 3 g(x) ! #x "2 3

2

#4 #2 #2 #4

4

x 3

h(x) ! √ x "1

x left 1 unit.

Exercise Set 1.7 71. Think of the graph of f (x) = |x|. Since g(x) = f (3x), the graph of g(x) = |3x| is the graph of f (x) = |x| shrunk horizontally by dividing each x! 1" . coordinate by 3 or multiplying each x-coordinate by 3 √ 1 72. Think of the graph of g(x) = 3 x. Since f (x) = g(x), the 2 √ 1√ graph of f (x) = 3 x is the graph of g(x) = 3 x shrunk 2 1 vertically by multiplying each y-coordinate by . 2 1 73. Think of the graph of f (x) = . Since h(x) = 2f (x), x 2 1 the graph of h(x) = is the graph of f (x) = stretched x x vertically by multiplying each y-coordinate by 2. 74. Think of the graph of g(x) = |x|. Since f (x) = g(x−3)−4, the graph of f (x) = |x − 3| − 4 is the graph of g(x) = |x| shifted right 3 units and down 4 units. √ 75. Think of the graph of g(x) √ = x. Since f (x) = 3g(x) −√5, the graph of f (x) = 3 x − 5 is the graph of g(x) = x stretched vertically by multiplying each y-coordinate by 3 and then shifted down 5 units. 1 . Since f (x) = 5 − g(x), or x 1 f (x) = −g(x) + 5, the graph of f (x) = 5 − is the graph x 1 of g(x) = reflected across the x-axis and then shifted up x 5 units.

76. Think of the graph of g(x) =

77. Think of the graph of f (x) = |x|. Since g(x) = % % # $ %1 % 1 x − 4, the graph of g(x) = %% x%% − 4 is the graph of f 3 3

f (x) = |x| stretched horizontally by multiplying each xcoordinate by 3 and then shifted down 4 units.

78. Think of the graph of g(x) = x3 . Since 2 2 f (x) = g(x) − 4, the graph of f (x) = x3 − 4 is the 3 3 graph of g(x) = x3 shrunk vertically by multiplying each 2 y-coordinate by and then shifted down 4 units. 3 1 79. Think of the graph of g(x) = x2 . Since f (x) = − g(x − 5), 4 1 the graph of f (x) = − (x − 5)2 is the graph of g(x) = x2 4 shifted right 5 units, shrunk vertically by multiplying each 1 y-coordinate by , and reflected across the x-axis. 4 80. Think of the graph of g(x) = x3 . Since f (x) = g(−x) − 5, the graph of f (x) = (−x)3 − 5 is the graph of g(x) = x3 reflected across the y-axis and shifted down 5 units. 1 . Since f (x) = x 1 + 2 is the graph g(x + 3) + 2, the graph of f (x) = x+3 1 of g(x) = shifted left 3 units and up 2 units. x

81. Think of the graph of g(x) =

99 √ 82. Think of the graph of√f (x) = x. Since g(x) = f (−x) + √5, the graph of g(x) = −x + 5 is the graph of f (x) = x reflected across the y-axis and shifted up 5 units. 83. Think of the graph of f (x) = x2 . Since h(x) = −f (x−3)+ 5, the graph of h(x) = −(x − 3)2 + 5 is the graph of f (x) = x2 shifted right 3 units, reflected across the x-axis, and shifted up 5 units. 84. Think of the graph of g(x) = x2 . Since f (x) = 3g(x + 4)− 3, the graph of f (x) = 3(x+4)2 −3 is the graph of g(x) = x2 shifted left 4 units, stretched vertically by multiplying each y-coordinate by 3, and then shifted down 3 units. 85. The graph of y = g(x) is the graph of y = f (x) shrunk 1 vertically by a factor of . Multiply the y-coordinate by 2 1 : (−12, 2). 2 86. The graph of y = g(x) is the graph of y = f (x) shifted right 2 units. Add 2 to the x-coordinate: (−10, 4). 87. The graph of y = g(x) is the graph of y = f (x) reflected across the y-axis, so we reflect the point across the y-axis: (12, 4). 88. The graph of y = g(x) is the graph of y = f (x) shrunk 1 the horizontally. The x-coordinates of y = g(x) are 4 corresponding x-coordinates of y = f (x), so we divide the ! 1" x-coordinate by 4 or multiply it by : (−3, 4). 4 89. The graph of y = g(x) is the graph of y = f (x) shifted down 2 units. Subtract 2 from the y-coordinate: (−12, 2). 90. The graph of y = g(x) is the graph of y = f (x) stretched horizontally. The x-coordinates of y = g(x) are twice the corresponding x-coordinates of y = f (x), so we multiply ! 1" : (−24, 4). the x-coordinate by 2 or divide it by 2 91. The graph of y = g(x) is the graph of y = f (x) stretched vertically by a factor of 4. Multiply the y-coordinate by 4: (−12, 16). 92. The graph of y = g(x) is the graph y = f (x) reflected across the x-axis. Reflect the point across the x-axis: (−12, −4). 93. g(x) = x2 + 4 is the function f (x) = x2 + 3 shifted up 1 unit, so g(x) = f (x) + 1. Answer B is correct. 94. If we substitute 3x for x in f , we get 9x2 + 3, so g(x) = f (3x). Answer D is correct. 95. If we substitute x − 2 for x in f , we get (x − 2)3 + 3, so g(x) = f (x − 2). Answer A is correct. 96. If we multiply x2 + 3 by 2, we get 2x2 + 6, so g(x) = 2f (x). Answer C is correct. 97. Shape: h(x) = x2 Turn h(x) upside-down (that is, reflect it across the xaxis): g(x) = −h(x) = −x2 Shift g(x) right 8 units: f (x) = g(x − 8) = −(x − 8)2

100

Chapter 1: Graphs, Functions, and Models

98. Shape: h(x) =

√

x

Shift h(x) left 6 units: g(x) = h(x + 6) =

√

Shift g(x) down 5 units: f (x) = g(x) − 5 =

x+6 √ x+6−5

99. Shape: h(x) = |x|

Shift h(x) left 7 units: g(x) = h(x + 7) = |x + 7|

107. Each y-coordinate is multiplied by −2. We plot and connect (−4, 0), (−3, 4), (−1, 4), (2, −6), and (5, 0).

(5, 0) 2

!4 !2

6 x

!4

Turn h(x) upside-down (that is, reflect it across the xaxis): g(x) = −h(x) = −x3 Shift g(x) right 5 units: f (x) = g(x − 5) = −(x − 5)3 1 x

1 ! that is, Shrink h(x) vertically by a factor of 2 " 1 : multiply each function value by 2 1 1 1 1 g(x) = h(x) = · , or 2 2 x 2x 1 Shift g(x) down 3 units: f (x) = g(x) − 3 = −3 2x

!6

(2, !6)

1 108. Each y-coordinate is multiplied by . We plot and connect 2 (−4, 0), (−3, −1), (−1, −1), (2, 1.5), and (5, 0). y 4

(!4, 0)

2 (2, 1.5) 2

!4 !2

(5, 0) 4

x

(!3, !1) (!1, !1) !4

102. Shape: h(x) = x2 Shift h(x) right 6 units: g(x) = h(x − 6) = (x − 6)2

Shift g(x) up 2 units: f (x) = g(x) + 2 = (x − 6)2 + 2 103. Shape: m(x) = x2 Turn m(x) upside-down (that is, reflect it across the xaxis): h(x) = −m(x) = −x2

y

Shift h(x) right 3 units: g(x) = h(x − 3) = −(x − 3)

Shift g(x) up 4 units: f (x) = g(x) + 4 = −(x − 3) + 4 2

g(x) " qf(x)

109. The graph is reflected across the y-axis and stretched horizontally by a factor of 2. That is, each x-coordinate is ! 1" multiplied by −2 or divided by − . We plot and con2 nect (8, 0), (6, −2), (2, −2), (−4, 3), and (−10, 0).

2

y " f !!q x"

4

(!4, 3)

2

(!10, 0)

(8, 0) 2

!10 !8 !6 !4 !2 !2

!

Stretch h(x) horizontally by a factor of 2 that is, multiply # $ % % %1 % 1" 1 each x-value by : g(x) = h x = %% x%% 2 2 2 % % %1 % Shift g(x) down 5 units: f (x) = g(x) − 5 = %% x%% − 5 2 √ 105. Shape: m(x) = x √ Reflect m(x) across the y-axis: h(x) = m(−x) = −x & Shift h(x) left 2 units: g(x) = h(x + 2) = −(x + 2) Shift g(x) down 1 unit: f (x) = g(x) − 1 = & −(x + 2) − 1

1 106. Shape: h(x) = x

4

!2

100. Shape: h(x) = x3

104. Shape: h(x) = |x|

y " !2f(x)

4

(!4, 0)

Shift g(x) up 2 units: f (x) = g(x) + 2 = |x + 7| + 2

101. Shape: h(x) =

y

(!1, 4) (!3, 4)

1 Reflect h(x) across the x-axis: g(x) = −h(x) = − x 1 Shift g(x) up 1 unit: f (x) = g(x) + 1 = − + 1 x

(2, !2)

4

6

8

x

(6, !2)

!4

110. The graph is shrunk horizontally by a factor of 2. That ! 1" is, each x-coordinate is divided by 2 or multiplied by . 2 We plot and connect (−2, 0), (−1.5, −2), (−0.5, −2), (1, 3), and (2.5, 0). y 4

(1, 3)

2

(2.5, 0)

(!2, 0)

2

!4

(1.5, !2)

!4

4

x

(!0.5, !2) g(x) " f(2x)

111. The graph is shifted right 1 unit so each x-coordinate is increased by 1. The graph is also reflected across the xaxis, shrunk vertically by a factor of 2, and shifted up 3 1 units. Thus, each y-coordinate is multiplied by − and 2

Exercise Set 1.7

101

then increased by 3. We plot and connect (−3, 3), (−2, 4), (0, 4), (3, 1.5), and (6, 3).

(!3, 3)

6

(3, 1.5) 4

2

y " !q f(x ! 1) # 3

112. The graph is shifted left 1 unit so each x-coordinate is decreased by 1. The graph is also reflected across the x-axis, stretched vertically by a factor of 3, and shifted down 4 units. Thus, each y-coordinate is multiplied by −3 and then decreased by 4. We plot and connect (−5, −4), (−4, 2), (−2, 2), (1, −13), and (4, −4). y (!2, 2)

4

4 (0, 0)

(!2, 2)

(2, 2)

(5, 2) (7, 0) 2 4 6 8 x 1 (1, !) 2

2

(1, !13)

!2

1 (!1, !) 2 !4

x

(4, !4)

!12

x

y (!7, 3)

(!5, !1)

(!5, !4) (!1, !4)

6

116. The graph is reflected across the y-axis so each x-coordinate is replaced with its opposite. It is also shrunk 1 vertically by a factor of , so each y-coordinate is multi2 1 plied by (or divided by 2). 2

!8 !6 !4 !2

4 2

4

(5, !5)

!6

6 x

!4

(!4, 2)

(!1, 0)

(!7, !3) (!4, !3) !4 (0, !3)

2

!2

(3, 3)

!2

(1, 3) (6, 3)

2

4

!10 !8 !6 !4 !2

(0, 4)

!4 !2

y

(!2, 1) 2 (!3, 0)

(!9, 1)

y (!2, 4)

h(x) " !g(x # 2) # 1

1

h(x) " !g(!x) 2

117. The graph is shrunk horizontally. The x-coordinates of y = h(x) are one-half the corresponding x-coordinates of y = g(x).

g(x) " !3f(x # 1) ! 4

y

113. The graph is reflected across the y-axis so each x-coordinate is replaced by its opposite.

g(x) " f (!x )

4

(!2, 3)

2

(!5, 0)

2 !2

4

6

x

(3, !2) (1, !2)

!4

114. The graph is reflected across the x-axis so each y-coordinate is replaced by its opposite. (!1, 2) (!3, 2) (!4, 0)

y 4

g(x) " !f (x )

2

4

4

6

1

2

1 !2 (!!, 2 1)

(0, 0)

y 10

(!1, 5) (!4, 5)

115. The graph is shifted left 2 units so each x-coordinate is decreased by 2. It is also reflected across the x-axis so each y-coordinate is replaced with its opposite. In addition, the graph is shifted up 1 unit, so each y-coordinate is then increased by 1.

x

118. The graph is shifted right 1 unit, so each x-coordinate is increased by 1. It is also stretched vertically !by a factor of 2, so each y-coordinate is multiplied by 2 or divided 1" . In addition, the graph is shifted down 3 units, so by 2 each y-coordinate is decreased by 3.

x

(2, !3)

4

5 (!, 2 !2)

!4

!2 !4

(!, 2 6) (!, 2 1)

2

7

(!!, 2 0)

(5, 0) 2

!6 !4 !2

(1, 4)

!4 !2

(4, 0)

!6 !4 !2

7

6

5 (!!, 2 4)

y

h(x) " g(2x)

8

(!1, 4)

h(x) " 2g (x ! 1) ! 3 (8, 9)

8 6

(3, 5)

4 2 2

!6 !4 !2

4

(0, !1) !2 (!6, !3)

!4 !6 !8

(1, !3)

6

8

x

(2, !1)

(6, !7)

102

Chapter 1: Graphs, Functions, and Models

119. g(x) = f (−x) + 3 The graph of g(x) is the graph of f (x) reflected across the y-axis and shifted up 3 units. This is graph (f). 120. g(x) = f (x) + 3 The graph of g(x) is the graph of f (x) shifted up 3 units. This is graph (h). 121. g(x) = −f (x) + 3

The graph of g(x) is the graph of f (x) reflected across the x-axis and shifted up 3 units. This is graph (f).

122. g(x) = −f (−x)

The graph of g(x) is the graph of f (x) reflected across the x-axis and the y-axis. This is graph (a).

131. The graph of f (x) = x3 − 3x2 is shifted left 1 unit. A formula for the transformed function is k(x) = f (x + 1), or k(x) = (x + 1)3 − 3(x + 1)2 . 132. The graph of f (x) = x3 − 3x2 is shifted right 2 units and up 1 unit. A formula for the transformed function is t(x) = f (x − 2) + 1, or t(x) = (x − 2)3 − 3(x − 2)2 + 1. 133. The graph of f (x) = 0 is symmetric with respect to the x-axis, the y-axis, and the origin. This function is both even and odd. 134. For every point (x, y) on the graph of y = f (x), its reflection across the y-axis (−x, y) is on the graph of y = f (−x). 135. f (x) = 5x2 − 7

a) f (−3) = 5(−3)2 − 7 = 5 · 9 − 7 = 45 − 7 = 38

1 123. g(x) = f (x − 2) 3 The graph of g(x) ! is the graph of f (x) shrunk vertically by a factor of 3 that is, each y-coordinate is multiplied 1" and then shifted right 2 units. This is graph (d). by 3

b) f (3) = 5 · 32 − 7 = 5 · 9 − 7 = 45 − 7 = 38 c) f (a) = 5a2 − 7

d) f (−a) = 5(−a)2 − 7 = 5a2 − 7 136. f (x) = 4x3 − 5x

a) f (2) = 4 · 23 − 5 · 2 = 4 · 8 − 5 · 2 = 32 − 10 = 22

1 f (x) − 3 3 The graph of g(x) ! is the graph of f (x) shrunk vertically by a factor of 3 that is, each y-coordinate is multiplied 1" and then shifted down 3 units. This is graph (e). by 3

124. g(x) =

1 f (x + 2) 3 The graph of g(x) ! is the graph of f (x) shrunk vertically

125. g(x) =

b) f (−2) = 4(−2)3 − 5(−2) = 4(−8) − 5(−2) = −32 + 10 = −22

c) f (a) = 4a3 − 5a

d) f (−a) = 4(−a)3 − 5(−a) = 4(−a3 ) − 5(−a) = −4a3 + 5a 137. First find the slope of the given line. 8x − y = 10

8x = y + 10

by a factor of 3 that is, each y-coordinate is multiplied 1" and then shifted left 2 units. This is graph (c). by 3

8x − 10 = y

The slope of the given line is 8. The slope of a line perpendicular to this line is the opposite of the reciprocal of 1 8, or − . 8 y − y1 = m(x − x1 ) 1 y − 1 = − [x − (−1)] 8 1 y − 1 = − (x + 1) 8 1 1 y−1 = − x− 8 8 7 1 y = − x+ 8 8

126. g(x) = −f (x + 2)

The graph of g(x) is the graph f (x) reflected across the x-axis and shifted left 2 units. This is graph (b).

127. f (−x) = 2(−x)4 − 35(−x)3 + 3(−x) − 5 = 2x4 + 35x3 − 3x − 5 = g(x) 128. f (−x) =

1 1 (−x)4 + (−x)3 − 81(−x)2 − 17 = 4 5

1 4 1 3 x − x − 81x2 − 17 #= g(x) 4 5

129. The graph of f (x) = x3 − 3x2 is shifted up 2 units. A formula for the transformed function is g(x) = f (x) + 2, or g(x) = x3 − 3x2 + 2. 130. Each y-coordinate of the graph of f (x) = x3 − 3x2 is mul1 tiplied by . A formula for the transformed function is 2 1 1 h(x) = f (x), or h(x) = (x3 − 3x2 ). 2 2

138.

2x − 9y + 1 = 0

2x + 1 = 9y 2 1 x+ = y 9 9 # $ 1 2 Slope: ; y-intercept: 0, 9 9

Exercise Set 1.7

103

139. Each point for which f (x) < 0 is reflected across the x-axis. y

144.

f (x) = f (−x) =

4

(1, 2) (!1, 2) 2 (!4, 0) 2

!4 !2

−f (x) = −

x

!2

Since f (−x) #= −f (x), f is not odd.

|f(x)|

!4

x2 + 1 (−x)2 + 1 = 3 (−x) + 1 −x3 + 1

x2 + 1 x3 + 1 Since f (x) #= f (−x), f is not even.

(4, 2) 4

x2 + 1 x3 + 1

Thus, f (x) = 140. The graph of y = f (|x|) consists of the points of y = f (x) for which x ≥ 0 along with their reflections across the y-axis. y 4

(!1, 2) (!4, 2)

2

(1, 2) 2

!4 !2

(4, 2) 4

x

!2

f(|x|)

!4

141. The graph of y = g(|x|) consists of the points of y = g(x) for which x ≥ 0 along with their reflections across the y-axis. y

(2, 1) 2

!4 !2

(4, 0) x

4

!2

g (|x|)

!4

142. Each point for which g(x) < 0 is reflected across the x-axis.

4

(2, 1) 2

!4 !2

(4, 0) x

4

!2 !4

143.

If the graph were folded on the y-axis, the parts to the left and right of the y-axis would not coincide, so the graph is not symmetric with respect to the y-axis. If the graph were rotated 180◦ , the resulting graph would not coincide with the original graph, so it is not symmetric with respect to the origin. 146. If the graph were folded on the x-axis, the parts above and below the x-axis would not coincide, so the graph is not symmetric with respect to the x-axis. If the graph were folded on the y-axis, the parts to the left and right of the y-axis would not coincide, so the graph is not symmetric with respect to the y-axis.

147. If the graph were folded on the x-axis, the parts above and below the x-axis would coincide, so the graph is symmetric with respect to the x-axis. If the graph were folded on the y-axis, the parts to the left and right of the y-axis would not coincide, so the graph is not symmetric with respect to the y-axis. If the graph were rotated 180◦ , the resulting graph would not coincide with the original graph, so it is not symmetric with respect to the origin.

y

(!2, 1) 2 (!4, 0)

145. If the graph were folded on the x-axis, the parts above and below the x-axis would coincide, so the graph is symmetric with respect to the x-axis.

If the graph were rotated 180◦ , the resulting graph would coincide with the original graph, so it is symmetric with respect to the origin.

4

(!2, 1) 2 (!4, 0)

x2 + 1 is neither even nor odd. x3 + 1

|g(x)|

√ f (x) = x 10 − x2 & √ f (−x) = −x 10 − (−x)2 = −x 10 − x2 √ −f (x) = −x 10 − x2

Since f (−x) = −f (x), f is odd.

148. Call the transformed function g(x). Then

g(5) = 4 − f (−3) = 4 − f (5 − 8), g(8) = 4 − f (0) = 4 − f (8 − 8),

and g(11) = 4 − f (3) = 4 − f (11 − 8).

Thus g(x) = 4 − f (x − 8), or g(x) = 4 − |x − 8|. 149. f (2 − 3) = f (−1) = 5, so b = 5.

(The graph of y = f (x − 3) is the graph of y = f (x) shifted right 3 units, so the point (−1, 5) on y = f (x) is transformed to the point (−1 + 3, 5), or (2, 5) on y = f (x − 3).)

150. Let f (x) = g(x) = x. Now f and g are odd functions, but (f g)(x) = x2 = (f g)(−x). Thus, the product is even, so the statement is false.

104

Chapter 1: Graphs, Functions, and Models For (0, −9):

151. Let f (x) and g(x) be even functions. Then by definition, f (x) = f (−x) and g(x) = g(−x). Thus, (f + g)(x) = f (x) + g(x) = f (−x) + g(−x) = (f + g)(−x) and f + g is even. The statement is true. 152. Let f (x) be an even function, and let g(x) be an odd function. By definition f (x) = f (−x) and g(−x) = −g(x), or g(x) = −g(−x). Then f g(x) = f (x) · g(x) = f (−x) · [−g(−x)] = −f (−x) · g(−x) = −f g(−x), and f g is odd. The statement is true.

2(0) − 9(−9) ?% −18 % 0 + 81 % % 81 % −18 FALSE

(0, −9) is not a solution.

8. For (0, 7):

y=7

7 ? 7 TRUE (0, 7) is a solution.

153. See the answer section in the text. f (−x) − f (x) f (−x) − f (−(−x)) = , 2 2 f (x) − f (−x) f (−x) − f (x) −O(x) = − = . Thus, 2 2 O(−x) = −O(x) and O is odd.

2x − 9y = −18

For (7, 1):

154. O(−x) =

y=7 1 ? 7 FALSE

(7, 1) is not a solution. 9.

155. a), b) See the answer section in the text.

Chapter 1 Review Exercises 1. Because the line passes through the origin, its x-intercept is (0, 0). Thus, the statement is false. 2. The statement is true. See page 77 in the text.

10.

3. First we solve each equation for y. ax + y = c

x − by = d

y = −ax + c

−by = −x + d d 1 y = x− b b If the lines are perpendicular, the product of their slopes is a a 1 −1, so we have −a · = −1, or − = −1, or = 1. The b b b statement is true. 4. The line parallel to the y-axis that passes through (−5, 25) is x = −5, so the statement is false.

11.

1 and x = −5, the x-coordinate of the 2 1 point of intersection is −5 and the y-coordinate is , so 2 the statement is true. & √ 3 − (−3) 6 = , so −3 is in the domain of 6. f (−3) = −3 −3 f (x). Thus, the statement is false. 5. For the lines y =

7. For

#

3,

$ 24 : 9

2x − 9y = −18

24 ? −18 9 %% % 6 − 24 % % −18 % −18 TRUE

12.

= =

2·3−9·

#

3,

24 9

$

is a solution.

d=

13.

&

(x1 − x2 )2 + (y1 − y2 )2

(3 − (−2))2 + (7 − 4)2 √ 52 + 32 = 34 ≈ 5.831 $ # x1 + x2 y1 + y2 , 2 2 $ # 3 + (−2) 7 + 4 , 2 2 $ # 1 11 , 2 2

& √

m= = =

Chapter 1 Review Exercises 14.

15.

(x − h)2 + (y − k)2 = r2 √ [x − (−2)]2 + (y − 6)2 = ( 13)2

105 27. Find the inputs that make the denominator zero: x2 − 6x + 5 = 0

(x + 2)2 + (y − 6)2 = 13

(x − 1)(x − 5) = 0

(x + 1)2 + (y − 3)2 = 16

x = 1 or

x − 1 = 0 or x − 5 = 0

[x − (−1)2 ] + (y − 3)2 = 42

The domain is {x|x #= 1 and x #= 5}, or (−∞, 1) ∪ (1, 5) ∪ (5, ∞).

Center: (−1, 3); radius: 4 16.

(x − 3)2 + (y + 5)2 = 1

28. Find the inputs that make the denominator zero:

(x − 3)2 + [y − (−5)]2 = 12

|16 − x2 | = 0

Center: (3, −5); radius: 1

17. The center is the midopint of the diameter: $ # $ # 4 8 −3 + 7 5 + 3 , = , = (2, 4) 2 2 2 2 Use the center and either endpoint of the diameter to find the radius. We use the point (7, 3). & & 2)2 + (3 − 4)2 = 52 + (−1)2 = r = (7 − √ √ 25 + 1 = 26 √ The equation of the circle is (x−2)2 + (y−4)2 = ( 26)2 , 2 2 or (x − 2) + (y − 4) = 26.

16 − x2 = 0

(4 + x)(4 − x) = 0 4+x = 0

x = −4 or

y

4 3 2 1 -5 -4 -3 -2 -1

2

3

4

5

-3 -4 -5

The inputs on the x axis extend from −4 to 4, so the domain is [−4, 4]. The outputs on the y-axis extend from 0 to 4, so the range is [0, 4]. 30.

21. This is the graph of a function, because there is no vertical line that crosses the graph more than once.

y 10

22. This is not the graph of a function, because we can find a vertical line that crosses the graph more than once.

26. The input 0 results in a denominator of zero. Thus, the domain is {x|x #= 0}, or (−∞, 0) ∪ (0, ∞).

x 1

-1 -2

19. The relation is a function, because no two ordered pairs have the same first coordinate and different second coordinates. The domain is the set of first coordinates: {−2, 0, 1, 2, 7}. The range is the set of second coordinates: {−7, −4, −2, 2, 7}.

25. We can substitute any real number for x. Thus, the domain is the set of all real numbers, or (−∞, ∞).

f (x ) = 16 – x 2

5

Domain: {3, 5, 7}

24. From the graph we see that when the input is 2, the output is −1, so f (2) = −1. When the input is −4, the output is −3, so f (−4) = −3. When the input is 0, the output is −1, so f (0) = −1.

4=x

29.

Range: {1, 3, 5, 7}

23. This is the graph of a function, because there is no vertical line that crosses the graph more than once.

or 4 − x = 0

The domain is {x|x #= −4 and x #= 4}, or (−∞, −4) ∪ (−4, 4) ∪ (4, ∞).

18. The relation is not a function, because the ordered pairs (3, 1) and (3, 5) have the same first coordinate and different second coordinates.

20. This is not the graph of a function, because we can find a vertical line that crosses the graph more than once.

x=5

g (x )

8

= |x – 5|

6 4 2 -10 -8 -6 -4 -2

-2

x 2

4

6

8 10

-4 -6 -8 -10

Each point on the x-axis corresponds to a point on the graph, so the domain is the set of all real numbers, or (−∞, ∞).

The number 0 is the smallest output on the y-axis and every number greater than 0 is also an output, so the range is [0, ∞).

106

Chapter 1: Graphs, Functions, and Models Then f (x + h) − f (x) h x2 + 2xh + h2 − x − h − 3 − (x2 − x − 3) = h x2 + 2xh + h2 − x − h − 3 − x2 + x + 3 = h 2xh + h2 − h = h h(2x + h − 1) = = 2x + h − 1 h

31. We graph y = x3 − 7 in the window [−10, 10, −15, 15]. y 10 8 6

x3 – 7

f (x ) =

4 2 -10 -8 -6 -4 -2

x 2

-2

4

6

8 10

-4 -6 -8 -10

Every point on the x-axis corresponds to a point on the graph, so the domain is the set of all real numbers, or (−∞, ∞). Each point on the y-axis also corresponds to a point on the graph, so the range is the set of all real numbers, or (−∞, ∞). 32. We graph y = x4 + x2 in the window [−5, 5, −5, 5].

34. a) Yes. Each input is 1 more than the one that precedes it. b) No. The change in the output varies. c) No. Constant changes in inputs do not result in constant changes in outputs. 35. a) Yes. Each input is 10 more than the one that precedes it. b) Yes. Each output is 12.4 more than the one that precedes it.

y

c) Yes. Constant changes in inputs result in constant changes in outputs.

5 4 3 2 h (x ) =

1 -5 -4 -3 -2 -1

1

-1

2

x4 + x2

3

4

x

36.

5

-2 -3 -4

y2 − y1 x2 − x1 −6 − (−11) 5 = = 5−2 3

m=

37. m =

-5

Each point on the x-axis corresponds to a point on the graph, so the domain is the set of all real numbers, or (−∞, ∞).

The number 0 is the smallest output on the y-axis and every number greater than 0 is also an output, so the range is [0, ∞). 33. f (x) = x − x − 3 2

a) f (0) = 02 − 0 − 3 = −3

b) f (−3) = (−3) − (−3) − 3 = 9 + 3 − 3 = 9 2

c) f (a − 1) = (a − 1) − (a − 1) − 3 2

= a2 − 2a + 1 − a + 1 − 3 = a2 − 3a − 1

d) First we find f (x + h).

f (x + h) = (x + h)2 − (x + h) − 3

= x + 2xh + h − x − h − 3 2

2

4−4 0 = =0 −3 − 5 −8

0−3 −3 = 1 1 0 − 2 2 The slope is not defined.

38. m =

39.

−2x − y = 7

−y = 2x + 7

y = −2x − 7

Slope: −2; y-intercept: (0, −7) 40. We use the points (1994, 254, 000) and (2004, 629, 000) and find the average rate of change, or slope. 375, 000 629, 000 − 254, 000 = = 37, 500 2004 − 1994 10 The average rate of change over the 10-year period was 37,500 travelers per year. 1 41. Graph y = − x + 3. 4 Plot the y-intercept, (0, 3). We can think of the slope as −1 . Start at (0, 3) and find another point by moving down 4 1 unit and right 4 units. We have the point (4, 2). 1 We could also think of the slope as . Then we can start −4 at (0, 3) and find another point by moving up 1 unit and

Chapter 1 Review Exercises

107

left 4 units. We have the point (−4, 4). Connect the three points and draw the graph.

42.

43.

y = mx + b 2 y = − x−4 3

Substituting −

2 for m and −4 for b 3

y − y1 = m(x − x1 )

y − (−1) = 3(x − (−2)) y + 1 = 3(x + 2) y + 1 = 3x + 6 y = 3x + 5 44. First we find the slope. −2 1 −1 − 1 = = m= −2 − 4 −6 3 Use the point-slope equation: 1 Using (4, 1): y − 1 = (x − 4) 3 1 Using (−2, −1): y − (−1) = (x − (−2)), or 3 1 y + 1 = (x + 2) 3 1 1 In either case, we have y = x − . 3 3 45. 3x − 2y = 8 6x − 4y = 2 3 1 3 y = x− y = x−4 2 2 2 3 The lines have the same slope, , and different 2$ # 1 y-intercepts, (0, −4) and 0, − , so they are parallel. 2 46.

y − 2x = 4

2y − 3x = −7 3 7 y = 2x + 4 y = x− 2 2 3 The lines have different slopes, 2 and , so they are not 2 3 parallel. The product of the slopes, 2 · , or 3, is not −1, so 2 the lines are not perpendicular. Thus the lines are neither parallel nor perpendicular.

2 The slope of a line parallel to the given line is − . 3 We use the point-slope equation. y − y1 = m(x − x1 ) 2 y − (−1) = − (x − 1) 3 2 1 y = − x− 3 3 49. From Exercise 48 we know that the slope of the given line 2 is − . The slope of a line perpendicular to this line is the 3 2 3 negative reciprocal of − , or . 3 2 We use the slope-intercept equation to find the y-intercept. y = mx + b 3 −1 = · 1 + b 2 3 −1 = + b 2 5 − =b 2 3 5 Then the equation of the desired line is y = x − . 2 2 50. Let t = number of months of basic service. C(t) = 25 + 20t C(6) = 25 + 20(6) = $145 51. a) T (d) = 10d + 20 T (5) = 10(5) + 20 = 70◦ C T (20) = 10(20) + 20 = 220◦ C T (1000) = 10(1000) + 20 = 10, 020◦ C b) 5600 km is the maximum depth. Domain: [0, 5600]. 52. Answers will vary depending on the data points used. We will use (3, 524) and (6, 440). −84 440 − 524 = = −28 m= 6−3 3 Use the point-slope equation with (3, 524). y − 524 = −28(x − 3) y = −28x + 608

In 2009, y = −28(14) + 608 = 216 drive-in movie sites.  for x ≤ −4,  −x, 1 53. f (x) =  2 x + 1, for x > −4

We create the graph in two parts. Graph f (x) = −x for 1 inputs less than or equal to −4. Then graph f (x) = x + 1 2 for inputs greater than −4.

3 2 3 47. The slope of y = x + 7 is and the slope of y = − x − 4 2 3 #2 $ 3 2 2 is − . Since − = −1, the lines are perpendicular. 3 2 3 48.

2x + 3y = 4 3y = −2x + 4 4 2 2 y = − x+ ; m=− 3 3 3

y 4 2 2

!4 !2 !2 !4

4

x

108

Chapter 1: Graphs, Functions, and Models

 3 for x < −2,  x , |x|, for −2 ≤ x ≤ 2, 54. f (x) =  √ x − 1, for x > 2

We create the graph in three parts. Graph f (x) = x3 for inputs less than −2. Then graph f (x) = |x| for inputs greater than or equal √to −2 and less than or equal to 2. Finally graph f (x) = x − 1 for inputs greater than 2.

 3  x , |x|, 58. f (x) =  √ x − 1, Since −1 is in the

for x < −2, for −2 ≤ x ≤ 2,

for x > 2 interval [−2, 2], f (−1) = | − 1| = 1. √ √ Since 5 > 2, f (5) = 5 − 1 = 4 = 2.

Since −2 is in the interval [−2, 2], f (−2) = | − 2| = 2. Since −3 < −2, f (−3) = (−3)3 = −27.

59. The length of the rectangle is 2x. The width is the second coordinate of the point (x, y) on the circle. The circle has center (0,√0) and radius 2, so its equation is x2 + y 2 = 4 and y = 4 −√x2 . Thus the area of the rectangle is given by A(x) = 2x 4 − x2 . 60. a) Let h = the height of the box. Since the volume is 108 in3 , we have: 108 = x · x · h

 2  x − 1 , for x #= −1, 55. f (x) = x+1  3, for x = −1

x2 − 1 x+1 for inputs not equal to −1. Then graph f (x) = 3 for x = −1. We create the graph in two parts. Graph f (x) =

56. f (x) = [[x]]. See Example 8 on page 127 of the text.

108 = x2 h 108 =h x2 Now find the surface area. S = x2 + 4 · x · h 108 S(x) = x2 + 4 · x · 2 x 432 2 S(x) = x + x b) x must be positive, so the domain is (0, ∞). c) From the graph, we see that the minimum value of the function occurs when x = 6 in. For this value of x, 108 108 108 = 3 in. h= 2 = 2 = x 6 36 61. (f − g)(6) = f (6) − g(6) √ = 6 − 2 − (62 − 1) √ = 4 − (36 − 1) = 2 − 35

62.

= −33

(f g)(2) = f (2) · g(2) √ = 2 − 2 · (22 − 1) = 0 · (4 − 1) =0

63.

57. f (x) = [[x − 3]]

(f + g)(−1) = f (−1) + g(−1) √ = −1 − 2 + ((−1)2 − 1) √ = −3 + (1 − 1) √ Since −3 is not a real number, (f +g)(−1) does not exist. 4 , g(x) = 3 − 2x x2 a) Division by zero is undefined, so the domain of f is {x|x #= 0}. The domain of g is the set of all real numbers. The domain # $of f + g, f − g and f g is 3 = 0, the domain of f /g is {x|x #= 0}. Since g 2 + % , % 3 x%%x #= 0 and x #= . 2

64. f (x) =

Chapter 1 Review Exercises

109

4 4 + (3 − 2x) = 2 + 3 − 2x x2 x 4 4 (f − g)(x) = 2 − (3 − 2x) = 2 − 3 + 2x x x # $ 8 12 4 (3 − 2x) = 2 − (f g)(x) = x2 x x 4 2 4 x = 2 (f /g)(x) = 3 − 2x x (3 − 2x)

b) (f + g)(x) =

65. a) The domain of f , g, # f$+ g, f − g, and f g is all real 1 numbers. Since g = 0, the domain of f /g is 2 , + % % 1 . x%%x #= 2

4 b) f ◦ g(x) = f (3 − 2x) = (3 − 2x)2 # $ # $ 4 4 8 g ◦ f (x) = g 2 = 3 − 2 2 = 3 − 2 x x x 69. a) Domain of f = domain of g = all real numbers, so domain of f ◦g = domain of g ◦f = all real numbers. b) f ◦ g(x) = f (2x − 1)

= 3(2x − 1)2 + 4(2x − 1)

= 3(4x2 − 4x + 1) + 4(2x − 1) = 12x2 − 12x + 3 + 8x − 4 = 12x2 − 4x − 1

(g ◦ f )(x) = g(3x2 + 4x)

= 2(3x2 + 4x) − 1

b) (f + g)(x) = (3x2 + 4x) + (2x − 1) = 3x2 + 6x − 1

(f −g)(x) = (3x2 +4x)−(2x−1) = 3x2+4x−2x+1 =

x, g(x) = 5x + 2. Answers may vary.

3x2 + 2x + 1

70. f (x) =

(f g)(x) = (3x2 +4x)(2x−1) = 6x3 −3x2 +8x2 −4x =

71. f (x) = 4x2 + 9, g(x) = 5x − 1. Answers may vary.

6x + 5x − 4x 3

66.

√

= 6x2 + 8x − 1

2

72. x2 + y 2 = 4

3x2 + 4x (f /g)(x) = 2x − 1

y

P (x) = R(x) − C(x)

5 4

= (120x − 0.5x2 ) − (15x + 6)

2 1

= −0.5x2 + 105x − 6 67. f (x) = 3 − x2

x2 + y2 = 4

3

= 120x − 0.5x2 − 15x − 6

-5 -4 -3 -2 -1

-1

x 1

2

3

4

5

-2

f (x + h) = 3 − (x + h) = 3 − (x + 2xh + h ) = 2

2

2

3 − x2 − 2xh − h2 f (x + h) − f (x) h 3 − x2 − 2xh − h2 − (3 − x2 ) = h 3 − x2 − 2xh − h2 − 3 + x2 = h −2xh − h2 = h h(−2x − h) = h = −2x − h

68. a) The domain of f is {x|x #= 0} and the domain of g is the set of all real numbers. To find the domain of f ◦ g, we find the values of x for which g(x) = 0. 3 Since 3 − 2x = 0 when x = , the domain of f ◦ g is 2 , + % % 3 . Since any real number can be an input x%%x #= 2 for g, the domain of g ◦ f is the same as the domain of f , {x|x #= 0}.

-3 -4 -5

The graph is symmetric with respect to the x-axis, the y-axis, and the origin. Replace y with −y to test algebraically for symmetry with respect to the x-axis. x2 + (−y)2 = 4 x2 + y 2 = 4 The resulting equation is equivalent to the original equation, so the graph is symmetric with respect to the x-axis. Replace x with −x to test algebraically for symmetry with respect to the y-axis. (−x)2 + y 2 = 4 x2 + y 2 = 4 The resulting equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Replace x and −x and y with −y to test for symmetry with respect to the origin. (−x)2 + (−y)2 = 4 x2 + y 2 = 4 The resulting equation is equivalent to the original equation, so the graph is symmetric with respect to the origin.

110

Chapter 1: Graphs, Functions, and Models Replace x with −x to test algebraically for symmetry with respect to the y-axis.

73. y 2 = x2 + 3 y

−x + y = 3

5

The resulting equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis.

4 3 2

y2 = x2 + 3

1 -5 -4 -3 -2 -1

x 1

-1

2

3

4

5

-2

Replace x and −x and y with −y to test for symmetry with respect to the origin. −x − y = 3

-3

x + y = −3

-4 -5

The graph is symmetric with respect to the x-axis, the y-axis, and the origin. Replace y with −y to test algebraically for symmetry with respect to the x-axis.

The resulting equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. 75. y = x2 y

(−y)2 = x2 + 3

5

y 2 = x2 + 3

4 3

The resulting equation is equivalent to the original equation, so the graph is symmetric with respect to the x-axis. Replace x with −x to test algebraically for symmetry with respect to the y-axis.

1 -5 -4 -3 -2 -1

2

3

4

5

-4

y = x +3

-5

The resulting equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Replace x and −x and y with −y to test for symmetry with respect to the origin. (−y)2 = (−x)2 + 3 y 2 = x2 + 3 The resulting equation is equivalent to the original equation, so the graph is symmetric with respect to the origin. 74. x + y = 3

Replace y with −y to test algebraically for symmetry with respect to the x-axis. −y = x2

y = −x2

The resulting equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis.

y = (−x)2

4

x+y=3

3

y = x2

2 1 -1

The graph is symmetric with respect to the y-axis. It is not symmetric with respect to the x-axis or the origin.

Replace x with −x to test algebraically for symmetry with respect to the y-axis.

y 5

x 1

2

3

4

5

-2 -3 -4 -5

The graph is not symmetric with respect to the x-axis, the y-axis, or the origin. Replace y with −y to test algebraically for symmetry with respect to the x-axis. x−y =3

x 1

-3

2

-5 -4 -3 -2 -1

-1 -2

y 2 = (−x)2 + 3 2

y = x2

2

The resulting equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis.

The resulting equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Replace x and −x and y with −y to test for symmetry with respect to the origin. −y = (−x)2

−y = x2

y = −x2

The resulting equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin.

Chapter 1 Review Exercises

111

76. y = x3

The resulting equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis.

y 10 8 6 4 2 !5 !4 !3 !2

!2

Replace x with −x to test algebraically for symmetry with respect to the y-axis.

!4 !6 !8 !10

y = (−x)4 − (−x)2

x

1 2 3 4 5

y = x4 − x2

y ! x3

The resulting equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Replace x and −x and y with −y to test for symmetry with respect to the origin.

The graph is symmetric with respect to the origin. It is not symmetric with respect to the x-axis or the y-axis.

−y = (−x)4 − (−x)2

Replace y with −y to test algebraically for symmetry with respect to the x-axis.

−y = x4 − x2

y = −x4 + x2

−y = x3

The resulting equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin.

y = −x3

The resulting equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with −x to test algebraically for symmetry with respect to the y-axis. y = (−x)3 y = −x3

The resulting equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Replace x and −x and y with −y to test for symmetry with respect to the origin. −y = (−x)3

78. The graph is symmetric with respect to the y-axis, so the function is even. 79. The graph is symmetric with respect to the y-axis, so the function is even. 80. The graph is symmetric with respect to the origin, so the function is odd. 81. The graph is symmetric with respect to the y-axis, so the function is even. 82.

f (−x) = 9 − (−x2 ) = 9 − x2

−y = −x

3

f (x) = f (−x), so f is even.

y = x3

The resulting equation is equivalent to the original equation, so the graph is symmetric with respect to the origin.

83.

f (x) = x3 − 2x + 4

f (−x) = (−x)3 − 2(−x) + 4 = −x3 + 2x + 4

f (x) #= f (−x), so f is not even.

77. y = x4 − x2

−f (x) = −(x3 − 2x + 4) = −x3 + 2x − 4 f (−x) #= −f (x), so f is not odd.

y 5

Thus, f (x) = x3 − 2x + 4 is neither even or odd.

4 3 2

y=

x4 – x2

2

3

1 -5 -4 -3 -2 -1

f (x) = 9 − x2

-1

84.

f (−x) = (−x)7 − (−x)5 = −x7 + x5

x 1

4

f (x) #= f (−x), so f is not even.

5

−f (x) = −(x7 − x5 ) = −x7 + x5

-2 -3 -4 -5

85.

−y = x4 − x2

y = −x4 + x2

f (−x) = −f (x), so f is odd. f (x) = |x|

f (−x) = | − x| = |x|

The graph is symmetric with respect to the y-axis. It is not symmetric with respect to the x-axis or the origin. Replace y with −y to test algebraically for symmetry with respect to the x-axis.

f (x) = x7 − x5

86.

f (x) = f (−x), so f is even. √ f (x) = 16 − x2 & √ f (−x) = 16 − (−x2 ) = 16 − x2 f (x) = f (−x), so f is even.

112 87.

Chapter 1: Graphs, Functions, and Models 10x x2 + 1 10x 10(−x) =− 2 f (−x) = (−x)2 + 1 x +1 f (x) #= f (−x), so f (x) is not even. 10x −f (x) = − 2 x +1 f (−x) = −f (x), so f is odd. f (x) =

94. Each y-coordinate is increased by 3. We plot and connect (−5, 6), (−3, 3), (0, 4) and (4, 1).

88. Shape: g(x) = x2

Shift g(x) left 3 units: f (x) = g(x + 3) = (x + 3)2 √ 89. Shape: t(x) = x Turn t(x) upside down √ (that is, reflect it across the x-axis): h(x) = −t(x) = − x. √ Shift h(x) right 3 units: g(x) = h(x − 3) = − x − 3. √ Shift g(x) up 4 units: f (x) = g(x) + 4 = − x − 3 + 4.

90. Shape: h(x) = |x|

Stretch h(x) vertically by a factor of 2 (that is, multiply each function value by 2): g(x) = 2h(x) = 2|x|.

Shift g(x) right 3 units: f (x) = g(x − 3) = 2|x − 3|.

91. The graph is shifted right 1 unit so each x-coordinate is increased by 1. We plot and connect (−4, 3), (−2, 0), (1, 1) and (5, −2).

x+3 8 − 4x When x = 2, the denominator is 0, so 2 is not in the domain of the function. Thus, the domain is (−∞, 2) ∪ (2, ∞) and answer B is correct.

95. f (x) =

96.

(x − 1)2 + y 2 = 9

(x − 1)2 + (y − 0)2 = 32

The center is (1, 0), so answer B is correct. 97. a) Enter the data in a graphing calculator and use the linear regression option. We get y = −21.9x + 588.3.

In 2009, x = 2009 − 1995 = 14. Then we have y = −21.9(14) + 588.3 ≈ 282.

b) The correlation coefficient is approximately −0.9903. This indicates that the line fits the data well. 98. f (x) = 4x3 − 2x + 7

a) f (x) + 2 = 4x3 − 2x + 7 + 2 = 4x3 − 2x + 9

92. The graph is shrunk horizontally by a factor of 2. That is, $ # $is divided by 2. We plot and connect # each x-coordinate 3 5 − , 3 , − , 0 , (0, 1) and (2, −2). 2 2

b) f (x + 2) = 4(x + 2)3 − 2(x + 2) + 7

= 4(x3 + 6x2 + 12x + 8) − 2(x + 2) + 7 = 4x3 + 24x2 + 48x + 32 − 2x − 4 + 7

= 4x3 + 24x2 + 46x + 35

c) f (x) + f (2) = 4x3 − 2x + 7 + 4 · 23 − 2 · 2 + 7 = 4x3 − 2x + 7 + 32 − 4 + 7 = 4x3 − 2x + 42

f (x) + 2 adds 2 to each function value; f (x + 2) adds 2 to each input before the function value is found; f (x) + f (2) adds the output for 2 to the output for x.

93. Each y-coordinate is multiplied by −2. We plot and connect (−5, −6), (−3, 0), (0, −2) and (4, 4).

99. In the graph of y = f (cx), the constant c stretches or shrinks the graph of y = f (x) horizontally. The constant c in y = cf (x) stretches or shrinks the graph of y = f (x) vertically. For y = f (cx), the x-coordinates of y = f (x) are divided by c; for y = cf (x), the y-coordinates of y = f (x) are multiplied by c. 100. a) To draw the graph of y2 from the graph of y1 , reflect across the x-axis the portions of the graph for which the y-coordinates are negative. b) To draw the graph of y2 from the graph of y1 , draw the portion of the graph y1 to the right of the y-axis; then draw its reflection across the y-axis.

Chapter 1 Test

113

√

1−x x − |x| We cannot find the square root of a negative number, so x ≤ 1. Division by zero is undefined, so x < 0.

101. f (x) =

7. f (x) = 2x2 − x + 5

a) f (−1) = 2(−1)2 − (−1) + 5 = 2 + 1 + 5 = 8

b) f (a + 2) = 2(a + 2)2 − (a + 2) + 5

= 2(a2 + 4a + 4) − (a + 2) + 5

Domain of f is {x|x < 0}, or (−∞, 0). 102. f (x) = (x − 9x−1 )−1 =

= 2a2 + 8a + 8 − a − 2 + 5

1

9 x− x Division by zero is undefined, so x #= 0. Also, note that we x , so x #= −3, 0, 3. can write the function as f (x) = 2 x −9 Domain of f is {x|x #= −3 and x #= 0 and x #= 3}, or (−∞, −3) ∪ (−3, 0) ∪ (0, 3) ∪ (3, ∞).

= 2a2 + 7a + 11

8. a)

103. Let f (x) and g(x) be odd functions. Then by definition, f (−x) = −f (x), or f (x) = −f (−x), and g(−x) = −g(x), or g(x) = −g(−x). Thus (f + g)(x) = f (x) + g(x) = −f (−x) + [−g(−x)] = −[f (−x) + g(−x)] = −(f + g)(−x) and f + g is odd.

b) Each point on the x-axis corresponds to a point on the graph, so the domain is the set of all real numbers, or (−∞, ∞). c) The number 3 is the smallest output on the y-axis and every number greater than 3 is also an output, so the range is [3, ∞).

104. Reflect the graph of y = f (x) across the x-axis and then across the y-axis.

Chapter 1 Test

9. The input 4 results in a denominator of 0. Thus the domain is {x|x #= 4}, or (−∞, 4) ∪ (4, ∞).

10. We can substitute any real number for x. Thus the domain is the set of all real numbers, or (−∞, ∞).

1.

11. We cannot find the square root of a negative number. Thus 25 − x2 ≥ 0 and the domain is {x| − 5 ≤ x ≤ 5}, or [−5, 5]. 12. a) This is not the graph of a function, because we can find a vertical line that crosses the graph more than once. b) This is the graph of a function, because there is no vertical line that crosses the graph more than once.

& √ √ 2. d = (5 − (−1))2 + (8 − 5)2 = 62 + 32 = 36 + 9 = √ 45 ≈ 6.708 $ # $ # $ # −6 9 9 −2 + (−4) 6 + 3 , , = = − 3, 3. m = 2 2 2 2 2 √ 2 2 2 4. [x − (−1)] + (y − 2) = ( 5) 5.

(x + 1)2 + (y − 2)2 = 5

(x + 4)2 + (y − 5)2 = 36

[x − (−4)]2 + (y − 5)2 = 62

Center: (−4, 5); radius: 6

6. a) The relation is a function, because no two ordered pairs have the same first coordinate and different second coordinates. b) The domain is the set of first coordinates: {−4, 0, 1, 3}.

c) The range is the set of second coordinates: {0, 5, 7}.

13 2 3 = 3 13. m = −2 − (−2) 0 The slope is not defined. 5−

22 11 12 − (−10) = =− −8 − 4 −12 6 0 6−6 = =0 15. m = 3 23 − (−5) 4 4 14. m =

16. We have the points (1995, 5.3) and (2004, 6.8). 1.5 6.8 − 5.3 = ≈ 0.167 m= 2004 − 1995 9 The average rate of change in weekend attendance from 1995 to 2004 was about 0.167 million per year, or about 167,000 per year. 17.

−3x + 2y = 5

2y = 3x + 5 5 3 y = x+ 2 2 # $ 5 3 Slope: ; y-intercept: 0, 2 2

114 18.

Chapter 1: Graphs, Functions, and Models y = mx + b 5 y = − x−5 8

19. First we find the slope: −6 3 −2 − 4 = =− m= 3 − (−5) 8 4 Use the point-slope equation. 3 Using (−5, 4): y − 4 = − (x − (−5)), or 4 3 y − 4 = − (x + 5) 4 3 Using (3, −2): y − (−2) = − (x − 3), or 4 3 y + 2 = − (x − 3) 4 3 1 In either case, we have y = − x + . 4 4 20. First find the slope of the given line. x + 2y = −6

2y = −x − 6 1 1 y = − x − 3; m = − 2 2

1 A line parallel to the given line has slope − . We use the 2 point-slope equation. 1 y − 3 = − (x − (−1)) 2 1 y − 3 = − (x + 1) 2 1 1 y−3 = − x− 2 2 5 1 y = − x+ 2 2 21.

2x + 3y = −12 2y − 3x = 8 2 3 y = − x−4 y = x+4 3 2 2 3 m1 = − , m2 = ; m1 m2 = −1. 3 2 The lines are perpendicular.

22. Answers will vary depending on the data points used. We will use (4, 5.06) and (24, 12.03). 6.97 12.03 − 5.06 = = 0.3485 m= 24 − 4 20 Now we use the point-slope equation with (4, 5.06). y − 5.06 = 0.3485(x − 4)

y − 5.06 = 0.3485x − 1.394 y = 0.3485x + 3.666

In 2010, x = 2010 − 1972 = 38. Then we have y = 0.3485(38) + 3.666 ≈ $16.91.

 2 for x < −1,  x , for −1 ≤ x ≤ 1, 23. f (x) = |x|,  √ x − 1, for x > 1

% $ % # % 7% 7 7 7 = %% − %% = . ≤ 1, f − 8 8 8 8 √ √ Since 5 > 1, f (5) = 5 − 1 = 4 = 2.

24. Since −1 ≤ −

Since −4 < −1, f (−4) = (−4)2 = 16.

25. (f − g)(−1) = f (−1) − g(−1) = & (−1)2 − 4(−1) + 3 − 3 − (−1) = √ 1+4+3− 4=1+4+3−2=6

26. a) We can substitute any real number for x, so the domain of f is the set of all real numbers, or (−∞, ∞). b) We must have x − 3 ≥ 0, or x ≥ 3, so the domain of g is [3, ∞). √ c) (f − g)(x) = f (x) − g(x) = x2 − x − 3 √ d) (f g)(x) = f (x) · g(x) = x2 x − 3 e) g(3) = 0, so the domain of (f /g)(x) is (3, ∞). (See parts (a) and (b).)

27. Answers may vary. f (x) = x4 , g(x) = 2x − 7 28. f (x) = x2 + 4

f (x + h) = (x + h)2 + 4 = x2 + 2xh + h2 + 4 f (x + h) − f (x) h x2 + 2xh + h2 + 4 − (x2 + 4) = h x2 + 2xh + h2 + 4 − x2 − 4 = h 2xh + h2 = h h(2x + h) = 2x + h = h √ 29. (f ◦ g)(x) = f (g(x)) = f (x2 + 1) = x2 + 1 − 5 = √ x2 − 4 √ √ (g ◦ f )(x) = g(f (x)) = g( x − 5) = ( x − 5)2 + 1 = x−5+1=x−4

30. The inputs for f (x) must be such that x − 5 ≥ 0, or x ≥ 5. Then for (f ◦g)(x) we must have g(x) ≥ 5, or x2 +1 ≥ 5, or x2 ≥ 4. Then the domain of (f ◦g)(x) is (−∞, −2]∪[2, ∞).

Since we can substitute any real number for x in g, the domain of (g ◦ f )(x) is the same as the domain of f (x), [5, ∞).

Chapter 1 Test 31. y = x4 − 2x2

Replace y with −y to test for symmetry with respect to the x-axis. −y = x4 − 2x2 y = −x4 + 2x2

The resulting equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Replace x with −x to test for symmetry with respect to the y-axis. y = (−x)4 − 2(−x)2 y = x4 − 2x2

The resulting equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Replace x with −x and y with −y to test for symmetry with respect to the origin. −y = (−x)4 − 2(−x)2 −y = x4 − 2x2

y = −x4 + 2x2

The resulting equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. 2x x2 + 1 2x 2(−x) =− 2 f (−x) = (−x)2 + 1 x +1 f (x) #= f (−x), so f is not even. 2x −f (x) = − 2 x +1 f (−x) = −f (x), so f is odd.

32. f (x) =

33. Shape: h(x) = x2

Shift h(x) right 2 units: g(x) = h(x − 2) = (x − 2)2 Shift g(x) down 1 unit: f (x) = (x − 2)2 − 1

34. Shape: h(x) = x2

Shift h(x) left 2 units: g(x) = h(x + 2) = (x + 2)2 Shift g(x) down 3 units: f (x) = (x + 2)2 − 3

1 35. Each y-coordinate is multiplied by − . We plot and con2 nect (−5, 1), (−3, −2), (1, 2) and (4, −1).

36. Each x-coordinate on the graph of y = f (x) is $ by # divided −3 , 1 , or 3 on the graph of y = f (3x). Thus the point 3 (−1, 1) is on the graph of f (3x).

115

Chapter 2

Functions, Equations, and Inequalities 9.

Exercise Set 2.1 1.

1 = y−7

x=4

Dividing by 4 on both sides

10.

2y − 1 = 3

The solution is

y=2 The solution is 2.

11.

The solution is 3 − x = 12 −x = 9

Dividing by 4 on both sides 12.

3x = 9

The solution is 3. 16 . 3

13.

4 − x = −5 −x = −9

x=6

14.

The solution is −5.

−3 = 3x −1 = x

The solution is −1.

Adding 3 on both sides 16.

5x + 1 = 9x − 7 8 = 4x 2=x The solution is 2.

9 = 4x − 8

The solution is

4x − 5 = 7x − 2 −5 = 3x − 2

8 = 5x − 3

17 = 4x 17 =x 4

4x + 3 = 2x − 7 x = −5

The solution is 9.

8.

Adding 5 on both sides

2x = −10

15.

11 . 5

Subtracting 2x on both sides

The solution is 6.

x=9

Dividing by 5 on both sides

3x − 5 = 2x + 1 x−5 = 1

The solution is −9.

The solution is

5x − 4 = 2x + 5

x=3

Multiplying (or dividing) by −1 on both sides

11 = 5x 11 =x 5

Subtracting 7 on both sides

3x − 4 = 5

Subtracting 3 on both sides

x = −9

Subtracting x on both sides

The solution is −4.

3x − 16 = 0

3x = 16 16 x= 3

7.

2x + 7 = x + 3 x = −4

Subtracting 3 on both sides

3 The solution is − . 4

6.

18 . 5

x+7 = 3

4x + 3 = 0 4x = −3 3 x=− 4

5.

5 − 4x = x − 13 18 = 5x 18 =x 5

2y = 4

4.

Adding 7 on both sides

The solution is 8.

The solution is 4.

3.

Subtracting y on both sides

8=y

4x + 5 = 21 4x = 16 Subtracting 5 on both sides

2.

y + 1 = 2y − 7

17 . 4

Subtracting 4x on both sides Adding 2 on both sides Dividing by 3 on both sides

118 17.

Chapter 2: Functions, Equations, and Inequalities 22.

5x − 2 + 3x = 2x + 6 − 4x

12x + 8 − 7 = 3x − 6

8x − 2 = 6 − 2x Collecting like terms

8x + 2x = 6 + 2 10x = 8 8 x= 10 4 5 4 The solution is . 5 18.

12x + 1 = 3x − 6

Adding 2x and 2 on both sides Collecting like terms

9x + 1 = −6

9x = −7 7 x=− 9 7 The solution is − . 9

Dividing by 10 on both sides

x=

Simplifying 23.

12x = 6 1 x= 2

−2x = 16

x = −8

The solution is −8.

The solution is

7(3x + 6) = 11 − (x + 2) 21x + 42 = 11 − x − 2 21x + 42 = 9 − x

21x + x = 9 − 42 22x = −33 33 x=− 22 3 2

x=−

Using the distributive property Collecting like terms Adding x and subtracting 42 on both sides Collecting like terms Dividing by 22 on both sides Simplifying

3 The solution is − . 2 20.

14y = −27 27 y=− 14 27 The solution is − . 14

3(x + 1) = 5 − 2(3x + 4) 3x + 3 = 5 − 6x − 8 3x + 3 = −6x − 3 9x + 3 = −3

9x = −6 2 x=− 3

2 The solution is − . 3

24.

Adding 10x and 8 on both sides Dividing by 12 on both sides

1 . 2

3(2x − 5) + 4 = 2(4x + 3) 6x − 15 + 4 = 8x + 6 6x − 11 = 8x + 6

−2x = 17 17 x=− 2 17 The solution is − . 2

25. Familiarize. Let w = the wholesale sales of bottled water in 2001, in billions of dollars. An increase of 45% over this amount is 45% · w, or 0.45w. Translate. Wholesale sales in 2001 "# $ !  & w

4(5y + 3) = 3(2y − 5) 20y + 12 = 6y − 15

21.

2(x − 4) = 3 − 5(2x + 1)

2x − 8 = 3 − 10x − 5 Using the distributive property 2x − 8 = −10x − 2 Collecting like terms

5x − 17 − 2x = 6x − 1 − x 3x − 17 = 5x − 1

19.

4(3x + 2) − 7 = 3(x − 2)

increase in sales  ! "#  $ & & + 0.45w

plus

Wholesale sales in 2005. "# $ !  & & = 10 is

Carry out. We solve the equation. w + 0.45w = 10 Removing parentheses Collecting like terms Adding 6x Subtracting 3 Dividing by 9

1.45w = 10 10 w= 1.45 w ≈ 6.9

Check. 45% of 6.9 is 0.45(6.9) or about 3.1, and 6.9+3.1 = 10. The answer checks. State. Wholesale sales of bottled water in 2001 were about $6.9 billion. 26. Let d = the daily global demand for oil in 2005, in millions of barrels. Solve: d + 0.23d = 103 d ≈ 84 million barrels per day

Exercise Set 2.1

119

27. Familiarize. Let d = the average credit card debt per household in 1990. Translate. debt$ is !debt in 2004.$ Debt in 1990$ plus !additional "# "# "# !      & & & & & d + 6346 = 9312

Carry out. We solve the equation. d + 6346 = 9312 d = 2966

Subtracting 6346

Check. $2966 + $6346 = $9312, so the answer checks. State. The average credit card debt in 1990 was $2966 per household. 28. Let n = the number of nesting pairs of bald eagles in the lower 48 states in 1963. Solve: n + 6649 = 7066

32. Let m = the number of calories in a Big Mac. Solve: m + (m + 20) = 1200 m = 590, so a Big Mac has 590 calories and an order of Super-Size fries has 590 + 20, or 610 calories. 33. Familiarize. Let v = the number of ABC viewers, in millions. Then v + 1.7 = the number of CBS viewers and v − 1.7 = the number of NBC viewers. Translate. ABC CBS NBC total plus plus is viewers viewers viewers viewers. ! "#  $  ! "#  $  ! "#  $  ! "#  $ & & & & & & & v + (v + 1.7) + (v − 1.7) = 29.1

Carry out.

v + (v + 1.7) + (v − 1.7) = 29.1 3v = 29.1

n = 417 pairs of bald eagles

v = 9.7

29. Familiarize. Let d = the number of gigabytes of digital data stored in a typical household in 2004. Translate. Data stored in 2004 "# $ !  & d

additional data "# $  !  & & + 4024

plus

Carry out. We solve the equation.

data stored in 2010. "# $ !  & & = 4430 is

d + 4024 = 4430 d = 406

Subtracting 4024

Check. 406 + 4024 = 4430, so the answer checks. State. In 2004, 406 GB of digital data were stored in a typical household. 30. Let s = the amount of a student’s expenditure for books that goes to the college store. Solve: s = 0.232(501) s ≈ $116.23

31. Familiarize. Let c = the average daily calorie requirement for many adults.

Then v+1.7 = 9.7+1.7 = 11.4 and v−1.7 = 9.7−1.7 = 8.0. Check. 9.7 + 11.4 + 8.0 = 29.1, so the answer checks. State. ABC had 9.7 million viewers, CBS had 11.4 million viewers, and NBC had 8.0 million viewers. 34. Let h = the number of households represented by the rating. Solve: h = 11.0(1, 102, 000) h = 12, 122, 000 households 35. Familiarize. Let P = the amount Tamisha borrowed. We will use the formula I = P rt to find the interest owed. For r = 5%, or 0.05, and t = 1, we have I = P (0.05)(1), or 0.05P . Translate. Amount borrowed$ plus interest is $1365. "# !      & & & & & P + 0.05P = 1365 Carry out. We solve the equation. P + 0.05P = 1365 1.05P = 1365 P = 1300

Translate. 3 the average daily requirement of 1560 calories$ is "# ! for many adults. 4 "# $     !  & & & & & 3 · c 1560 = 4 Carry out. We solve the equation. 3 1560 = · c 4 4 4 · 1560 = c Multiplying by 3 3 2080 = c 3 Check. · 2080 = 1560, so the answer checks. 4 State. The average daily calorie requirement for many adults is 2080 calories.

Adding Dividing by 1.05

Check. The interest due on a loan of $1300 for 1 year at a rate of 5% is $1300(0.05)(1), or $65, and $1300 + $65 = $1365. The answer checks. State. Tamisha borrowed $1300. 36. Let P = the amount invested. Solve: P + 0.04P = $1560 P = $1500 37. Familiarize. Let s = Ryan’s sales for the month. Then his commission is 8% of s, or 0.08s. Translate. Base salary plus commission is total pay. "# $  !    ! "#  $ & & & & & 1500 + 0.08s = 2284

120

Chapter 2: Functions, Equations, and Inequalities Carry out. We solve the equation. 1500 + 0.08s = 2284 0.08s = 784

Subtracting 1500

x + 5x + x − 2 = 180 7x − 2 = 180

7x = 182

s = 9800 Check. 8% of $9800, or 0.08($9800), is $784 and $1500 + $784 = $2284. The answer checks. State. Ryan’s sales for the month were $9800. 38. Let s = the amount of sales for which the two choices will be equal. Solve: 1800 = 1600 + 0.04s s = $5000 39. Familiarize. Let d = the number of miles Diego traveled in the cab.

x = 26 If x = 26, then 5x = 5 · 26, or 130, and x − 2 = 26 − 2, or 24. Check. The measure of angle B, 130◦ , is five times the measure of angle A, 26◦ . The measure of angle C, 24◦ , is 2◦ less than the measure of angle A, 26◦ . The sum of the angle measures is 26◦ + 130◦ + 24◦ , or 180◦ . The answer checks. State. The measure of angles A, B, and C are 26◦ , 130◦ , and 24◦ , respectively. 42. Let x = the measure of angle A.

Translate. Pickup fee ! "# $  & 1.75

Carry out. We solve the equation.

cost number per times of miles is $19.75. mile traveled  ! "#  $  ! "#  $  & & & & & & + 1.50 · d = 19.75

plus

Carry out. We solve the equation. 1.75 + 1.50 · d = 19.75 1.5d = 18 d = 12

Subtracting 1.75 Dividing by 1.5

Check. If Diego travels 12 mi, his fare is $1.75 + $1.50 · 12, or $1.75 + $18, or $19.75. The answer checks. State. Diego traveled 12 mi in the cab. 40. Let w = Soledad’s regular hourly wage. She worked 48 − 40, or 8 hr, of overtime.

Solve: x + 2x + x + 20 = 180 x = 40◦ , so the measure of angle A is 40◦ ; the measure of angle B is 2 · 40◦ , or 80◦ ; and the measure of angle C is 40◦ + 20◦ , or 60◦ . 43. Familiarize. Using the labels on the drawing in the text, we let w = the width of the test plot and w + 25 = the length, in meters. Recall that for a rectangle, Perimeter = 2 · length + 2 · width. Translate. + 2 · width Perimeter ! "# $ = 2! · length "# $ ! "# $ 322 = 2(w + 25) + 2 · w

Carry out. We solve the equation. 322 = 2(w + 25) + 2 · w 322 = 2w + 50 + 2w

Solve: 40w + 8(1.5w) = 442

322 = 4w + 50

w = $8.50

272 = 4w

41. Familiarize. We make a drawing.

! ! ! ! !x A

B "" !5x

68 = w When w = 68, then w + 25 = 68 + 25 = 93. Check. The length is 25 m more than the width: 93 = 68 + 25. The perimeter is 2 · 93 + 2 · 68, or 186 + 136, or 322 m. The answer checks.

"

"

"

" x − 2"" C

We let x = the measure of angle A. Then 5x = the measure of angle B, and x − 2 = the measure of angle C. The sum of the angle measures is 180◦ . Translate. Measure Measure Measure of of of + + = !"#$ 180. angle B angle C angle A ! "# $ ! "# $ ! "# $ x + 5x + x−2 = 180

State. The length is 93 m; the width is 68 m. 44. Let w = the width of the garden. Solve: 2 · 2w + 2 · w = 39

w = 6.5, so the width is 6.5 m, and the length is 2(6.5), or 13 m. 45. Familiarize. Let l = the length of the soccer field and l − 35 = the width, in yards.

Translate. We use the formula for the perimeter of a rectangle. We substitute 330 for P and l − 35 for w. P = 2l + 2w

330 = 2l + 2(l − 35)

Exercise Set 2.1

121

Carry out. We solve the equation.

d

330 = 2l + 2(l − 35) 330 = 2l + 2l − 70

=

r

·

t

Distance

Rate

Time

330 = 4l − 70

Freight train

d

60

t+1

100 = l

Passenger train

d

80

t

400 = 4l

If l = 100, then l − 35 = 100 − 35 = 65.

Check. The width, 65 yd, is 35 yd less than the length, 100 yd. Also, the perimeter is 2 · 100 yd + 2 · 65 yd = 200 yd + 130 yd = 330 yd.

Translate. Using the formula d = rt in each row of the table, we get two equations. d = 60(t + 1) and d = 80t.

The answer checks.

Since the distances are the same, we have the equation

State. The length of the field is 100 yd, and the width is 65 yd.

Carry out. We solve the equation.

2 46. Let h = the height of the poster and h = the width, in 3 inches. 2 Solve: 100 = 2 · h + 2 · h 3 2 h = 30, so the height is 30 in. and the width is · 30, or 3 20 in. 47. Familiarize. Let w = the number of pounds of Kimiko’s body weight that is water. Translate. 50% of body weight is water. "# $ ! ↓ ↓ ↓ ↓ ↓ 0.5 × 135 = w 0.5 × 135 = w 67.5 = w

Check. Since 50% of 138 is 67.5, the answer checks. State. 67.5 lb of Kimiko’s body weight is water. 48. Let w = the number of pounds of Emilio’s body weight that is water. Solve: 0.6 × 186 = w w = 111.6 lb

49. Familiarize. We make a drawing. Let t = the number of hours the passenger train travels before it overtakes the freight train. Then t + 1 = the number of hours the freight train travels before it is overtaken by the passenger train. Also let d = the distance the trains travel.

Passenger train ! 80 mph

60(t + 1) = 80t 60t + 60 = 80t 60 = 20t 3=t When t = 3, then t + 1 = 3 + 1 = 4. Check. In 4 hr the freight train travels 60 · 4, or 240 mi. In 3 hr the passenger train travels 80 · 3, or 240 mi. Since the distances are the same, the answer checks. State. It will take the passenger train 3 hr to overtake the freight train. 50. Let t = the time the private airplane travels.

Carry out. We solve the equation.

Freight train ! 60 mph

60(t + 1) = 80t.

t + 1 hr

d

t hr

d

# #

We can also organize the information in a table.

Distance

Rate

Time

Private airplane

d

180

t

Jet

d

900

t−2

From the table we have the following equations: d = 180t

and d = 900(t − 2)

Solve: 180t = 900(t − 2)

t = 2.5

In 2.5 hr the private airplane travels 180(2.5), or 450 km. This is the distance from the airport at which it is overtaken by the jet. 51. Familiarize. Let t = the number of hours it takes the kayak to travel 36 mi upstream. The kayak travels upstream at a rate of 12 − 4, or 8 mph. Translate. We use the formula d = rt. 36 = 8 · t

Carry out. We solve the equation. 36 = 8 · t

4.5 = t

Check. At a rate of 8 mph, in 4.5 hr the kayak travels 8(4.5), or 36 mi. The answer checks. State. It takes the kayak 4.5 hr to travel 36 mi upstream.

122

Chapter 2: Functions, Equations, and Inequalities

52. Let t = the number of hours it will take Angelo to travel 20 km downstream. The kayak travels downstream at a rate of 14 + 2, or 16 km/h. Solve: 20 = 16t

56. Let x = the amount borrowed at 5%. Then 9000 − x = the amount invested at 6%. Solve: 0.05x + 0.06(9000 − x) = 492

x = 4800, so $4800 was borrowed at 5% and $9000 − $4800 = $4200 was borrowed at 6%.

t = 1.25 hr 53. Familiarize. Let t = the number of hours it will take the plane to travel 1050 mi into the wind. The speed into the headwind is 450 − 30, or 420 mph. Translate. We use the formula d = rt. 1050 = 420 · t

Carry out. We solve the equation. 1050 = 420 · t 2.5 = t

57. Familiarize. Let c = the calcium content of the cheese, in mg. Then 2c + 4 = the calcium content of the yogurt. Translate. Calcium calcium content plus content of cheese of yogurt ! "# $  ! "# $   & & & c + (2c + 4)

total calcium content. ! "# $   & & = 676 is

Check. At a rate of 420 mph, in 2.5 hr the plane travels 420(2.5), or 1050 mi. The answer checks.

Carry out. We solve the equation.

State. It will take the plane 2.5 hr to travel 1050 mi into the wind.

3c + 4 = 676

54. Let t = the number of hours it will take the plane to travel 700 mi with the wind. The speed with the wind is 375+25, or 400 mph.

c = 224

Solve: 700 = 400t

c + (2c + 4) = 676 3c = 672 Then 2c + 4 = 2 · 224 + 4 = 448 + 4 = 452.

Check. 224 + 452 = 676, so the answer checks. State. The cheese contains 224 mg of calcium, and the yogurt contains 452 mg.

t = 1.75 hr 55. Familiarize. Let x = the amount invested at 3% interest. Then 5000 − x = the amount invested at 4%. We organize the information in a table, keeping in mind the simple interest formula, I = P rt. Amount Interest Amount invested rate Time of interest 3% 3%, or x(0.03)(1), investx 1 yr ment 0.03 or 0.03x 4% 4%, or (5000−x)(0.04)(1), invest- 5000−x 1 yr ment 0.04 or 0.04(5000−x) Total 5000 176 Translate. interest on Interest on plus 4% investment 3% investment "# $  ! "# $ !   & & & 0.03x + 0.04(5000 − x)

Carry out. We solve the equation. 0.03x + 0.04(5000 − x) = 176

58. Let p = the number of working pharmacists in the United States in 1975. Solve: 224, 500 = 1.84p p ≈ 122, 011 pharmacists 59. Familiarize. Let s = the total holiday spending in 2004, in billions of dollars. Translate. $29.4 billion$ "# !  & 29.4

is 7% of total spending. "# $    !  & & & & = 0.07 · s

Carry out. We solve the equation. is $176.  & =

 & 176

0.03x + 200 − 0.04x = 176 −0.01x + 200 = 176

−0.01x = −24

x = 2400

If x = 2400, then 5000 − x = 5000 − 2400 = 2600.

Check. The interest on $2400 at 3% for 1 yr is $2400(0.03)(1) = $72. The interest on $2600 at 4% for 1 yr is $2600(0.04)(1) = $104. Since $72 + $104 = $176, the answer checks. State. $2400 was invested at 3%, and $2600 was invested at 4%.

29.4 = 0.07 · s 420 = s

Dividing by 0.07

Check. 7% of 420 is 0.07 · 420, or 29.4, so the answer checks. State. Total holiday spending in 2004 was $420 billion. 60. Let b = the number of high-speed Internet customers in 2000, in millions. Solve: 5b + 2.4 = 37.9 b = 7.1 million customers 61. Familiarize. Let n = the number of inches the volcano rises in a year. We will express one-half mile in inches: 5280 ft 12 in. 1 mi × × = 31, 680 in. 2 1 mi 1 ft

Exercise Set 2.1

123 71.

Translate. number of inches 50,000 yr times per year ! "# $ "# $ !    & & & 50, 000 · n

Carry out. We solve the equation. 50, 000n = 31, 680

is 31,680 in. ! "# $   & & = 31, 680

n = 0.6336

6=x

The zero of the function is 6. 72.

Setting f (x) = 0 Subtracting 5 on both sides

20 − x = 0

The zero of the function is 20. 74.

−3x + 13 = 0

−3x = −13 13 x= , or 4.3 3

75.

x−6 = 0

x=6

64.

5x + 20 = 0

76.

65.

−x + 18 = 0

3x − 9 = 0

x=3 77.

Setting f (x) = 0

The zero of the function is 18.

67.

Setting f (x) = 0

16 − x + x = 0 + x Adding x on both sides 16 = x

The zero of the function is 16. 68.

69.

x + 12 − 12 = 0 − 12 x = −12

4=x

79. a) The graph crosses the x-axis at (4, 0). This is the x-intercept.

80. a) (5, 0) b) 5 81. a) The graph crosses the x-axis at (−2, 0). This is the x-intercept.

82. a) (2, 0) b) 2 Setting f (x) = 0 Subtracting 12 on both sides

The zero of the function is −12. 70.

4−x = 0

b) The zero of the function is the first coordinate of the x-intercept. It is −2.

−2x + 7 = 0

−2x = −7 7 x= 2 x + 12 = 0

78.

b) The zero of the function is the first coordinate of the x-intercept. It is 4.

8+x = 0 16 − x = 0

Setting f (x) = 0

15 = x Adding x on both sides

18 = x

x = −8

−x + 15 = 0

The zero of the function is 15.

−x + 18 + x = 0 + x Adding x on both sides

66.

Adding 6 on both sides

3x = 9

5x = −20 x = −4

Setting f (x) = 0

The zero of the function is 6.

x = −5

The zero of the function is −5.

Setting f (x) = 0

20 = x

x = 660 yr x+5 = 0

x = −4 20 − x + x = 0 + x Adding x on both sides

State. On average, the volcano rises 0.6336 in. in a year.

x+5−5 = 0−5

4+x = 0

73.

62. Let x = the number of years it will take Horseshoe Falls to migrate one-fourth mile upstream. We will express onefourth mile in feet: 5280 ft 1 mi × = 1320 ft 4 1 mi Solve: 2x = 1320

Setting f (x) = 0

−x + 6 + x = 0 + x Adding x on both sides

Check. Rising at a rate of 0.6336 in. per year, in 50,000 yr the volcano will rise 50, 000(0.6336), or 31,680 in. The answer checks.

63.

−x + 6 = 0

8x + 2 = 0

8x = −2 1 x = − , or − 0.25 4

83. a) The graph crosses the x-axis at (−4, 0). This is the x-intercept. b) The zero of the function is the first coordinate of the x-intercept. It is −4. 84. a) (−2, 0) b) −2

124 85.

86.

87.

Chapter 2: Functions, Equations, and Inequalities 1 bh 2 2A = bh Multiplying by 2 on both sides 2A =b Dividing by h on both sides h A=

A = πr2 A =π r2 Subtracting 2l on both sides

P − 2l =w 2 88.

Ax = C − By C − By A= x 96.

97.

Dividing by 2 on both sides

A = P + P rt

1 h(b1 + b2 ) 2 2A = h(b1 + b2 ) Multiplying by 2 on both sides 2A = h Dividing by b1 + b2 on both sides b1 + b 2 A=

90.

A=

98.

1 h(b1 + b2 ) 2

93.

94.

4 3 πr 3 3V = 4πr3 Multiplying by 3 on both sides 3V =π Dividing by 4r3 on both sides 4r3 V =

4 V = πr3 3 3V = r3 4π 9 F = C + 32 5 9 F − 32 = C Subtracting 32 on both sides 5 5 5 (F − 32) = C Multiplying by on both sides 9 9 Ax + By = C By = C − Ax C − Ax y= B

2w + 2h + l = p Subtracting 2w and l

p − 2w − l 2

Dividing by 2

3x + 4y = 12 4y = 12 − 3x 12 − 3x y= 4

99.

2x − 3y = 6

−3y = 6 − 2x 6 − 2x , or y= −3 2x − 6 3

100.

T =

Subtracting 2x Dividing by −3

3 (I − 12, 000) 10

10 T = I − 12, 000 3

2A − b1 = b2 , or h 2A − b1 h = b2 h

92.

Dividing by x on both sides

2w + 2h + l = p

h=

2A = b1 + b 2 h

91.

Subtracting By on both sides

2h = p − 2w − l

A − P = P rt A−P =r Pt 89.

Ax + By = C

2w = p − 2h − l p − 2h − l w= 2

P = 2l + 2w P − 2l = 2w

95.

10 T + 12, 000 = I, or 3 10T + 36, 000 =I 3 101.

a = b + bcd a = b(1 + cd) Factoring a =b Dividing by 1 + cd 1 + cd

102.

103.

104.

q = p − np

q = p(1 − n) q =p 1−n

z = xy − xy 2

z = x(y − y 2 ) Factoring z =x Dividing by y − y 2 y − y2 st = t − 4

st − t = −4

t(s − 1) = −4 4 −4 t= , or s−1 1−s

105. Left to the student 106. Left to the student

Exercise Set 2.1 107. The graph of f (x) = mx + b, m %= 0, is a straight line that is not horizontal. The graph of such a line intersects the x-axis exactly once. Thus, the function has exactly one zero.

125 119.

2x − {x − [3x − (6x + 5)]} = 4x − 1 2x − {x − [3x − 6x − 5]} = 4x − 1 2x − {x − [−3x − 5]} = 4x − 1 2x − {x + 3x + 5} = 4x − 1

108. If a person wanted to convert several Fahrenheit temperatures to Celsius, it would be useful to solve the formula for C and then use the formula in that form.

2x − {4x + 5} = 4x − 1 2x − 4x − 5 = 4x − 1 −2x − 5 = 4x − 1

109. First find the slope of the given line. 3x + 4y = 7 4y = −3x + 7 7 3 y = − x+ 4 4 3 The slope is − . Now write a slope-intersect equation of 4 3 the line containing (−1, 4) with slope − . 4 3 y − 4 = − [x − (−1)] 4 3 y − 4 = − (x + 1) 4 3 3 y−4 = − x− 4 4 3 13 y = − x+ 4 4 4 − (−2) 6 3 = =− −5 − 3 −8 4 3 y − 4 = − (x − (−5)) 4 15 3 y−4 = − x− 4 4 3 1 y = − x+ 4 4

110. m =

111. The domain of f is the set of all real numbers as is the domain of g, so the domain of f + g is the set of all real numbers, or (−∞, ∞). 112. The domain of f is the set of all real numbers as is the domain of g. When x = −2, g(x) = 0, so the domain of f /g is (−∞, −2) ∪ (−2, ∞).

113. 114.

(f − g)(x) = f (x) − g(x) = 2x − 1 − (3x + 6) = 2x − 1 − 3x − 6 = −x − 7

f g(−1) = f (−1) · g(−1) = [2(−1) − 1][3(−1) + 6] =

−3 · 3 = −9

3 3 115. f (x) = 7 − x = − x + 7 2 2 The function can be written in the form y = mx + b, so it is a linear function. 3 + 5 cannot be written in the form f (x) = mx + 116. f (x) = 2x b, so it is not a linear function. 117. f (x) = x2 +1 cannot be written in the form f (x) = mx+b, so it is not a linear function. 3 118. f (x) = x − (2.4)2 is in the form f (x) = mx + b, so it is 4 a linear function.

−6x − 5 = −1 −6x = 4

2 The solution is − . 3 120.

x=−

2 3

14 − 2[3 + 5(x − 1)] = 3{x − 4[1 + 6(2 − x)]} 14 − 2[3 + 5x − 5] = 3{x − 4[1 + 12 − 6x]} 14 − 2[5x − 2] = 3{x − 4[13 − 6x]} 14 − 10x + 4 = 3{x − 52 + 24x} 18 − 10x = 3{25x − 52} 18 − 10x = 75x − 156 174 = 85x 174 =x 85

121. The size of the cup was reduced 8 oz − 6 oz, or 2 oz, and 2 oz = 0.25, so the size was reduced 25%. The price per 8 oz 89/ c , or 11.25/ c/oz. The price ounce of the 8 oz cup was 8 oz 71/ c per ounce of the 6 oz cup is , or 11.83/ c/oz. Since the 6 oz price per ounce was not reduced, it is clear that the price per ounce was not reduced by the same percent as the size c, or of the cup. The price was increased by 11.83 − 11.125/ 0.7083/ c 0.7083/ c per ounce. This is an increase of ≈ 0.064, 11.83/ c or about 6.4% per ounce. 122. The size of the container was reduced 100 oz − 80 oz, or 20 oz = 0.2, so the size of the container was 20 oz, and 100 oz reduced 20%. The price per ounce of the 100-oz container $6.99 , or $0.0699/oz. The price per ounce of the was 100 oz $5.75 , or $0.071875. Since the price per 80-oz container is 80 oz ounce was not reduced, it is clear that the price per ounce was not reduced by the same percent as the size of the container. The price increased by $0.071875 − $0.0699, or $0.001975 $0.001975. This is an increase of ≈ 0.028, or $0.0699 about 2.8% per ounce. 123. We use a proportion to determine the number of calories c burned running for 75 minutes, or 1.25 hr. c 720 = 1 1.25 720(1.25) = c 900 = c

126

Chapter 2: Functions, Equations, and Inequalities Next we use a proportion to determine how long the person would have to walk to use 900 calories. Let t represent this time, in hours. We express 90 min as 1.5 hr. t 1.5 = 480 900 900(1.5) =t 480 2.8125 = t Then, at a rate of 4 mph, the person would have to walk 4(2.8125), or 11.25 mi.

124. Let x = the number of copies of The DaVinci Code that were sold. Then 3570−x = the number of copies of Marley and Me that were sold. 10 x = Solve: 3570 − x 1.9 x = 3000, so 3000 copies of The DaVinci Code were sold and 3570 − x = 3570 − 3000 = 570 copies of Marley and Me were sold.

15.

= (12 − 8) + (3i + 5i) = 4 + 8i

16. (−11 + 4i) + (6 + 8i) = (−11 + 6) + (4i + 8i) = −5 + 12i 17.

= −4 − 2i

18. (−5 − i) + (6 + 2i) = (−5 + 6) + (−i + 2i) = 1 + i √ √ 19. (3 + −16) + (2 + −25) = (3 + 4i) + (2 + 5i) = (3 + 2) + (4i + 5i) = 5 + 9i 20. 21.

1. 2. 3. 4. 5. 6. 7. 8. 9.

10. 11.

−36) + (2 +

√

−9) = (7 − 6i) + (2 + 3i) =

(7 + 2) + (−6i + 3i) = 9 − 3i (10 + 7i) − (5 + 3i)

= 5 + 4i √

√

(−5 + 3i) + (7 + 8i)

22. 23.

25.

= 5 + (−9 − 3)i = 5 − 12i

14. (7 − 2i) + (4 − 5i) = (7 + 4) + (−2i − 5i) = 11 − 7i

−10 + 18i

(6 − 4i) − (−5 + i)

26. (8 − 3i) − (9 − i) = (8 − 9) + [−3i − (−i)] = −1 − 2i 27.

(−5 + 2i) − (−4 − 3i)

= [−5 − (−4)] + [2i − (−3i)]

= (−5 + 4) + (2i + 3i)

29.

= −1 + 5i

(−6 + 7i)−(−5 − 2i) = [−6 − (−5)] + [7i−(−2i)] = −1 + 9i

(4 − 9i) − (2 + 3i)

= (4 − 2) + (−9i − 3i)

(4 − 9i) + (1 − 3i)

and the imaginary parts

(−7 + 12i) − (3 − 6i) = (−7 − 3) + [12i − (−6i)] =

= 11 − 5i

= 2 + 11i

= (4 + 1) + (−9i − 3i) Collecting the real parts

The 8 and the 2i are both being subtracted.

= (6 + 5) + (−4i − i)

28.

12. (−6 − 5i) + (9 + 2i) = (−6 + 9) + (−5i + 2i) = 3 − 3i

(13 + 9i) − (8 + 2i)

= [6 − (−5)] + (−4i − i)

and the imaginary parts = 2 + (3 + 8)i

−11 − 3i

= 5 + 7i 24.

The 5 and the 3i are both being subtracted.

(−3 − 4i) − (8 − i) = (−3 − 8) + [−4i − (−i)] =

= (13 − 8) + (9i − 2i)

= (−5 + 7) + (3i + 8i) Collecting the real parts

13.

(7 −

√

= (10 − 5) + (7i − 3i)

√ √ −3 = −1 · 3 = −1 · 3 = i 3, or 3i √ √ √ √ −21 = −1 · 21 = i 21, or 21i √ √ √ √ −25 = −1 · 25 = −1 · 25 = i · 5 = 5i √ √ −100 = −1 · 100 = i · 10 = 10i √ √ √ √ √ √ − −33 = − −1 · 33 = − −1 · 33 = −i 33, or − 33i √ √ √ √ − −59 = − −1 · 59 = −i 59, or − 59i √ √ √ √ − −81 = − −1 · 81 = − −1 · 81 = −i · 9 = −9i √ √ √ √ − −9 = − −1 · 9 = − −1 · 9 = −i · 3 = −3i √ √ √ √ √ −98 = −1 · 98 = −1 · 98 = i 49 · 2 = √ √ √ i · 7 2 = 7i 2, or 7 2i √ √ √ √ √ √ −28 = −1 · 28 = i 28 = i 4 · 7 = 2i 7, or 2 7i √

(−1 − i) + (−3 − i)

= (−1 − 3) + (−i − i)

Exercise Set 2.2 √

(12 + 3i) + (−8 + 5i)

30. 31.

= 2 − 12i

(10 − 4i) − (8 + 2i) = (10 − 8) + (−4i − 2i) = 2 − 6i

7i(2 − 5i)

= 14i − 35i2

Using the distributive law

= 35 + 14i

Writing in the form a + bi

= 14i + 35

i2 = −1

Exercise Set 2.2

127

32. 3i(6 + 4i) = 18i + 12i2 = 18i − 12 = −12 + 18i

46. (5 + 9i)(5 − 9i) = 25 − 81i2 = 25 + 81 = 106

33.

47.

−2i(−8 + 3i)

= 16i − 6i2

Using the distributive law

= 6 + 16i

Writing in the form a + bi

= 16i + 6

i2 = −1

34. −6i(−5 + i) = 30i − 6i2 = 30i + 6 = 6 + 30i 35.

(1 + 3i)(1 − 4i)

= 1 − 4i + 3i − 12i2

= 9 − 6i + 6i + 4

= 13 48. 49.

Using FOIL

38. 39.

= 49 + 35i − 35i + 25

(1 − 2i)(1 + 3i) = 1 + 3i − 2i − 6i2 = 1 + i + 6 =

50. 51.

(2 + 3i)(2 + 5i) = 4 + 10i + 6i + 15i2

Using FOIL

= 4 + 10i + 6i − 15

i2 = −1

= −11 + 16i

(−4 + i)(3 − 2i)

= −10 + 11i

43.

52. 53.

54. (−3 + 2i)2 = 9 − 12i + 4i2 = 9 − 12i − 4 = 5 − 12i

i = −1 2

55.

(7 − 4i)(−3 − 3i) = −21 − 21i + 12i + 12i =

−21 − 21i + 12i − 12 = −33 − 9i √ √ (3 + −16)(2 + −25)

= 1 − 6i + 9i2 = 1 − 6i − 9

56.

= 6 + 15i + 8i + 20i

i2 = −1

= −14 + 23i √ √ (7 − −16)(2 + −9) = (7 − 4i)(2 + 3i) =

57.

= −8 − 6i

(2 − 5i)2 = 4 − 20i + 25i2 = 4 − 20i − 25 =

−21 − 20i

(−1 − i)2

= 1 + 2i + i2 = 1 + 2i − 1

26 + 13i

= 2i 58.

i = −1 2

i2 = −1

(−4 − 2i)2 = 16 + 16i + 4i2 = 16 + 16i − 4 = 12 + 16i

2

i2 = −1

= (−1)2 − 2(−1)(i) + i2

14 + 21i − 8i − 12i2 = 14 + 21i − 8i + 12 =

= 25 + 16

(1 − 3i)2

= 12 − 2 · 1 · (3i) + (3i)2

2

= 41

i2 = −1

= −45 − 28i

(8 − 3i)(−2 − 5i)

(5 − 4i)(5 + 4i) = 52 − (4i)2

Recall (A + B)2 = A2 + 2AB + B 2

= 4 − 28i − 49

−5 + 5i + 2i + 2 = −3 + 7i

= 25 − 16i

(−2 + 7i)2

= 4 − 28i + 49i

(5 − 2i)(−1 + i) = −5 + 5i + 2i − 2i =

2

45.

9 − 40i

2

= (3 + 4i)(2 + 5i)

44.

(5 − 4i)2 = 25 − 40i + 16i2 = 25 − 40i − 16 =

= (−2)2 + 2(−2)(7i) + (7i)2 2

= 6 + 15i + 8i − 20

Recall (A + B)2 = A2 + 2AB + B 2 i2 = −1

= 12 + 16i

i2 = −1

= −31 − 34i

(4 + 2i)2

= 16 + 16i − 4

Using FOIL

= −16 − 40i + 6i − 15

36 + 48i − 48i + 64 = 100

= 16 + 16i + 4i2

24 − 6i − 40i − 10 = 14 − 46i

i2 = −1

(6 − 8i)(6 + 8i) = 36 + 48i − 48i − 64i2 =

= 16 + 2 · 4 · 2i + (2i)2

(3 − 5i)(8 − 2i) = 24 − 6i − 40i + 10i2 =

= −16 − 40i + 6i + 15i2

42.

(7 − 5i)(7 + 5i)

= 74

= −12 + 8i + 3i + 2

41.

64 − 8i + 8i + 1 = 65

= 13 − i

= −12 + 8i + 3i − 2i2

40.

(8 + i)(8 − i) = 64 − 8i + 8i − i2 =

= 49 + 35i − 35i − 25i2

7+i 37.

i2 = −1

= 1 − 4i + 3i − 12(−1) i2 = −1 = 1 − i + 12

36.

(3 + 2i)(3 − 2i)

= 9 − 6i + 6i − 4i2

128

Chapter 2: Functions, Equations, and Inequalities

59.

64.

(3 + 4i)2 = 9 + 2 · 3 · 4i + (4i)2 = 9 + 24i + 16i2 = 9 + 24i − 16

60.

= −7 + 24i

i2 = −1

(6 + 5i)2 = 36 + 60i + 25i2 = 36 + 60i − 25 =

11 + 60i 61.

3 5 − 11i 5 + 11i 3 = · 5 − 11i 5 + 11i = = = = =

62.

65.

5 − 11i is the conjugate of 5 + 11i.

3(5 + 11i) (5 − 11i)(5 + 11i) 15 + 33i 25 − 121i2 15 + 33i i2 = −1 25 + 121 15 + 33i 146 33 15 + i Writing in the form a + bi 146 146

−12 + 5i + 2i2 9 − 4i2 −12 + 5i − 2 = i2 = −1 9+4 −14 + 5i = 13 5 14 Writing in the form a + bi =− + i 13 13

5 2 + 3i 2 − 3i 5 = · 2 + 3i 2 − 3i

67.

= = = =

(4 + i)(−3 + 2i) (−3 − 2i)(−3 + 2i)

−3 + 2i is the conjugate of the divisor.

=

66.

=

4+i −3 − 2i −3 + 2i 4+i · = −3 − 2i −3 + 2i =

i i 2−i = · 2+i 2+i 2−i

5−i 5−i −7 − 2i = · −7 + 2i −7 + 2i −7 − 2i −35 − 3i + 2i2 49 − 4i2 −35 − 3i − 2 = 49 + 4 3 37 =− − i 53 53

=

2i − i2 = 4 − i2 2i + 1 = 4+1 1 2 = + i 5 5

63.

−3 −3 4 + 5i = · 4 − 5i 4 − 5i 4 + 5i −12 − 15i = 16 − 25i2 −12 − 15i = 16 + 25 12 15 =− − i 41 41

2 − 3i is the conjugate of 2 + 3i.

5(2 − 3i) (2 + 3i)(2 − 3i) 10 − 15i 4 − 9i2 10 − 15i i2 = −1 4+9 10 − 15i 13 10 15 − i Writing in the form a + bi 13 13

5 − 3i 4 + 3i 5 − 3i 4 − 3i = · 4 + 3i 4 − 3i =

(5 − 3i)(4 − 3i) (4 + 3i)(4 − 3i)

4 − 3i is the conjugate of 4 + 3i.

20 − 27i + 9i2 16 − 9i2 20 − 27i − 9 = i2 = −1 16 + 9 11 − 27i = 25 11 27 − i Writing in the form a + bi = 25 25 =

68.

6 + 5i 6 + 5i 3 + 4i = · 3 − 4i 3 − 4i 3 + 4i

18 + 39i + 20i2 9 − 16i2 18 + 39i − 20 = 9 + 16 39 2 =− + i 25 25

=

Exercise Set 2.2 69.

70.

129

√ 2 + 3i 5 − 4i √ 2 + 3i 5 + 4i = · 5 + 4i is the conjugate 5 − 4i 5 + 4i of the divisor. √ (2 + 3i)(5 + 4i) = (5 − 4i)(5 + 4i) √ √ 10 + 8i + 5 3i + 4 3i2 = 25 − 16i2 √ √ 10 + 8i + 5 3i − 4 3 = i2 = −1 25 + 16 √ √ 10 − 4 3 + (8 + 5 3)i = 41 √ √ 10 − 4 3 8 + 5 3 = + i Writing in the 41 41 form a + bi √ √ 5 + 3i 5 + 3i 1 + i = · 1−i 1−i 1+i √ √ 5 + 5i + 3i + 3i2 = 1 − i2 √ √ 5 + 5i + 3i − 3 = 1+1 √ √ 5−3 5+3 + i = 2 2

71. = = = = =

1+i (1 − i)2 1+i 1 − 2i + i2 1+i 1 − 2i − 1 1+i −2i 1 + i 2i · −2i 2i (1 + i)(2i) (−2i)(2i)

2i + 2i2 −4i2 2i − 2 = 4 2 2 =− + i 4 4 1 1 =− + i 2 2

72.

1−i 1−i = (1 + i)2 1 + 2i + i2 1−i = 1 + 2i − 1 1−i = 2i 1 − i −2i · = 2i −2i −2i + 2i2 −4i2 −2i − 2 = 4 1 1 =− − i 2 2 =

73.

4 − 2i 1+i 6 − 7i = 1+i 6 − 7i = 1+i =

+

2 − 5i 1+i Adding

·

1−i 1−i

1 − i is the conjugate of 1 + i.

(6 − 7i)(1 − i) (1 + i)(1 − i)

6 − 13i + 7i2 1 − i2 6 − 13i − 7 = 1+1 −1 − 13i = 2 1 13 =− − i 2 2 =

i2 = −1 74. 2i is the conjugate of −2i.

i2 = −1

9 + 4i 3 + 2i 6 + 2i + = 1−i 1−i 1−i 9 + 4i 1 + i = · 1−i 1+i 9 + 13i + 4i2 1 − i2 9 + 13i − 4 = 1+1 5 13 = + i 2 2

=

=

i2 = −1

75. i11 = i10 · i = (i2 )5 · i = (−1)5 · i = −1 · i = −i 76. i7 = i6 · i = (i2 )3 · i = (−1)3 · i = −1 · i = −i 77. i35 = i34 · i = (i2 )17 · i = (−1)17 · i = −1 · i = −i 78. i24 = (i2 )12 = (−1)12 = 1 79. i64 = (i2 )32 = (−1)32 = 1 80. i42 = (i2 )21 = (−1)21 = −1 81.

(−i)71 = (−1 · i)71 = (−1)71 · i71 = −i70 · i =

−(i2 )35 · i = −(−1)35 · i = −(−1)i = i 82. (−i)6 = i6 = (i2 )3 = (−1)3 = −1

130

Chapter 2: Functions, Equations, and Inequalities

83. (5i)4 = 54 · i4 = 625(i2 )2 = 625(−1)2 = 625 · 1 = 625

84. (2i) = 32i = 32·i ·i = 32(i ) ·i = 32(−1) ·i = 32·1·i = 32i 5

5

4

2 2

2

85. Left to the student 86. Left to the student 87. The sum of two imaginary numbers is not always an imaginary number. For example, (2 + i) + (3 − i) = 5, a real number.

97. (a + bi)(c + di) = (ac − bd) + (ad + bc)i. The conjugate of the product is (ac − bd) − (ad + bc)i = (a − bi)(c − di), the product of the conjugates of the individual complex numbers. Thus, the statement is true. 98.

99. zz = (a + bi)(a − bi) = a2 − b2 i2 = a2 + b2

−6y = −3x + 7 7 1 y = x− 2 6 1 The slope is . The slope of the desired line is the oppo2 1 site of the reciprocal of , or −2. Write a slope-intercept 2 equation of the line containing (3, −5) with slope −2.

a + bi + 6(a − bi) = 7

a + bi + 6a − 6bi = 7

7a − 5bi = 7 Then 7a = 7, so a = 1, and −5b = 0, so b = 0. Thus, z = 1.

Exercise Set 2.3 1. (2x − 3)(3x − 2) = 0

2x − 3 = 0 or 3x − 2 = 0 Using the principle of zero products 2x = 3 or 3x = 2 2 3 or x= x= 2 3 3 2 The solutions are and . 2 3

y − (−5) = −2(x − 3) y + 5 = −2x + 6 y = −2x + 1

90. The domain of f is the set of all real numbers as is the domain of g. Then the domain of (f − g)(x) is the set of all real numbers, or (−∞, ∞). 91. The domain of f is the set of all real numbers as is the 5 domain of g. When x = − , g(x) = 0, so the domain of 3 ' ( 5( ' 5 ∪ − ,∞ . f /g is − ∞, − 3 3

2. (5x − 2)(2x + 3) = 0 3 2 x = or x = − 5 2 3 2 The solutions are and − . 5 2 x2 − 8x − 20 = 0 3. (x − 10)(x + 2) = 0 x − 10 = 0

92. (f − g)(x) = f (x) − g(x) = x2 + 4 − (3x + 5) = x2 − 3x − 1 93. (f /g)(2) = 94. = = = = =

22 + 4 4+4 8 f (2) = = = g(2) 3·2+5 6+5 11

f (x + h) − f (x) h (x + h)2 − 3(x + h) + 4 − (x2 − 3x + 4) h x2 + 2xh + h2 − 3x − 3h + 4 − x2 + 3x − 4 h 2xh + h2 − 3h h h(2x + h − 3) h 2x + h − 3

95. (a+bi)+(a−bi) = 2a, a real number. Thus, the statement is true. 96. (a + bi) + (c + di) = (a + c) + (b + d)i. The conjugate of this sum is (a+c)−(b+d)i = a+c−bi−di = (a−bi)+(c−di), the sum of the conjugates of the individual complex numbers. Thus, the statement is true.

z + 6z = 7

100.

88. The product of two imaginary numbers is not always an imaginary number. For example, i · i = i2 = −1, a real number. 89. First find the slope of the given line. 3x − 6y = 7

1 1 a − bi a −b = · = 2 + 2 i z a + bi a − bi a + b2 a + b2

4.

Factoring

or x + 2 = 0

x = 10 or x = −2 The solutions are 10 and −2.

Using the principle of zero products

x2 + 6x + 8 = 0

(x + 2)(x + 4) = 0 x = −2 or x = −4 5.

The solutions are −2 and −4. 3x2 + x − 2 = 0

(3x − 2)(x + 1) = 0

Factoring

3x − 2 = 0 or x + 1 = 0 x=

2 or 3

x = −1

2 and −1. 3 10x2 − 16x + 6 = 0

The solutions are 6.

2(5x − 3)(x − 1) = 0 3 or x = 1 x= 5 3 The solutions are and 1. 5

Using the principle of zero products

Exercise Set 2.3 7.

131

4x2 − 12 = 0

16.

4x2 = 12

9x(2 + x) = 0

x2 = 3 √ √ x = 3 or x = − 3 The solutions are 8.

9.

6x2 = 36

√

Using the principle of square roots

√ 3 and − 3.

x = 0 or x = −2

The solutions are −2 and 0. 17.

y(3y + 1)(y − 2) = 0

y = 0 or 3y + 1 = 0 y = 0 or

18.

t = 0 or t = 1 or t = The solutions are 0, 19.

15.

7x3 + x2 − 7x − 1 = 0

(x2 − 1)(7x + 1) = 0

x+1 = 0

x2 = −2 √ √ x = 2i or x = − 2i √ √ The solutions are 2i and − 2i. 4x2 + 12 = 0

2x2 − 34 = 0

x2 = 17 √ √ x = 17 or x = − 17 √ √ The solutions are 17 and − 17. 3x2 = 33

x2 = 11 √ √ x = 11 or x = − 11 √ √ The solutions are 11 and − 11. 2x2 = 6x

Subtracting 6x on both sides

2x(x − 3) = 0

2x = 0 or x − 3 = 0

x = 0 or x=3 The solutions are 0 and 3.

or x − 1 = 0 or 7x + 1 = 0

x = −1 or

x = 1 or

1 The solutions are −1, − , and 1. 7 20.

x2 = −3 √ √ x = 3i or x = − 3i √ √ The solutions are 3i and − 3i.

2x2 − 6x = 0

2 , and 1. 3

(x + 1)(x − 1)(7x + 1) = 0

2x2 = 34

14.

2 3

x (7x + 1) − (7x + 1) = 0 2

5x2 + 10 = 0

4x2 = −12

13.

3t3 + 2t = 5t2 t(t − 1)(3t − 2) = 0

5x2 = −10

12.

y=2

3t − 5t2 + 2t = 0

2x2 = 20

11.

1 or 3

3

2x2 − 20 = 0

x2 = 10 √ √ x = 10 or x = − 10 √ √ The solutions are 10 and − 10.

y=−

or y − 2 = 0

1 The solutions are − , 0 and 2. 3

3x2 = 21

Using the principle of square roots √ √ The solutions are 7 and − 7.

3y 3 − 5y 2 − 2y = 0

y(3y 2 − 5y − 2) = 0

x2 = 6 √ √ x = 6 or x = − 6 √ √ The solutions are 6 and − 6. x2 = 7 √ √ x = 7 or x = − 7

10.

18x + 9x2 = 0

x=−

1 7

3x3 + x2 − 12x − 4 = 0

x2 (3x + 1) − 4(3x + 1) = 0 (3x + 1)(x2 − 4) = 0

(3x + 1)(x + 2)(x − 2) = 0

1 or x = −2 or x = 2 3 1 The solutions are −2, − , and 2. 3 x=−

21. a) The graph crosses the x-axis at (−4, 0) and at (2, 0). These are the x-intercepts. b) The zeros of the function are the first coordinates of the x-intercepts of the graph. They are −4 and 2. 22. a) (−1, 0), (2, 0) b) −1, 2 23. a) The graph crosses the x-axis at (−1, 0) and at (3, 0). These are the x-intercepts. b) The zeros of the function are the first coordinates of the x-intercepts of the graph. They are −1 and 3. 24. a) (−3, 0), (1, 0) b) −3, 1

132

Chapter 2: Functions, Equations, and Inequalities

25. a) The graph crosses the x-axis at (−2, 0) and at (2, 0). These are the x-intercepts.

31.

b) The zeros of the function are the first coordinates of the x-intercepts of the graph. They are −2 and 2.

x2 + 8x + 25 = 0 x2 + 8x = −25

Subtracting 25

x + 8x + 16 = −25 + 16 2

1 2

26. a) (−1, 0), (1, 0) (x + 4) = −9

27.

x + 4 = ±3i

x2 + 6x = 7 x + 6x + 9 = 7 + 9

Completing the square: 1 2 2 · 6 = 3 and 3 = 9 Factoring

2

(x + 3)2 = 16 x + 3 = ±4 x = −3 ± 4

Using the principle of square roots

x = −4 ± 3i

32.

x2 +8x+16 = −15+16 (x + 4)2 = 1

( 12 ·8 = 4 and 42 = 16)

x = −4 − 1 or x = −4 + 1 or x = −3

The solutions are −5 and −3. x2 = 8x − 9

x2 − 8x = −9

x2 − 8x + 16 = −9 + 16

Subtracting 8x Completing the square:

1 2 (−8)

(x − 4)2 = 7

√ x−4 = ± 7 x = 4±

√

= −4 and (−4)2 = 16

Factoring Using the principle of square roots

7 √ √ √ The solutions are 4 − 7 and 4 + 7, or 4 ± 7. 30.

x2 = 22 + 10x x − 10x = 22 2

x − 10x + 25 = 22 + 25 2

2

* · 6 = 3 and 32 = 9

(x + 3)2 = −4

x = −3 ± 2i

x = −4 ± 1

29.

)1

x + 3 = ±2i

x + 4 = ±1

x = −5

x2 + 6x = −13

x + 6x + 9 = −13 + 9

or x = 1

x2 + 8x = −15

x2 + 6x + 13 = 0 2

The solutions are −7 and 1. 28.

Using the principle of square roots

The solutions are −4 − 3i and −4 + 3i, or −4 ± 3i.

x = −3 − 4 or x = −3 + 4 x = −7

· 8 = 4 and 42 = 16 Factoring

2

b) −1, 1

Completing the square:

( 12 (−10) = −5 and (−5)2 = 25)

(x − 5)2 = 47 √ x − 5 = ± 47 √ x = 5 ± 47 √ √ √ The solutions are 5 − 47 and 5 + 47, or 5 ± 47.

The solution are −3 − 2i and −3 + 2i, or −3 ± 2i. 33.

3x2 + 5x − 2 = 0

3x2 + 5x = 2 2 5 x2 + x = 3 3 5 25 2 25 2 x + x+ = + 3 36 3 36 1 2

+

x+

5 6

,2

=

49 36

·

5 3

Adding 2 Dividing by 3 Completing the square: =

5 6

and ( 56 )2 =

25 36

Factoring and simplifying

5 7 =± Using the principle 6 6 of square roots 5 7 x=− ± 6 6 5 7 5 7 x=− − or x = − + 6 6 6 6 2 12 or x = x=− 6 6 1 x = −2 or x = 3 1 The solutions are −2 and . 3 x+

Exercise Set 2.3 34.

133

2x2 − 5x = 3 3 5 x2 − x = 2 2 5 25 3 25 x2 − x + = + 2 16 2 16 +

5 4

,2

x=

( 12 (− 52 ) = − 54 and

(− 54 )2 =

25 16 )

49 16 7 5 x− = ± 4 4 5 7 x= ± 4 4 5 7 5 7 x= − or x = + 4 4 4 4 1 x=− or x = 3 2 1 The solutions are − and 3. 2 35.

x−

=

x2 − 2x = 15

x − 2x − 15 = 0 2

(x − 5)(x + 3) = 0

Factoring

x − 5 = 0 or x + 3 = 0 x = 5 or

41. x2 + x + 2 = 0 We use the quadratic formula. Here a = 1, b = 1, and c = 2. √ −b ± b2 − 4ac x= 2a √ −1 ± 12 − 4 · 1 · 2 Substituting = 2·1 √ −1 ± −7 = 2 √ √ 1 −1 ± 7i 7 =− ± i = 2 2 2 √ √ 1 7 1 7 i and − + i, or The solutions are − − 2 2 2 2 √ 7 1 i. − ± 2 2

x = −3

x2 + 4x = 5 x2 + 4x − 5 = 0

(x + 5)(x − 1) = 0 x+5 = 0

or x − 1 = 0

x = −5 or

x=1

The solutions are −5 and 1. 37.

5m2 + 3m = 2 5m2 + 3m − 2 = 0

(5m − 2)(m + 1) = 0 Factoring

5m − 2 = 0 or m + 1 = 0 2 or m = −1 m= 5 2 The solutions are and −1. 5 38.

2y 2 − 3y − 2 = 0

(2y + 1)(y − 2) = 0 2y + 1 = 0

or y − 2 = 0 1 y = − or y=2 2 1 The solutions are − and 2. 2 39.

3x2 + 6 = 10x 3x − 10x + 6 = 0 2

We use the quadratic formula. Here a = 3, b = −10, and c = 6.

−b ±

40. 3t2 + 8t + 3 = 0 √ −8 ± 82 − 4 · 3 · 3 t= 2·3 √ √ −8 ± 2 7 −8 ± 28 = = 6 6 √ √ 2(−4 ± 7) −4 ± 7 = = 2·3 3 √ √ −4 + 7 −4 − 7 and , or The solutions are 3 3 √ −4 ± 7 . 3

The solutions are 5 and −3. 36.

√

b2 − 4ac 2a −(−10)± (−10)2 −4·3·6 = Substituting 2·3 √ √ 10 ± 28 10 ± 2 7 = = 6 6 √ √ 2(5 ± 7) 5± 7 = = 2·3 3 √ √ √ 5+ 7 5± 7 5− 7 and , or . The solutions are 3 3 3

2x2 − 5x − 3 = 0

42.

x2 + 1 = x x −x+1 = 0 −(−1) ± (−1)2 − 4 · 1 · 1 x= 2·1 √ √ 1 ± −3 1 ± 3i = = 2 2 √ 3 1 = ± i 2 2 √ √ 3 3 1 1 i and + i, or The solutions are − 2 2 2 2 √ 3 1 ± i. 2 2 2

134 43.

Chapter 2: Functions, Equations, and Inequalities 5±

5t2 − 8t = 3

5t2 − 8t − 3 = 0 We use the quadratic formula. Here a = 5, b = −8, and c = −3. √ −b ± b2 − 4ac t= 2a −(−8) ± (−8)2 − 4 · 5(−3) = 2·5 √ √ 8 ± 124 8 ± 2 31 = = 10 10 √ √ 2(4 ± 31) 4 ± 31 = = 2·5 √5 √ 4 − 31 4 + 31 The solutions are and , or 5 5 √ 4 ± 31 . 5 5x2 + 2 = x 44. 5x2 − x + 2 = 0 −(−1) ± (−1)2 − 4 · 5 · 2 x= 2·5 √ √ 1 ± 39i 1 ± −39 = = 10 10 √ 39 1 = ± i 10 10 √ √ 1 39 1 39 The solutions are − i and + i, or 10 10 10 10 √ 39 1 ± i. 10 10 3x2 + 4 = 5x 45. 3x2 − 5x + 4 = 0 We use the quadratic formula. Here a = 3, b = −5, and c = 4. √ −b ± b2 − 4ac x= 2a −(−5) ± (−5)2 − 4 · 3 · 4 = 2·3 √ √ 5 ± 23i 5 ± −23 = = 6 6 √ 5 23 i = ± 6 6 √ √ 5 5 23 23 i and + i, or The solutions are − 6 6 6 6 √ 23 5 ± i. 6 6 2 2t − 5t = 1 46. 2t2 − 5t − 1 = 0 −(−5) ± (−5)2 − 4 · 2(−1) t= 2·2 √ 5 ± 33 = 4 √ √ 5 + 33 5 − 33 and , or The solutions are 4 4

√ 4

33

.

47. x2 − 8x + 5 = 0

We use the quadratic formula. Here a = 1, b = −8, and c = 5. √ −b ± b2 − 4ac x= 2a −(−8) ± (−8)2 − 4 · 1 · 5 = 2·1 √ √ 8 ± 2 11 8 ± 44 = = 2 2 √ √ 2(4 ± 11) = = 4 ± 11 2 √ √ √ The solutions are 4 − 11 and 4 + 11, or 4 ± 11.

48. x2 − 6x + 3 = 0 −(−6) ± (−6)2 − 4 · 1 · 3 x= 2·1 √ √ 6 ± 24 6±2 6 = = 2 2 √ √ 2(3 ± 6) = =3± 6 2 √ √ √ The solutions are 3 − 6 and 3 + 6, or 3 ± 6. 49.

3x2 + x = 5 3x2 + x − 5 = 0

We use the quadratic formula. We have a = 3, b = 1, and c = −5. √ −b ± b2 − 4ac x= 2a −1 ± 12 − 4 · 3 · (−5) = 2·3 √ −1 ± 61 = 6 √ √ −1 − 61 −1 + 61 The solutions are and , or 6 6 √ −1 ± 61 . 6 50.

5x2 + 3x = 1 5x2 + 3x − 1 = 0 −3 ± 32 − 4 · 5 · (−1) x= 2·5 √ −3 ± 29 = 10 √ √ −3 − 29 −3 + 29 and , or The solutions are 10 10 √ −3 ± 29 . 10

Exercise Set 2.3 51.

135 Since b2 − 4ac > 0, there are two different real-number solutions.

2x2 + 1 = 5x 2x2 − 5x + 1 = 0

We use the quadratic formula. We have a = 2, b = −5, and c = 1. √ −b ± b2 − 4ac x= 2a √ −(−5) ± (−5)2 − 4 · 2 · 1 5 ± 17 = = 2·2 4 √ √ √ 5 + 17 5 ± 17 5 − 17 and , or . The solutions are 4 4 4 52.

56.

b2 − 4ac = (−12)2 − 4 · 4 · 9 = 0

There is one real-number solution. 57.

b2 − 4ac = 32 − 4 · 1 · 4 = −7

Since b2 − 4ac < 0, there are two different imaginarynumber solutions.

4x + 3 = x

53.

58.

There are two different imaginary-number solutions. 59.

b2 − 4ac = (−7)2 − 4 · 5 · 0 = 49

Since b2 − 4ac > 0, there are two different real-number solutions. 60.

3x2 + 3x = −4

3x2 + 3x + 4 = 0 √ −3 ± 32 − 4 · 3 · 4 x= 2·3 √ √ −3 ± 39i −3 ± −39 = = 6 6 √ 1 39 i =− ± 2 6 √ √ 1 39 1 39 i and − + i or The solutions are − − 2 6 2 6 √ 39 1 − ± i. 2 6 55.

4x2 = 8x + 5 4x2 − 8x − 5 = 0

a = 4, b = −8, c = −5

b2 − 4ac = (−8)2 − 4 · 4(−5) = 144

5t2 − 4t = 11

5t − 4t − 11 = 0 2

b − 4ac = (−4)2 − 4 · 5(−11) = 236 > 0 2

There are two different real-number solutions. 61.

x2 + 6x + 5 = 0 (x + 5)(x + 1) = 0 x+5 = 0

Setting f (x) = 0 Factoring

or x + 1 = 0

x = −5 or

=

54.

5t2 − 7t = 0

a = 5, b = −7, c = 0

5x2 + 2x + 2 = 0

2(−1 ± 3i) −1 ± 3i = 2·5 5 1 3 =− ± i 5 5 1 3 1 3 1 3 The solutions are − − i and − + i, or − ± i. 5 5 5 5 5 5

x2 − 2x + 4 = 0

b2 − 4ac = (−2)2 − 4 · 1 · 4 = −12 < 0

5x2 + 2x = −2

We use the quadratic formula. We have a = 5, b = 2, and c = 2. √ −b ± b2 − 4ac x= 2a √ −2 ± 22 − 4 · 5 · 2 = 2·5 √ −2 ± 6i −2 ± −36 = = 10 10

x2 + 3x + 4 = 0 a = 1, b = 3, c = 4

2

4x2 − x + 3 = 0 −(−1) ± (−1)2 − 4 · 4 · 3 x= 2·4 √ √ √ 1 ± −47 1 ± 47i 1 47 = = = ± i 8 8 8 8 √ √ √ 1 47 1 47 1 47 The solutions are − i and + i, or ± i. 8 8 8 8 8 8

4x2 − 12x + 9 = 0

x = −1

The zeros of the function are −5 and −1. 62.

x2 − x − 2 = 0

(x + 1)(x − 2) = 0 x+1 = 0

or x − 2 = 0

x = −1 or

x=2

The zeros of the function are −1 and 2. 63. x2 − 3x − 3 = 0

a = 1, b = −3, c = −3 √ −b ± b2 − 4ac x= 2a −(−3) ± (−3)2 − 4 · 1 · (−3) = 2·1 √ 3 ± 9 + 12 = 2 √ 3 ± 21 = 2 √ √ 3 + 21 3 − 21 and , or The zeros of the function are 2 2 √ 3 ± 21 . 2

136

Chapter 2: Functions, Equations, and Inequalities

64. 3x2 + 8x + 2 = 0 √ −8 ± 82 − 4 · 3 · 2 x= 2·3 √ √ −8 ± 2 10 −8 ± 40 = = 6 6 √ −4 ± 10 = 3 √ −4 − 10 The zeros of the function are and 3 √ √ −4 ± 10 −4 + 10 , or . 3 3

The zeros of the function are √ 1 ± 17 . 2 69.

66. x2 − 3x − 7 = 0 −(−3) ± (−3)2 − 4 · 1 · (−7) x= 2·1 √ 3 ± 37 = 2 √ √ 3 − 37 3 + 37 The zeros of the function are and , or 2 2 √ 3 ± 37 . 2 67. x2 + 2x − 5 = 0

a = 1, b = 2, c = −5 √ −b ± b2 − 4ac x= 2a −2 ± 22 − 4 · 1 · (−5) = 2·1 √ √ −2 ± 24 −2 ± 4 + 20 = = 2 2 √ √ −2 ± 2 6 = = −1 ± 6 2 √ √ The zeros √ of the function are −1 + 6 and −1 − 6, or −1 ± 6.

68. x2 − x − 4 = 0 x= =

−(−1) ±

1±

√ 2

17

(−1)2 − 4 · 1 · (−4) 2·1

√ 2

17

or

1−

√ 2

17

, or

2x2 − x + 4 = 0

a = 2, b = −1, c = 4 √ −b ± b2 − 4ac x= 2a −(−1) ± (−1)2 − 4 · 2 · 4 = 2·2 √ √ 1 ± 31i 1 ± −31 = = 4 4 √ 1 31 = ± i 4 4 √ √ 31 1 31 1 The zeros of the function are − i and + i, or 4 4 4 4 √ 31 1 ± i. 4 4

65. x2 − 5x + 1 = 0

a = 1, b = −5, c = 1 √ −b ± b2 − 4ac x= 2a −(−5) ± (−5)2 − 4 · 1 · 1 = 2·1 √ 5 ± 25 − 4 = 2 √ 5 ± 21 = 2 √ √ 5 + 21 5 − 21 and , or The zeros of the function are 2 2 √ 5 ± 21 . 2

1+

70. 2x2 + 3x + 2 = 0 √ −3 ± 32 − 4 · 2 · 2 x= 2·2 √ √ −3 ± −7 −3 ± 7i = = 4 4 √ 7 3 i =− ± 4 4

√ √ 3 7 3 7 The zeros of the function are − − i and − + i, 4 4 4 4 √ 3 7 i. or − ± 4 4

71.

3x2 − x − 1 = 0

a = 3, b = −1, c = −1 √ −b ± b2 − 4ac x= 2a −(−1) ± (−1)2 − 4 · 3 · (−1) = 2·3 √ 1 ± 13 = 6 √ √ 1 − 13 1 + 13 and , or The zeros of the function are 6 6 √ 1 ± 13 . 6 72. 3x2 + 5x + 1 = 0 √ −5 ± 52 − 4 · 3 · 1 x= 2·3 √ −5 ± 13 = 6

√ √ −5 − 13 −5 + 13 The zeros of the function are and , 6 6 √ −5 ± 13 . or 6

Exercise Set 2.3 73.

137

a = 5, b = −2, c = −1 √ −b ± b2 − 4ac x= 2a −(−2) ± (−2)2 − 4 · 5 · (−1) = 2·5 √ √ 2 ± 24 2±2 6 = = 10 10 √ √ 1± 6 2(1 ± 6) = = 2·5 5 √ √ 1− 6 1+ 6 The zeros of the function are and , or 5 5 √ 1± 6 . 5 74. 4x2 − 4x − 5 = 0 −(−4) ± (−4)2 − 4 · 4 · (−5) x= 2·4 √ √ 4±4 6 4 ± 96 = = 8 8 √ √ 1± 6 4(1 ± 6) = = 4·2 2 √ √ 1+ 6 1− 6 The zeros of the function are and , or 2 2 √ 1± 6 . 2 75.

u − 1 = 0 or u − 2 = 0

5x2 − 2x − 1 = 0

u = 1 or

Now substitute x for u and solve for x. x2 = 1 or x2 = 2 √ x = ±1 or x = ± 2 √ √ The solutions are −1, 1, − 2, and 2. 78.

The zeros of the function are √ −3 ± 57 or . 8

Let u = x2 . u2 − 4u + 3 = 0

u − 1 = 0 or u − 3 = 0 u = 1 or

u=3

Substitute x for u and solve for x. x2 = 1 or x2 = 3 √ x = ±1 or x = ± 3 √ √ The solutions are −1, 1, − 3, and 3. 2

79.

x4 + 3x2 = 10 x4 + 3x2 − 10 = 0

Let u = x2 . u2 + 3u − 10 = 0

Substituting u for x2

(u + 5)(u − 2) = 0 u+5 = 0

or u − 2 = 0

u=2

Now substitute x for u and solve for x. or x2 = 2 x2 = −5 √ √ x = ± 5i or x = ± 2 √ √ √ √ The solutions are − 5i, 5i, − 2, and 2. 80. √

√

−3 + 57 −3 − 57 and , 8 8

77. x4 − 3x2 + 2 = 0

(u − 1)(u − 2) = 0

Substituting u for x2

(u − 1)(u − 3) = 0

u = −5 or

76. x2 + 6x − 3 = 0 −6 ± 62 − 4 · 1 · (−3) x= 2·1 √ √ −6 ± 48 −6 ± 4 3 = = 2 2 √ √ 2(−3 ± 2 3) = −3 ± 2 3 = 2 √ √ The zeros √ of the function are −3 − 2 3 and −3 + 2 3, or −3 ± 2 3. Let u = x2 . u2 − 3u + 2 = 0

x4 + 3 = 4x2 x4 − 4x2 + 3 = 0

4x2 + 3x − 3 = 0

a = 4, b = 3, c = −3 √ −b ± b2 − 4ac x= 2a −3 ± 32 − 4 · 4 · (−3) = 2·4 √ −3 ± 57 = 8

u=2

2

2

x4 − 8x2 = 9

x4 − 8x2 − 9 = 0

Let u = x2 . u2 − 8u − 9 = 0

Substituting u for x2

(u − 9)(u + 1) = 0

u − 9 = 0 or u + 1 = 0 u = 9 or

u = −1

Now substitute x for u and solve for x. x2 = 9 or x2 = −1 2

x = ±3 or

x = ±i

The solutions are −3, 3, i, and −i. 81. y 4 + 4y 2 − 5 = 0 Let u = y 2 .

u2 + 4u − 5 = 0

Substituting u for y 2

(u − 1)(u + 5) = 0 Substituting u for x2

u − 1 = 0 or u + 5 = 0 u = 1 or

u = −5

138

Chapter 2: Functions, Equations, and Inequalities Now substitute m1/3 for u and solve for m. m1/3 = −2 or m1/3 = 4

Substitute y 2 for u and solve for y. y2 = 1

or y 2 = −5 √ y = ±1 or y = ± 5i

(m1/3 )3 = (−2)3 or (m1/3 )3 = 43 Using the principle of powers m = −8 or m = 64

√ √ The solutions are −1, 1, − 5i, and 5i. 82. y 4 − 15y 2 − 16 = 0

The solutions are −8 and 64.

Let u = y 2 .

u − 15u − 16 = 0

Substituting u for y

2

2

86. t2/3 + t1/3 − 6 = 0

Let u = t1/3 . u2 + u − 6 = 0

(u − 16)(u + 1) = 0

u − 16 = 0

or u + 1 = 0

u = 16 or

(u + 3)(u − 2) = 0

u = −1

u+3 = 0

Now substitute y 2 for u and solve for y.

u = −3 or

y 2 = 16 or y 2 = −1 y = ±4 or

u − 3u − 4 = 0

Substituting u for

2

t = −27 or

√

x

87. x1/2 − 3x1/4 + 2 = 0 Let u = x1/4 . u2 − 3u + 2 = 0

u = −1 or u=4 √ Now substitute x for u and solve for x. √ √ x = −1 or x=4

No solution x = 16 √ √ Note that x must be nonnegative, so x = −1 has no solution. The number 16 checks and is the solution. The solution is 16. √ 84. 2x − 9 x + 4 = 0 √ Let u = x. √ 2u2 − 9u + 4 = 0 Substituting u for x (2u − 1)(u − 4) = 0

Both numbers check. The solutions are

u+2 = 0

or u − 4 = 0

u = −2 or

u=4

u = 1 or

u=2

Now substitute x1/4 for u and solve for x. x1/4 = 2 x1/4 = 1 or (x1/4 )4 = 14 or (x1/4 )4 = 24 x=1

x = 16

or

The solutions are 1 and 16. 88.

x1/2 − 4x1/4 = −3

x1/2 − 4x1/4 + 3 = 0

Let u = x1/4 . u2 − 4u + 3 = 0

u − 1 = 0 or u − 3 = 0 u = 1 or

u=3

for u and solve for x. Substitute x x1/4 = 1 or x1/4 = 3 1/4

x = 1 or 1 and 16. 4

Substituting u for m1/3

(u + 2)(u − 4) = 0

u − 1 = 0 or u − 2 = 0

(u − 1)(u − 3) = 0

85. m2/3 − 2m1/3 − 8 = 0 Let u = m1/3 . u2 − 2u − 8 = 0

Substituting u for x1/4

(u − 1)(u − 2) = 0

or u − 4 = 0

2u − 1 = 0 or u − 4 = 0 1 or u=4 u= 2 √ Substitute x for u and solve for u. √ √ 1 x= or x=4 2 1 x= or x = 16 4

t=8

The solutions are −27 and 8.

(u + 1)(u − 4) = 0

u+1 = 0

u=2

Substitute t1/3 for u and solve for t. t1/3 = −3 or t1/3 = 2

y = ±i

The solutions are −4, 4, −i and i. √ 83. x − 3 x − 4 = 0 √ Let u = x.

or u − 2 = 0

x = 81

The solutions are 1 and 81. 89. (2x − 3)2 − 5(2x − 3) + 6 = 0 Let u = 2x − 3. u2 − 5u + 6 = 0

(u − 2)(u − 3) = 0

Substituting u for 2x − 3

u − 2 = 0 or u − 3 = 0 u = 2 or

u=3

Now substitute 2x − 3 for u and solve for x.

Exercise Set 2.3

139

2x − 3 = 2 or 2x − 3 = 3 2x = 5 or 5 x= or 2

5±

2x = 6

5 and 3. 2

Carry out. We solve the equation. 2120 = 16t2

Substituting u for 3x + 2

(u + 8)(u − 1) = 0

u+8 = 0

or u − 1 = 0

3x = −10 or 10 or x=− 3

3x = −1 1 x=− 3 10 1 The solutions are − and − . 3 3 91. (2t2 + t)2 − 4(2t2 + t) + 3 = 0

Substituting u for 2t2 + t

(u − 1)(u − 3) = 0

u − 1 = 0 or u − 3 = 0 u = 1 or

u=3

Now substitute 2t2 + t for u and solve for t. 2t2 + t = 1 or 2t2 + t = 3 2t2 + t − 1 = 0 or

2t2 + t − 3 = 0

(2t − 1)(t + 1) = 0 or (2t + 3)(t − 1) = 0

2t−1=0 or t+1=0 or 2t+3=0 or t−1 = 0 3 1 t=−1 or t=− or t=1 t= or 2 2 3 1 The solutions are , −1, − and 1. 2 2 92. 12 = (m2 − 5m)2 + (m2 − 5m) 0 = (m2 − 5m)2 + (m2 − 5m) − 12

Let u = m2 − 5m. 0 = u2 + u − 12

0 = (u + 4)(u − 3)

u+4 = 0

Substituting u for m2 − 5m

Dividing by 16 on both sides

11.5 ≈ t

Taking the square root on both sides

State. It would take an object about 11.5 sec to reach the ground. 94.

Solve: 2063 = 16t2 t ≈ 11.4 sec

95. Substitute 9.7 for w(x) and solve for x. −0.01x2 + 0.27x + 8.60 = 9.7 −0.01x2 + 0.27x − 1.1 = 0

a = −0.01, b = 0.27, c = −1.1 √ −b ± b2 − 4ac x= 2a −0.27 ± (0.27)2 − 4(−0.01)(−1.1) = 2(−0.01) √ −0.27 ± 0.0289 = −0.02 x ≈ 5 or x ≈ 22

There were 9.7 million self-employed workers in the United States 5 years after 1980 and also 22 years after 1980, or in 1985 and in 2002. 96. Solve: −0.01x2 + 0.27x + 8.60 = 9.1

x ≈ 2 or x ≈ 25, so there were or will be 9.1 self-employed workers in the United States 2 years after 1980, or in 1982, and 25 years after 1980, or in 2005.

97. Familiarize. Let w = the width of the rug. Then w + 1 = the length. Translate. We use the Pythagorean equation. w2 + (w + 1)2 = 52

or u − 3 = 0

u = −4 or

132.5 = t2

Check. When t = 11.5, s = 16(11.5)2 = 2116 ≈ 2120. The answer checks.

u = −8 or u=1 Substitute 3x + 2 for u and solve for x. 3x + 2 = −8 or 3x + 2 = 1

Let u = 2t2 + t. u2 − 4u + 3 = 0

Carry out. We solve the equation.

u=3

Substitute m − 5m for u and solve for m. m2 −5m = −4 or m2 −5m = 3

w2 + (w + 1)2 = 52

2

m −5m+4 = 0 or m −5m−3 = 0 2

2

(m−1)(m−4) = 0 or

-

(−5)2 −4·1(−3) 2·1 √ 5 ± 37 m = 1 or m = 4 or m = 2 √ √ 5 + 37 5 − 37 , and , or 1, 4, and The solutions are 1, 4, 2 2 m=

.

2120 = 16t2

90. (3x + 2)2 + 7(3x + 2) − 8 = 0 Let u = 3x + 2. u2 + 7u − 8 = 0

37

2

93. Familiarize and Translate. We will use the formula s = 16t2 , substituting 2120 for s.

x=3

The solutions are

√

−(−5) ±

w + w2 + 2w + 1 = 25 2

2w2 + 2w + 1 = 25 2w2 + 2w − 24 = 0

2(w + 4)(w − 3) = 0

w+4 = 0

or w − 3 = 0

w = −4 or

w=3

Since the width cannot be negative, we consider only 3. When w = 3, w + 1 = 3 + 1 = 4.

140

Chapter 2: Functions, Equations, and Inequalities Check. The length, 4 ft, is 1 ft more than the width, 3 ft. The length of√a diagonal of √ with width 3 ft and √ a rectangle length 4 ft is 32 + 42 = 9 + 16 = 25 = 5. The answer checks. State. The length is 4 ft, and the width is 3 ft.

98. Let x = the length of the longer leg. 2

2

Only 12 has meaning in the original problem. The length of one leg is 12 cm, and the length of the other leg is 12−7, or 5 cm. 99. Familiarize. Let n = the smaller number. Then n + 5 = the larger number. Translate. The product of the numbers is 36. "# $ ! ↓ ↓ ↓ n(n + 5) = 36

or x − 2 = 0

x = 13 or

x=2

Check. When x = 13, both 20 − 2x and 10 − 2x are negative numbers, so we only consider x = 2. When x = 2, then 20 − 2x = 20 − 2 · 2 = 16 and 10 − 2x = 10 − 2 · 2 = 6, and the area of the base is 16 · 6, or 96 cm2 . The answer checks.

Solve: (32 − 2w)(28 − 2w) = 192 w = 8 or w = 22

n2 + 5n = 36

Only 8 has meaning in the original problem. The width of the frame is 8 cm.

n2 + 5n − 36 = 0

(n + 9)(n − 4) = 0

103. Familiarize. We have P = 2l + 2w, or 28 = 2l + 2w. Solving for w, we have

or n − 4 = 0

n=4

28 = 2l + 2w

If n = −9, then n + 5 = −9 + 5 = −4. If n = 4, then n + 5 = 4 + 5 = 9. Check. The number −4 is 5 more than −9 and (−4)(−9) = 36, so the pair −9 and −4 check. The number 9 is 5 more than 4 and 9 · 4 = 36, so the pair 4 and 9 also check. State. The numbers are −9 and −4 or 4 and 9. 100. Let n = the larger number. Solve: n(n − 6) = 72

14 = l + w

Dividing by 2

14 − l = w.

Then we have l = the length of the rug and 14 − l = the width, in feet. Recall that the area of a rectangle is the product of the length and the width. Translate. The area ! "#  $ & l(14 − l)

is !48"#ft2$.   & & = 48

Carry out. We solve the equation.

n = −6 or n = 12

When n = −6, then n − 6 = −6 − 6 = −12, so one pair of numbers is −6 and −12. When n = 12, then n − 6 = 12 − 6 = 6, so the other pair of numbers is 6 and 12. 101. Familiarize. We add labels to the drawing in the text. We let x represent the length of a side of the square in each corner. Then the length and width of the resulting base are represented by 20 − 2x and 10 − 2x, respectively. Recall that for a rectangle, Area = length × width. x

x

x

x

x

10 − 2x 20 − 2x x

x

x − 13 = 0

102. Let w = the width of the frame.

n(n + 5) = 36

10 cm

(x − 13)(x − 2) = 0

State. The length of the sides of the squares is 2 cm.

Carry out.

n = −9 or

200 − 60x + 4x2 = 96 x2 − 15x + 26 = 0

x = −5 or x = 12

n+9 = 0

Carry out. We solve the equation. 4x2 − 60x + 104 = 0

Solve: x + (x − 7) = 13 2

Translate. The area "# of the base$ is 96 cm2$. ! ! "# (20 − 2x)(10 − 2x) = 96

x

20 cm

l(14 − l) = 48

14l − l2 = 48

0 = l2 − 14l + 48

0 = (l − 6)(l − 8)

l − 6 = 0 or l − 8 = 0 l = 6 or

l=8

If l = 6, then 14 − l = 14 − 6 = 8. If l = 8, then 14 − l = 14 − 8 = 6.

In either case, the dimensions are 8 ft by 6 ft. Since we usually consider the length to be greater than the width, we let 8 ft = the length and 6 ft = the width. Check. The perimeter is 2·8 ft+2·6 ft = 16 ft +12 ft = 28 ft. The answer checks. State. The length of the rug is 8 ft, and the width is 6 ft.

Exercise Set 2.3

141

104. We have 170 = 2l + 2w, so w = 85 − l. Solve: l(85 − l) = 1750 l = 35 or l = 50

Choosing the larger number to be the length, we find that the length of the petting area is 50 m, and the width is 35 m. 105. f (x) = 4 − 5x = −5x + 4

The function can be written in the form y = mx + b, so it is a linear function.

The solutions are −8 and −0.4. 113. Graph y = 7x2 − 43x + 6 and use the Zero feature twice.

106. f (x) = 4 − 5x2 = −5x2 + 4

The function can be written in the form f (x) = ax2 +bx+c, a %= 0, so it is a quadratic function.

107. f (x) = 7x2 The function is in the form f (x) = ax2 + bx + c, a %= 0, so it is a quadratic function. 108. f (x) = 23x + 6 The function is in the form f (x) = mx + b, so it is a linear function. 109. f (x) = 1.2x − (3.6)2

The function is in the form f (x) = mx + b, so it is a linear function.

110. f (x) = 2 − x − x2 = −x2 − x + 2

The function can be written in the form f (x) = ax2 +bx+c, a %= 0, so it is a quadratic function.

One solution is approximately 0.143 and the other is 6. 114. Graph y = 10x2 − 23x + 12 and use the Zero feature twice.

111. Graph y = x2 − 8x + 12 and use the Zero feature twice.

The solutions are 0.8 and 1.5.

The solutions are 2 and 6. 112. Graph y = 5x2 + 42x + 16 and use the Zero feature twice.

115. Graph y1 = 6x + 1 and y2 = 4x2 and use the Intersect feature twice.

142

Chapter 2: Functions, Equations, and Inequalities

The solutions are approximately −0.151 and 1.651. 116. Graph y1 = 3x2 + 5x and y2 = 3 and use the Intersect feature twice.

The solutions are approximately −2.135 and 0.468. 117. Graph y = 2x2 − 5x − 4 and use the Zero feature twice.

The zeros are approximately −0.637 and 3.137. 118. Graph y = 4x2 − 3x − 2 and use the Zero feature twice.

The zeros are approximately −0.425 and 1.175. 119. Graph y = 3x2 + 2x − 4 and use the Zero feature twice.

The zeros are approximately −1.535 and 0.869. 120. Graph y = 9x2 − 8x − 7 and use the Zero feature twice.

The zeros are approximately −0.543 and 1.432. 121. Graph y = 5.02x2 − 4.19x − 2.057 and use the Zero feature twice.

Exercise Set 2.3

143 The last equation is equivalent to the original equation, so the equation is symmetric with respect to the y-axis. Test for symmetry with respect to the origin: 3x2 + 4y 2 = 5 Original equation 3(−x) + 4(−y)2 = 5 Replacing x by −x and y by −y 3x2 + 4y 2 = 5 Simplifying 2

The zeros are approximately −0.347 and 1.181. 122. Graph y = 1.21x2 − 2.34x − 5.63 and use the Zero feature twice.

The last equation is equivalent to the original equation, so the equation is symmetric with respect to the origin. 128. Test for symmetry with respect to the x-axis: y 3 = 6x2

Original equation

(−y)3 = 6x2

Replacing y by −y

−y = 6x

Simplifying

3

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis: y 3 = 6x2

Original equation

y 3 = 6(−x)2

Replacing x by −x

y = 6x 3

Simplifying

2

The last equation is equivalent to the original equation, so the equation is symmetric with respect to the y-axis. Test for symmetry with respect to the origin: y 3 = 6x2

The zeros are approximately −1.397 and 3.331. 123. No; consider the quadratic formula √ −b −b ± b2 − 4ac . If b2 − 4ac = 0, then x = , so x= 2a 2a √ 2 2 there is one real zero. If b − 4ac > 0, then b − 4ac is a real number and there are two real zeros. If b2 − 4ac < 0, √ then b2 − 4ac is an imaginary number and there are two imaginary zeros. Thus, a quadratic function cannot have one real zero and one imaginary zero. 124. Use the discriminant. If b − 4ac < 0, there are no xintercepts. If b2 − 4ac = 0, there is one x-intercept. If b2 − 4ac > 0, there are two x-intercepts.

126. 2010 − 1980 = 30

a(30) = 9096(30) + 387, 725 = 660, 605 associate’s degrees

127. Test for symmetry with respect to the x-axis: 3x2 + 4y 2 = 5 3x2 + 4(−y)2 = 5 3x2 + 4y 2 = 5

Original equation Replacing y by −y Simplifying

The last equation is equivalent to the original equation, so the graph is symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis: 3x + 4y = 5 2

2

3(−x)2 + 4y 2 = 5 3x + 4y = 5 2

2

Original equation Replacing x by −x Simplifying

2

−y 3 = 6x2

Replacing x by −x y by −y Simplifying

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. 129.

f (x) = 2x3 − x

f (−x) = 2(−x)3 − (−x) = −2x3 + x

−f (x) = −2x3 + x

2

125. 1998 − 1980 = 18, so we substitute 18 for x. a(18) = 9096(18) + 387, 725 = 551, 453 associate’s degrees

Original equation

(−y) = 6(−x) 3

f (x) %= f (−x) so f is not even

130.

f (−x) = −f (x), so f is odd. f (x) = 4x2 + 2x − 3

f (−x) = 4(−x)2 + 2(−x) − 3 = 4x2 − 2x − 3

−f (x) = −4x2 − 2x + 3

f (x) %= f (−x) so f is not even

f (−x) %= −f (x), so f is not odd.

Thus f (x) = 4x2 + 2x − 3 is neither even nor odd. 131. a)

kx2 − 17x + 33 = 0

k(3)2 − 17(3) + 33 = 0 Substituting 3 for x 9k − 51 + 33 = 0

9k = 18 k=2

144

Chapter 2: Functions, Equations, and Inequalities b)

2x2 − 17x + 33 = 0

b) x2 − (6 + 3i)x + 9 + 9i = 0 −[−(6+3i)] ± [−(6+3i)]2 −4(1)(9+9i) x= 2·1 √ 6 + 3i ± 36 + 36i − 9 − 36 − 36i x= 2 √ 6 + 3i ± −9 6 + 3i ± 3i x= = 2 2 6 + 3i − 3i 6 + 3i + 3i or x = x= 2 2 6 6 + 6i or x = x= 2 2 x = 3 + 3i or x = 3

Substituting 2 for k

(2x − 11)(x − 3) = 0

2x − 11 = 0 or x − 3 = 0 11 x= or x=3 2 11 . The other solution is 2 132. a)

kx2 − 2x + k = 0

k(−3) − 2(−3) + k = 0 2

9k + 6 + k = 0

Substituting −3 for x

10k = −6 3 k=− 5

The other solution is 3 + 3i.

3 3 3 b) − x2 − 2x − = 0 Substituting − for k 5 5 5 3x2 + 10x + 3 = 0 Multiplying by −5

135.

(x − 2)3 = x3 − 2

x − 6x + 12x − 8 = x3 − 2 3

2

0 = 6x2 − 12x + 6

0 = 6(x2 − 2x + 1)

(3x + 1)(x + 3) = 0

3x + 1 = 0

3x = −1 or 1 x = − or 3

x − 1 = 0 or x − 1 = 0

x = −3

x = 1 or

x = −3

1 The other solution is − . 3 133. a)

0 = 6(x − 1)(x − 1)

or x + 3 = 0

(1 + i)2 − k(1 + i) + 2 = 0 1 + 2i − 1 − k − ki + 2 = 0

136.

x3 + 3x2 + 3x + 1 = x3 − 3x2 + 3x + 25 6x2 − 24 = 0

2 + 2i = k + ki

6(x2 − 4) = 0

2(1 + i) = k(1 + i)

6(x + 2)(x − 2) = 0

2=k

x+2 = 0

b) x2 − 2x + 2 = 0

134. a)

x2 − (6 + 3i)x + k = 0

3 − (6 + 3i) · 3 + k = 0 Substituting 3 for x 2

9 − 18 − 9i + k = 0

k = 9 + 9i

(x + 1)3 = (x − 1)3 + 26

x3 + 3x2 + 3x + 1 = x3 − 3x2 + 3x − 1 + 26

Substituting 1 + i for x

Substituting 2 for k −(−2) ± (−2)2 − 4 · 1 · 2 x= 2·1 √ 2 ± −4 = 2 2 ± 2i = =1±i 2 The other solution is 1 − i.

x=1

The solution is 1.

or x − 2 = 0

x = −2 or

x=2

The solutions are −2 and 2. 137.

(6x3 + 7x2 − 3x)(x2 − 7) = 0 x(6x2 + 7x − 3)(x2 − 7) = 0

x(3x − 1)(2x + 3)(x2 − 7) = 0

x=0 or 3x − 1=0 or 2x + 3=0 or x2 − 7 = 0 √ 3 1 x=− or x = 7 or x=0 or x= or 3 2 √ x=− 7 √ √ 3 1 The exact solutions are − 7, − , 0, , and 7. 2 3 ,+ , + ,+ , + 1 1 1 1 2 2 138. x − + x− x + =0 x− 5 4 5 8 ,+ , + 1 1 x− 2x2 − =0 5 8 , + ,+ , + 1 1 1 (2) x + x− =0 x− 5 4 4 1 1 1 x= or x = − or x = 5 4 4 1 1 1 The solutions are − , , and . 4 5 4

Exercise Set 2.3 139. x2 + x −

145

√

2=0 √ −b ± b2 − 4ac x= 2a . √ √ −1 ± 12 − 4 · 1(− 2) −1 ± 1 + 4 2 = = 2·1 2 √ −1 ± 1 + 4 2 The solutions are . 2 √ √ 140. x2 + 5x − 3 = 0 √ Use the √ quadratic formula. Here a = 1, b = 5, and c = − 3. √ −b ± b2 − 4ac x= 2a .√ √ √ − 5 ± ( 5)2 − 4 · 1(− 3) = 2·1 √ √ − 5± 5+4 3 = 2 √ √ − 5± 5+4 3 The solutions are . 2 141.

2t2 + (t − 4)2 = 5t(t − 4) + 24

2t + t2 − 8t + 16 = 5t2 − 20t + 24 2

0 = 2t2 − 12t + 8 0 = t2 − 6t + 4

Dividing by 2

Use the quadratic formula. √ −b ± b2 − 4ac t= 2a −(−6) ± (−6)2 − 4 · 1 · 4 = 2·1 √ √ 6±2 5 6 ± 20 = = 2 2 √ √ 2(3 ± 5) = =3± 5 2 √ The solutions are 3 ± 5. 142.

9t(t + 2) − 3t(t − 2) = 2(t + 4)(t + 6) 9t2 + 18t − 3t2 + 6t = 2t2 + 20t + 48 4t2 + 4t − 48 = 0

4(t + 4)(t − 3) = 0

t+4 = 0

or t − 3 = 0

t = −4 or

t=3

The solutions are −4 and 3. √ √ 143. x − 3 − 4 x − 3 = 2 √ Substitute u for 4 x − 3. u2 − u − 2 = 0

(u − 2)(u + 1) = 0

u − 2 = 0 or u + 1 = 0 u = 2 or

u = −1

√ Substitute 4 x − 3 for u and solve for x. √ √ 4 x − 3 = 2 or 4 x − 3 = 1 x − 3 = 16

No solution

x = 19

The value checks. The solution is 19. 144. x6 − 28x3 + 27 = 0

Substitute u for x3 . u2 − 28u + 27 = 0

(u − 27)(u − 1) = 0

u = 27 or u = 1

Substitute x3 for u and solve for x. x3 = 27 or x − 27 = 0 or

x3 = 1

3

x −1 = 0 3

(x−3)(x +3x+9) = 0 or (x−1)(x +x+1) = 0 2

2

Using the principle of zero products and, where necessary, the quadratic formula, we √ √ find that the solutions are 1 3 3 3 3 i, 1, and − ± i. 3, − ± 2 2 2 2 ,2 + 6 2 145. y+ + 3y + = 4 y y ,2 + , + 2 2 +3 y+ y+ −4 = 0 y y 2 Substitute u for y + . y u2 + 3u − 4 = 0

(u + 4)(u − 1) = 0

u = −4 or u = 1 2 Substitute y + for u and solve for y. y 2 2 y+ =1 y + = −4 or y y y 2 + 2 = −4y or y2 + 2 = y

or y 2 − y + 2 = 0 y 2 + 4y + 2 = 0 √ −4± 42 −4·1·2 or y= 2·1 −(−1)± (−1)2 −4·1·2 y= 2·1 √ √ 1 ± −7 −4 ± 8 y= or y = 2 2 √ √ 1 ± 7i −4 ± 2 2 or y = y= 2 2 √ √ 1 7 i y = −2 ± 2 or y = ± 2 2 √ √ 7 1 i. The solutions are −2 ± 2 and ± 2 2

146 146.

Chapter 2: Functions, Equations, and Inequalities √ x2 + 3x + 1 − x2 + 3x + 1 = 8 √ x2 + 3x + 1 − x2 + 3x + 1 − 8 = 0

u2 − u − 8 = 0 √ √ 1 + 33 1 − 33 u= or u = 2 2 √ 1 + 33 x2 + 3x + 1 = or 2 √ 1 − 33 x2 + 3x + 1 = 2 √ 34 + 2 33 x2 + 3x + 1 = or 4 √ 34 − 2 33 2 x + 3x + 1 = 4 √ −15 − 33 x2 + 3x + = 0 or 2 √ −15 + 33 =0 x2 + 3x + 2 √ −3 ± 39 + 2 33 x= or 2 √ −3 ± 39 − 2 33 x= 2 √ −3 ± 39 + 2 33 Only checks. The solutions are 2 √ −3 ± 39 + 2 33 . 2 147.

1 at + v0 t + x0 = 0 2

1 Use the quadratic formula. Here a = a, b = v0 , and 2 c = x0 . / 1 −v0 ± (v0 )2 − 4 · a · x0 2 t= 1 2· a 2 2 −v0 ± v0 − 2ax0 t= a

Exercise Set 2.4 value 1. a) The minimum function , occurs at the vertex, + 1 9 so the vertex is − , − . 2 4 b) The axis of symmetry is a vertical line through the 1 vertex. It is x = − . 2 9 c) The minimum value of the function is − . 4 + , 1 25 2. a) − , 2 4 1 b) Axis of symmetry: x = − 2 25 c) Maximum: 4

3.

f (x) = x2 − 8x + 12

16 completes the square for x2 − 8x. = x2 − 8x + 16 − 16 + 12 Adding 16−16 on the right side = (x2 − 8x + 16) − 16 + 12 = (x − 4)2 − 4

Factoring and simplifying

= (x − 4)2 + (−4)

Writing in the form f (x) = a(x − h)2 + k

a) Vertex: (4, −4)

b) Axis of symmetry: x = 4 c) Minimum value: −4

d) We plot the vertex and find several points on either side of it. Then we plot these points and connect them with a smooth curve. x f (x)

4.

4

−4

2

0

1

5

5

−3

6

0

y 4 2 !2

2

4

6

!4

x

f(x) ! x 2 " 8x # 12

g(x) = x2 + 7x − 8 + 49 49 1 = x2 + 7x + − −8 ·7= 4 4 2 + ,2 7 = 2 ,2 + 81 7 − = x+ 2 4 0 + ,12 + , 81 7 + − = x− − 2 4 , + 7 81 a) Vertex: − ,− 2 4 7 b) Axis of symmetry: x = − 2 81 c) Minimum value: − 4 d) y 10 !8

8

!2

!4

4

6

!10 !20 !30

g(x) ! x 2 # 7x " 8

x

7 and 2 , 49 4

Exercise Set 2.4 5.

f (x) = x2 − 7x + 12

147 49 completes the 4 square for x2 − 7x.

d)

y 4

49 49 − + 12 Adding 4 4 49 49 − on the right side 4 4 + , 49 49 = x2 − 7x + − + 12 4 4 ,2 + 1 7 − Factoring and = x− 2 4 simplifying ,2 + , + 7 1 = x− + − Writing in the 2 4

= x2 − 7x +

a) b) c) d)

!2

−

4

0

5

2

3

0

7.

1

6

f (x) = x2 + 4x + 5

4 completes the square for x2 + 4x = x2 + 4x + 4 − 4 + 5 Adding 4 − 4 on the right side = (x + 2)2 + 1 Factoring and simplifying = [x − (−2)]2 + 1

a) Vertex: (−2, 1) c) Minimum value: 1

d) We plot the vertex and find several points on either side of it. Then we plot these points and connect them with a smooth curve.

6 4 2 2

4

6

x

f(x) ! x 2 " 7x # 12

g(x) = x − 5x + 6 + 25 25 1 5 = x2 − 5x + − +6 (−5) = − 4 4 2 2 ,2 , + 25 5 = and − 2 4 + ,2 1 5 = x− − 2 4 ,2 + , + 1 5 + − = x− 2 4 , + 5 1 a) Vertex: ,− 2 4 5 b) Axis of symmetry: x = 2 1 c) Minimum value: − 4

Writing in the form f (x) = a(x − h)2 + k

b) Axis of symmetry: x = −2

8

!2

x

g(x) ! x 2 " 5x # 6

y

1 4

4

!4

form f (x) = a(x − h)2 + k , 7 1 ,− Vertex: 2 4 7 Axis of symmetry: x = 2 1 Minimum value: − 4 We plot the vertex and find several points on either side of it. Then we plot these points and connect them with a smooth curve. 7 2

2 !2

+

x f (x)

6.

2

8.

x

f (x)

y

−2

1

16

−1

2

12

0

5

−3

2

−4

5

8 4 !6

!4

f(x) !

!2

x2

2

x

# 4x # 5

f (x) = x2 + 2x + 6

2

= x2 + 2x + 1 − 1 + 6

+

= (x + 1)2 + 5

= [x − (−1)]2 + 5

a) Vertex: (−1, 5)

b) Axis of symmetry: x = −1 c) Minimum value: 5 d)

y 16 12 8 4 !4

!2

f(x) !

2

x2

4

# 2x # 6

x

, 1 · 2 = 1 and 12 = 1 2

148 9.

Chapter 2: Functions, Equations, and Inequalities x2 + 4x + 6 2 1 = (x2 + 8x) + 6 2

11.

g(x) =

= 2(x2 + 3x) + 8

Factoring 2 out of the first two terms , + 9 9 +8 Adding = 2 x2 + 3x + − 4 4 9 9 − inside the parentheses 4 4 , + 9 9 2 − 2 · + 8 Removing = 2 x + 3x + 4 4 9 − from within the parentheses 4 ,2 + 7 3 = 2 x+ + Factoring and 2 2 simplifying

1 out of the 2 first two terms

Factoring

1 2 (x +8x+16−16)+6 Adding 16 − 16 inside 2 the parentheses 1 1 2 = (x +8x+16)− ·16+6 Removing −16 from 2 2 within the parentheses 1 2 = (x + 4) − 2 Factoring and simplifying 2 1 = [x − (−4)]2 + (−2) 2 a) Vertex: (−4, −2) =

b) Axis of symmetry: x = −4

,12 0 + 7 3 + = 2 x− − 2 2 + , 3 7 a) Vertex: − , 2 2

c) Minimum value: −2

d) We plot the vertex and find several points on either side of it. Then we plot these points and connect them with a smooth curve.

10.

b) Axis of symmetry: x = −

y

x

g(x)

−4

−2

8

−2

0

4

0

6

−6

0

!8

−8

6

2

!8

!4

7 2 d) We plot the vertex and find several points on either side of it. Then we plot these points and connect them with a smooth curve.

4

8

x

!4

x g(x) ! " # 4x # 6 2

x − 2x + 1 3 1 = (x2 − 6x) + 1 3 1 2 = (x − 6x + 9 − 9) + 1 3 1 1 2 = (x − 6x + 9) − · 9 + 1 3 3 1 = (x − 3)2 − 2 3 1 = (x − 3) + (−2) 3 a) Vertex: (3, −2)

b) Axis of symmetry: x = 3 c) Minimum value: −2 d)

y 8 4 !8

!4

4

8

!4 !8

x2 g(x) ! " " 2x # 1 3

x

3 2

c) Minimum value:

2

g(x) =

g(x) = 2x2 + 6x + 8

12.

x

f (x)

y

3 − 2

7 2

16

−1

4

0

8

−2

4

!4

−3

8

g(x) ! 2x 2 # 6x # 8

f (x) = 2x2 − 10x + 14

12 8 4 !2

2

= 2(x2 − 5x) + 14 , + 25 25 = 2 x2 − 5x + − + 14 4 4 , + 25 25 −2· + 14 = 2 x2 − 5x + 4 4 + ,2 3 5 = 2 x− + 2 2 , + 5 3 , a) Vertex: 2 2 5 b) Axis of symmetry: x = 2 3 c) Minimum value: 2

4

x

Exercise Set 2.4 d)

149 d)

y

y

16

20

12

10

8

!8

!4

!2

2

4

8

x

f(x) ! "x 2 " 8x # 5

15.

f (x) = −x2 − 6x + 3

= −(x2 + 6x) + 3 9 completes the square for x2 + 6x. = −(x2 + 6x + 9 − 9) + 3 = −(x + 3)2 − (−9) + 3 = −(x + 3)2 + 9 + 3

g(x) = −2x2 + 2x + 1

= −2(x2 − x) + 1

Removing −9 from the parentheses

= −[x − (−3)]2 + 12

a) Vertex: (−3, 12)

b) Axis of symmetry: x = −3 c) Maximum value: 12

d) We plot the vertex and find several points on either side of it. Then we plot these points and connect them with a smooth curve.

14.

4

!20

x

f(x) ! 2x 2 " 10x # 14

13.

!4 !10

4

a)

y

x

f (x)

−3

12

8

0

3

4

1

−4

−6

3

−7

−4

b)

!8

!4

c) 4

8

d)

x

!4 !8

f(x) ! "x 2 " 6x # 3

f (x) = −x2 − 8x + 5

= −(x + 8x) + 5 2

= −(x2 + 8x + 16 − 16) + 5 '1 ( · 8 = 4 and 42 = 16 2 = −(x2 + 8x + 16) − (−16) + 5 = −(x2 + 8x + 16) + 21 = −[x − (−4)]2 + 21

a) Vertex: (−4, 21)

b) Axis of symmetry: x = −4 c) Maximum value: 21

16.

Factoring −2 out of the

first two terms , + 1 1 1 1 +1 Adding − = −2 x2 −x+ − 4 4 4 4 inside the parentheses , + , + 1 1 = −2 x2 −x+ −2 − +1 4 4 1 Removing − from within the parentheses 4 ,2 + 3 1 + = −2 x − 2 2 , + 1 3 , Vertex: 2 2 1 Axis of symmetry: x = 2 3 Maximum value: 2 We plot the vertex and find several points on either side of it. Then we plot these points and connect them with a smooth curve. x

f (x)

1 2

3 2

1

1

2

−3

0

1

−1

−3

f (x) = −3x2 − 3x + 1

y 4 2 !4

!2

2

4

!2 !4

g(x) ! "2x 2 # 2x # 1

= −3(x2 + x) + 1 + , 1 1 2 = −3 x + x + − +1 4 4 , + , + 1 1 −3 − +1 = −3 x2 + x + 4 4 ,2 + 7 1 + = −3 x + 2 4 ,12 0 + 7 1 + = −3 x − − 2 4

x

150

Chapter 2: Functions, Equations, and Inequalities

a) Vertex:

+

1 7 − , 2 4

,

b) Axis of symmetry: x = − 7 4

c) Maximum value: d)

1 2

2 2

4

d) Increasing: (−2, ∞); decreasing: (−∞, −2) a) The x-coordinate of the vertex is 4 b =− = −1. − 2a 2·2 Since f (−1) = 2(−1)2 + 4(−1) − 16 = −18, the vertex is (−1, −18).

4

!2

c) Range: [−9, ∞)

27. f (x) = 2x2 + 4x − 16

y

!4

b) Since a = 1 > 0, the graph opens up. The minimum value of f (x) is −9.

x

b) Since a = 2 > 0, the graph opens up so the second coordinate of the vertex, −18, is the minimum value of the function.

!2 !4

c) The range is [−18, ∞).

f(x) ! "3x 2 " 3x # 1

17. The graph of y = (x + 3)2 has vertex (−3, 0) and opens up. It is graph (f). 18. The graph of y = −(x − 4)2 + 3 has vertex (4, 3) and opens down. It is graph (e).

d) Since the graph opens up, function values decrease to the left of the vertex and increase to the right of the vertex. Thus, f (x) is increasing on (−1, ∞) and decreasing on (−∞, −1).

22. The graph of y = (x − 3)2 has vertex (3, 0) and opens up. It is graph (a).

5 1 2 x − 3x + 2 2 b −3 a) − =3 =− 1 2a 2· 2 5 1 f (3) = · 32 − 3 · 3 + = −2 2 2 The vertex is (3, −2). 1 b) Since a = > 0, the graph opens up. The minimum 2 value of f (x) is −2.

23. The graph of y = −(x + 3) + 4 has vertex (−3, 4) and opens down. It is graph (c).

d) Increasing: (3, ∞); decreasing: (−∞, 3)

19. The graph of y = 2(x − 4)2 − 1 has vertex (4, −1) and opens up. It is graph (b). 20. The graph of y = x2 − 3 has vertex (0, −3) and opens up. It is graph (g). 1 21. The graph of y = − (x + 3)2 + 4 has vertex (−3, 4) and 2 opens down. It is graph (h).

2

24. The graph of y = 2(x − 1)2 − 4 has vertex (1, −4) and opens up. It is graph (d). 25. f (x) = x2 − 6x + 5

a) The x-coordinate of the vertex is −6 b =− = 3. − 2a 2·1 Since f (3) = 32 −6·3+5 = −4, the vertex is (3, −4).

b) Since a = 1 > 0, the graph opens up so the second coordinate of the vertex, −4, is the minimum value of the function. c) The range is [−4, ∞).

d) Since the graph opens up, function values decrease to the left of the vertex and increase to the right of the vertex. Thus, f (x) is increasing on (3, ∞) and decreasing on (−∞, 3).

26. f (x) = x + 4x − 5 4 b a) − =− = −2 2a 2·1 f (−2) = (−2)2 + 4(−2) − 5 = −9 2

The vertex is (−2, −9).

28. f (x) =

c) Range: [−2, ∞)

1 29. f (x) = − x2 + 5x − 8 2 a) The x-coordinate of the vertex is b 5 , = 5. − =− + 1 2a 2 − 2 9 1 Since f (5) = − · 52 + 5 · 5 − 8 = , the vertex is 2 2 , + 9 . 5, 2 1 b) Since a = − < 0, the graph opens down so the sec2 9 ond coordinate of the vertex, , is the maximum 2 value of the function. 1 + c) The range is − ∞, 9 . 2 d) Since the graph opens down, function values increase to the left of the vertex and decrease to the right of the vertex. Thus, f (x) is increasing on (−∞, 5) and decreasing on (5, ∞).

Exercise Set 2.4 30. f (x) = −2x2 − 24x − 64 b −24 a) − =− = −6. 2a 2(−2) f (−6) = −2(−6)2 − 24(−6) − 64 = 8

The vertex is (−6, 8).

b) Since a = −2 < 0, the graph opens down. The maximum value of f (x) is 8. c) Range: (−∞, 8] d) Increasing: (−∞, −6); decreasing: (−6, ∞) 31. f (x) = 3x2 + 6x + 5 a) The x-coordinate of the vertex is 6 b =− = −1. − 2a 2·3 Since f (−1) = 3(−1)2 + 6(−1) + 5 = 2, the vertex is (−1, 2). b) Since a = 3 > 0, the graph opens up so the second coordinate of the vertex, 2, is the minimum value of the function. c) The range is [2, ∞).

d) Since the graph opens up, function values decrease to the left of the vertex and increase to the right of the vertex. Thus, f (x) is increasing on (−1, ∞) and decreasing on (−∞, −1). 32. f (x) = −3x2 + 24x − 49 b 24 a) − =− = 4. 2a 2(−3) f (4) = −3 · 42 + 24 · 4 − 49 = −1 The vertex is (4, −1).

b) Since a = −3 < 0, the graph opens down. The maximum value of f (x) is −1. c) Range: (−∞, −1]

d) Increasing: (−∞, 4); decreasing: (4, ∞) 33. g(x) = −4x2 − 12x + 9

a) The x-coordinate of the vertex is −12 3 b − =− =− . 2a 2(−4) 2 ' 3 (2 ' 3( ' 3( = −4 − + 9 = 18, Since g − −12 − 2 ' 2 2 ( 3 the vertex is − , 18 . 2 b) Since a = −4 < 0, the graph opens down so the second coordinate of the vertex, 18, is the maximum value of the function.

c) The range is (−∞, 18]. d) Since the graph opens down, function values increase to the left of the vertex and decrease to the right of the vertex. Thus, g(x) is increasing on ' 3 ( ' 3( and decreasing on − , ∞ . − ∞, − 2 2

151 34. g(x) = 2x2 − 6x + 5 b −6 3 =− = a) − 2a 2·2 2 + ,2 + , + , 3 3 1 3 g =2 −6 +5= 2 2 2 2 + , 3 1 , . The vertex is 2 2 b) Since a = 2 > 0, the graph opens up. The minimum 1 value of g(x) is . 0 ,2 1 c) Range: ,∞ 2 + , + , 3 3 d) Increasing: , ∞ ; decreasing: − ∞, 2 2 35. Familiarize and Translate. The function s(t) = −16t2 + 20t + 6 is given in the statement of the problem. Carry out. The function s(t) is quadratic and the coefficient of t2 is negative, so s(t) has a maximum value. It occurs at the vertex of the graph of the function. We find the first coordinate of the vertex. This is the time at which the ball reaches its maximum height. 20 b =− = 0.625 t=− 2a 2(−16) The second coordinate of the vertex gives the maximum height. s(0.625) = −16(0.625)2 + 20(0.625) + 6 = 12.25

Check. Completing the square, we write the function in the form s(t) = −16(t − 0.625)2 + 12.25. We see that the coordinates of the vertex are (0.625, 12.25), so the answer checks. State. The ball reaches its maximum height after 0.625 seconds. The maximum height is 12.25 ft. 36. Find the first coordinate of the vertex: 60 = 1.875 t=− 2(−16) Then s(1.875) = −16(1.875)2 + 60(1.875) + 30 = 86.25. Thus the maximum height is reached after 1.875 sec. The maximum height is 86.25 ft. 37. Familiarize and Translate. The function s(t) = −16t2 + 120t + 80 is given in the statement of the problem. Carry out. The function s(t) is quadratic and the coefficient of t2 is negative, so s(t) has a maximum value. It occurs at the vertex of the graph of the function. We find the first coordinate of the vertex. This is the time at which the rocket reaches its maximum height. 120 b =− = 3.75 t=− 2a 2(−16) The second coordinate of the vertex gives the maximum height. s(3.75) = −16(3.75)2 + 120(3.75) + 80 = 305

Check. Completing the square, we write the function in the form s(t) = −16(t − 3.75)2 + 305. We see that the

152

Chapter 2: Functions, Equations, and Inequalities When b = 10, then 20 − b = 20 − 10 = 10, and the area is 1 · 10 · 10 = 50 cm2 . 2 Check. As a partial check, we can find A(b) for a value of b less than 10 and for a value of b greater than 10. For instance, V (9.9) = 49.995 and V (10.1) = 49.995. Since both of these values are less than 50, our result appears to be correct.

coordinates of the vertex are (3.75, 305), so the answer checks. State. The rocket reaches its maximum height after 3.75 seconds. The maximum height is 305 ft. 38. Find the first coordinate of the vertex: 150 = 4.6875 t=− 2(−16) Then s(4.6875) = −16(4.6875)2 + 150(4.6875) + 40 = 391.5625. Thus the maximum height is reached after 4.6875 sec. The maximum height is 391.5625 ft. 39. Familiarize. Using the label in the text, we let x = the height of the file. Then the length = 10 and the width = 18 − 2x.

Translate. Since the volume of a rectangular solid is length × width × height we have V (x) = 10(18 − 2x)x, or −20x2 + 180x.

Carry out. Since V (x) is a quadratic function with a = −20 < 0, the maximum function value occurs at the vertex of the graph of the function. The first coordinate of the vertex is 180 b =− = 4.5. − 2a 2(−20) Check. When x = 4.5, then 18 − 2x = 9 and V (x) = 10 · 9(4.5), or 405. As a partial check, we can find V (x) for a value of x less than 4.5 and for a value of x greater than 4.5. For instance, V (4.4) = 404.8 and V (4.6) = 404.8. Since both of these values are less than 405, our result appears to be correct.

State. The area is a maximum when the base and the height are both 10 cm. 42. Let b = the length of the base. Then 69 − b = the height and A(b) = b(69−b), or −b2 +69b. The maximum function value occurs at the vertex of the graph of A(b). The first coordinate of the vertex is 69 b =− = 34.5. − 2a 2(−1) When b = 34.5, then 69−b = 34.5. The area is a maximum when the base and height are both 34.5 cm. 43. C(x) = 0.1x2 − 0.7x + 2.425

Since C(x) is a quadratic function with a = 0.1 > 0, a minimum function value occurs at the vertex of the graph of C(x). The first coordinate of the vertex is −0.7 b =− = 3.5. − 2a 2(0.1) Thus, 3.5 hundred, or 350 bicycles should be built to minimize the average cost per bicycle.

44.

P (x) = 5x − (0.001x2 + 1.2x + 60)

State. The file should be 4.5 in. tall in order to maximize the volume. 40. Let w = the width of the garden. Then the length = 32 − 2w and the area is given by A(w) = (32 − 2w)w, or −2w2 + 32w. The maximum function value occurs at the vertex of the graph of A(w). The first coordinate of the vertex is 32 b =− = 8. − 2a 2(−2) When w = 8, then 32 − 2w = 16 and the area is 16 · 8, or 128 ft2 . A garden with dimensions 8 ft by 16 ft yields this area. 41. Familiarize. Let b = the length of the base of the triangle. Then the height = 20 − b. 1 Translate. Since the area of a triangle is × base×height, 2 we have 1 1 A(b) = b(20 − b), or − b2 + 10b. 2 2 Carry out. Since A(b) is a quadratic function with 1 a = − < 0, the maximum function value occurs at the 2 vertex of the graph of the function. The first coordinate of the vertex is 10 b , = 10. =− + − 1 2a 2 − 2

P (x) = R(x) − C(x)

P (x) = −0.001x2 + 3.8x − 60

Since P (x) is a quadratic function with a = −0.001 < 0, a maximum function value occurs at the vertex of the graph of the function. The first coordinate of the vertex is 3.8 b =− = 1900. − 2a 2(−0.001) P (1900) = −0.001(1900)2 + 3.8(1900) − 60 = 3550

Thus, the maximum profit is $3550. It occurs when 1900 units are sold. 45.

P (x) = R(x) − C(x)

P (x) = (50x − 0.5x2 ) − (10x + 3) P (x) = −0.5x2 + 40x − 3

Since P (x) is a quadratic function with a = −0.5 < 0, a maximum function value occurs at the vertex of the graph of the function. The first coordinate of the vertex is 40 b =− = 40. − 2a 2(−0.5) P (40) = −0.5(40)2 + 40 · 40 − 3 = 797

Thus, the maximum profit is $797. It occurs when 40 units are sold.

Exercise Set 2.4 46.

153 t1 + t2 = 3,

P (x) = R(x) − C(x)

s = 16t21 ,

P (x) = −0.1x + 16x − 2 2

Since P (x) is a quadratic function with a = −0.1 < 0, a maximum function value occurs at the vertex of the graph of the function. The first coordinate of the vertex is 16 b =− = 80. − 2a 2(−0.1) P (80) = −0.1(80)2 + 16(80) − 2 = 638

Thus, the maximum profit is $638. It occurs when 80 units are sold.

u2 + 275u − 5500 = 0 √ −b + b2 − 4ac We only want the u= 2a positive solution. 2 −275 + 275 − 4 · 1(−5500) = 2·1 √ −275 + 97, 625 ≈ 18.725 = 2 √ Since u ≈ 18.725, we have s = 18.725, so s ≈ 350.6. √ √ s 350.6 Check. If s ≈ 350.6, then t1 = = ≈ 4 4 s 350.6 = ≈ 0.32, so t1 + t2 = 4.68 + 4.68 and t2 = 1100 1100 0.32 = 5. The result checks. State. The elevator shaft is about 350.6 ft tall. 48. Let s = the height of the cliff, t1 = the time it takes the balloon to hit the ground, and t2 = the time it takes for the sound to reach the top of the cliff. Then we have

or t1 =

49. Familiarize. Using the labels on the drawing in the text, we let x = the width of each corral and 240 − 3x = the total length of the corrals. Translate. Since the area of a rectangle is length × width, we have

47. Familiarize. We let s = the height of the elevator shaft, t1 = the time it takes the screwdriver to reach the bottom of the shaft, and t2 = the time it takes the sound to reach the top of the shaft. Translate. We know that t1 + t2 = 5. Using the information in Example 4 we also know that √ s and s = 16t21 , or t1 = 4 s s = 1100t2 , or t2 = . 1100 √ s s + = 5. Then 4 1100 Carry out. We solve the last equation above. √ s s + =5 4 1100 √ 275 s + s = 5500 Multiplying by 1100 √ s + 275 s − 5500 = 0 √ Let u = s and substitute.

√

s , and 4 s s = 1100t2 , or t2 = , so 1100 √ s s + = 3. 4 1100 Solving the last equation, we find that s ≈ 132.7 ft.

P (x) = 20x − 0.1x − (4x + 2) 2

A(x) = (240 − 3x)x = −3x2 + 240x.

Carry out. Since A(x) is a quadratic function with a = −3 < 0, the maximum function value occurs at the vertex of the graph of A(x). The first coordinate of the vertex is 240 b =− = 40. − 2a 2(−3) A(40) = −3(40)2 + 240(40) = 4800

Check. As a partial check we can find A(x) for a value of x less than 40 and for a value of x greater than 40. For instance, A(39.9) = 4799.97 and A(40.1) = 4799.97. Since both of these values are less than 4800, our result appears to be correct. State. The largest total area that can be enclosed is 4800 yd2 . 50.

πx 1 · 2πx + 2x + 2y = 24, so y = 12 − − x. 2 2 , + 1 πx −x A(x) = · πx2 + 2x 12 − 2 2 A(x) =

πx2 + 24x − πx2 − 2x2 2

A(x) = 24x −

πx2 − 2x2 , or 24x − 2

+

, π + 2 x2 2

Since A(x) is , a quadratic function with + π a=− + 2 < 0, the maximum function value occurs 2 at the vertex of the graph of A(x). The first coordinate of the vertex is 24 24 −b ,1 = =− 0 + . π 2a π+4 2 − +2 2 24 24 , then y = . Thus, the maxiWhen x = π+4 π+4 mum amount of light will enter when the dimensions of the rectangular part of the window are 2x by y, or 48 24 ft by ft, or approximately 6.72 ft by 3.36 ft. π+4 π+4 51. Left to the student 52. Left to the student

154

Chapter 2: Functions, Equations, and Inequalities 62. f (x) = −4x2 + bx + 3

53. Left to the student

b b , or . The x-coordinate of the vertex of f (x) is − 2(−4) 8 + , b = 50. Now we find b such that f 8 + ,2 b b −4 + b · + 3 = 50 8 8

54. Left to the student 55. Answers will vary. The problem could be similar to Examples 5 and 6 or Exercises 35 through 50. 56. Completing the square was used in Section 2.3 to solve quadratic equations. It was used again in this section to write quadratic functions in the form f (x) = a(x − h)2 + k.

−

57. The x-intercepts of g(x) are also (x1 , 0) and (x2 , 0). This is true because f (x) and g(x) have the same zeros. Consider g(x) = 0, or −ax2 − bx − c = 0. Multiplying by −1 on both sides, we get an equivalent equation ax2 + bx + c = 0, or f (x) = 0. 58.

3(x + h) − 7 − (3x − 7) f (x + h) − f (x) = h h 3x + 3h − 7 − 3x + 7 = h 3h = =3 h

b2 = 47 16 2 b = 16 · 47 √ b = ± 16 · 47 √ b = ±4 47 63. f (x) = −0.2x2 − 3x + c

b The x-coordinate of the vertex of f (x) is − = 2a −3 = −7.5. Now we find c such that − 2(−0.2) f (−7.5) =

59. f (x) = 2x2 − x + 4

f (x + h) = 2(x + h)2 − (x + h) + 4

−225. −0.2(−7.5)2 − 3(−7.5) + c = −225

= 2(x2 + 2xh + h2 ) − (x + h) + 4

= 2x2 + 4xh + 2h2 − x − h − 4

= = = = 60.

f (x + h) − f (x) h 2x2 + 4xh + 2h2 − x − h − 4 − (2x2 − x + 4) h 2x2 + 4xh + 2h2 − x − h − 4 − 2x2 + x − 4 h 4xh + 2h2 − h h(4x + 2h − 1) = h h 4x + 2h − 1

−11.25 + 22.5 + c = −225

c = −236.25

64. f (x) = a(x − h)2 + k

6 . Then 1 = a(−3 − 4)2 − 5, so a = 49 6 2 f (x) = (x − 4) − 5. 49

65.

y " (|x | ! 5)2 ! 3 y

y 4

(!1.5, 2)

24

g(x) " f(2x)

8

4

!4 !2 !2 !4

x

(1.5, !2)

y 4

(3, 4)

2

(5, 0) 2

!4 !2

4

!2

(!3, !4)

(0, !4)

g(x) " !2f (x)

!8

!4

4

8

x

!8

61. The graph of f (x) is stretched vertically and reflected across the x-axis.

(!5, 0)

16

(0, 2) (2.5, 0)

(!2.5, 0)

b2 b2 + + 3 = 50 16 8

x

66. First we find the radius r of a circle with circumference x: 2πr = x x r= 2π Then we find the length s of a side of a square with perimeter 24 − x: 4s = 24 − x 24 − x s= 4

Exercise Set 2.5

155

Then S = area of circle + area of square

3.

+ s πr + ,2 + ,2 x 24 − x S(x) = π + 2π 4 , + 1 1 S(x) = + x2 − 3x + 36 4π 16 S=

2

2

Since S(x) is a quadratic function with 1 1 + > 0, the minimum function value occurs at a= 4π 16 the vertex of the graph of S(x). The first coordinate of the vertex is 24π −3 b ,= =− + . − 1 1 2a 4+π + 2 4π 16 24π in., Then the string should be cut so that one piece is 4+π 24π , or or about 10.56 in. The other piece will be 24 − 4+π 96 in., or about 13.44 in. 4+π

5x + 10 − 4x + 4 = 300 x + 14 = 300

x = 286 The solution is 286. 4.

1.

Check: 1 1 1 + = 4 5 t

t = −1

The solution is −1. 5.

3x − 2x = 18 − 12 x=6

Check: 1 3 1 2 + = + 2 x 3 x 1 2 1 3 + ? + 2 6 3 6 2| 1 1 22 1 1 + + 2 3 2 3 2 The solution is 6.

9t = 20 20 t= 9

1 1 5 − = , 3 6 x 2x − 5x = 6

6.

TRUE

1 1 1 + + = 5, LCD is 6t t 2t 3t 6 + 3 + 2 = 30t Multiplying by 6t 11 = 30t 11 =t 30

11 11 checks. The solution is . 30 30

TRUE

LCD is 6x Multiplying by 6x

−3x = 6

x = −2

1 3 1 2 + = + , LCD is 6x 2 x 3 x '1 2( '1 3( 6x + = 6x + 2 x 3 x 3x + 12 = 2x + 18

1 = , LCD is 20t t 1 = 20t · t 1 = 20t · t = 20

1 1 1 + ? 4 5 2 20 2 9 2 4 22 9 5 1· + 20 20 22 20 9 22 9 20 2 20 20 . The solution is 9 2.

t+1 t−1 − = 1, LCD is 6 3 2 2t + 2 − 3t + 3 = 6 Multiplying by 6 −t = 1

Exercise Set 2.5 1 1 + 4 5 '1 1( 20t + 4 5 1 1 20t · + 20t · 4 5 5t + 4t

x+2 x−1 − = 15, LCD is 20 4 5 'x + 2 x − 1( − = 20 · 15 20 4 5 5(x + 2) − 4(x − 1) = 300

−2 checks. The solution is −2.

7.

5 3 = , LCD is 2x(3x + 2) 3x + 2 2x 5 3 2x(3x + 2) · = 2x(3x + 2) · 3x + 2 2x 2x · 5 = 3(3x + 2) 10x = 9x + 6 x=6 6 checks, so the solution is 6.

156 8.

Chapter 2: Functions, Equations, and Inequalities 2 3 = , LCD is (x − 1)(x + 2) x−1 x+2 2(x + 2) = 3(x − 1) 2x + 4 = 3x − 3

m = 4 or m = −2

Only 4 checks. The solution is 4. 13.

7=x

The answer checks. The solution is 7. 9.

6 = 5, LCD is x x , + 6 x x+ = x·5 x x2 + 6 = 5x x+

(2x − 1)(x − 5) = 0

2x − 1 = 0 or x − 5 = 0

(x − 2)(x − 3) = 0

2x = 1 or 1 or x= 2

x − 2 = 0 or x − 3 = 0

x=3

12 = 1, LCD is x x x2 − 12 = x x−

2x − 3x − 35 = 0 2

(2x + 7)(x − 5) = 0

Both numbers check. The solutions are 4 and −3. 2 5y − 3 6 + = 2 y+3 y y −9 6 2 5y − 3 + = , y+3 y (y + 3)(y − 3)

LCD is y(y+3)(y−3) + , 6 2 5y−3 y(y+3)(y−3) + = y(y+3)(y−3)· y+3 y (y+3)(y−3) 6y(y−3)+2(y+3)(y−3) = y(5y − 3)

x=−

15.

8y 2 − 18y − 18 = 5y 2 − 3y

3x − 5 = 16

y 2 − 5y − 6 = 0

3x = 21

(y − 6)(y + 1) = 0

x=7

y − 6 = 0 or y + 1 = 0

2 4m − 4 3 + = 2 m+2 m m −4 3 2 4m − 4 + = , m+2 m (m + 2)(m − 2)

LCD is m(m + 2)(m − 2)

3m2 − 6m + 2m2 − 8 = 4m2 − 4m

Multiplying by m(m + 2)(m − 2)

m2 − 2m − 8 = 0

(m − 4)(m + 2) = 0

1 16 2 + = 2 x+5 x−5 x − 25 1 16 2 + = , x+5 x−5 (x + 5)(x − 5)

2x − 10 + x + 5 = 16

3y 2 − 15y − 18 = 0

Both numbers check. The solutions are 6 and −1.

7 and 5. 2

LCD is (x + 5)(x − 5) + , 1 16 2 + = (x+5)(x−5)· (x+5)(x−5) x+5 x−5 (x+5)(x−5) 2(x − 5) + x + 5 = 16

6y 2 − 18y + 2y 2 − 18 = 5y 2 − 3y

y = −1

7 or x = 5 2

Both numbers check. The solutions are −

6y 2 − 18y + 2(y 2 − 9) = 5y 2 − 3y

12.

1 and 5. 2

2x2 + 2x = 5x + 35

(x − 4)(x + 3) = 0 x = 4 or x = −3

y = 6 or

x=5

5 2x = , LCD is (x + 7)(x + 1) x+7 x+1 2x(x + 1) = 5(x + 7)

14.

x2 − x − 12 = 0

11.

x=5

Both numbers check. The solutions are

Both numbers check. The solutions are 2 and 3. 10.

2x2 − 6x = 5x − 5

2x − 11x + 5 = 0 2

x2 − 5x + 6 = 0

x = 2 or

2x 5 = , LCD is (x − 1)(x − 3) x−1 x−3 5 2x = (x − 1)(x − 3) · (x − 1)(x − 3) · x−1 x−3 2x(x − 3) = 5(x − 1)

7 checks, so the solution is 7. 16.

5 3 2 + = x2 − 9 x − 3 x+3 2 5 3 + = , LCD is (x + 3)(x−3) (x + 3)(x−3) x−3 x+3 2 + 5(x + 3) = 3(x − 3) 2 + 5x + 15 = 3x − 9

5x + 17 = 3x − 9 2x = −26 x = −13

The answer checks. The solution is −13.

Exercise Set 2.5 17.

157

3x 6 + x+2 x 6 3x + x+2 x ! 3x 6" + x(x + 2) x+2 x 3x · x + 6(x + 2)

12 x2 + 2x 12 = , LCD is x(x+2) x(x+2) 12 = x(x + 2) · x(x + 2) = 12 =

21.

3x2 + 6x + 12 = 12

2 3 4 − = 5x + 5 x2 − 1 x−1 3 4 2 − = , 5(x + 1) (x + 1)(x − 1) x−1 LCD is # $ 5(x + 1)(x − 1) 3 4 2 − = 5(x+1)(x−1)· 5(x+1)(x−1) 5(x+1) (x+1)(x−1) x−1 2(x − 1) − 5 · 3 = 20(x + 1) 2x − 2 − 15 = 20x + 20

3x2 + 6x = 0

2x − 17 = 20x + 20

3x(x + 2) = 0

−18x − 17 = 20

3x = 0 or x + 2 = 0 x = 0 or

−18x = 37 37 x=− 18 37 37 − checks, so the solution is − . 18 18

x = −2

Neither 0 nor −2 checks, so the equation has no solution. 18.

y+4 y+1 3y + 5 + = y 2 + 5y y + 5 y y+4 y+1 3y + 5 + = , LCD is y(y + 5) y(y + 5) y + 5 y 3y + 5 + y 2 + 4y = y 2 + 6y + 5 y=0

22.

Multiplying by y(y + 5)

0 does not check. There is no solution. 19.

1 3 1 − = 5x + 20 x2 − 16 x−4 1 3 1 − = , 5(x + 4) (x + 4)(x − 4) x−4 LCD is $ 5(x + 4)(x − 4) # 1 1 3 5(x+4)(x−4) − = 5(x+4)(x−4)· 5(x+4) (x+4)(x−4) x−4 x − 4 − 5 = 15(x + 4) x − 9 = 15x + 60

−14x − 9 = 60

−14x = 69 69 x=− 14 69 69 − checks, so the solution is − . 14 14 20.

1 5 1 − = 4x + 12 x2 − 9 x−3 1 1 5 − = , 4(x + 3) (x + 3)(x − 3) x−3 LCD is 4(x + 3)(x − 3) x − 3 − 4 = 20x + 60

−19x = 67 67 x=− 19 67 67 checks. The solution is − . − 19 19

1 3 1 − = 3x + 6 x2 − 4 x−2 1 3 1 − = , 3(x + 2) (x + 2)(x − 2) x−2 LCD is 3(x + 2)(x − 2) x − 2 − 3 = 9x + 18 x − 5 = 9x + 18

−8x = 23 23 x=− 8 23 23 − checks. The solution is − . 8 8 23.

x 24 8 = + , x2 − 2x + 4 x + 2 x3 + 8 LCD is (x + 2)(x2 − 2x + 4) 8 (x+2)(x2 −2x+4) · 2 = x −2x+4 # $ x 24 + (x+2)(x2− 2x+4) x+2 (x+2)(x2 −2x+4) 8(x + 2) = x(x2 −2x+4)+24 8x + 16 = x3 −2x2 +4x+24 0 = x3 −2x2 −4x+8

0 = x2 (x−2) − 4(x−2) 0 = (x−2)(x2 −4)

0 = (x−2)(x+2)(x−2) x − 2 = 0 or x + 2 = 0 x = 2 or

or x − 2 = 0

x = −2 or

Only 2 checks. The solution is 2.

x=2

158 24.

Chapter 2: Functions, Equations, and Inequalities 18 x 81 − = 3 , x2 − 3x + 9 x + 3 x + 27 LCD is (x + 3)(x2 − 3x + 9)

18x+54−x3 +3x2 −9x = 81 Multiplying by (x + 3)(x2 − 3x + 9) 3 2 −x + 3x + 9x − 27 = 0 −x2 (x − 3) + 9(x − 3) = 0 (x − 3)(9 − x ) = 0 2

(x − 3)(3 + x)(3 − x) = 0

x = 3 or x = −3

Only 3 checks. The solution is 3. 25.

4 32 x − = 2 x−4 x+4 x − 16 4 32 x − = , x−4 x+4 (x+4)(x−4) LCD is (x+4)(x−4) ! x 4 " 32 − = (x+4)(x−4)· (x+4)(x−4) x−4 x+4 (x+4)(x−4) x(x + 4) − 4(x − 4) = 32 x2 + 4x − 4x + 16 = 32 x2 + 16 = 32

x2 = 16 x = ±4

Neither 4 nor −4 checks, so the equation has no solution. 26.

1 2 x − = 2 x−1 x+1 x −1 1 2 x − = , x−1 x+1 (x + 1)(x − 1) LCD is (x + 1)(x − 1)

x +x−x+1 = 2 2

1 1 15 − = 2 x − 15 x x − 15x 1 15 1 − = , LCD is x(x − 15) x − 15 x x(x − 15) x − (x − 15) = 15 x − x + 15 = 15 15 = 15

We get an equation that is true for all real numbers. Note, however, that when x = 0 or x = 15, division by 0 occurs in the original equation. Thus, the solution set is {x|x is a real number and x "= 0 and x "= 15}, or (−∞, 0) ∪ (0, 15) ∪ (15, ∞). √ 29. 3x − 4 = 1 √ ( 3x − 4)2 = 12 3x − 4 = 1

3x = 5 5 x= 3

Check: √ 3x − 4 % 5 3· −4 3 √ 5−4 √ 1 1

=1 ? 1 & && & & & & & & 1

TRUE

5 The solution is . 3 √ 30. 4x + 1 = 3 4x + 1 = 9 4x = 8 x=2

x2 = 1 x = ±1

Neither 1 nor −1 checks. There is no solution. 27.

28.

1 6 1 − = 2 x−6 x x − 6x 1 6 1 − = , LCD is x(x−6) x−6 x x(x−6) $ # 1 1 6 − = x(x−6) · x(x−6) x−6 x x(x−6) x − (x − 6) = 6 x−x+6 = 6 6=6

We get an equation that is true for all real numbers. Note, however, that when x = 6 or x = 0, division by 0 occurs in the original equation. Thus, the solution set is {x|x is a real number and x "= 6 and x "= 0}, or (−∞, 0) ∪ (0, 6) ∪ (6, ∞).

The answer checks. The solution is 2. √ 31. 2x − 5 = 2 √ ( 2x − 5)2 = 22 2x − 5 = 4

2x = 9 9 x= 2

Check: √ 2x − 5 % 9 2· −5 2 √ 9−5 √ 4 2

=2 ? 2 & && & & & & & & 2

9 The solution is . 2

TRUE

Exercise Set 2.5 32.

√

159 Check: √ 3

3x + 2 = 6 3x + 2 = 36

' 3

3x = 34 34 x= 3 The answer checks. The solution is 33.

√ 7−x = 2 √ ( 7 − x)2 = 22

34 . 3

5(−5) − 2 √ 3 −25 − 2 √ 3 −27 −3

2x = −126 x = −63

x=3

The answer checks. The solution is −63. √ 4 39. x2 − 1 = 1 √ 4 ( x2 − 1)4 = 14

=2

x2 = 2

Check:

5−x = 1

4=x

1 − 2x = 9 −2x = 8

x = −4

TRUE

2 − 7x = 4

x=−

2 7

2 The answer checks. The solution is − . 7 √ 3 37. 5x − 2 = −3 √ 3 ( 5x − 2)3 = (−3)3 5x − 2 = −27 5x = −25 x = −5

x2 − 1 = 1 ? 1 & & & & & & & & 1

TRUE √ The solutions are ± 2. √ 40. 5 3x + 4 = 2 3x = 28 28 x= 3

The solution is −4. √ 36. 2 − 7x = 2 −7x = 2

√ 4

3x + 4 = 32

1 − 2x = 3 ?& 3 & & & & & & 3

√ x=± 2

√ (± 2)2 − 1 √ 4 2−1 √ 4 1 1

( 4

The answer checks. The solution is 4. √ 35. 1 − 2x = 3 √ ( 1 − 2x)2 = 32

' 1 − 2(−4) √ 1+8 √ 9 3

x2 − 1 = 1

TRUE

The solution is 3. √ 34. 5−x = 1

Check: √

TRUE

2x + 1 = −125

−x = −3

?& 2 & & & & 2

?& −3 & & & & & & −3

The solution is −5. √ 38. 3 2x + 1 = −5

7−x = 4

Check: √ 7−x √ 7−3 √ 4 2

5x − 2 = −3

28 The answer checks. The solution is . 3 √ 41. y−1+4 = 0 √ y − 1 = −4

The principal square root is never negative. Thus, there is no solution. If we do not observe the above fact, we can continue and reach the same answer. √ ( y − 1)2 = (−4)2 y − 1 = 16

y = 17

Check: √ y−1+4 √ 17 − 1 + 4 √ 16 + 4 4+4 8

=0 ?& 0 & & & & & & 0

FALSE

Since 17 does not check, there is no solution.

160 42.

Chapter 2: Functions, Equations, and Inequalities √

Check: √ 2x + 1 − 3 % 35 2· +1−3 2 √ 35 + 1 − 3 √ 36 − 3

m+1−5 = 8 √ m + 1 = 13 m + 1 = 169

? 3 && & & & & & & 6−3 & & 3 & 3

m = 168 The answer checks. The solution is 168. √ 43. b+3−2 = 1 √ b+3 = 3 ' 2 ( b + 3) = 32 b+3 = 9

b=6 Check: √ b+3−2 √ 6+3−2 √ 9−2 3−2 1

=1 ?& 1 & & & & & & 1

The answer checks. The solution is 20. √ 45. z+2+3 = 4 √ z+2 = 1 √ ( z + 2)2 = 12 z+2 = 1 z = −1 =4

TRUE

The solution is −1. √ 46. y−5−2 = 3 √ y−5 = 5 y − 5 = 25

y = 30

The answer checks. The solution is 30. √ 47. 2x + 1 − 3 = 3 √ 2x + 1 = 6 √ ( 2x + 1)2 = 62 2x = 35 35 x= 2

The answer checks. The solution is 49.

x = 20

2x + 1 = 36

3x = 26 26 x= 3

TRUE

x − 4 = 16

?& 4 & & & & & & 4

TRUE

35 The solution is . 2 √ 48. 3x − 1 + 2 = 7 √ 3x − 1 = 5 3x − 1 = 25

The solution is 6. √ 44. x−4+1 = 5 √ x−4 = 4

Check: √ z+2+3 √ −1 + 2 + 3 √ 1+3 1+3 4

=3

√

26 . 3

2−x−4 = 6 √ 2 − x = 10 √ ( 2 − x)2 = 102

2 − x = 100 −x = 98

Check: √

x = −98

2−x−4 ' 2 − (−98) − 4 √ 100 − 4 10 − 4 6

=6 ?& 6 & & & & & & 6

TRUE

The solution is −98. √ 50. 5−x+2 = 8 √ 5−x = 6 5 − x = 36 −x = 31

x = −31

The answer checks. The solution is −31. √ 51. 3 6x + 9 + 8 = 5 √ 3 6x + 9 = −3 √ 3 ( 6x + 9)3 = (−3)3 6x + 9 = −27 6x = −36 x = −6

Exercise Set 2.5 Check: √ 3 ' 3

161 55.

6x + 9 + 8 = 5

6(−6) + 9 + 8 √ 3 −27 + 8

?& 5 & & & −3 + 8 && 5 & 5

0 = x2 − 11x + 28

0 = (x − 4)(x − 7)

TRUE

x − 4 = 0 or x − 7 = 0 x = 4 or

2x = 35 35 x= 2

The answer checks. The solution is √

35 . 2

x+4+2 = x √ x+4 = x−2 √ ( x + 4)2 = (x − 2)2

x + 4 = x2 − 4x + 4 0 = x2 − 5x

0 = x(x − 5)

x = 0 or x − 5 = 0

x=5

Check: For 0: √ x+4+2 = x √ 0 + 4 + 2 ?& 0 & 2+2 & & 4 & 0 For 5: √ x+4+2 √ 5+4+2 √ 9+2

For 4: √ x−3+5 √ 4−3+5 √ 1+5 1+5 6 For 7: √ x−3+5 √ 7−3+5 √ 4+5 2+5 7

=x ?& 4 & & & & & & 4

FALSE

=x ?& 7 & & & & & & 7

TRUE

The number 7 checks but 4 does not. The solution is 7. √ 56. x+3−1 = x √ x+3 = x+1 x + 3 = x2 + 2x + 1 0 = x2 + x − 2

0 = (x + 2)(x − 1)

FALSE

=x

?& 5 & & & 3 + 2 && 5 & 5

x=7

Check:

2x − 3 = 32

x = 0 or

x−3+5 = x √ x−3 = x−5 √ ( x − 3)2 = (x − 5)2

x − 3 = x2 − 10x + 25

The solution is −6. √ 52. 5 2x − 3 − 1 = 1 √ 5 2x − 3 = 2

53.

√

x = −2 or x = 1

Only 1 checks. The solution is 1. √ 57. x+7 = x+1 √ ( x + 7)2 = (x + 1)2 x + 7 = x2 + 2x + 1 0 = x2 + x − 6

TRUE

The number 5 checks but 0 does not. The solution is 5. √ 54. x+1+1 = x √ x+1 = x−1 x + 1 = x2 − 2x + 1 0 = x2 − 3x

0 = x(x − 3)

x = 0 or x = 3

Only 3 checks. The solution is 3.

0 = (x + 3)(x − 2)

x+3 = 0

or x − 2 = 0

x = −3 or

x=2

Check:

For −3: √ x+7 √ −3 + 7 √ 4 2

= x+1 ?& −3 + 1 & & −2 & & −2

FALSE

162

Chapter 2: Functions, Equations, and Inequalities For 2: √ x+7 √ 2+7 √ 9

Check: = x+1

?& 2 + 1 & & 3 & 3 & 3

TRUE

The number 2 checks but −3 does not. The solution is 2. √ 58. 6x + 7 = x + 2 6x + 7 = x2 + 4x + 4 0 = x2 − 2x − 3

0 = (x − 3)(x + 1)

x = 3 or x = −1

Both values check. The solutions are 3 and −1. √ 59. 3x + 3 = x + 1 √ ( 3x + 3)2 = (x + 1)2 3x + 3 = x2 + 2x + 1 0 = (x − 2)(x + 1)

x − 2 = 0 or x + 1 = 0 x = 2 or

For 2: √ 3x + 3 √ 3·2+3 √ 9 3

x = −1

= x+1 ?& 2 + 1 & & 3 & & 3

For −1: √ 3x + 3 ' 3(−1) + 3 √ 0

TRUE

= x+1

?& −1 + 1 & 0 & & 0 & 0

TRUE

0 = x2 − 12x + 20

0 = (x − 2)(x − 10)

x = 2 or x = 10

Only 10 checks. The solution is 10. √ 61. 5x + 1 = x − 1 √ ( 5x + 1)2 = (x − 1)2

0 = x(x − 7)

x = 0 or x − 7 = 0 x = 0 or

x=7

FALSE

?& 7 − 1 & & 6 & 6 & 6

TRUE

For 7: √ 5x + 1 √ 5·7+1 √ 36

= x−1

The number 7 checks but 0 does not. The solution is 7. √ 62. 7x + 4 = x + 2 7x + 4 = x2 + 4x + 4 0 = x2 − 3x x = 0 or x = 3

Both numbers check. The solutions are 0 and 3. √ √ 63. x−3+ x+2 = 5 √ √ x+2 = 5− x−3 √ √ ( x + 2)2 = (5 − x − 3)2 √ x + 2 = 25 − 10 x − 3 + (x − 3) √ x + 2 = 22 − 10 x − 3 + x √ 10 x − 3 = 20 √ x−3 = 2 √ ( x − 3)2 = 22 x=7

2x + 5 = x2 − 10x + 25

0 = x2 − 7x

?& 0 − 1 & & −1 & & −1

x−3 = 4

Both numbers check. The solutions are 2 and −1. √ 60. 2x + 5 = x − 5

5x + 1 = x2 − 2x + 1

= x−1

0 = x(x − 3)

0 = x2 − x − 2

Check:

For 0: √ 5x + 1 √ 5·0+1 √ 1 1

Check: √ √ x−3+ x+2 √ √ 7−3+ 7+2 √ √ 4+ 9 2+3 5

=5 ? 5 &| & & & & & 5

TRUE

The solution is 7. √ √ 64. x− x−5 = 1 √ √ x = x−5+1 √ x = x−5+2 x−5+1 √ 4 = 2 x−5 √ 2 = x−5 4 = x−5 9=x

The answer checks. The solution is 9.

Exercise Set 2.5 √ 3x − 5 + 2x + 3 + 1 = 0 √ √ 3x − 5 + 2x + 3 = −1 The principal square root is never negative. Thus the sum of two principal square roots cannot equal −1. There is no solution. √ √ 66. 2m − 3 = m + 7 − 2 √ 2m − 3 = m + 7 − 4 m + 7 + 4 √ m − 14 = −4 m + 7 65.

√

m2 − 28m + 196 = 16m + 112 m2 − 44m + 84 = 0

(m − 2)(m − 42) = 0

m = 2 or m = 42

Only 2 checks. The solution is 2. √ √ 67. x − 3x − 3 = 1 √ √ x = 3x − 3 + 1 √ √ 2 ( x) = ( 3x − 3 + 1)2 √ x = (3x − 3) + 2 3x − 3 + 1 √ 2 − 2x = 2 3x − 3 √ 1 − x = 3x − 3 √ (1 − x)2 = ( 3x − 3)2 1 − 2x + x2 = 3x − 3

x2 − 5x + 4 = 0

(x − 4)(x − 1) = 0 x = 4 or x = 1 The number 4 does not check, but 1 does. The solution is 1. √ √ 68. 2x + 1 − x = 1 √ √ 2x + 1 = x + 1 √ 2x + 1 = x + 2 x + 1 √ x=2 x x2 = 4x x − 4x = 0 2

x(x − 4) = 0 x = 0 or x = 4

Both values check. The solutions are 0 and 4. √ √ 2y − 5 − y − 3 = 1 69. √ √ 2y − 5 = y − 3 + 1 √ √ ( 2y − 5)2 = ( y − 3 + 1)2 √ 2y − 5 = (y − 3) + 2 y − 3 + 1 √ y−3 = 2 y−3 √ (y − 3)2 = (2 y − 3)2 y 2 − 6y + 9 = 4(y − 3) y 2 − 6y + 9 = 4y − 12

y − 10y + 21 = 0 2

(y − 7)(y − 3) = 0 y = 7 or y = 3 Both numbers check. The solutions are 7 and 3.

163 70.

√

√ 4p + 5 + p + 5 = 3 √ √ 4p + 5 = 3 − p + 5 √ 4p + 5 = 9 − 6 p + 5 + p + 5 √ 3p − 9 = −6 p + 5 √ p − 3 = −2 p + 5 p2 − 6p + 9 = 4p + 20

p − 10p − 11 = 0 2

(p − 11)(p + 1) = 0

p = 11 or p = −1

Only −1 checks. The solution is −1. √ √ 71. y+4− y−1 = 1 √ √ y+4 = y−1+1 √ √ ( y + 4)2 = ( y − 1 + 1)2 √ y+4 = y−1+2 y−1+1 √ 4 = 2 y−1 √ 2 = y−1 Dividing by 2 √ 22 = ( y − 1)2 4 = y−1 5=y

The answer checks. The solution is 5. √ √ 72. y + 7 + y + 16 = 9 √ √ y + 7 = 9 − y + 16 √ y + 7 = 81 − 18 y + 16 + y + 16 √ −90 = −18 y + 16 √ 5 = y + 16 25 = y + 16 9=y The answer checks. The solution is 9. √ √ 73. x+5+ x+2 = 3 √ √ x+5 = 3− x+2 √ √ ( x + 5)2 = (3 − x + 2)2 √ x+5 = 9−6 x+2+x+2 √ −6 = −6 x + 2 √ Dividing by −6 1 = x+2 √ 2 2 1 = ( x + 2) 1 = x+2 −1 = x

The answer checks. The solution is −1. √ √ 74. 6x + 6 = 5 + 21 − 4x √ 6x + 6 = 25 + 10 21 − 4x + 21 − 4x √ 10x − 40 = 10 21 − 4x √ x − 4 = 21 − 4x x2 − 8x + 16 = 21 − 4x x2 − 4x − 5 = 0

(x − 5)(x + 1) = 0

164

Chapter 2: Functions, Equations, and Inequalities 86. |5x| = 4

x = 5 or x = −1

Only 5 checks. The solution is 5. 75. (x

x1/3 = −2

) = (−2)3

(x1/3 =

1/3 3

√ 3

x)

x = −8

The value checks. The solution is −8. 76.

5x = −4 or 5x = 4 4 4 x = − or x = 5 5 4 4 The solutions are − and . 5 5

87. |x| = 0

The distance of 0 from 0 on a number line is 0. That is,

t1/5 = 2

x = 0.

t = 32

The solution is 0.

The value checks. The solution is 32. 77.

t1/4 = 3 (t1/4 )4 = 34

(t1/4 =

√ 4

88. |6x| = 0

6x = 0

t)

x=0

t = 81

The solution is 0.

The value checks. The solution is 81.

89. |3x + 2| = 1

78. m1/2 = −7

3x + 2 = −1 or 3x + 2 = 1

The principal square root is never negative. There is no solution.

3x = −3 or

3x = −1 1 x = −1 or x=− 3 1 The solutions are −1 and − . 3

79. |x| = 7

The solutions are those numbers whose distance from 0 on a number line is 7. They are −7 and 7. That is, x = −7 or x = 7.

The solutions are −7 and 7. 80. |x| = 4.5

x = −4.5 or x = 4.5

The solutions are −4.5 and 4.5.

81. |x| = −10.7

The absolute value of a number is nonnegative. Thus, the equation has no solution. 3 5 The absolute value of a number is nonnegative. Thus, there is no solution.

82. |x| = −

83. |x − 1| = 4

x − 1 = −4 or x − 1 = 4 x = −3 or

x=5

The solutions are −3 and 5. 84. |x − 7| = 5

x − 7 = −5 or x − 7 = 5 x=2

or

x = 12

The solutions are 2 and 12. 85. |3x| = 1

3x = −1 or 3x = 1 1 1 x = − or x = 3 3 1 1 The solutions are − and . 3 3

90. |7x − 4| = 8

7x − 4 = −8 or 7x − 4 = 8 7x = −4 or 4 x = − or 7

7x = 12 12 x= 7 12 4 The solutions are − and . 7 7 & &1 & & 91. & x − 5& = 17 2 1 1 x − 5 = −17 or x − 5 = 17 2 2 1 1 x = −12 or x = 22 2 2 x = −24 or x = 44

The solutions are −24 and 44. & &1 & & 92. & x − 4& = 13 3 1 1 x − 4 = −13 or x − 4 = 13 3 3 1 1 x = −9 or x = 17 3 3 x = −27 or x = 51 The solutions are −27 and 51.

93.

|x − 1| + 3 = 6

|x − 1| = 3

x − 1 = −3 or x − 1 = 3 x = −2 or

x=4

The solutions are −2 and 4.

Exercise Set 2.5 94.

165 101.

|x + 2| − 5 = 9

−|x + 6| = −7

|x + 2| = 14

|x + 6| = 7

x + 2 = −14 or x + 2 = 14 x = −16 or

x + 6 = −7

x = 12

The solutions are −16 and 12. 95.

|x + 3| − 2 = 8

x=0

x = −2 or

103.

x = 10

The solutions are −2 and 10. 97.

3x + 1 = −3 or 3x + 1 = 3

3x = 2 2 x= 3 4 2 The solutions are − and . 3 3 |2x − 1| − 5 = −3

105.

2x = −1 or 1 x = − or 2

2x = 3 3 x= 2 1 3 The solutions are − and . 2 2

100.

|5x + 4| + 2 = 5

|5x + 4| = 3

5x + 4 = −3 or 5x + 4 = 3 5x = −7 or 7 x = − or 5

5x = −1 1 x=− 5 1 7 The solutions are − and − . 5 5

1 1 1 = + R R1 R2 # $ 1 1 1 RR1 R2 · = RR1 R2 + R R1 R2 Multiplying by RR1 R2 on both sides R1 R2 − RR2 = RR1

Subtracting RR2 on both sides Factoring

R2 (R1 − R) = RR1 RR1 R2 = R1 − R

|4x − 3| = 6

4x = 9 9 x= 4 3 9 The solutions are − and . 4 4

1 1 1 = + F m p

R1 R2 = RR2 + RR1

|4x − 3| + 1 = 7

4x = −3 or 3 x = − or 4

Dividing by P2 V2 on both sides

mp = F (p + m) mp =F p+m

2x − 1 = −2 or 2x − 1 = 2

4x − 3 = −6 or 4x − 3 = 6

Multiplying by T1 T2 on both sides

mp = F p + F m

|2x − 1| = 2

99.

P1 V1 P2 V2 = T1 T2 P1 V1 T2 = P2 V2 T1

104.

3x = −4 or 4 x = − or 3

98.

x=4

P1 V 1 T 2 = T1 P2 V 2

|3x + 1| − 4 = −1 |3x + 1| = 3

or

The solutions are 0 and 4.

|x − 4| + 3 = 9

x − 4 = −6 or x − 4 = 6

9 − |x − 2| = 7

x − 2 = −2 or x − 2 = 2

x=7

|x − 4| = 6

x=1

2 = |x − 2|

The solutions are −13 and 7. 96.

or x + 6 = 7

The solutions are −13 and 1.

x + 3 = −10 or x + 3 = 10 x = −13 or

Multiplying by −1

x = −13 or

102.

|x + 3| = 10

12 − |x + 6| = 5

A = P (1 + i)2 A = (1 + i)2 P

106.

%

%

A = 1+i P

A −1 = i P

Dividing by R1 − R on both sides

166 107.

Chapter 2: Functions, Equations, and Inequalities 1 1 1 = + F m p !1 1 1" F mp · = F mp + Multiplying by F m p F mp on both sides Subtracting F p on both sides Factoring

p(m − F ) = F m Fm Dividing by m − F on p= m−F both sides 108. Left to the student 109. Left to the student 110. When both sides of an equation are multiplied by the LCD, the resulting equation might not be equivalent to the original equation. One or more of the possible solutions of the resulting equation might make a denominator of the original equation 0. 111. When both sides of an equation are raised to an even power, the resulting equation might not be equivalent to the original equation. For example, the solution set of x = −2 is {−2}, but the solution set of x2 = (−2)2 , or x2 = 4, is {−2, 2}. 112.

−3x + 9 = 0

−3x = −9 x=3

The zero of the function is 3. 113.

15 − 2x = 0

Setting f (x) = 0

15 = 2x 15 = x, or 2 7.5 = x

The zero of the function is

15 , or 7.5. 2

114. Let d = the number of acres Disneyland occupies. Solve: (d + 11) + d = 181 d = 85, so Disneyland occupies 85 acres and the Mall of America occupies 85 + 11, or 96 acres. 115. Familiarize. Let p = the number of prescriptions for sleeping pills filled in 2000, in millions. Then the number of prescriptions filled in 2005 was p + 60% · p, or p + 0.6p, or 1.6p. Translate. Number of prescriptions was )42 filled in 2005 *+ , )   . . 1.6p =

Carry out. We solve the equation. 1.6p = 42 p = 26.25

State. 26.25 million prescriptions for sleeping pills were filled in 2000. 116.

mp = F p + F m mp − F p = F m

Check. 60% of 26.25 is 0.6(26.25), or 15.75, and 26.25 + 15.75 = 42. The answer checks.

million. *+ ,  . 42

x+3 x+4 x+5 x+6 − = − , x+2 x+3 x+4 x+5 LCD is (x + 2)(x + 3)(x + 4)(x + 5) x4 + 15x3 + 83x2 + 201x + 180 − x4 − 15x3 −

82x2 − 192x − 160 = x4 + 15x3 + 81x2 + 185x+ 150 − x4 − 15x3 − 80x2 − 180x − 144 x2 + 9x + 20 = x2 + 5x + 6

4x = −14 7 x=− 2 7 7 The number − checks. The solution is − . 2 2 117. (x − 3)2/3 = 2 [(x − 3)2/3 ]3 = 23 (x − 3)2 = 8

x2 − 6x + 9 = 8

x2 − 6x + 1 = 0 a = 1, b = −6, c = 1 √ −b ± b2 − 4ac x= 2a ' −(−6) ± (−6)2 − 4 · 1 · 1 = 2·1 √ √ 6 ± 32 6±4 2 = = 2 2 √ √ 2(3 ± 2 2) =3±2 2 = 2 √ Both values check. The solutions are 3 ± 2 2. ' √ 118. 15 + 2x + 80 = 5 "2 !' √ 15 + 2x + 80 = 52 √ 15 + 2x + 80 = 25 √ 2x + 80 = 10 √ ( 2x + 80)2 = 102 2x + 80 = 100 2x = 20 x = 10 This number checks. The solution is 10. √ √ 6 119. x+5+1 = √ , LCD is x + 5 x+5 √ √ Multiplying by x + 5 x+5+ x+5 = 6 √ x+5 = 1−x x + 5 = 1 − 2x + x2

0 = x2 − 3x − 4

0 = (x − 4)(x + 1) x = 4 or x = −1

Only −1 checks. The solution set is −1.

Exercise Set 2.6

167

x2/3 = x

120. (x

4.

5y − 5 + y ≤ 2 − 6y − 8 6y − 5 ≤ −6y − 6

) =x

2/3 3

3

12y ≤ −1 1 y≤− 12

x2 = x3 0 = x −x 3

2

0 = x2 (x − 1)

# 2 / & 10 1 & The solution set is y &y ≤ − , or − ∞, − , . The 12 12 graph is shown below.

x = 0 or x − 1 = 0 2

x = 0 or x=1 Both numbers check. The solutions are 0 and 1.

1 !! ! 12

Exercise Set 2.6 1.

0

x + 6 < 5x − 6

6 + 6 < 5x − x Subtracting x and adding 6 on both sides 12 < 4x 12 <x Dividing by 4 on both sides 4 3<x This inequality could also be solved as follows: x + 6 < 5x − 6

5.

14 + 8 ≤ 8y + 5y

22 ≤ 13y 22 ≤y 13 This inequality could also be solved as follows: 14 − 5y ≤ 8y − 8

−5y − 8y ≤ −8 − 14 −13y ≤ −22 22 y≥ 13

x − 5x < −6 − 6 Subtracting 5x and 6 on both sides −4x < −12 −12 Dividing by −4 on both sides and x> −4 reversing the inequality symbol x>3

Dividing by −13 on both sides and reversing the inequality symbol 4 1 $ 3 & & 22 22 , or , ∞ . The graph The solution set is y &&y ≥ 13 13 is shown below.

The solution set is {x|x > 3}, or (3, ∞). The graph is shown below. 0

2.

3

3 − x < 4x + 7 −5x < 4

x>−

0

6.

22 ! ! 13

8x − 7 < 6x + 3 2x < 10 x<5

4 5

The solution set is {x|x < 5}, or (−∞, 5). The graph is shown below.

! 4 " / & 40 & The solution set is x&x > − , or − , ∞ . The graph 5 5 is shown below. 4 !! 5

0

7. 0

3.

14 − 5y ≤ 8y − 8

3x − 3 + 2x ≥ 1 − 7x − 9

5x − 3 ≥ −7x − 8 Collecting like terms 5x + 7x ≥ −8 + 3 Adding 7x and 3 on both sides 12x ≥ −5 5 Dividing by 12 on both sides x≥− 12 1 $ / & 50 5 & The solution set is x&x ≥ − , or − , ∞ . The 12 12 graph is shown below. 5 !! ! 12

0

3 − x 4 5 8 5 8 5 8 12 5 · 17 8 15 34

5

5 2 ≥− + x 8 3 3 2 ≥ x+ x 4 3 9 8 ≥ x+ x 12 12 17 ≥ x 12 12 17 ≥ · x 17 12 ≥x

4 # 2 3 & & 15 15 , or − ∞, . x&&x ≤ 34 34 graph is shown below. The solution set is

The

168

Chapter 2: Functions, Equations, and Inequalities 13.

15 ! ! 34

8 ≤ x ≤ 10

0

8.

3 8 5 − x≤ + x 6 4 3 21 3 − x≤ 6 4 3 x≥− 14

0

14.

8

−1 < x − 4 < 7 (3, 11) 0

15.

3

−3 ≤ x + 4 ≤ 3

4x2 − 8x < 4x2 − 14x + 6 −8x < −14x + 6

!7

16.

!1

6x < 6 6 x< 6 x<1

−5 < x + 2 < 15 (−7, 13) !7

The solution set is {x|x < 1}, or (−∞, 1). The graph is shown below.

17.

0

−2 < 2x + 1 < 5

x2 + 3x + 2 > x2 + x 2x > −2 The solution set is {x|x > −1}, or (−1, ∞). The graph is shown below.

Adding −1

1 Multiplying by 2 # $ 3 The solution set is − , 2 . The graph is shown below. 2

(x + 1)(x + 2) > x(x + 1)

x > −1

13

−3 < 2x < 4 3 − <x<2 2

0 1

3 !! 2

18.

0

2

−3 ≤ 5x + 1 ≤ 3

−4 ≤ 5x ≤ 2 2 4 − ≤x≤ 5 5 2 1 4 2 − , 5 5

0

−3 ≤ x < 3

0

−7 < x < 13

−8x + 14x < 6

−2 ≤ x + 1 < 4

Subtracting 4

The solution set is [−7, −1]. The graph is shown below.

4x(x − 2) < 2(2x − 1)(x − 3)

4x(x − 2) < 2(2x2 − 7x + 3)

11.

11

−7 ≤ x ≤ −1

0

!1

10

3 < x < 11

3 !! ! 14

10.

Adding 3

The solution set is [8, 10]. The graph is shown below.

4 1 $ 3 & & 3 3 , or − , ∞ . The The solution set is x&&x ≥ − 14 14 graph is shown below.

9.

5 ≤ x−3≤7

Subtracting 1

The solution set is [−3, 3). The graph is shown below. 4 !! 5 !3

12.

0

3

19.

! 5

−4 ≤ 6 − 2x < 4

−3 < x + 2 ≤ 5

−10 ≤ −2x < −2

−5 < x ≤ 3

5≥x>1

(−5, 3] !5

0 2

or 0

3

Adding −6 Multiplying by −

1<x≤5

1 2

The solution set is (1, 5]. The graph is shown below. 0 1

5

Exercise Set 2.6 20.

169 26.

−3 < 1 − 2x ≤ 3 −4 < −2x ≤ 2

3x < −4 or 3x > 6 4 x < − or x>2 3 $ # 4 ∪ (2, ∞) − ∞, − 3

2 > x ≥ −1

[−1, 2)

!1

21.

0

3x − 1 < −5 or 3x − 1 > 5

2

1 (3x + 1) < 7 2 −10 < 3x + 1 < 14 Multiplying by 2

4 !! 3

−5 <

−11 < 3x < 13 13 11 <x< − 3 3

Adding −1

27.

0

2

2x − 20 < −0.8 or 2x − 20 > 0.8 2x < 19.2 or

1 Multiplying by 3 # $ 11 13 The solution set is − , . The graph is shown be3 3 low.

x < 9.6

22.

13 ! ! 3

0

23.

28.

7 ! 4

5x + 11 ≤ −4 or 5x + 11 ≥ 4 x ≤ −3 or

!3

29.

13 ! ! 6

x>1

The solution set is (−∞, −2] ∪ (1, ∞). The graph is shown below.

24.

x < 4 or

30.

x ≥7

(−∞, 4) ∪ [7, ∞)

25.

2

3

)

4

5

6

[

7

8

9

10

2x + 3 ≤ −4 or 2x + 3 ≥ 4 2x ≤ −7 or 7 x ≤ − or 2

The solution set is shown below. 7 !! 2

0

1 1 or x + 14 ≥ 4 4 55 57 or x≥− x≤− 4 4 # 2 1 $ 57 55 The solution set is − ∞, − ∪ − ,∞ . 4 4 graph is shown below.

The

0

55 !! ! 4

0 1

2x < 8 or x + 3 ≥ 10

1

7 !! 5

x + 14 ≤ −

57 !! ! 4 !2

5x ≥ −7 7 x≥− 5

$ 1 7 (−∞, −3] ∪ − , ∞ 5

3x ≤ −6 or x − 1 > 0 x ≤ −2 or

9.6 10.4

5x ≤ −15 or

4 2 ≤ − (x − 3) < 1 3 5 5 5 − ≥ x−3>− 6 4 13 7 ≥x> 6 4 2 # 7 13 , 4 6 0

x > 10.4

The solution set is (−∞, 9.6) ∪ (10.4, ∞). The graph is shown below. 0

11 !! ! 3

2x > 20.8

or

1 ! 2

0

#

2x ≥ 1 1 x≥ 2 − ∞, −

1 or x − 9 > 2 17 or x> x< 2 # $ # $ 17 19 − ∞, ∪ ,∞ 2 2 x−9 < −

0

2 1 $ 1 7 ∪ , ∞ . The graph is 2 2

31. |x| < 7

1 2 19 2

17 19 ! ! ! 2 ! 2

To solve we look for all numbers x whose distance from 0 is less than 7. These are the numbers between −7 and 7. That is, −7 < x < 7. The solution set and its graph are as follows:

170

Chapter 2: Functions, Equations, and Inequalities 38. |x + 6| > 10

(−7, 7)

(

x + 6 < −10 or x + 6 > 10

) 0

!7

x>4

x < −16 or

7

The solution set is (−∞, −16) ∪ (4, ∞).

32. |x| ≤ 4.5

−4.5 ≤ x ≤ 4.5

!16

The solution set is [−4.5, 4.5]. 39. 0

!4.5

4.5

1 2 1 − 4

{x|x ≤ −4.5 or x ≥ 4.5}, or (−∞, −4.5] ∪ [4.5, ∞) !4.5

[ 0

1 0 !! 4

3 ! 4

0

4.5

40.

34. |x| > 7

|x − 0.5| ≤ 0.2

−0.2 ≤ x − 0.5 ≤ 0.2

x < −7 or x > 7

The solution set is (−∞, −7) ∪ (7, ∞). !7

& 1 && 1 < 4& 2 1 1 <x− < 4 2 3 <x< 4

1 Adding 4 # $ 1 3 The solution set is − , . The graph is shown below. 4 4

To solve we look for all numbers x whose distance from 0 is greater than or equal to 4.5. That is, x ≤ −4.5 or x ≥ 4.5. The solution set and its graph are as follows.

]

4

& & &x − &

−

33. |x| ≥ 4.5

0

0.3 ≤ x ≤ 0.7

The solution set is [0.3,0.7]. 0.3

7

0

0.7

0

35.

|x + 8| < 9

−9 < x + 8 < 9 −17 < x < 1

41. |3x| < 1 Subtracting 8

The solution set is (−17, 1). The graph is shown below. !17

36.

01

−1 < 3x < 1 1 1 Dividing by 3 − <x< 3 3 $ # 1 1 . The graph is shown below. The solution set is − , 3 3

|x + 6| < 10

1 0 1 !! ! 3 3

−10 ≤ x + 6 ≤ 10

−16 ≤ x ≤ 4

42. |5x| ≤ 4

The solution set is [−16, 4]. !16

37. |x + 8| ≥ 9

x + 8 ≤ −9

0

4

or x + 8 ≥ 9

x ≤ −17 or

−4 ≤ 5x ≤ 4 4 4 − ≤x≤ 5 5

x≥1

The solution set is

01

2 4 4 − , . 5 5

Subtracting 8

The solution set is (−∞, −17]∪[1, ∞). The graph is shown below. !17

1

4 !! 5

4 ! 5

0

43. |2x + 3| ≤ 9

−9 ≤ 2x + 3 ≤ 9

−12 ≤ 2x ≤ 6

Subtracting 3

−6 ≤ x ≤ 3

Dividing by 2

The solution set is [−6, 3]. The graph is shown below. !6

0

3

Exercise Set 2.6

171

44. |3x + 4| < 13

−13 < 3x + 4 < 13

17 !! ! 3

#

$ 17 ,3 . 3

−

& 3 && 1 < 4& 4

3

7 !! 3

45. |x − 5| > 0.1

x − 5 < −0.1 or x − 5 > 0.1 x > 5.1

or

Adding 5

The solution set is (−∞, 4.9) ∪ (5.1, ∞). The graph is shown below. 0

& & 50. &&x +

−

0

x < 4.9

& 2 && 5 ≤ 3& 3

5 2 5 ≤ x+ ≤ 3 3 3 7 2 − ≤x≤1 Subtracting 3 3 2 1 7 The solution set is − , 1 . The graph is shown below. 3

−17 < 3x < 9 17 <x<3 − 3

The solution set is

& & 49. &&x +

4.9 5.1

0 1

1 3 1 < x+ < 4 4 4 1 −1 < x < − 2

−

The solution set is

46. |x − 7| ≥ 0.4

x − 7 ≤ −0.4 or x − 7 ≥ 0.4 x ≤ 6.6

x ≥ 7.4

or

The solution set is (−∞, 6.6] ∪ [7.4, ∞). 0

47. |6 − 4x| ≤ 8

−8 ≤ 6 − 4x ≤ 8

−14 ≤ −4x ≤ 2 Subtracting 6 2 14 ≥x≥− Dividing by −4 and revers4 4 ing the inequality symbols 1 7 ≥x≥− Simplifying 2 2 1 2 1 7 The solution set is − , . The graph is shown below. 2 2 7 ! 2

0

48. |5 − 2x| > 10

5 − 2x < −10 or 5 − 2x > 10 −2x < −15 or 15 or x> 2

−2x > 5

5 2 $ # $ # 15 5 ∪ ,∞ . The solution set is − ∞, − 2 2 5 !! 2

0

1 !! 2

− 1, −

$ 1 . 2

0

& & & 2x + 1 & &>5 51. && 3 &

2x + 1 2x + 1 < −5 or >5 3 3 2x + 1 < −15 or 2x + 1 > 15 Multiplying by 3

6.6 7.4

1 !! 2

!1

#

x<−

15 ! ! 2

2x < −16 or

2x > 14 Subtracting 1

x > 7 Dividing by 2

x < −8 or

The solution set is {x|x < −8 or x > 7}, or (−∞, −8) ∪ (7, ∞). The graph is shown below. !8

0

& & & 2x − 1 & 5 &≥ & 52. & 3 & 6 2x − 1 5 ≤− 3 6 5 2x − 1 ≤ − 2 3 2x ≤ − 2 3 x≤− 4

7

or or or or

The solution set is

3 !! 4

53. |2x − 4| < −5

0

2x − 1 5 ≥ 3 6 5 2x − 1 ≥ 2 7 2x ≥ 2 7 x≥ 4 2 1 $ # 7 3 − ∞, − ∪ , ∞ . 4 4 7 ! 4

Since |2x − 4| ≥ 0 for all x, there is no x such that |2x − 4| would be less than −5. There is no solution.

172

Chapter 2: Functions, Equations, and Inequalities

54. |3x + 5| < 0

Translate. Interest at 4%, plus )interest at 5%, is least $325. ) at*+ ) *+ *+     ,  . . . . . 0.04x + 0.05(7500 − x) ≥ 325

|3x + 5| ≥ 0 for all x, so there is no solution.

55. Familiarize and Translate. Spending is given by the equation y = 12.7x + 15.2. We want to know when the spending will be more than $66 billion, so we have 12.7x + 15.2 > 66. Carry out. We solve the inequality. 12.7x + 15.2 > 66 12.7x > 50.8

Carry out. We solve the inequality. 0.04x + 0.05(7500 − x) ≥ 325 0.04x + 375 − 0.05x ≥ 325 −0.01x + 375 ≥ 325

−0.01x ≥ −50

x ≤ 5000

x>4 Check. When x = 4, the spending is 12.7(4) + 15.2 = 66. As a partial check, we could try a value of x less than 4 and one greater than 4. When x = 3.9, we have y = 12.7(3.9) + 15.2 = 64.73 < 66; when x = 4.1, we have y = 12.7(4.1) + 15.2 = 67.27 > 66. Since y = 66 when x = 4 and y > 66 when x = 4.1 > 4, the answer is probably correct. State. The spending will be more than $66 billion more than 4 yr after 2002. 56. Solve: 5x + 5 ≥ 20

x ≥ 3, so 3 or more y after 2002, or in 2005 and later, there will be at least 20 million homes with devices installed that receive and manage broadband TV and Internet content.

57. Familiarize. Let t = the number of hours worked. Then Acme Movers charge 100 + 30t and Hank’s Movers charge 55t. Translate. less than, Acme’s charge. Hank’s charge is *+ *+ , ) *+ , ) )    . . . 55t < 100 + 30t

Carry out. We solve the inequality. 55t < 100 + 30t 25t < 100 t<4

Check. When t = 4, Hank’s Movers charge 55 · 4, or $220 and Acme Movers charge 100 + 30 · 4 = 100 + 120 = $220, so the charges are the same. As a partial check, we find the charges for a value of t < 4. When t = 3.5, Hank’s Movers charge 55(3.5) = $192.50 and Acme Movers charge 100 + 30(3.5) = 100 + 105 = $205. Since Hank’s charge is less than Acme’s, the answer is probably correct. State. For times less than 4 hr it costs less to hire Hank’s Movers. 58. Let x = the amount invested at 4%. Then 12, 000 − x = the amount invested at 6%. Solve: 0.04x + 0.06(12, 000 − x) ≥ 650

x ≤ 3500, so at most $3500 can be invested at 4%. 59. Familiarize. Let x = the amount invested at 4%. Then 7500 − x = the amount invested at 5%. Using the simpleinterest formula, I = P rt, we see that in one year the 4% investment earns 0.04x and the 5% investment earns 0.05(7500 − x).

Check. When $5000 is invested at 4%, then $7500−$5000, or $2500, is invested at 5%. In one year the 4% investment earns 0.04($5000), or $200, in simple interest and the 5% investment earns 0.05($2500), or $125, so the total interest is $200 + $125, or $325. As a partial check, we determine the total interest when an amount greater than $5000 is invested at 4%. Suppose $5001 is invested at 4%. Then $2499 is invested at 5%, and the total interest is 0.04($5001) + 0.05($2499), or $324.99. Since this amount is less than $325, the answer is probably correct. State. The most that can be invested at 4% is $5000. 60. Let c = the number of checks written per month. Solve: 0.20c < 6 + 0.05c c < 40, so the Smart Checking plan will cost less than the Consumer Checking plan when fewer than 40 checks are written per month. 61. Familiarize. Let c = the number of checks written per month. Then the No Frills plan costs 0.35c per month and the Simple Checking plan costs 5 + 0.10c per month. Translate. less than, No cost., Simple Checking cost is *+ ) Frills *+ *+ , ) )    . . . 5 + 0.10c < 0.35c

Carry out. We solve the inequality. 5 + 0.10c < 0.35c 5 < 0.25c 20 < c

Check. When 20 checks are written the No Frills plan costs 0.35(20), or $7 per month and the Simple Checking plan costs 5 + 0.10(20), or $7, so the costs are the same. As a partial check, we compare the cost for some number of checks greater than 20. When 21 checks are written, the No Frills plan costs 0.35(21), or $7.35 and the Simple Checking plan costs 5+0.10(21), or $7.10. Since the Simple Checking plan costs less than the No Frills plan, the answer is probably correct. State. The Simple Checking plan costs less when more than 20 checks are written per month. 62. Let s = the monthly sales. Solve: 750 + 0.1s > 1000 + 0.08(s − 2000)

s > 4500, so Plan A is better for monthly sales greater than $4500.

Exercise Set 2.6

173

63. Familiarize. Let s = the monthly sales. Then the amount of sales in excess of $8000 is s − 8000. Translate. is greater income from Income from than plan A. plan B *+ , ) *+ ,) *+ , )    . . . 1200 + 0.15(s − 8000) > 900 + 0.1s

Carry out. We solve the inequality.

1200 + 0.15(s − 8000) > 900 + 0.1s 1200 + 0.15s − 1200 > 900 + 0.1s 0.15s > 900 + 0.1s 0.05s > 900 s > 18, 000 Check. For sales of $18,000 the income from plan A is $900+0.1($18, 000), or $2700, and the income from plan B is 1200 + 0.15(18, 000 − 8000), or $2700 so the incomes are the same. As a partial check we can compare the incomes for an amount of sales greater than $18,000. For sales of $18,001, for example, the income from plan A is $900 + 0.1($18, 001), or $2700.10, and the income from plan B is $1200 + 0.15($18, 001 − $8000), or $2700.15. Since plan B is better than plan A in this case, the answer is probably correct. State. Plan B is better than plan A for monthly sales greater than $18,000. 64. Solve: 200 + 12n > 20n n < 25

−2x ≤ −2

4x ≤ 4

and

x≥1

x≤1

and

The solution is 1. 79. |3x − 1| > 5x − 2

3x − 1 < −(5x − 2) or 3x − 1 > 5x − 2 3x − 1 < −5x + 2

1 > 2x 1 8x < 3 or >x 2 1 3 or >x x< 8 2 $ # $ # 1 3 ∪ − ∞, . This is equivThe solution set is − ∞, 8 2 # $ 1 alent to − ∞, . 2 or

80. |x + 2| ≤| x − 5|

Divide the set of real numbers into three intervals: (−∞, −2), [−2, 5), and [5, ∞, ).

Find the solution set of |x + 2| ≤| x − 5| in each interval. Then find the union of the three solution sets. If x < −2, then |x + 2| = −(x + 2) and |x − 5| = −(x − 5). Solve: x < −2 and −(x + 2) ≤ −(x − 5) x < −2 and x < −2 and

−x − 2 ≤ −x + 5 −2 ≤ 5

The solution set for this interval is (−∞, −2).

65. Left to the student

If −2 ≤ x < 5, then |x + 2| = x + 2 and |x − 5| = −(x − 5).

66. Left to the student

Solve: −2 ≤ x < 5 and x + 2 ≤ −(x − 5)

67. Absolute value is nonnegative. 68. |x| ≥ 0 > p for any real number x. 69. y-intercept

−2 ≤ x < 5 and x + 2 ≤ −x + 5 −2 ≤ x < 5 and

2x ≤ 3

−2 ≤ x < 5 and

x≤

3 2 1

2 3 . 2 If x ≥ 5, then |x + 2| = x + 2 and |x − 5| = x − 5.

70. distance formula

The solution set for this interval is

71. relation 72. function

− 2,

Solve: x ≥ 5 and x + 2 ≤ x − 5 x ≥ 5 and

73. horizontal line

2 ≤ −5

The solution set for this interval is ∅.

74. parallel 75. decreasing 76. symmetric with respect to the y-axis 77. 2x ≤ 5 − 7x < 7 + x 2x ≤ 5 − 7x and 5 − 7x < 7 + x 9x ≤ 5 5 x≤ 9

78. x ≤ 3x − 2 ≤ 2 − x x ≤ 3x − 2 and 3x − 2 ≤ 2 − x

and

−8x < 2

1 and x>− 4 2 # 1 5 The solution set is − , . 4 9

The union2 of the above three solution set is # 3 − ∞, . This is the solution set of |x + 2| ≤| x − 5|. 2 81. |p − 4| + |p + 4| < 8

If p < −4, then |p − 4| = −(p − 4) and |p + 4| = −(p + 4). Solve: −(p − 4) + [−(p + 4)] < 8 −p + 4 − p − 4 < 8 −2p < 8

p > −4

Since this is false for all values of p in the interval (−∞, −4) there is no solution in this interval.

174

Chapter 2: Functions, Equations, and Inequalities 5 2 5 x<− 2 5 x<− 2 5 x<− 2

If p ≥ −4, then |p + 4| = p + 4.

Solve: x < −

Solve: |p − 4| + p + 4 < 8

|p − 4| < 4 − p

p − 4 > −(4 − p) and p − 4 < 4 − p p−4 > p−4 −4 > −4

and

2p < 8

and

p<4

Thus, |p − 4| + |p + 4| < 8 has no solution.

82. |x| + |x + 1| < 10

If x < −1, then |x| = −x and |x + 1| = −(x + 1) and we have: x < −1 and −x + [−(x + 1)] < 10 x < −1 and

−x − x − 1 < 10 −2x − 1 < 10

−2x < 11 11 x < −1 and x>− 2 $ # 11 The solution set for this interval is − , −1 . 2 x < −1 and

The union of the above solution sets is $ # 8 − ∞, − ∪ (−2, ∞). This is the solution set of 3 |x − 3| + |2x + 5| > 6.

x ≥ 0 and x + x + 1 < 10

x ≥ 0 and

2x < 9 9 x< 2

1 $ 9 The solution set for this interval is 0, . 2 The union of the three solution sets above is $ # 11 9 . This is the solution set of − , 2 2 |x| + |x + 1| < 10. 83. |x − 3| + |2x + 5| > 6

Divide the set of real numbers into three intervals: $ 1 $ # 5 5 − ∞, − , − , 3 , and [3, ∞). 2 2

Find the solution set of |x − 3| + |2x + 5| > 6 in each interval. Then find the union of the three solution sets. 5 If x < − , then |x−3| = −(x−3) and |2x+5| = −(2x+5). 2

and

3x > 4 4 x ≥ 3 and x> 3 The solution set in this interval is [3, ∞).

1 < 10

2x + 1 < 10

−3x > 8

x ≥ 3 and

If x ≥ 0, then |x| = x and |x + 1| = x + 1 and we have:

x ≥ 0 and

and

Solve: x ≥ 3 and x − 3 + 2x + 5 > 6

The solution set for this interval is [−1, 0].

x ≥ 0 and

−x + 3 − 2x − 5 > 6

If x ≥ 3, then |x − 3| = x − 3 and |2x + 5| = 2x + 5.

If −1 ≤ x < 0, then |x| = −x and |x + 1| = x + 1 and we have: −1 ≤ x and −x + x + 1 < 10 −1 ≤ x and

and

8 x<− 3 $ # 8 The solution set in this interval is − ∞, − . 3 5 If − ≤ x < 3, then |x−3| = −(x−3) and |2x+5| = 2x+5. 2 5 Solve: − ≤ x < 3 and −(x − 3) + 2x + 5 > 6 2 5 − ≤ x < 3 and −x + 3 + 2x + 5 > 6 2 5 − ≤ x < 3 and x > −2 2 The solution set in this interval is (−2, 3).

Since −4 > −4 is false for all values of p, there is no solution in the interval [−4, ∞).

x < −1 and

and −(x − 3) + [−(2x + 5)] > 6

Chapter 2 Review Exercises 1. The statement is false. We find the zeros of a function f (x) by finding the values of x for which f (x) = 0. 2. The statement is true. See page 186 in the text. 3. The statement is true. See page 205 in the text. 4. The statement is true. See page 217 in the text. 5. The statement is false. For example, 32 = (−3)2 , but 3 "= −3. 6. If a < b and c < 0 are true, then ac > bc is true. Thus, the statement is false. 7.

4y − 5 = 1

4y = 6 3 y= 2

The solution is 8.

3 . 2

3x − 4 = 5x + 8 −12 = 2x −6 = x

Chapter 2 Review Exercises 9.

5(3x + 1) = 2(x − 4)

175 19.

−2x = −8

15x + 5 = 2x − 8

x=4

13x = −13

The zero of the function is 4.

x = −1

The solution is −1. 10.

20.

x−1 = 0

2n − 6 = 3n + 15

11. (2y + 5)(3y − 1) = 0 2y + 5 = 0

x=1

21.

2y = −5 or 5 y = − or 2

12.

x2 + 4x − 5 = 0

(x + 5)(x − 1) = 0

x = −5 or x = 1 13.

3x2 + 2x = 8 3x2 + 2x − 8 = 0

x+5 = 0

x = −5 or

23. 3x2 + 2x − 3 = 0

a = 3, b = 2, c = −3 √ −b ± b2 − 4ac x= 2a ' −2 ± 22 − 4 · 3 · (−3) x= 2·3 √ √ √ −2 ± 2 10 −1 ± 10 −2 ± 40 = = = 6 6 3 √ −1 ± 10 The zeros of the function are . 3

or 3x − 4 = 0

3x = 4 4 x = −2 or x= 3 4 The solutions are −2 and . 3 5x2 = 15 x2 = 3 √ √ x = − 3 or x = 3 15.

24.

x2 − 10 = 0

7x − 27 = 0

7x = 27 27 x= 7 This number checks.

6x − 18 = 0

6x = 18 x=3

17.

x−4 = 0

x=4

The zero of the function is 4. 18.

2 − 10x = 0

−10x = −2 x=

1 5

1 5 + = 0, LCD is (2x + 3)(x − 6) 2x + 3 x − 6 5(x − 6) + 2x + 3 = 0 5x − 30 + 2x + 3 = 0

x2 = 10 √ √ x = − 10 or x = 10 √ √ The solutions are − 10 and 10. 16.

x=3

22. 2x2 − x − 5 = 0 ' −(−1) ± (−1)2 − 4 · 2 · (−5) x= 2·2 √ 1 ± 41 = 4

x = −2 or

14.

or x − 3 = 0

The zeros of the function are −5 and 3.

(x + 2)(3x − 4) = 0 x+2 = 0

x + 2x − 15 = 0 2

(x + 5)(x − 3) = 0

or 3y − 1 = 0

3y = 1 1 y= 3 5 1 The solutions are − and . 2 3

x2 − 2x + 1 = 0

(x − 1)2 = 0

2(n − 3) = 3(n + 5) −21 = n

8 − 2x = 0

25.

3 8 + =1 8x + 1 2x + 5 LCD is$(8x+1)(2x+5) # 8 3 + = (8x+1)(2x+5)·1 (8x+1)(2x+5) 8x+1 2x+5 3(2x + 5) + 8(8x + 1) = (8x+1)(2x+5) 6x + 15 + 64x + 8 = 16x2 + 42x + 5 70x + 23 = 16x2 + 42x + 5 0 = 16x2 −28x−18

0 = 2(8x2 −14x−9)

0 = 2(2x+1)(4x−9)

176

Chapter 2: Functions, Equations, and Inequalities 2x + 1 = 0

or 4x − 9 = 0

2x = −1 or 1 x = − or 2

4x = 9 9 x= 4

Both numbers check. The solutions are − √

26.

√

5x + 1 − 1 =

√

31.

2 < 5x ≤ 10 2 <x≤2 5

3x

32.

5x + 1 = x2 + 2x + 1 0 = x2 − 3x

Both numbers check. √ √ 27. x−1− x−4 = 1 √ √ x−1 = x−4+1 √ √ ( x − 1)2 = ( x − 4 + 1)2 √ x−1 = x−4+2 x−4+1 √ x−1 = x−3+2 x−4 √ 2 = 2 x−4 √ Dividing by 2 1 = x−4 √ 12 = ( x − 4)2 1 = x−4 5=x

Every real number is less than −

33.

34.

x=7

or

2y = 2 y=1

2 5 ∪ [1, ∞). 3

|6x − 1| < 5

x + 4 ≤ −2 or x + 4 ≥ 2 x ≤ −6 or

−3 ≤ 3x + 1 ≤ 5

−4 ≤ 3x ≤ 4 4 4 − ≤x≤ 3 3 2 1 4 4 − , 3 3

− ∞, −

35. |x + 4| ≥ 2

The solutions are −8 and 1. 30.

x≥1

or 2y + 7 = 9

2y = −16 or y = −8

#

−4 < 6x < 6 2 − <x<1 3 $ # 2 − ,1 3

The solutions are 1 and 7. 2y + 7 = −9

2x ≥ 2

−5 < 6x − 1 < 5

x − 4 = −3 or x − 4 = 3

29. |2y + 7| = 9

or 2x + 3 ≥ 5

The solution set is

28. |x − 4| = 3

or

3x + 7 ≤ 2

3x ≤ −5 or 5 x ≤ − or 3

This number checks. The solution is 5.

x=1

2x < −1 or x + 3 > 0 1 x > −3 x < − or 2

1 or greater than −3, so 2 the solution set is the set of all real numbers, or (−∞, ∞).

0 = x(x − 3)

x = 0 or x = 3

Adding 4

Dividing by 5 2 # 2 ,2 . The solution set is 5

1 9 and . 2 4

5x + 1 − 2 5x + 1 + 1 = 3x √ −2 5x + 1 = −2x − 2 √ 5x + 1 = x + 1

−2 < 5x − 4 ≤ 6

x ≥ −2

The solution is (−∞, −6] ∪ [−2, ∞).

36.

37.

V = lwh V =h lw M = n + 0.3s M − n = 0.3s M −n =s 0.3

Chapter 2 Review Exercises 38.

v=

√

177 3 9 3 9 − or x = + 2 2 2 2 x = −3 or x = 6

2gh

x=

v 2 = 2gh v2 =h 2g 39.

The solutions are −3 and 6.

1 1 1 + = , LCD is abt a b t # $ 1 1 1 + = abt · abt a b t bt + at = ab

t(b + a) = ab ab t= a+b √ √ √ √ √ 40. − −40 = − −1 · 4 · 10 = −2 10i √ √ √ √ √ √ √ √ 41. −12 · −20 = −1 · 4 · 3 · −1 · 4 · 5 √ √ = 2i 3 · 2i 5 √ = 4i2 3 · 5 √ = −4 15 √ 7i −49 7 42. √ = =− −8i 8 − −64 43.

(6 + 2i)(−4 − 3i) = −24 − 18i − 8i − 6i2 = −24 − 26i + 6 = −18 − 26i

44.

45. 46.

2 − 3i 1 + 3i 2 − 3i = · 1 − 3i 1 − 3i 1 + 3i 2 + 3i − 9i2 = 1 − 9i2 2 + 3i + 9 = 1+9 11 + 3i = 10 3 11 + i = 10 10 (3 − 5i) − (2 − i) = (3 − 2) + [−5i − (−i)] = 1 − 4i

(6 + 2i) + (−4 − 3i) = (6 − 4) + (2i − 3i) = 2−i

47. i23 = (i2 )11 · i = (−1)11 · i = −1 · i = −i 48. 49.

(−3i)28 = (−3)28 · i28 = 328 · (i2 )14 =

328 · (−1)14 = 328 x2 − 3x = 18

9 = 4 $2 # 3 = x− 2 3 x− = 2

x2 −3x +

18+

9 4

81 4 9 ± 2 3 9 x= ± 2 2

# # $2 $ 9 1 3 3 = (−3) = − and − 2 2 2 4

50.

3x2 − 12x − 6 = 0

3x2 − 12x = 6 x2 − 4x = 2

x2 − 4x + 4 = 2+4 (x − 2)2 = 6

# $ 1 (−4) = −2 and (−2)2 = 4 2

√ x−2 = ± 6 √ x = 2± 6

3x2 + 10x = 8

51.

3x2 + 10x − 8 = 0

(x + 4)(3x − 2) = 0 x+4 = 0

or 3x − 2 = 0

x = −4 or

3x = 2 2 x = −4 or x= 3 2 The solutions are −4 and . 3 52. r2 − 2r + 10 = 0 ' −(−2) ± (−2)2 − 4 · 1 · 10 r= 2·1 √ 2 ± 6i 2 ± −36 = = 2 2 = 1 ± 3i x2 = 10 + 3x

53.

x − 3x − 10 = 0 2

(x + 2)(x − 5) = 0 x+2 = 0

or x − 5 = 0

x = −2 or

x=5

The solutions are −2 and 5. √ 54. x = 2 x−1 √ x−2 x+1 = 0 √ Let u = x. u2 − 2u + 1 = 0

(u − 1)2 = 0 u−1 = 0

Substitute √ x=1

u=1 √ x for u and solve for x.

x=1 55. y 4 − 3y 2 + 1 = 0 Let u = y 2 .

u2 − 3u + 1 = 0 ' √ −(−3) ± (−3)2 − 4 · 1 · 1 3± 5 = u= 2·1 2

178

Chapter 2: Functions, Equations, and Inequalities Substitute y 2 for u and solve for y. √ 3± 5 y2 = 2 5 √ 3± 5 y=± 2 5 √ 3± 5 . The solutions are ± 2

d) Range:

#

− ∞, −

7 16

2

e) Since the graph opens down, function values increase to the left of the vertex and decrease to the right of the (x) is $ increasing on $ vertex. Thus, f # # 3 3 and decreasing on ,∞ . − ∞, 8 8

f)

56. (x2 − 1)2 − (x2 − 1) − 2 = 0 Let u = x2 − 1. u2 − u − 2 = 0

(u + 1)(u − 2) = 0

u+1 = 0

or u − 2 = 0

u = −1 or

u=2

Substitute x2 − 1 for u and solve for x. x2 − 1 = −1 or x2 − 1 = 2 x =0

or

x=0

or

2

x =3 2

p − 3 = 0 or 3p + 2 = 0

3p = −2 or 2 p = 3 or p = − or 3 2 The solutions are −2, − and 3. 3

= 5(x2 − 2x + 1) − 5 · 1 + 3 p = −2 p = −2

(x + 5)(x2 − 4) = 0

59.

or x + 2 = 0

a) Vertex: (1, −2)

b) Axis of symmetry: x = 1 c) Minimum value: −2 e) Since the graph opens up, function values decrease to the left of the vertex and increase to the right of the vertex. Thus, f (x) is increasing on (1, ∞) and decreasing on (−∞, 1).

x2 (x + 5) − 4(x + 5) = 0 (x + 5)(x + 2)(x − 2) = 0

= 5(x − 1)2 − 2

d) Range: [−2, ∞)

x3 + 5x2 − 4x − 20 = 0

x = −5 or

= 5(x2 − 2x) + 3

= 5(x2 − 2x + 1 − 1) + 3

or p + 2 = 0

p = 3 or

x+5 = 0

f (x) = 5x2 − 10x + 3

√ x=± 3

57. (p − 3)(3p + 2)(p + 2) = 0

58.

60.

f) or x − 2 = 0

x = −2 or

x=2

f (x) = −4x2 + 3x − 1 # $ 3 2 = −4 x − x − 1 4 $ # 9 9 3 − −1 = −4 x2 − x + 4 64 64 # $ # $ 3 9 9 2 = −4 x − x + −4 − −1 4 64 64 $ # 3 9 9 = −4 x2 − x + + −1 4 64 16 $2 # 7 3 − = −4 x − 8 16 $ # 3 7 ,− a) Vertex: 8 16 3 b) Axis of symmetry: x = 8 7 c) Maximum value: − 16

61. The graph of y = (x − 2)2 has vertex (2, 0) and opens up. It is graph (d). 62. The graph of y = (x + 3)2 − 4 has vertex (−3, −4) and opens up. It is graph (c). 63. The graph of y = −2(x + 3)2 + 4 has vertex (−3, 4) and opens down. It is graph (b). 1 64. The graph of y = − (x − 2)2 + 5 has vertex (2, 5) and 2 opens down. It is graph (a). 65. Familiarize. Using the labels in the textbook, the legs of the right triangle are represented by x and x + 10.

Chapter 2 Review Exercises

179

Translate. We use the Pythagorean theorem. x2 + (x + 10)2 = 502 Carry out. We solve the equation. x2 + (x + 10)2 = 502 x + x2 + 20x + 100 = 2500 2

2x2 + 20x − 2400 = 0

2(x2 + 10x − 1200) = 0 2(x + 40)(x − 30) = 0

x + 40 = 0

or x − 30 = 0

x = −40 or

x = 30

Check. Since the length cannot be negative, we need to check only 30. If x = 30, then x+10 = 30+10 = 40. Since 302 + 402 = 900 + 1600 = 2500 = 502 , the answer checks. State. The lengths of the legs are 30 ft and 40 ft. 66. Let r = the speed of the boat in still water. 8 8 + =3 Solve: r−2 r+2 2 r = − or r = 6 3 Only 6 has meaning in the original problem. The speed of the boat in still water is 6 mph. 67. Familiarize. Let r = the speed of the second train, in km/h. After 1 hr the first train has traveled 60 km, and the second train has traveled r km, and they are 100 km apart. We make a drawing. ! !

r

! ! ! !

100 km ! ! ! ! !! 60 km

Translate. We use the Pythagorean theorem. 602 + r2 = 1002 Carry out. We solve the equation. 602 + r2 = 1002 3600 + r2 = 10, 000 r2 = 6400 r = ±80

Check. Since the speed cannot be negative, we need to check only 80. We see that 602 + 802 = 3600 + 6400 = 10, 000 = 1002 , so the answer checks. State. The second train is traveling 80 km/h. 68. Let w = the width of the sidewalk. 2 Solve: (80 − 2w)(60 − 2w) = · 80 · 60 3 √ w = 35 ± 5 33

√ If w = 35 + 5 33 ≈ 64, both the new length, 80 − 2w, √ and the new width, 60 − 2w, would be negative, so 35 + 5 33 cannot be a solution. √ The other number, 35 − 5 33 ft ≈ 6.3 ft, checks in the original problem. 69. Familiarize. Let l = the length of the toy corral, in ft. 24 − 2l , or 12 − l. The height of the Then the width is 2 corral is 2 ft. Translate. We use the formula for the volume of a rectangular solid, V = lwh. V (l) = l(12 − l)(2) = 24l − 2l2

= −2l2 + 24l

Carry out. Since V (l) is a quadratic function with a = −2 < 0, the maximum function value occurs at the vertex of the graph of the function. The first coordinate of the vertex is 24 b =− =6 − 2a 2(−2) When l = 6, then 12 − l = 12 − 6 = 6.

Check. The volume of a corral with length 6 ft, width 6 ft, and height 2 ft is 6 · 6 · 2, or 72 ft3 . As a partial check, we can find V (l) for a value of l less than 6 and for a value of l greater than 6. For instance, V (5.9) = 71.98 and V (6.1) = 71.98. Since both of these values are less than 72, our result appears to be correct. State. The dimensions of the corral should be 6 ft by 6 ft.

70. Using the labels in the textbook, let x = the length of the sides of the squares, in cm. Solve: (20 − 2x)(10 − 2x) = 90 √ 15 ± 115 x= 2 √ 15 + 115 If x = ≈ 12.9, both the length of the base, 2 20 − √ 2x, and the width 10 − 2x, would be negative, so 15 + 115 cannot be a solution. 2 √ 15 − 115 cm ≈ 2.1 cm, checks in the The other number, 2 original problem. 71. Familiarize and Translate. The number of faculty, in thousands, is given by the equation y = 6x+121. We want to know when this number will exceed 325 thousand, so we have 6x + 121 > 325. Carry out. We solve the inequality. 6x + 121 > 325 6x > 204 x > 34 Check. When x = 34. the number of faculty is 6 · 34 + 121 = 325. As a partial check, we could try a value of x less than 34 and one greater than 34. When x = 33.9 we have y = 6(33.9) + 121 = 324.4 < 325; when x = 34.1

180

Chapter 2: Functions, Equations, and Inequalities we have y = 6(34.1) + 121 = 325.6 > 325. Since y = 325 when x = 34 and y > 325 when x > 34, the answer is probably correct.

81.

x $2 x '√ x 6'√ 72 x √ x √ 2 ( x)

#(

State. There will be more than 325 thousand faculty members more than 34 years after 1970, or in years after 2004. 5 72. Solve: (F − 32) < 45 9 F < 113, so Celsius temperature is lower than 45◦ for Fahrenheit temperatures less than 113◦ . 73.

2x2 − 5x + 1 = 0

a = 2, b = −5, c = 1 √ −b ± b2 − 4ac x= 2a ' √ −(−5) ± (−5)2 − 4 · 2 · 1 5 ± 25 − 8 = = 2·2 4 √ 5 ± 17 = 4 Answer B is correct. √ √ 74. 4x + 1 + 2x = 1 √ √ 4x + 1 = 1 − 2x √ √ ( 4x + 1)2 = (1 − 2x)2 √ 4x + 1 = 1 − 2 2x + 2x √ 2x = −2 2x √ x = − 2x √ x2 = (− 2x)2 x2 = 2x x2 − 2x = 0

(' √

'√

=2 = 22 =4 = 42 = 16 = 162

x = 256 The answer checks. The solution is 256. 82.

(x − 1)2/3 = 4

(x − 1)2 = 43 √ x − 1 = ± 64 x − 1 = ±8

x − 1 = −8 or x − 1 = 8 x = −7 or

x=9

Both numbers check. 83.

(t − 4)4/5 = 3 8 95 (t − 4)4/5 = 35

(t − 4)4 = 243 √ t − 4 = ± 4 243 √ t = 4 ± 4 243

√ √ The exact solutions are 4 + 4 243 and 4 − 4 243. The approximate solutions are 7.948 and 0.052. √ √ 84. x + 2 + 4 x + 2 − 2 = 0 √ √ √ Let u = 4 x + 2, so u2 = ( 4 x + 2)2 = x + 2. u2 + u − 2 = 0

x(x − 2) = 0

(u + 2)(u − 1) = 0

x = 0 or x = 2

u = −2 or u = 1 √ Substitute 4 x + 2 for u and solve for x. √ √ 4 x + 2 = −2 or 4 x + 2 = 1

Only 0 checks, so answer B is correct. 75. Left to the student 76. Left to the student

No real solution

x+2 = 1

77. Left to the student This number checks.

78. Left to the student 79. If an equation contains no fractions, using the addition principle before using the multiplication principle eliminates the need to add or subtract fractions. 80. You can conclude that |a1 | = |a2 | since these constants determine how wide the parabolas are. Nothing can be concluded about the h’s and the k’s.

x = −1

85.

(2y − 2)2 + y − 1 = 5

4y − 8y + 4 + y − 1 = 5 2

4y 2 − 7y + 3 = 5

4y 2 − 7y − 2 = 0

(4y + 1)(y − 2) = 0

4y + 1 = 0

or y − 2 = 0

4y = −1 or 1 y = − or 4

The solutions are −

y=2 y=2

1 and 2. 4

Chapter 2 Test

181

86. The maximum value occurs at the vertex. The first # coordi$ b b b b =− = and f = 2. nate of the vertex is − 2a 2(−3) 6 6 # $2 # $ b b −3 +b −1 = 2 6 6 b2 b2 −1 = 2 − + 12 6 2 2 −b + 2b − 12 = 24

4. (2x − 1)(x + 5) = 0

2x − 1 = 0 or x + 5 = 0 2x = 1 or 1 x= or 2

The solutions are 5.

b = 36 2

6.

r = 0.09 or r = −3.23 Check. Since the interest rate cannot be negative, we need to check only 0.09. At 9%, the $3500 deposit would grow to $3500(1 + 0.09)2 , or $4158.35. The $4000 deposit would grow to $4000(1+0.09), or $4360. Since $4158.35+$4360 = $8518.35, the answer checks. State. The interest rate was 9%.

Chapter 2 Test 1.

6x + 7 = 1 6x = −6

2.

x = −1 The solution is −1. 3y − 4 = 5y + 6 −4 = 2y + 6

−10 = 2y

3.

−5 = y The solution is −5.

2(4x + 1) = 8 − 3(x − 5) 8x + 2 = 8 − 3x + 15 8x + 2 = 23 − 3x

11x + 2 = 23

11x = 21 21 x= 11 21 . The solution is 11

6x2 − 36 = 0

x2 + 4 = 0

x2 = −4 √ x = ± −4 x = −2i or x = 2i

3500(1 + r)2 + 4000(1 + r) = 8518.35.

3500u2 + 4000u − 8518.35 = 0 Using the quadratic formula, we find that u = 1.09 or u ≈ −2.23. Substitute 1 + r for u and solve for r. 1 + r = 1.09 or 1 + r = −2.23

1 and −5. 2

x2 = 6 √ √ x = − 6 or x = 6 √ √ The solutions are − 6 and 6.

Translate. The amount in the account at the end of 2 years was $8518.35, so we have Carry out. We solve the equation. Let u = 1 + r. 3500u2 + 4000u = 8518.35

x = −5

6x2 = 36

b = ±6

87. Familiarize. When principal P is deposited in an account at interest rate r, compounded annually, the amount A to which it grows in t years is given by A = P (1 + r)t . In 2 years the $3500 deposit had grown to $3500(1 + r)2 . In one year the $4000 deposit had grown to $4000(1 + r).

x = −5

The solutions are −2i and 2i. 7.

x2 − 2x − 3 = 0

(x + 1)(x − 3) = 0 x+1 = 0

or x − 3 = 0

x = −1 or

x=3

The solutions are −1 and 3.

8. x2 − 5x + 3 = 0

a = 1, b = −5, c = 3 √ −b ± b2 − 4ac x= 2a ' −(−5) ± (−5)2 − 4 · 1 · 3 x= 2·1 √ 5 ± 13 = 2 √ √ 5 + 13 5 − 13 The solutions are and . 2 2

9. 2t2 − 3t + 4 = 0

a = 2, b = −3, c = 4 √ −b ± b2 − 4ac x= 2a ' −(−3) ± (−3)2 − 4 · 2 · 4 x= 2·2 √ √ 3 ± i 23 3 ± −23 = = 4 4 √ 3 23 = ± i 4 4 √ √ 3 23 3 23 The solutions are + i and − i. 4 4 4 4 √ 10. x + 5 x − 36 = 0 √ Let u = x. u2 + 5u − 36 = 0

(u + 9)(u − 4) = 0

182

Chapter 2: Functions, Equations, and Inequalities u+9 = 0

or u − 4 = 0

u = −9 or u=4 √ Substitute x for u and solve for x. √ √ x = −9 or x=4 No solution

16.

2x ≤ 4 or

3 2 + = 2, LCD is (3x+4)(x−1) 3x + 4 x−1 $ # 3 2 (3x+4)(x−1) + = (3x+4)(x−1)(2) 3x+4 x−1 3(x − 1) + 2(3x + 4) = 2(3x2 + x − 4)

5x ≥ 20

x ≤ 2 or

x = 16

x≥4

The solution set is (−∞, 2] ∪ [4, ∞).

The number 16 checks. It is the solution. 11.

2x − 1 ≤ 3 or 5x + 6 ≥ 26

17.

|x + 3| ≤ 4

−4 ≤ x + 3 ≤ 4 −7 ≤ x ≤ 1

3x − 3 + 6x + 8 = 6x2 + 2x − 8

The solution set is [−7, 1].

9x + 5 = 6x2 + 2x − 8

0 = 6x2 − 7x − 13

0 = (x + 1)(6x − 13)

x+1 = 0

or 6x − 13 = 0

x = −1 or

18. |x + 5| > 2

x + 5 < −2 or x + 5 > 2

6x = 13 13 x= 6

x = −1 or

13 Both numbers check. The solutions are −1 and . 6 √ 12. x+4−2 = 1 √ x+4 = 3 √ ( x + 4)2 = 32

x < −7 or

The solution set is (−∞, −7) ∪ (−3, ∞).

19.

x+4 = 9 x=5 This number checks. √ √ 13. x+4− x−4 = √ x+4 = √ ( x + 4)2 = x+4 = 4= 1= 1 = 2

The solution is 5. 20.

2 √

x−4+2 √ ( x − 4 + 2)2 √ x−4+4 x−4+4 √ 4 x−4 √ x−4 √ ( x − 4)2

1 = x−4 5=x

This number checks. The solution is 5. 14. |4y − 3| = 5

4y − 3 = −5 or 4y − 3 = 5 4y = −2 or 1 y = − or 2

The solutions are − 15.

−7 < 2x + 3 < 9

−10 < 2x < 6 −5 < x < 3

4y = 8 y=2

1 and 2. 2 Subtracting 3 Dividing by 2

The solution set is (−5, 3).

x > −3

V =

2 2 πr h 3

3 3V = πr2 h Multiplying by 2 2 3V =h Dividing by πr2 2πr2 √ R = 3np √ R2 = ( 3np)2 R2 = 3np R2 =n 3p

21.

x2 + 4x = 1 #

x + 4x + 4 = 1 + 4 2

(x + 2)2 = 5

√ x+2 = ± 5

x = −2 ±

√

The solutions are −2 +

$ 1 2 (4) = 2 and 2 = 4 2

5 √

5 and −2 −

√

5.

3 22. Familiarize. Let l = the length, in meters. Then l = 4 the width. Recall that the formula for the perimeter P of a rectangle with length l and width w is P = 2l + 2w. Translate. The perimeter *+ , )  . 3 2l + 2 · l 4

is 210 m.  ) *+  , . .

=

210

Chapter 2 Test

183 Translate. We express 25/c as $0.25.

Carry out. We solve the equation. 3 2l + 2 · l = 210 4 3 2l + l = 210 2 7 l = 210 2 l = 60

Wholesale price  . p

23. Familiarize. Let c = the speed of the current, in km/h. The boat travels downstream at a speed of 12 + c and upstream at a speed of 12 − c. Using the formula d = rt in d the form t = , we see that the travel time downstream is r 45 45 and the time upstream is . 12 + c 12 − c Translate. hr. Total travel time, is )8 *+ *+ )   , . . . 45 45 + = 8 12 + c 12 − c Carry out. We solve the equation. First we multiply both sides by the LCD, (12 + c)(12 − c). 45 45 + =8 12 + c 12 − c $ # 45 45 + = (12+c)(12−c)(8) (12+c)(12− c) 12+c 12−c 45(12 − c) + 45(12 + c) = 8(144 − c2 )

p = 1.8 Check. 50% of $1.80 is $0.90 and $1.80 + $0.90 + $0.25 = $2.95, so the answer checks. State. The wholesale price of a bottle of juice is $1.80. √ √ √ √ √ 25. −43 = −1 · 43 = i 43, or 43i √ √ √ 26. − −25 = − −1 · 25 = −5i 27. 28.

(5 − 2i) − (2 + 3i) = (5 − 2) + (−2i − 3i) = 3 − 5i

(3 + 4i)(2 − i) = 6 − 3i + 8i − 4i2 = 6 + 5i + 4

= 10 + 5i 29.

= = = =

0 = 72 − 8c

=

0 = 8(9 − c2 )

Check. Since the speed of the current cannot be negative, we need to check only 3. If the speed of the current is 3 km/h, then the boat’s speed downstream is 12 + 3, or 15 km/h, and the speed upstream is 12 − 3, or 9 km/h. 45 , or 3 hr, to travel 45 km At 15 km/h, it takes the boat 15 45 downstream. At 9 km/h, it takes the boat , or 5 hr, to 9 travel 45 km upstream. The total travel time is 3 hr + 5 hr, or 8 hr, so the answer checks. State. The speed of the current is 3 km/h. 24. Familiarize. Let p = the wholesale price of the juice.

(i2 = −1)

1 − i 6 − 2i 1−i = · 6 + 2i 6 + 2i 6 − 2i

2

3=c

 . 2.95

1.5p = 2.7

1080 = 1152 − 8c2

0 = 8(3 + c)(3 − c)

  . . 0.25 =

1.5p + 0.25 = 2.95

540 − 45c + 540 + 45c = 1152 − 8c2

or 3 − c = 0

 . +

p + 0.5p + 0.25 = 2.95

State. The length is 60 m and the width is 45 m.

c = −3 or

 . +

plus $0.25 is $2.95.

Carry out. We solve the equation.

3 3 If l = 60, then l = · 60 = 45. 4 4 Check. The width, 45 m, is three-fourths of the length, 60 m. Also, 2 · 60 m + 2 · 45 m = 210 m, so the answer checks.

3+c = 0

50% of wholesale price  . 0.5p

plus

6 − 2i − 6i + 2i2 36 − 4i2 6 − 8i − 2 36 + 4 4 − 8i 40 4 8 − i 40 40 1 1 − i 10 5

30. i33 = (i2 )16 · i = (−1)16 · i = 1 · i = i 31.

3x + 9 = 0 3x = −9 x = −3

The zero of the function is −3. 32.

4x2 − 11x − 3 = 0

(4x + 1)(x − 3) = 0 4x + 1 = 0

or x − 3 = 0

4x = −1 or 1 x = − or 4

x=3 x=3

The zeros of the functions are −

1 and 3. 4

184

Chapter 2: Functions, Equations, and Inequalities

33. 2x2 − x − 7 = 0

If w = 20, then 80 − 2w = 80 − 2 · 20 = 40.

a = 2, b = −1, c = −7 √ −b ± b2 − 4ac x= 2a ' −(−1) ± (−1)2 − 4 · 2 · (−7) x= 2·2 √ 1 ± 57 = 4 √ √ 1 − 57 1 + 57 and . The solutions are 4 4

34.

f (x) = −x2 + 2x + 8

= −(x2 − 2x) + 8

= −(x2 − 2x + 1 − 1) + 8

= −(x2 − 2x + 1) − (−1) + 8 = −(x2 − 2x + 1) + 1 + 8 = −(x − 1)2 + 9

Check. The area of a rectangle with length 40 ft and width 20 ft is 40 · 20, or 800 ft2 . As a partial check, we can find A(w) for a value of w less than 20 and for a value of w greater than 20. For instance, A(19.9) = 799.98 and A(20.1) = 799.98. Since both of these values are less than 800, the result appears to be correct. State. The dimensions for which the area is a maximum are 20 ft by 40 ft. 36. Familiarize. Let t = the number of hours a move requires. Then Morgan Movers charges 90+25t to make a move and McKinley Movers charges 40t. Translate. McKinley Movers’ Morgan Movers’ is ) less *+than, charge. charge *+ , ) *+ , )    . . . 90 + 25t < 40t

Carry out. We solve the inequality.

a) Vertex: (1, 9)

90 + 25t < 40t

b) Axis of symmetry: x = 1

90 < 15t

c) Maximum value: 9

6
d) Range: (−∞, 9] e) The graph opens down, so function values increase to the left of the vertex and decrease to the right of the vertex. Thus, f (x) is increasing on (−∞, 1) and decreasing on (1, ∞). f)

Check. For t = 6, Morgan Movers charge 90 + 25 · 6, or $240, and McKinley Movers charge 40 · 6, or $240, so the charge is the same for 6 hours. As a partial check, we can find the charges for a value of t greater than 6. For instance, for 6.5 hr Morgan Movers charge 90 + 25(6.5), or $252.50, and McKinley Movers charge 40(6.5), or $260. Since Morgan Movers cost less for a value of t greater than 6, the answer is probably correct. State. It costs less to hire Morgan Movers when a move takes more than 6 hr.

35. Familiarize. We make a drawing, letting w = the width of the rectangle, in ft. This leaves 80 − w − w, or 80 − 2w ft of fencing for the length. House w

w 80 − 2w

Translating. The area of a rectangle is given by length times width. A(w) = (80 − 2w)w = 80w − 2w2 , or − 2w2 + 80w

Carry out. This is a quadratic function with a < 0, so it has a maximum value that occurs at the vertex of the graph of the function. The first coordinate of the vertex is 80 b =− = 20. w=− 2a 2(−2)

37. The maximum value occurs at the vertex. The first # $coordi(−4) 2 2 b =− = and f = 12. nate of the vertex is − 2a 2a a a Then we have: # $ # $2 2 2 + 3 = 12 −4 a a a 8 4 a · 2 − + 3 = 12 a a 4 8 − + 3 = 12 a a 4 − + 3 = 12 a 4 − =9 a −4 = 9a 4 − =a 9

Chapter 3

Polynomial and Rational Functions 10. f (x) = 12 + x = x + 12

Exercise Set 3.1 1 3 x − 10x + 8 2 The degree of the polynomial is 3, so the polynomial is 1 cubic. The leading term is x3 and the leading coefficient 2 1 is . 2

1. g(x) =

The degree of the polynomial is 1, so the polynomial is linear. The leading term is x and the leading coefficient is 1 (x = 1 · x). 11. f (x) = −3x3 − x + 4

The leading term is −3x3 . The degree, 3, is odd and the leading coefficient, −3, is negative. Thus the end behavior of the graph is like that of (d).

1 4 1 3 x + x − 6x2 + x − 5 4 2 1 The leading term is x4 . The degree, 4, is even and the 4 1 leading coefficient, , is positive. Thus the end behavior 4 of the graph is like that of (a).

2. f (x) = 15x2 − 10 + 0.11x4 − 7x3 = 0.11x4 − 7x3 + 15x2 − 10

12. f (x) =

3. h(x) = 0.9x − 0.13

3 13. f (x) = −x6 + x4 4 The leading term is −x6 . The degree, 6, is even and the leading coefficient, −1, is negative. Thus the end behavior of the graph is like that of (b).

The degree of the polynomial is 4, so the polynomial is quartic. The leading term is 0.11x4 and the leading coefficient is 0.11.

The degree of the polynomial is 1, so the polynomial is linear. The leading term is 0.9x and the leading coefficient is 0.9.

4. f (x) = −6 = −6x0

The degree of the polynomial is 0, so the polynomial is constant. The leading term and leading coefficient are both −6.

5. g(x) = 305x4 + 4021 The degree of the polynomial is 4, so the polynomial is quartic. The leading term is 305x4 and the leading coefficient is 305. 7 8 The degree of the polynomial is 3, so the polynomial is cubic. The leading term is 2.4x3 and the leading coefficient is 2.4.

6. h(x) = 2.4x3 + 5x2 − x +

7. h(x) = −5x2 + 7x3 + x4 = x4 + 7x3 − 5x2

The degree of the polynomial is 4, so the polynomial is quartic. The leading term is x4 and the leading coefficient is 1 (x4 = 1 · x4 ).

8. f (x) = 2 − x2 = −x2 + 2

The degree of the polynomial is 2, so the polynomial is quadratic. The leading term is −x2 and the leading coefficient is −1.

1 9. g(x) = 4x3 − x2 + 8 2 The degree of the polynomial is 3, so the polynomial is cubic. The leading term is 4x3 and the leading coefficient is 4.

1 2 5 x − 2x4 + x3 − x + 3 5 2 2 5 The leading term is x . The degree, 5, is odd and the 5 2 leading coefficient, , is positive. Thus the end behavior 5 of the graph is like that of (c).

14. f (x) =

15. f (x) = −3.5x4 + x6 + 0.1x7 = 0.1x7 + x6 − 3.5x4

The leading term is 0.1x7 . The degree, 7, is odd and the leading coefficient, 0.1, is positive. Thus the end behavior of the graph is like that of (c).

16. f (x) = −x3 + x5 − 0.5x6 = −0.5x6 + x5 − x3

The leading term is −0.5x6 . The degree, 6, is even and the leading coefficient, −0.5, is negative. Thus the end behavior of the graph is like that of (b).

1 4 2 3 1 4 2 3 x − x = x − x + 10 10 5 10 5 1 4 x . The degree, 4, is even and the The leading term is 10 1 leading coefficient, , is positive. Thus the end behavior 10 of the graph is like that of (a).

17. f (x) = 10 +

18. f (x) = 2x + x3 − x5 = −x5 + x3 + 2x

The leading term is −x5 . The degree, 5, is odd and the leading coefficient, −1, is negative. Thus the end behavior of the graph is like that of (d).

186

Chapter 3: Polynomial and Rational Functions

19. f (x) = −x6 + 2x5 − 7x2

The leading term is −x6 . The degree, 6, is even and the leading coefficient, −1, is negative. Thus, (c) is the correct graph.

20. f (x) = 2x4 − x2 + 1

The leading term is 2x . The degree, 4, is even and the leading coefficient, 2, is positive. Thus, (b) is the correct graph. 4

1 x−3 10 The leading term is x5 . The degree, 5, is odd and the leading coefficient, 1, is positive. Thus, (d) is the correct graph.

21. f (x) = x5 +

22. f (x) = −x3 + x2 − 2x + 4

The leading term is −x3 . The degree, 3, is odd and the leading coefficient, −1, is negative. Thus, (a) is the correct graph.

23. f (x) = x3 − 9x2 + 14x + 24

f (4) = 43 − 9 · 42 + 14 · 4 + 24 = 0

Since f (4) = 0, 4 is a zero of f (x). f (5) = 53 − 9 · 52 + 14 · 5 + 24 = −6

Since f (5) "= 0, 5 is not a zero of f (x).

f (−2) = (−2)3 − 9(−2)2 + 14(−2) + 24 = −48

Since f (−2) "= 0, −2 is not a zero of f (x). 24. f (x) = 2x − 3x + x + 6 3

2

f (2) = 2 · 2 − 3 · 2 + 2 + 6 = 12 3

2

f (2) "= 0, so 2 is not a zero of f (x). f (3) = 2 · 33 − 3 · 32 + 3 + 6 = 36

f (3) "= 0, so 3 is not a zero of f (x).

f (−1) = 2(−1) − 3(−1) + (−1) + 6 = 0 3

2

f (−1) = 0, so −1 is a zero of f (x). 25. g(x) = x − 6x + 8x + 6x − 9 4

3

2

g(2) = 2 − 6 · 2 + 8 · 2 + 6 · 2 − 9 = 3 4

3

2

Since g(2) "= 0, 2 is not a zero of g(x).

g(3) = 34 − 6 · 33 + 8 · 32 + 6 · 3 − 9 = 0 Since g(3) = 0, 3 is a zero of g(x).

g(−1) = (−1) − 6(−1) + 8(−1) + 6(−1) − 9 = 0 4

3

2

Since g(−1) = 0, −1 is a zero of g(x). 26. g(x) = x4 − x3 − 3x2 + 5x − 2

g(1) = 14 − 13 − 3 · 12 + 5 · 1 − 2 = 0

Since g(1) = 0, 1 is a zero of g(x).

g(−2) = (−2)4 − (−2)3 − 3(−2)2 + 5(−2) − 2 = 0

Since g(−2) = 0, −2 is a zero of g(x).

g(3) = 34 − 33 − 3 · 32 + 5 · 3 − 2 = 40

Since g(3) "= 0, 3 is not a zero of g(x).

27. f (x) = (x + 3)2 (x − 1) = (x + 3)(x + 3)(x − 1)

To solve f (x) = 0 we use the principle of zero products, solving x + 3 = 0 and x − 1 = 0. The zeros of f (x) are −3 and 1. The factor x + 3 occurs twice. Thus the zero −3 has a multiplicity of two. The factor x − 1 occurs only one time. Thus the zero 1 has a multiplicity of one.

28. f (x) = (x + 5)3 (x − 4)(x + 1)2

−5, multiplicity 3; 4, multiplicity 1; −1, multiplicity 2

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

To solve f (x) = 0 we use the principle of zero products, solving x − 4 = 0 and x + 6 = 0. The zeros of f (x) are 4 and −6. The factor x − 4 occurs three times. Thus the zero 4 has a multiplicity of 3.

The factor x + 6 occurs only one time. Thus the zero −6 has a multiplicity of 1. " ! " ! 1 1 30. f (x) = x+ (x+7)(x+7)(x+5) = x+ (x+7)2 (x+5) 2 2 1 − , multiplicity 1; −7, multiplicity 2; 2 −5, multiplicity 1 31. f (x) = (x2 + 9)3 = [(x + 3i)(x − 3i)]3 = (x + 3i)3 (x − 3i)3

To solve f (x) = 0 we use the principle of zero products, solving x + 3i = 0 and x − 3i = 0. The zeros of f (x) are −3i and 3i. The factors x + 3i and x − 3i each occur three times so each zero has a multiplicity of 3.

32. f (x) = (x2 − 4)2 = [(x + 2)(x − 2)]2 = (x + 2)2 (x − 2)2 −2, multiplicity 2; 2, multiplicity 2

33. f (x) = x3 (x − 1)2 (x + 4)

To solve f (x) = 0 we use the principle of zero products, solving x = 0, x − 1 = 0, and x + 4 = 0. The zeros of f (x) are 0, 1, and −4. The factor x occurs three times. Thus the zero 0 has a multiplicity of three.

The factor x − 1 occurs twice. Thus the zero 1 has a multiplicity of two. The factor x + 4 occurs only one time. Thus the zero −4 has a multiplicity of one. 34. f (x) = x2 (x + 3)2 (x − 4)(x + 1)4

0, multiplicity 2; −3, multiplicity 2; 4, multiplicity 1; −1, multiplicity 4

35. f (x) = −8(x − 3)2 (x + 4)3 x4

To solve f (x) = 0 we use the principle of zero products, solving x − 3 = 0, x + 4 = 0, and x = 0. The zeros of f (x) are 3, −4, and 0.

Exercise Set 3.1

187

The factor x − 3 occurs twice. Thus the zero 3 has a multiplicity of 2. The factor x + 4 occurs three times. Thus the zero −4 has a multiplicity of 3. The factor x occurs four times. Thus the zero 0 has a multiplicity of 4. 36.

f (x) = (x2 − 5x + 6)2

= [(x − 3)(x − 2)]

2

= (x − 3)2 (x − 2)2

3, multiplicity 2; 2, multiplicity 2 37. f (x) = x4 − 4x2 + 3

We factor as follows: f (x) = (x2 − 3)(x2 − 1) √ √ = (x − 3)(x + 3)(x − 1)(x + 1) √ √ The zeros of the function are 3, − 3, 1, and −1. Each has a multiplicity of 1.

38.

f (x) = x4 − 10x2 + 9

= (x2 − 9)(x2 − 1)

= (x + 3)(x − 3)(x + 1)(x − 1)

±3, ±1; each has a multiplicity of 1.

39. f (x) = x + 3x − x − 3 3

2

We factor by grouping: f (x) = x2 (x + 3) − (x + 3) = (x − 1)(x + 3) 2

= (x − 1)(x + 1)(x + 3)

The zeros of the function are 1, −1, and −3. Each has a multiplicity of 1. 40.

f (x) = x3 − x2 − 2x + 2

= x2 (x − 1) − 2(x − 1)

= (x − 2)(x − 1) √ √ = (x − 2)(x + 2)(x − 1) √ √ 2, − 2, 1; each has a multiplicity of 1. 2

41.

f (x) = 2x3 − x2 − 8x + 4

= x2 (2x − 1) − 4(2x − 1)

= (2x − 1)(x2 − 4)

= (2x − 1)(x + 2)(x − 2) 1 The zeros of the function are , −2, and 2. Each has a 2 multiplicity of 1. 42.

f (x) = 3x3 + x2 − 48x − 16

= x2 (3x + 1) − 16(3x + 1)

= (3x + 1)(x2 − 16)

= (3x + 1)(x + 4)(x − 4)

1 − , −4, 4; each has a multiplicity of 1 3 43. Graphing the function, we see that the graph touches the x-axis at (3, 0) but does not cross it, so the statement is false.

44. Graphing the function, we see that the statement is true. 45. Graphing the function, we see that the statement is true. 46. The number 4 is not a zero of the polynomial, so the statement is false. 47. In 1960, x = 60: f (60) = 0.0013924589(60)4 − 0.2121392316(60)3 + 3.414011377(60)2 + 310.7243489(60) + 16, 209.42591 ≈ 19, 368 There were about 19,368 thousand, or 19,368,000, milk cows in 1960. In 1980, x = 80: f (80) = 0.0013924589(80)4 − 0.2121392316(80)3 + 3.414011377(80)2 + 310.7243489(80) + 16, 209.42591 ≈ 11, 337 There were about 11,337 thousand, or 11,337,000, milk cows in 1980. 48. Solve 294 = 4.9t2 + 34.3t. t = 5 sec 49. In 2002, x = 1: f (1) = 2.375(1)2 − 7.395(1) + 28.205 ≈ 23.2

There were about 23.2 million real Christmas trees in 2002. In 2004, x = 3: f (3) = 2.375(3)2 − 7.395(3) + 28.205 ≈ 27.4

There were about 27.4 million real Christmas trees in 2004. 50. f (95) ≈ 250, 869 mi

f (129) ≈ 221, 035 mi f (162) ≈ 147, 527 mi

51. First find the number of games played. N (x) = x2 − x

N (9) = 92 − 9 = 72

Now multiply the number of games by the cost per game to find the total cost. 72 · 45 = 3240

It will cost $3240 to play the entire schedule. 52. a) P (15) = 0.015(15)3 = 50.625 watts per hour b) Solve 120 = 0.015v 3 . v = 20 mph 53. In 1996, x = 1: f (1) = −0.2607808858(1)4 + 5.281857032(1)3 − 30.77185315(1)2 + 60.89063714(1) + 663.8111888 ≈ $699 In 1999, x = 4:

f (4) = −0.2607808858(4)4 + 5.281857032(4)3 − 30.77185315(4)2 + 60.89063714(4) + 663.8111888 ≈ $686 In 2004, x = 9:

f (9) = −0.2607808858(9)4 + 5.281857032(9)3 − 30.77185315(9)2 + 60.89063714(9) + 663.8111888 ≈ $859

188 54.

Chapter 3: Polynomial and Rational Functions Relative minimum: 0.494 when x ≈ 0.632

f (7) ≈ 15.5% f (16) = 24%

Range: (−∞, ∞)

f (24) ≈ 41.7% 55. A = P (1 + i)

t

a)

4368.10 = 4000(1 + i)2 Substituting 4368.10 = (1 + i)2 4000 ±1.045 = 1 + i Taking the square root on both sides −1 ± 1.045 = i −1 − 1.045 = i or −1 + 1.045 = i −2.045 = i or

0.045 = i

Only the positive result has meaning in this application. The interest rate is 0.045, or 4.5%. b)

13, 310 = 10, 000(1 + i)3 Substituting 13, 310 = (1 + i)3 10, 000 1.10 ≈ 1 + i Taking the cube root on both sides 0.10 ≈ i

The interest rate is 0.1, or 10%. 56. 5 ft, 7 in. = 67 in. "3 ! 67 ≈ 162 lb W (67) = 12.3

57. Graph y = x3 − 3x − 1 and use the Zero feature three times. The real-number zeros are about −1.532, −0.347, and 1.879. 58. Graph y = x3 +3x2 −9x−13 and use the Zero feature three times. The real-number zeros are about −4.378, −1.167, and 2.545. 59. Graph y = x4 − 2x2 and use the Zero feature three times. The real-number zeros are about −1.414, 0, and 1.414. 60. Graph y = x4 − 2x3 − 5.6 and use the Zero feature twice. The real-number zeros are about −1.205 and 2.403. 61. Graph y = x3 − x and use the Zero feature three times. The real-number zeros are −1, 0, and 1. 62. Graph y = 2x3 −x2 −14x−10 and use the Zero feature three times. The real-number zeros are about −1.831, −0.856, and 3.188. 63. Graph y = x8 + 8x7 − 28x6 − 56x5 + 70x4 + 56x3 − 28x2 − 8x + 1 and use the Zero feature eight times. The realnumber zeros are about −10.153, −1.871, −0.821, −0.303, 0.098, 0.535, 1.219, and 3.297. 64. Graph y = x6 − 10x5 + 13x3 − 4x2 − 5 and use the Zero feature twice. The real-number zeros are −1.281 and 9.871. 65. g(x) = x3 − 1.2x + 1

Graph the function and use the Zero, Maximum, and Minimum features. Relative maximum: 1.506 when x ≈ −0.632

1 66. h(x) = − x4 + 3x3 − 5x2 + 3x + 6 2 Graph the function and use the Zero, Maximum, and Minimum features. Relative maxima: 6.59375 when x = 0.5, 10.5 when x = 3 Relative minimum: 6.5 when x = 1 Range: (−∞, 10.5] 67. f (x) = x6 − 3.8

Graph the function and use the Zero and Minimum features. There is no relative maximum. Relative minimum: −3.8 when x = 0 Range: [−3.8, ∞)

68. h(x) = 2x3 − x4 + 20

Graph the function and use the Zero and Maximum features. Relative maximum: 21.688 when x = 1.5 There is no relative minimum. Range: (−∞, 21.688]

69. f (x) = x2 + 10x − x5

Graph the function and use the Zero, Maximum, and Minimum features. Relative maximum: 11.012 when x ≈ 1.258

Relative minimum: −8.183 when x ≈ −1.116 Range: (−∞, ∞)

70. f (x) = 2x4 − 5.6x2 + 10

Graph the function and use the Zero, Maximum, and Minimum features. Relative maximum: 10 when x = 0 Relative minima: 6.08 when x ≈ −1.183 and when x ≈ 1.183 Range: [6.08, ∞)

71. The sales drop and then rise. This suggests that a quadratic function that opens up might fit the data. Thus, we choose (b). 72. The sales rise steadily. Thus, we choose (a), a linear function. 73. The sales rise and then drop. This suggests that a quadratic function that opens down might fit the data. Thus, we choose (c). 74. The sales rise, then fall, and then rise again. This suggests that a polynomial function that is neither linear nor quadratic fits the data. Thus, we choose (d). 75. The sales fall steadily. Thus, we choose (a), a linear function.

Exercise Set 3.1

189

76. The sales rise at an increasing rate. This suggests that a quadratic function that opens up might fit the data. Thus, we choose (b). (It is also possible that there is no polynomial function that fits the data well.)

80. a) Linear: y = 4.550243651x + 10.30607732 Cubic: y = 0.0102548814x3 − 0.3401188859x2 +

7.397498939x + 6.618361098

Since R2 is higher for the cubic function, it has the better fit.

77. a) For each function, x represents the number of years after 1900. Cubic: y = 0.000053215724x3 − 0.0059922487x2 + 0.0288132843x + 14.25265259 Quartic: y = −0.0000007674506x4 + 0.0002128619x3 −0.0163703506x2 +0.2450880151x+ 13.59031965 b) Evaluate each function for x = 2010 − 1900 = 110. Cubic: y ≈ 15.7%

Quartic: y ≈ 13.4%

Given the trend shown in the table, many would say that the quartic function gives the more realistic estimate. Answers will vary. 78. a) For each function, x represents the number of years after 1900. Quadratic: y = 0.0125777765x2 + 2.198116867x + 111.0690528 Cubic: y = −0.001201986x3 + 0.2045619198x2 − 5.604327275x + 164.1011737 Quartic: y = −0.00001673164x + 0.002317775x − 0.0275590018x2 − 0.6827647773x + 148.6082941 4

b) f (5) ≈ 36 yr; f (10) ≈ 57 yr; f (15) ≈ 76 yr 81. The range of a polynomial function with an odd degree is (−∞, ∞). The range of a polynomial function with an even degree is [s, ∞) for some real number s if an > 0 and is (−∞, s] for some real number s if an < 0. 82. In terms of the domain, “continuous” means that all real numbers are in the domain. In terms of the graphs, “continuous” means that there are no holes or breaks and no sharp corners. # 83. d = (x2 − x1 )2 + (y2 − y1 )2 # = [−1 − (−5)]2 + (0 − 3)2 # √ = 42 + (−3)2 = 16 + 9 √ = 25 = 5 # √ √ √ 84. d = (−2 − 4)2 + (−4 − 2)2 = 36 + 36 = 72 = 6 2

85.

(x − 3)2 + [y − (−5)]2 = 72

3

b) For 2010:

Quadratic: y ≈ 505 Cubic: y ≈ 423

Quartic: y ≈ 375

Center: (3, −5); radius: 7

86. The center of the circle is the midpoint of a segment that is a diameter: ! " −6 + (−2) 5 + 1 , = (−4, 3). 2 2 The length of a radius is the distance from the center to one of the endpoints of the diameter: # r = [−6 − (−4)]2 + (5 − 3)2 √ √ = 4+4= 8 √ =2 2

For 2015:

Quadratic: y ≈ 530 Cubic: y ≈ 399

Quartic: y ≈ 304

Given the numbers and the trend shown in the table, many would say that the quadratic function provides the most realistic estimates. Answers will vary.

87.

Quartic: y = −0.2966200466x4 + 4.341491841x3 − 9.860722611x2 − 39.24009324x + 399.986014

b) Evaluate each function for x = 2008 − 1996 = 12. Cubic: y ≈ 231 thousand

Quartic: y ≈ −139 thousand

Since the quartic function gives a negative value, the cubic function provides a more realistic estimate.

2y − 3 ≥ 1 − y + 5 2y − 3 ≥ 6 − y 3y − 3 ≥ 6

3y ≥ 9

79. a) For each function, x represents the number of years after 1996. Cubic: y = −1.590909091x3 + 27.21678322x2 − 113.3951049x + 421.3426573

(x − 3)2 + (y + 5)2 = 49

y≥3

Collecting like terms Adding y Adding 3 Dividing by 3

The solution set is {y|y ≥ 3}, or [3, ∞). 88.

(x − 2)(x + 5) > x(x − 3) x2 + 3x − 10 > x2 − 3x 6x − 10 > 0

6x > 10 5 x> 3 % & ! " $ % 5 5 % , or ,∞ . The solution set is x%x > 3 3

190

Chapter 3: Polynomial and Rational Functions

89. |x + 6| ≥ 7

x + 6 ≤ −7

or x + 6 ≥ 7

x ≤ −13 or

x≥1

The solution set is {x|x ≤ −13 or x ≥ 1}, or (−∞, −13] ∪ [1, ∞). % % % 1 %% 2 % 90. %x + % ≤ 4 3 2 1 2 − ≤ x+ ≤ 3 4 3 11 5 − ≤x≤ 12 12 $ % & % 11 5 ≤x≤ , or The solution set is x%% − 12 12 ' ( 11 5 . − , 12 12 91. Familiarize. We will use the compound interest formula. The $2000 deposit is invested for two years and grows to an amount A1 given by A1 = 2000(1 + i)2 . The $1200 deposit is invested for one year and grows to an amount A2 given by A2 = 1200(1 + i). Translate. There is a total of $3573.80 in both accounts at the end of the second year, so we have A1 + A2 = 3573.80, or 2000(1 + i)2 + 1200(1 + i) = 3573.80. Carry out. We solve the equation. 2000(1 + i)2 + 1200(1 + i) − 3573.8 = 0

Substitute u for 1 + i.

2000u2 + 1200u − 3573.8 = 0

Using the quadratic formula we find that u = 1.07 or u = −3.34. Only the positive value has meaning in this application. Then since u = 1.07, we have 1 + i = 1.07, or i = 0.07. Check. At an interest rate of 0.07, or 7%, in two years $2000 would grow to 2000(1 + 0.07)2 , or $2289.80. In one year $1200 would grow to 1200(1 + 0.07), or $1284. Now $2289.80 + $1284 = $3573.80, so the result checks. State. The interest rate is 7%.

Exercise Set 3.2 1. f (x) = x − x + 6 a) This function has degree 5, so its graph can have at most 5 real zeros. b) This function has degree 5, so its graph can have at most 5 x-intercepts. c) This function has degree 5, so its graph can have at most 5 − 1, or 4, turning points. 5

2

2. f (x) = −x2 + x4 − x6 + 3 = −x6 + x4 − x2 + 3 a) This function has degree 6, so its graph can have at most 6 real zeros. b) This function has degree 6, so its graph can have at most 6 x-intercepts. c) This function has degree 6, so its graph can have at most 6 − 1, or 5, turning points.

3. f (x) = x10 − 2x5 + 4x − 2 a) This function has degree 10, at most 10 real zeros. b) This function has degree 10, at most 10 x-intercepts. c) This function has degree 10, at most 10 − 1, or 9, turning

so its graph can have so its graph can have so its graph can have points.

1 3 x + 2x2 4 a) This function has degree 3, so its graph can have at most 3 real zeros. b) This function has degree 3, so its graph can have at most 3 x-intercepts. c) This function has degree 3, so its graph can have at most 3 − 1, or 2, turning points.

4. f (x) =

5. f (x) = −x − x3 = −x3 − x a) This function has degree 3, so its graph can have at most 3 real zeros. b) This function has degree 3, so its graph can have at most 3 x-intercepts. c) This function has degree 3, so its graph can have at most 3 − 1, or 2, turning points.

6. f (x) = −3x4 + 2x3 − x − 4 a) This function has degree 4, so its graph can have at most 4 real zeros. b) This function has degree 4, so its graph can have at most 4 x-intercepts. c) This function has degree 4, so its graph can have at most 4 − 1, or 3, turning points. 7. f (x) =

1 2 x −5 4

1 The leading term is x2 . The sign of the leading coef4 1 ficient, , is positive and the degree, 2, is even, so we 4 would choose either graph (b) or graph (d). Note also that f (0) = −5, so the y-intercept is (0, −5). Thus, graph (d) is the graph of this function. 8. f (x) = −0.5x6 − x5 + 4x4 − 5x3 − 7x2 + x − 3

The leading term is −0.5x6 . The sign of the leading coefficient, −0.5, is negative and the degree, 6, is even. Thus, graph (a) is the graph of this function.

9. f (x) = x5 − x4 + x2 + 4

The leading term is x5 . The sign of the leading coefficient, 1, is positive and the degree, 5, is odd. Thus, graph (f) is the graph of this function.

1 10. f (x) = − x3 − 4x2 + 6x + 42 3 1 The leading term is − x3 . The sign of the leading coef3 1 ficient, − , is negative and the degree, 3, is odd, so we 3 would choose either graph (c) or graph (e). Note also that f (0) = 42, so the y-intercept is (0, 42). Thus, graph (c) is the graph of this function.

Exercise Set 3.2

191

11. f (x) = x4 − 2x3 + 12x2 + x − 20

14.

y

The leading term is x . The sign of the leading coefficient, 1, is positive and the degree, 4, is even, so we would choose either graph (b) or graph (d). Note also that f (0) = −20, so the y-intercept is (0, −20). Thus, graph (b) is the graph of this function. 4

12. f (x) = −0.3x7 + 0.11x6 − 0.25x5 + x4 + x3 − 6x − 5

The leading term is −0.3x7 . The sign of the leading coefficient, −0.3, is negative and the degree, 7, is odd, so we would choose either graph (c) or graph (e). Note also that f (0) = −5, so the y-intercept is (0, −5). Thus, graph (e) is the graph of this function.

13. f (x) = −x3 − 2x2

1. The leading term is −x3 . The degree, 3, is odd and the leading coefficient, −1, is negative so as x → ∞, f (x) → −∞ and as x → −∞, f (x) → ∞.

2. We solve f (x) = 0.

2

!4 !2

4

x

!2

g(x ) " x 4 ! 4x 3 # 3x 2 !4

15. h(x) = x5 − 4x3

1. The leading term is x5 . The degree, 5, is odd and the leading coefficient, 1, is positive so as x → ∞, h(x) → ∞ and as x → −∞, h(x) → −∞.

2. We solve h(x) = 0.

x5 − 4x3 = 0

x (x2 − 4) = 0 x (x + 2)(x − 2) = 0 3

3

x = 0 or

−x2 (x + 2) = 0

or x − 2 = 0

x = −2 or

x=2

The zeros of the function are 0, −2, and 2 so the xintercepts of the graph are (0, 0), (−2, 0), and (2, 0).

−x2 = 0 or x + 2 = 0 x = 0 or

2

x3 = 0 or x + 2 = 0

−x3 − 2x2 = 0

x2 = 0 or

4

x = −2 x = −2

The zeros of the function are 0 and −2, so the xintercepts of the graph are (0, 0) and (−2, 0). 3. The zeros divide the x-axis into 3 intervals, (−∞, −2), (−2, 0), and (0, ∞). We choose a value for x from each interval and find f (x). This tells us the sign of f (x) for all values of x in that interval. In (−∞, −2), test −3: f (−3) = −(−3)3 − 2(−3)2 = 9 > 0

3. The zeros divide the x-axis into 4 intervals, (−∞, −2), (−2, 0), (0, 2), and (2, ∞). We choose a value for x from each interval and find h(x). This tells us the sign of h(x) for all values of x in that interval. In (−∞, −2), test −3: h(−3) = (−3)5 − 4(−3)3 = −135 < 0

In (−2, 0), test −1:

h(−1) = (−1)5 − 4(−1)3 = 3 > 0 In (0, 2), test 1:

In (−2, 0), test −1:

h(1) = 15 − 4 · 13 = −3 < 0

In (0, ∞), test 1:

h(3) = 35 − 4 · 33 = 135 > 0 Thus the graph lies below the x-axis on (−∞, −2) and on (0, 2). It lies above the x-axis on (−2, 0) and on (2, ∞). We also know the points (−3, −135), (−1, 3), (1, −3), and (3, 135) are on the graph.

In (2, ∞), test 3:

f (−1) = −(−1)3 − 2(−1)2 = −1 < 0 f (1) = −13 − 2 · 12 = −3 < 0 Thus the graph lies above the x-axis on (−∞, −2) and below the x-axis on (−2, 0) and (0, ∞). We also know the points (−3, 9), (−1, −1), and (1, −3) are on the graph. 4. From Step 2 we see that the y-intercept is (0, 0). 5. We find additional points on the graph and then draw the graph. x

f (x)

−2.5

3.125

−1.5 −1.125 1.5

−7.875

y 4 2

5. We find additional points on the graph and then draw the graph. x

y

h(x)

8

−2.5 −35.2 f(x) " !x 3 ! 2x 2 2

!4 !2

4. From Step 2 we see that the y-intercept is (0, 0).

4

x

!2 !4

6. Checking the graph as described on page 272 in the text, we see that it appears to be correct.

−1.5

5.9

1.5

−5.9

2.5

35.2

4 2

!4 !2

4

x

!4 !8

h(x) " x 5 ! 4x 3

6. Checking the graph as described on page 272 in the text, we see that it appears to be correct.

192

Chapter 3: Polynomial and Rational Functions

16.

18.

y 2

8

1

4 1

!2 !1

2

2

!6 !4 !2

x

!1

!4

!2

!8

x

1 3 5 2 f(x) " !x # !x 2 2

f(x) " x 3 ! x

17. h(x) = x(x − 4)(x + 1)(x − 2)

1. The leading term is x · x · x · x, or x4 . The degree, 4, is even and the leading coefficient, 1, is positive so as x → ∞, h(x) → ∞ and as x → −∞, h(x) → ∞.

2. We see that the zeros of the function are 0, 4, −1, and 2 so the x-intercepts of the graph are (0, 0), (4, 0), (−1, 0), and (2, 0).

3. The zeros divide the x-axis into 5 intervals, (−∞, −1), (−1, 0), (0, 2), (2, 4), and (4, ∞). We choose a value for x from each interval and find h(x). This tells us the sign of h(x) for all values of x in that interval. In (−∞, −1), test −2: h(−2) = −2(−2 − 4)(−2 + 1)(−2 − 2) = 48 > 0 In (−1, 0), test −0.5:

h(−0.5) = (−0.5)(−0.5 − 4)(−0.5 + 1)(−0.5 − 2) = −2.8125 < 0 In (0, 2), test 1:

2. We solve g(x) = 0. −x4 − 2x3 = 0

−x3 (x + 2) = 0

−x3 = 0 or x + 2 = 0 x = 0 or

h(3) = 3(3 − 4)(3 + 1)(3 − 2) = −12 < 0

3. The zeros divide the x-axis into 3 intervals, (−∞, −2), (−2, 0), and (0, ∞). We choose a value for x from each interval and find g(x). This tells us the sign of g(x) for all values of x in that interval. In (−∞, −2), test −3:

In (0, ∞), test 1:

h(5) = 5(5 − 4)(5 + 1)(5 − 2) = 90 > 0

Thus the graph lies above the x-axis on (−∞, −1), (0, 2), and (4, ∞). It lies below the x-axis on (−1, 0) and on (2, 4). We also know the points (−2, 48), (−0.5, −2.8125), (1, 6), (3, −12), and (5, 90) are on the graph.

4. From Step 2 we see that the y-intercept is (0, 0). 5. We find additional points on the graph and then draw the graph. y

g(1) = −14 − 2 · 13 = −3 < 0 Thus the graph lies below the x-axis on (−∞, −2) and (0, ∞) and above the x-axis on (−2, 0). We also know the points (−3, −27), (−1, 1), and (1, −3) are on the graph. 4. From Step 2 we see that the y-intercept is (0, 0). 5. We find additional points on the graph and then draw the graph. x

−2.5 −7.8

4

−1.5

1.7

0.5

−0.3

2

−32

1.5

4.7

2.5

−6.6

!8

4.5

30.9

!12

2

4

6 x

!4

h(x) " x(x ! 4)(x # 1)(x ! 2)

6. Checking the graph as described on page 272 in the text, we see that it appears to be correct.

y

g(x)

8

!4 !2

x = −2

The zeros of the function are 0 and −2, so the xintercepts of the graph are (0, 0) and (−2, 0).

g(−1) = −(−1)4 − 2(−1)3 = 1 > 0

In (4, ∞), test 5:

−1.5 14.4

1. The leading term is −x4 . The degree, 4, is even and the leading coefficient, −1, is negative so as x → ∞, g(x) → −∞ and as x → −∞, g(x) → −∞.

In (−2, 0), test −1:

In (2, 4), test 3:

h(x)

19. g(x) = −x4 − 2x3

g(−3) = −(−3)4 − 2(−3)3 = −27 < 0

h(1) = 1(1 − 4)(1 + 1)(1 − 2) = 6 > 0

x

y

4 2 2

!4 !2

4

x

!2 !4

g(x) " !x 4 ! 2x 3

6. Checking the graph as described on page 272 in the text, we see that it appears to be correct.

Exercise Set 3.2 20.

193 22.

y

y 30

8

10

4 2

!4 !2

4

x

!4

!30

!8

!50

!12

!70

1 21. f (x) = − (x − 2)(x + 1)2 (x − 1) 2 1 1 1. The leading term is − · x · x · x · x, or − x4 . The 2 2 1 degree, 4, is even and the leading coefficient, − , is 2 negative so as x → ∞, f (x) → −∞ and as x → −∞, f (x) → −∞. 2. We see that the zeros of the function are 2, −1, and 1 so the x-intercepts of the graph are (2, 0), (−1, 0), and (1, 0). 3. The zeros divide the x-axis into 4 intervals, (−∞, −1), (−1, 1), (1, 2), and (2, ∞). We choose a value for x from each interval and find f (x). This tells us the sign of f (x) for all values of x in that interval. In (−∞, −1), test −2: 1 f (−2) = − (−2 − 2)(−2 + 1)2 (−2 − 1) = −6 < 0 2 In (−1, 1), test 0: 1 f (0) = − (0 − 2)(0 + 1)2 (0 − 1) = −1 < 0 2 In (1, 2), test 1.5: 1 g(1.5) = − (1.5−2)(1.5+1)2 (1.5−1) = 0.78125 > 0 2 In (2, ∞), test 3: 1 f (3) = − (3 − 2)(3 + 1)2 (3 − 1) = −16 < 0 2 Thus the graph lies below the x-axis on (−∞, −1), (−1, 1), and (2, ∞) and above the x-axis on (1, 2). We also know that the points (−2, −6), (0, −1), (1.5, 0.78125), and (3, −16) are on the graph. 4. From Step 3 we see that f (0) = −1, so the y-intercept is (0, −1). 5. We find additional points on the graph and then draw the graph. f (x)

−3

−40

−1.5 −1.1 −0.5 −0.5 0.5

−0.8

4

x

g(x) " (x ! 2)3 (x # 3)

h(x) " x 3 ! 3x 2

x

2

!4 !2!10

y 4 2

1. The leading term is −x · x · x · x · x, or −x5 . The degree, 5, is odd and the leading coefficient, −1, is negative so as x → ∞, g(x) → −∞ and as x → −∞, g(x) → ∞.

2. We see that the zeros of the function are 0, 1, and −4 so the x-intercepts are (0, 0), (1, 0), and (−4, 0). 3. The zeros divide the x-axis into 4 intervals, (−∞, −4), (−4, 0), (0, 1), and (1, ∞). We choose a value of x from each interval and find g(x). This tells us the sign of g(x) for all values of x in that interval. In (−∞, −4), test −5: g(−5) = −(−5)(−5 − 1)2 (−5 + 4)2 = 180 > 0 In (−4, 0), test −1:

g(−1) = −(−1)(−1 − 1)2 (−1 + 4)2 = 36 > 0

In (0, 1), test 0.5:

g(0.5) = −0.5(0.5 − 1)2 (0.5 + 4)2 = −2.53125 < 0 In (1, ∞), test 2:

g(2) = −2(2 − 1)2 (2 + 4)2 = −72 < 0 Thus the graph lies above the x-axis on (−∞, −4) and (−4, 0) and below the x-axis on (0, 1) and (1, ∞). We also know that the points (−5, 180), (−1, 36), (0.5, −2.53125), and (2, −72) are on the graph.

4. From Step 2 we know that the y-intercept is (0, 0).

5. We find additional points on the graph and then draw the graph. x

g(x)

−3

48

−2

72

−0.5

13.8

1.5

−11.3

y 80 40 !6 !4 !2 !40

2

4 x

!80

g(x) " !x(x ! 1)2 (x # 4)2 2

!4 !2

23. g(x) = −x(x − 1)2 (x + 4)2

4

x

!2 !4

1 f(x) " !!(x ! 2)(x # 1)2 (x ! 1) 2

6. Checking the graph as described on page 272 in the text, we see that it appears to be correct.

6. Checking the graph as described on page 272 in the text, we see that it appears to be correct.

194

Chapter 3: Polynomial and Rational Functions

24.

26.

y

y 30

40

10

20 !4 !2 !20

2

4

x

!70

g(x) " x 4 ! 9x 2

25. f (x) = (x − 2)2 (x + 1)4

1. The leading term is x · x · x · x · x · x, or x6 . The degree, 6, is even and the leading coefficient, 1, is positive so as x → ∞, f (x) → ∞ and as x → −∞, f (x) → ∞. We see that the zeros of the function are 2 and −1 so 2. the x-intercepts of the graph are (2, 0) and (−1, 0).

27. g(x) = −(x − 1)4

1. The leading term is −1 · x · x · x · x, or −x4 . The degree, 4, is even and the leading coefficient, −1, is negative so as x → ∞, g(x) → −∞ and as x → −∞, g(x) → −∞. 2. We see that the zero of the function is 1, so the xintercept is (1, 0).

3. The zeros divide the x-axis into 3 intervals, (−∞, −1), (−1, 2), and (2, ∞). We choose a value for x from each interval and find f (x). This tells us the sign of f (x) for all values of x in that interval. In (−∞, −1), test −2:

3. The zero divides the x-axis into 2 intervals, (−∞, 1) and (1, ∞). We choose a value for x from each interval and find g(x). This tells us the sign of g(x) for all values of x in that interval. In (−∞, 1), test 0:

f (−2) = (−2 − 2)2 (−2 + 1)4 = 16 > 0

g(0) = −(0 − 1)4 = −1 < 0

In (−1, 2), test 0:

In (1, ∞), test 2:

f (0) = (0 − 2)2 (0 + 1)4 = 4 > 0

g(2) = −(2 − 1)4 = −1 < 0 Thus the graph lies below the x-axis on both intervals. We also know the points (0, −1) and (2, −1) are on the graph.

In (2, ∞), test 3:

f (3) = (3 − 2)2 (3 + 1)4 = 256 > 0 Thus the graph lies above the x-axis on all 3 intervals. We also know the points (−2, 16), (0, 4), and (3, 256) are on the graph.

4. From Step 3 we know that g(0) = −1 so the yintercept is (0, −1).

4. From Step 3 we know that f (0) = 4 so the y-intercept is (0, 4).

5. We find additional points on the graph and then draw the graph.

5. We find additional points on the graph and then draw the graph.

−1.5

0.8

−0.5

0.4

1

16

1.5

9.8

x

!50

h (x ) " !x (x ! 3)(x ! 3)(x # 2)

f (x)

4

!30

!40

x

2

!4 !2!10

y 30

g(x)

−1

−16

y 4

10 2

4

x

0.1

3

−16

4

x

!4 !8 !12 !14

g(x) " !(x ! 1)4

f(x) " (x ! 2)2 (x # 1)4

6. Checking the graph as described on page 272 in the text, we see that it appears to be correct.

1.5

2

!4 !2

−0.5 −5.1

20

!4 !2 !10

x

6. Checking the graph as described on page 272 in the text, we see that it appears to be correct. 28.

y 8 6 4 2 2

!6 !4 !2 !2

h(x) " (x # 2)3

x

Exercise Set 3.2

195

29. h(x) = x3 + 3x2 − x − 3

30.

1. The leading term is x . The degree, 3, is odd and the leading coefficient, 1, is positive so as x → ∞, h(x) → ∞ and as x → −∞, h(x) → −∞.

2. We solve h(x) = 0.

x3 + 3x2 − x − 3 = 0

(x + 3)(x2 − 1) = 0

(x + 3)(x + 1)(x − 1) = 0

or x + 1 = 0

x = −3 or

or x − 1 = 0

x = −1 or

x=1

h(−4) = (−4)3 + 3(−4)2 − (−4) − 3 = −15 < 0 In (−3, −1), test −2:

h(−2) = (−2)3 + 3(−2)2 − (−2) − 3 = 3 > 0

In (−1, 1), test 0:

h(0) = 03 + 3 · 02 − 0 − 3 = −3 < 0

In (1, ∞), test 2:

h(2) = 23 + 3 · 22 − 2 − 3 = 15 > 0 Thus the graph lies below the x-axis on (−∞, −3) and on (−1, 1) and above the x-axis on (−3, −1) and on (1, ∞). We also know the points (−4, −15), (−2, 3), (0, −3), and (2, 15) are on the graph.

4. From Step 3 we know that h(0) = −3 so the yintercept is (0, −3).

5. We find additional points on the graph and then draw the graph. y 4

−4.5 −28.9 −2.5

2.6

0.5

−2.6

2.5

28.9

!4 !2 !10

2

4

x

31. f (x) = 6x3 − 8x2 − 54x + 72

3. The zeros divide the x-axis into 4 intervals, (−∞, −3), (−3, −1), (−1, 1), and (1, ∞). We choose a value for x from each interval and find h(x). This tells us the sign of h(x) for all values of x in that interval. In (−∞, −3), test −4:

h(x)

10

g(x) " !x 3 # 2x 2 # 4x ! 8

The zeros of the function are −3, −1, and 1 so the xintercepts of the graph are (−3, 0), (−1, 0), and (1, 0).

x

20

!20

x2 (x + 3) − (x + 3) = 0

x+3 = 0

y

3

2 2

!4 !2

4

x

!2 !4

h (x ) " x 3 # 3x 2 ! x ! 3

6. Checking the graph as described on page 272 in the text, we see that it appears to be correct.

1. The leading term is 6x3 . The degree, 3, is odd and the leading coefficient, 6, is positive so as x → ∞, f (x) → ∞ and as x → −∞, f (x) → −∞.

2. We solve f (x) = 0.

6x3 − 8x2 − 54x + 72 = 0

2(3x3 − 4x2 − 27x + 36) = 0

2[x2 (3x − 4) − 9(3x − 4)] = 0 2(3x − 4)(x2 − 9) = 0

2(3x − 4)(x + 3)(x − 3) = 0

3x − 4 = 0 or x + 3 = 0 or x − 3 = 0 4 or x = −3 or x=3 x= 3 4 The zeros of the function are , −3, and 3, so the x! 3" 4 intercepts of the graph are , 0 , (−3, 0), and (3, 0). 3 The zeros"divide the 3. ! ! " x-axis into 4 intervals, (−∞, −3), 4 4 − 3, , , 3 , and (3, ∞). We choose a value for 3 3 x from each interval and find f (x). This tells us the sign of f (x) for all values of x in that interval. In (−∞, −3), test −4: f (−4) = 6(−4)3 − 8(−4)2 − 54(−4) + 72 = −224 < 0 ! " 4 In − 3, , test 0: 3 f (0) = 6 · 03 − 8 · 02 − 54 · 0 + 72 = 72 > 0 " ! 4 , 3 , test 2: In 3 f (2) = 6 · 23 − 8 · 22 − 54 · 2 + 72 = −20 < 0

In (3, ∞), test 4:

f (4) = 6 · 43 − 8 · 42 − 54 · 4 + 72 = 112 > 0 Thus on (−∞,"−3) and ! the"graph lies below the x-axis ! 4 4 , 3 and above the x-axis on − 3, and on on 3 3 (3, ∞). We also know the points (−4, −224), (0, 72), (2, −20), and (4, 112) are on the graph.

4. From Step 3 we know that f (0) = 72 so the yintercept is (0, 72).

196

Chapter 3: Polynomial and Rational Functions 5. We find additional points on the graph and then draw the graph. x

112

1

16

By the intermediate value theorem, since f (4) and f (5) have opposite signs then f (x) has a zero between 4 and 5.

100 50 !4 !2 !50

3.5 42.25

40. 2

4

f (x ) " 6x 3 ! 8x 2 ! 54x # 72

6. Checking the graph as described on page 272 in the text, we see that it appears to be correct. y 4 2 1

!2 !1

2

x

!2 !4

h(x) " x 5 ! 5x 3 # 4x

33. f (−5) = (−5)3 + 3(−5)2 − 9(−5) − 13 = −18 f (−4) = (−4)3 + 3(−4)2 − 9(−4) − 13 = 7

By the intermediate value theorem, since f (−5) and f (−4) have opposite signs then f (x) has a zero between −5 and −4.

34. f (1) = 13 + 3 · 12 − 9 · 1 − 13 = −18 f (2) = 23 + 3 · 22 − 9 · 2 − 13 = −11

Since both f (1) and f (2) are negative, we cannot use the intermediate value theorem to determine if there is a zero between 1 and 2. 35. f (−3) = 3(−3)2 − 2(−3) − 11 = 22 f (−2) = 3(−2)2 − 2(−2) − 11 = 5

Since both f (−3) and f (−2) are positive, we cannot use the intermediate value theorem to determine if there is a zero between −3 and −2. 41. The zeros of a polynomial function are the first coordinates of the points at which the graph of the function crosses or is tangent to the x-axis. 42. Since we can find f (0) for any polynomial function f (x), it is not possible for the graph of a polynomial function to have no y-intercept. It is possible for a polynomial function to have no x-intercepts. For instance, a function of the form f (x) = x2 + a, a > 0, has no x-intercepts. There are other examples as well. 43. The graph of y = x, or y = x + 0, has y-intercept (0, 0), so (d) is the correct answer. 44. The graph of x = −4 is a vertical line 4 units to the left of the y-axis, so (f) is the correct answer. 45. The graph of y − 2x = 6, or y = 2x + 6, has y-intercept (0, 6), so (e) is the correct answer. 46. When y = 0, x = −2 and when x = 0, y = −3 so the intercepts of the graph are (0, −2) and (−3, 0). Thus (a) is the correct answer. 47. The graph of y = 1 − x, or y = −x + 1, has y-intercept (0, 1), so (b) is the correct answer. 48. The graph of y = 2 is a horizontal line 2 units above the x-axis, so (c) is the correct answer. 49.

Since both f (−3) and f (−2) are positive, we cannot use the intermediate value theorem to determine if there is a zero between −3 and −2.

36. f (2) = 3 · 22 − 2 · 2 − 11 = −3 f (3) = 3 · 32 − 2 · 3 − 11 = 10

By the intermediate value theorem, since f (2) and f (3) have opposite signs then f (x) has a zero between 2 and 3. 37. f (2) = 24 − 2 · 22 − 6 = 2

f (3) = 34 − 2 · 32 − 6 = 57

Since both f (2) and f (3) are positive, we cannot use the intermediate value theorem to determine if there is a zero between 2 and 3. 38. f (1) = 2 · 15 − 7 · 1 + 1 = −4 f (2) = 2 · 25 − 7 · 2 + 1 = 51

By the intermediate value theorem, since f (1) and f (2) have opposite signs then f (x) has a zero between 1 and 2.

f (−3) = (−3)4 − 3(−3)2 + (−3) − 1 = 50

f (−2) = (−2)4 − 3(−2)2 + (−2) − 1 = 1

x

!100

32.

f (4) = 43 − 5 · 42 + 4 = −12

f (5) = 53 − 5 · 52 + 4 = 4

y

f (x)

−1

39.

50.

1 = 4 − 3x 2 1 5x − = 4 Adding 3x 2 1 9 5x = Adding 2 2 1 9 1 x= · Multiplying by 5 2 5 9 x= 10 9 . The solution is 10 2x −

x3 − x2 − 12x = 0

x(x2 − x − 12) = 0

x(x − 4)(x + 3) = 0

x = 0 or x − 4 = 0 or x + 3 = 0 x = 0 or

x = 4 or

x = −3

Exercise Set 3.3 51.

197

6x2 − 23x − 55 = 0

2. a)

(3x + 5)(2x − 11) = 0 3x + 5 = 0

or 2x − 11 = 0

3x = −5 or 5 x = − or 3

2x = 11 11 x= 2 5 11 The solutions are − and . 3 2 52.

1 3 x + 10 = + 2x 4 5 15x + 200 = 4 + 40x Multiplying by 20

Since the remainder is not 0, x + 5 is not a factor of h(x). b)

196 = 25x 196 =x 25

Exercise Set 3.3 1. a)

2

Since the remainder is not 0, x + 1 is not a factor of f (x). b)

c)

x2 − 2x − 15 x + 1 x3 − x2 − 17x − 15 x3 + x2 − 2x2 − 17x − 2x2 − 2x − 15x − 15 − 15x − 15 0 Since the remainder is 0, x + 1 is a factor of h(x).

x − 7x + 8x + 16 x + 1 x4 − 6x3 + x2 + 24x − 20 x4 + x3 − 7x3 + x2 − 7x3 − 7x2 8x2 + 24x 8x2 + 8x 16x − 20 16x + 16 − 4 3

x2 − 6x + 13 x + 5 x3 − x2 − 17x − 15 x3 + 5x2 − 6x2 − 17x − 6x2 − 30x 13x − 15 13x + 65 − 80

x3 − 4x2 − 7x + 10 x − 2 x4 − 6x3 + x2 + 24x − 20 x4 − 2x3 − 4x3 + x2 − 4x3 + 8x2 − 7x2 + 24x − 7x2 + 14x 10x − 20 10x − 20 0

c)

Since the remainder is 0, x + 3 is a factor of h(x). 3. a)

b)

Since the remainder is 0, x − 2 is a factor of f (x). x3 − 11x2 + 56x − 256 x + 5 x4 − 6x3 + x2 + 24x − 20 x4 + 5x3 − 11x3 + x2 − 11x3 − 55x2 56x2 + 24x 56x2 + 280x − 256x − 20 − 256x − 1280 1260

Since the remainder is not 0, x + 5 is not a factor of f (x).

x2 − 4x − 5 x + 3 x3 − x2 − 17x − 15 x3 + 3x2 − 4x2 − 17x − 4x2 − 12x − 5x − 15 − 5x − 15 0

x2 + 2x − 3 x − 4 x3 − 2x2 − 11x + 12 x3 − 4x2 2x2 − 11x 2x2 − 8x − 3x + 12 − 3x + 12 0 Since the remainder is 0, x − 4 is a factor of g(x). x2 + x − 8 x − 3 x3 − 2x2 − 11x + 12 x3 − 3x2 x2 − 11x x2 − 3x − 8x + 12 − 8x + 24 − 12

Since the remainder is not 0, x − 3 is not a factor of g(x). c)

x2 − x − 12 x − 1 x3 − 2x2 − 11x + 12 x3 − x2 − x2 − 11x − x2 + x − 12x + 12 − 12x + 12 0 Since the remainder is 0, x − 1 is a factor of g(x).

198

Chapter 3: Polynomial and Rational Functions

4. a)

x3 + 2x2 − 7x + 4 x + 6 x4 + 8x3 + 5x2 − 38x + 24 x4 + 6x3 2x3 + 5x2 2x3 + 12x2 − 7x2 − 38x − 7x2 − 42x 4x + 24 4x + 24 0 Since the remainder is 0, x + 6 is a factor of f (x).

b)

x3 + 7x2 − 2x − 36 x + 1 x4 + 8x3 + 5x2 − 38x + 24 x4 + x3 7x3 + 5x2 7x3 + 7x2 − 2x2 − 38x − 2x2 − 2x − 36x + 24 − 36x − 36 60

7.

x3 + 6x2 − 25x + 18 = (x + 9)(x2 − 3x + 2) + 0 8.

x3 + 12x2 + 53x + 174 x − 4 x4 + 8x3 + 5x2 − 38x + 24 x4 − 4x3 12x3 + 5x2 12x3 − 48x2 53x2 − 38x 53x2 − 212x 174x + 24 174x − 696 720

9.

x2 − 2x + 4 x + 2 x3 + 0x2 + 0x − 8 x3 + 2x2 − 2x2 + 0x − 2x2 − 4x 4x − 8 4x + 8 −16

x3 − 8 = (x + 2)(x2 − 2x + 4) − 16 6.

2x2 + 3x + x − 3 2x3 − 3x2 + 2x3 − 6x2 3x2 + 3x2 −

10 x − 1

x 9x 10x − 1 10x − 30 29

2x3 − 3x2 + x − 1 = (x − 3)(2x2 + 3x + 10) + 29

x3 − 2x2 + 2x − 4 x + 2 x4 + 0x3 − 2x2 + 0x + 3 x4 + 2x3 − 2x3 − 2x2 − 2x3 − 4x2 2x2 + 0x 2x2 + 4x − 4x + 3 − 4x − 8 11

x4 − 2x2 + 3 = (x + 2)(x3 − 2x2 + 2x − 4) + 11 10.

Since the remainder is not 0, x − 4 is not a factor of f (x). 5.

x2 − 4x − 5 x − 5 x3 − 9x2 + 15x + 25 x3 − 5x2 − 4x2 + 15x − 4x2 + 20x − 5x + 25 − 5x + 25 0

x3 − 9x2 + 15x + 25 = (x − 5)(x2 − 4x − 5) + 0

Since the remainder is not 0, x + 1 is not a factor of f (x). c)

x2 − 3x + 2 x + 9 x3 + 6x2 − 25x + 18 x3 + 9x2 − 3x2 − 25x − 3x2 − 27x 2x + 18 2x + 18 0

x3 + 7x2 + 7x + 7 x − 1 x4 + 6x3 + 0x2 + 0x + 0 x4 − x3 7x3 + 0x2 7x3 − 7x2 7x2 + 0x 7x2 − 7x 7x + 0 7x − 7 7

x4 + 6x3 = (x − 1)(x3 + 7x2 + 7x + 7) + 7 11.

(2x4 + 7x3 + x − 12) ÷ (x + 3)

= (2x4 + 7x3 + 0x2 + x − 12) ÷ [x − (−3)] % −3 % 2 7 0 1 −12 −6 −3 9 −30 2 1 −3 10 −42

The quotient is 2x3 + x2 − 3x + 10. The remainder is −42. % 12. 2 % 1 −7 13 3 2 −10 6 1 −5 3 9 Q(x) = x2 − 5x + 3, R(x) = 9

Exercise Set 3.3 13.

199

(x3 − 2x2 − 8) ÷ (x + 2)

= (x − 2x + 0x − 8) ÷ [x − (−2)] % −2 % 1 −2 0 −8 −2 8 −16 1 −4 8 −24 3

2

The quotient is x2 − 4x + 8. The remainder is −24. % 14. 2 % 1 0 −3 10 2 4 2 1 2 1 12 Q(x) = x2 + 2x + 1, R(x) = 12 15.

(3x3 − x2 + 4x − 10) ÷ (x + 1)

= (3x3 − x2 + 4x − 10) ÷ [x − (−1)] % −1 % 3 −1 4 −10 −3 4 −8 3 −4 8 −18

The quotient is 3x2 − 4x + 8. The remainder is −18. % 0 0 −2 5 16. −3 % 4 −12 36 −108 330 4 −12 36 −110 335 17.

Q(x) = 4x3 − 12x2 + 36x − 110, R(x) = 335 (x5 + x3 − x) ÷ (x − 3)

= (x5 + 0x4 + x3 + 0x2 − x + 0) ÷ (x − 3) % 3% 1 0 1 0 −1 0 3 9 30 90 267 1 3 10 30 89 267 The quotient is x4 + 3x3 + 10x2 + 30x + 89. The remainder % 18. −1 % 1 −1 −1 1 −2 19.

is 267. 1 −1 0 0 0 2 2 −3 4 −4 4 −4 3 −4 4 −4 4 −2

Q(x) = x6 − 2x5 + 3x4 − 4x3 + 4x2 − 4x + 4, R(x) = −2 (x4 − 1) ÷ (x − 1)

= (x4 + 0x3 + 0x2 + 0x − 1) ÷ (x − 1) % 1 % 1 0 0 0 −1 1 1 1 1 1 1 1 1 0 The quotient is % 0 20. −2 % 1 −2 1 −2

x3 + x2 + x + 1. The remainder is 0. 0 0 0 32 4 −8 16 −32 4 −8 16 0

Q(x) = x − 2x + 4x − 8x + 16, R(x) = 0 ! " 1 4 2 21. (2x + 3x − 1) ÷ x − 2 " ! 1 4 3 2 (2x + 0x + 3x + 0x − 1) ÷ x − 2 % 1 % 2 0 3 0 −1 2 7 1 12 47 8 4

2

1

7 2

3

2

7 4

− 18

7 1 7 The quotient is 2x3 + x2 + x + . The remainder is − . 2 4 8

22.

1 4

% %

3

0

3

3 4 3 4

−2

3 16 29 − 16

0 2 29 − 29 64 − 256 − 29 64

483 256

3 29 29 483 Q(x) = 3x3 + x2 − x − , R(x) = 4 16 64 256 23. f (x) = x3 − 6x2 + 11x − 6 Find f (1). % 1 % 1 −6 11 −6 1 −5 6 1 −5 6 0 f (1) = 0

Find f (−2). % −2 % 1 −6 11 −6 −2 16 −54 1 −8 27 −60

f (−2) = −60

Find f (3). % 3 % 1 −6 11 −6 3 −9 6 1 −3 2 0

f (3) = 0 % 24. −3 % 1

7 −12 −3 −3 −12 72 1 4 −24 69

f (−3) = 69 % −2 % 1 7 −12 −3 −2 −10 44 1 5 −22 41 f (−2) = 41 % 1 % 1 7 −12 −3 1 8 −4 1 8 −4 −7 f (1) = −7

25. f (x) = x4 − 3x3 + 2x + 8

Find f (−1). % −1 % 1 −3 0 2 8 −1 4 −4 2 1 −4 4 −2 10

f (−1) = 10

Find f (4). % 4 % 1 −3 0 2 8 4 4 16 72 1 1 4 18 80 f (4) = 80 Find f (−5). % −5 % 1 −3 −5 1 −8

f (−5) = 998

0 2 8 40 −200 990 40 −198 998

200 26.

Chapter 3: Polynomial and Rational Functions % −10 %

2 2

0 1 −10 1 −20 200 −2010 20, 200 −20 201 −2020 20, 201

f (−10) = 20, 201 % 2 % 2 0 1 −10 1 4 8 18 16 2 4 9 8 17 f (2) = 17 % 3% 2 0 6 2 6

1 −10 1 18 57 141 19 47 142

f (−2) = 0 (See % 3% 1 0 0 3 9 1 3 9

27. f (x) = 2x5 − 3x4 + 2x3 − x + 8 −1 8 296, 800 5, 935, 980 296, 799 5, 935, 988

f (20) = 5, 935, 988 Find f (−3). % −3 % 2 −3 −6 2 −9

f (−3) = −772 % 28. −10 % 1 −10 −10 1 −20

2 0 −1 8 27 −87 261 −780 29 −87 260 −772 20 200 220

0 −2200 −2200

−5 −100 22, 000 −219, 950 21, 995 −220, 050

f (−10) = −220, 050 % 5 % 1 −10 20 0 −5 −100 5 −25 −25 −125 −650 1 −5 −5 −25 −130 −750 f (5) = −750

29. f (x) = x4 − 16

Find f (2). % 2 % 1 0 0 0 −16 2 4 8 16 1 2 4 8 0

f (2) = 0 Find f (−2). % −2 % 1 0 0 0 −16 −2 4 −8 16 1 −2 4 −8 0

f (−2) = 0

Find f (3). % 3% 1 0 0 3 9 1 3 9 f (3) = 65

0 −16 27 81 27 65

0√ −16√ 7−5√2 17−12√ 2 7−5 2 1−12 2

f (2) = 64

f (3) = 142 Find f (20). % 2 0 20 % 2 −3 40 740 14, 840 2 37 742 14, 840

√ Find f (1 − 2). √ % 0√ 0√ 1− 2 % 1 1− √2 3−2√2 1 1− 2 3−2 2 √ √ f (1 − 2) = 1 − 12 2 % 30. 2 % 1 0 0 0 0 32 2 4 8 16 32 1 2 4 8 16 64

Exercise 20.) 0 0 32 27 81 243 27 81 275

f (3) = 275 % 2+3i % 1

0 0 0 0 32 2+3i −5+12i −46+9i −119−120i 122−597i 1 2+3i −5+12i −46+9i −119−120i 154−597i

f (2 + 3i) = 154 − 597i

31. f (x) = 3x3 + 5x2 − 6x + 18

If −3 is a zero of f (x), then f (−3) = 0. Find f (−3) using synthetic division. % −3 % 3 5 −6 18 −9 12 −18 3 −4 6 0

Since f (−3) = 0, −3 is a zero of f (x).

If 2 is a zero of f (x), then f (2) = 0. Find f (2) using synthetic division. % 2 % 3 5 −6 18 6 22 32 3 11 16 50 Since f (2) "= 0, 2 is not a zero of f (x). % 32. −4 % 3 11 −2 8 −12 4 −8 3 −1 2 0 f (−4) = 0, so −4 % 2 % 3 11 −2 6 34 3 17 32

is a zero of f (x). 8 64 72

f (2) "= 0, so 2 is not a zero of f (x). 33. h(x) = x4 + 4x3 + 2x2 − 4x − 3

If −3 is a zero of h(x), then h(−3) = 0. Find h(−3) using synthetic division. % −3 % 1 4 2 −4 −3 −3 −3 3 3 1 1 −1 −1 0

Since h(−3) = 0, −3 is a zero of h(x).

If 1 is a zero of h(x), then h(1) = 0. Find h(1) using synthetic division.

Exercise Set 3.3 % 1%

1

4 2 1 5 5 7

1

201 −4 −3 7 3 3 0

Since h(1) = 0, 1 is a zero of h(x). % 34. 2 % 1 −6 1 24 −20 2 −8 −14 20 1 −4 −7 10 0 g(2) = 0, so 2 is a zero of g(x). % −1 % 1 −6 1 24 −20 −1 7 −8 −16 1 −7 8 16 −36

g(−1) "= 0, so −1 is not a zero of g(x).

35. g(x) = x3 − 4x2 + 4x − 16

If i is a zero of g(x), then g(i) = 0. Find g(i) using synthetic division. Keep in mind that i2 = −1. % i% 1 −4 4 −16 i −4i − 1 3i + 4 1 −4 + i 3 − 4i −12 + 3i

Since g(i) "= 0, i is not a zero of g(x).

If −2i is a zero of g(x), then g(−2i) = 0. Find g(−2i) using synthetic division. Keep in mind that i2 = −1. % −2i % 1 −4 4 −16 −2i 8i − 4 16 1 −4 − 2i 8i 0

Since g(−2i) = 0, −2i is a zero of g(x). % % 1 36. 13 % 1 −1 − 19 9 1 2 1 −9 −9 3 1 − 23 − 13 0 ! " 1 1 h = 0, so is a zero of h(x). 3 3 % % 2 % 1 −1 − 19 91 2 2 34 9 1

1

17 9

35 9

h(2) "= 0, so 2 is not a zero of h(x).

7 3 37. f (x) = x3 − x2 + x − 2 2 If −3 is a zero of f (x), then f (−3) = 0. Find f (−3) using synthetic division. % % 1 − 32 −3 % 1 − 72 −3 39 − 123 2 2 1

− 13 2

41 2

−63

Since f (−3) "= 0, −3 is not a zero of f (x). ! " 1 1 If is a zero of f (x), then f = 0. 2 2 ! " 1 using synthetic division. Find f 2

1 2

% % %

1 − 72 1 2

1

1 − 32 − 14

− 32 − 12

−3 − 74 ! " 1 1 Since f "= 0, is not a zero of f (x). 2 2 % 38. i % 1 2 1 2 i −1 + 2i −2 1 2+i 2i 0 f (i) = 0, so i is a zero of f (x). % 2 1 2 −i % 1 −i −1 − 2i −2 1 2−i −2i 0

f (−i) = 0, so −i is a zero of f (x). % −2 % 1 2 1 2 −2 0 −2 1 0 1 0 f (−2) = 0, so −2 is a zero of f (x). 39. f (x) = x3 + 4x2 + x − 6

Try x − 1. Use synthetic division to see whether f (1) = 0. % 1 % 1 4 1 −6 1 5 6 1 5 6 0 Since f (1) = 0, x − 1 is a factor of f (x). Thus f (x) = (x − 1)(x2 + 5x + 6). Factoring the trinomial we get f (x) = (x − 1)(x + 2)(x + 3).

To solve the equation f (x) = 0, use the principle of zero products. (x − 1)(x + 2)(x + 3) = 0

x − 1 = 0 or x + 2 = 0 x = 1 or

or x + 3 = 0

x = −2 or

The solutions are 1, −2, and −3. % 40. 2 % 1 5 −2 −24 2 14 24 1 7 12 0

x = −3

f (x) = (x − 2)(x2 + 7x + 12) = (x − 2)(x + 3)(x + 4)

The solutions of f (x) = 0 are 2, −3, and −4. 41. f (x) = x3 − 6x2 + 3x + 10

Try x − 1. Use synthetic division to see whether f (1) = 0. % 1 % 1 −6 3 10 1 −5 −2 1 −5 −2 8 Since f (1) "= 0, x − 1 is not a factor of P (x).

Try x+1. Use synthetic division to see whether f (−1) = 0. % −1 % 1 −6 3 10 −1 7 −10 1 −7 10 0

202

Chapter 3: Polynomial and Rational Functions Since f (−1) = 0, x + 1 is a factor of f (x). Thus f (x) = (x + 1)(x − 7x + 10). 2

Factoring the trinomial we get f (x) = (x + 1)(x − 2)(x − 5).

To solve the equation f (x) = 0, use the principle of zero products. (x + 1)(x − 2)(x − 5) = 0 x+1 = 0

or x − 2 = 0 or x − 5 = 0

x = −1 or

x = 2 or

x=5

f (x) = (x + 2)(x − 3)(x − 3)2 , or f (x) = (x + 2)(x − 3)3 .

To solve the equation f (x) = 0, use the principle of zero products.

f (x) = (x − 1)(x + 3x − 10)

(x + 2)(x − 3)(x − 3)(x − 3) = 0

= (x − 1)(x − 2)(x + 5)

The solutions of f (x) = 0 are 1, 2, and −5. 43. f (x) = x3 − x2 − 14x + 24

Try x + 1, x − 1, and x + 2. Using synthetic division we find that f (−1) "= 0, f (1) "= 0 and f (−2) "= 0. Thus x + 1, x − 1, and x + 2, are not factors of f (x). Try x − 2. Use synthetic division to see whether f (2) = 0. % 2 % 1 −1 −14 24 2 2 −24 1 1 −12 0

Since f (2) = 0, x − 2 is a factor of f (x). Thus f (x) = (x − 2)(x2 + x − 12). Factoring the trinomial we get

To solve the equation f (x) = 0, use the principle of zero products. or x − 3 = 0

x = −4 or

x=3

The solutions are 2, −4, and 3. % 44. 2 % 1 −3 −10 24 2 −2 −24 1 −1 −12 0 f (x) = (x − 2)(x2 − x − 12)

= (x − 2)(x − 4)(x + 3)

The solutions of f (x) = 0 are 2, 4, and −3. 45. f (x) = x4 − 7x3 + 9x2 + 27x − 54

x+2 = 0

or x − 3 = 0 or x − 3 = 0 or x − 3 = 0

x = −2 or

x = 3 or

x = 3 or

x=3

The solutions are −2 and 3. % 46. 1 % 1 −4 −7 34 −24 1 −3 −10 24 1 −3 −10 24 0 % 2 % 1 −3 −10 24 2 −2 −24 1 −1 −12 0

f (x) = (x − 1)(x − 2)(x2 − x − 12)

= (x − 1)(x − 2)(x − 4)(x + 3)

The solutions of f (x) = 0 are 1, 2, 4, and −3.

f (x) = (x − 2)(x + 4)(x − 3)

x = 2 or

Since g(3) = 0, x − 3 is a factor of x3 − 9x2 + 27x − 27. Factoring the trinomial we get

2

x − 2 = 0 or x + 4 = 0

We continue to use synthetic division to factor g(x) = x3 − 9x2 + 27x − 27. Trying x + 2 again and x − 2 we find that g(−2) "= 0 and g(2) "= 0. Thus x + 2 and x − 2 are not factors of g(x). Try x − 3. % 27 −27 3 % 1 −9 3 −18 27 1 −6 9 0

Thus f (x) = (x + 2)(x − 3)(x2 − 6x + 9).

The solutions are −1, 2, and 5. % 42. 1 % 1 2 −13 10 1 3 −10 1 3 −10 0

(x − 2)(x + 4)(x − 3) = 0

Since f (−2) = 0, x + 2 is a factor of f (x). Thus f (x) = (x + 2)(x3 − 9x2 + 27x − 27).

Try x + 1 and x − 1. Using synthetic division we find that f (−1) "= 0 and f (1) "= 0. Thus x + 1 and x − 1 are not factors of f (x). Try x + 2. Use synthetic division to see whether f (−2) = 0. % −2 % 1 −7 9 27 −54 −2 18 −54 54 1 −9 27 −27 0

47. f (x) = x4 − x3 − 19x2 + 49x − 30

Try x − 1. Use synthetic division to see whether f (1) = 0. % 1 % 1 −1 −19 49 −30 1 0 −19 30 1 0 −19 30 0

Since f (1) = 0, x − 1 is a factor of f (x). Thus f (x) = (x − 1)(x3 − 19x + 30).

We continue to use synthetic division to factor g(x) = x3 − 19x + 30. Trying x − 1, x + 1, and x + 2 we find that g(1) "= 0, g(−1) "= 0, and g(−2) "= 0. Thus x − 1, x + 1, and x + 2 are not factors of x3 − 19x + 30. Try x − 2. % 2 % 1 0 −19 30 2 4 −30 1 2 −15 0

Since g(2) = 0, x − 2 is a factor of x3 − 19x + 30.

Thus f (x) = (x − 1)(x − 2)(x2 + 2x − 15). Factoring the trinomial we get

f (x) = (x − 1)(x − 2)(x − 3)(x + 5).

To solve the equation f (x) = 0, use the principle of zero products. (x − 1)(x − 2)(x − 3)(x + 5) = 0

Exercise Set 3.3

203 In (1, 3), test 2:

x−1 = 0 or x−2 = 0 or x−3 = 0 or x+5 = 0 x = 1 or

x = 2 or

x = 3 or

f (2) = 24 − 23 − 7 · 22 + 2 + 6 = −12 < 0

x = −5

The solutions are 1, 2, 3, and −5. % 48. −1 % 1 11 41 61 30 −1 −10 −31 −30 1 10 31 30 0 % −2 % 1 10 31 30 −2 −16 −30 1 8 15 0

In (3, ∞), test 4:

f (4) = 44 − 43 − 7 · 42 + 4 + 6 = 90 > 0 Thus the graph lies above the x-axis on (−∞, −2), on (−1, 1), and on (3, ∞). It lies below the x-axis on (−2, −1) and on (1, 3). We also know the points (−3, 48), (−1.5, −2.8125), (0, 6), (2, −12), and (4, 90) are on the graph. 4. From Step 3 we see that f (0) = 6 so the y-intercept is (0, 6).

f (x) = (x + 1)(x + 2)(x2 + 8x + 15)

5. We find additional points on the graph and draw the graph.

= (x + 1)(x + 2)(x + 3)(x + 5) The solutions of f (x) = 0 are −1, −2, −3, and −5. 49. f (x) = x − x − 7x + x + 6 4

3

2

1. The leading term is x . The degree, 4, is even and the leading coefficient, 1, is positive so as x → ∞, f (x) → ∞ and as x → −∞, f (x) → ∞. 4

2. Find the zeros of the function. We first use synthetic division to determine if f (1) = 0. % 1 % 1 −1 −7 1 6 1 0 −7 −6 1 0 −7 −6 0 1 is a zero of the function and we have f (x) = (x − 1)(x3 − 7x − 6). Synthetic division shows that −1 is a zero of g(x) = x3 − 7x − 6. % −1 % 1 0 −7 −6 −1 1 6 1 −1 −6 0

Then we have f (x) = (x − 1)(x + 1)(x − x − 6). To find the other zeros we solve the following equation: x2 − x − 6 = 0 2

(x − 3)(x + 2) = 0

x − 3 = 0 or x + 2 = 0 x = 3 or

x = −2

The zeros of the function are 1, −1, 3, and −2 so the x-intercepts of the graph are (1, 0), (−1, 0), (3, 0), and (−2, 0). 3. The zeros divide the x-axis into five intervals, (−∞, −2), (−2, −1), (−1, 1), (1, 3), and (3, ∞). We choose a value for x from each interval and find f (x). This tells us the sign of f (x) for all values of x in the interval. In (−∞, −2), test −3: f (−3) = (−3)4 − (−3)3 − 7(−3)2 + (−3) + 6 = 48 > 0

In (−2, −1), test −1.5:

f (−1.5) = (−1.5)4 −(−1.5)3 −7(−1.5)2 +(−1.5)+6 =

−2.8125 < 0

In (−1, 1), test 0: f (0) = 0 − 0 − 7 · 0 + 0 + 6 = 6 > 0 4

3

2

x

f (x)

−2.5

14.3

−0.5

3.9

0.5

4.7

2.5

−11.8

y 8 4 2

!4 !2

4

x

!4 !8 !12

f(x) " x 4 ! x 3 ! 7x 2 # x # 6

6. Checking the graph as described on page 272 in the text, we see that it appears to be correct. 50.

y 8 4 2

!4 !2

4

x

!4 !8 !12

f (x ) " x 4 # x 3 ! 3x 2 ! 5x ! 2

51. f (x) = x3 − 7x + 6

1. The leading term is x3 . The degree, 3, is odd and the leading coefficient, 1, is positive so as x → ∞, f (x) → ∞ and as x → −∞, f (x) → −∞.

2. Find the zeros of the function. We first use synthetic division to determine if f (1) = 0. % 1 % 1 0 −7 6 1 1 −6 1 1 −6 0 1 is a zero of the function and we have f (x) = (x − 1)(x2 + x − 6). To find the other zeros we solve the following equation. x2 + x − 6 = 0

(x + 3)(x − 2) = 0 x+3 = 0

or x − 2 = 0

x = −3 or

x=2

The zeros of the function are 1, −3, and 2 so the xintercepts of the graph are (1, 0), (−3, 0), and (2, 0).

204

Chapter 3: Polynomial and Rational Functions 1 is a zero of the function and we have f (x) = (x − 1)(−x2 + 2x + 8). To find the other zeros we solve the following equation.

3. The zeros divide the x-axis into four intervals, (−∞, −3), (−3, 1), (1, 2), and (2, ∞). We choose a value for x from each interval and find f (x). This tells us the sign of f (x) for all values of x in the interval.

−x2 + 2x + 8 = 0

In (−∞, −3), test −4:

x2 − 2x − 8 = 0

f (−4) = (−4) − 7(−4) + 6 = −30 < 0 3

(x − 4)(x + 2) = 0

In (−3, 1), test 0:

x − 4 = 0 or x + 2 = 0

f (0) = 03 − 7 · 0 + 6 = 6 > 0 In (1, 2), test 1.5:

x = 4 or

f (1.5) = (1.5)3 − 7(1.5) + 6 = −1.125 < 0 In (2, ∞), test 3: f (3) = 33 − 7 · 3 + 6 = 12 > 0

Thus the graph lies below the x-axis on (−∞, −3) and on (1, 2). It lies above the x-axis on (−3, 1) and on (2, ∞). We also know the points (−4, −30), (0, 6), (1.5, −1.125), and (3, 12) are on the graph. 4. From Step 3 we see that f (0) = 6 so the y-intercept is (0, 6).

f (−3) = (−3)3 + 3(−3)2 + 6(−3) − 8 = 28 > 0

In (−2, 1), test 0:

f (0) = −03 + 3 · 02 + 6 · 0 − 8 = −8 < 0

5. We find additional points on the graph and draw the graph. x

12

2.5

4.1

4

42

f (2) = −23 + 3 · 22 + 6 · 2 − 8 = 8 > 0 In (4, ∞), test 5:

20

−3.5 −12.4 −2

In (1, 4), test 2:

y

f (x)

f (5) = −53 + 3 · 52 + 6 · 5 − 8 = −28 < 0 Thus the graph lies above the x-axis on (−∞, −2) and on (1, 4). It lies below the x-axis on (−2, 1) and on (4, ∞). We also know the points (−3, 28), (0, −8), (2, 8), and (5, −28) are on the graph. 4. From Step 3 we see that f (0) = −8 so the y-intercept is (0, −8).

10 !4 !2 !10

2

4

x

!20

f(x) " x 3 ! 7x # 6

5. We find additional points on the graph and draw the graph.

6. Checking the graph as described on page 272 in the text, we see that it appears to be correct. 52.

x

f (x)

y

−2.5

11.4

30

−1

−10

3

10

4.5

−11.4

20 10 !4 !2 !10

2

4

x = −2

The zeros of the function are 1, 4, and −2 so the xintercepts of the graph are (1, 0), (4, 0), and (−2, 0). 3. The zeros divide the x-axis into four intervals, (−∞, −2), (−2, 1), (1, 4), and (4, ∞). We choose a value for x from each interval and find f (x). This tells us the sign of f (x) for all values of x in the interval. In (−∞, −2), test −3:

x

1. The leading term is −x3 . The degree, 3, is odd and the leading coefficient, −1, is negative so as x → ∞, f (x) → −∞ and as x → −∞, f (x) → ∞.

2. Find the zeros of the function. We first use synthetic division to determine if f (1) = 0. % 1 % −1 3 6 −8 −1 2 8 −1 2 8 0

8 4 2

!4 !2

4

x

!4 !8

f (x ) " !x 3 # 3x 2 # 6x ! 8

6. Checking the graph as described on page 272 in the text, we see that it appears to be correct.

f(x) " x 3 ! 12x # 16

53. f (x) = −x3 + 3x2 + 6x − 8

y

54.

y 4 2

!4 !2

4

x

!4 !8 !12 !14

f (x ) " !x 4 # 2x 3 # 3x 2 ! 4x ! 4

Exercise Set 3.3

205 x+3 = 0

55. Left to the student 56. No; the polynomial cannot have more than n linear factors. 57. We can write P (x) = p(x) · q(x), where q(x) is also a factor of P (x). If p(2) = 0, then P (2) = p(2) · q(2) = 0 · q(2) = 0, so P (2) = 0. If P (2) = 0 it does not necessarily follow that p(2) = 0. Consider the function P (x) = p(x)·q(x) where p(x) = x−1 and q(x) = x − 2. Then P (2) = p(2) · q(2) = (2 − 1)(2 − 2) = 1 · 0 = 0, but p(2) = 1 "= 0. 58.

2x − 7 = 5x + 8 −7 = 3x + 8

−15 = 3x −5 = x

Subtracting 2x on both sides Subtracting 8 on both sides Dividing by 3 on both sides

The solution is −5. 59.

2x − 5x + 12 = 0

a = 2, b = −5, c = 12 √ −b ± b2 − 4ac x= 2a ! −(−5) ± (−5)2 − 4 · 2 · 12 x= 2·2 √ 5 ± −71 = 4 √ √ 5 ± i 71 5 71 = = ± i 4 4 4 √ √ √ 71 5 71 5 71 5 i and − i, or ± i. The solutions are + 4 4 4 4 4 4 7x2 + 4x = 3 (7x − 3)(x + 1) = 0

7x − 3 = 0 or x + 1 = 0

x = −1

x = −1

3 The solutions are and −1. 7

61. We substitute −14 for g(x) and solve for x. −14 = x2 + 5x − 14 0 = x2 + 5x

0 = x(x + 5) x = 0 or x + 5 = 0 x = 0 or

When the output is −20, the input is −3 or −2. 64. We use the points (0, 27.2) and (12, 72) to find m. 72 − 27.2 44.8 448 56 m= = = = 12 − 0 12 120 15 Since we have the point (0, 27.2), we know that b = 27.2, 56 so we have f (x) = x + 27.2, where x is the number of 15 years since 1989 and f (x) is in billions of dollars. 56 · 11 + 27.2 ≈ $68.3 billion For 2000, f (11) = 15 56 For 2005, f (16) = · 16 + 27.2 ≈ $86.9 billion 15 56 For 2010, f (21) = · 21 + 27.2 = $105.6 billion 15

b + h = 30, so b = 30 − h. 1 1 1 A = bh = (30 − h)h = − h2 + 15h 2 2 2 Find the value of h for which A is a maximum: −15 h= = 15 2(−1/2) When h = 15, b = 30 − 15 = 15.

The area is a maximum when the base and the height are each 15 in.

66. a) −5, −3, 4, 6, and 7 are zeros of the function, so x + 5, x + 3, x − 4, x − 6, and x − 7 are factors. b) We first write the product of the factors:

F (x) = (x + 5)(x + 3)(x − 4)(x − 6)(x − 7)

7x2 + 4x − 3 = 0

7x = 3 or 3 or x= 7

x = −2

65. Let b and h represent the length of the base and the height of the triangle, respectively.

2x2 + 12 = 5x 2

60.

or x + 2 = 0

x = −3 or

x = −5

When the output is −14, the input is 0 or −5. 62. g(3) = 32 + 5 · 3 − 14 = 10 63. We substitute −20 for g(x) and solve for x. −20 = x + 5x − 14 2

0 = x2 + 5x + 6

0 = (x + 3)(x + 2)

Note that F (0) = 5 · 3(−4)(−6)(−7) < 0, but that the y-intercept of the graph is positive. Thus we must reflect F (x) across the x-axis to obtain a function P (x) with the given graph. We have: P (x) = −F (x)

P (x) = −(x + 5)(x + 3)(x − 4)(x − 6)(x − 7)

c) Yes; two examples are f (x) = c · P (x), for any nonzero constant c, and g(x) = (x − a)P (x).

d) No; only the function in part (b) has the given graph. 67. a) −4, −3, 2, and 5 are zeros of the function, so x + 4, x + 3, x − 2, and x − 5 are factors. b) We first write the product of the factors: P (x) = (x + 4)(x + 3)(x − 2)(x − 5)

Note that P (0) = 4 · 3(−2)(−5) > 0 and the graph shows a positive y-intercept, so this function is a correct one. c) Yes; two examples are f (x) = c · P (x) for any nonzero constant c and g(x) = (x − a)P (x).

d) No; only the function in part (b) has the given graph.

206

Chapter 3: Polynomial and Rational Functions Using synthetic division and several trials, we find that −3 is a factor of f (x) = x3 + 5x2 − 18: " 5 0 −18 −3 " 1 −3 −6 18 1 2 −6 0

68. Divide x2 + kx + 4 by x − 1. " 1" 1 k 4 1 k+1 1 k+1 k+5 The remainder is k + 5.

Then we have:

Divide x2 + kx + 4 by x + 1. " k 4 −1 " 1 −1 −k + 1 1 k − 1 −k + 5

(x + 3)(x2 + 2x − 6) = 0 x+3 = 0

x = −3 or √ Only −1 ± 7 check.

The remainder is −k + 5.

Let k + 5 = −k + 5 and solve for k. k + 5 = −k + 5 2k = 0 k=0

69. Divide x3 − kx2 + 3x + 7k by x + 2. " −2 " 1 −k 3 7k −2 2k + 4 −4k − 14 1 −k − 2 2k + 7 3k − 14 Thus P (−2) = 3k − 14.

We know that if x + 2 is a factor of f (x), then f (−2) = 0. We solve 0 = 3k − 14 for k. 0 = 3k − 14 14 =k 3 1 1 3 70. y = x − x 13 14 $ # 1 1 2 x − y=x 13 14 We use the principle of zero products to find the zeros of the polynomial. 1 1 2 x = 0 or x − =0 13 14 1 1 2 x = x = 0 or 13 14 13 x = 0 or x2 = 14 % 13 x = 0 or x=± 14 x = 0 or x ≈ ±0.9636 Only 0 and 0.9636 are in the interval [0, 2].

71.

2x2 4 32 + = 2 , −1 x+3 3x − x − 3 LCM is (x + 1)(x − 1)(x + 3) ' & 4 2x2 + = (x + 1)(x − 1)(x + 3) (x + 1)(x − 1) x + 3 32 (x + 1)(x − 1)(x + 3) · (x + 1)(x − 1)(x + 3) x2

2x2 (x + 3) + 4(x + 1)(x − 1) = 32 2x3 + 6x2 + 4x2 − 4 = 32 2x3 + 10x2 − 36 = 0 x3 + 5x2 − 18 = 0

or x2 + 2x − 6 = 0

72.

x = −1 ±

√

7

60 1 6x2 + = , + 11 x3 − 7x2 + 11x − 77 x−7 LCM is (x2 + 11)(x − 7) & ' 6x2 60 (x2 + 11)(x − 7) 2 + 2 = x + 11 (x + 11)(x − 7) 1 (x2 + 11)(x − 7) · x−7 6x2 (x − 7) + 60 = x2 + 11 x2

6x3 − 42x2 + 60 = x2 + 11 6x3 − 43x2 + 49 = 0

Use synthetic division to find factors of f (x) = 6x3 −43x2 + 49. " −1 " 6 −43 0 49 −6 49 −49 6 −49 49 0 Then we have: (x + 1)(6x2 − 49x + 49) = 0

(x + 1)(6x − 7)(x − 7) = 0

x = −1 or x = Only −1 and

7 or x = 7 6

7 check. 6

73. Answers may vary. One possibility is P (x) = x15 − x14 . " 74. 3 + 2i " 1 −4 −2 3 + 2i −7 + 4i 1 −1 + 2i −9 + 4i

The answer is x − 1 + 2i, R −9 + 4i. " 75. i " 1 −3 7 (i2 = −1) i −3i − 1 1 −3 + i 6 − 3i The answer is x − 3 + i, R 6 − 3i.

Exercise Set 3.4

207

Exercise Set 3.4 1. Find a polynomial function of degree 3 with −2, 3, and 5 as zeros.

= (x2 − 2x + 1 − 3)(x + 2) = (x2 − 2x − 2)(x + 2)

Such a function has factors x + 2, x − 3, and x − 5, so we have f (x) = an (x + 2)(x − 3)(x − 5).

The number an can be any nonzero number. The simplest polynomial will be obtained if we let it be 1. Multiplying the factors, we obtain f (x) = (x + 2)(x − 3)(x − 5)

= x3 − 2x2 − 2x + 2x2 − 4x − 4

8.

3. Find a polynomial function of degree 3 with −3, 2i, and −2i as zeros.

= x3 + 2x2 − 12x − 16

9. Find a polynomial function of degree 3 with 1 + 6i, 1 − 6i, and −4 as zeros.

Such a function has factors x − (1 + 6i), x − (1 − 6i), and x + 4, so we have

Such a function has factors x + 3, x − 2i, and x + 2i, so we have f (x) = an (x + 3)(x − 2i)(x + 2i).

f (x) = an [x − (1 + 6i)][x − (1 − 6i)](x + 4).

The number an can be any nonzero number. The simplest polynomial will be obtained if we let it be 1. Multiplying the factors, we obtain

The number an can be any nonzero number. The simplest polynomial will be obtained if we let it be 1. Multiplying the factors, we obtain f (x) = (x + 3)(x − 2i)(x + 2i)

f (x) = [x − (1 + 6i)][x − (1 − 6i)](x + 4) = [(x − 1) − 6i][(x − 1) + 6i](x + 4)

= (x + 3)(x + 4) 2

= [(x − 1)2 − (6i)2 ](x + 4)

= x + 3x + 4x + 12. 3

4.

2

= (x2 − 2x + 1 + 36)(x + 4)

f (x) = (x − 2)(x − i)(x + i)

= (x2 − 2x + 37)(x + 4)

= (x − 2)(x2 + 1)

= x3 − 2x2 + 37x + 4x2 − 8x + 148

= x3 − 2x2 + x − 2

√ √ 5. Find a polynomial function of degree 3 with 2, − 2, and 3 as zeros. √ √ Such a function has factors √ x − √2, x + 2, and x − 3, so we have f (x) = an (x − 2)(x + 2)(x − 3).

The number an can be any nonzero number. The simplest polynomial will be obtained if we let it be 1. Multiplying the factors, we obtain √ √ f (x) = (x − 2)(x + 2)(x − 3) = (x2 − 2)(x − 3)

6.

= x3 − 3x2 − 2x + 6. √ √ f (x) = (x + 5)(x − 3)(x + 3) = (x + 5)(x2 − 3)

= x3 + 5x2 − 3x − 15

√ √ 7. Find a polynomial function of degree 3 with 1− 3, 1+ 3, and −2 as zeros. √ √ Such a function has factors x − (1 − 3), x − (1 + 3), and x + 2, so we have √ √ f (x) = an [x − (1 − 3)][x − (1 + 3)](x + 2).

The number an can be any nonzero number. The simplest polynomial will be obtained if we let it be 1. Multiplying the factors, we obtain

√

= x3 − 2x2 − 4x + 4x2 − 8x − 16

f (x) = (x + 1)(x)(x − 4) = x3 − 3x2 − 4x

√ 5)][x − (1 + 5)] √ √ = (x + 4)[(x − 1) + 5][(x − 1) − 5]

f (x) = (x + 4)[x − (1 −

= (x + 4)(x2 − 2x − 4)

= x3 − 6x2 − x + 30. = (x2 + x)(x − 4)

= x3 − 6x − 4.

= (x + 4)(x2 − 2x + 1 − 5)

= (x2 − x − 6)(x − 5)

2.

√

√ 3)][x − (1 + 3)](x + 2) √ √ = [(x − 1) + 3][(x − 1) − 3](x + 2) √ = [(x − 1)2 − ( 3)2 ](x + 2)

f (x) = [x − (1 −

= x3 + 2x2 + 29x + 148. 10.

f (x) = [x − (1 + 4i)][x − (1 − 4i)](x + 1) = (x2 − 2x + 17)(x + 1) = x3 − x2 + 15x + 17

1 11. Find a polynomial function of degree 3 with − , 0, and 2 3 as zeros. 1 Such a function has factors x + , x − 0 (or x), and x − 2 3 so we have # $ 1 (x)(x − 2). f (x) = an x + 3 The number an can be any nonzero number. The simplest polynomial will be obtained if we let it be 1. Multiplying the factors, we obtain $ # 1 (x)(x − 2) f (x) = x + 3 $ # 1 = x2 + x (x − 2) 3 5 2 2 3 = x − x − x. 3 3

208 12.

Chapter 3: Polynomial and Rational Functions # $ 1 f (x) = (x + 3)(x) x − 2 # $ 1 2 = (x + 3x)( x − 2 5 2 3 3 =x + x − x 2 2

13. A polynomial function of degree 5 has at most 5 zeros. Since 5 zeros are given, these are all of the zeros of the desired function. We proceed as in Exercises 1-11, letting an = 1. f (x) = (x + 1)3 (x − 0)(x − 1) = (x3 + 3x2 + 3x + 1)(x2 − x)

14.

= x5 + 2x4 − 2x2 − x

f (x) = (x + 2)(x − 3)2 (x + 1)

= x4 − 3x3 − 7x2 + 15x + 18

15. A polynomial function of degree 4 has at most 4 zeros. Since 4 zeros are given, these are all of the zeros of the desired function. We proceed as in Exercises 1-11, letting an = 1. f (x) = (x + 1)3 (x − 0) = (x3 + 3x2 + 3x + 1)(x)

16.

= x4 + 3x3 + 3x2 + x $2 # 1 (x − 0)(x − 1)2 f (x) = x + 2 3 1 1 = x5 − x4 − x3 + x2 + x 4 2 4

17. A polynomial function of degree 4 can have at most 4 zeros. Since f (x) has rational coefficients, in addition√to the three √ zeros given, the other zero is the conjugate of 3, or − 3. 18. A polynomial function of degree 4 can have at most 4 zeros. Since f (x) has rational coefficients, in addition to √ √the three zeros given, the other zero is the conjugate of − 2, or 2.

25. A polynomial function f (x) of degree 5 has at most 5 zeros. Since f (x) has rational coefficients, in addition to √ the 3 given zeros, √ the other zeros are the conjugates of 5 and −4i, or − 5 and 4i. 26. A polynomial function f (x) of degree 5 has at most 5 zeros. Since f (x) has rational coefficients, in addition to√ the 3 given zeros, the other zeros are the conjugates of − 3 and √ 2i, or 3 and −2i. 27. A polynomial function f (x) of degree 5 has at most 5 zeros. Since f (x) has rational coefficients, the other zero is the conjugate of 2 − i, or 2 + i. 28. A polynomial function f (x) of degree 5 has at most 5 zeros. Since f (x) has rational coefficients, the other zero is the conjugate of 1 − i, or 1 + i. 29. A polynomial function f (x) of degree 5 has at most 5 zeros. Since f (x) has rational coefficients, in addition to the 3 given zeros, √ conjugates of −3 + 4i √ the other zeros are the and 4 − 5, or −3 − 4i and 4 + 5. 30. A polynomial function f (x) of degree 5 has at most 5 zeros. Since f (x) has rational coefficients, in addition to the 3 given zeros, √ conjugates of −3 − 3i √ the other zeros are the and 2 + 13, or −3 + 3i and 2 − 13. 31. A polynomial function f (x) of degree 5 has at most 5 zeros. Since f (x) has rational coefficients, the other zero is the conjugate of 4 − i, or 4 + i. 32. A polynomial function f (x) of degree 5 has at most 5 zeros. Since f (x) has rational coefficients, √ the other zero is the √ conjugate of −3 + 2, or −3 − 2. 33. Find a polynomial function of lowest degree with rational coefficients that has 1 + i and 2 as some of its zeros. 1 − i is also a zero. Thus the polynomial function is f (x) = an (x − 2)[x − (1 + i)][x − (1 − i)].

19. A polynomial function of degree 4 can have at most 4 zeros. Since f (x) has rational coefficients, the other zeros√are the conjugates of the given zeros. They are i and 2 + 5.

If we let an = 1, we obtain

f (x) = (x − 2)[(x − 1) − i][(x − 1) + i]

20. A polynomial function of degree 4 can have at most 4 zeros. Since f (x) has rational coefficients, the other zeros are√the conjugates of the given zeros. They are −i and −3 − 3. 21. A polynomial function of degree 4 can have at most 4 zeros. Since f (x) has rational coefficients, in addition to the three zeros given, the other zero is the conjugate of 3i, or −3i. 22. A polynomial function of degree 4 can have at most 4 zeros. Since f (x) has rational coefficients, in addition to the three zeros given, the other zero is the conjugate of −2i, or 2i. 23. A polynomial function of degree 4 can have at most 4 zeros. Since f (x) has rational coefficients, the other zeros are √ the conjugates of the given zeros. They are −4+3i and 2+ 3. 24. A polynomial function of degree 4 can have at most 4 zeros. Since f (x) has rational coefficients, the other zeros are √ the conjugates of the given zeros. They are 6+5i and −1− 7.

= (x − 2)[(x − 1)2 − i2 ]

= (x − 2)(x2 − 2x + 1 + 1) = (x − 2)(x2 − 2x + 2)

34.

= x3 − 4x2 + 6x − 4.

f (x) = [x − (2 − i)][x − (2 + i)](x + 1) = x3 − 3x2 + x + 5

35. Find a polynomial function of lowest degree with rational coefficients that has 4i as one of its zeros. −4i is also a zero. Thus the polynomial function is f (x) = an (x − 4i)(x + 4i).

If we let an = 1, we obtain

f (x) = (x − 4i)(x + 4i) = x2 + 16. 36. f (x) = (x + 5i)(x − 5i) = x2 + 25

Exercise Set 3.4

209

37. Find a polynomial function of lowest degree with rational coefficients that has −4i and 5 as some of its zeros. 4i is also a zero.

If we let an = 1, we obtain = (x − 5)(x2 + 16)

= x3 − 5x2 + 16x − 80

38. f (x) = (x − 3)(x + i)(x − i) = x3 − 3x2 + x − 3 39. Find a polynomial function of lowest √ degree with rational coefficients√that has 1 − i and − 5 as some of its zeros. 1 + i and 5 are also zeros. Thus the polynomial function is f (x) = an [x − (1 − i)][x − (1 + i)](x +

= (x − 2x + 1 + 1)(x − 5) 2

2

√

√

√

5)(x −

5)(x − 5)(x −

√

√ √

5) 5)

= x4 − 2x3 + 2x2 − 5x2 + 10x − 10

= x − 2x − 3x + 10x − 10 √ √ f (x) = [x−(2− 3)][x−(2+ 3)][x−(1+i)][x−(1−i)]

40.

3

2

= x − 6x + 11x − 10x + 2 4

3

2

41. Find a polynomial function of lowest degree with rational √ coefficients that has 5 and −3i as some of its zeros. √ − 5 and 3i are also zeros. Thus the polynomial function is √ √ f (x) = an (x − 5)(x + 5)(x + 3i)(x − 3i).

If we let an = 1, we obtain f (x) = (x − 5)(x + 9) 2

42.

2

= x4 + 4x2 − 45 √ √ f (x) = (x + 2)(x − 2)(x − 4i)(x + 4i) = x4 + 14x2 − 32

43. f (x) = x3 + 5x2 − 2x − 10

Since −5 is a zero of f (x), we have f (x) = (x + 5) · Q(x). We use synthetic division to find Q(x). " −5 " 1 5 −2 −10 −5 0 10 1 0 −2 0

Then f (x) = (x + 5)(x2 − 2). To find the other zeros we solve x2 − 2 = 0. x2 − 2 = 0 x2 = 2

√ x=± 2

√ √ The other zeros are − 2 and 2.

1

x2 + 1

x2 − x4 − x4

5).

= (x2 − 2x + 2)(x2 − 5) 4

−1 1 −1 1 0 1 0 1 0

45. If −i is a zero of f (x) = x4 − 5x3 + 7x2 − 5x + 6, i is also a zero. Thus x + i and x − i are factors of the polynomial. Since (x + i)(x − i) = x2 + 1, we know that f (x) = (x2 + 1) · Q(x). Divide x4 − 5x3 + 7x2 − 5x + 6 by x2 + 1.

f (x) = (x − 5)[x2 − (4i)2 ]

= [(x − 1) + i][(x − 1) − i](x +

1

Solving x2 + 1 = 0, we find that the other zeros are −i and i.

f (x) = an (x − 5)(x + 4i)(x − 4i).

f (x) = [x − (1 − i)][x − (1 + i)](x +

" 1"

f (x) = (x − 1)(x2 + 1)

Thus the polynomial function is

If we let an = 1, we obtain

44.

5x + 6 5x3 + 7x2 − 5x + 6 + x2 3 −5x + 6x2 − 5x − 5x −5x3 6x2 +6 6x2 +6 0

Thus x4 −5x3 +7x2 −5x+6 = (x+i)(x−i)(x2 −5x+6)

= (x+i)(x−i)(x−2)(x−3)

Using the principle of zero products we find the other zeros to be i, 2, and 3. 46. (x − 2i) and (x + 2i) are both factors of P (x) = x4 − 16. (x − 2i)(x + 2i) = x2 + 4 x2 + 4

x2 − x4 + x4 +

4 0x2 − 16 4x2 −4x2 − 16 −4x2 − 16 0

(x − 2i)(x + 2i)(x2 − 4) = 0

(x − 2i)(x + 2i)(x + 2)(x − 2) = 0

The other zeros are −2i, −2, and 2. 47. x3 − 6x2 + 13x − 20 = 0

If 4 is a zero, then x − 4 is a factor. Use synthetic division to "find another factor. 4 " 1 −6 13 −20 4 −8 20 1 −2 5 0 (x − 4)(x2 − 2x + 5) = 0

x−4 = 0 or x2 −2x+5 = 0

Principle of zero products

√ 2± 4−20 2 Quadratic formula 2±4i = 1 ± 2i x = 4 or x= 2 The other zeros are 1 + 2i and 1 − 2i. x = 4 or

x=

210 48.

Chapter 3: Polynomial and Rational Functions " 2"

1 1

55. f (x) = x3 + 3x2 − 2x − 6 Possibilities for p ±1, ±2, ±3, ±6 a) : Possibilities for q ±1 Possibilities for p/q: 1, −1, 2, −2, 3, −3, 6, −6

0 0 −8 2 4 8 2 4 0

(x − 2)(x2 + 2x + 4) = 0 √ x = 2 or x = −1 ± 3i

The other zeros are −1 +

√

3i and −1 −

√

3i.

49. f (x) = x5 − 3x2 + 1

According to the rational zeros theorem, any rational zero of f must be of the form p/q, where p is a factor of the constant term, 1, and q is a factor of the coefficient of x5 , 1. ±1 Possibilities for p : Possibilities for q ±1 Possibilities for p/q: 1, −1

50. f (x) = x + 37x − 6x + 12 ±1, ±2, ±3, ±4, ±6, ±12 Possibilities for p : Possibilities for q ±1 7

5

2

Possibilities for p/q: 1, −1, 2, −2, 3, −3, 4, −4, 6, −6, 12, −12

51. f (x) = 2x4 − 3x3 − x + 8

According to the rational zeros theorem, any rational zero of f must be of the form p/q, where p is a factor of the constant term, 8, and q is a factor of the coefficient of x4 , 2. ±1, ±2, ±4, ±8 Possibilities for p : Possibilities for q ±1, ±2 1 1 Possibilities for p/q: 1, −1, 2, −2, 4, −4, 8, −8, , − 2 2

52. f (x) = 3x3 − x2 + 6x − 9 Possibilities for p ±1, ±3, ±9 : Possibilities for q ±1, ±3 Possibilities for p/q: 1, −1, 3, −3, 9, −9,

Then we have f (x) = (x + 3)(x2 − 2). We find the other zeros: x2 − 2 = 0 x2 = 2

√ x = ± 2.

There is only √ one rational zero, −3. The other zeros are ± 2. (Note that we could have used factoring by grouping to find this result.) √ √ b) f (x) = (x + 3)(x − 2)(x + 2)

56. f (x) = x3 − x2 − 3x + 3 ±1, ±3 Possibilities for p a) : Possibilities for q ±1 Possibilities for p/q: 1, −1, 3, −3

From the graph of y = x3 − x2 − 3x + 3, we see that, of the possibilities above, only 1 might be a zero. " 1 " 1 −1 −3 3 1 0 −3 1 0 −3 0 f (x) = (x − 1)(x2 − 3)

1 1 ,− 3 3

53. f (x) = 15x6 + 47x2 + 2 According to the rational zeros theorem, any rational zero of f must be of the form p/q, where p is a factor of 2 and q is a factor of 15. ±1, ±2 Possibilities for p : Possibilities for q ±1, ±3, ±5, ±15 1 1 2 2 1 Possibilities for p/q: 1, −1, 2, −2, , − , , − , , 3 3 3 3 5 1 2 2 1 1 2 2 − , ,− , ,− , ,− 5 5 5 15 15 15 15 54. f (x) = 10x25 + 3x17 − 35x + 6 ±1, ±2, ±3, ±6 Possibilities for p : Possibilities for q ±1, ±2, ±5, ±10

From the graph of y = x3 + 3x2 − 2x − 6, we see that, of the possibilities above, only −3 might be a zero. We use synthetic division to determine whether −3 is indeed a zero. " −3 " 1 3 −2 −6 −3 0 6 1 0 −2 0

Possibilities for p/q: 1, −1, 2, −2, 3, −3, 6, −6, 1 1 3 3 1 1 2 2 ,− , ,− , ,− , ,− , 2 2 2 2 5 5 5 5 1 3 3 3 6 6 1 ,− , ,− , ,− , , 5 5 5 5 10 10 10 3 − 10

√ Now x2 − 3 = 0 for x = ± 3. Thus, there is only √ one rational zero, 1. The other zeros are ± 3. (Note that we would have used factoring by grouping to find this result.) √ √ b) f (x) = (x − 1)(x − 3)(x + 3)

57. f (x) = x3 − 3x + 2 Possibilities for p ±1, ±2 a) : Possibilities for q ±1 Possibilities for p/q: 1, −1, 2, −2

From the graph of y = x3 − 3x + 2, we see that, of the possibilities above, −2 and 1 might be a zeros. We use synthetic division to determine whether −2 is a zero. " −2 " 1 0 −3 2 −2 4 −2 1 −2 1 0

Then we have f (x) = (x + 2)(x2 − 2x + 1) = (x + 2)(x − 1)2 .

Now (x − 1)2 = 0 for x = 1. Thus, the rational zeros are −2 and 1. (The zero 1 has a multiplicity of 2.) These are the only zeros.

b) f (x) = (x + 2)(x − 1)2

Exercise Set 3.4

211

58. f (x) = x3 − 2x + 4 a)

Possibilities for p ±1, ±2, ±4 : Possibilities for q ±1 Possibilities for p/q: 1, −1, 2, −2, 4, −4

From the graph of y = x3 −2x+4, we see that, of the possibilities above, only −2 might be a zero. " −2 " 1 0 −2 4 −2 4 −4 1 −2 2 0 f (x) = (x + 2)(x2 − 2x + 2)

Using the quadratic √ formula, we find that the other −3 ± 41 zeros are . The only rational zero is −2. 4 √ −3 ± 41 . The other zeros are 4 √ $# √ $ # −3+ 41 −3− 41 b) f (x) = (x+2) x− x− 4 4 61. f (x) = 5x4 − 4x3 + 19x2 − 16x − 4 a)

b) f (x) = (x + 2)[x − (1 + i)][x − (1 − i)] = (x + 2)(x − 1 − i)(x − 1 + i)

59. f (x) = x3 − 5x2 + 11x + 17

Possibilities for p ±1, ±17 : Possibilities for q ±1 Possibilities for p/q: 1, −1, 17, −17

From the graph of y = x3 −5x2 +11x+17, we see that, of the possibilities above, we see that only −1 might be a zero. We use synthetic division to determine whether −1 is indeed a zero. " −1 " 1 −5 11 17 −1 6 −17 1 −6 17 0

Then we have f (x) = (x − 1)(5x3 + x2 + 20x + 4)

= (x − 1)[x2 (5x + 1) + 4(5x + 1)] = (x − 1)(5x + 1)(x2 + 4).

We find the other zeros: 5x + 1 = 0 or x2 + 4 = 0

Then we have f (x) = (x + 1)(x2 − 6x + 17). We use the quadratic formula to find the other zeros.

5x = −1 or 1 x = − or 5

x2 − 6x + 17 = 0

! (−6)2 − 4 · 1 · 17 x= 2·1 √ √ 6 ± 4 2i 6 ± −32 = = 2√ 2 = 3 ± 2 2i −(−6) ±

The only√rational zero is −1. The other zeros are 3 ± 2 2i. √ √ b) f (x) = (x + 1)[x − (3 + 2 2i)][x − (3 − 2 2i)] √ √ = (x + 1)(x − 3 − 2 2i)(x − 3 + 2 2i) 60. f (x) = 2x + 7x + 2x − 8 3

a)

2

Possibilities for p : Possibilities for q

±1, ±2, ±4, ±8 ±1, ±2

Possibilities for p/q: 1, −1, 2, −2, 4, −4, 8, −8 1 1 ,− 2 2 From the graph of y = 2x3 + 7x2 + 2x − 8, we see that, of the possibilities above, only −2 and 1 might be a zeros. " −2 " 2 7 2 −8 −4 −6 8 2 3 −4 0 f (x) = (x + 2)(2x2 + 3x − 4)

±1, ±2, ±4 ±1, ±5

1 1 Possibilities for p/q: 1, −1, 2, −2, 4, −4, , − 5 5 2 2 4 4 ,− , ,− 5 5 5 5 From the graph of y = 5x4 − 4x3 + 19x2 − 16x − 4, we see that, of the possibilities above, only 2 1 − , − and 1 might be zeros. We use synthetic 5 5 division to determine whether 1 is a zero. " 1 " 5 −4 19 −16 −4 5 1 20 4 5 1 20 4 0

Using the quadratic formula, we find that the other zeros are 1 ± i. The only rational zero is −2. The other zeros are 1 ± i.

a)

Possibilities for p : Possibilities for q

x2 = −4 x = ±2i

1 The rational zeros are − and 1. The other zeros 5 are ±2i.

b) From part (a) we see that

f (x) = (5x + 1)(x − 1)(x + 2i)(x − 2i). 62. f (x) = 3x4 − 4x3 + x2 + 6x − 2 a)

Possibilities for p : Possibilities for q

±1, ±2 ±1, ±3

1 1 Possibilities for p/q: 1, −1, 2, −2, , − 3 3 2 2 ,− 3 3 From the graph of y = 3x4 − 4x3 + x2 + 6x − 2, we see that, of the possibilities above, only −1 1 and might be zeros. "3 −1 " 3 −4 1 6 −2 −3 7 −8 2 3 −7 8 −2 0 " 1 " 8 −2 3 " 3 −7 1 −2 2 3 −6 6 0

212

Chapter 3: Polynomial and Rational Functions $ 1 (3x2 − 6x + 6) 3 $ 1 (3)(x2 − 2x + 2) 3 Using the quadratic formula, we find that the other zeros are 1 ± i. 1 The rational zeros are −1 and . The other zeros 3 are 1 ± i. $ # 1 b) f (x) = 3(x+1) x− [x−(1+i)][x−(1−i)] 3

" 2"

# f (x) = (x + 1) x − # = (x + 1) x −

= (x + 1)(3x − 1)(x − 1 − i)(x − 1 + i) 63. f (x) = x4 − 3x3 − 20x2 − 24x − 8 a)

Possibilities for p ±1, ±2, ±4, ±8 : Possibilities for q ±1 Possibilities for p/q: 1, −1, 2, −2, 4, −4, 8, −8

Using the quadratic √ formula, we find that the other zeros are −4 ± 21.

The rational √ zeros are 1 and 2. The other zeros are −4 ± 21. √ √ b) f (x) = (x−1)(x−2)[x−(−4+ 21)][x−(−4− 21)] √ √ = (x−1)(x−2)(x+4− 21)(x+4+ 21) 65. f (x) = x3 − 4x2 + 2x + 4 a)

Then we have f (x) = (x − 2)(x2 − 2x − 2). Use the quadratic formula to find the other zeros. x2 − 2x − 2 = 0

! (−2)2 − 4 · 1 · (−2) 2·1 √ √ 2±2 3 2 ± 12 = = 2 2 √ = 1± 3

x=

Then we have f (x) = (x + 2)(x + 1)(x2 − 6x − 4).

Use the quadratic formula to find the other zeros. ! (−6)2 − 4 · 1 · (−4) x= 2·1 √ √ 6 ± 2 13 6 ± 52 = = 2 2 √ = 3 ± 13

The rational √ zeros are −2 and −1. The other zeros are 3 ± 13. √ √ b) f (x) = (x+2)(x+1)[x−(3+ 13)][x−(3− 13)] √ √ = (x+2)(x+1)(x−3− 13)(x−3+ 13) 64. f (x) = x4 + 5x3 − 27x2 + 31x − 10 a)

Possibilities for p ±1, ±2, ±5, ±10 : Possibilities for q ±1 Possibilities for p/q: 1, −1, 2, −2, 5, −5, 10, −10

From the graph of y = x4 +5x3 −27x2 +31x−10, we see that, of the possibilities above, only 1 and 2 might be zeros. " 1 " 1 5 −27 31 −10 1 6 −21 10 1 6 −21 10 0

±1, ±2, ±4 Possibilities for p : Possibilities for q ±1 Possibilities for p/q: 1, −1, 2, −2, 4, −4

From the graph of y = x3 − 4x2 + 2x + 4, we see that, of the possibilities above, only −1, 1, and 2 might be zeros. Synthetic division shows that neither −1 nor 1 is a zero. Try 2. " 2 " 1 −4 2 4 2 −4 −4 1 −2 −2 0

We see that −2 is a zero. Now we determine whether −1 is a zero. " −1 " 1 −5 −10 −4 −1 6 4 1 −6 −4 0

−(−6) ±

−21 10 16 −10 −5 0

f (x) = (x − 1)(x − 2)(x2 + 8x − 5)

From the graph of y = x4 − 3x3 − 20x2 − 24x − 8, we see that, of the possibilities above, only −2 and −1 might be zeros. We use synthetic division to determine if −2 is a zero. " −2 " 1 −3 −20 −24 −8 −2 10 20 8 1 −5 −10 −4 0

x2 − 6x − 4 = 0

1 6 2 1 8

−(−2) ±

The√only rational zero is 2. The other zeros are 1 ± 3. √ √ b) f (x) = (x − 2)[x − (1 + 3)][x − (1 − 3)] √ √ = (x − 2)(x − 1 − 3)(x − 1 + 3) 66. f (x) = x3 − 8x2 + 17x − 4 a)

Possibilities for p ±1, ±2, ±4 : Possibilities for q ±1 Possibilities for p/q: 1, −1, 2, −2, 4, −4

From the graph of y = x3 − 8x2 + 17x − 4, we see that, of the possibilities above, only 4 might be "a zero. 4 " 1 −8 17 −4 4 −16 4 1 −4 1 0 f (x) = (x − 4)(x2 − 4x + 1)

Using the quadratic√ formula, we find that the other zeros are 2 ± 3.

The√only rational zero is 4. The other zeros are 2 ± 3. √ √ b) f (x) = (x − 4)[x − (2 + 3)][x − (2 − 3)] √ √ = (x − 4)(x − 2 − 3)(x − 2 + 3)

Exercise Set 3.4

213 From the graph of y = 2x3 −3x2 −x+1, we see that, 1 1 of the possibilities above, only − and might be 2 2 1 zeros. Synthetic division shows that − is not a 2 1 zero. Try . 2 " 1 " 1 2 " 2 −3 −1 1 −1 −1 2 −2 −2 0 $ # 1 (2x2 − 2x − 2) = We have g(x) = x − 2 # $ 1 (2)(x2 − x − 1). Use the quadratic forx− 2 mula to find the other zeros. x2 − x − 1 = 0 ! −(−1) ± (−1)2 − 4 · 1 · (−1) x= 2·1 √ 1± 5 = 2 1 The only rational zero is . The other zeros are 2 √ 1± 5 . 2 1 b) f (x) = g(x) 6 √ '& √ ' # $ & 1 1+ 5 1− 5 1 x− (2) x− x− = 6 2 2 2 √ $# √ $ # $# 1 1+ 5 1− 5 1 = x− x− x− 3 2 2 2

67. f (x) = x3 + 8 a)

Possibilities for p ±1, ±2, ±4, ±8 : Possibilities for q ±1 Possibilities for p/q: 1, −1, 2, −2, 4, −4, 8, −8

From the graph of y = x3 + 8, we see that, of the possibilities above, only −2 might be a zero. We use synthetic division to see if it is. " 0 0 8 −2 " 1 −2 4 −8 1 −2 4 0 We have f (x) = (x + 2)(x2 − 2x + 4). Use the quadratic formula to find the other zeros. x2 − 2x + 4 = 0

! (−2)2 − 4 · 1 · 4 x= 2·1 √ √ 2 ± 2 3i 2 ± −12 = = 2 2 √ = 1 ± 3i −(−2) ±

The only √ rational zero is −2. The other zeros are 1 ± 3i. √ √ b) f (x) = (x + 2)[x − (1 + 3i)][x − (1 − 3i)] √ √ = (x + 2)(x − 1 − 3i)(x − 1 + 3i) 68. f (x) = x3 − 8 a) As in Exercise 67, the possibilities for p/q are 1, −1, 2, −2, 4, −4, 8, and −8.

From the graph of y = x3 − 8, we see that, of the possibilities above, only 2 might be a zero. " 2 " 1 0 0 −8 2 4 8 1 2 4 0 f (x) = (x − 2)(x2 + 2x + 4)

Using the quadratic √ formula, we find that the other zeros are −1 ± 3i.

The only √ rational zero is 2. The other zeros are −1 ± 3i. √ √ b) f (x) = (x−2)[x−(−1+ 3i)][x−(−1− 3i)] √ √ = (x − 2)(x + 1 − 3i)(x + 1 + 3i)

69.

1 3 1 2 1 1 x − x − x+ 3 2 6 6 1 = (2x3 − 3x2 − x + 1) 6 a) The second form of the equation is equivalent to the first and has the advantage of having integer coefficients. Thus, we can use the rational zeros theorem for g(x) = 2x3 − 3x2 − x + 1. The zeros of g(x) are the same as the zeros of f (x). We find the zeros of g(x). ±1 Possibilities for p : Possibilities for q ±1, ±2 1 1 Possibilities for p/q: 1, −1, , − 2 2 f (x) =

70.

2 3 1 2 2 1 x − x + x− 3 2 3 2 1 = (4x3 − 3x2 + 4x − 3) 6 a) Find the zeros of g(x) = 4x3 − 3x2 + 4x − 3. Possibilities for p ±1, ±3 : Possibilities for q ±1, ±2, ±4 1 1 3 3 Possibilities for p/q: 1, −1, 3, −3, , − , , − 2 2 2 2 1 1 3 3 ,− , ,− 4 4 4 4 From the graph of y = 4x3 − 3x2 + 4x − 3, we see 1 3 that, of the possibilities above, only , , and 1 2 4 1 might be zeros. Synthetic division shows that is 2 3 not a zero. Try . 4 " 3 " 4 " 4 −3 4 −3 3 0 3 4 0 4 0 $ # $ # 3 3 (4x2 +4) = x− (4)(x2 +1) g(x) = x− 4 4 f (x) =

214

Chapter 3: Polynomial and Rational Functions Now x2 + 1 = 0 when x = ±i. Thus, the only 3 rational zero is . The other zeros are ±i. (Note 4 that we could have used factoring by grouping to find this result.) 1 g(x) 6 # 1 x− = 6 # 2 x− = 3

$ 3 (4)(x + i)(x − i) 4 $ 3 (x + i)(x − i) 4

71. f (x) = x4 + 2x3 − 5x2 − 4x + 6

According to the rational zeros theorem, the possible rational zeros are ±1, ±2 , ±3, and ±6. Synthetic division shows that only 1 and −3 are zeros.

72. f (x) = x4 − 3x3 − 9x2 − 3x − 10

Possible rational zeros: ±1, ±2, ±5, ±10

Synthetic division shows that only −2 and 5 are zeros.

73. f (x) = x3 − x2 − 4x + 3

According to the rational zeros theorem, the possible rational zeros are ±1 and ±3. Synthetic division shows that none of these is a zero. Thus, there are no rational zeros.

74. f (x) = 2x3 + 3x2 + 2x + 3 1 3 Possible rational zeros: ±1, ±3, ± , ± 2 2 3 Synthetic division shows that only − is a zero. 2 75. f (x) = x4 + 2x3 + 2x2 − 4x − 8

According to the rational zeros theorem, the possible rational zeros are ±1, ±2, ±4, and ±8. Synthetic division shows that none of the possibilities is a zero. Thus, there are no rational zeros.

76. f (x) = x + 6x + 17x + 36x + 66 3

= −3x5 − 2x2 − x − 1

There are no variations in sign in f (−x), so there are 0 negative real zeros. 80. g(x) = 5x6 − 3x3 + x2 − x

b) f (x) =

4

f (−x) = 3(−x)5 − 2(−x)2 + (−x) − 1

2

Possible rational zeros: ±1, ±2, ±3, ±6, ±11, ±22, ±33, ±66

Synthetic division showns that none of these is a zero. Thus, there are no rational zeros. 77. f (x) = x5 − 5x4 + 5x3 + 15x2 − 36x + 20

According to the rational zeros theorem, the possible rational zeros are ±1, ±2, ±4, ±5, ±10, and ±20. Synthetic division shows that only −2, 1 and 2 are zeros.

78. f (x) = x5 − 3x4 − 3x3 + 9x2 − 4x + 12

Possible rational zeros: ±1, ±2, ±3, ±4, ±6, ±12. Synthetic division shows that only −2, 2 and 3 are zeros.

79. f (x) = 3x5 − 2x2 + x − 1

The number of variations in sign in f (x) is 3. Then the number of positive real zeros is either 3 or less than 3 by 2, 4, 6, and so on. Thus, the number of positive real zeros is 3 or 1.

The number of variations in sign in g(x) is 3. Then the number of positive real zeros is either 3 or less than 3 by 2, 4, 6, and so on. Thus, the number of positive real zeros is 3 or 1. g(−x) = 5(−x)6 − 3(−x)3 + (−x)2 − (−x) = 5x6 + 3x3 + x2 + x

There are no variations in sign in g(−x), so there are 0 negative real zeros. 81. h(x) = 6x7 + 2x2 + 5x + 4 There are no variations in sign in h(x), so there are 0 positive real zeros. h(−x) = 6(−x)7 + 2(−x)2 + 5(−x) + 4 = −6x7 + 2x2 − 5x + 4

The number of variations in sign in h(−x) is 3. Thus, there are 3 or 1 negative real zeros. 82. P (x) = −3x5 − 7x3 − 4x − 5

There are no variations in sign in P (x), so there are 0 positive real zeros. P (−x) = −3(−x)5 − 7(−x)3 − 4(−x) − 5 = 3x5 + 7x3 + 4x − 5

There is 1 variation in sign in P (−x), so there is 1 negative real zero. 83. F (p) = 3p18 + 2p4 − 5p2 + p + 3

There are 2 variations in sign in F (p), so there are 2 or 0 positive real zeros. F (−p) = 3(−p)18 + 2(−p)4 − 5(−p)2 + (−p) + 3 = 3p18 + 2p4 − 5p2 − p + 3

There are 2 variations in sign in F (−p), so there are 2 or 0 negative real zeros. 84. H(t) = 5t12 − 7t4 + 3t2 + t + 1

There are 2 variations in sign in H(t), so there are 2 or 0 positive real zeros. H(−t) = 5(−t)12 − 7(−t)4 + 3(−t)2 + (−t) + 1 = 5t12 − 7t4 + 3t2 − t + 1

There are 4 variations in sign in H(−t), so there are 4, 2, or 0 negative real zeros. 85. C(x) = 7x6 + 3x4 − x − 10

There is 1 variation in sign in C(x), so there is 1 positive real zero. C(−x) = 7(−x)6 + 3(−x)4 − (−x) − 10 = 7x6 + 3x4 + x − 10

There is 1 variation in sign in C(−x), so there is 1 negative real zero.

Exercise Set 3.4 86. g(z) = −z 10 + 8z 7 + z 3 + 6z − 1

There are 2 variations in sign in g(z), so there are 2 or 0 positive real zeros. g(−z) = −(−z)10 + 8(−z)7 + (−z)3 + 6(−z) − 1 = −z 10 − 8z 7 − z 3 − 6z − 1

There are no variations in sign in g(−z), so there are 0 negative real zeros. 87. h(t) = −4t5 − t3 + 2t2 + 1

215 93. R(x) = 3x5 − 5x3 − 4x

There is 1 variation in sign in R(x), so there is 1 positive real zero. R(−x) = 3(−x)5 − 5(−x)3 − 4(−x) = −3x5 + 5x3 + 4x

There is 1 variation in sign in R(−x), so there is 1 negative real zero. 94. f (x) = x4 − 9x2 − 6x + 4

There is 1 variation in sign in h(t), so there is 1 positive real zero. h(−t) = −4(−t)5 − (−t)3 + 2(−t)2 + 1

There are 2 variations in sign in f (x), so there are 2 or 0 positive real zeros. f (−x) = (−x)4 − 9(−x)2 − 6(−x) + 4

There are no variations in sign in h(−t), so there are 0 negative real zeros.

There are 2 variations in sign in f (−x), so there are 2 or 0 negative real zeros.

= 4t5 + t3 + 2t2 + 1

88. P (x) = x6 + 2x4 − 9x3 − 4

There is 1 variation in sign in P (x), so there is 1 positive real zero. P (−x) = (−x)6 + 2(−x)4 − 9(−x)3 − 4 = x6 + +2x4 + 9x3 − 4

There is 1 variation in sign in P (−x), so there is 1 negative real zero. 89. f (y) = y 4 + 13y 3 − y + 5

There are 2 variations in sign in f (y), so there are 2 or 0 positive real zeros. f (−y) = (−y)4 + 13(−y)3 − (−y) + 5 = y 4 − 13y 3 + y + 5

There are 2 variations in sign in f (−y), so there are 2 or 0 negative real zeros. 90. Q(x) = x4 − 2x2 + 12x − 8

There are 3 variations in sign in Q(x), so there are 3 or 1 positive real zeros. Q(−x) = (−x)4 − 2(−x)2 + 12(−x) − 8 = x4 − 2x2 − 12x − 8

There is 1 variation in sign in Q(−x), so there is 1 negative real zero. 91. r(x) = x4 − 6x2 + 20x − 24

There are 3 variations in sign in r(x), so there are 3 or 1 positive real zeros. r(−x) = (−x)4 − 6(−x)2 + 20(−x) − 24 = x4 − 6x2 − 20x − 24

There is 1 variation in sign in r(−x), so there is 1 negative real zero. 92. f (x) = x5 − 2x3 − 8x

There is 1 variation in sign in f (x), so there is 1 positive real zero. f (−x) = (−x)5 − 2(−x)3 − 8(−x) = −x5 + 2x3 + 8x

There is 1 variation in sign in f (−x), so there is 1 negative real zero.

= x4 − 9x2 + 6x + 4

95. f (x) = 4x3 + x2 − 8x − 2 1. The leading term is 4x3 . The degree, 3, is odd and the leading coefficient, 4, is positive so as x → ∞, f (x) → ∞ and x → −∞, f (x) → −∞. 2. We find the rational zeros p/q of f (x). ±1, ±2 Possibilities for p : Possibilities for q ±1, ±2, ±4 1 1 1 1 Possibilities for p/q: 1, −1 , 2, −2, , − , , − 2 2 4 4

1 Synthetic division shows that − is a zero. 4 " − 14 " 4 1 −8 −2 −1 0 2 4 0 −8 0 $ # 1 (4x2 − 8) = We have f (x) = x + 4 $ # 1 (x2 − 2). Solving x2 − 2 = 0 we get 4 x+ 4 √ 1 x = ± 2. Thus the zeros of the function are − , 4 √ √ − $ 2 so the x-intercepts of the graph are # 2, and √ √ 1 − , 0 , (− 2, 0), and ( 2, 0). 4 3. The zeros divide # the x-axis $ into # 4 intervals, $ √ √ 1 1 √ (−∞, − 2), − 2, − , − , 2 , and 4 4 √ ( 2, ∞). We choose a value for x from each interval and find f (x). This tells us the sign of f (x) for all values of x in that interval. √ In (−∞, − 2), test −2: f (−2) = 4(−2)3 + (−2)2 − 8(−2) − 2 = −14 < 0 # $ √ 1 In − 2, − , test −1: 4 f (−1) = 4(−1)3 + (−1)2 − 8(−1) − 2 = 3 > 0 # $ 1 √ In − , 2 , test 0: 4 f (0) = 4 · 03 + 02 − 8 · 0 − 2 = −2 < 0 √ In ( 2, ∞), test 2: f (2) = 4 · 23 + 22 − 8 · 2 − 2 = 18 > 0

216

Chapter 3: Polynomial and Rational Functions √ Thus the#graph lies$below the x-axis on (−∞, − 2) 1 √ and on − , 2 . It lies above the x-axis on $4 # √ √ 1 and on ( 2, ∞). We also know the − 2, − 4 points (−2, −14), (−1, 3), (0, −2), and (2, 18) are on the graph. 4. From Step 3 we see that f (0) = −2 so the y-intercept is (0, −2).

5. We find additional points on the graph and then draw the graph. x

y

f (x)

8

−1.5 −1.25 −0.5 1 1.5

4

1.75

2

#4 #2

4

x

#4

−5

#8

1.75

f (x ) ! 4x 3 " x 2 # 8x # 2

6. Checking the graph as described on page 272 in the text, we see that it appears to be correct. 96.

3. The zeros divide#the x-axis into (−∞, −1), $ # 5 intervals, $ 3 3 , and , ∞ . We choose a (−1, 0), (0, 1), 1, 2 2 value for x from each interval and find f (x). This tells us the sign of f (x) for all values of x in that interval. In (−∞, −1), test −2: f (−2) = 2(−2)4 − 3(−2)3 − 2(−2)2 + 3(−2) = 42 > 0

In (−1, 0), test −0.5:

f (−0.5) = 2(−0.5)4 −3(−0.5)3 −2(−0.5)2 +3(−0.5) =

−1.5 < 0

In (0, 1), test 0.5: f (0.5) = 2(0.5)4 −3(0.5)3 −2(0.5)2 +3(0.5) = 0.75 > 0 $ # 3 , test 1.25: In 1, 2 f (1.25) = 2(1.25)4 − 3(1.25)3 − 2(1.25)2 + 3(1.25) =

y 4 2 2

#4 #2

Then f (x) = x(x−1)(2x2 −x−3). Using the principle 3 of zero products to solve 2x2 −x−3 = 0, we get x = 2 or x = −1. 3 Thus the zeros of the function are 0, 1, , and −1 so 2 # $ 3 ,0 , the x-intercepts of the graph are (0, 0), (1, 0), 2 and (−1, 0).

4

x

#2 #4

f (x ) ! 3x 3 # 4x 2 # 5x " 2

97. f (x) = 2x4 − 3x3 − 2x2 + 3x

1. The leading term is 2x4 . The degree, 4, is even and the leading coefficient, 2, is positive so as x → ∞, f (x) → ∞ and as x → −∞, f (x) → ∞.

2. We find the rational zeros p/q of f (x). First note that f (x) = x(2x3 − 3x2 − 2x + 3), so 0 is a zero. Now consider g(x) = 2x3 − 3x2 − 2x + 3. ±1, ±3 Possibilities for p : Possibilities for q ±1, ±2 1 1 3 3 Possibilities for p/q: 1, −1 , 3, −3, , − , , − 2 2 2 2 We try 1. " 1 " 2 −3 −2 3 2 −1 −3 2 −1 −3 0

−0.3515625 < 0 $ # 3 , ∞ , test 2: In 2

f (2) = 2 · 24 − 3 · 23 − 2 · 22 + 3 · 2 = 6 > 0

Thus the graph lies $ the x-axis on (−∞, −1), # above 3 , ∞ . It lies below the x-axis on (0, 1), and on #2 $ 3 on (−1, 0) and on 1, . We also know the points 2 (−2, 42), (−0.5, −1.5), (0.5, 0.75), (1.25, −0.3515625), and (2, 6) are on the graph. 4. From Step 2 we know that f (0) = 0 so the y-intercept is (0, 0). 5. We find additional points on the graph and then draw the graph. x

y

f (x)

4

−1.5 11.25 2.5 3

26.25 72

2 2

#4 #2

4

x

#2 #4

f(x) ! 2x 4 # 3x 3 # 2x 2 " 3x

6. Checking the graph as described on page 272 in the text, we see that it appears to be correct.

Exercise Set 3.4 98.

217 106. f (x) = −x2 − 3x + 6

y

The degree is 2, so the function is quadratic.

100

Leading term: −x2 ; leading coefficient: −1

50 2

#4 #2 #50

4

Since the degree is even and the leading coefficient is negative, as x → ∞, f (x) → −∞, and as x → −∞, f (x) → −∞.

x

#100

107. g(x) = −x3 − 2x2

The degree is 3, so the function is cubic.

f(x) ! 4x 4 # 37x 2 " 9

Leading term: −x3 ; leading coefficient: −1

99. Left to the student 100. Yes; let c be a zero of P (x). Then P (c) = 0, so −P (c) = Q(c) = 0. Thus, every zero of P (x) is a zero of Q(x). Now let r be a zero of Q(x). Then Q(r) = 0, so −P (r) = 0 and P (r) = 0. Every zero of Q(x) is also a zero of P (x). Thus, P (x) and Q(x) have the same zeros. 101. No; since imaginary zeros of polynomials with rational coefficients occur in conjugate pairs, a third-degree polynomial with rational coefficients can have at most two imaginary zeros. Thus, there must be at least one real zero. 102. f (x) = 3x2 − 6x − 1 b −6 =− =1 a) − 2a 2·3 f (1) = 3 · 12 − 6 · 1 − 1 = −4 The vertex is (1, −4).

Since the degree is odd and the leading coefficient is negative, as x → ∞, g(x) → −∞ and as x → −∞, g(x) → ∞. 108. h(x) = x − 2

The degree is 1, so the function is linear. Leading term: x; leading coefficient: 1 Since the degree is odd and the leading coefficient is positive, as x → ∞, h(x) → ∞ and as x → −∞, h(x) → −∞.

4 9 The degree is 0, so this is a constant function. 4 4 Leading term: − ; leading coefficient: − ; 9 9 4 for all x, f (x) = − 9

109. f (x) = −

1 110. h(x) = x3 + x2 − 4x − 3 2 The degree is 3, so the function is cubic.

b) x = 1

c) Minimum: −4 at x = 1

Leading term: x3 ; leading coefficient: 1

103. f (x) = x2 − 8x + 10 b −8 =− = −(−4) = 4 a) − 2a 2·1 2 f (4) = 4 − 8 · 4 + 10 = −6

Since the degree is odd and the leading coefficient is positive, as x → ∞, h(x) → ∞ and as x → −∞, h(x) → −∞. 111. g(x) = x4 − 2x3 + x2 − x + 2

The degree is 4, so the function is quartic.

The vertex is (4, −6).

Leading term: x4 ; leading coefficient: 1

b) The axis of symmetry is x = 4. c) Since the coefficient of x2 is positive, there is a minimum function value. It is the second coordinate of the vertex, −6. It occurs when x = 4. 104.

x2 − 8x − 33 = 0

(x − 11)(x + 3) = 0

Q(x) = x2 + 2ix + (2 − 4i), R(x) = −6 − 2i

x = 11 or x = −3

The zeros are −3 and 11. 105.

4 − x+8 5 4 − x 5 # $ 5 4 − − x 4 5 x

113.

= −8

Subtracting 8

5 = − (−8) 4 = 10

Multiplying by −

(x4 − y 4 ) ÷ (x − y)

= (x4 + 0x3 + 0x2 + 0x − y 4 ) ÷ (x − y) " y " 1 0 0 0 −y 4 y y2 y3 y4 2 3 1 y y y 0

=0

The zero is 10.

Since the degree is even and the leading coefficient is positive, as x → ∞, g(x) → ∞ and as x → −∞, g(x) → ∞. " 112. −i " 1 3i −4i −2 −i 2 −4 − 2i 1 2i 2 − 4i −6 − 2i

5 4

The quotient is x3 + x2 y + xy 2 + y 3 . The remainder is 0. 114. By the rational zeros theorem, only ±1, ±2, ±3, ±4, ±6, and ±12 can be rational solutions of x4 − 12 = 0. Since none of them is√a solution, the equation has no rational √ solutions. But 4 12 is a solution of the equation, so 4 12 must be irrational.

218

Chapter 3: Polynomial and Rational Functions

115. f (x) = 2x3 − 5x2 − 4x + 3

a) 2x3 − 5x2 − 4x + 3 = 0 Possibilities for p ±1, ±3 : Possibilities for q ±1, ±2

1 3 3 1 ,− , ,− 2 2 2 2 The first possibility that is a solution of f (x) = 0 is −1: " 3 −1 " 2 −5 −4 −2 7 −3 2 −7 3 0 Possibilities for p/q: 1, −1, 3, −3,

Thus, −1 is a solution. Then we have: (x + 1)(2x2 − 7x + 3) = 0 (x + 1)(2x − 1)(x − 3) = 0

1 and 3. 2 b) The graph of y = f (x − 1) is the graph of y = f (x) shifted 1 unit right. Thus, we add 1 to each solution of f (x) = 0 to find the solutions of f (x − 1) = 0. The solutions are −1 + 1, or 0; 3 1 + 1, or ; and 3 + 1, or 4. 2 2 c) The graph of y = f (x + 2) is the graph of y = f (x) shifted 2 units left. Thus, we subtract 2 from each solution of f (x) = 0 to find the solutions of f (x + 2) = 0. The solutions are 1 3 −1 − 2, or −3; − 2, or − ; and 3 − 2, or 1. 2 2 d) The graph of y = f (2x) is a horizontal shrinking of the graph of y = f (x) by a factor of 2. We divide each solution of f (x) = 0 by 2 to find the −1 or solutions of f (2x) = 0. The solutions are 2 1 1 1/2 − ; , or ; 2 2 4 3 and . 2 The other solutions are

116. P (x) = x − x − 72x − 81x + 486x + 5832 6

5

4

2

a) x6 − 6x5 − 72x4 − 81x2 + 486x + 5832 = 0

Synthetic division shows that we can factor as follows: P (x) = (x−3)(x+3)(x+6)(x3 −12x2 +9x−108) = (x−3)(x+3)(x+6)[x2 (x−12) + 9(x−12)] = (x−3)(x+3)(x+6)(x−12)(x2 +9) The rational zeros are 3, −3, −6, and 12.

117. P (x) = 2x5 − 33x4 − 84x3 + 2203x2 − 3348x − 10, 080 a) 2x5 − 33x4 − 84x3 + 2203x2 − 3348x − 10, 080 = 0 Trying some of the many possibilities for p/q, we find that 4 is a zero. " 4 " 2 −33 −84 2203 −3348 −10, 080 8 −100 −736 5868 10, 080 2 −25 −184 1467 2520 0

Then we have:

(x − 4)(2x4 − 25x3 − 184x2 + 1467x + 2520) = 0

We now use the fourth degree polynomial above to find another zero. Synthetic division shows that 4 is not a double zero, but 7 is a zero. " 1467 2520 7 " 2 −25 −184 14 −77 −1827 −2520 2 −11 −261 −360 0

Now we have:

(x − 4)(x − 7)(2x3 − 11x2 − 261x − 360) = 0 Use the third degree polynomial above to find a third zero. Synthetic division shows that 7 is not a double zero, but 15 is a zero. " 15 " 2 −11 −261 −360 30 285 360 2 19 24 0 We have: P (x) = (x − 4)(x − 7)(x − 15)(2x2 + 19x + 24) = (x − 4)(x − 7)(x − 15)(2x + 3)(x + 8)

3 The rational zeros are 4, 7, 15, − , and −8. 2

Exercise Set 3.5 8 . x2 − 4 x2 − 4 = 0 when x = ±2, so x = −2 and x = 2 are vertical asymptotes.

1. Graph (d) is the graph of f (x) =

The x-axis, y = 0, is the horizontal asymptote because the degree of the numerator is less than the degree of the denominator. There is no oblique asymptote. 8 . x2 + 4 x2 + 4 = 0 has no real solutions, so there is no vertical asymptote.

2. Graph (f) is the graph of f (x) =

The x-axis, y = 0, is the horizontal asymptote because the degree of the numerator is less than the degree of the denominator. There is no oblique asymptote. 8x . x2 − 4 As in Exercise 1, x = −2 and x = 2 are vertical asymptotes.

3. Graph (e) is the graph of f (x) =

The x-axis, y = 0, is the horizontal asymptote because the degree of the numerator is less than the degree of the denominator. There is no oblique asymptote. 8x2 . −4 As in Exercise 1, x = 2 and x = −2 are vertical asymptotes.

4. Graph (a) is the graph of f (x) =

x2

The numerator and denominator have the same degree, so y = 8/1, or y = 8, is the horizontal asymptote. There is no oblique asymptote.

Exercise Set 3.5

219

8x3 . −4 As in Exercise 1, x = −2 and x = 2 are vertical asymptotes.

5. Graph (c) is the graph of f (x) =

x2

The degree of the numerator is greater than the degree of the denominator, so there is no horizontal asymptote but there is an oblique asymptote. To find it we first divide to find an equivalent expression.

32x 8x3 = 8x + 2 x2 − 4 x −4 Now we multiply by 1, using (1/x2 )/(1/x2 ). 1 32 32x x2 = x · 4 x2 − 4 1 1 − x2 x2 As |x| becomes very large, each expression with x in the denominator tends toward zero. Then, as |x| →∞ , we have 32 0 x , or 0. → 4 1−0 1− 2 x Thus, as |x| becomes very large, the graph of f (x) gets very close to the graph of y = 8x, so y = 8x is the oblique asymptote. 8x3 . +4 As in Exercise 2, there is no vertical asymptote. x2

The degree of the numerator is greater than the degree of the denominator, so there is no horizontal asymptote but there is an oblique asymptote. To find it we first divide to find an equivalent expression. 32x 8x3 = 8x − 2 +4 x +4 32 32x x and, as |x| →∞ , Now 2 = 4 x +4 1+ 2 x 32 0 x , or 0. → 4 1+0 1+ 2 x Thus, as y = 8x is the oblique asymptote. x2

1 x2 The zero of the denominator is 0, so the vertical asymptote is x = 0.

7. g(x) =

4 x + 10 x + 10 = 0 when x = −10, so the vertical asymptote is x = −10.

8. f (x) =

x4 + 2 x The zero of the denominator is 0, so the vertical asymptote is x = 0.

10. g(x) =

3−x (x − 4)(x + 6) The zeros of the denominator are 4 and −6, so the vertical asymptotes are x = 4 and x = −6.

11. f (x) =

8x x2 − 4 8x3 8x3 −32x 32x

6. Graph (b) is the graph of f (x) =

x+7 2−x 2 − x = 0 when x = 2, so the vertical asymptote is x = 2.

9. h(x) =

x2 + 4 x(x + 5)(x − 2) The zeros of the denominator are 0, −5, and 2, so the vertical asymptotes are x = 0, x = −5, and x = 2.

12. h(x) =

x2 x2 = −x−3 (2x − 3)(x + 1) 3 The zeros of the denominator are and −1, so the vertical 2 3 asymptotes are x = and x = −1. 2

13. g(x) =

2x2

x+5 x+5 = x2 + 4x − 32 (x + 8)(x − 4) The zeros of the denominator are −8 and 4, so the vertical asymptotes are x = −8 and x = 4.

14. f (x) =

3x2 + 5 4x2 − 3 The numerator and denominator have the same degree and 3 3 the ratio of the leading coefficients is , so y = is the 4 4 horizontal asymptote.

15. f (x) =

x+6 x3 + 2x2 The degree of the numerator is less than the degree of the denominator, so y = 0 is the horizontal asymptote.

16. g(x) =

x2 − 4 2x4 + 3 The degree of the numerator is less than the degree of the denominator, so y = 0 is the horizontal asymptote.

17. h(x) =

x5 +x The numerator and denominator have the same degree and the ratio of the leading coefficients is 1, so y = 1 is the horizontal asymptote.

18. f (x) =

x5

x3 − 2x2 + x − 1 x2 − 16 The degree of the numerator is greater than the degree of the denominator, so there is no horizontal asymptote.

19. g(x) =

8x4 + x − 2 2x4 − 10 The numerator and denominator have the same degree and the ratio of the leading coefficients is 4, so y = 4 is the horizontal asymptote.

20. h(x) =

220

Chapter 3: Polynomial and Rational Functions

21. g(x) =

x2 + 4x − 1 x+3

x+ 1 x + 3 x2 + 4x − 1 x2 + 3x x − 1 x + 3 −4 Then g(x) = x + 1 + y = x + 1. 22. f (x) =

−4 . The oblique asymptote is x+3

y = x − 1.

−5 . The oblique asymptote is x−5

x4 − 2 x3 + 1

Then h(x) = 5x + 4 + is y = 5x + 4.

2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. There are no oblique asymptotes. 3. The numerator has no zeros, so there is no xintercept.

5. Find other function values to determine the shape of the graph and then draw it. y 4

y!0 #4

12x3 − x 6x2 + 4

y = 2x.

#2

2

4

x

x!0

1 x2 1. 0 is the zero of the denominator, so the domain excludes 0. It is (−∞, 0) ∪ (0, ∞). The line x = 0, or the y-axis, is the vertical asymptote.

28. g(x) =

−9x . The oblique asymptote is 6x2 + 4

x− 3 x2 + 2x − 1 x3 − x2 + x − 4 x3 + 2x2 − x − 3x2 + 2x − 4 − 3x2 − 6x + 3 8x − 7 Then f (x) = x − 3 +

f (x) ! 1 x

#4

x3 − x2 + x − 4 x2 + 2x − 1

is y = x − 3.

2

#2

2x 6x2 + 4 12x3 − x 12x3 + 8x − 9x Then g(x) = 2x +

−5x − 9 . The oblique asymptote x2 − x + 2

4. Since 0 is not in the domain of the function, there is no y-intercept.

x x3 + 1 x4 + 0x3 + 0x2 + 0x + x x4 − x −x . The oblique asymptote is Then h(x) = x + 3 x +1 y = x.

25. f (x) =

5x + 4 x2 − x + 2 5x3 − x2 + x − 1 5x3 − 5x2 + 10x 4x2 − 9x − 1 4x2 − 4x + 8 − 5x − 9

27. f (x) =

Then f (x) = x − 1 +

24. g(x) =

5x3 − x2 + x − 1 x2 − x + 2

1 x 1. 0 is the zero of the denominator, so the domain excludes 0. It is (−∞, 0) ∪ (0, ∞). The line x = 0, or the y-axis, is the vertical asymptote.

x2 − 6x x−5

x− 1 x − 5 x2 − 6x + 0 x2 − 5x − x +0 − x +5 −5

23. h(x) =

26. h(x) =

8x − 7 . The oblique asymptote x2 + 2x − 1

2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. There is no oblique asymptote. 3. The numerator has no zeros, so there is no xintercept. 4. Since 0 is not in the domain of the function, there is no y-intercept. 5. Find other function values to determine the shape of the graph and then draw it.

Exercise Set 3.5

221 y 4 2

#4

#2

y!0

g (x) ! 12 x 2

#2

x2 − 4x + 3 x+1 1. The denominator, x + 1, is 0 when x = −1, so the domain excludes −1. It is (−∞, −1)∪(−1, ∞). The line x = −1 is the vertical asymptote.

31. g(x) =

4

x

x!0

2. The degree of the numerator is 1 greater than the degree of the denominator, so we divide to find the oblique asymptote.

#4

4 x2 1. 0 is the zero of the denominator, so the domain excludes 0. It is (−∞, 0) ∪ (0, ∞). The line x = 0, or the y-axis, is the vertical asymptote.

29. h(x) = −

2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. There is no oblique asymptote. 3. The numerator has no zeros, so there is no xintercept. 4. Since 0 is not in the domain of the function, there is no y-intercept. 5. Find other function values to determine the shape of the graph and then draw it.

x− 5 x + 1 x2 − 4x + 3 x2 + x − 5x + 3 − 5x − 5 8 The oblique asymptote is y = x − 5. There is no horizontal asymptote. 3. x2 − 4x + 3 = (x − 1)(x − 3), so the zeros of the numerator are 1 and 3. Thus the x-intercepts are (1, 0) and (3, 0). 02 − 4 · 0 + 3 = 3, so the y-intercept is (0, 3). 0+1 5. Find other function values to determine the shape of the graph and then draw it. 4. g(0) =

y

y 4

y!0 2

4

x ! #1 x!0

#8

#4

4 #4

#4

#2

2

4

x

#8

#2 #4

h (x) ! # 42 x

6 x 1. 0 is the zero of the denominator, so the domain excludes 0. It is (−∞, 0) ∪ (0, ∞). The line x = 0, or the y-axis, is the vertical asymptote.

30. f (x) = −

2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. There is no oblique asymptote. 3. The numerator has no zeros, so there is no xintercept. 4. Since 0 is not in the domain of the function, there is no y-intercept. 5. Find other function values to determine the shape of the graph and then draw it. y 4 2 #4

#2

y!0

x!0 f(x) ! # 6 x 2

#2

4

x

8 x y!x #5

#12

2 g (x) ! x # 4x " 3 x"1

#16

2x2 − x − 3 x−1 1. The denominator is 0 when x = 1, so the domain excludes 1. It is (−∞, 1) ∪ (1, ∞). The line x = 1 is the vertical asymptote.

32. h(x) =

2. The degree of the numerator is 1 greater than the degree of the denominator, so we divide to find the oblique asymptote. 2x + 1 x − 1 2x2 − x − 3 2x2 − 2x x −3 x −1 −2

The oblique asymptote is y = 2x + 1. There is no horizontal asymptote. 3. 2x2 − x − 3 = (2x − 3)(x + 1), so the zeros of the 3 numerator are and −1. Thus the x-intercepts are 2 # $ 3 , 0 and (−1, 0). 2

#4

4. h(0) =

2 · 02 − 0 − 3 = 3, so the y-intercept is (0, 3). 0−1

222

Chapter 3: Polynomial and Rational Functions y

5. Find other function values to determine the shape of the graph and then draw it.

f(x) !

4 2

y 4

y!0 4

y ! 2x " 1

6

#2

2

1 x#5

8

x

x!5

#4 #4

#2

2 #2

4

x

x!1

−2 x−5 1. 5 is the zero of the denominator, so the domain excludes 5. It is (−∞, 5) ∪ (5, ∞). The line x = 5 is the vertical asymptote.

35. f (x) =

#4

2 h (x) ! 2x # x # 3 x#1

1 x+3 1. −3 is the zero of the denominator, so the domain excludes −3. It is (−∞, −3) ∪ (−3, ∞). The line x = −3 is the vertical asymptote.

33. f (x) =

2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. There is no oblique asymptote. 3. The numerator has no zeros, so there is no xintercept. # $ 1 1 1 4. f (0) = = , so 0, is the y-intercept. 0+3 3 3 5. Find other function values to determine the shape of the graph and then draw it.

2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. There is no oblique asymptote. 3. The numerator has no zeros, so there is no xintercept. # $ 2 2 −2 4. f (0) = = , so 0, is the y-intercept. 0−5 5 5 5. Find other function values to determine the shape of the graph and then draw it. y 8

4 2

f (x) !

2

x ! #3

#2

2

#2

1 x"3

y!0

#2

x

y!0

#4

1 x−5 1. 5 is the zero of the denominator, so the domain is (−∞, 5)∪(5, ∞) and x = 5 is the vertical asymptote.

34. f (x) =

2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. There is no oblique asymptote. 3. The numerator has no zeros, so there is no xintercept. # $ 1 1 1 4. f (0) = = − , so 0, − is the y0−5 5 5 intercept. 5. Find other function values to determine the shape of the graph and then draw it.

4

6

x

x!5

#4 2

#2

#2 x#5

6

y 4

f(x) !

3 3−x 1. 3 is the zero of the denominator, so the domain is (−∞, 3)∪(3, ∞) and x = 3 is the vertical asymptote.

36. f (x) =

2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. There is no oblique asymptote.

3. The numerator has no zeros, so there is no xintercept. 3 4. f (0) = = 1, so (0, 1) is the y-intercept. 3−0 5. Find other function values to determine the shape of the graph and then draw it. y f(x) !

4 2

y!0 2

#2 #4

3 3#x

x!3

4

6

x

Exercise Set 3.5

223

2x + 1 x 1. 0 is the zero of the denominator, so the domain excludes 0. It is (−∞, 0) ∪ (0, ∞). The line x = 0, or the y-axis, is the vertical asymptote.

37. f (x) =

2. The numerator and denominator have the same degree, so the horizontal asymptote is determined by the ratio of the leading coefficients, 2/1, or 2. Thus, y = 2 is the horizontal asymptote. There is no oblique asymptote. 3. The zero of the numerator is the # solution $ of 2x+1 = 1 1 0, or − . The x-intercept is − , 0 . 2 2 4. Since 0 is not in the domain of the function, there is no y-intercept.

2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. There is no oblique asymptote. 3. The numerator has no zeros, so there is no xintercept. # $ 1 1 1 4. f (0) = = , so 0, is the y(0 − 2)2 4 4 intercept. 5. Find other function values to determine the shape of the graph and then draw it. y 3

5. Find other function values to determine the shape of the graph and then draw it.

2

!0, ~"1

y 6

f(x) !

4

2x " 1 x

#1 #2

2

4

!#q, 0"

x!0

x

3x − 1 x 1. 0 is the zero of the denominator, so the domain is (−∞, 0) ∪ (0, ∞). The line x = 0, or the y-axis, is the vertical asymptote.

38. f (x) =

2. The numerator and denominator have the same degree, so the horizontal asymptote is determined by the ratio of the leading coefficients, 3/1, or 3. Thus, y = 3 is the horizontal asymptote. There is no oblique asymptote. 1 3. The zero of the numerator is , so the x-intercept 3 $ # 1 is ,0 . 3 4. Since 0 is not in the domain of the function, there is no y-intercept. 5. Find other function values to determine the shape of the graph and then draw it.

f(x) !

4

x

1 f(x) ! (x # 2)2

−2 (x − 3)2 1. 3 is the zero of the denominator, so the domain is (−∞, 3)∪(3, ∞) and x = 3 is the vertical asymptote.

2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. There is no oblique asymptote. 3. The numerator has no zeros, so there is no xintercept. # $ 2 2 −2 4. f (0) = = − , so 0, − is the (0 − 3)2 9 9 y-intercept. 5. Find other function values to determine the shape of the graph and then draw it. y x!3

4 2

y!0 2

#2

4

x

#2

3x # 1 x

2 2

4

x

#2 x ! 0

1 (x − 2)2 1. 2 is the zero of the denominator, so the domain excludes 2. It is (−∞, 2) ∪ (2, ∞). The line x = 2 is the vertical asymptote.

39. f (x) =

3

#4

4

#4 #2

2

40. f (x) =

y

y!3

y!0 1

#1

y!2 #4 #2

x!2

4

f(x) !

#2 (x # 3)2

1 x2 1. 0 is the zero of the denominator, so the domain excludes 0. It is (−∞, 0) ∪ (0, ∞). The line x = 0, or the y-axis, is the vertical asymptote.

41. f (x) = −

2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. There is no oblique asymptote.

224

Chapter 3: Polynomial and Rational Functions 3. The numerator has no zeros, so there is no xintercept.

5. Find other function values to determine the shape of the graph and then draw it.

4. Since 0 is not in the domain of the function, there is no y-intercept.

y

!0, a"

5. Find other function values to determine the shape of the graph and then draw it.

0.5 0.5

#0.5

y x!0

y!0 y!0

1 1

#2 #1

#0.5

2

x

#1

−1 x2 + 2 1. The denominator has no real-number zeros, so the domain is (−∞, ∞) and there is no vertical asymptote.

44. f (x) =

#2 #3

x

1 f(x) ! 2 x "3

f(x) ! #

1 x2

#4

1 42. f (x) = 2 3x 1. 0 is the zero of the denominator, so the domain is (−∞, 0) ∪ (0, ∞). The line x = 0, or the y-axis, is the vertical asymptote. 2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. There is no oblique asymptote. 3. The numerator has no zeros, so there is no xintercept.

2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. There is no oblique asymptote. 3. The numerator has no zeros, so there is no xintercept. # $ 1 1 −1 4. f (0) = = − , so 0, − is the y02 + 2 2 2 intercept. 5. Find other function values to determine the shape of the graph and then draw it. y

4. Since 0 is not in the domain of the function, there is no y-intercept.

0.5

5. Find other function values to determine the shape of the graph and then draw it.

y!0

f(x) !

0.25

#1 x2 " 2

2

#2

x

y f (x) !

#0.5

1 3x 2 y!0 2

#4 #2 #2

4

x

x!0

#4

1 x2 + 3 1. The denominator has no real-number zeros, so the domain is (−∞, ∞) and there is no vertical asymptote.

43. f (x) =

(x + 2)(x − 2) x2 − 4 = = x + 2, x "= 2 x−2 x−2 The graph is the same as the graph of f (x) = x + 2 except at x = 2, where there is a hole. The zero of f (x) = x + 2 is −2, so the x-intercept is (−2, 0); f (0) = 2, so the yintercept is (0, 2).

45. f (x) =

y 4

2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. There is no oblique asymptote. 3. The numerator has no zeros, so there is no xintercept. # $ 1 1 1 4. f (0) = = , so 0, is the y-intercept. 02 + 3 3 3

2

#4 #2 #4

4

x

x2 # 4 f (x) ! x#2

x2 − 9 (x + 3)(x − 3) = = x − 3, x "= −3 x+3 x+3 The zero of f (x) = x − 3 is 3, so the x-intercept is (3, 0); f (0) = −3, so the y-intercept is (0, −3).

46. f (x) =

Exercise Set 3.5

225 y

y

2

4 2

!4 !2

4

2

x

!2

y#1 2

!4 !2

!6

x−1 x+2 1. −2 is the zero of the denominator, so the domain excludes −2. It is (−∞, −2) ∪ (−2, ∞). The line x = −2 is the vertical asymptote.

3. The zero of the numerator is 1, so the x-intercept is (1, 0). ! " 1 1 0−1 4. f (0) = = − , so 0, − is the y0+2 2 2 intercept. 5. Find other function values to determine the shape of the graph and then draw it. y 4 2

y#1 2

!4

x # !2 !4

4

f(x) #

x

f(x) #

x+3 2x2 − 5x − 3 1. The zeros of the denominator are the solutions of 2x2 − 5x − 3 = 0. Since 2x2 − 5x − 3 = 1 (2x+1)(x−3), the zeros are − and 3. Thus, the do2 " ! " ! 1 1 main is − ∞, − ∪ − , 3 ∪ (3, ∞) and the 2 2 1 lines x = − and x = 3 are vertical asymptotes. 2 2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. There is no oblique asymptote. 3. −3 is the zero of the numerator, so (−3, 0) is the x-intercept. 0+3 4. f (0) = = −1, so (0, −1) is the y2 2·0 −5·0−3 intercept. 5. Find other function values to determine the shape of the graph and then draw it. y y#0

1 !4 !2

x−2 x+1 1. −1 is the zero of the denominator, so the domain is (−∞, −1) ∪ (−1, ∞) and x = −1 is the vertical asymptote.

48. f (x) =

3. The zero of the numerator is 2, so the x-intercept is (2, 0). 0−2 4. f (0) = = −2, so (0, −2) is the y-intercept. 0+1 5. Find other function values to determine the shape of the graph and then draw it.

x!2 x"1

49. f (x) =

x!1 x"2

2. The numerator and denominator have the same degree, so the horizontal asymptote is determined by the ratio of the leading coefficients, 1/1, or 1. Thus, y = 1 is the horizontal asymptote. There is no oblique asymptote.

x

x # !1

47. f (x) =

2. The numerator and denominator have the same degree, so the horizontal asymptote is determined by the ratio of the leading coefficients, 1/1, or 1. Thus, y = 1 is the horizontal asymptote. There is no oblique asymptote.

4

!2

x2 ! 9 f(x) # x"3

2

4

x # !q

x#3

f(x) #

x"3 2x 2 ! 5x ! 3

x

3x x2 + 5x + 4 1. x2 + 5x + 4 = (x + 4)(x + 1), so the domain excludes −4 and −1. It is (−∞, −4)∪(−4, −1)∪(−1, ∞) and the lines x = −4 and x = −1 are vertical asymptotes.

50. f (x) =

2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. There is no oblique asymptote. 3. 0 is the zero of the numerator, so (0, 0) is the xintercept. 4. From part (3) we see that (0, 0) is the y-intercept. 5. Find other function values to determine the shape of the graph and then draw it.

226

Chapter 3: Polynomial and Rational Functions y

y x # !1

x # !4 2

2

y#0 4

!8 !4

x#1

4

8

4

!4

x

x

!2 !4

y#x"1 f (x) #

3x x 2 " 5x " 4

f(x) #

x2 − 9 x+1 1. −1 is the zero of the denominator, so the domain excludes −1. It is (−∞, −1) ∪ (−1, ∞). The line x = −1 is the vertical asymptote.

51. f (x) =

2. Because the degree of the numerator is one greater than the degree of the denominator, there is an oblique asymptote. Using division, we find that −8 x2 − 9 =x−1+ . As |x| becomes very large, x+1 x+1 the graph of f (x) gets close to the graph of y = x−1. Thus, the line y = x − 1 is the oblique asymptote.

3. Since x2 − 9 = (x + 3)(x − 3), the zeros of the numerator are −3 and 3. Thus, the x-intercepts are (−3, 0) and (3, 0). 2 4. f (0) = 0 − 9 = −9, so (0, −9) is the y0+1 intercept.

5. Find other function values to determine the shape of the graph and then draw it. y x # !1

x2 + x − 2 2x2 + 1 1. The denominator has no real-number zeros, so the domain is (−∞, ∞) and there is no vertical asymptote.

53. f (x) =

2. The numerator and denominator have the same degree, so the horizontal asymptote is determined by the ratio of the leading coefficients, 1/2. Thus, y = 1/2 is the horizontal asymptote. There is no oblique asymptote. 3. Since x2 + x − 2 = (x + 2)(x − 1), the zeros of the numerator are −2 and 1. Thus, the x-intercepts are (−2, 0) and (1, 0). 2 4. f (0) = 0 + 0 − 2 = −2, so (0, −2) is the 2 · 02 + 1 y-intercept. 5. Find other function values to determine the shape of the graph and then draw it. y 4

y#x!1

4

y#q

2

!4 !2

!4 !2

f (x) #

x2 ! 4 x!1

2

4

x

x2 ! 9 x"1

x2 − 4 x−1 1. 1 is the zero of the denominator, so the domain is (−∞, 1)∪(1, ∞) and x = 1 is the vertical asymptote. 2 2. x − 4 = x + 1 + −3 , so y = x + 1 is the oblique x−1 x−1 asymptote.

52. f (x) =

3. x2 −4 = (x+2)(x−2), so the x-intercepts are (−2, 0) and (2, 0). 2 4. f (0) = 0 − 4 = 4, so (0, 4) is the y-intercept. 0−1 5. Find other function values to determine the shape of the graph and then draw it.

2 2 !4

4

f(x) #

x x2 " x ! 2 2x 2 " 1

x2 − 2x − 3 3x2 + 2 1. The denominator has no real-number zeros, so the domain is (−∞, ∞) and there is no vertical asymptote.

54. f (x) =

2. The numerator and denominator have the same degree, so the horizontal asymptote is determined by the ratio of the leading coefficients, 1/3. Thus, y = 1/3 is the horizontal asymptote. There is no oblique asymptote. 3. x2 − 2x − 3 = (x + 1)(x − 3), so the x-intercepts are (−1, 0) and (3, 0). ! " 2 4. f (0) = 0 − 2 · 0 − 3 = − 3 , so 0, − 3 is the y3 · 02 + 2 2 2 intercept. 5. Find other function values to determine the shape of the graph and then draw it.

Exercise Set 3.5

227 2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. There is no oblique asymptote.

y 1

y#a 4

!4 !2

3. 1 is the zero of the numerator, so (1, 0) is the xintercept. ! " 0−1 1 1 4. f (0) = = , so 0, is the 02 − 2 · 0 − 3 3 3 y-intercept.

x

!1.5

x 2 ! 2x ! 3 3x 2 " 2

f (x) #

55. g(x) = x $= 1

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

5. Find other function values to determine the shape of the graph and then draw it. y

The graph is the same as the graph of g(x) = 3x+2 except at x = 1, where there is a hole. 2 The zero of g(x) = 3x + 2 is − , so the x-intercept is 3 ! " 2 − , 0 ; g(0) = 2, so the y-intercept is (0, 2). 3

4

x!1 x 2 ! 2x ! 3

x+2 x2 + 2x − 15 1. x2 +2x−15 = (x+5)(x−3), so the domain excludes −5 and 3. It is (−∞, −5) ∪ (−5, 3) ∪ (3, ∞) and the lines x = −5 and x = 3 are vertical asymptotes.

58. f (x) = 2

!4 !2 !4

x

x#3

f(x) #

2

!2

4

!4

y 4

y#0

x # !1 2

4

x

3x 2 ! x ! 2 g(x) # x!1

2x + 1 2x + 1 1 56. f (x) = 2 = = , 2x − 5x − 3 (2x + 1)(x − 3) x−3 1 x $= − 2 1 has no zero, so there is no x-intercept; f (x) = x−3 ! " 1 1 f (0) = − , so the y-intercept is 0, − . 3 3 y

2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. There is no oblique asymptote. 3. −2 is the zero of the numerator, so (−2, 0) is the x-intercept. ! " 2 2 0+2 4. f (0) = = − , so 0, − is the 02 + 2 · 0 − 15 15 15 y-intercept. 5. Find other function values to determine the shape of the graph and then draw it.

4

y

2 2

!4 !2 !4

4

2

x

1

x#3 2

!4 !2

4

x

!1

f (x) #

2x " 1 2x 2 ! 5x ! 3

x−1 − 2x − 3 1. The zeros of the denominator are the solutions of x2 − 2x − 3 = 0. Since x2 − 2x − 3 = (x + 1)(x − 3), the zeros are −1 and 3. Thus, the domain is (−∞, −1) ∪ (−1, 3) ∪ (3, ∞) and the lines x = −1 and x = 3 are the vertical asymptotes.

57. f (x) =

!2

x # !5 f(x) #

x2

x#3

x"2 x 2 " 2x ! 15

x−3 (x + 1)3 1. −1 is the zero of the denominator, so the domain excludes −1. It is (−∞, −1) ∪ (−1, ∞). The line x = −1 is the vertical asymptote.

59. f (x) =

228

Chapter 3: Polynomial and Rational Functions 2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. There is no oblique asymptote. 3. 3 is the zero of the numerator, so (3, 0) is the xintercept. 0−3 4. f (0) = = −3, so (0, −3) is the y(0 + 1)3 intercept.

4. Since 0 is not in the domain of the function, there is no y-intercept. 5. Find other function values to determine the shape of the graph and then draw it. y 4 2 2

!4 !2

5. Find other function values to determine the shape of the graph and then draw it.

f (x) #

y 4

2

!4 !2

x # !1

4

x

x!3 f (x) # (x " 1)3

x+2 (x − 1)3 1. 1 is the zero of the denominator, so the domain is (−∞, 1)∪(1, ∞) and x = 1 is the vertical asymptote.

60. f (x) =

2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. There is no oblique asymptote.

2. Because the degree of the numerator is more than one greater than the degree of the denominator, there is no horizontal or oblique asymptote. 3. The real-number zero of the numerator is 1, so the x-intercept is (1, 0). 4. Since 0 is not in the domain of the function, there is no y-intercept. 5. Find other function values to determine the shape of the graph and then draw it. y

!4

f(x) #

x x3

!1 x

x + 2x2 − 15x x2 − 5x − 14 1. The zeros of the denominator are the solutions of x2 −5x−14 = 0. Since x2 −5x−14 = (x+2)(x−7), the zeros are −2 and 7. Thus, the domain is (−∞, −2) ∪ (−2, 7) ∪ (7, ∞) and the lines x = −2 and x = 7 are the vertical asymptotes.

63. f (x) =

8 4 2

f (x) #

4

!2

x#0

y

!8

2

!4 !2

5. Find other function values to determine the shape of the graph and then draw it.

!4

x3 " 1 x

x3 − 1 x 1. 0 is the zero of the denominator, so the domain excludes 0. It is (−∞, 0) ∪ (0, ∞). The line x = 0, or the y-axis, is the vertical asymptote.

3. −2 is the zero of the numerator, so (−2, 0) is the x-intercept. 0+2 4. f (0) = = −2, so (0, −2), is the y(0 − 1)3 intercept.

!4 !2

x

62. f (x) =

2

y#0

4

x#0

4

x

x#1

x"2 (x ! 1)3

x3 + 1 x 1. 0 is the zero of the denominator, so the domain excludes 0. It is (−∞, 0) ∪ (0, ∞). The line x = 0, or the y-axis, is the vertical asymptote.

61. f (x) =

2. Because the degree of the numerator is more than one greater than the degree of the denominator, there is no horizontal or oblique asymptote. 3. The real-number zero of the numerator is −1, so the x-intercept is (−1, 0).

3

2. Because the degree of the numerator is one greater than the degree of the denominator, there is an oblique asymptote. Using division, we find that 34x + 98 x3 + 2x2 − 15x =x+7+ 2 . As |x| bex2 − 5x − 14 x − 5x − 14 comes very large, the graph of f (x) gets close to the graph of y = x + 7. Thus, the line y = x + 7 is the oblique asymptote. 3. The zeros of the numerator are the solutions of x3 + 2x2 − 15x = 0. Since x3 + 2x2 − 15x = x(x + 5)(x − 3), the zeros are 0, −5, and 3. Thus, the x-intercepts are (−5, 0), (0, 0), and (3, 0). 4. From part (3) we see that (0, 0) is the y-intercept.

Exercise Set 3.5

229 y

5. Find other function values to determine the shape of the graph and then draw it.

y#5

4

y

2

50

x#7

x # !2

2

!4 !2

4

x

5x 4 f(x) # 4 x "1

y#x"7

x+1 x2 + x − 6 1. x2 + x − 6 = (x + 3)(x − 2), so the domain excludes −3 and 2. It is (−∞, −3) ∪ (−3, 2) ∪ (2, ∞) and the lines x = −3 and x = 2 are vertical asymptotes.

66. f (x) = 20

!20

f (x) #

x

x 3 " 2x 2 ! 15x x 2 ! 5x ! 14

x + 2x − 3x x2 − 25 1. The zeros of the denominator are −5 and 5, so the domain excludes −5 and 5. It is (−∞, −5)∪(−5, 5)∪ (5, ∞) and the lines x = −5 and x = 5 are the vertical asymptotes. 3 2 2. x + 2x − 3x = x + 2 + −28x + 50 , so y = x + 2 2 x − 25 x2 − 25 is the oblique asymptote.

64. f (x) =

3

2

3. x3 + 2x2 − 3x = x(x + 3)(x − 1), so the x-intercepts are (0, 0), (−3, 0), and (1, 0).

2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. There is no oblique asymptote. 3. The zero of the numerator is −1, so the x-intercept is (−1, 0). ! " 1 1 0+1 4. f (0) = = − , so 0, − is the 02 + 0 − 6 6 6 y-intercept. 5. Find other function values to determine the shape of the graph and then draw it. y

4. From part (3) we see that (0, 0) is the y-intercept.

2

5. Find other function values to determine the shape of the graph and then draw it.

x # !3

1 2

!4 !2

x # !5

y

!1

16

!2

x

x#2

y#x"2

8 6

!12

4

12

f (x) #

x

x"1 x2 " x ! 6

!8 !16

f (x) #

x#5

x 3 " 2x 2 !3x x 2 ! 25

5x4 +1 1. The denominator has no real-number zeros, so the domain is (−∞, ∞) and there is no vertical asymptote.

65. f (x) =

x4

2. The numerator and denominator have the same degree, so the horizontal asymptote is determined by the ratio of the leading coefficients, 5/1, or 5. Thus, y = 5 is the horizontal asymptote. There is no oblique asymptote. 3. The zero of the numerator is 0, so (0, 0) is the xintercept. 4. From part (3) we see that (0, 0) is the y-intercept. 5. Find other function values to determine the shape of the graph and then draw it.

x2 x2 − x − 2 1. The zeros of the denominator are the solutions of x2 − x − 2 = 0. Since x2 − x − 2 = (x + 1)(x − 2), the zeros are −1 and 2. Thus, the domain is (−∞, −1)∪ (−1, 2) ∪ (2, ∞) and the lines x = −1 and x = 2 are the vertical asymptotes.

67. f (x) =

2. The numerator and denominator have the same degree, so the horizontal asymptote is determined by the ratio of the leading coefficients, 1/1, or 1. Thus, y = 1 is the horizontal asymptote. There is no oblique asymptote. 3. The zero of the numerator is 0, so the x-intercept is (0, 0). 4. From part (3) we see that (0, 0) is the y-intercept. 5. Find other function values to determine the shape of the graph and then draw it.

230

Chapter 3: Polynomial and Rational Functions

y 4

y#1

2

4

!4 !2

x # !1

x

73. a) The horizontal asymptote of N (t) is the ratio of the leading coefficients of the numerator and denominator, 0.8/5, or 0.16. Thus, N (t) → 0.16 as t → ∞.

x#2

x2 x !x!2

f(x) #

72. Answers may vary. The degree of the numerator must be 1 greater than the degree of the denominator and the quotient, when long division is performed, must be x − 1. If we let the remainder be 1, a function that satisfies these 1 x2 − x + 1 conditions is f (x) = x − 1 + , or f (x) = . x x

b) The medication never completely disappears from the body; a trace amount remains.

2

x2 − x − 2 x+2 1. −2 is the zero of the denominator, so the domain is (−∞, −2) ∪ (−2, ∞) and the line x = −2 is the vertical asymptote. 4 x2 − x − 2 2. =x−3+ , so y = x − 3 is the x+2 x+2 oblique asymptote.

68. f (x) =

3. x − x − 2 = (x + 1)(x − 2), so the x-intercepts are (−1, 0) and (2, 0). 02 − 0 − 2 4. f (0) = = −1, so (0, −1) is the 0+2 y-intercept. 2

5. Find other function values to determine the shape of the graph and then draw it.

74. a) A(x) → 2/1, or 2 as x → ∞.

b) As more DVDs are produced, the average cost approaches $2.

75. a) P (0) = 0; P (1) = 45.455 thousand, or 45, 455; P (3) = 55.556 thousand, or 55, 556; P (8) = 29.197 thousand, or 29, 197 b) The degree of the numerator is less than the degree of the denominator, so the x-axis is the horizontal asymptote. Thus, P (t) → 0 as t → ∞. c) Eventually, no one will live in Lordsburg. 76. a) S = area of base + 4 · area of a side = x2 + 4xy

Now we express y in terms of x:

y

Volume = 108 = x2 y 108 =y x2 ! " 108 , or Thus, S = x2 + 4x x2 432 S = x2 + . x

12

x # !2

6 4

!4 !2 !6

y#x!3

x

!12

f (x) #

x2 ! x ! 2 x"2

69. Answers may vary. The numbers −4 and 5 must be zeros of the denominator. A function that satisfies these conditions is 1 1 , or f (x) = 2 . f (x) = (x + 4)(x − 5) x − x − 20

70. Answers may vary. The numbers −4 and 5 must be zeros of the denominator and −2 must be a zero of the numerator. x+2 x+2 , or f (x) = 2 . f (x) = (x + 4)(x − 5) x − x − 20

71. Answers may vary. The numbers −4 and 5 must be zeros of the denominator and −2 must be a zero of the numerator. In addition, the numerator and denominator must have the same degree and the ratio of their leading coefficients must be 3/2. A function that satisfies these conditions is 3x2 + 6x 3x(x + 2) , or f (x) = 2 . f (x) = 2(x + 4)(x − 5) 2x − 2x − 40 Another function that satisfies these conditions is 3x2 + 12x + 12 3(x + 2)2 , or g(x) = . g(x) = 2(x + 4)(x − 5) 2x2 − 2x − 40

b)

y # x2 "

432 x

600

0

0

12

c) Using the Minimum feature on a graphing calculator, we estimate that the minimum surface area is 108 cm2 . This occurs when x = 6 cm. 77. Using the Maximum feature on a graphing calculator, we find that the maximum population is about 58.926 thousand, or 58,926. This occurs when t ≈ 2.12 months. x3 + 4 4 = x2 + x x 4 As |x| →∞ , → 0 and the value of y1 → x2 . Thus, x the parabola y2 = x2 can be thought of as a nonlinear asymptote for y1 . The graph confirms this.

78. y1 =

79. The denominator has no real-number zeros.

Exercise Set 3.6

231

80. A horizontal asymptote occurs when the degree of the numerator of a rational function is less than or equal to the degree of the denominator. An oblique asymptote occurs when the degree of the numerator is 1 greater than the degree of the denominator. Thus, a rational function cannot have both a horizontal asymptote and an oblique asymptote. 81. slope

y = x2−4x−21 we see that the domain is {x|x < −3 or x > 7}, or (−∞, −3) ∪ (7, ∞). √ 94. f (x) = x2 − 4x − 21

The radicand must be nonnegative. By inspecting the graph of y = x2 − 4x − 21 we see that the domain is {x|x ≤ −3 or x ≥ 7}, or (−∞, −3] ∪ [7, ∞).

Exercise Set 3.6

82. slope-intercept equation

1.

83. point-slope equation

(x + 5)(x − 3) = 0 x+5 = 0

84. domain, range, domain, range

2. Solve x2 + 2x − 15 < 0.

87. midpoint formula

From Exercise 1 we know the solutions of the related equation are −5 and 3. These numbers divide the x-axis into the intervals (−∞, −5), (−5, 3), and (3, ∞). We test a value in each interval.

88. vertical lines −8 x5 + 2x3 + 4x2 = x3 + 4 + 2 x2 + 2 x +2 −8 → 0 and the value of f (x) → x3 + 4. As |x| →∞ , 2 x +2 Thus, the nonlinear asymptote is y = x3 + 4.

89. f (x) =

−2 x + 3x = x2 + 2 + 2 x2 + 1 x +1 −2 → 0 and the value of f (x) → x2 + 2. As |x| →∞ , 2 x +1 Thus, the nonlinear asymptote is y = x2 + 2.

91.

(−∞, −5): f (−6) = 9 > 0 (−5, 3): f (0) = −15 < 0

(3, ∞): f (4) = 9 > 0

Function values are negative only in the interval (−5, 3). The solution set is (−5, 3).

2

y

x # !3

3. Solve x2 + 2x − 15 ≤ 0.

From Exercise 2 we know the solution set of x2 +2x−15 < 0 is (−5, 3). The solution set of x2 +2x−15 ≤ 0 includes the endpoints of this interval. Thus the solution set is [−5, 3].

x#3

4. Solve x2 + 2x − 15 > 0.

4

From our work in Exercise 2 we see that the solution set is (−∞, −5) ∪ (3, ∞).

y#2 4

!4

x

5. Solve x2 + 2x − 15 ≥ 0.

x # !1 f (x) #

2x 3 " x 2 ! 8x ! 4 x 3 " x 2 ! 9x ! 9

92.

15

y#x"5 4

8

x

x#2

x 3 " 4x 2 " x ! 6 x2 ! x ! 2

72 − 4x − 21 The radicand must be nonnegative and the denominator must be nonzero. Thus, the values of x for which x2 −4x− 21 > 0 comprise the domain. By inspecting the graph of

93. f (x) =

x2

x5 − 9x3 = 0

x (x2 − 9) = 0

x # !1

#

6.

From Exercise 4 we know the solution set of x2 +2x−15 > 0 is (−∞, −5) ∪ (3, ∞). The solution set of x2 + 2x − 15 ≥ 0 includes the endpoints −5 and 3. Thus the solution set is (−∞, −5] ∪ [3, ∞). 3

y

f (x) #

x=3

The solution set is {−5, 3}.

86. x-intercept

90. f (x) =

or x − 3 = 0

x = −5 or

85. f (−x) = −f (x)

4

x2 + 2x − 15 = 0

x3 (x + 3)(x − 3) = 0

The solution set is {0, −3, 3}.

7. Solve x5 − 9x3 < 0.

From Exercise 6 we know the solutions of the related equation are −3, 0, and 3. These numbers divide the x-axis into the intervals (−∞, −3), (−3, 0), (0, 3), and (3, ∞). We test a value in each interval. (−∞, −3): g(−4) = −448 < 0 (−3, 0): g(−1) = 8 > 0 (0, 3): g(1) = −8 < 0

(3, ∞): g(4) = 448 > 0

Function values are negative on (−∞, −3) and on (0, 3). The solution set is (−∞, −3) ∪ (0, 3).

232

Chapter 3: Polynomial and Rational Functions

8. Solve x5 − 9x3 ≤ 0.

From Exercise 7 we know the solution set of x − 9x < 0 is (−∞, −3) ∪ (0, 3). The solution set of x5 − 9x3 ≤ 0 includes the endpoints −3, 0, and 3. Thus the solution set is (−∞, −3] ∪ [0, 3]. 5

(1, ∞): f (2) = 6 > 0

Function values are negative only in the interval (−4, 1). The solution set is (−4, 1).

3

16. (x + 3)(x − 5) < 0

The related equation is (x + 3)(x − 5) = 0. Its solutions are −3 and 5. These numbers divide the x-axis into the intervals (−∞, −3), (−3, 5), and (5, ∞). Let f (x) = (x + 3)(x − 5) and test a value in each interval.

9. Solve x5 − 9x3 > 0.

From our work in Exercise 7 we see that the solution set is (−3, 0) ∪ (3, ∞).

(−∞, −3): f (−4) = 9 > 0

10. Solve x5 − 9x3 ≥ 0.

From Exercise 9 we know the solution set of x5 − 9x3 > 0 is (−3, 0) ∪ (3, ∞). The solution set of x5 − 9x3 ≥ 0 includes the endpoints −3, 0, and 3. Thus the solution set is [−3, 0] ∪ [3, ∞).

11. First we find an equivalent inequality with 0 on one side. x + 6x < x + 30 3

2

(−3, 5): f (0) = −15 < 0 (5, ∞): f (6) = 9 > 0

Function values are negative only in the interval (−3, 5). The solution set is (−3, 5).

17. (x − 4)(x + 2) ≥ 0

The related equation is (x − 4)(x + 2) = 0. Using the principle of zero products, we find that the solutions of the related equation are 4 and −2. These numbers divide the x-axis into the intervals (−∞, −2), (−2, 4), and (4, ∞). We let f (x) = (x − 4)(x + 2) and test a value in each interval.

x + 6x − x − 30 < 0 3

2

From the graph we see that the x-intercepts of the related function occur at x = −5, x = −3, and x = 2. They divide the x-axis into the intervals (−∞, −5), (−5, −3), (−3, 2), and (2, ∞). From the graph we see that the function has negative values only on (−∞, −5) and (−3, 2). Thus, the solution set is (−∞, −5) ∪ (−3, 2). 12. From the graph we see that the x-intercepts of the related function occur at x = −4, x = −3, x = 2, and x = 5. They divide the x-axis into the intervals (−∞, −4), (−4, −3), (−3, 2), (2, 5), and (5, ∞). The function has positive values only on (−∞, −4), (−3, 2), and (5, ∞). Since the inequality symbol is ≥, the endpoints of the intervals are included in the solution set. It is (−∞, −4] ∪ [−3, 2] ∪ [5, ∞).

(−∞, −2): f (−3) = 7 > 0 (−2, 4): f (0) = −8 < 0 (4, ∞): f (5) = 7 > 0

Function values are positive on (−∞, −2) and (4, ∞). Since the inequality symbol is ≥, the endpoints of the intervals must be included in the solution set. It is (−∞, −2] ∪ [4, ∞).

18. (x − 2)(x + 1) ≥ 0

The related equation is (x − 2)(x + 1) = 0. Its solutions are 2 and −1. These numbers divide the x-axis into the intervals (−∞, −1), (−1, 2), and (2, ∞). We let f (x) = (x − 2)(x + 1) and test a value in each interval.

13. By observing the graph or the denominator of the function, we see that the function is not defined for x = −2 or x = 2. We also see that 0 is a zero of the function. These numbers divide the x-axis into the intervals (−∞, −2), (−2, 0), (0, 2), and (2, ∞). From the graph we see that the function has positive values only on (−2, 0) and (2, ∞). Since the inequality symbol is ≥, 0 must be included in the solution set. It is (−2, 0] ∪ (2, ∞). 14. By observing the graph or the denominator of the function, we see that the function is not defined for x = −2 or x = 2. We also see that the function has no zeros. Thus, the numbers divide the x-axis into the intervals (−∞, −2), (−2, 2), and (2, ∞). From the graph we see that the function has negative values only on (−2, 2). Thus, the solution set is (−2, 2). 15. (x − 1)(x + 4) < 0

The related equation is (x − 1)(x + 4) = 0. Using the principle of zero products, we find that the solutions of the related equation are 1 and −4. These numbers divide the x-axis into the intervals (−∞, −4), (−4, 1), and (1, ∞). We let f (x) = (x − 1)(x + 4) and test a value in each interval. (−∞, −4): f (−5) = 6 > 0 (−4, 1): f (0) = −4 < 0

(−∞, −1): f (−2) = 4 > 0 (−1, 2): f (0) = −2 < 0 (2, ∞): f (3) = 4 > 0

19.

Function values are positive on (−∞, −1) and (2, ∞). Since the inequality symbol is ≥, the endpoints of the intervals must be included in the solution set. It is (−∞, −1] ∪ [2, ∞). x2 + x − 2 > 0

x2 + x − 2 = 0

(x + 2)(x − 1) = 0

Polynomial inequality

Related equation Factoring

Using the principle of zero products, we find that the solutions of the related equation are −2 and 1. These numbers divide the x-axis into the intervals (−∞, −2), (−2, 1), and (1, ∞). We let f (x) = x2 + x − 2 and test a value in each interval. (−∞, −2): f (−3) = 4 > 0 (−2, 1): f (0) = −2 < 0 (1, ∞): f (2) = 4 > 0

Function values are positive on (−∞, −2) and (1, ∞). The solution set is (−∞, −2) ∪ (1, ∞).

Exercise Set 3.6 20.

x2 − x − 6 > 0

x −x−6 = 0 2

(x − 3)(x + 2) = 0

233 Polynomial inequality

(−2, 2): f (0) = 4 > 0

Related equation

(2, ∞): f (3) = −5 < 0

Factoring

The solutions of the related equation are 3 and −2. These numbers divide the x-axis into the intervals (−∞, −2), (−2, 3), and (3, ∞). We let f (x) = x2 − x − 6 and test a value in each interval. (−∞, −2): f (−3) = 6 > 0

Function values are negative on (−∞, −2) and (2, ∞). Since the inequality symbol is ≤, the endpoints of the intervals must be included in the solution set. It is (−∞, −2] ∪ [2, ∞). 24.

11 − x2 = 0

(−2, 3): f (0) = −6 < 0

Function values are positive on (−∞, −2) and (3, ∞). The solution set is (−∞, −2) ∪ (3, ∞). x2 > 25

Polynomial inequality

x − 25 > 0

Equivalent inequality with 0 on one side Related equation

2

x2 − 25 = 0

(x + 5)(x − 5) = 0

Factoring

Using the principle of zero products, we find that the solutions of the related equation are −5 and 5. These numbers divide the x-axis into the intervals (−∞, −5), (−5, 5), and (5, ∞). We let f (x) = x2 − 25 and test a value in each interval.

25.

Using the principle of zero products, we find that the solution of the related equation is 3. This number divides the x-axis into the intervals (−∞, 3) and (3, ∞). We let f (x) = 6x − 9 − x2 and test a value in each interval.

Function values are positive on (−∞, −5) and (5, ∞). The solution set is (−∞, −5) ∪ (5, ∞).

x2 − 1 = 0

(x + 1)(x − 1) = 0

(−∞, 3): f (−4) = −49 < 0

Polynomial inequality

(3, ∞): f (4) = −1 < 0

Equivalent inequality with 0 on one side Related equation Factoring

The solutions of the related equation are −1 and 1. These numbers divide the x-axis into the intervals (−∞, −1), (−1, 1), and (1, ∞). We let f (x) = x2 − 1 and test a value in each interval.

Function values are negative on both intervals. The solution set is (−∞, 3) ∪ (3, ∞). 26.

(x + 1)(x + 1) = 0

(−1, ∞): f (0) = 1 > 0

Function values are negative only on (−1, 1). Since the inequality symbol is ≤, the endpoints of the interval must be included in the solution set. It is [−1, 1].

(2 + x)(2 − x) = 0

Polynomial inequality Related equation Factoring

Using the principle of zero products, we find that the solutions of the related equation are −2 and 2. These numbers divide the x-axis into the intervals (−∞, −2), (−2, 2), and (2, ∞). We let f (x) = 4 − x2 and test a value in each interval. (−∞, −2): f (−3) = −5 < 0

Factoring

(−∞, −1): f (−2) = 1 > 0

(1, ∞): f (2) = 3 > 0

4 − x2 = 0

Polynomial inequality Related equation

The solution of the related equation is −1. This number divides the x-axis into the intervals (−∞, −1), and (−1, ∞). We let f (x) = x2 + 2x + 1 and test a value in each interval.

(−1, 1): f (0) = −1 < 0

4 − x2 ≤ 0

x2 + 2x + 1 ≤ 0 x2 + 2x + 1 = 0

(−∞, −1): f (−2) = 3 > 0

23.

Factoring out −1 and rearranging Factoring

−(x − 3)(x − 3) = 0

(5, ∞): f (6) = 11 > 0

x2 − 1 ≤ 0

Polynomial inequality Related equation

−(x2 − 6x + 9) = 0

(−5, 5): f (0) = −25 < 0

x2 ≤ 1

6x − 9 − x2 < 0 6x − 9 − x2 = 0

(−∞, −5): f (−6) = 11 > 0

22.

Polynomial inequality Related equation

√ The solutions of the related equation are ± 11. These √ numbers into the intervals (−∞, − 11), √ √divide the x-axis √ (− 11, 11), and ( 11, ∞). We let f (x) = 11 − x2 and test a value in each interval. √ (−∞, − 11): f (−4) = −5 < 0 √ √ (− 11, 11): f (0) = 11 > 0 √ ( 11, ∞): f (4) = −5 < 0 √ √ Function values are positive only on (− 11, 11). Since the inequality symbol is ≥, the endpoints of√the √ interval must be included in the solution set. It is [− 11, 11].

(3, ∞): f (4) = 6 > 0

21.

11 − x2 ≥ 0

Function values are negative in neither interval. The function is equal to 0 when x = −1. Thus, the solution set is {−1}. 27.

x2 + 12 < 4x Polynomial inequality x2 − 4x + 12 < 0 x2 − 4x + 12 = 0

Equivalent inequality with 0 on one side Related equation

Using the quadratic formula, we find that the related equation has no real-number solutions. The graph lies entirely above the x-axis, so the inequality has no solution. We could determine this algebraically by letting f (x) = x2 − 4x + 12 and testing any real number (since

234

28.

Chapter 3: Polynomial and Rational Functions The solutions of the related equation divide the x-axis into four intervals. We let f (x) = 2x3 −√ x2 − 5x and test a 1 − 41 value in each interval. Note that ≈ −1.3508 and 4 √ 1 + 41 ≈ 1.8508. 4 √ " ! 1 − 41 − ∞, : f (−2) = −10 < 0 4 √ " ! 1 − 41 , 0 : f (−1) = 2 > 0 4 √ " ! 1 + 41 : f (1) = −4 < 0 0, 4 √ ! " 1 + 41 , ∞ : f (2) = 2 > 0 4 √ " ! 1 − 41 Function values are negative on − ∞, and 4 √ " ! 1 + 41 . The solution set is 0, 4 √ √ " " ! ! 1 − 41 1 + 41 − ∞, ∪ 0, , or 4 4 approximately (−∞, −1.3508) ∪ (0, 1.8508).

there are no real-number solutions of f (x) = 0 to divide the x-axis into intervals). For example, f (0) = 12 > 0, so we see algebraically that the inequality has no solution. The solution set is ∅. x2 − 8 > 6x Polynomial inequality

x − 6x − 8 > 0

Equivalent inequality with 0 on one side Related equation x2 − 6x − 8 = 0 Using the quadratic formula, we √ solutions √ find that the of the related equation are 3 − 17 and 3 + 17. These √ numbers √ intervals (−∞, 3− 17), √ divide√the x-axis into the (3 − 17, 3 + 17), and (3 + 17, ∞). We let f (x) = x2 − 6x − 8 and test a value in each interval. √ (−∞, 3 − 17): f (−2) = 8 > 0 √ √ (3 − 17, 3 + 17): f (0) = −8 < 0 √ (3 + 17, ∞): f (8) = 8 > 0 √ Function √ values are positive on (−∞, 3 − 17) and ∞). The solution set is (3 + 17, √ √ (−∞, 3 − 17) ∪ (3 + 17, ∞) or approximately (−∞, −1.123) ∪ (7.123, ∞). 2

29.

4x3 − 7x2 ≤ 15x Polynomial inequality

4x3 − 7x2 − 15x ≤ 0

Equivalent inequality with 0 on one side Related equation

4x − 7x − 15x = 0 3

2

31.

" 5 and (0, 3). 4 Since the inequality symbol is ≤, the endpoints of the intervals must be included in the solution set. It is $ ! 5 − ∞, − ∪ [0, 3]. 4 Function values are negative on

30.

− ∞, −

2x3 − x2 < 5x Polynomial inequality

2x − x2 − 5x < 0 3

!

2x3 − x2 − 5x = 0

Equivalent inequality with 0 on one side Related equation

x(2x2 − x − 5) = 0

x = 0 or 2x2 − x − 5 = 0

x = 0 or

x=

1±

x3 + 3x2 − x − 3 = 0 Related equation

x (x + 3) − (x + 3) = 0 Factoring 2

x(4x + 5)(x − 3) = 0 Factoring Using the principle of zero products, we find that the so5 lutions of the related equation are 0, − , and 3. These 4 ! " 5 numbers divide the x-axis into the intervals − ∞, − , 4 " ! 5 3 − , 0 , (0, 3), and (3, ∞). We let f (x) = 4x − 7x2 − 4 15x and test " a value in each interval. ! 5 − ∞, − : f (−2) = −30 < 0 4 " ! 5 − , 0 : f (−1) = 4 > 0 4 (0, 3) f (1) = −18 < 0 (3, ∞): f (4) = 84 > 0

x3 + 3x2 − x − 3 ≥ 0 Polynomial inequality

(x2 − 1)(x + 3) = 0

(x + 1)(x − 1)(x + 3) = 0

Using the principle of zero products, we find that the solutions of the related equation are −1, 1, and −3. These numbers divide the x-axis into the intervals (−∞, −3), (−3, −1), (−1, 1), and (1, ∞). We let f (x) = x3 +3x2 −x−3 and test a value in each interval. (−∞, −3): f (−4) = −15 < 0 (−3, −1): f (−2) = 3 > 0 (−1, 1): f (0) = −3 < 0 (1, ∞): f (2) = 15 > 0

Function values are positive on (−3, −1) and (1, ∞). Since the inequality symbol is ≥, the endpoints of the intervals must be included in the solution set. It is [−3, −1]∪[1, ∞). 32.

x3 + x2 − 4x − 4 ≥ 0

Polynomial inequality

x3 + x2 − 4x − 4 = 0

Related equation

(x + 2)(x − 2)(x + 1) = 0

Factoring

The solutions of the related equation are −2, 2, and −1. These numbers divide the x-axis into the intervals (−∞, −2), (−2, −1), (−1, 2), and (2, ∞). We let f (x) = x3 + x2 − 4x − 4 and test a value in each interval. (−∞, −2): f (−3) = −10 < 0

√ 4

41

(−2, −1): f (−1.5) = 0.875 > 0 (−1, 2): f (0) = −4 < 0

Exercise Set 3.6

235 √ √ √ √ (−∞, − 2), (− 2, −1), (−1, 2), and ( 2, ∞). We let f (x) = x5 − 2x3 + x2 − 2 and test a value in each interval. √ (−∞, − 2): f (−2) = −14 < 0 √ (− 2, −1): f (−1.3) ≈ 0.37107 > 0 √ (−1, 2): f (0) = −2 < 0 √ ( 2, ∞): f (2) = 18 > 0 √ Function values are positive on (− 2, −1) and √ ( 2, ∞). Since the inequality symbol is ≥, the endpoints of √ the intervals √ must be included in the solution set. It is [− 2, −1] ∪ [ 2, ∞).

(2, ∞): f (3) = 20 > 0

Function values are positive only on (−2, −1) and (2, ∞). Since the inequality symbol is ≥, the endpoints of the interval must be included in the solution set. It is [−2, −1] ∪ [2, ∞).

33.

x3 − 2x2 < 5x − 6 x3 − 2x2 − 5x + 6 < 0 x3 − 2x2 − 5x + 6 = 0

Polynomial inequality Equivalent inequality with 0 on one side Related equation

Using the techniques of Section 3.3, we find that the solutions of the related equation are −2, 1, and 3. They divide the x-axis into the intervals (−∞, −2), (−2, 1), (1, 3), and (3, ∞). Let f (x) = x3 − 2x2 − 5x + 6 and test a value in each interval.

36.

3

Function values are negative on (−∞, −2) and (1, 3). The solution set is (−∞, −2) ∪ (1, 3). Polynomial inequality Equivalent inequality with 0 on one side Related equation

Using the techniques of Section 3.3, we find that the solutions of the related equation are −3, −2, and 1. They divide the x-axis into the intervals (−∞, −3), (−3, −2), (−2, 1), and (1, ∞). Let f (x) = x3 + 4x2 + x − 6 and test a value in each interval.

(2, ∞): f (3) = 114 > 0

√ √ Function values are positive on √ (−√ 3, 3) and (2, ∞). The solution set is (− 3, 3) ∪ (2, ∞). 37.

2x3 + 6 ≤ 5x2 + x Polynomial inequality

(−∞, −3): f (−4) = −10 < 0

2x3 − 5x2 − x + 6 ≤ 0

(−2, 1): f (0) = −6 < 0

2x3 − 5x2 − x + 6 = 0

(−3, −2): f (−2.5) = 0.875 > 0 (1, ∞): f (2) = 20 > 0

Function values are negative on (−∞, −3) and (−2, 1). Since the inequality symbol is ≤, the endpoints of the intervals must be included in the solution set. It is (−∞, −3] ∪ [−2, 1]. 35.

x5 + x2 ≥ 2x3 + 2 x5 − 2x3 + x2 − 2 ≥ 0 x5 − 2x3 + x2 − 2 = 0 x3 (x2 − 2) + x2 − 2 = 0

(x + 1)(x − 2) = 0 3

Factoring

Using the principle of zero products, we find that √ the real-number solutions of the related equation are 2, − 3, √ and 3. These √ numbers √ √divide√the x-axis into the intervals (−∞, − 3), (− 3, 3), ( 3, 2), and (2, ∞). We let f (x) = x5 −3x3 −8x2 +24 and test a value in each interval. √ (−∞, − 3): f (−2) = −16 < 0 √ √ (− 3, 3): f (0) = 24 > 0 √ ( 3, 2): f (1.8) ≈ −0.5203 < 0

(3, ∞): f (4) = 18 > 0

x3 + 4x2 + x − 6 = 0

Equivalent inequality with 0 on one side Related equation

(x3 − 8)(x2 − 3) = 0

(1, 3): f (2) = −4 < 0

x3 + 4x2 + x − 6 ≤ 0

x5 − 3x3 − 8x2 + 24 = 0

x (x2 − 3) − 8(x2 − 3) = 0

(−2, 1): f (0) = 6 > 0

x3 + x ≤ 6 − 4x2

Polynomial inequality

x5 − 3x3 − 8x2 + 24 > 0

(−∞, −2): f (−3) = −24 < 0

34.

x5 + 24 > 3x3 + 8x2

Polynomial inequality Related inequality with 0 on one side Related equation Factoring

2

Using the principle of zero products, we find that the √ realnumber solutions of the related equation are −1, − 2, and √ 2. These numbers divide the x-axis into the intervals

Equivalent inequality with 0 on one side Related equation

Using the techniques of Section 3.3, we find that the solu3 tions of the related equation are −1, , and 2. We can also 2 use the graph of y = 2x3 − 5x2 − x + 6 to find these solutions. They " !divide " the x-axis into the intervals (−∞, −1), ! 3 3 , , 2 , and (2, ∞). Let f (x) = 2x3 −5x2 −x+6 − 1, 2 2 and test a value in each interval. (−∞, −1): f (−2) = −28 < 0 " ! 3 : f (0) = 6 > 0 − 1, 2 ! " 3 , 2 : f (1.6) = −0.208 < 0 2 (2, ∞): f (3) = 12 > 0

Function " values are negative in (−∞, −1) and ! 3 , 2 . Since the inequality symbol is ≤, the endpoints 2

236

38.

Chapter 3: Polynomial and Rational Functions of the intervals!must be included in $ the solution set. The $ % 3 solution set is − ∞, −1 ∪ , 2 . 2

40.

x3 − 3x2 − 9x + 27 ≥ 0

2x3 + x2 < 10 + 11x Polynomial inequality 2x3 + x2 − 11x − 10 < 0 2x3 + x2 − 11x − 10 = 0

x3 − 3x2 − 9x + 27 = 0

Equivalent inequality with 0 on one side Related equation

x (x − 3) − 9(x − 3) = 0 2

(x + 3)(x − 3)(x − 3) = 0

(−∞, −3): f (−4) = −49 < 0 (−3, 3): f (0) = 27 > 0

Function values are positive only on (−3, 3) and (3, ∞). Since the inequality symbol is ≥, the endpoints of the intervals must be included in the solution set. It is [−3, 3] ∪ [3, ∞), or [−3, ∞).

Function"values are negative on (−∞, −2) and ! 5 . The solution set is − 1, 2 ! " 5 (−∞, −2) ∪ − 1, . 2

x + 5x − 25x − 125 = 0 3

2

x2 (x + 5) − 25(x + 5) = 0

41.

Polynomial inequality Equivalent inequality with 0 on one side Related equation Factoring

(x − 25)(x + 5) = 0

(x + 5)(x − 5)(x + 5) = 0

Using the principle of zero products, we find that the solutions of the related equation are −5 and 5. These numbers divide the x-axis into the intervals (−∞, −5), (−5, 5), and (5, ∞). We let f (x) = x3 + 5x2 − 25x − 125 and test a value in each interval. (−∞, −5): f (−6) = −11 < 0 (−5, 5): f (0) = −125 < 0

Function values are negative on (−∞, −5) and (−5, 5). Since the inequality symbol is ≤, the endpoints of the intervals must be included in the solution set. It is (−∞, −5] ∪ [−5, 5] or (−∞, 5].

1 > 0 Rational inequality x+4 1 = 0 Related equation x+4 1 is 0 when x = −4, so the The denominator of f (x) = x+4 function is not defined for x = −4. The related equation has no solution. Thus, the only critical value is −4. It divides the x-axis into the intervals (−∞, −4) and (−4, ∞). We test a value in each interval. (−∞, −4): f (−5) = −1 < 0 1 (−4, ∞): f (0) = > 0 4 Function values are positive on (−4, ∞). This can also be 1 . The solution set determined from the graph of y = x+4 is (−4, ∞).

2

(5, ∞): f (6) = 121 > 0

Factoring

(3, ∞): f (4) = 7 > 0

(−2, −1): f (−1.5) = 2 > 0 " ! 5 : f (0) = −10 < 0 − 1, 2 " ! 5 , ∞ : f (3) = 20 > 0 2

x3 + 5x2 − 25x − 125 ≤ 0

Equivalent inequality with 0 on one side Related equation

The solutions of the related equation are −3 and 3. These numbers divide the x-axis into the intervals (−∞, −3), (−3, 3), and (3, ∞). We let f (x) = x3 − 3x2 − 9x + 27 and test a value in each interval.

(−∞, −2): f (−3) = −22 < 0

x3 + 5x2 − 25x ≤ 125

Polynomial inequality

(x2 − 9)(x − 3) = 0

Using the techniques of Section 3.3, we find that the realnumber solutions of the related equation are −2, −1, and 5 . These numbers divide the x-axis into the intervals 2 ! " ! " 5 5 (−∞, −2), (−2, −1), − 1, , and , ∞ . We let 2 2 f (x) = 2x3 +x2 −11x−10 and test a value in each interval.

39.

x3 − 9x + 27 ≥ 3x2

42.

1 ≤ 0 Rational inequality x−3 1 = 0 Related equation x−3 1 is 0 when x = 3, so the The denominator of f (x) = x−3 function is not defined for x = 3. The related equation has no solution. Thus, the only critical value is 3. It divides the x-axis into the intervals (−∞, 3) and (3, ∞). We test a value in each interval. 1 (−∞, 3): f (0) = − < 0 3 (3, ∞): f (4) = 1 > 0

Function values are negative on (−∞, 3). Note that since 3 is not in the domain of f (x), it cannot be included in the solution set. It is (−∞, 3).

Exercise Set 3.6 43.

44.

237

−4 < 0 Rational inequality 2x + 5 −4 = 0 Related equation 2x + 5 −4 5 The denominator of f (x) = is 0 when x = − , 2x + 5 2 5 so the function is not defined for x = − . The related 2 equation has no solution. Thus, the only critical value " is ! 5 5 − . It divides the x-axis into the intervals − ∞, − 2 ! 2 " 5 and − , ∞ . We test a value in each interval. 2 " ! 5 − ∞, − : f (−3) = 4 > 0 2 ! " 4 5 − , ∞ : f (0) = − < 0 2 5 ! " 5 Function values are negative on − , ∞ . The solution 2 " ! 5 set is − , ∞ . 2 −2 ≥ 0 Rational inequality 5−x −2 = 0 Related equation 5−x −2 The denominator of f (x) = is 0 when x = 5, so the 5−x function is not defined for x = 5. The related equation has no solution. Thus, the only critical value is 5. It divides the x-axis into the intervals (−∞, 5) and (5, ∞). We test a value in each interval. 2 (−∞, 5): f (0) = − < 0 5 (5, ∞): f (6) = 2 > 0 Function values are positive on (5, ∞). Note that since 5 is not in the domain of f (x), it cannot be included in the solution set. It is (5, ∞).

45.

x−4 x+2 − ≤0 x+3 x−1

x−4 x+2 − is 0 when x = The denominator of f (x) = x+3 x−1 −3 or x = 1, so the function is not defined for these values of x. We solve the related equation f (x) = 0. x−4 x+2 − =0 x+3 x−1 ! " x−4 x+2 (x+3)(x−1) − = (x+3)(x−1)·0 x+3 x−1 (x − 1)(x − 4) − (x + 3)(x + 2) = 0 x2 − 5x + 4 − (x2 + 5x + 6) = 0

x=−

(−∞, −3): f (−4) = 7.6 > 0 " ! 1 − 3, − : f (−1) = −2 < 0 5 ! " 1 2 − , 1 : f (0) = > 0 5 3 (1, ∞): f (2) = −4.4 < 0

" 1 and (1, ∞). 5 Note " since the inequality symbol is ≤ and ! that 1 1 = 0, then − must be included in the solution f − 5 5 set. Note also that since neither −3 nor 1 is in the domain of $ are not included in the solution set. It is ! f (x), they 1 − 3, − ∪ (1, ∞). 5 Function values are negative on

46.

!

− 3, −

x+1 x−3 + <0 x−2 x−1

x+1 x−3 + is 0 when x = 2 x−2 x−1 or x = 1, so the function is not defined for these values of x. We solve the related equation f (x) = 0. x+1 x−3 + =0 x−2 x−1 (x − 1)(x + 1) + (x − 2)(x − 3) = 0 The denominator of f (x) =

Multiplying by (x − 2)(x − 1)

x − 1 + x2 − 5x + 6 = 0 2

2x2 − 5x + 5 = 0

This equation has no real-number solutions, so the critical values are 1 and 2. They divide the x-axis into the intervals (−∞, 1), (1, 2), and (2, ∞). We test a value in each interval. (−∞, 1): f (0) = 2.5 > 0 (1, 2): f (1.5) = −8 < 0 (2, ∞): f (3) = 4 > 0

Function values are negative on (1, 2). The solution set is (1, 2). x+1 2x − 1 47. ≥ Rational inequality x+3 3x + 1 x+1 2x − 1 − ≥0 Equivalent inequality x+3 3x + 1 with 0 on one side The denominator of f (x) =

2x − 1 x+1 − is 0 when x = x+3 3x + 1

1 −3 or x = − , so the function is not defined for these 3 values of x. We solve the related equation f (x) = 0.

−10x − 2 = 0

−10x = 2

1 The critical values are −3, − , and 1. They divide the 5 " ! " ! 1 1 x-axis into the intervals (−∞, −3), 3, − , − , 1 , 5 5 and (1, ∞). We test a value in each interval.

1 5

238

Chapter 3: Polynomial and Rational Functions x + 5 3x + 2 − =0 x − 4 2x + 1 (2x + 1)(x + 5) − (x − 4)(3x + 2) = 0

2x − 1 x+1 − =0 x+3 3x + 1 " ! x+1 2x − 1 − = (x + 3)(3x + 1) x+3 3x + 1 (x + 3)(3x + 1) · 0

Multiplying by (x − 4)(2x + 1)

2x2 + 11x + 5 − (3x2 − 10x − 8) = 0

(3x + 1)(2x − 1) − (x + 3)(x + 1) = 0

−x2 + 21x + 13 = 0

6x − x − 1 − (x + 4x + 3) = 0 2

2

Using the √ quadratic formula we find that 21 ± 493 x= . Then the critical values are 2 √ √ 1 21 + 493 21 − 493 , − , 4, and . They divide 2 2 √ " !2 21 − 493 , the x-axis into the intervals − ∞, 2 √ √ ! " ! " ! " 21 − 493 1 1 21 + 493 , − , − , 4 , 4, , 2 2 2 2 √ ! " 21 + 493 and , ∞ . We test a value in each interval. 2 √ " ! 21 − 493 : f (−1) = −1.8 < 0 − ∞, 2 √ " ! 21 − 493 1 − : f (−0.55) = 2.522 > 0 2 2 " ! 1 − , 4 : f (0) = −3.25 < 0 2 √ " ! 21 + 493 4, : f (5) ≈ 8.4545 > 0 2 √ " ! 21 + 493 , ∞ : f (22) ≈ −0.0111 < 0 2 √ " ! 21 − 493 1 Function values are positive on ,− and 2 2 √ " ! 21 + 493 . The solution set is 4, 2 √ √ " ! " ! 21 − 493 1 21 + 493 ,− ∪ 4, . 2 2 2

5x2 − 5x − 4 = 0

Using the√quadratic formula we find that √ 5 − 105 5 ± 105 . Then the critical values are −3, , x= 10 10 √ 1 5 + 105 − , and . They divide the x-axis into the in3 10 √ √ " ! " ! 5 − 105 1 5 − 105 , ,− , tervals (−∞, −3), − 3, 10 10 3 √ ! " 1 5 + 105 , and − , 3 10 √ " ! 5 + 105 , ∞ . We test a value in each interval. 10 (−∞, −3): f (−4) ≈ 8.7273 > 0 √ " ! 5 − 105 − 3, : f (−2) = −5.2 < 0 10 √ " ! 5 − 105 1 , − : f (−0.4) ≈ 2.3077 > 0 10 3 √ " ! 1 5 + 105 : f (0) ≈ −1.333 < 0 − , 3 10 √ " ! 5 + 105 , ∞ : f (2) ≈ 0.1714 > 0 10

Function are positive on −3), √ values " √ (−∞, " ! ! 5 − 105 1 5 + 105 ,− and , ∞ . Note that 10 3 10 since symbol is√ ≥ and √ inequality " ! the 5 ± 105 5 ± 105 f = 0, then 10 10 must be included in the solution set. Note also that since 1 neither −3 nor − is in the domain of f (x), they are not 3 included in the solution set. It is √ √ " % " % 5 + 105 5 − 105 1 ,− ∪ ,∞ . (−∞, −3) ∪ 10 3 10 48.

x+5 3x + 2 > Rational inequality x−4 2x + 1 x + 5 3x + 2 − >0 Equivalent inequality x − 4 2x + 1 with 0 on one side

The denominator of f (x) = 1 x + 5 3x + 2 − is 0 when x = 4 or x = − , so the function x − 4 2x + 1 2 is not defined for these values of x. We solve the related equation f (x) = 0.

49.

x+1 ≥ 3 Rational inequality x−2 x+1 − 3 ≥ 0 Equivalent inequality x−2 with 0 on one side

x+1 − 3 is 0 when x = 2, so x−2 the function is not defined for this value of x. We solve the related equation f (x) = 0. x+1 −3 = 0 x−2 " ! x+1 − 3 = (x − 2) · 0 (x − 2) x−2 x + 1 − 3(x − 2) = 0 The denominator of f (x) =

x + 1 − 3x + 6 = 0 −2x + 7 = 0

−2x = −7 7 x= 2

Exercise Set 3.6

239 1 = x " ! 1 = x x−2− x 1 x2 − 2x − x · = x x2 − 2x − 1 =

7 The critical values are 2 and . They divide the x-axis 2 " ! " ! 7 7 , and , ∞ . We test into the intervals (−∞, 2), 2, 2 2 a value in each interval. (−∞, 2): f (0) = −3.5 < 0 " ! 7 : f (3) = 1 > 0 2, 2 ! " 7 , ∞ : f (4) = −0.5 < 0 2

x−2−

50.

x <2 x−5

!

Rational inequality

x −2 < 0 x−5

Equivalent inequality with 0 on one side x − 2 is 0 when x = 5, so The denominator of f (x) = x−5 the function is not defined for this value of x. We solve the related equation f (x) = 0. x −2 = 0 x−5 x − 2(x − 5) = 0 Multiplying by x − 5

52.

x = 10

(−∞, 0): f (−1) = 9 > 0 (0, 2): f (1) = −1 < 0 1 (2, ∞): f (3) = − < 0 3 Function values are positive on (−∞, 0). Note that since the inequality symbol is ≥ and f (2) = 0, then 2 must be included in the solution set. Note also that since 0 is not in the domain of f (x), it is not included in the solution set. It is (−∞, 0) ∪ {2}.

Rational inequality Equivalent inequality with 0 on one side

1 The denominator of f (x) = x − 2 − is 0 when x = 0, so x the function is not defined for this value of x. We solve the related equation f (x) = 0.

4 −x ≥ 0 x

x=2

(−∞, 5): f (0) = −2 < 0

1 >0 x

4 + x Rational inequality x

The critical values are 0 and 2. They divide the x-axis into the intervals (−∞, 0), (0, 2), and (2, ∞). We test a value in each interval.

(5, 10): f (6) = 4 > 0 1 (10, ∞): f (11) = − < 0 6 Function values are negative on (−∞, 5) and (10, ∞). The solution set is (−∞, 5) ∪ (10, ∞).

x−2−

0

−(x − 2)2 = 0

The critical values are 5 and 10. They divide the x-axis into the intervals (−∞, 5), (5, 10), and (10, ∞). We test a value in each interval.

1 x

0

Equivalent inequality with 0 on one side 4 The denominator of f (x) = 4 − − x is 0 when x = 0, so x the function is not defined for this value of x. We solve the related equation f (x) = 0. 4 4− −x = 0 x 4x − 4 − x2 = 0 Multiplying by x

−x + 10 = 0

x−2 >

4≥ 4−

x − 2x + 10 = 0

51.

x·0

√ Using the quadratic formula √ we find that√x = 1 ± 2. The critical values are 1 − 2, 0, and 1 + √2. They √ divide the x-axis (−∞, 1 − 2), (1 − 2, 0), √ into the intervals √ (0, 1 + 2), and (1 + 2, ∞). We test a value in each interval. √ (−∞, 1 − 2): f (−1) = −2 < 0 √ (1 − 2, 0): f (−0.1) = 7.9 > 0 √ (0, 1 + 2): f (1) = −2 < 0 √ 2 (1 + 2, ∞): f (3) = > 0 3 √ Function √ values are positive on (1 − 2, 0) and solution set is (1 + √2, ∞). The √ (1 − 2, 0) ∪ (1 + 2, ∞).

" 7 2, . Note that since ! 2" 7 7 = 0, then must the inequality symbol is ≥ and f 2 2 be included in the solution set. Note also that since 2 is not in the!domain $ of f (x), it is not included in the solution 7 set. It is 2, . 2 Function values are positive on

0

53.

5 2 ≤ 2 x2 − 4x + 3 x −9 5 2 − ≤0 x2 − 4x + 3 x2 − 9 2 5 − ≤0 (x − 1)(x − 3) (x + 3)(x − 3) The denominator of f (x) = 5 2 − is 0 when x = 1, 3, or −3, (x − 1)(x − 3) (x + 3)(x − 3)

240

Chapter 3: Polynomial and Rational Functions 25 . They divide the 2 x-axis into the intervals (−∞, −5), " ! " ! 25 25 , and , ∞ . We test a value (−5, −2), (−2, 2), 2, 2 2 in each interval.

so the function is not defined for these values of x. We solve the related equation f (x) = 0. 2 5 − =0 (x − 1)(x − 3) (x + 3)(x − 3) " ! 5 2 − (x−1)(x−3)(x+3) (x−1)(x−3) (x+3)(x−3)

The critical values are −5, −2, 2, and

(−∞, −5): f (−6) ≈ −1.156 < 0

= (x − 1)(x − 3)(x + 3) · 0

(−5, −2): f (−3) = 3.1 > 0

2(x + 3) − 5(x − 1) = 0

(−2, 2): f (0) = −1.25 < 0 " ! 25 : f (3) = 0.475 > 0 2, 2 " ! 25 , ∞ : f (13) ≈ −0.0003 < 0 2 Function" values are negative on (−∞, −5), (−2, 2), and ! 25 , ∞ . Note that since the inequality symbol is ≤ and !2 " 25 25 = 0, then must be included in the solution set. f 2 2 Note also that since −5, −2, and 2 are not in the domain of f (x), they are not%included " in the solution set. It is 25 ,∞ . (−∞, −5) ∪ (−2, 2) ∪ 2

2x + 6 − 5x + 5 = 0 −3x + 11 = 0

−3x = −11 11 x= 3

11 . They divide the x3 " ! 11 axis into the intervals (−∞, −3), (−3, 1), (1, 3), 3, , 3 " ! 11 , ∞ . We test a value in each interval. and 3 The critical values are −3, 1, 3, and

(−∞, −3): f (−4) ≈ −0.6571 < 0 (−3, 1): f (0) ≈ 1.2222 > 0

(1, 3): f (2) = −1 < 0 " ! 11 : f (3.5) ≈ 0.6154 > 0 3, 3 " ! 11 , ∞ : f (4) ≈ −0.0476 < 0 3

55.

6 3 − has no realx2 + 1 5x2 + 2 number zeros. We solve the related equation f (x) = 0. 6 3 − =0 x2 + 1 5x2 + 2 ! " 3 6 (x2 + 1)(5x2 + 2) 2 − = x + 1 5x2 + 2 2 (x + 1)(5x2 + 2) · 0 The denominator of f (x) =

Function" values are negative on (−∞, −3), (1, 3), and ! 11 , ∞ . Note that since the inequality symbol is ≤ and !3 " 11 11 f = 0, then must be included in the solution set. 3 3 Note also that since −3, 1, and 3 are not in the domain of f (x), they are not " in the solution set. It is % included 11 ,∞ . (−∞, −3) ∪ (1, 3) ∪ 3 54.

6 3 ≥ 2 +1 5x + 2 3 6 − ≥0 x2 + 1 5x2 + 2 x2

3(5x2 + 2) − 6(x2 + 1) = 0 15x2 + 6 − 6x2 − 6 = 0

9x2 = 0

5 3 ≤ 2 x2 − 4 x + 7x + 10

x2 = 0 x=0

5 3 − ≤0 (x + 2)(x − 2) (x + 2)(x + 5) The denominator of f (x) = 5 3 − is 0 when x = −2, 2, or (x + 2)(x − 2) (x + 2)(x + 5) −5, so the function is not defined for these values of x. We solve the related equation f (x) = 0. 3 5 − =0 (x+2)(x−2) (x+2)(x+5) 3(x + 5) − 5(x − 2) = 0 Multiplying

The only critical value is 0. It divides the x-axis into the intervals (−∞, 0) and (0, ∞). We test a value in each interval. (−∞, 0): f (−1) ≈ 0.64286 > 0 (0, ∞): f (1) ≈ 0.64286 > 0

Function values are positive on both intervals. Note that since the inequality symbol is ≥ and f (0) = 0, then 0 must be included in the solution set. It is (−∞, 0] ∪ [0, ∞), or (−∞, ∞).

by (x+2)(x−2)(x+5)

3x + 15 − 5x + 10 = 0

−2x + 25 = 0 25 x= 2

56.

3 4 < 2 x2 − 9 x − 25 4 3 − <0 x2 − 9 x2 − 25 3 4 − <0 (x + 3)(x − 3) (x + 5)(x − 5)

Exercise Set 3.6

241 !

4x2 − 100 − 3x2 + 27 = 0

x2 − 73 = 0

√ x = ± 73

√ √ The critical values are − 73, −5, −3, 3, 5, and √ 73. They √ divide the x-axis into the intervals √ (−∞, − 73), (− 73, −5), (−5, −3), (−3, 3), (3, 5), (5, 73), and √ ( 73, ∞). We test a value in each interval. √ (−∞, − 73): f (−9) ≈ 0.00198 > 0 √ (− 73, −5): f (−6) ≈ −0.1246 < 0 (−5, −3): f (−4) ≈ 0.90476 > 0

(−3, 3): f (0) ≈ −0.3244 < 0

√ Function values are negative on (− √73, −5), (−3, 3), and √ (5, √73). The solution set is (− 73, −5) ∪ (−3, 3) ∪ (5, 73). 57.

58.

3 5 < x2 + 3x 2x + 1 3 5 − <0 x2 + 3x 2x + 1 5 3 − <0 x(x + 3) 2x + 1 The denominator of f (x) =

5(2x + 1) − 3x(x + 3) = 0

10x + 5 − 3x2 − 9x = 0 −3x2 + x + 5 = 0

Using the√quadratic formula we find that √ 1 − 61 1 1 ± 61 . The critical values are −3, , − , x= 6 √ 6 2 1 + 61 0, and . They divide the x-axis into 6 √ " ! 1 − 61 , the intervals (−∞, −3), − 3, 6

61

,−

3 2 > x2 + 3 5 + 4x2 3 2 − >0 x2 + 3 5 + 4x2

3 2 − has no realx2 + 3 5 + 4x2 number zeros. We solve the related equation f (x) = 0. 3 2 =0 − x2 + 3 5 + 4x2 2 2 2(5 + 4x ) − 3(x + 3) = 0 Multiplying by (x2 + 3)(5 + 4x2 ) 2 2 10 + 8x − 3x − 9 = 0 The denominator of f (x) =

3 5 − is 0 when x(x + 3) 2x + 1

1 x = 0, −3, or − , so the function is not defined for these 2 values of x. We solve the related equation f (x) = 0. 3 5 − =0 x(x + 3) 2x + 1 ! " 5 3 x(x+3)(2x+1) − = x(x+3) 2x+1 x(x+3)(2x+1)·0

1−

(−∞, −3): f (−4) ≈ 1.6786 > 0 √ " ! 1 − 61 : f (−2) = −1.5 < 0 − 3, 6 √ " ! 1 − 61 1 , − : f (−1) = 0.5 > 0 6 2 " ! 1 − , 0 : f (−0.1) ≈ −20.99 < 0 2 √ " ! 1 + 61 0, : f (1) = 0.25 > 0 6 √ " ! 1 + 61 , ∞ : f (2) = −0.1 < 0 6 √ " ! 1 − 61 Function values are negative on − 3, , 6 √ " ! " ! 1 + 61 1 , ∞ . The solution set is − , 0 and 2 6 √ √ " ! " ! " ! 1 1 + 61 1 − 61 − 3, ∪ − ,0 ∪ ,∞ . 6 2 6

Multiplying by (x + 3)(x − 3)(x + 5)(x − 5)

(3, 5): f (4) ≈ 0.90476 > 0 √ (5, 73): f (6) ≈ −0.1246 < 0 √ ( 73, ∞): f (9) ≈ 0.00198 > 0

√

√ " " ! " ! 1 1 1 + 61 , − , 0 , 0, , and 6√ 2 2 6 " ! 1 + 61 ,∞ . 6 We test a value in each interval.

The denominator of f (x) = 3 4 − is 0 when x = −3, 3, −5, (x + 3)(x − 3) (x + 5)(x − 5) or 5, so the function is not defined for these values of x. We solve the related equation f (x) = 0. 4 3 − =0 (x + 3)(x − 3) (x + 5)(x − 5) 4(x + 5)(x − 5) − 3(x + 3)(x − 3) = 0

5x2 + 1 = 0

This equation has no real-number solutions. Thus, there are no critical values. We test a value in (−∞, ∞): 1 > 0. The function is positive on (−∞, ∞). This f (0) = 15 is the solution set. 59.

x 5x > 7x − 2 x+1 5x x − >0 7x − 2 x + 1

The denominator of f (x) =

x 5x − is 0 7x − 2 x + 1

2 or x = −1, so the function is not defined for 7 these values of x. We solve the related equation f (x) = 0.

when x =

242

Chapter 3: Polynomial and Rational Functions 5x x − =0 7x − 2 x + 1 " ! 5x x (7x−2)(x+1) − = (7x−2)(x+1) · 0 7x−2 x+1 5x(x + 1) − x(7x − 2) = 0

61.

5x2 + 5x − 7x2 + 2x = 0 −2x2 + 7x = 0

7 x = 0 or x = 2

−x(2x − 7) = 0

2 7 , and . They divide the x7 2 ! " ! " 2 7 2 , , , axis into the intervals (−∞, −1), (−1, 0), 0, 7 7 2 " ! 7 , ∞ . We test a value in each interval. and 2 The critical values are −1, 0,

Multiplying by (x + 5)(x − 1)(x − 5)

(−∞, −1): f (−2) = −1.375 < 0 (−1, 0): f (−0.5) ≈ 1.4545 > 0 " ! 2 : f (0.1) ≈ −0.4755 < 0 0, 7 ! " 2 7 , : f (1) = 0.5 > 0 7 2 " ! 7 , ∞ : f (4) ≈ −0.0308 < 0 2

Function values are positive on (−1, 0) and " ! 2 7 , . solution set is (−1, 0) ∪ 7 2 60.

x2 − 5x + 3x − 3 − 2x2 − 10x = 0 −x2 − 12x − 3 = 0 x2 + 12x + 3 = 0

!

√ Using the quadratic formula,√we find that x√= −6 ± 33. The critical values are −6 − 33, −5, −6 + 33, 1, and √ 5. 33), They divide the x-axis into the intervals (−∞, −6 − √ √ √ (−6 − 33, −5), (−5, −6 + 33), (−6 + 33, 1), (1, 5), and (5, ∞). We test a value in each interval. √ (−∞, −6 − 33): f (−12) ≈ 0.00194 > 0 √ (−6 − 33, −5): f (−6) ≈ −0.4286 < 0 √ (−5, −6 + 33): f (−1) ≈ 0.16667 > 0 √ (−6 + 33, 1): f (0) = −0.12 < 0

" 2 7 , . The 7 2

x2 − x − 2 <0 x2 + 5x + 6

(1, 5): f (2) ≈ 1.4762 > 0

x2 − x − 2 <0 (x + 3)(x + 2)

(5, ∞): f (6) ≈ −2.018 < 0

√ Function √ values are negative on (−6 − 33, −5), (−6 + 33, 1), and (5, ∞).√Note that since the inequality √ symbol √ is ≤ and f (−6 ± 33) = 0, then −6 − 33 and −6 + 33 must be included in the solution set. Note also that since −5, 1, and 5 are not in the domain of√ f (x), they are not√included in the solution set. It is [−6 − 33, −5) ∪ [−6 + 33, 1) ∪ (5, ∞).

x2 − x − 2 is 0 when (x + 3)(x + 2) x = −3 or x = −2, so the function is not defined for these values of x. We solve the related equation f (x) = 0. The denominator of f (x) =

x2 − x − 2 =0 (x + 3)(x + 2) (x − 2)(x + 1) = 0

x = 2 or x = −1

The critical values are −3, −2, −1, and 2. They divide the x-axis into the intervals (−∞, −3), (−3, −2), (−2, −1), (−1, 2), and (2, ∞). We test a value in each interval. (−∞, −3): f (−4) = 9 > 0

(−3, −2): f (−2.5) = −27 < 0

(−2, −1): f (−1.5) ≈ 2.3333 > 0 (−1, 2): f (0) ≈ −0.3333 < 0 (2, ∞): f (3) ≈ 0.13333 > 0

x 3 + ≤ x2 +4x−5 x2 −25 2x x2 −6x+5 x 3 2x + − ≤0 x2 +4x−5 x2 −25 x2 −6x+5 x 3 2x + − ≤0 (x+5)(x−1) (x+5)(x−5) (x−5)(x−1) The denominator of x 3 2x f (x) = + − (x+5)(x−1) (x+5)(x−5) (x−5)(x−1) is 0 when x = −5, 1, or 5, so the function is not defined for these values of x. We solve the related equation f (x) = 0. 3 2x x + − =0 (x+5)(x−1) (x+5)(x−5) (x−5)(x−1) x(x − 5) + 3(x − 1) − 2x(x + 5) = 0

Function values are negative on (−3, −2) and (−1, 2). The solution set is (−3, −2) ∪ (−1, 2).

62.

x 2x + ≥ x2 −9 x2 +x−12 3x x2 +7x+12 x 3x 2x + − ≥0 (x+3)(x−3) (x+4)(x−3) (x+4)(x+3) The denominator of f (x) = 2x x 3x + − (x + 3)(x − 3) (x + 4)(x − 3) (x + 4)(x + 3) is 0 when x = −3, 3, or −4, so the function is not defined for these values of x. We solve the related equation f (x) = 0.

Exercise Set 3.6

243

2x x 3x + − =0 (x+3)(x−3) (x+4)(x−3) (x+4)(x+3) 2x(x + 4) + x(x + 3) − 3x(x − 3) = 0 2x + 8x + x + 3x − 3x + 9x = 0 2

2

2

20x = 0 x=0

The critical values are −4, −3, 0, and 3. They divide the x-axis into the intervals (−∞, −4), (−4, −3), (−3, 0), (0, 3), and (3, ∞). We test a value in each interval.

500t − 40 = 0 2t2 + 9 500t − 80t2 − 360 = 0

Multiplying by 2t2 + 9

Using the√quadratic formula, we find that 25 ± 337 ; that is, t ≈ 0.830 or t ≈ 5.420. These t= 8 numbers divide the t-axis into the intervals (−∞, 0.830), (0.830, 5.420), and (5.420, ∞). We test a value in each interval.

(−∞, −4): f (−5) = 6.25 > 0

(−∞, 0.830): f (0) = −40 < 0

(−3, 0): f (−1) ≈ 0.83333 > 0

(5.420, ∞): f (6) ≈ −2.963 < 0

(−4, −3): f (−3.5) ≈ −43.08 < 0 (0, 3): f (1) = −0.5 < 0

(3, ∞): f (4) ≈ 1.4286 > 0

Function values are positive on (−∞, −4), (−3, 0), and (3, ∞). Note that since the inequality symbol is ≥ and f (0) = 0, then 0 must be included in the solution set. Note also that since −4, −3, and 3 are not in the domain of f (x), they are not included in the solution set. It is (−∞, −4) ∪ (−3, 0] ∪ (3, ∞). 63. We write and solve a rational inequality. 4t + 98.6 > 100 t2 + 1 4t − 1.4 > 0 2 t +1 4t The denominator of f (t) = 2 − 1.4 has no realt +1 number zeros. We solve the related equation f (t) = 0. 4t − 1.4 = 0 t2 + 1 4t − 1.4(t2 + 1) = 0 Multiplying by t2 + 1 4t − 1.4t2 − 1.4 = 0

Using the √ quadratic formula, we find that 4 ± 8.16 t= ; that is, t ≈ 0.408 or t ≈ 2.449. These 2.8 numbers divide the t-axis into the intervals (−∞, 0.408), (0.408, 2.449), and (2.449, ∞). We test a value in each interval. (−∞, 0.408): f (0) = −1.4 < 0 (0.408, 2.449): f (1) = 0.6 > 0 (2.449, ∞): f (3) = −0.2 < 0

Function values are positive on (0.408, 2.449). The solution set is (0.408, 2.449). 64. We write and solve a rational inequality. 500t ≥ 40 2t2 + 9 500t − 40 ≥ 0 2t2 + 9 500t − 40 has no realThe denominator of f (t) = 2 2t + 9 number zeros. We solve the related equation f (t) = 0.

(0.830, 5.420): f (1) ≈ 5.4545 > 0 Function values are positive on (0.830, 5.420). The solution set is [0.830, 5.420]. 65. a) We write and solve a polynomial inequality. −3x2 + 630x − 6000 > 0

(x ≥ 0)

We first solve the related equation. −3x2 + 630x − 6000 = 0

x2 − 210x + 2000 = 0

Dividing by −3

(x − 10)(x − 200) = 0

Factoring

Using the principle of zero products or by observing the graph of y = −3x2 + 630 − 6000, we see that the solutions of the related equation are 10 and 200. These numbers divide the x-axis into the intervals (−∞, 10), (10, 200), and (200, ∞). Since we are restricting our discussion to nonnegative values of x, we consider the intervals [0, 10), (10, 200), and (200, ∞). We let f (x) = −3x2 + 630x − 6000 and test a value in each interval. [0, 10): f (0) = −6000 < 0

(10, 200): f (11) = 567 > 0 (200, ∞): f (201) = −573 < 0 Function values are positive only on (10, 200). The solution set is {x|10 < x < 200}, or (10, 200).

b) From part (a), we see that function values are negative on [0, 10) and (200, ∞). Thus, the solution set is {x|0 < x < 10 or x > 200}, or (0, 10) ∪ (200, ∞). 66. a) We write and solve a polynomial inequality. −16t2 + 32t + 1920 > 1920 −16t2 + 32t > 0

−16t2 + 32t = 0 −16t(t − 2) = 0

Related equation Factoring

The solutions of the related equation are 0 and 2. This could also be determined from the graph of y = −16x2 + 32x. These numbers divide the taxis into the intervals (−∞, 0), (0, 2), and (2, ∞). Since only nonnegative values of t have meaning in this application, we restrict our discussion to the intervals (0, 2) and (2, ∞). We let f (x) = −16t2 + 32t and test a value in each interval.

244

Chapter 3: Polynomial and Rational Functions (0, 2): f (1) = 16 > 0

n = −20 or n = 23

(2, ∞): f (3) = −48 < 0

Function values are positive on (0, 2). The solution set is {t|0 < t < 2}, or (0, 2).

b) We write and solve a polynomial inequality. 2

−16t + 32t + 1280 = 0 2

Related equation

2

t2 − 2t − 80 = 0

(t − 10)(t + 8) = 0

The solutions of the related equation are 10 and −8. These numbers divide the t-axis into the intervals (−∞, −8), (−8, 10), and (10, ∞). As in part (a), we will not consider negative values of t. In addition, note that the nonnegative solution of S(t) = 0 is 12. This means that the object reaches the ground in 12 sec. Thus, we also restrict our discussion to values of t such that t ≤ 12. We consider the intervals [0, 10) and (10, 12]. We let f (x) = −16t2 + 32t + 1280 and test a value in each interval. [0, 10): f (0) = 1280 > 0 (10, 12]: f (11) = −304 < 0

Function values are negative on (10, 12]. The solution set is {x|10 < x ≤ 12}, or (10, 12]. 67. We write an inequality. n(n − 3) ≤ 230 27 ≤ 2 54 ≤ n(n − 3) ≤ 460 Multiplying by 2 54 ≤ n2 − 3n ≤ 460

We write this as two inequalities. 54 ≤ n2 − 3n

and

Solve each inequality.

n2 − 3n ≤ 460

n2 − 3n ≥ 54

n − 3n − 54 ≥ 0 2

n2 − 3n − 54 = 0

Related equation

Since only positive values of n have meaning in this application, we consider the intervals (0, 9) and (9, ∞). Let f (n) = n2 − 3n − 54 and test a value in each interval.

(0, 9): f (1) = −56 < 0

(9, ∞): f (10) = 16 > 0

Function values are positive on (9, ∞). Since the inequality symbol is ≥, 9 must also be included in the solution set for this portion of the inequality. It is {n|n ≥ 9}. Now solve the second inequality. n2 − 3n ≤ 460

(n + 20)(n − 23) = 0

The solution set of the original inequality is {n|n ≥ 9 and 0 < n ≤ 23}, or {n|9 ≤ n ≤ 23}. 68. We write an inequality. n(n − 1) ≤ 300 66 ≤ 2 132 ≤ n2 − n ≤ 600

We write this as two inequalities. 132 ≤ n2 − n and n2 − n ≤ 600

Solve each inequality.

n2 − n ≥ 132

n − n − 132 ≥ 0 2

n2 − n − 132 = 0

Related equation

(n + 11)(n − 12) = 0

n = −11 or n = 12

Since only positive values of n have meaning in this application, we consider the intervals (0, 12) and (12, ∞). Let f (n) = n2 − n − 132 and test a value in each interval. (0, 12): f (1) = −132 < 0 (12, ∞): f (13) = 24 > 0

Function values are positive on (12, ∞). Since the inequality symbol is ≥, 12 must also be included in the solution set for this portion of the inequality. It is {n|n ≥ 12}. n2 − n ≤ 600

n2 − n − 600 ≤ 0 n2 − n − 600 = 0

n = −6 or n = 9

n2 − 3n − 460 = 0

Function values are negative on (0, 23). Since the inequality symbol is ≤, 23 must also be included in the solution set for this portion of the inequality. It is {n|0 < n ≤ 23}.

Now solve the second inequality.

(n + 6)(n − 9) = 0

n2 − 3n − 460 ≤ 0

(0, 23): f (1) = −462 < 0 (23, ∞): f (24) = 44 > 0

−16t + 32t + 1920 < 640 2

−16t + 32t + 1280 < 0

We consider only positive values of n as above. Thus, we consider the intervals (0, 23) and (23, ∞). Let f (n) = n2 − 3n − 460 and test a value in each interval.

Related equation

Related equation

(n + 24)(n − 25) = 0

n = −24 or n = 25

We consider only positive values of n as above. Thus, we consider the intervals (0, 25) and (25, ∞). Let f (n) = n2 − n − 600 and test a value in each interval. (0, 25): f (1) = −600 < 0 (25, ∞): f (26) = 50 > 0

Function values are negative on (0, 25). Since the inequality symbol is ≤, 25 must also be included in the solution set for this portion of the inequality. It is {n|0 < n ≤ 25}. The solution set of the original inequality is {n|n ≥ 12 and 0 < n ≤ 25}, or {n|12 ≤ n ≤ 25}. 69. Left to the student

Exercise Set 3.6

245

70. We need to know these values because they cannot be included in the solution set. The sign of the function often changes at these values as well. 71. A quadratic inequality ax2 +bx+c ≤ 0, a > 0, or ax2 +bx+ c ≥ 0, a < 0, has a solution set that is a closed interval. 7 7/2 = 2 4

72. r =

! "2 7 (x − 0) + [y − (−3)] = 4 49 x2 + (y + 3)2 = 16 2

73.

2

(x − h)2 + (y − k)2 = r2

[x − (−2)]2 + (y − 4)2 = 32 (x + 2)2 + (y − 4)2 = 9

74. g(x) = x2 − 10x + 2 −10 b =− =5 a) − 2a 2·1 g(5) = 52 − 10 · 5 + 2 = −23 The vertex is (5, −23).

b) Minimum: −23 at x = 5

c) [−23, ∞)

75. h(x) = −2x2 + 3x − 8 b 3 3 − =− = a) 2a 2(−2) 4 ! " ! "2 3 3 55 3 = −2 +3· −8=− h 4 4 4 8 ! " 3 55 The vertex is ,− . 4 8

76.

b) The coefficient of x2 is negative, so there is a maximum value. It is the second coordinate of 55 3 the vertex, − . It occurs at x = . 8 4 ! $ 55 c) The range is − ∞, − . 8 x4 − 6x2 + 5 > 0

x4 − 6x2 + 5 = 0 Related equation

(x − 1)(x2 − 5) = 0 √ x = ±1 or x = ± 5 2

Let f (x) = x4 − 6x2 + 5 and test a value in each of the intervals determined by the solutions of the related equation. √ (−∞, − 5): f (−3) = 32 > 0 √ (− 5, −1): f (−2) = −3 < 0 (−1, 1): f (0) = 5 > 0 √ (1, 5): f (2) = −3 < 0 √ ( 5, ∞): f (3) = 32 > 0

√ √ The solution set is (−∞, − 5) ∪ (−1, 1) ∪ ( 5, ∞).

77.

x4 + 3x2 > 4x − 15

x4 + 3x2 − 4x + 15 > 0

The graph of y = x4 + 3x2 − 4x + 15 lies entirely above the x-axis. Thus, the solution set is the set of all real numbers, or (−∞, ∞). & & &x + 3& & & 78. &<2 & &x − 4& x+3 −2 < <2 x−4 x+3 x+3 and <2 −2 < x−4 x−4 x+3 First solve −2 < . x−4 x+3 +2>0 x−4 x+3 + 2 is 0 when x = 4, so The denominator of f (x) = x−4 the function is not defined for this value of x. Now solve the related equation. x+3 +2 = 0 x−4 x + 3 + 2(x − 4) = 0 Multiplying by x − 4 x + 3 + 2x − 8 = 0

3x − 5 = 0 5 x= 3 5 The critical values are and 4. Test a value in each of the 3 intervals determined by them. " ! 5 : f (0) = 1.25 > 0 − ∞, 3 " ! 5 , 4 : f (2) = −0.5 < 0 3 (4, ∞): f (5) = 10 > 0

The solution set for this portion of the inequality is " ! 5 ∪ (4, ∞). − ∞, 3 x+3 x+3 < 2, or − 2 < 0. Next solve x−4 x−4 x+3 − 2 is 0 when x = 4, so The denominator of f (x) = x−4 the function is not defined for this value of x. Now solve the related equation. x+3 −2 = 0 x−4 x + 3 − 2(x − 4) = 0 Multiplying by x − 4 x + 3 − 2x + 8 = 0 −x + 11 = 0

x = 11

The critical values are 4 and 11. Test a value in each of the intervals determined by them. (−∞, 4): f (0) = −2.75 < 0 (4, 11): f (5) = 6 > 0

246

Chapter 3: Polynomial and Rational Functions (11, ∞): f (12) = −0.125 < 0

The solution set for this portion of the inequality is (−∞, 4) ∪ (11, ∞).

The solution set of the original inequality is ( '! " ) * 5 ∪ (4, ∞) and (−∞, 4) ∪ (11, ∞) , or − ∞, 3 ! " 5 ∪ (11, ∞). − ∞, 3

79. |x2 − 5| = |5 − x2 | = 5 − x2 when 5 − x2 ≥ 0. Thus we solve 5 − x2 ≥ 0. 5 − x2 ≥ 0 5 − x2 = 0

Related equation

5 = x2

√ ± 5=x

80.

Let f (x) = 5 − x2 and test a value in each of the intervals determined by the solutions of the related equation. √ (−∞, − 5): f (−3) = −4 < 0 √ √ (− 5, 5): f (0) = 5 > 0 √ ( 5, ∞): f (3) = −4 < 0 √ √ Function values are positive on (− 5, 5). Since the inequality symbol is ≥, the endpoints +of the must √ interval √ , be included in the solution set. It is − 5, 5 . (7 − x)−2 < 0 1 <0 (7 − x)2

1 >0 (7 − x)2 1 . Thus, for all values of x in the domain of f (x) = (7 − x)2 the solution set is ∅. Since (7−x)2 ≥ 0 for all real numbers x, then

81.

2|x|2 − |x| + 2 ≤ 5 2|x|2 − |x| − 3 ≤ 0

2|x|2 − |x| − 3 = 0 Related equation

(2|x| − 3)(|x| + 1) = 0 Factoring 2|x| − 3 = 0 or |x| + 1 = 0 3 or |x| = −1 |x| = 2

3 3 or x = . 2 2 The second equation has no solution. Let f (x) = 2|x|2 − |x| − 3 and test a value in each interval determined by the solutions of the related equation. ! " 3 − ∞, − : f (−2) = 3 > 0 2 " ! 3 3 : f (0) = −3 < 0 − , 2 2 ! " 3 , ∞ : f (2) = 3 > 0 2 ! " 3 3 . Since the inFunction values are negative on − , 2 2 equality symbol is ≤, the endpoints of the $ must % interval 3 3 also be included in the solution set. It is − , . 2 2 The solution of the first equation is x = −

& & & & & &1& 1& 82. &&2 − && ≤ 2 + && && x x 1 Note that is not defined when x = 0. Thus, x $= 0. Also x 1 1 1 note that 2 − = 0 when x = . The numbers 0 and x 2 2 " ! 1 , and divide the x-axis into the intervals (−∞, 0), 0, 2 ! " 1 , ∞ . Find the solution set of the inequality for each 2 interval. Then find the union of the three solution sets. 1 1 If x < 0, then 2 − > 0 and < 0, so x x & & & & & & & & &2 − 1 & = 2 − 1 and & 1 & = − 1 . & &x& x& x x 1 1 Solve: x < 0 and 2 − ≤ 2 − x x x < 0 and 2≤2

The solution set for this interval is (−∞, 0). 1 1 1 If 0 < x < , then 2 − < 0 and > 0, so 2 x x & & & & " ! & & & & &2 − 1 & = − 2 − 1 = −2 + 1 and & 1 & = 1 . & &x& x & x x x 1 1 1 and −2 + ≤ 2 + Solve: 0 < x < 2 x x 1 0<x< and −2 ≤ 2 2 " ! 1 . The solution set for this interval is 0, 2 1 1 1 If x ≥ , then 2 − > 0 and > 0, so 2 x x & & & & & & & & &2 − 1 & = 2 − 1 and & 1 & = 1 . &x& x & x& x 1 1 1 Solve: x ≥ and 2 − ≤ 2 + 2 x x 1 1 1 x≥ and − ≤ 2 x x 1 x≥ and −1 ≤ −1 2 " % 1 ,∞ . Then the solution set for this interval is 2 Then the solution set of the original inequality is " % " ! 1 1 ∪ , ∞ , or (−∞, 0) ∪ (0, ∞). (−∞, 0) ∪ 0, 2 2 & & & 1 && & 83. &1 + & < 3 & x& 1 −3 < 1 + < 3 x 1 1 −3 < 1 + and 1 + < 3 x x 1 First solve −3 < 1 + . x 1 1 +4>0 0 < 4 + , or x x

Exercise Set 3.6

247

1 The denominator of f (x) = + 4 is 0 when x = 0, so the x function is not defined for this value of x. Now solve the related equation. 1 +4 = 0 x 1 + 4x = 0 Multiplying by x 1 x=− 4 1 The critical values are − and 0. Test a value in each of 4 the intervals determined by them. " ! 1 ∞, − : f (−1) = 3 > 0 4 " ! 1 − , 0 : f (−0.1) = −6 < 0 4

Let f (x) = 6 + 5x − x2 and test a value in each of the intervals determined by the solution of the related equation. (−∞, −1): f (−2) = −8 < 0 (−1, 6): f (0) = 6 > 0

(6, ∞): f (7) = −8 < 0

The solution set for this portion of the inequality is (−∞, −1] ∪ [6, ∞). Next solve 1 + 5x − x2 ≥ 5. 1 + 5x − x2 ≥ 5 −4 + 5x − x2 ≥ 0

x2 − 5x + 4 ≤ 0 x2 − 5x + 4 = 0

(x − 1)(x − 4) = 0

x = 1 or x = 4

(0, ∞): f (1) = 5 > 0

The solution " set for this portion of the inequality is ! 1 ∪ (0, ∞). − ∞, − 4 1 1 Next solve 1 + < 3, or − 2 < 0. The denominator of x x 1 f (x) = − 2 is 0 when x = 0, so the function is not dex fined for this value of x. Now solve the related equation. 1 −2 = 0 x 1 − 2x = 0 Multiplying by x 1 x= 2 1 The critical values are 0 and . Test a value in each of the 2 intervals determined by them. (−∞, 0): f (−1) = −3 < 0 " ! 1 : f (0.1) = 8 > 0 0, 2 " ! 1 , ∞ : f (1) = −1 < 0 2 The solution ! set"for this portion of the inequality is 1 (−∞, 0) ∪ ,∞ . 2 The solution set of the original inequality is ( ' '! " ! "( 1 1 ∪ (0, ∞) and (−∞, 0) ∪ ,∞ , − ∞, − 4 2 ! " ! " 1 1 ∪ ,∞ . or − ∞, − 4 2 84. |1 + 5x − x2 | ≥ 5

1 + 5x − x2 ≤ −5 or 1 + 5x − x2 ≥ 5

First solve 1 + 5x − x2 ≤ −5. 1 + 5x − x2 ≤ −5 6 + 5x − x ≤ 0 2

6 + 5x − x2 = 0

(6 − x)(1 + x) = 0

x = 6 or x = −1

Related equation

Related equation

Let f (x) = x2 − 5x + 4 and test a value in each of the intervals determined by the solution of the related equation. (−∞, 1): f (0) = 4 > 0 (1, 4): f (2) = −2 < 0 (4, ∞): f (5) = 4 > 0

The solution set for this portion of the inequality is [1, 4]. The solution set of *the original inequality is ) (−∞, −1] ∪ [6, ∞) or [1, 4], or (−∞, −1] ∪ [1, 4] ∪ [6, ∞). 85.

|x2 + 3x − 1| < 3

−3 < x2 + 3x − 1 < 3

−3 < x2 + 3x − 1 and x2 + 3x − 1 < 3

First solve −3 < x2 + 3x − 1, or x2 + 3x − 1 > −3. x2 + 3x − 1 > −3 x2 + 3x + 2 > 0 x2 + 3x + 2 = 0

Related equation

(x + 2)(x + 1) = 0 x = −2 or x = −1

Let f (x) = x2 + 3x + 2 and test a value in each of the intervals determined by the solution of the related equation. (−∞, −2): f (−3) = 2 > 0

(−2, −1): f (−1.5) = −0.25 < 0 (−1, ∞): f (0) = 2 > 0

The solution set for this portion of the inequality is (−∞, −2) ∪ (−1, ∞). Next solve x2 + 3x − 1 < 3. x2 + 3x − 1 < 3 x2 + 3x − 4 < 0 x2 + 3x − 4 = 0

Related equation

(x + 4)(x − 1) = 0

x = −4 or x = 1

Let f (x) = x2 + 3x − 4 and test a value in each of the intervals determined by the solution of the related equation. (−∞, −4): f (−5) = 6 > 0

248

Chapter 3: Polynomial and Rational Functions (−4, 1): f (0) = −4 < 0

4.

(1, ∞): f (2) = 6 > 0

The solution set for this portion of the inequality is (−4, 1). The solution set of the ) * original inequality is (−∞, −2) ∪ (−1, ∞) and (−4, 1), or (−4, −2) ∪ (−1, 1).

86. First find the polynomial with solutions −4, 3, and 7. (x + 4)(x − 3)(x − 7) = 0

5.

2

(−4, 3): (0 + 4)(0 − 3)(0 − 7) = 84 > 0

The variation constant is 4. The equation of variation is y = 4x. 6.

(7, ∞): (8 + 4)(8 − 3)(8 − 7) = 60 > 0

Then a polynomial inequality for which the solution set is [−4, 3] ∪ [7, ∞) is x3 − 6x2 − 19x + 84 ≥ 0. Answers may vary. 87. First find a quadratic equation with solutions −4 and 3. (x + 4)(x − 3) = 0

7.

x2 + x − 12 = 0 Test a point in each of the three intervals determined by −4 and 3. (−∞, −4): (−5 + 4)(−5 − 3) = 8 > 0 (−4, 3): (0 + 4)(0 − 3) = −12 < 0 (3, ∞): (4 + 4)(4 − 3) = 8 > 0

Then a quadratic inequality for which the solution set is (−4, 3) is x2 + x − 12 < 0. Answers may vary.

Exercise Set 3.7 1.

9 The variation constant is , or 4.5. The equation of vari2 9 ation is y = x, or y = 4.5x. 2 y = kx 2. 0.1 = k(0.2) 1 = k Variation constant 2 1 Equation of variation: y = x, or y = 0.5x. 2 k 3. y = x k 3= 12 36 = k The variation constant is 36. The equation of variation is 36 . y= x

k x k 0.1 = 0.5 0.05 = k Variation constant 0.05 Equation of variation: y = x y=

k x k 32 = 1 8 1 · 32 = k 8 4=k y=

The variation constant is 4. The equation of variation is 4 y= . x 8.

y = kx 3 = k · 33 1 = k Variation constant 11 1 Equation of variation: y = x 11

y = kx 54 = k · 12 9 54 = k, or k = 12 2

1 4

4=k

(−∞, −4): (−5 + 4)(−5 − 3)(−5 − 7) = −96 < 0

(3, 7): (4 + 4)(4 − 3)(4 − 7) = −24 < 0

y = kx 1 = k·

x − 6x − 19x + 84 = 0 Test a point in each of the four intervals determined by −4, 3, and 7. 3

k x k 12 = 5 60 = k Variation constant 60 Equation of variation: y = x y=

9.

y 3 4 1 3 · 2 4 3 8

= kx = k·2 =k =k

3 The variation constant is . The equation of variation is 8 3 y = x. 8 10.

k x 1 k = 5 35 7 = k Variation constant 7 Equation of variation: y = x y=

Exercise Set 3.7 11.

249

k x k 1.8 = 0.3 0.54 = k

16.

y=

The variation constant is 0.54. The equation of variation 0.54 . is y = x 12.

y = kx

0.9 = k(0.4) 9 = k Variation constant 4 9 Equation of variation: y = x, or y = 2.25x 4 k 13. T = T varies inversely as P . P k Substituting 5= 7 35 = k Variation constant 35 P 35 T = 10 T = 3.5

T =

9600 L 9600 W = 14 5 W = 685 7 W =

Equation of variation Substituting

It will take 10 bricklayers 3.5 hr to complete the job. 14.

17.

1 w 2 1 F = · 180 2

Substituting Variation constant Equation of variation Substituting

5 kg, or about 686 kg. 7

Substituting 29 =k Variation constant 19, 227, 088 29 P N= 19, 227, 088 29 N = · 8, 541, 221 Substituting 19, 227, 088 N ≈ 13

North Carolina has 13 representatives. 19. F varies directly as w. Substituting Solving for k Variation constant

20.

F = 90 The maximum daily fat intake for a person weighing 180 lb is 90 g.

M varies directly as E.

A 100-lb person would weigh 40 lb on Mars.

Equation of variation Substituting

M = kE

38 = k · 95 Substituting 2 =k Variation constant 5 2 M = E Equation of variation 5 2 M = · 100 Substituting 5 M = 40

15. Let F = the number of grams of fat and w = the weight.

F =

W varies inversely as L.

29 = k · 19, 227, 088

A=

60 = k · 120 60 = k, or 120 1 =k 2

5 hr 7

N = kP

18.

A = kG

F = kw

or 5

A 14-m beam can support 685

9.66 = k · 9 161 =k 150 161 G 150 161 A= ·4 150 A ≈ $4.29

k r k 5= 80 400 = r 400 t= r 400 t= 70 40 , t= 7 k W = L k 1200 = 8 9600 = k t=

t= 45 = 27, 000 = t= t= t=

k r k 600 k 27, 000 r 27, 000 1000 27 min

250 21.

Chapter 3: Polynomial and Rational Functions d = km

d varies directly as m.

40 = k · 3 Substituting 40 =k Variation constant 3 40 m Equation of variation 3 40 200 d= ·5= Substituting 3 3 2 d = 66 3

The equation of variation is y = 26.

d=

27.

f = 0.042(80)

28.

Variation constant

550W = 1056 W =

1056 550

W = 1.92

y= 29.

Multiplying by W Dividing by 550 Simplifying

25.

B = kP 4445 = k · 9880 889 =k 1976 889 B= P 1976 889 B= · 74, 650 1976 B ≈ 33, 585 lb k y= 2 x k 0.15 = Substituting (0.1)2 k 0.15 = 0.01 0.15(0.01) = k 0.0015 = k

y = kxz Substituting

56 = 56k 1=k

The equation of variation is y = xz. kx 30. y = z k · 12 4= 15 5=k

A tone with a pitch of 550 vibrations per second has a wavelength of 1.92 ft. 24.

2 2 x 3

56 = k · 7 · 8

Equation of variation Substituting

y = kx2 6 = k · 32 2 =k 3

f = 3.36 k P varies inversely as W . P = W k Substituting 330 = 3.2

1056 W 1056 550 = W

Substituting

The equation of variation is y = 15x2 .

f = 0.042F

P =

y = kx2 0.15 = 0.01k 0.15 =k 0.01 15 = k

6.3 = k · 150

1056 = k

54 x2

0.15 = k(0.1)2

0.042 = k

23.

k x2 k 6= 2 3 54 = k y=

y=

2 A 5-kg mass will stretch the spring 66 cm. 3 f = kF 22.

0.0015 . x2

y= 31.

5x z y = kxz 2

105 = k · 14 · 52

105 = 350k 105 =k 350 3 =k 10

Substituting

3 2 xz . The equation of variation is y = 10 xz 32. y = k · w 2·3 3 = k· 2 4 1=k y=

xz w

Exercise Set 3.7 33.

251

xz wp 3 · 10 3 =k 28 7·8 3 30 = k· 28 56 3 56 · =k 28 30 1 =k 5 y=k

Substituting

The equation of variation is y = 34.

xz w2 12 16 · 3 = k· 2 5 5 5 =k 4

36 = r

xz . 5wp

A car can travel 36 mph and still stop in 72 ft. k 38. W = 2 d k 220 = (3978)2 3, 481, 386, 480 = k

y = k·

y= 35.

Substitute 72 for d and find r. 1 2 72 = r 18 1296 = r2

3, 481, 386, 480 d2 3, 481, 386, 480 W = (3978 + 200)2 W ≈ 199 lb W =

5xz 4w2

kR I We first find k.

39. E =

k d2 k 90 = 2 5 k 90 = 25 2250 = k I =

k · 59 5.69 = 93.3 " ! 93.3 =k 5.69 59 9≈k

Substituting

d = 7.5 40.

222 = k · 37.8 · 40 37 =k 252 37 Av 252 37 · 51v 430 = 252 v ≈ 57.4 mph

110T P 110 · 30 V = 15 V = 220 cm3

D=

37.

Curt Schilling would have given up about 126 earned runs if he had pitched 200 innings. kT P k · 42 231 = 20 110 = k

D = kAv

V =

V =

d = kr2

200 = k · 602

93.3 59

9R . I Substitute 5.69 for E and 200 for I and solve for R. 9R 5.69 = 200 - 200 . 200 =R Multiplying by 5.69 9 9 126 ≈ R

d2 = 56.25

36.

Multiplying by

The equation of variation is E =

2250 The equation of variation is I = 2 . d Substitute 40 for I and find d. 2250 40 = 2 d 40d2 = 2250

The distance from 5 m to 7.5 m is 7.5 − 5, or 2.5 m, so it is 2.5 m further to a point where the intensity is 40 W/m2 .

Substituting

41. Let y(x) = kx2 . Then y(2x) = k(2x)2 = k · 4x2 = 4 · kx2 = 4 · y(x). Thus, doubling x causes y to be quadrupled.

Substituting

200 = 3600k 200 =k 3600 1 =k 18

The equation of variation is d =

k2 k2 k1 k2 , 42. Let y = k1 x and x = . Then y = k1 · , or y = z z z so y varies inversely as z.

1 2 r . 18

252

Chapter 3: Polynomial and Rational Functions

43.

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis.

y 4

Test for symmetry with respect to the y-axis:

2 2

!4 !2

4

x

!2

−2x − 5y = 0 Simplifying

!4

44. Test for symmetry with respect to the x-axis. y = 3x4 − 3

Original equation

−y = 3x4 − 3

Replacing y by −y

y = −3x4 + 3 Simplifying

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis. y = 3x4 − 3

Original equation

y = 3x4 − 3

Simplifying

y = 3(−x)4 − 3 Replacing x by −x

The last equation is equivalent to the original equation, so the graph is symmetric with respect to the y-axis. Test for symmetry with respect to the origin: y = 3x4 − 3

−y = 3(−x)4 − 3 Replacing x by −x and y by −y −y = 3x4 − 3 y = −3x4 + 3

Simplifying

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. 45. Test for symmetry with respect to the x-axis. y 2 = x Original equation (−y)2 = x Replacing y by −y y 2 = x Simplifying

The last equation is equivalent to the original equation, so the graph is symmetric with respect to the x-axis. Test for symmetry with respect to the y-axis: y2 = x

2x − 5y = 0 Original equation

2(−x) − 5y = 0 Replacing x by −x

Original equation

y 2 = −x Replacing x by −x

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Test for symmetry with respect to the origin: y2 = x

Original equation

(−y)2 = −x Replacing x by −x and y by −y y 2 = −x Simplifying

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the origin. 46. Test for symmetry with respect to the x-axis: 2x − 5y = 0 Original equation

2x − 5(−y) = 0 Replacing y by −y 2x + 5y = 0 Simplifying

The last equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis. Test for symmetry with respect to the origin: 2x − 5y = 0 Original equation

2(−x) − 5(−y) = 0 Replacing x by −x and y by −y −2x + 5y = 0 2x − 5y = 0 Simplifying

The last equation is equivalent to the original equation, so the graph is symmetric with respect to the origin. 47. Let V represent the volume and p represent the price of a jar of peanut butter. V = kp ! "2 3 π (5) = k(1.8) 2 6.25π = k V = 6.25πp π(1.625) (5.5) = 6.25πp 2

V varies directly as p. Substituting Variation constant Equation of variation Substituting

2.32 ≈ p

If cost is directly proportional to volume, the larger jar should cost $2.32. Now let W represent the weight and p represent the price of a jar of peanut butter. W = kp 18 = k(1.8)

Substituting

10 = k

Variation constant

W = 10p

Equation of variation

28 = 10p

Substituting

2.8 = p If cost is directly proportional to weight, the larger jar should cost $2.80. 48. a) 7xy = 14 2 y= x Inversely b) x − 2y = 12 x y = −6 2 Neither c) −2x + y = 0

y = 2x

Directly

Chapter 3 Review Exercises 3 y 4 4 y= x 3 Directly

1 3 x − 2x + 3 3 The degree of the polynomial is 3, so the polynomial is 1 cubic. The leading term is x3 and the leading coefficient 3 1 is . 3

d) x =

9. f (x) =

e) x = 2 y 1 y= x 2 Directly 49. We are told A = kd2 , and we know A = πr2 so we have: kd2 = πr2 kd2 = π

- d .2

πd 4 π k= 4

kd2 =

2

253

r=

d 2

2

Variation constant

kp2 q3 Q varies directly as the square of p and inversely as the cube of q.

50. Q =

Chapter 3 Review Exercises 1. f (−b) = (−b+a)(−b+b)(−b−c) = (−b+a)·0·(−b−c) = 0, so the statement is true. 2. The statement is true. See page 308 in the text.

1 10. f (x) = − x4 + 3x2 + x − 6 2 1 The leading term is − x4 . The degree, 4, is even and the 2 1 leading coefficient, − , is negative. As x → ∞, 2 f (x) → −∞, and as x → −∞, f (x) → −∞. 11. f (x) = x5 + 2x3 − x2 + 5x + 4

The leading term is x5 . The degree, 5, is odd and the leading coefficient, 1, is positive. As x → ∞, f (x) → ∞, and as x → −∞, f (x) → −∞. " ! 2 (x + 2)3 (x − 5)2 12. g(x) = x − 3 2 , multiplicity 1; 3 −2, multiplicity 3; 5, multiplicity 2

13.

f (x) = x4 − 26x2 + 25

= (x2 − 1)(x2 − 25)

= (x + 1)(x − 1)(x + 5)(x − 5)

±1, ±5; each has multiplicity 1 14.

h(x) = x3 + 4x2 − 9x − 36

= x2 (x + 4) − 9(x + 4) = (x + 4)(x2 − 9)

3. In addition to the given possibilities, 9 and −9 are also possible rational zeros. The statement is false. 4. The degree of P is 8, so the graph of P (x) has at most 8 x-intercepts. The statement is false. 5. The domain of the function is the set of all real numbers except −2 and 3, or {x|x $= −2 and x $= 3}. The statement is false. 6.

= (x + 4)(x + 3)(x − 3)

−4, ±3; each has multiplicity 1 15. A = P (1 + r)t a)

f (x) = 7x2 − 5 + 0.45x4 − 3x3

−1 ± 1.04 = r

= 0.45x4 − 3x3 + 7x2 − 5

−2.04 = r or 0.04 = r

The degree of the polynomial is 4, so the polynomial is quartic. The leading term is 0.45x4 and the leading coefficient is 0.45. 7. h(x) = −25

The degree of the polynomial is 0, so the polynomial is constant. The leading term is −25 and the leading coefficient is −25.

8.

g(x) = 6 − 0.5x

= −0.5x + 6

The degree of the polynomial is 1, so the polynomial is linear. The leading term is −0.5x and the leading coefficient is −0.5.

6760 = 6250(1 + r)2 6760 = (1 + r)2 6250 ±1.04 = 1 + r

Only 0.04 has meaning in this application. The interest rate is 0.04 or 4%. b)

1, 215, 506.25 = 1, 000, 000(1 + r)4 1, 215, 506.25 = (1 + r)4 1, 000, 000 ±1.05 = 1 + r −1 ± 1.05 = r

−2.05 = r or 0.05 = r

Only 0.05 has meaning in this application. The interest rate is 0.05 or 5%.

254

Chapter 3: Polynomial and Rational Functions

16. f (x) = −x4 + 2x3

1. The leading term is −x . The degree, 4, is even and the leading coefficient, −1, is negative so as x → ∞, f (x) → −∞ and as x → −∞, f (x) → −∞. 4

2. We solve f (x) = 0.

g(−3) = (−3 − 1)3 (−3 + 2)2 = −64 < 0

In (−2, 1), test 0:

g(0) = (0 − 1)3 (0 + 2)2 = −4 < 0 In (1, ∞), test 2:

−x4 + 2x3 = 0

−x3 (x − 2) = 0

−x3 = 0 or x − 2 = 0 x = 0 or

In (−∞, −2), test −3:

x=2

The zeros of the function are 0 and 2, so the xintercepts of the graph are (0, 0) and (2, 0). 3. The zeros divide the x-axis into 3 intervals, (−∞, 0), (0, 2), and (2, ∞). We choose a value for x from each interval and find f (x). This tells us the sign of f (x) for all values of x in that interval. In (−∞, 0), test −1:

g(2) = (2 − 1)3 (2 + 2)2 = 16 > 0 Thus the graph lies below the x-axis on (−∞, −2) and on (−2, 1) and above the x-axis on (1, ∞). We also know that the points (−3, −64), (0, −4), and (2, 16) are on the graph. 4. From Step 3 we know that g(0) = −4, so the yintercept is (0, −4).

5. We find additional points on the graph and then draw the graph. x

g(x)

f (−1) = −(−1)4 + 2(−1)3 = −3 < 0

−2.5 −10.7

f (1) = −1 + 2 · 1 = 1 > 0

−1

−8

−0.5

−7.6

0.5

−0.8

In (0, 2), test 1: 4

3

In (2, ∞), test 3:

f (3) = −34 + 2 · 33 = −27 < 0 Thus the graph lies below the x-axis on (−∞, 0) and on (2, ∞) and above the x-axis on (0, 2). We also know the points (−1, −3), (1, 1), and (3, −27) are on the graph. 4. From Step 2 we know that the y-intercept is (0, 0). 5. We find additional points on the graph and then draw the graph. x

f (x)

−2

−32

−0.5 −0.3125 0.5

0.1875

1.5

1.6875

6. Checking the graph as described on page 272 in the text, we see that it appears to be correct. 18. h(x) = x3 + 3x2 − x − 3

1. The leading term is x3 . The degree, 3, is odd and the leading coefficient, 1, is positive so as x → ∞, h(x) → ∞ and as x → −∞, h(x) → −∞.

2. We solve h(x) = 0.

x3 + 3x2 − x − 3 = 0

x (x + 3) − (x + 3) = 0 2

(x + 3)(x2 − 1) = 0

(x + 3)(x + 1)(x − 1) = 0

x+3 = 0 6. Checking the graph as described on page 272 in the text, we see that it appears to be correct. 17. g(x) = (x − 1)3 (x + 2)2

1. The leading term is x · x · x · x · x, or x5 . The degree, 5, is odd and the leading coefficient, 1, is positive so as x → ∞, g(x) → ∞ and as x → −∞, g(x) → −∞.

2. We see that the zeros of the function are 1 and −2, so the x-intercepts of the graph are (1, 0) and (−2, 0). 3. The zeros divide the x-axis into 3 intervals, (−∞, −2), (−2, 1), and (1, ∞). We choose a value for x from each interval and find g(x). This tells us the sign of g(x) for all values of x in that interval.

or x + 1 = 0

x = −3 or

or x − 1 = 0

x = −1 or

x=1

The zeros of the function are −3, −1, and 1, so the x-intercepts of the graph are (−3, 0), (−1, 0), and (1, 0). 3. The zeros divide the x-axis into 4 intervals, (−∞, −3), (−3, −1), (−1, 1), and (1, ∞). We choose a value for x from each interval and find h(x). This tells us the sign of h(x) for all values of x in that interval. In (−∞, −3), test −4: h(−4) = (−4)3 + 3(−4)2 − (−4) − 3 = −15 < 0 In (−3, −1), test −2:

h(−2) = (−2)3 + 3(−2)2 − (−2) − 3 = 3 > 0 In (−1, 1), test 0:

h(0) = 03 + 3 · 02 − 0 − 3 = −3 < 0

Chapter 3 Review Exercises In (1, ∞), test 2:

h(2) = 23 + 3 · 22 − 2 − 3 = 15 > 0 Thus the graph lies below the x-axis on (−∞, −3) and on (−1, 1) and above the x-axis on (−3, −1) and on (1, ∞). We also know the points (−4, −15), (−2, 3), (0, −3), and (2, 15) are on the graph.

4. From Step 3 we know that h(0) = −3, so the yintercept is (0, −3).

5. We find additional points on the graph and then draw the graph.

255 Thus the graph lies above the x-axis on (−∞, −1) and on (2, ∞) and below the x-axis on (−1, 2). We also know that the points (−2, 64), (0, −8), and (3, 4) are on the graph. 5. We find additional points on the graph and then draw the graph. x

f (x)

−1.5 21.4 −0.5 −7.8

x

h(x)

1

−2

−2.5

2.625

4

40

−0.5 −1.875 0.5

−1.875

1.5

5.625

6. Checking the graph as described on page 272 in the text, we see that it appears to be correct. 19. f (x) = x4 − 5x3 + 6x2 + 4x − 8

1. The leading term is x4 . The degree, 4, is even and the leading coefficient, 1, is positive so as x → ∞, f (x) → ∞ and as x → −∞, f (x) → −∞.

2. We solve f (x) = 0, or x4 − 5x3 + 6x2 + 4x − 8 = 0. The possible rational zeros are ±1, ±2, ±4, and ±8. We try −1. & −1 & 1 −5 6 4 −8 −1 6 −12 8 1 −6 12 −8 0

Now we have (x + 1)(x3 − 6x2 + 12x − 8) = 0. We use synthetic division to determine if 2 is a zero of x3 − 6x2 + 12x − 8 = 0. & 2 & 1 −6 12 −8 2 −8 8 1 −4 4 0 We have (x + 1)(x − 2)(x2 − 4x + 4) = 0, or (x + 1)(x − 2)(x − 2)2 = 0. Thus the zeros of f (x) are −1 and 2 and the x-intercepts of the graph are (−1, 0) and (2, 0).

3. The zeros divide the x-axis into 3 intervals, (−∞, −1), (−1, 2), and (2, ∞). We choose a value for x from each interval and find f (x). This tells us the sign of f (x) for all values of x in that interval. In (−∞, −1), test −2: f (−2) = (−2)4 − 5(−2)3 + 6(−2)2 + 4(−2) − 8 = 64 > 0

In (−1, 2), test 0: f (0) = 04 − 5 · 03 + 6 · 02 + 4 · 0 − 8 = −8 < 0 In (2, ∞), test 3:

f (3) = 34 − 5 · 33 + 6 · 32 + 4 · 3 − 8 = 4 > 0

6. Checking the graph as described on page 272 in the text, we see that it appears to be correct. 20. g(x) = 2x3 + 7x2 − 14x + 5

1. The leading term is x3 . The degree, 3, is odd and the leading coefficient, 2, is positive so as x → ∞, g(x) → ∞ and as x → −∞, h(x) → −∞. 2. We solve g(x) = 0, or 2x3 + 7x2 − 14x + 5 = 0. The 5 1 possible rational zeros are ±1, ±5, ± , and ± . 2 2 We try 1. & 1 & 2 7 −14 5 2 9 −5 2 9 −5 0 Now we have:

(x − 1)(2x2 + 9x − 5) = 0

(x − 1)(2x − 1)(x + 5) = 0

x − 1 = 0 or 2x − 1 = 0 or x + 5 = 0 x = 1 or x = 1 or

2x = 1 or 1 or x= 2

x = −5

x = −5

1 The zeros of the function are 1, , and −5, so the " 2 ! 1 x-intercepts are (1, 0), , 0 , and (−5, 0). 2 the"x-axis 3. The zeros divide ! into " 4 intervals, ! 1 1 , , 1 , and (1, ∞). We (−∞, −5), − 5, 2 2 choose a value for x from each interval and find g(x). This tells us the sign of g(x) for all values of x in that interval. In (−∞, −5), test −6: g(−6) = 2(−6)3 + 7(−6)2 − 14(−6) + 5 = −91 < 0 " ! 1 In − 5, , test 0: 2

g(0) = 2 · 03 + 7 · 02 − 14 · 0 + 5 = 5 > 0 " ! 3 1 , 1 , test : In 2 4 ! "3 ! " 2 ! " 3 3 3 3 =2 g +7 −14· +5 = −0.71875 < 0 4 4 4 4

256

Chapter 3: Polynomial and Rational Functions In (1, ∞), test 2:

g(2) = 2 · 2 + 7 · 2 − 14 · 2 + 5 = 21 > 0 Thus the (−∞, −5) " lies below the x-axis on ! " ! graph 1 1 and on , 1 and above the x-axis on − 5, 2 2 and on (1, ∞). We also know the points (−6, −91), (0, 5), (0.75, −0.71875), and (2, 21) are on the graph. 3

2

4. From Step 3 we know that g(0) = 5, so the yintercept is (0, 5).

5. We find additional points on the graph and then draw the graph. x

g(x)

−3

56

−1

24

1.5

6.5

3

80

f (1) = 4 · 12 − 5 · 1 − 3 = −4

f (2) = 4 · 22 − 5 · 2 − 3 = 3

By the intermediate value theorem, since f (1) and f (2) have opposite signs, f (x) has a zero between 1 and 2. 1 22. f (−1) = (−1) − 4(−1) + (−1) + 2 = −3.5 2 1 f (1) = 13 − 4 · 12 + · 1 + 2 = −0.5 2 Since f (−1) and f (1) have the same sign, the intermediate value theorem does not allow us to determine if there is a real zero between −1 and 1. 3

23.

2

6x2 + 16x + 52 x − 3 6x3 − 2x2 + 4x − 1 6x3 − 18x2 16x2 + 4x 16x2 − 48x 52x − 1 52x − 156 155 Q(x) = 6x2 + 16x + 52; R(x) = 155; P (x) = (x − 3)(6x2 + 16x + 52) + 155

24.

x3 − 3x2 + 3x − 2 x + 1 x − 2x3 + 0x2 + x + 5 x4 + x3 − 3x3 + 0x2 − 3x3 − 3x2 3x2 + x 3x2 + 3x − 2x + 5 − 2x − 2 7 4

P (x) = (x + 1)(x3 − 3x2 + 3x − 2) + 7 & 25. 5 & 1 2 −13 10 5 35 110 1 7 22 120 The quotient is x2 + 7x + 22; the remainder is 120. 26. (x4 + 3x3 + 3x2 + 3x + 2) ÷ (x + 2) =

(x4 + 3x3 + 3x2 + 3x + 2) ÷ [x − (−2)] & −2 & 1 3 3 3 2 −2 −2 −2 −2 1 1 1 1 0

The quotient is & 27. −1 & 1 0 −1 1 −1

6. Checking the graph as described on page 272 in the text, we see that it appears to be correct. 21.

Q(x) = x3 − 3x2 + 3x − 2; R(x) = 7;

x3 + x2 + x + 1; the remainder is 0. 0 0 −2 0 1 −1 1 1 1 −1 −1 1

The quotient is x4 − x3 + x2 − x − 1; the remainder is 1. & 28. −2 & 1 2 −13 10 −2 0 26 1 0 −13 36 f (−2) = 36 & 29. −2 & 1 0 0 0 −16 −2 4 −8 16 1 −2 4 −8 0

f (−2) = 0 & 30. −10 & 1 1

−4 1 −1 −10 140 −1410 −14 141 −1411

f (−10) = −141, 220 & 31. −i & 1 −5 1 −5 −i −1 + 5i 5 1 −5 − i 5i 0

2 −100 14, 110 −141, 120 14, 112 −141, 220

f (−i) = 0, so −i is a zero of f (x). & −5 & 1 −5 1 −5 −5 50 −255 1 −10 51 −260

f (−5) $= 0, so −5 is not a zero of f (x). & 32. −1 & 1 −4 −3 14 −8 −1 5 −2 −12 1 −5 2 12 −20 f (−1) $= 0, so −1 is not a zero of f (x). & −2 & 1 −4 −3 14 −8 −2 12 −18 8 1 −6 9 −4 0

f (−2) = 0, so −2 is a zero of f (x).

Chapter 3 Review Exercises 33.

1 3

& &

1

− 43

− 53

1 3

2 3 − 23

− 13

37. f (x) = x4 − 4x3 − 21x2 + 100x − 100

1 −1 −2 0 ! " 1 1 f = 0, so is a zero of f (x). 3 3 & 2 1 & 1 − 43 − 53 3 1

1

− 13 −2

− 13

−2 − 43

f (1) $= 0, so 1 is not a zero of f (x). & 34. 2 & 1 0 −5 0 6 2 4 −2 −4 1 2 −1 −2 2

Q(2) $= 0, so 2 is not a zero of f (x). √ & − 3&

1

6 √0 −5 √0 −√3 3 2√3 −6 1 − 3 −2 2 3 0 √ √ f (− 3) = 0, so − 3 is a zero of f (x). 35. f (x) = x3 + 2x2 − 7x + 4

Try x + 1 and x + 2. Using synthetic division we find that f (−1) $= 0 and f (−2) $= 0. Thus x + 1 and x + 2 are not factors of f (x). Try x + 4. & −4 & 1 2 −7 4 −4 8 −4 1 −2 1 0 Since f (−4) = 0, x + 4 is a factor of f (x). Thus f (x) = (x + 4)(x2 − 2x + 1) = (x + 4)(x − 1)2 .

Now we solve f (x) = 0. x+4 = 0

or (x − 1)2 = 0

x = −4 or

x−1 = 0

x = −4 or

x=1

The solutions of f (x) = 0 are −4 and 1. 36. f (x) = x3 + 4x2 − 3x − 18

Try x + 1, x − 1, and x + 2. Using synthetic division we find that f (−1) $= 0, f (1) $= 0, and f (−2) $= 0. Thus x + 1, x − 1, and x + 2 are not factors of f (x). Try x − 2. & 2 & 1 4 −3 −18 2 12 18 1 6 9 0 Since f (2) = 0, x − 2 is a factor of f (x). Thus f (x) = (x − 2)(x2 + 6x + 9) = (x − 2)(x + 3)2 .

Now solve f (x) = 0. (x − 2)(x + 3)2 = 0

x − 2 = 0 or (x + 3)2 = 0 x = 2 or

x = 2 or

257

x+3 = 0

x = −3

The solutions of f (x) = 0 are 2 and −3.

Using synthetic division we find that f (2) = 0: & 2 & 1 −4 −21 100 −100 2 −4 −50 100 1 −2 −25 50 0 Then we have: f (x) = (x − 2)(x3 − 2x2 − 25x + 50) = (x − 2)[x2 (x − 2) − 25(x − 2)] = (x − 2)(x − 2)(x2 − 25)

= (x − 2)2 (x + 5)(x − 5)

Now solve f (x) = 0.

(x − 2)2 = 0 or x + 5 = 0 x − 2 = 0 or x = 2 or

or x − 5 = 0

x = −5 or x = −5 or

x=5 x=5

The solutions of f (x) = 0 are 2, −5, and 5.

38. f (x) = x4 − 3x2 + 2 = x4 + 0 · x3 − 3x2 + 0 · x + 2 Try x + 1. & −1 & 1

0 −3 0 2 −1 1 2 −2 1 −1 −2 2 0

Since f (−1) = 0, (x + 1) is a factor of f (x). We have: f (x) = (x + 1)(x3 − x2 − 2x + 2)

= (x + 1)[x2 (x − 1) − 2(x − 1)]

= (x + 1)(x − 1)(x2 −√2) √ = (x + 1)(x − 1)(x + 2)(x − 2)

Now solve f (x) = 0. √ √ (x + 1)(x − 1)(x + 2)(x − 2) = 0 √ √ x+1=0 or x−1=0 or x+ 2=0 or x− 2=0 √ √ x= 2 x=−1 or x=1 or x=− 2 or √ √ The solutions of f (x) = 0 are −1, 1, − 2, and 2.

39. A polynomial function of degree 3 with −4, −1, and 2 as zeros has factors x + 4, x + 1, and x − 2 so we have f (x) = an (x + 4)(x + 1)(x − 2). The simplest polynomial is obtained if we let an = 1. f (x) = (x + 4)(x + 1)(x − 2) = (x2 + 5x + 4)(x − 2)

= x3 − 2x2 + 5x2 − 10x + 4x − 8

= x3 + 3x2 − 6x − 8

40. A polynomial function of degree 3 with −3, 1 + i, and 1 − i as zeros has factors x + 3, x − (1 + i), and x − (1 − i) so we have f (x) = an (x + 3)[x − (1 + i)][x − (1 − i)]. The simplest polynomial is obtained if we let an = 1. f (x) = (x + 3)[x − (1 + i)][x − (1 − i)] = (x + 3)[(x − 1) − i][(x − 1) + i)

= (x + 3)[(x − 1)2 − i2 ]

= (x + 3)(x2 − 2x + 1 + 1) = (x + 3)(x2 − 2x + 2)

= x3 − 2x2 + 2x + 3x2 − 6x + 6

= x3 + x2 − 4x + 6

258

Chapter 3: Polynomial and Rational Functions

√ 1 41. A polynomial function of degree 3 with , 1 − 2, and 2 √ √ 1 1 + 2 as zeros has factors x − , x − (1 − 2), and 2 √ x − (1 + !2) so we " have √ √ 1 f (x) = an x − [x − (1 − 2)][x − (1 + 2)]. 2 Let an = 1. ! " √ √ 1 f (x) = x − [x − (1 − 2)][x − (1 + 2)] 2 ! " √ √ 1 [(x − 1) + 2][(x − 1) − 2] = x− 2 " ! 1 = x− (x2 − 2x + 1 − 2) 2 ! " 1 = x− (x2 − 2x − 1) 2 1 1 = x3 − 2x2 − x − x2 + x + 2 2 5 1 = x3 − x2 + 2 2 If we let an = 2, we obtain f (x) = 2x3 − 5x2 + 1. 42. A polynomial function of degree 4 has at most 4 real zeros. Since 4 zeros are given, these are all of the zeros of the desired function. We proceed as in Exercise 39 above, letting an = 1. ! " 1 f (x) = (x + 5)3 x − 2 " ! 1 3 2 = (x + 15x + 75x + 125) x − 2 1 3 15 2 75 125 4 3 2 = x +15x +75x +125x− x − x − x− 2 2 2 2 29 135 175 125 = x4 + x3 + x2 + x− 2 2 2 2 If we had let an = 2, the result would have been f (x) = 2x4 + 29x3 + 135x2 + 175x − 125. 43. A polynomial function of degree 5 has at most 5 real zeros. Since 5 zeros are given, these are all of the zeros of the desired function. We proceed as in Exercise 39 above, letting an = 1. f (x) = (x + 3)2 (x − 2)(x − 0)2

= (x2 + 6x + 9)(x3 − 2x2 )

= x5 + 6x4 + 9x3 − 2x4 − 12x3 − 18x2 = x5 + 4x4 − 3x3 − 18x2

44. A polynomial function of degree 5 can have at most 5 zeros. Since f (x) has rational coefficients, in addition to √ the 3 given zeros, the other zeros are the conjugates of 5 and √ i, or − 5 and −i. 45. A polynomial function of degree 5 can have at most 5 zeros. Since f (x) has rational coefficients, in addition to the√3 given zeros, the other √ zeros √ are the conjugates of 1 + 3 √ and − 3, or 1 − 3 and 3. 46. A polynomial function of degree 5 can have at most 5 zeros. Since f (x) has rational coefficients, in addition √4 √ to the given zeros, the other zero is the conjugate of − 2, or 2.

√ 47. − 11 is also a zero. √ √ f (x) = (x − 11)(x + 11) = x2 − 11

48. i is also a zero. f (x) = (x + i)(x − i)(x − 6) = (x2 + 1)(x − 6)

= x3 − 6x2 + x − 6

49. 1 − i is also a zero. f (x) = (x + 1)(x − 4)[x − (1 + i)][x − (1 − i)] = (x2 − 3x − 4)(x2 − 2x + 2)

= x4 − 2x3 + 2x2 − 3x3 + 6x2 − 6x − 4x2 + 8x − 8

= x4 − 5x3 + 4x2 + 2x − 8 √ 50. − 5 and 2i are also zeros. √ √ f (x) = (x − 5)(x + 5)(x + 2i)(x − 2i) = (x2 − 5)(x2 + 4)

51.

= x4 − x2 − 20 " ! 1 f (x) = x − (x − 0)(x + 3) 3 " ! 1 = x2 − x (x + 3) 3 8 = x3 + x2 − x 3

52. h(x) = 4x5 − 2x3 + 6x − 12 Possibilities for p ±1, ±2, ±3, ±4, ±6, ±12 : Possibilities for q ±1, ±2, ±4

Possibilities for p/q: ±1, ±2, ±3, ±4, ±6, ±12, 1 3 1 3 ± ,± ,± ,± 2 2 4 4

53. g(x) = 3x4 − x3 + 5x2 − x + 1 Possibilities for p ±1 : Possibilities for q ±1, ±3 1 Possibilities for p/q: ±1, ± 3 54. f (x) = x3 − 2x2 + x − 24 Possibilities for p ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24 : Possibilities for q ±1 Possibilities for p/q: ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24

55. f (x) = 3x5 + 2x4 − 25x3 − 28x2 + 12x a) We know that 0 is a zero since f (x) = x(3x4 + 2x3 − 25x2 − 28x + 12).

Now consider g(x) = 3x4 + 2x3 − 25x2 − 28x + 12.

Possibilities for p/q: ±1, ±2, ±3, ±4, ±6, ±12, 1 2 4 ± ,± ,± 3 3 3 From the graph of y = 3x4 +2x3 −25x2 −28x+12, we 1 see that, of all the possibilities above, only −2, , 3 2 , and 3 might be zeros. We use synthetic division 3 to determine if −2 is a zero.

Chapter 3 Review Exercises & −2 &

3 3

2 −25 −28 12 −6 8 34 −12 −4 −17 6 0

Now try 3 in the quotient above. & 3 & 3 −4 −17 6 9 15 −6 3 5 −2 0

We have f (x) = (x + 2)(x − 3)(3x2 + 5x − 2).

We find the other zeros. 3x3 + 5x − 2 = 0

(3x − 1)(x + 2) = 0

3x − 1 = 0 or x + 2 = 0

3x = 1 or x = −2 1 or x = −2 x= 3 The rational zeros of g(x) = 3x4 +2x3 −25x2 −28x+ 1 12 are −2, 3, and . Since 0 is also a zero of f (x), 3 1 the zeros of f (x) are −2, 3, , and 0. (The zero −2 3 has multiplicity 2.) These are the only zeros. b) From our work above we see f (x) = x(x + 2)(x − 3)(3x − 1)(x + 2), or x(x + 2)2 (x − 3)(3x − 1).

56. f (x) = x3 − 2x2 − 3x + 6

a) Possibilities for p/q: ±1, ±2, ±3, ±6

From the graph of f (x), we see that −2 and 2 might be zeros. Synthetic division shows that −2 is not a zero. Try 2. & 2 & 1 −2 −3 6 2 0 −6 1 0 −3 0 f (x) = (x − 2)(x2 − 3) We find the other zeros. x2 − 3 = 0 x2 = 3

√ x=± 3

There√is only one rational zero, 2. The other zeros are ± 3. (Note that we could have used factoring by grouping to find this result.) √ √ b) f (x) = (x − 2)(x + 3)(x − 3) 57. f (x) = x4 − 6x3 + 9x2 + 6x − 10

a) Possibilities for p/q: ±1, ±2, ±5, ±10

From the graph of f (x), we see that −1 and 1 might be zeros. & −1 & 1 −6 9 6 −10 −1 7 −16 10 1 −7 16 −10 0 & & 1 1 −7 16 −10 1 −6 10 1 −6 10 0

f (x) = (x + 1)(x − 1)(x2 − 6x + 10)

259 Using the quadratic formula, we find that the other zeros are 3 ± i. The rational zeros are −1 and 1. The other zeros are 3 ± i.

b) f (x) = (x + 1)(x − 1)[x − (3 + i)][x − (3 − i)] = (x + 1)(x − 1)(x − 3 − i)(x − 3 + i)

58. f (x) = x3 + 3x2 − 11x − 5

a) Possibilities for p/q: ±1, ±5

From the graph of f (x), we see that −5 might be a zero.& −5 & 1 3 −11 −5 −5 10 5 1 −2 −1 0

f (x) = (x + 5)(x2 − 2x − 1) Use the quadratic formula to find the other zeros. x2 − 2x − 1 = 0 / −(−2) ± (−2)2 − 4 · 1 · (−1) x= 2·1 √ √ 2±2 2 2± 8 = = 2√ 2 = 1± 2

√ The rational zero is −5. The other zeros are 1 ± 2. √ √ b) f (x) = (x + 5)[x − (1 + 2)][x − (1 − 2)] √ √ = (x + 5)(x − 1 − 2)(x − 1 + 2) 59. f (x) = 3x3 − 8x2 + 7x − 2

1 2 a) Possibilities for p/q: ±1, ±2, ± , ± 3 3

2 From the graph of f (x), we see that and 1 might 3 be &zeros. 1 & 3 −8 7 −2 3 −5 2 3 −5 2 0 We have f (x) = (x − 1)(3x2 − 5x + 2). We find the other zeros. 3x2 − 5x + 2 = 0 (3x − 2)(x − 1) = 0

3x − 2 = 0 or x − 1 = 0 2 or x=1 x= 3

2 . (The zero 1 has 3 multiplicity 2.) These are the only zeros. The rational zeros are 1 and

b) f (x) = (x − 1)2 (3x − 2) 60. f (x) = x5 − 8x4 + 20x3 − 8x2 − 32x + 32

a) Possibilities for p/q: ±1, ±2, ±4, ±8, ±16, ±32

From the graph of f (x), we see that −1, 2, and 4 might be zeros. Synthetic division shows that −1 is not a zero. Try 2.

260

Chapter 3: Polynomial and Rational Functions & 2&

1 1

−8 2 −6

20 −8 −32 32 −12 16 16 −32 8 8 −16 0

62. f (x) = 2x5 − 13x4 + 32x3 − 38x2 + 22x − 5

1 5 a) Possibilities for p/q: ±1, ±5, ± , ± 2 2 From the graph of y = 2x5−13x4+32x3−38x2+22x−5, 5 we see that 1 and might be zeros. We try 1. 2 & 32 −38 22 −5 1 & 2 −13 2 −11 21 −17 5 2 −11 21 −17 5 0

We try 2 again. & 8 8 −16 2 & 1 −6 2 −8 0 16 1 −4 0 8 0 We try 2 a third time. & 2 & 1 −4 0 8 2 −4 −8 1 −2 −4 0

We have f (x) = (x − 2)3 (x2 − 2x − 4). Use the quadratic formula to find the other zeros.

We try 1 again. & 5 1 & 2 −11 21 −17 2 −9 12 −5 2 −9 12 −5 0

x2 − 2x − 4 = 0 / −(−2) ± (−2)2 − 4 · 1 · (−4) x= 2·1 √ √ 2±2 5 2 ± 20 = = 2 2 √ = 1± 5

We have f (x) = (x − 1)(x − 1)(x − 1)(2x2 − 7x + 5). We find the other zeros. 2x2 − 7x + 5 = 0

The rational zero is 2. (It is √ a zero of multiplicity 3.) The other zeros are 1 ± 5.

√ √ b) f (x) = (x − 2)3 [x − (1 + 5)][x − (1 − 5)] √ √ = (x − 2)3 (x − 1 − 5)(x − 1 + 5) 61. f (x) = x + x − 28x − 16x + 192x 6

5

4

3

2

f (x) = x2 (x4 + x3 − 28x2 − 16x + 192).

Consider g(x) = x4 + x3 − 28x2 − 16x + 192.

Possibilities for p/q: ±1, ±2, ±3, ±4, ±6, ±8, ±12,

±16, ±24, ±32, ±48, ±64, ±96, ±192

From the graph of y = g(x), we see that −4, 3 and 4 might be zeros. & −4 & 1 1 −28 −16 192 −4 12 64 −192 1 −3 −16 48 0 We have f (x) = x · g(x) = x2 (x + 4)(x3 − 3x2 − 16x + 48). We find the other zeros. x3 − 3x2 − 16x + 48 = 0 2

2x = 5 or 5 or x= 2

x=1 x=1

5 . (The number 1 is a 2 zero of multiplicity 4.) These are the only zeros. b) f (x) = (x − 1)4 (2x − 5) 63. f (x) = 2x6 − 7x3 + x2 − x

There are 3 variations in sign in f (x), so there are 3 or 1 positive real zeros. f (−x) = 2(−x)6 − 7(−x)3 + (−x)2 − (−x) = 2x6 + 7x3 + x2 + x

There are no variations in sign in f (−x), so there are no negative real zeros. 64. h(x) = −x8 + 6x5 − x3 + 2x − 2

h(−x) = −(−x)8 + 6(−x)5 − (−x)3 + 2(−x) − 2 = −x8 − 6x5 + x3 − 2x − 2

There are 2 variations in sign in h(−x), so there are 2 or 0 negative real zeros.

(x − 3)(x2 − 16) = 0

x = 3 or

2x − 5 = 0 or x − 1 = 0

There are 4 variations in sign in h(x), so there are 4 or 2 or 0 positive real zeros.

x2 (x − 3) − 16(x − 3) = 0

x − 3 = 0 or x + 4 = 0

(2x − 5)(x − 1) = 0

The rational zeros are 1 and

a) We know that 0 is a zero since

(x − 3)(x + 4)(x − 4) = 0

We try 1 a third time. & −5 1 & 2 −9 12 2 −7 5 2 −7 5 0

or x − 4 = 0

x = −4 or

x=4

The rational zeros are 0, −4, 3, and 4. (The zeros 0 and −4 each have multiplicity 2.) These are the only zeros. b) f (x) = x2 (x + 4)2 (x − 3)(x − 4)

65. g(x) = 5x5 − 4x2 + x − 1

There are 3 variations in sign in g(x), so there are 3 or 1 positive real zeros. g(−x) = 5(−x)5 − 4(−x)2 + (−x) − 1 = −5x5 − 4x2 − x − 1

There is no variation in sign in g(−x), so there are 0 negative real zeros.

Chapter 3 Review Exercises x2 − 5 x+2 1. The denominator is zero when x = −2, so the domain excludes −2. It is (−∞, −2) ∪ (−2, ∞). The line x = −2 is the vertical asymptote.

66. f (x) =

2. The degree of the numerator is 1 greater than the degree of the denominator, so we divide to find the oblique asymptote. x −2 x + 2 x2 + 0x − 5 x2 + 2x − 2x − 5 − 2x − 4 −1

The oblique asymptote is y = x − 2. There is no horizontal asymptote. √ 3. The numerator√is zero when√ x ± 5, so the xintercepts are ( 5, 0) and (− 5, 0). ! " 5 5 02 − 5 4. f (0) = = − , so the y-intercept is 0, − . 0+2 2 2 Find other function values to determine the shape 5. of the graph and then draw it.

5 67. f (x) = (x − 2)2 1. The denominator is zero when x = 2, so the domain excludes 2. It is (−∞, 2) ∪ (2, ∞). The line x = 2 is the vertical asymptote.

261 x2 + x − 6 (x + 3)(x − 2) = x2 − x − 20 (x + 4)(x − 5) 1. The denominator is zero when x = −4, or x = 5, so the domain excludes −4 and 5. It is (−∞, −4) ∪ (−4, 5) ∪ (5, ∞). The lines x = −4 and x = 5 are vertical asymptotes.

68. f (x) =

2. The numerator and the denominator have the same degree, so the horizontal asymptote is determined by the ratio of the leading coefficients, 1/1, or 1. Thus, y = 1 is the horizontal asymptote. There is no oblique asymptote. 3. The numerator is zero when x = −3 or x = 2, so the x-intercepts are (−3, 0) and (2, 0). 3 02 + 0 − 6 4. f (0) = = , so the y-intercept is 02 − 0 − 20 10 ! " 3 0, . 10 5. Find other function values to determine the shape of the graph and then draw it.

x−2 x−2 = − 2x − 15 (x + 3)(x − 5) 1. The denominator is zero when x = −3, or x = 5, so the domain excludes −3 and 5. It is (−∞, −3) ∪ (−3, 5) ∪ (5, ∞). The lines x = −3 and x = 5 are vertical asymptotes.

69. f (x) =

x2

2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. There is no oblique asymptote.

2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. There is no oblique asymptote.

3. The numerator has no zeros, so there is no xintercept. ! " 5 5 5 4. f (0) = , so the y-intercept is 0, . = (0 − 2)2 4 4 5. Find other function values to determine the shape of the graph and then draw it.

3. The numerator is zero when x = 2, so the xintercept is (2, 0). 2 0−2 = , so the y-intercept is 4. f (0) = 2 15 "0 − 2 · 0 − 15 ! 2 . 0, 15 5. Find other function values to determine the shape of the graph and then draw it.

262

Chapter 3: Polynomial and Rational Functions !

" 1 − , 2 : f (0) = −2 < 0 2

(2, ∞): f (3) = 7 > 0

75.

" ! 1 and (2, ∞). Function values are positive on − ∞, − 2 ! " 1 The solution set is − ∞, − ∪ (2, ∞). 2 (1 − x)(x + 4)(x − 2) ≤ 0 (1 − x)(x + 4)(x − 2) = 0

We let f (x) = (1 − x)(x + 4)(x − 2) and test a value in each interval. (−∞, −4) : f (−5) = 42 > 0

71. Answers may vary. The numbers −2 and 3 must be zeros of the denominator, and −3 must be zero of the numerator. In addition, the numerator and denominator must have the same degree and the ratio of the leading coefficients must be 4. 4x2 + 12x 4x(x + 3) , or f (x) = 2 f (x) = (x + 2)(x − 3) x −x−6

b) The medication never completely disappears from the body; a trace amount remains.

73.

x2 − 9 < 0 x −9 = 0 2

(x + 3)(x − 3) = 0

Polynomial inequality Related equation Factoring

The solutions of the related equation are −3 and 3. These numbers divide the x-axis into the intervals (−∞, −3), (−3, 3), and (3, ∞).

We let f (x) = (x + 3)(x − 3) and test a value in each interval. (−∞, −3): f (−4) = 7 > 0

(−3, 3): f (0) = −9 < 0

Function values are negative only on (−3, 3). The solution set is (−3, 3). 2x2 > 3x + 2 2x − 3x − 2 > 0 2

2x2 − 3x − 2 = 0

(2x + 1)(x − 2) = 0

(−4, 1) : f (0) = −8 < 0 ! " 11 3 = >0 (1, 2) : f 2 8 (2, ∞) : f (3) = −14 < 0

Function values are negative on (−4, 1) and (2, ∞). Since the inequality symbol is ≤, the endpoints of the intervals must be included in the solution set. It is [−4, 1] ∪ [2, ∞). 76.

x−2 < 4 Rational inequality x+3 x−2 − 4 < 0 Equivalent inequality x+3 x−2 − 4 = 0 Related equation x+3 x−2 − 4 is zero when The denominator of f (x) = x+3 x = −3, so the function is not defined for this value of x. We solve the related equation f (x) = 0. x−2 −4 = 0 x+3 " ! x−2 − 4 = (x + 3) · 0 (x + 3) x+3 x − 2 − 4(x + 3) = 0 x − 2 − 4x − 12 = 0

(3, ∞): f (4) = 7 > 0

74.

Related equation

The solutions of the related equation are 1, −4 and 2. These numbers divide the x-axis into the intervals (−∞, −4), (−4, 1), (1, 2) and (2, ∞).

70. Answers may vary. The numbers −2 and 3 must be zeros of the denominator. 1 1 , or f (x) = 2 f (x) = (x + 2)(x − 3) x −x−6

72. a) The horizontal asymptote of N (t) is the ratio of the leading coefficients of the numerator and denominator, 0.7/8, or 0.0875. Thus, N (t) → 0.0875 as t → ∞.

Polynomial inequality

Polynomial inequality Equivalent inequality Related equation Factoring

1 and 2. 2 These numbers " !divide the " x-axis into the intervals ! 1 1 − ∞, − , − , 2 , and (2, ∞). We let f (x) = 2 2 (2x + 1)(x − 2) and test a value in each interval. " ! 1 − ∞, − : f (−1) = 3 > 0 2 The solutions of the related equation are −

−3x − 14 = 0

−3x = 14 14 x=− 3 14 The critical values are − and −3. They divide the !3 " ! " 14 14 x-axis into the intervals − ∞, − , − , −3 and 3 3 (−3, ∞). We test a value in each interval. " ! 14 : f (−8) = −2 < 0 − ∞, − 3 " ! 14 − , −3 : f (−4) = 2 > 0 3 (−3, ∞) : f (2) = −4 < 0

Chapter 3 Review Exercises

263

" ! 14 Function values are negative for − ∞, − and ! "3 14 ∪ (−3, ∞). (−3, ∞). The solution set is − ∞, − 3 77. a) We write and solve a polynomial equation. −16t2 + 80t + 224 = 0

−16(t2 − 5t − 14) = 0 −16(t + 2)(t − 7) = 0

The solutions are t = −2 and t = 7. Only t = 7 has meaning in this application. The rocket reaches the ground at t = 7 seconds. b) We write and solve a polynomial inequality. −16t2 + 80t + 224 > 320 −16t2 + 80t − 96 > 0 −16t + 80t − 96 = 0 2

Polynomial inequality Equivalent inequality Related equation

−16(t2 − 5t + 6) = 0

−16(t − 2)(t − 3) = 0

The solutions of the related equation are 2 and 3. These numbers divide the t-axis into the intervals (−∞, 2), (2, 3) and (3, ∞). We restrict our discussion to values of t such that 0 ≤ t ≤ 7 since we know from part (a) the rocket is in the air for 7 sec. We consider the intervals [0, 2), (2, 3) and (3, 7]. We let f (t) = −16t2 + 80t − 96 and test a value in each interval. [0, 2): f (1) = −32 < 0 ! " 5 =4>0 (2, 3): f 2

79.

4=k

Variation constant

Equation of variation: y = 4x 80.

Function values are positive on (2, 3). The solution set is (2, 3)

k x k 100 = 25 2500 = k Variation constant y=

Equation of variation: y =

78. We write and solve a rational inequality. 8000t ≥ 400 4t2 + 10 8000t − 400 ≥ 0 4t2 + 10 8000t − 400 has no real The denominator of f (t) = 2 4t + 10 number zeros. We solve the related equation f (t) = 0. Keep in mind that the formula gives the population in thousands. 8000t − 400 = 0 4t2 + 10 ! " 8000t (4t2 + 10) − 400 = (4t2 + 10) · 0 4t2 + 10 8000t − 1600t2 − 4000 = 0 −1600t2 + 8000t − 4000 = 0

Using the quadratic formula, we find that t =

y = kx 100 = k · 25

(3, 7]: f (4) = −32 < 0

−800(2t2 − 10t + 5) = 0

These numbers divide the t-axis into the intervals √ " ! √ √ " ! 5 − 15 5 − 15 5 + 15 − ∞, , , and 2 2 2 √ ! " 5 + 15 ,∞ . 2 We test a value in each interval, restricting our discussion to values of t ≥ 0. √ " % 400 5 − 15 : f (0.5) = − <0 0, 2 11 √ " √ ! 1200 5 − 15 5 + 15 , : f (1) = >0 2 2 7 √ ! " 5 + 15 400 , ∞ : f (5) = − <0 2 11 √ " √ ! 5 − 15 5 + 15 , . Function values are positive on 2 2 Since the inequality symbol is ≥, the endpoints of the interval must be included in the solution set. It is √ √ $ % 5 − 15 5 + 15 , . 2 2

81.

t= 35 = 28, 000 = t= t= t=

82.

k r k 800 k 28, 000 r 28, 000 1400 20 min

5±

2

15

.

Variation constant Variation equation

P = kC 2 180 = k · 62 5=k

Variation constant

P = 5C 2

Variation equation

P = 5 · 102

P = 500 watts

√

2500 x

264 83.

Chapter 3: Polynomial and Rational Functions N = ka 87 = k · 28 87 =k Variation constant 28 87 a Variation equation N = 28 87 N = · 25 28 N ≈ 78

84.

85.

k x2 k 12 = 2 2 48 = k 48 y= 2 x y=

90. f (x) = x4 − 3x3 + 2x2

Graph the function and use the Zero, Maximum and Minimum features. a) Zeros: 0, 1, 2 b) Relative maximum: 0.202 at x = 0.610 c) Relative minima: 0 at x = 0, −0.620 at x = 1.640 d) Domain: all real numbers; range: [−0.620, ∞)

91. a) Linear: f (x) = 0.5408695652x − 30.30434783; quadratic: f (x) = 0.0030322581x2 − 0.5764516129x + 57.53225806; cubic: f (x) = 0.0000247619x3 − 0.0112857143x2 + 2.002380952x − 82.14285714

b) For the linear function, f (300) ≈ 132; for the quadratic function, f (300) ≈ 158; for the cubic function, f (300) ≈ 171. Since the cubic function yields the result that is closest to the actual number of 180, it makes the best prediction.

kxz 2 w ! " 2 1 k(16) 2 2= 0.2 2 = 20k 1 =k 10 xz 2 1 xz 2 · , or y= 10 w 10w y=

c) Using the cubic function, we have: f (350) ≈ 298 and f (400) ≈ 498.

92. A polynomial function is a function that can be defined by a polynomial expression. A rational function is a function that can be defined as a quotient of two polynomials.

x−4 (x + 1)(x − 2)(x + 4) The zeros of the denominator are −1, 2, and −4, so the vertical asymptotes are x = −1, x = 2, and x = 4. Answer C is correct.

86. f (x) =

x + 2x − 3 x + 2x − 3 = x2 − 5x + 6 (x − 2)(x − 3) The values of x that make the denominator 0 are 2 and 3, so the domain is (−∞, 2) ∪ (2, 3) ∪ (3, ∞). Answer A is correct.

87. g(x) =

2

2

88. f (x) = −2x2 − 3x + 6

Graph the function and use the Zero, Maximum and Minimum features. a) Zeros: −2.637, 1.137

b) Relative maximum: 7.125 at x = −0.75

c) There is no relative minimum.

d) Domain: all real numbers; range: (−∞, 7.125] 89. f (x) = x3 + 3x2 − 2x − 6

Graph the function and use the Zero, Maximum and Minimum features. a) Zeros: −3, −1.414, 1.414

b) Relative maximum: 2.303 at x = −2.291

93. Vertical asymptotes occur at any x-values that make the denominator zero. The graph of a rational function does not cross any vertical asymptotes. Horizontal asymptotes occur when the degree of the numerator is less than or equal to the degree of the denominator. Oblique asymptotes occur when the degree of the numerator is greater than the degree of the denominator. Graphs of rational functions may cross horizontal or oblique asymptotes. 94.

x2 ≥ 5 − 2x

x + 2x − 5 ≥ 0 2

x2 + 2x − 5 = 0

Polynomial inequality Equivalent inequality Related equation

√ Using the quadratic formula, we find that x = −1 ± 6; that is, x ≈ −3.449 or x ≈ 1.449. These √ divide √ numbers the √ x-axis into the √ intervals (−∞, −1− 6),2(−1− 6, −1+ 6) and (−1 + 6, ∞). We let f (x) = x + 2x − 5 and test a value in each interval. √ (−∞, −1 − 6) : f (−4) = 3 > 0 √ √ (−1 − 6, −1 + 6) : f (0) = −5 < 0 √ (−1 + 6, ∞) : f (2) = 3 > 0 √ Function √ values are positive on (−∞, −1 − 6) and (−1 + 6, ∞). Since the inequality symbol is ≥, the endpoints of the intervals must be√included in the solution set. √ It is (−∞, −1 − 6] ∪ [−1 + 6, ∞). & & & & &1 − 1 & < 3 95. & x2 & 1 −3 < 1 − 2 < 3 x

c) Relative minimum: −6.303 at x = 0.291

−3 <

d) Domain: all real numbers; range: all real numbers −3 <

x2 − 1 <3 x2

(x + 1)(x − 1) <3 x2

Chapter 3 Review Exercises

265

(x + 1)(x − 1) (x + 1)(x − 1) and <3 x2 x2 First, solve (x + 1)(x − 1) −3 < x2 (x + 1)(x − 1) 0< +3 x2 (x + 1)(x − 1) +3 is zero when The denominator of f (x) = x2 x = 0, so the function is not defined for this value of x. Solve the related equation. (x + 1)(x − 1) +3 = 0 x2 (x + 1)(x − 1) + 3x2 = 0 Multiplying by x2 −3 <

x2 − 1 + 3x2 = 0

4x2 = 1 1 x2 = 4 x=±

1 2

1 1 The critical values are − , 0 and . Test a value in each 2 2 of the intervals determined by them. " ! 1 : f (−1) = 3 > 0 − ∞, − 2 " ! " ! 1 1 = −12 < 0 − ,0 : f − 2 4 " ! " ! 1 1 : f = −12 < 0 0, 2 4 " ! 1 , ∞ : f (1) = 3 > 0 2 The solution " set ! for "this portion of the inequality is ! 1 1 ,∞ . ∪ − ∞, − 2 2 Next, solve (x + 1)(x − 1) <3 x2 (x + 1)(x − 1) −3 < 0 x2 (x + 1)(x − 1) The denominator of f (x) = − 3 is zero x2 when x = 0, so the function is not defined for this value of x. Now solve the related equation. (x + 1)(x − 1) −3 = 0 x2 (x + 1)(x − 1) − 3x2 = 0 Multiplying by x2 x2 − 1 − 3x2 = 0

2x2 = −1 1 x2 = − 2 There are no real solutions for this portion of the inequality. " ! set of"the original inequality is ! The solution 1 1 ∪ ,∞ . − ∞, − 2 2

96. x4 − 2x3 + 3x2 − 2x + 2 = 0

The only possible rational solutions are ±1, ±2. Synthetic division shows that none are solutions. Try complex numbers. & −2 3 −2 2 i& 1 i −1 − 2i 2 + 2i −2 1 −2 + i 2 − 2i 2i 0 & −i & 1 −2 + i 2 − 2i 2i −i 2i −2i 1 −2 2 0

We can use the quadratic formula to find that the solutions of x2 − 2x + 2 = 0 are 1 + i and 1 − i. The solutions are 1 + i, 1 − i, i, −i. 97.

(x − 2)−3 < 0 1 <0 (x − 2)3

1 is zero when x = 2, (x − 2)3 so the function is not defined for this value of x. The 1 = 0 has no solution, so 2 is the related equation (x − 2)3 only critical point. Test a value in each of the intervals determined by this critical point. The denominator of f (x) =

(−∞, 2) : f (1) = −1 < 0 (2, ∞) : f (3) = 1 > 0

Function values are negative on (−∞, 2). The solution set is (−∞, 2). 98. f (x) = x3 − 1

The possibilities for p/q are ±1. From the graph of y = x3 − 1 we see that 1 might be a zero. & 1 & 1 0 0 −1 1 1 1 1 1 1 0 f (x) = (x − 1)(x2 + x + 1)

Using the √ quadratic formula, we find that the other zeros 3 1 i. are − ± 2 2 √ "$% ! √ "$ % ! 1 1 3 3 i x− − − i f (x) = (x−1) x − − + 2 2 2 2 √ "! √ " ! 1 1 3 3 = (x − 1) x + − i x+ + i 2 2 2 2 99. Divide x3 + kx2 + kx − 15 by x + 3. & −3 & 1 k k −15 −3 9 − 3k −27 + 6k 1 −3 + k 9 − 2k −42 + 6k Thus f (−3) = −42 + 6k.

We know that if x + 3 is a factor of f (x), then f (−3) = 0. We solve −42 + 6k = 0 for k. −42 + 6k = 0 6k = 42 k=7

266

Chapter 3: Polynomial and Rational Functions 5 − |7x + 2| > 0

100. Divide x2 − 4x + 3k by x + 5. & −5 & 1 −4 3k −5 45 1 −9 45 + 3k

−5 < 7x + 2 < 5

−7 < 7x < 3 3 −1 < x < 7 ! " 3 The solution set is − 1, . 7

Since the remainder is 33, we solve 45 + 3k = 33. 45 + 3k = 33 3k = −12

101. f (x) =

√

k = −4

x2 + 3x − 10

Since we cannot take the square root of a negative number, then x2 + 3x − 10 ≥ 0. x2 + 3x − 10 ≥ 0 Polynomial inequality x + 3x − 10 = 0 2

(x + 5)(x − 2) = 0

Related equation Factoring

The solutions of the related equation are −5 and 2. These numbers divide the x-axis into the intervals (−∞, −5), (−5, 2) and (2, ∞).

We let g(x) = (x + 5)(x − 2) and test a value in each interval. (−∞, −5) : g(−6) = 8 > 0 (−5, 2) : g(0) = −10 < 0

(2, ∞) : g(3) = 8 > 0

Functions values are positive on (−∞, −5) and (2, ∞). Since the equality symbol is ≥, the endpoints of the intervals must be included in the solution set. It is (−∞, −5] ∪ [2, ∞). √ 102. f (x) = x2 − 3.1x + 2.2 + 1.75 Since we cannot take the square root of a negative number, then x2 − 3.1x + 2.2 ≥ 0. x2 − 3.1x + 2.2 ≥ 0 Polynomial inequality x2 − 3.1x + 2.2 = 0

(x − 1.1)(x − 2) = 0

Related equation Factoring

The solutions of the related equation are 1.1 and 2. These numbers divide the x-axis into the intervals (−∞, 1.1), (1.1, 2), and (2, ∞).

We let g(x) = (x − 1.1)(x − 2) and test a value in each interval. (−∞, 1.1) : g(0) = 2.2 > 0 (1.1, 2) : g(1.5) = −0.2 < 0

(2, ∞) : g(3) = 1.9 > 0

Function values are positive on (−∞, 1.1) and (2, ∞). Since the inequality symbol is ≥, the endpoints of the intervals must be included in the solution set. It is (−∞, 1.1] ∪ [2, ∞).

1 103. f (x) = / 5 − |7x + 2| We cannot take the square root of a negative number; neither can the denominator be zero. Thus we have 5 − |7x + 2| > 0.

Polynomial inequality

|7x + 2| < 5

Chapter 3 Test 1.

f (x) = 2x3 + 6x2 − x4 + 11

= −x4 + 2x3 + 6x2 + 11

The degree of the polynomial is 4, so the polynomial is quartic. The leading term is −x4 and the leading coefficient is −1. 2. h(x) = −4.7x + 29

The degree of the polynomial is 1, so the polynomial is linear. The leading term is −4.7x and the leading coefficient is −4.7.

3. f (x) = x(3x − 5)(x − 3)2 (x + 1)3 5 The zeros of the function are 0, , 3, and −1. 3 The factors x and 3x − 5 each occur once, so the zeros 0 5 and have multiplicity 1. 3 The factor x − 3 occurs twice, so the zero 3 has multiplicity 2. The factor x + 1 occurs three times, so the zero −1 has multiplicity 3. 4. For 1986, x = 1986 − 1980 = 6.

f (6) = 0.152(6)4 − 9.786(6)3 + 173.317(6)2 − 831.220(6) + 4053.134 ≈ 3388 In 1986, 3388 young people were killed by gunfire. For 1993, x = 1993 − 1980 = 13.

f (13) = 0.152(13)4 − 9.786(13)3 + 173.317(13)2 − 831.220(13) + 4053.134 ≈ 5379

In 1993, 5379 young people were killed by gunfire. For 1999, x = 1999 − 1980 = 19.

f (19) = 0.152(19)4 − 9.786(19)3 + 173.317(19)2 − 831.220(19) + 4053.134 ≈ 3514

In 1999, 3514 young people were killed by gunfire. 5. f (x) = x3 − x2 − 6x

1. The leading term is x3 . The degree, 3, is odd and the leading coefficient, 1, is positive so as x → ∞, f (x) → ∞ and as x → −∞, f (x) → −∞.

Chapter 3 Test

267 We have f (x) = (x + 1)(−2x3 + 3x2 + 8x − 12). Find the other zeros. −2x3 + 3x2 + 8x − 12 = 0

2. We solve f (x) = 0. x3 − x2 − 6x = 0

x(x2 − x − 6) = 0

2x3 − 3x2 − 8x + 12 = 0

x(x − 3)(x + 2) = 0

x2 (2x − 3) − 4(2x − 3) = 0

x = 0 or x − 3 = 0 or x + 2 = 0

x = 0 or

x = 3 or

(2x − 3)(x2 − 4) = 0

x = −2

The zeros of the function are −2, 0, and 3, so the x-intercepts are (−2, 0), (0, 0), and (3, 0). 3. The zeros divide the x-axis in 4 intervals, (−∞, −2), (−2, 0), (0, 3), and (3, ∞). We choose a value for x in each interval and find f (x). This tells us the sign of f (x) for all values of x in that interval. In (−∞, −2), test −3:

f (−3) = (−3)3 − (−3)2 − 6(−3) = −18 < 0 In (−2, 0), test −1:

f (−1) = (−1)3 − (−1)2 − 6(−1) = 4 > 0 In (0, 3), test 1:

f (1) = 13 − 12 − 6 · 1 = −6 < 0 In (3, ∞), test 4:

f (4) = 43 − 42 − 6 · 4 = 24 > 0 Thus the graph lies below the x-axis on (−∞, −2) and on (0, 3) and above the x-axis on (−2, 0) and (3, ∞). We also know the points (−3, −18), (−1, 4), (1, −6), and (4, 24) are on the graph.

4. From Step 2 we know that the y-intercept is (0, 0).

5. We find additional points on the graph and draw the graph. y

x

f (x)

−1.5

3.375

−0.5

2.625

0.5

−3.125

2.5

−5.625

8 4 2

!4 !2

4

x

!4 !8

f (x ) # x 3 ! x 2 ! 6x

6. Checking the graph as described on page 272 in the text, we see that it appears to be correct. 6. f (x) = −2x + x + 11x − 4x − 12 4

3

Multiplying by −1

2

1. The leading term is −2x . The degree, 4, is even and the leading coefficient, −2, is negative so x → ∞, f (x) → −∞ and as x → −∞, f (x) → −∞. 4

2. We solve f (x) = 0.

The possible rational zeros are ±1, ±2, ±3, ±4, ±6, 3 1 ±12, ± , and ± . We try −1. 2 2 & −1 & −2 1 11 −4 −12 2 −3 −8 12 −2 3 8 −12 0

(2x − 3)(x + 2)(x − 2) = 0

2x − 3 = 0 or x + 2 = 0 2x = 3 or 3 or x= 2

or x − 2 = 0

x = −2 or

x=2

x = −2 or

x=2

3 , −2, and 2, so 2 ! " 3 , 0 , and the x-intercepts are (−2, 0), (−1, 0), 2 (2, 0). The zeros of the function are −1,

3. The zeros divide the x-axis into " 5! intervals, " ! 3 3 , , 2 , and (−∞, −2), (−2, −1), − 1, 2 2 (2, ∞). We choose a value for x in each interval and find f (x). This tells us the sign of f (x) for all values of x in that interval. In (−∞, −2), test −3: f (−3) = −90 < 0 In (−2, −1), test −1.5: f (−1.5) = 5.25 > 0 " ! 3 , test 0: f (0) = −12 < 0 In − 1, 2 " ! 3 In , 2 , test 1.75: f (1.75) ≈ 1.29 > 0 2 In (2, ∞), test 3: f (3) = −60 < 0

Thus " lies below the x-axis on (−∞, −2), ! the graph 3 , and on (2, ∞) and above the x-axis on on − 1, 2 " ! 3 , 2 . We also know the points (−2, −1) and on 2 (−3, −90), (−1.5, 5.25), (0, −12), (1.75, 1.29), and (3, −60) are on the graph.

4. From Step 3 we know that f (0) = −12, so the yintercept is (0, −12).

5. We find additional points on the graph and draw the graph. x

f (x)

−0.5

−7.5

0.5

−11.25

2.5

−15.75

y 4 2

!4 !2

4

x

!4 !8 !12

f (x) # !2x 4 " x 3 " 11x 2 ! 4x ! 12

6. Checking the graph as described on page 272 in the text, we see that it appears to be correct.

268 7.

Chapter 3: Polynomial and Rational Functions f (0) = −5 · 02 + 3 = 3

f (2) = −5 · 2 + 3 = −17 2

By the intermediate value theorem, since f (0) and f (2) have opposite signs, f (x) has a zero between 0 and 2. 8.

g(−2) = 2(−2)3 + 6(−2)2 − 3 = 5

g(−1) = 2(−1)3 + 6(−1)2 − 3 = 1

Since both g(−2) and g(−1) are positive, we cannot use the intermediate value theorem to determine if there is a zero between −2 and −1. 9.

x3 + 4x2 + 4x + 6 x − 1 x4 + 3x3 + 0x2 + 2x − 5 x4 − x3 4x3 + 0x2 4x3 − 4x2 4x2 + 2x 4x2 − 4x 6x − 5 6x − 6 1 The quotient is x3 + 4x2 + 4x + 6; the remainder is 1.

P (x) = (x − 1)(x3 + 4x2 + 4x + 6) + 1 & 10. 5 & 3 0 −12 7 15 75 315 3 15 63 322 Q(x) = 3x2 + 15x + 63; R(x) = 322 & 11. −3 & 2 −6 1 −4 −6 36 −111 2 −12 37 −115

P (−3) = −115 & 12. −2 & 1 4 1 −6 −2 −4 6 1 2 −3 0

f (−2) = 0, so −2 is a zero of f (x).

13. The function can be written in the form f (x) = an (x + 3)2 (x)(x − 6).

The simplest polynomial is obtained if we let an = 1. f (x) = (x + 3)2 (x)(x − 6)

= (x2 + 6x + 9)(x2 − 6x)

= x4 + 6x3 + 9x2 − 6x3 − 36x2 − 54x

= x4 − 27x2 − 54x

14. A polynomial function of degree 5 can have at most 5 zeros. Since f (x) has rational coefficients, in addition to √ the 3 given zeros,√the other zeros are the conjugates of 3 and 2 − i, or − 3 and 2 + i. 15. −3i is also a zero. f (x) = (x + 10)(x − 3i)(x + 3i) = (x + 10)(x2 + 9)

= x3 + 10x2 + 9x + 90

16.

√

3 and 1 + i are also zeros. √ √ f (x) = (x − 0)(x + 3)(x − 3)[x − (1 − i)][x − (1 + i)] = x(x2 − 3)[(x − 1) + i][(x − 1) − i] = (x3 − 3x)(x2 − 2x + 1 + 1)

= (x3 − 3x)(x2 − 2x + 2)

= x5 − 2x4 + 2x3 − 3x3 + 6x2 − 6x

= x5 − 2x4 − x3 + 6x2 − 6x

17. f (x) = 2x3 + x2 − 2x + 12 ±1, ±2, ±3, ±4, ±6, ±12 Possibilities for p : Possibilities for q ±1, ±2

1 3 Possibilities for p/q: ±1, ±2, ±3, ±4, ±6, ±12, ± , ± 2 2

18. h(x) = 10x4 − x3 + 2x − 5 Possibilities for p ±1, ±5 : Possibilities for q ±1, ±2, ±5, ±10 1 1 5 1 Possibilities for p/q: ±1, ±5, ± , ± , ± , ± 2 2 5 10 19. f (x) = x3 + x2 − 5x − 5

a) Possibilities for p/q: ±1, ±5

From the graph of y = f (x), we see that −1 might be a zero. & −1 & 1 1 −5 −5 −1 0 5 1 0 −5 0

We have f (x) = (x + 1)(x2 − 5). We find the other zeros. x2 − 5 = 0 x2 = 5

√ x=± 5

√ The rational zero is −1. The other zeros are ± 5. √ √ b) f (x) = (x + 1)(x − 5)(x + 5) 20. g(x) = 2x4 − 11x3 + 16x2 − x − 6

1 3 a) Possibilities for p/q: ±1, ±2, ±3, ±6, ± , ± 2 2 1 From the graph of y = g(x), we see that − , 1, 2, 2 1 and 3 might be zeros. We try − . 2 & − 12 & 2 −11 16 −1 −6 −1 6 −11 6 2 −12 22 −12 0

Now we try 1. & 1 & 2 −12 22 −12 2 −10 12 2 −10 12 0 " ! 1 (x−1)(2x2 −10x + 12) = We have g(x) = x + 2 " ! 1 (x − 1)(x2 − 5x + 6). We find the other 2 x+ 2 zeros.

Chapter 3 Test

269

x2 − 5x + 6 = 0

(x − 2)(x − 3) = 0

x − 2 = 0 or x − 3 = 0 x = 2 or

x=3

1 The rational zeros are − , 1, 2, and 3. These are 2 the only zeros. " ! 1 b) g(x) = 2 x + (x − 1)(x − 2)(x − 3) 2 = (2x + 1)(x − 1)(x − 2)(x − 3) 21. h(x) = x3 + 4x2 + 4x + 16 a) Possibilities for p/q: ±1, ±2, ±4, ±8, ±16

From the graph of h(x), we see that −4 might be a zero.& 4 4 16 −4 & 1 −4 0 −16 1 0 4 0

We have h(x) = (x + 4)(x2 + 4). We find the other zeros. x2 + 4 = 0

! " 2 b) f (x) = 3 x − (x − 1)(x − 1)(x − 1) 3 = (3x − 2)(x − 1)3 23. g(x) = −x8 + 2x6 − 4x3 − 1

There are 2 variations in sign in g(x), so there are 2 or 0 positive real zeros. g(−x) = −(−x)8 + 2(−x)6 − 4(−x)3 − 1 = −x8 + 2x6 + 4x3 − 1

There are 2 variations in sign in g(−x), so there are 2 or 0 negative real zeros.

1.

2.

3. 4.

x2 = −4

x = ±2i

2 (x − 3)2 The denominator is zero when x = 3, so the domain excludes 3. it is (−∞, 3) ∪ (3, ∞). The line x = 3 is the vertical asymptote. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. The numerator has no zeros, so there is no xintercept. ! " 2 2 2 = , so the y-intercept is 0, . f (0) = (0 − 3)2 9 9 Find other function values to determine the shape of the graph and then draw it.

24. f (x) =

5.

The rational zero is −4. The other zeros are ±2i.

b) h(x) = (x + 4)(x + 2i)(x − 2i)

22. f (x) = 3x4 − 11x3 + 15x2 − 9x + 2

1 2 a) Possibilities for p/q: ±1, ±2, ± , ± 3 3

2 From the graph of f (x), we see that and 1 might 3 2 be zeros. We try . 3 & 2 & 3 −11 15 −9 2 3 2 −6 6 −2 3 −9 9 −3 0

Now we try 1. & 1 & 3 −9 9 −3 3 −6 3 3 −6 3 0 " ! 2 (x−1)(3x2 −6x + 3) = We have f (x) = x − 3 ! " 2 3 x− (x − 1)(x2 − 2x + 1). We find the other 3 zeros. x2 − 2x + 1 = 0 (x − 1)(x − 1) = 0

1.

2.

3. 4.

x − 1 = 0 or x − 1 = 0 x = 1 or

x=1

2 and 1. (The zero 1 has The rational zeros are 3 multiplicity 3.) These are the only zeros.

x+3 x+3 = x2 − 3x − 4 (x + 1)(x − 4) The denominator is zero when x = −1 or x = 4, so the domain excludes −1 and 4. It is (−∞, −1) ∪ (−1, 4) ∪ (4, ∞). The lines x = −1 and x = 4 are vertical asymptotes. Because the degree of the numerator is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. The numerator is zero at x = −3, so the x-intercept is (−3, 0). 3 0+3 = − , so the y-intercept is f (0) = (0 + 1)(0 − 4) 4 ! " 3 0, − . 4 Find other function values to determine the shape of the graph and then draw it.

25. f (x) =

5.

270

Chapter 3: Polynomial and Rational Functions 13 The critical values are 4 and . They divide the x-axis ! " !2 " 13 13 and , ∞ . We into the intervals (−∞, 4), 4, 2 2 test a value in each interval. (−∞, 4) : f (3) = −7 < 0 " ! 13 : f (5) = 3 > 0 4, 2 ! " 13 , ∞ : f (9) = −1 < 0 2 " ! 13 ,∞ . Function values are negative for (−∞, 4) and 2 Since the inequality symbol is ≤, the endpoint of the in" ! 13 terval , ∞ must be included in the solution set. It is 2% " 13 ,∞ . (−∞, 4) ∪ 2

26. Answers may vary. The numbers −1 and 2 must be zeros of the denominator, and −4 must be a zero of the numerator. x+4 x+4 f (x) = , or f (x) = 2 (x + 1)(x − 2) x −x−2

27.

2x2 > 5x + 3

2x − 5x − 3 > 0

Polynomial inequality Equivalent inequality

2

2x2 − 5x − 3 = 0

Related equation

(2x + 1)(x − 3) = 0

29. a) We write and solve a polynomial equation.

Factoring

−16t2 + 64t + 192 = 0

1 and 3. These 2! " 1 numbers divide the x-axis into the intervals − ∞, − , 2 ! " 1 − , 3 , and (3, ∞). 2

−16(t2 − 4t − 12) = 0

The solutions of the related equation are −

−16(t + 2)(t − 6) = 0

The solutions are t = −2 and t = 6. Only t = 6 has meaning in this application. The rocket reaches the ground at t = 6 seconds. b) We write and solve a polynomial inequality.

We let f (x) = (2x + 1)(x − 3) and test a value in each interval. " ! 1 : f (−1) = 4 > 0 − ∞, − 2 " ! 1 − , 3 : f (0) = −3 < 0 2 (3, ∞) : f (4) = 9 > 0 " ! 1 Function values are positive on − ∞, − and (3, ∞). 2 ! " 1 The solution set is − ∞, − ∪ (3, ∞). 2

28.

x+1 ≤3 x−4 x+1 −3 ≤ 0 x−4

−16t2 + 64t + 192 > 240 −16t2 + 64t − 48 > 0 −16t2 + 64t − 48 = 0

−16(t − 4t + 3) = 0

−16(t − 1)(t − 3) = 0

The solutions of the related equation are 1 and 3. These numbers divide the t-axis into the intervals (−∞, 1), (1, 3) and (3, ∞). Because the rocket returns to the ground at t = 6, we restrict our discussion to values of t such that 0 ≤ t ≤ 6. We consider the intervals [0, 1), (1, 3) and (3, 6]. We let f (t) = −16t2 + 64t − 48 and test a value in each interval. ! " 1 = −20 < 0 [0, 1) : f 2 (1, 3) : f (2) = 16 > 0

Rational inequality Equivalent inequality

x+1 − 3 is zero when x = 4, x−4 so the function is not defined for this value of x. We solve the related equation f (x) = 0. x+1 −3 = 0 x−4 " ! x+1 − 3 = (x − 4) · 0 (x − 4) x−4 x + 1 − 3(x − 4) = 0 The denominator of f (x) =

x + 1 − 3x + 12 = 0 −2x + 13 = 0

2x = 13 13 x= 2

Related equation

2

(3, 6] : f (4) = −48 < 0

Function values are positive on (1, 3). The solution set is (1, 3). 30.

k x k 5= 6 30 = k y=

Variation constant

Equation of variation: y =

30 x

Chapter 3 Test 31.

271

d = kr2 200 = k · 602 1 =k 18 1 2 r d= 18 1 d= · 302 18 d = 50 ft

32.

Variation constant Equation of variation

kxz 2 w k(0.1)(10)2 100 = 5 100 = 2k y=

50 = k

Variation constant

50xz Equation of variation w √ 33. f (x) = x2 + x − 12 y=

2

Since we cannot take the square root of a negative number, then x2 + x − 12 ≥ 0. x2 + x − 12 ≥ 0 Polynomial inequality x2 + x − 12 = 0 Related equation

(x + 4)(x − 3) = 0 Factoring

The solutions of the related equation are −4 and 3. These numbers divide the x-axis into the intervals (−∞, −4), (−4, 3) and (3, ∞).

We let f (x) = (x + 4)(x − 3) and test a value in each interval. (−∞, −4) : g(−5) = 8 > 0 (−4, 3) : g(0) = −12 < 0

(3, ∞) : g(4) = 8 > 0

Functions values are positive on (−∞, −4) and (3, ∞). Since the inequality symbol is ≥, the endpoints of the intervals must be included in the solution set. It is (−∞, −4] ∪ [3, ∞).

Exercise Set 4.5 82.

311 87.

log3 (log4 x) = 0 log4 x = 30

(5x − 1)(5x − 2) = 0

log4 x = 1

5x = 1

x = 41

or x log 5 = log 2 log 2 x=0 or x= ≈ 0.431 log 5 The solutions are 0 and 0.431. " ! 1 88. = ln 6 x ln 6 x(ln 1 − ln 6) = ln 6

(log3 x)2 − log3 x2 = 3

(log3 x)2 − 2 log3 x − 3 = 0

Let u = log3 x and substitute: u2 − 2u − 3 = 0

(u − 3)(u + 1) = 0 log3 x = 3

u = −1

or

−x ln 6 = ln 6

or log3 x = −1

x = 3−1 1 x= 3

x = 33 or

x = 27 or

Both answers check. The solutions are 84.

The solution is −1. 89.

Let u = log x and substitute. u2 − 2u − 3 = 0

|x| = 9

x = −9 or x = 9

Both answers check. The solutions are −9 and 9.

90.

x3 100 log x · log x = log x3 − log 100 (log x)2 = 3 log x − 2

(log x)2 − 3 log x + 2 = 0

Let u = log x and substitute. u2 − 3u + 2 = 0

ln x2 = (ln x)2

(u − 1)(u − 2) = 0

2 ln x = (ln x)2

u=1

0 = (ln x)2 − 2 ln x

log x = 1

Let u = ln x and substitute. 0 = u2 − 2u u = 0 or

u=2

x = 1 or

x = e ≈ 7.389 2

Both answers check. The solutions are 1 and e2 , or 1 and 7.389. 86.

2x

e

x

e =2 x

ln ex = ln 2 x = ln 2

or

e =7 x

or ln ex = ln 7 or

x ≈ 0.693 or

x = ln 7 x ≈ 1.946

The solutions are 0.693 and 1.946.

x = 102 = 100

ln xln x = 4 ln x · ln x = 4 (ln x)2 = 4 ln x = ±2

ln x = −2 x=e

− 9 · e + 14 = 0

(e − 2)(e − 7) = 0 x

u=2

or log x = 2

Both answers check. The solutions are 10 and 100.

−2

x

or

x = 10 or 91.

ln x = 0 or ln x = 2

x3 100

log xlog x = log

1 Both answers check. The solutions are and 1000. 10

0 = u(u − 2)

xlog x =

u=3

log x = −1 or log x = 3 1 or x = 1000 x= 10

85.

log3 |x| = 2

|x| = 32

1 and 27. 3

(log x)2 − 2 log x − 3 = 0

u = −1 or

(ln 1 = 0)

x = −1

(log x)2 − log x2 = 3

(u + 1)(u − 3) = 0

5x = 2

or

x log 5 = 0

The answer checks. The solution is 4.

u=3

This equation is quadratic in 5x .

log 5x = log 1 or log 5x = log 2

x=4 83.

52x − 3 · 5x + 2 = 0

or ln x = 2

or

x ≈ 0.135 or

x = e2 x ≈ 7.389

Both answers check. The solutions are e−2 and e2 , or 0.135 and 7.389.

312 92.

Chapter 4: Exponential and Logarithmic Functions (e3x+1 )2 = e10x e4

96.

2

2

2x − 8 = −3

e6x+2 = e10x e4 e6x−2 = e10x

or 2x − 8 = 3

2

2x = 5

2

2x = 11

or

2

2

log 2x = log 5

6x − 2 = 10x

or log 2x = log 11

x2 log 2 = log 5 or x2 log 2 = log 11 log 5 log 11 x2 = or x2 = log 2 log 2 x = ±1.524 or x = ±1.860

−2 = 4x 1 − =x 2

1 The solution is − . 2 # (e2x · e−5x )−4 93. = e7 ex ÷ e−x √ e12x = e7 ex−(−x)

The solutions are −1.860, −1.524, 1.524, and 1.860. 97.

a = log8 225, so 8a = 225 = 152 . b = log2 15, so 2b = 15. Then

8a = (2b )2 (23 )a = 22b 23a = 22b

e6x = e7 e2x e4x = e7

3a = 2b 2 a = b. 3

4x = 7 7 x= 4 7 The solution is . 4 94.

2

|2x − 8| = 3

1 , so a = 3 ! "3 1 1 is equivalent to a = = . Then 3 27

98. log5 125 = 3 and log125 5 = (log125 5)log5 125 log3 a = log3

4 5 ln ex < ln 0.8 ex <

1 = −3. 27

99. log2 [log3 (log4 x)] = 0 yields x = 64. log3 [log2 (log4 y)] = 0 yields y = 16.

x < −0.223

The solution set is (−∞, −0.223). 95. | log5 x| + 3 log5 |x| = 4

Note that we must have x > 0. First consider the case when 0 < x < 1. When 0 < x < 1, then log5 x < 0, so | log5 x| = − log5 x and |x| = x. Thus we have: − log5 x + 3 log5 x = 4

100. f (x) = ex − e−x

Replace f (x) with y: y = ex − e−x Interchange x and y: x = ey − e−y Solve for y: xey = e2y − 1

2 log5 x = 4

0=e

2y

log5 x2 = 4 x2 = 54 x = 52 x = 25

log4 [log3 (log2 z)] = 0 yields z = 8. Then x + y + z = 64 + 16 + 8 = 88.

(Recall that x > 0.)

25 cannot be a solution since we assumed 0 < x < 1. Now consider the case when x > 1. In this case log5 x > 0, so | log5 x| = log5 x and |x| = x. Thus we have: log5 x + 3 log5 x = 4 4 log5 x = 4 log5 x = 1 x=5 This answer checks. The solution is 5.

Multiplying by ey

− xe − 1 y

Using the quadratic formula with a = 1, b = −x, and y c = −1 and taking √ the positive square root (since e > 0), 2 x+ x +4 . Then we have we get ey = 2 √ ! " x + x2 + 4 y ln e = ln 2 √ " ! x + x2 + 4 y = ln 2 Replace y with f −1 (x): √ " ! x + x2 + 4 −1 . f (x) = ln 2

Exercise Set 4.6

313

Exercise Set 4.6

4. a) 1. a) Substitute 6.4 for P0 and 0.012 for k in P (t) = P0 ekt . We have:

ln 58.02 = ln e6k ln 58.02 = 6k ln 58.02 =k 6 0.6768 ≈ k

b) In 2008, t = 2008 − 2005 = 3.

P (3) = 6.4e0.012(3) ≈ 6.6 billion In 2010, t = 2010 − 2005 = 5.

D(t) = 0.5e0.6768t , where D(t) is in millions and t is the number of years after 1998.

P (5) = 6.4e0.012(5) ≈ 6.8 billion

c) Substitute 8 for P (t) and solve for t.

b) D(7) = 0.5e0.6768(7) ≈ 57.1 million

8 = 6.4e

0.012t

D(10) = 0.5e0.6768(10) ≈ 434.8 million

1.25 = e

0.012t

ln 1.25 = ln e0.012t ln 1.25 = 0.012t ln 1.25 =t 0.012 18.6 ≈ t

The world population will be 8 billion about 18.6 yr after 2005. ln 2 ≈ 57.8 yr d) T = 0.012 2. a) P (t) = 100e0.117t b) P (7) = 100e0.117(7) ≈ 227

Note that 2 weeks = 2 · 7 days = 14 days.

P (14) = 100e0.117(14) ≈ 514 ln 2 ≈ 5.9 days c) t = 0.117 ln 2 ≈ 28.9 yr 0.024 ln 2 k= ≈ 1.1% per yr 63 ln 2 T = ≈ 49.5 yr 0.014 ln 2 T = ≈ 346.6 yr 0.002 ln 2 k= ≈ 1.8% per yr 38.5 ln 2 ≈ 1.9% per yr k= 36.5 ln 2 k= ≈ 77.0 yr 0.009 ln 2 ≈ 1.0% per yr k= 69.3 ln 2 ≈ 57.8 yr k= 0.012

b) c) d) e) f) g) h) i)

29.01 = 0.5ek·6 58.02 = e6k

P (t) = 6.4e0.012t , where P (t) is in billions and t is the number of years after 2005.

3. a) T =

ln 2 ≈ 26.7 yr 0.026

j) k =

D(13) = 0.5e0.6768(13) ≈ 3312 million 5.

P (t) = P0 ekt 24, 839, 654, 400 24, 839, 654, 400 6, 276, 883 " ! 24, 839, 654, 400 ln 6, 276, 883 ! " 24, 839, 654, 400 ln 6, 276, 883 ! " 24, 839, 654, 400 ln 6, 276, 883 0.013 637

= 6, 276, 883e0.013t = e0.013t = ln e0.013t = 0.013t

=t ≈t

There will be one person for every square yard of land about 637 yr after 2005. 6. In 2010, t = 2010 − 1626 = 384.

P (384) = 24e0.06(384) ≈ $243, 419, 914, 000, or

about $243,000,000,000

7. a) Substitute 10,000 for P0 and 5.4%, or 0.054 for k. P (t) = 10, 000e0.054t b)

P (1) = 10, 000e0.054(1) ≈ $10, 555

P (2) = 10, 000e0.054(2) ≈ $11, 140 P (5) = 10, 000e0.054(5) ≈ $13, 100

P (10) = 10, 000e0.054(10) ≈ $17, 160 c) T =

ln 2 ≈ 12.8 yr 0.054

ln 2 ≈ 11.2 yr 0.062 P (5) = 35, 000e0.062(5) ≈ $47, 719.88

8. a) T =

314

Chapter 4: Exponential and Logarithmic Functions b)

7130.90 = 5000e5k 1.4618 = e5k ln 1.4618 = ln e5k ln 1.4618 = 5k ln 1.4618 =k 5 0.071 ≈ k 7.1% ≈ k

ln 2 ≈ 9.8 yr 0.071 11, 414.71 = P0 e0.084(5) 11, 414.71 = P0 e0.084(5) $7500 ≈ P0

T = c)

b) c) d) e) f) g)

12. a) t = 2004 − 1950 = 54

N (t) = N0 e−kt

2, 110, 000 2, 110, 000 5, 650, 000 " ! 2, 110, 000 ln 5, 650, 000 " ! 2, 110, 000 ln 5, 650, 000 " ! 2, 110, 000 ln 5, 650, 000 −54 0.018

ln 2 ≈ 8.3 yr 0.084 ln 2 d) k = ≈ 0.063, or 6.3% 11 17, 539.32 = P0 e0.063(5) 17, 539.32 = P0 e0.063(5) $12, 800 ≈ P0 T =

9. We use the function found in Example 5. If the mummy has lost 46% of its carbon-14 from an initial amount P0 , then 54%P0 , or 0.54P0 remains. We substitute in the function. 0.54P0 = P0 e−0.00012t 0.54 = e

−0.00012t

ln 0.54 = ln e−0.00012t ln 0.54 = −0.00012t ln 0.54 =t −0.00012 5135 ≈ t

The mummy is about 5135 years old. 10. Let x = the percent of carbon-14 remaining in the mummies. For 3300 yr: xP0 = P0 e−0.00012(3300) x = e−0.00012(3300) x ≈ 0.673, or 67.3%

If 63.7% of the carbon-14 remains, the mummies have lost 100% − 67.3%, or 32.7%, of their carbon-14. For 3500 yr: xP0 = P0 e

−0.00012(3500)

x = e−0.00012(3500) x ≈ 0.657, or 65.7%

If 65.7% of the carbon-14 remains, the mummies have lost 100% − 65.7%, or 34.3%, of their carbon-14. The mummies have lost about 32.7% to 34.3% of their carbon-14.

ln 2 ≈ 0.231, or 23.1% per min 3 ln 2 ≈ 0.0315, or 3.15% per yr k= 22 ln 2 T = ≈ 7.2 days 0.096 ln 2 ≈ 11 yr T = 0.063 ln 2 ≈ 0.028, or 2.8% per yr k= 25 ln 2 ≈ 0.00015, or 0.015% per yr k= 4560 ln 2 ≈ 0.00003, or 0.003% per yr k= 23, 105

11. a) K =

= 5, 650, 000e−k(54) = e−54k = ln e−54k = −54k =k ≈k

N (t) = 5, 650, 000e−0.018t b) In 2006, t = 2006 − 1950 = 56.

N (56) = 5, 650, 000e−0.018(56) ≈ 2, 061, 957 farms

In 2009, t = 2009 − 1950 = 59. c)

N (59) = 5, 650, 000e−0.018(59) ≈ 1, 953, 564 farms 1, 000, 000 1, 000, 000 5, 650, 000 ! " 1, 000, 000 ln 5, 650, 000 ! " 1, 000, 000 ln 5, 650, 000 " ! 1, 000, 000 ln 5, 650, 000 −0.018 96

= 5, 650, 000e−0.018t = e−0.018t = ln e−0.018t = −0.018t =t ≈t

Only 1,000,000 farms will remain about 96 years after 1950, or in 2046.

Exercise Set 4.6 13. a)

315

N (t) = N0 e−kt 4.8 4.8 125.0 4.8 ln 125.0 4.8 ln 125.0 4.8 ln 125.0 −20 0.1630

= 125.0e−k·20 = e−20k = ln e−20k = −20k =k ≈k

We have N (t) = 125.0e−0.1630t , where t is the number of years since 1983. b) In 1995, t = 1995 − 1983 = 12.

N (12) = 125.0e−0.1630(12) ≈ 17.7 students per computer In 2007, t = 2007 − 1983 = 24.

N (24) = 125.0e−0.1630(24) ≈ 2.5 students per computer c) Substitute 1 for N (t) and solve for t. 1 = 125.0e−0.1630t 0.008 = e−0.1630t ln 0.008 = ln e−0.1630t ln 0.008 = −0.1630t

ln 0.008 =t −0.1630 30 ≈ t

There will be a computer for each student about 30 yr after 1983, or in 2013. 14. a)

27, 000 = 800e

k·39

33.75 = e39k ln 33.75 = ln e

39k

ln 33.75 = 39k 0.0902 ≈ k

We have V (t) = 800e0.0902t , where t is the number of years after 1967. b) V (41) = 800e0.0902(41) ≈ $32, 300 ln 2 c) T = ≈ 7.7 yr 0.0902 d) 40, 000 = 800e0.0902t 50 = e

0.0902t

ln 50 = ln e0.0902t ln 50 = 0.0902t 43 ≈ t

The value of the car will be $40,000 about 43 yr after 1967, or in 2010.

3500 ≈ 167 1 + 19.9e−0.6(0) 3500 ≈ 500 b) N (2) = 1 + 19.9e−0.6(2) 3500 N (5) = ≈ 1758 1 + 19.9e−0.6(5) 3500 ≈ 3007 N (8) = 1 + 19.9e−0.6(8) 3500 ≈ 3449 N (12) = 1 + 19.9e−0.6(12) 3500 ≈ 3495 N (16) = 1 + 19.9e−0.6(16) c) As t → ∞, N (t) → 3500; the number of people infected approaches 3500 but never actually reaches it.

15. a) N (0) =

2500 = 400 1 + 5.25e−0.32(0) 2500 ≈ 520 P (1) = 1 + 5.25e−0.32(1) 2500 ≈ 1214 P (5) = 1 + 5.25e−0.32(5) 2500 ≈ 2059 P (10) = 1 + 5.25e−0.32(10) 2500 ≈ 2396 P (15) = 1 + 5.25e−0.32(15) 2500 ≈ 2478 P (20) = 1 + 5.25e−0.32(20)

16. P (0) =

17. To find k we substitute 105 for T1 , 0 for T0 , 5 for t, and 70 for T (t) and solve for k. 70 = 0 + (105 − 0)e−5k

70 70 105 70 ln 105 70 ln 105 70 ln 105 −5 0.081

= 105e−5k = e−5k = ln e−5k = −5k =k ≈k

The function is T (t) = 105e−0.081t . Now we find T (10). T (10) = 105e−0.081(10) ≈ 46.7 ◦ F

316

Chapter 4: Exponential and Logarithmic Functions

18. To find k we substitute 375 for T1 , 72 for T0 , 3 for t, and 365 for T (t). 365 = 72 + (375 − 72)e−3k

293 293 303 293 ln 303 293 ln 303 293 ln 303 −3 0.0112

= 303e−3k = e−3k = ln e−3k

98.6 = 70 + 24.6e−0.0008t 28.6 28.6 24.6 28.6 ln 24.6 28.6 ln 24.6

= 24.6e−0.0008t = e−0.0008t = ln e−0.0008t = −0.0008t

28.6 24.6 ≈ −188 t= −0.0008 The murder was committed at approximately 188 minutes, or about 3 hours, before 12:00 PM, or at about 9:00 AM. (Answers may vary slightly due to rounding differences.)

= −3k

ln

=k ≈k

The function is T (t) = 72 + 303e−0.0112t . T (15) = 72 + 303e

−0.0112(15)

≈ 328 F ◦

19. To find k we substitute 43 for T1 , 68 for T0 , 12 for t, and 55 for T (t) and solve for k.

21. a) y = 0.1367470864(1.024068023)x , where x is the number of years after 1900. b)

Since r2 ≈ 0.9688, the function is a good fit. 2

55 = 68 + (43 − 68)e−12k

−13 = −25e−12k 0.52 = e−12k

0

ln 0.52 = ln e−12k ln 0.52 = −12k

120

c) In 2007, x = 2007 − 1900 = 107.

0.0545 ≈ k

The function is T (t) = 68 − 25e

−0.0545t

y = 0.1367470864(1.024068023)107 ≈ 1.7%

.

In 2015, x = 2015 − 1900 = 115.

Now we find T (20). T (20) = 68 − 25e

−0.0545(20)

y = 0.1367470864(1.024068023)115 ≈ 2.1%

≈ 59.6 F ◦

20. To find k we substitute 94.6 for T1 , 70 for T0 , 60 for t (1 hr = 60 min), and 93.4 for T (t). 93.4 = 70 + |94.6 − 70|e−k(60)

23.4 23.4 24.6 23.4 ln 24.6 23.4 ln 24.6

0

= 24.6e−60k = e−60k

In 2020, x = 2020 − 1900 = 120.

y = 0.1367470864(1.024068023)120 ≈ 2.4% 22. a) Using the logarithmic regression feature on a graphing calculator, we get y = 84.94353992 − 0.5412834098 ln x. b) For x = 8, y ≈ 83.8%.

For x = 10, y ≈ 83.7%. For x = 24, y ≈ 83.2%.

= ln e

−60k

= −60k

23.4 24.6 ≈ 0.0008 k= −60 The function is T (t) = 70 + 24.6e−0.0008t .

For x = 36, y ≈ 83.0%.

c) 82 = 84.94353992 − 0.5412834098 ln x

Graph y1 = 84.94353992 − 0.5412834098 ln x and y2 = 82 and find the first coordinate of the point of intersection of the graphs. It is approximately 230, so test scores will fall below 82% after about 230 months.

ln

We substitute 98.6 for T (t) and solve for t. 23. a)

25

0

0

50

Exercise Set 4.6

317

b) Linear: y = 0.3920710572x + 0.695407279; r2 ≈ 0.9465

Quadratic: y = 0.0072218083x2 + 0.1172668587x + 1.973805022; r2 ≈ 0.9851

Exponential: y = 2.062091236(1.059437743)x ; r2 ≈ 0.9775

26. a) Using the logistic regression feature on a graphing calculator, we get 99.98884912 y= . 1 + 489.2438401e−0.1299899024x b) For x = 55, y ≈ 72.2%. For x = 100, y ≈ 99.9%.

The value of r2 is greatest for the quadratic function, so it provides the best fit.

c)

c) y = 100 is the horizontal asymptote; as more and more ads are run, the percent of people who bought the product approaches 100%.

25

27. Answers will vary.

0

0

50

d) In 2008, x = 2008 − 1967 = 41.

For the linear function, when x = 41, y ≈ 16.8%.

For the quadratic function, when x = 41, y ≈ 18.9%.

For the exponential function, when x = 41, y ≈ 22.0%.

Given the trend shown in the table, the quadratic model seems the most realistic. Answers may vary.

24. a) Using the exponential regression feature on a graphing calculator, we get y = 22.28866067(1.092917808)x , where x is the number of years after 1980 and y is in millions of dollars. b) In 2008, x = 2008 − 1980 = 28.

28. Measure the atmospheric pressure P at the top of the building. Substitute that value in the equation P = 14.7e−0.00005a , and solve for the height, or altitude, a, of the top of the building. Also measure the atmospheric pressure at the base of the building and solve for the altitude of the base. Then subtract to find the height of the building. 29. Multiplication principle for inequalities 30. Product rule 31. Principle of zero products 32. Principle of square roots 33. Power rule 34. Multiplication principle for equations 35.

480e−0.003p = 150e0.004p 480 e0.004p = −0.003p 150 e 3.2 = e0.007p

y = 22.28866067(1.092917808)28 ≈ $268.2 million In 2012, x = 2012 − 1980 = 32.

y = 22.28866067(1.092917808)32 ≈ $382.7 million

c) Graph y1 = 22.28866067(1.092917808)x and y2 = 546 and find the first coordinate of the point of intersection of the graphs. It is slightly less than 36, so the total cost will first exceed $546 million for the first convention year that is 36 yr after 1980, or 2016.

ln 3.2 = ln e0.007p

36.

25. a) y = 255.0890581(1.277632801)x , where x is the number of years after 1995.

In 2009, x = 2009 − 1995 = 14.

y = 255.0890581(1.277632801)14 ≈ 7877 surgeries

c) Graph y1 = 255.0890581(1.277632801)x and y2 = 12, 000 and find the first coordinate of the point of intersection of the graphs. It is approximately 16, so there will be 12,000 surgeries about 16 yr after 1995, or in 2011.

P (4000) = P0 e−0.00012(4000) = 0.619P0 , or 61.9%P0 Thus, about 61.9% of the carbon-14 remains, so about 38.1% has been lost.

b) In 2005, x = 2005 − 1995 = 10.

y = 255.0890581(1.277632801)10 ≈ 2956 surgeries

ln 3.2 = 0.007p ln 3.2 =p 0.007 $166.16 ≈ p

37.

P (t) = P0 ekt 50, 000 = P0 e0.07(18) 50, 000 = P0 e0.07(18) $14, 182.70 ≈ P0

38. a)

P = P0 ekt P = P0 , or ekt P e−kt = P0

b) P0 = 50, 000e−0.064(18) ≈ $15, 800.21

318 39.

Chapter 4: Exponential and Logarithmic Functions $ % V −(R/L)t i= 1−e R iR = 1 − e−(R/L)t V iR e−(R/L)t = 1 − V ! " iR −(R/L)t = ln 1 − ln e V ! " R iR − t = ln 1 − L V $ ! "% iR L ln 1 − t=− R V

5. The domain of all logarithmic functions is (0, ∞), so the statement is false. 6. The statement is true. See page 377 in the text. 7. We interchange the first and second coordinates of each pair to find the inverse of the relation. It is {(−2.7, 1.3), (−3, 8), (3, −5), (−3, 6), (−5, 7)}. 8. a) x = −2y + 3

b) x = 3y 2 + 2y − 1

c) 0.8y 3 − 5.4x2 = 3y 9. The graph of f (x) = −|x|+3 is shown below. The function is not one-to-one, because there are many horizontal lines that cross the graph more than once.

40. a) At 1 m: I = I0 e−1.4(1) ≈ 0.247I0 24.7% of I0 remains.

y

At 3 m: I = I0 e−1.4(3) ≈ 0.015I0

5

1.5% of I0 remains.

At 5 m: I = I0 e

−1.4(5)

≈ 0.0009I0

2 1

At 50 m: I = I0 e−1.4(50) ≈ (3.98 × 10−31 )I0 Now, 3.98 × 10 (3.98 × 10

−29

= (3.98 × 10

)% remains.

−29

) × 10

−2

-5 -4 -3 -2 -1

Thus, 0.00008% remains.

y = aex ln y = ln(aex )

-1

5

10. The graph of f (x) = x2 + 1 is shown below. The function is not one-to-one, because there are many horizontal lines that cross the graph more than once. y

Y = x + ln a

3

4

2

This function is of the form y = mx + b, so it is linear.

f (x ) =

x2 + 1

1 -5 -4 -3 -2 -1

x 1

-1

ln y = ln(axb )

-2

ln y = ln a + b ln x

-4

2

3

4

5

-3 -5

Y = ln a + bX

Chapter 4 Review Exercises

4

-5

5

This function is of the form y = mx + b, so it is linear.

3

-4

ln y = ln a + x

y = axb

2

-3

ln y = ln a + ln ex

42.

x 1

-2

, so

b) I = I0 e−1.4(10) ≈ 0.0000008I0 41.

3

3

0.09% of I0 remains. −31

f (x ) = – | x | +

4

3 11. The graph of f (x) = 2x − is shown below. The function 4 is one-to-one, because no horizontal line crosses the graph more than once. y

1. The statement is true. See page 351 in the text. 2. f (0) = e is false.

−0

= 1, so the y-intercept is (0, 1). The statement

5 4 3 f (x ) = 2 x – — 4

3 2 1

3. The graph of f −1 is a reflection of the graph of f across y = x, so the statement is false. 4. The statement is true. See page 352 in the text.

-5 -4 -3 -2 -1

-1 -2 -3 -4 -5

x 1

2

3

4

5

Chapter 4 Review Exercises

319

6 is shown below. The function x+1 is one-to-one, because no horizontal line crosses the graph more than once.

12. The graph of f (x) = −

x+2 x−1 y+2 Interchange x and y: x = y−1

b) Replace f (x) with y: y =

Solve for y: (y − 1)x = y + 2

y

xy − x = y + 2

10 8 4 2 -10 -8 -6 -4 -2

xy − y = x + 2

6 f (x ) = – —— x+1

6

y(x − 1) = x + 2 x+2 y= x−1

x 2

-2

4

6

8 10

-4

Replace y with f −1 (x): f −1 (x) =

-6 -8 -10

13. a) The graph of f (x) = 2−3x is shown below. It passes the horizontal-line test, so the function is one-toone.

x+2 x−1

√ 15. a) The graph of f (x) = x − 6 is shown below. It passes the horizontal-line test, so the function is oneto-one. y 10

y

8 6

5

f (x ) =

4

4

x–6

2

x

3 f (x ) = 2 – 3 x

2

-10 -8 -6 -4 -2

1 -5 -4 -3 -2 -1

-2

1

2

3

4

4

6

8 10

-4

x

-1

2

5

-6 -8

-2

-10

-3 -4

b) Replace f (x) with y: y =

-5

Interchange x and y: x =

b) Replace f (x) with y: y = 2 − 3x Interchange x and y: x = 2 − 3y −x + 2 Solve for y: y = 3

Replace y with f −1 (x): f −1 (x) =

Solve for y:

x2 = y − 6

−x + 2 3

Replace y with f −1 (x): f −1 (x) = x2 + 6, for all x in the range of f (x), or f −1 (x) = x2 + 6, x ≥ 0. 16. a) The graph of f (x) = x3 −8 is shown below. It passes the horizontal-line test, so the function is one-toone. y 10

y

8

5

6

4

4 2

3 -10 -8 -6 -4 -2

2 1 -5 -4 -3 -2 -1

-1

x−6 √ y−6

x2 + 6 = y

x+2 14. a) The graph of f (x) = is shown below. It passes x−1 the horizontal-line test, so the function is one-toone.

x+2 f (x ) = —— x–1

√

x 1

2

3

4

-2 -4

5

x 2

4

6

f (x ) =

8 10

x3 – 8

-6 -8

-2

-10

-3 -4 -5

b) Replace f (x) with y: y = x3 − 8 Interchange x and y: x = y 3 − 8 Solve for y:

√ 3

x + 8 = y3 x+8 = y

Replace y with f −1 (x): f −1 (x) =

√ 3

x+8

320

Chapter 4: Exponential and Logarithmic Functions

17. a) The graph of f (x) = 3x2 + 2x − 1 is shown below. It is not one-to-one since there are many horizontal lines that cross the graph more than once. The function does not have an inverse that is a function.

21. Replace f (x) with y: y = 2 − 5x Interchange x and y: x = 2 − 5y 2−x Solve for y: y = 5

2−x 5 The domain and range of f are (−∞, ∞), so the domain and range of f −1 are also (−∞, ∞).

y

Replace y with f −1 (x): f −1 (x) =

5 4 3 2 1 -5 -4 -3 -2 -1

-1

x 1

2

3

4

2 -2 f (x ) = 3 x + 2 x

5

—1

-3 -4 -5

18. a) The graph of f (x) = ex is shown below. It passes the horizontal-line test, so the function is one-toone.

x−3 x+2 y−3 Interchange x and y: x = y+2

22. Replace f (x) with y: y =

y

Solve for y: xy + 2x = y − 3

2x + 3 = y − xy

2x + 3 = y(1 − x) 2x + 3 =y 1−x

f (x) ! e x x

2x + 3 −2x − 3 , or 1−x x−1 The domain of f is (−∞, −2) ∪ (−2, ∞), and the range of f is (−∞, 1) ∪ (1, ∞). Thus the domain of f −1 is (−∞, 1) ∪ (1, ∞) and the range of f −1 is (−∞, −2) ∪ (−2, ∞). Replace y with f −1 (x): f −1 (x) =

b) Replace f (x) with y: y = ex Interchange x and y: x = ey Solve for y: y = ln x Replace y with f −1 (x): f −1 (x) = ln x 19. We find (f −1 ◦f )(x) and (f ◦f −1 )(x) and check to see that each is x. (f −1 ◦ f )(x) = f −1 (f (x)) = f −1 (6x − 5) = 6x (6x − 5) + 5 = =x 6 6 " ! x+5 (f ◦ f −1 )(x) = f (f −1 (x)) = f = 6 " ! x+5 −5=x+5−5=x 6 6 20.

" ! x+1 = (f −1 ◦ f )(x) = f −1 (f (x)) = f −1 x x 1 x " ! = = =x x+1 x+1−x 1 −1 x ! " 1 = (f ◦ f −1 )(x) = f (f −1 (x)) = f x−1 " ! 1 +1 1 + (x − 1) x x−1 = = =x 1 1 1 x−1

23. Since f (f −1 (x)) = x, then f (f −1 (657)) = 657. 24. Since f (f −1 (x)) = x, then f (f −1 (a)) = a. 25.

Chapter 4 Review Exercises 26.

321

34. y =

! "x 1 2

At x = 0, y =

! "0 1 =1 2

The only graph with y-intercept (0, 1) is (f). 35. f (x) = 3(1 − e−x ), x ≥ 0

This is the graph of f (x) = ex reflected across the y-axis, reflected across the x-axis, shifted up 1 unit, and stretched by a factor of 3. The correct choice is (e).

27.

36. f (x) = | ln(x − 4)|

This is the graph of f (x) = ln x shifted right 4 units. The absolute value reflects negative outputs across the x-axis. The line x = 4 is a vertical asymptote. The correct choice is (d).

37. log5 125 = 3 because the exponent to which we raise 5 to get 125 is 3.

28.

38. log 100, 000 = 5 because the exponent to which we raise 10 to get 100,000 is 5. 39. ln e = 1 because the exponent to which we raise e to get e is 1. 40. ln 1 = 0 because the exponent to which we raise e to get 1 is 0. 1 because the exponent to which we raise 10 4 1 to get 101/4 is . 4 √ 1 42. log3 3 = log3 31/2 = because the exponent to which we 2 1 raise 3 to get 31/2 is . 2 41. log 101/4 =

29.

43. log 1 = 0 because the exponent to which we raise 10 to get 1 is 0. 44. log 10 = 1 because the exponent to which we raise 10 to get 10 is 1.

30.

√ 3

1 2 = log2 21/3 = because the exponent to which we 3 1 1/3 is . raise 2 to get 2 3

45. log2

46. log 0.01 = −2 because the exponent to which we raise 10 to get 0.01 is −2. 31. f (x) = ex−3 This is the graph of f (x) = ex shifted right 3 units. The correct choice is (c). 32.

f (x) = log3 x f (1) = log3 (1) = 0 The only graph with x-intercept (1, 0) is (a).

33. f (x) = − log3 (x + 1)

This is the graph of log3 x shifted left 1 unit and reflected across the y-axis. The correct choice is (b).

47. log4 x = 2 ⇒ 42 = x 48. loga Q = k ⇒ ak = Q 49. 4−3 =

1 1 ⇒ log4 = −3 64 64

50. ex = 80 ⇒ ln 80 = x 51. log 11 ≈ 1.0414 52. log 0.234 ≈ −0.6308 53. ln 3 ≈ 1.0986 54. ln 0.027 ≈ −3.6119

322

Chapter 4: Exponential and Logarithmic Functions

55. log(−3) does not exist. (The calculator gives an error message.)

65.

loga

56. ln 0 does not exist. (The calculator gives an error message.) log 24 ≈ 1.9746 57. log5 24 = log 5 58. log8 3 = 59.

66.

log 3 ≈ 0.5283 log 8

1 logb z 2 = logb x3 − logb y 4 + logb z 1/2 √ x3 z 1/2 x3 z = logb , or log b y4 y4 3 logb x − 4 logb y +

60.

ln(x3 − 8) − ln(x2 + 2x + 4) + ln(x + 2) = ln

2

61.

62.

63.

ln

log

& 3

= ln(wr ) 1 = ln wr2 4 1 = (ln w + ln r2 ) 4 1 = (ln w + 2 ln r) 4 1 1 = ln w + ln r 4 2

M2

69.

N

= log

2

M N

"1/3

M2 1 = log 3 N 1 = (log M 2 − log N ) 3 1 = (2 log M − log N ) 3 2 1 = log M − log N 3 3 ! " 6 loga 3 = loga 2 = loga 6 − loga 2 ≈ 0.778 − 0.301

64.

≈ 0.477

loga 50 = loga (2 · 52 )

= loga 2 + loga 52 = loga 2 + 2 loga 5 ≈ 0.301 + 2(0.699) ≈ 1.699

(loga ax = x)

log4 x = 2 x = 42 = 16

70.

31−x = 92x 31−x = (32 )2x 31−x = 34x 1 − x = 4x 1 = 5x 1 =x 5

2 1/4

!

5 = loga 51/3 1 = loga 5 3 1 ≈ (0.699) 3 ≈ 0.233

68. log5 5−6t = −6t

= ln(x2 − 4) wr2

≈ −0.699

The solution is 16.

(x − 2)(x + 2x + 4)(x + 2) x2 + 2x + 4 = ln(x − 2)(x + 2) √ 4

√ 3

67. ln e−5k = −5k

(x3 − 8)(x + 2) x2 + 2x + 4

= ln

loga

1 = loga 5−1 5 = − loga 5

The solution is 71.

1 . 5

ex = 80 ln ex = ln 80 x = ln 80 x ≈ 4.382

The solution is 4.382. 72.

42x−1 − 3 = 61

42x−1 = 64 42x−1 = 43

2x − 1 = 3

2x = 4 x=2

The solution is 2. 73.

log16 4 = x 16x = 4 (42 )x = 41 42x = 41 2x = 1 1 x= 2 The solution is

1 . 2

Chapter 4 Review Exercises 74.

323

logx 125 = 3

82. k =

x3 = 125 √ x = 3 125

83.

P (t) = P0 ekt 0.73P0 = P0 e−0.00012t

x=5

0.73 = e−0.00012t

The solution is 5. 75.

ln 0.73 = ln e−0.00012t

log2 x + log2 (x − 2) = 3

ln 0.73 = −0.00012t ln 0.73 =t −0.00012 2623 ≈ t

log2 x(x − 2) = 3

x(x − 2) = 23 x2 − 2x = 8

The skeleton is about 2623 years old.

x − 2x − 8 = 0 2

(x + 2)(x − 4) = 0

x+2 = 0

84. pH = − log[2.3 × 10−6 ] ≈ −(−5.6) = 5.6

or x − 4 = 0

x = −2 or

85. R = log

x=4

The number 4 checks, but −2 does not. The solution is 4. 76.

log(x − 1) − log(x − 1) = 1 2

x2 − 1 =1 x−1 (x + 1)(x − 1) = 101 x−1 x + 1 = 10

86.

1000 · I0 I0 = 10 log 1000 = 10 · 3

= 30 decibels 87. a) We substitute 353.823 for P , since P is in thousands. W (353.823) = 0.37 ln 353.823 + 0.05

x=9 The answer checks. The solution is 9. log x2 = log x

≈ 2.2 ft/sec

b) We substitute 3.4 for W and solve for P . 3.4 = 3.7 ln P + 0.05

x2 = x

3.35 = 0.37 ln P 3.35 = ln P 0.37 e3.35/0.37 = P

x2 − x = 0

x(x − 1) = 0

x = 0 or x − 1 = 0 x = 0 or

P ≈ 8553.143

x=1

The population is about 8553.143 thousand, or 8,553,143. (Answers may vary due to rounding differences.)

The number 1 checks, but 0 does not. The solution is 1. 78.

e−x = 0.02 ln e−x = ln 0.02

88. a)

−x = ln 0.02

x = − ln 0.02

x ≈ 3.912

The answer checks. The solution is 3.912. 79. a) A(t) = 16, 000(1.0105)4t b)

A(0) = 16, 000(1.0105)4·0 = $16, 000 A(6) = 16, 000(1.0105)4·6 ≈ $20, 558.51

A(12) = 16, 000(1.0105)

4·12

A(18) = 16, 000(1.0105)

4·18

80.

≈ $26, 415.77 ≈ $33, 941.80

W (10) = 1665.945(1.087)10 ≈ 3837 cases W (20) = 1665.945(1.087)20 ≈ 8836 cases

W (27) = 1665.945(1.087)27 ≈ 15, 844 cases ln 2 ≈ 8.1 years 81. T = 0.086

106.3 · I0 = log 106.3 = 6.3 I0

L = 10 log

log

77.

ln 2 ≈ 0.023, or 2.3% 30

492 492 0.035 492 ln 0.035 492 ln 0.035 492 ln 0.035 64 0.1492

= 0.035ek·64 = e64k = ln e64k = 64k =k ≈k

b) We have S(t) = 0.035e0.1492t , where S(t) is in billions of dollars and t is the number of years after 1940. c) S(25) ≈ $1.459 billion S(55) ≈ $128.2 billion

S(75) ≈ $2534 billion, or $2.534 trillion

324

Chapter 4: Exponential and Logarithmic Functions 4 + 3x x+4 , y2 = , and y3 = x and obx−2 x−3 serve that the graphs of y1 and y2 are not reflections of each other across the third line, y = x. Thus f (x) and g(x) are not inverses.

d) 1 trillion is 1000 billion. 1000 = 0.035e0.1492t 1000 = e0.1492t 0.035 1000 ln = ln e0.1492t 0.035 1000 ln = 0.1492t 0.035 1000 ln 0.035 = t 0.1492 69 ≈ t

93. We graph y1 =

94. a)

Cash benefits will reach $1 trillion about 69 yr after 1940, or in 2009.

89. a) P (t) = 3.039e0.013t , where P (t) is in millions and t is the number of years after 2005. b) In 2009, t = 2009 − 2005 = 4.

P (4) = 3.039e0.013(4) ≈ 3.201 million

In 2015, t = 2015 − 2005 = 10. P (10) = 3.039e

0.013(10)

c)

10 10 3.039 " ! 10 ln 3.039 " ! 10 ln 3.039 ! " 10 ln 3.039 0.013 92

= 3.039e

≈ 3.461 million

0.013t

x = 4−3 or 1 x= or 64

= ln e0.013t = 0.013t

x = 43 x = 64

Both answers check. The solutions are

1 and 64. 64

98. log x = ln x

=t

Graph y1 = log x and y2 = ln x and find the first coordinates of the points of intersection of the graph. We see that the only solution is 1.

≈t

91. We must have 2x−3 > 0, or x >

3 , so answer A is correct. 2

92. a) y = 2.518123986(1.301660678)x , where y is in millions of dollars and x is the number of years after 2001. y ! 2.518123986(1.301660678)x 10

0

96. The inverse of a function f (x) is written f −1 (x), whereas 1 . [f (x)]−1 means f (x) log4 x = −3 or log4 x = 3

90. The graph of f (x) = ex−3 + 2 is a translation of the graph of y = ex right three units and up 2 units. The horizontal asymptote of y = ex is y = 0, so the horizontal asymptote of f (x) = ex−3 + 2 is translated up 2 units from y = 0. Thus, it is y = 2, and answer D is correct.

0

95. By the product rule, log2 x + log2 5 = log2 5x, not log2 (x+5). Also, substituting various numbers for x shows that the two sides of the inequality are indeed unequal. You could also graph each side and show that the graphs do not coincide.

97. | log4 x| = 3

= e0.013t

The population will be 10 million about 92 yr after 2005. ln 2 ≈ 53.3 yr d) T = 0.013

b)

b) Relative maximum: 0.486 at x = 1.763; no relative minimum

99.

5

√

x

√

x

= 625

5 = 54 √ x=4 x = 16 100. f (x) = log3 (ln x) ln x must be positive, so x > 1. The domain is (1, ∞).

Chapter 4 Test 1. We interchange the first and second coordinates of each pair to find the inverse of the relation. It is {(5, −2), (3, 4)(−1, 0), (−3, −6)}. 2. The function is not one-to-one, because there are many horizontal lines that cross the graph more than once.

5

c) For x = 9, y ≈ $27 million

3. The function is one-to-one, because no horizontal line crosses the graph more than once.

Chapter 4 Test

325

4. a) The graph of f (x) = x3 +1 is shown below. It passes the horizontal-line test, so the function is one-toone.

Solve for y: (2 − y)x = y

y

2x − xy = y

5 4

xy + y = 2x

3

x3 + 1

f (x ) =

2 1 -5 -4 -3 -2 -1

x 2−x y Interchange x and y: x = 2−y

b) Replace f (x) with y: y =

y(x + 1) = 2x 2x y= x+1

x 1

-1

2

3

4

5

-2

Replace y with f −1 (x): f −1 (x) =

-3 -4 -5

7. a) The graph of f (x) = x2 + x − 3 is shown below. It is not one-to-one since there are many horizontal lines that cross the graph more than once. The function does not have an inverse that is a function.

b) Replace f (x) with y: y = x3 + 1 Interchange x and y: x = y 3 + 1 Solve for y: y 3 = x − 1 √ y = 3x−1

2x x+1

y

Replace y with f −1 (x): f −1 (x) =

√ 3

5 4

x−1

5. a) The graph of f (x) = 1 − x is shown below. It passes the horizontal-line test, so the function is one-toone.

3 2 1 -5 -4 -3 -2 -1

-1

x 1

2

3

4

5

-2 y

-3 -4

5

f (x ) =

x2 + x — 3

-5

4 3 f (x ) = 1 –

2

x

1 -5 -4 -3 -2 -1

x 1

-1

2

3

4

5

(f −1 ◦ f )(x) = f −1 (f (x)) = f −1 (−4x + 3) = 3 − (−4x + 3) 3 + 4x − 3 4x = = =x 4 4 4 " ! 3−x = (f ◦ f −1 )(x) = f (f −1 (x)) = f 4 ! " 3−x −4 + 3 = −3 + x + 3 = x 4

-2 -3 -4 -5

b) Replace f (x) with y: y = 1 − x Interchange x and y: x = 1 − y Solve for y: y = 1 − x

Replace y with f −1 (x): f −1 (x) = 1 − x x is shown below. It passes 2−x the horizontal-line test, so the function is one-toone.

6. a) The graph of f (x) =

y

4 x f (x ) = —— 2–x

2 1 -5 -4 -3 -2 -1

-1 -2 -3 -4 -5

x 1

2

3

4

5

1 x−4 1 Interchange x and y: x = y−4

9. Replace f (x) with y: y =

Solve for y: x(y − 4) = 1

xy − 4x = 1

xy = 4x + 1 4x + 1 y= x

5 3

8. We find (f −1 ◦f )(x) and (f ◦f −1 )(x) and check to see that each is x.

4x + 1 x The domain of f (x) is (−∞, 4) ∪ (4, ∞) and the range of f (x) is (−∞, 0) ∪ (0, ∞). Thus, the domain of f −1 is (−∞, 0) ∪ (0, ∞) and the range of f −1 is (−∞, 4) ∪ (4, ∞). Replace y with f −1 (x): f −1 (x) =

326

Chapter 4: Exponential and Logarithmic Functions 19. 3x = 5.4 ⇒ x = log3 5.4

20. ln 16 ≈ 2.7726

21. log 0.293 ≈ −0.5331 22. log6 10 = 23. 10.

24.

11.

25.

log 10 ≈ 1.2851 log 6

1 loga z 2 = loga x2 − loga y + loga z 1/2 √ x2 z 1/2 x2 z , or loga = loga y y # 5 ln x2 y = ln(x2 y)1/5 1 = ln x2 y 5 1 = (ln x2 + ln y) 5 1 = (2 ln x + ln y) 5 1 2 = ln x + ln y 5 5 ! " 8 loga 4 = loga 2 = loga 8 − loga 2 2 loga x − loga y +

≈ 0.984 − 0.328 ≈ 0.656

12. 26. ln e

−4t

27.

= −4t

(loga ax = x)

log25 5 = x 25x = 5

(52 )x = 51 52x = 51 2x = 1 1 x= 2

13.

1 . 2 log3 x + log3 (x + 8) = 2

The solution is 28.

log3 x(x + 8) = 2 x(x + 8) = 32 x2 + 8x = 9 x + 8x − 9 = 0 2

14. log 0.00001 = −5 because the exponent to which we raise 10 to get 0.00001 is −5.

(x + 9)(x − 1) = 0 x = −9 or x = 1

15. ln e = 1 because the exponent to which we raise e to get e is 1. 16. ln 1 = 0 because the exponent to which we raise e to get 1 is 0. √ 1 5 17. log4 4 = log4 41/5 = because the exponent to which we 5 1 1/5 raise 4 to get 4 is . 5 18. ln x = 4 ⇒ x = e4

29.

The number 1 checks, but −9 does not. The solution is 1. 34−x = 27x

34−x = (33 )x 34−x = 33x 4 − x = 3x 4 = 4x

x=1 The solution is 1.

Chapter 4 Test 30.

ex = 65 ln ex = ln 65 x = ln 65 x ≈ 4.174

The solution is 4.174. 31. R = log

106.6 · I0 = log 106.6 = 6.6 I0

ln 2 ≈ 0.0154 ≈ 1.54% 45

32. k = 33. a)

1144.54 = 1000e3k 1.14454 = e3k ln 1.14454 = ln e3k ln 1.14454 = 3k ln 1.14454 =k 3 0.045 ≈ k

The interest rate is about 4.5%. b) P (t) = 1000e0.045t c) P (8) = 1000e0.045·8 ≈ $1433.33 ln 2 d) T = ≈ 15.4 yr 0.045 34.

4 (2 ) 2

2

2

√ 3

x

√ 3

=8

x

√ 3

= 23

x

= 23

√ 3

2 x=3 √ 3 3 x= 2 ! "3 3 x= 2 27 x= 8 27 The solution is . 8

327

Chapter 4

Exponential and Logarithmic Functions $ 2 ,0 . 3 Plot these points and draw the line. Then reflect the graph across the line y = x.

13. Graph y = 3x − 2. The intercepts are (0, −2) and

Exercise Set 4.1 1. We interchange the first and second coordinates of each ordered pair to find the inverse of the relation. It is {(8, 7), (8, −2), (−4, 3), (−8, 8)}.

y

2. {(1, 0), (6, 5), (−4, −2)}

4

3. We interchange the first and second coordinates of each ordered pair to find the inverse of the relation. It is

2

x " 3y ! 2 !4

{(−1, −1), (4, −3)}.

!2

2 !2

5. Interchange x and y. y = 4x−5   " " x = 4y−5

14.

4 2

!4

7. Interchange x and y. y = −5 x3  "" y 3 x = −5

!2

2 !2 !4

y"x

9. Interchange x and y. x = y2 −2y   " " " y = x2 −2x

y " !x!

y 4 2

11. Graph x = y − 3. Some points on the graph are (−3, 0), (−2, −1), (−2, 1), (1, −2), and (1, 2). Plot these points and draw the curve. Then reflect the graph across the line y = x. 2

y x2

!3

4

!4

x " !y! y

x

2 4

x

!2

x " y2 # 1

4 2

!4

!2

2 !2

x"y

4

!4

4

x " y2 ! 3

y

2

!4

y"x

16. 2

4

!2

y"x

!4

!4 !2

2

!4 !2

y # 2 " |x|

y"x

x

y " !x # 4

1 x+4 2

y " x2 # 1

4

15. Graph y = |x|. Some points on the graph are (0, 0), (−2, 2), (2, 2), (−5, 5), and (5, 5). Plot these points and draw the graph. Then reflect the graph across the line y = x.

8. x = 3y 2 − 5y + 9

12.

x

y x " !y # 4

6. 2y 2 + 5x2 = 4

y"

4

y " 3x ! 2

!4

y"x

4. {(3, −1), (5, 2), (5, −3), (0, 2)}

10. y =

#

!4

4

x

x # 2 " |y|

x

274

Chapter 4: Exponential and Logarithmic Functions

17. We show that if f (a) = f (b), then a = b. 1 1 a−6 = b−6 3 3 1 1 a= b Adding 6 3 3 a=b Multiplying by 3 Thus f is one-to-one.

33. The graph of f (x) = 5x − 8 is shown below. 10 8

4 − 2a = 4 − 2b

6 4 2

−2a = −2b a=b

f(x) = 5x – 8 x

-10 -8 -6 -4 -2 0 -2

Then f is one-to-one. 19. We show that if f (a) = f (b), then a = b. 1 1 a3 + = b3 + 2 2 1 3 3 Subtracting a =b 2 a=b Taking cube roots Thus f is one-to-one.

2 4 6 8 10

-4 -6 -8 -10

Since there is no horizontal line that crosses the graph more than once, the function is one-to-one. 34. The graph of f (x) = 3 + 4x is shown below.

20. Assume f (a) = f (b). √ √ 3 a= 3b

y

Using the principle of powers

Then f is one-to-one. 21. g(−1) = 1 − (−1) = 1 − 1 = 0 and g(1) = 1 − 12 = 1 − 1 = 0, so g(−1) = g(1) but −1 #= 1. Thus the function is not one-to-one. 2

22. g(−1) = 4 and g(1) = 4, so g(−1) = g(1) but −1 #= 1. Thus the function is not one-to-one. 23. f (−2) = (−2) − (−2) = 16 − 4 = 12 and f (2) = 24 − 22 = 16− 4 = 12, so f (−2) = f (2) but −2 #= 2. Thus the function is not one-to-one. 4

32. The function is one-to-one, because no horizontal line crosses the graph more than once.

y

18. Assume f (a) = f (b).

a=b

31. The function is one-to-one, because no horizontal line crosses the graph more than once.

10 8 6 4 2

f(x) = 3 + 4x x

-10 -8 -6 -4 -2 0 -2

2 4 6 8 10

-4 -6 -8 -10

2

24. g(−1) = 1 and g(1) = 1 so g(−1) = g(1) but −1 #= 1. Thus the function is not one-to-one. 25. The function is one-to-one, because no horizontal line crosses the graph more than once.

Since there is no horizontal line that crosses the graph more than once, the function is one-to-one. 35. The graph of f (x) = 1 − x2 is shown below. y 5 4

26. The function is one-to-one, because no horizontal line crosses the graph more than once.

3 2 1

f(x) = 1 – x 2

27. The function is not one-to-one, because there are many horizontal lines that cross the graph more than once.

-5 -4 -3 -2 -1 0 -1

1 2 3 4 5

28. The function is not one-to-one, because there are many horizontal lines that cross the graph more than once. 29. The function is not one-to-one, because there are many horizontal lines that cross the graph more than once. 30. The function is one-to-one, because no horizontal line crosses the graph more than once.

x

-2 -3 -4 -5

Since there are many horizontal lines that cross the graph more than once, the function is not one-to-one.

Exercise Set 4.1

275

36. The graph of f (x) = |x| − 2 is shown below.

39. The graph of f (x) = −

y

4 is shown below. x

y

5 4

f(x) = |x | – 2 4

f (x ) = - – x

3 2 1

5 4 3 2

x

-5 -4 -3 -2 -1 0 -1

1

1 2 3 4 5

x

–5 –4 –3 –2 –1 0 –1

-2 -3 -4

1

2

3

4

5

–2 –3 –4

-5

–5

Since there are many horizontal lines that cross the graph more than once, the function is not one-to-one. 37. The graph of f (x) = |x + 2| is shown below.

Since there is no horizontal line that crosses the graph more than once, the function is one-to-one. 2 is shown below. x+3

40. The graph of f (x) =

y y

5 4

f(x) = |x + 2 |

3 2 1

5 4 x

-5 -4 -3 -2 -1 0 -1

f(x) =

3 2 1

1 2 3 4 5

x

-5 -4 -3 -2 -1 0 -1

-2 -3 -4

2 x +3

1 2 3 4 5

-2 -3 -4

-5

-5

Since there are many horizontal lines that cross the graph more than once, the function is not one-to-one. 38. The graph of f (x) = −0.8 is shown below. y

Since there is no horizontal line that crosses the graph more than once, the function is one-to-one. 2 is shown below. 3

41. The graph of f (x) =

3 y

2

5 4

1

3

x -3

-2

-1

0 -1

1

2

3

f(x) = –0.8

-2

f (x ) = –23

2 1 –5 –4 –3 –2 –1 0 –1

x 1

2

3

4

5

–2 –3

-3

–4 –5

Since the horizontal line y = −0.8 crosses the graph more than once, the function is not one-to-one.

2 Since the horizontal line y = crosses the graph more than 3 once, the function is not one-to-one.

276

Chapter 4: Exponential and Logarithmic Functions 1 2 x + 3 is shown below. 2

42. The graph of f (x) =

45. a) The graph of f (x) = x + 4 is shown below. It passes the horizontal-line test, so it is one-to-one.

y

y

5

5 4

4 3

f (x ) = –12 x 2 + 3

2 1

x

–5 –4 –3 –2 –1 0 –1

1

2

3

4

5

f(x) = x + 4

3 2 1

x

-5 -4 -3 -2 -1 0 -1

–2

1 2 3 4 5

-2 -3 -4

–3 –4 –5

-5

Since there are many horizontal lines that cross the graph more than once, the function is not one-to-one. √ 43. The graph of f (x) = 25 − x2 is shown below.

b) Replace f (x) with y: y = x + 4 Interchange x and y: x = y + 4 Solve for y: x − 4 = y

Replace y with f −1 (x): f −1 (x) = x − 4

y 5

46. a) The graph of f (x) = 7 − x is shown below. It passes the horizontal-line test, so it is one-to-one.

4 3 2 1

–2

y

x

–5 –4 –3 –2 –1 0 –1

1

2

3

4

5

10 8

f (x ) = 25 – x 2

f(x) = 7 – x

6 4 2

–3 –4 –5

x

-10 -8 -6 -4 -2 0 -2

Since there are many horizontal lines that cross the graph more than once, the function is not one-to-one. 44. The graph of f (x) = −x3 + 2 is shown below. y

-10

Interchange x and y: x = 7 − y

4

Solve for y: y = 7 − x

3 2 1

x 1

2

3

4

5

–2

f (x ) = – x 3 + 2

-4 -6 -8

b) Replace f (x) with y: y = 7 − x

5

–5 –4 –3 –2 –1 0 –1

2 4 6 8 10

Replace y with f −1 (x): f −1 (x) = 7 − x 47. a) The graph of f (x) = 2x−1 is shown below. It passes the horizontal-line test, so it is one-to-one.

–3

y

–4 –5

Since there is no horizontal line that crosses the graph more than once, the function is one-to-one.

5 4 3 2 1 -5 -4 -3 -2 -1 0 -1 -2 -3 -4 -5

f(x) = 2x – 1 x 1 2 3 4 5

Exercise Set 4.1

277 3 50. a) The graph of f (x) = − is shown below. It passes x the horizontal-line test, so it is one-to-one.

b) Replace f (x) with y: y = 2x − 1 Interchange x and y: x = 2y − 1 x+1 =y Solve for y: 2

y

x+1 Replace y with f −1 (x): f −1 (x) = 2 48. a) The graph of f (x) = 5x+8 is shown below. It passes the horizontal-line test, so it is one-to-one.

-5

3 x 3 Interchange x and y: x = − y 3 Solve for y: y = − x

x

-10 -8 -6 -4 -2 0 -2

b) Replace f (x) with y: y = −

2 4 6 8 10

-4 -6 -8 -10

Replace y with f −1 (x): f −1 (x) = −

b) Replace f (x) with y: y = 5x + 8 Interchange x and y: x = 5y + 8 x−8 =y Solve for y: 5 Replace y with f −1 (x): f −1 (x) =

x−8 5

y

6 4 2 -10 -8 -6 -4 -2 0 -2

3 x

x+4 is shown below. It passes x−3 the horizontal-line test, so the function is one-toone.

51. a) The graph of f (x) =

49. a) The graph of f (x) = 4 is shown below. It passes x+7 the horizontal-line test, so the function is one-toone.

10 8

x 1 2 3 4 5

-2 -3 -4 f(x) = 5x + 8

6 4 2

33 f(x) = – — xx

3 2 1 -5 -4 -3 -2 -1 0 -1

y 10 8

5 4

y 10 8

x+4 f(x) = x – 3

6 4 2 -10 -8 -6 -4 -2 0 -2

x 2 4 6 8 10

-4 -6 -8

4 f(x) = x+7

-10 x 2 4 6 8 10

x+4 x−3 y+4 Interchange x and y: x = y−3

b) Replace f (x) with y: y =

-4 -6 -8 -10

Solve for y: (y − 3)x = y + 4

4 b) Replace f (x) with y: y = x+7 4 Interchange x and y: x = y+7 Solve for y: x(y + 7) = 4 4 y+7 = x 4 y = −7 x Replace y with f −1 (x): f −1 (x) =

xy − 3x = y + 4

xy − y = 3x + 4

y(x − 1) = 3x + 4 3x + 4 y= x−1

Replace y with f −1 (x): f −1 (x) = 4 −7 x

3x + 4 x−1

5x − 3 is shown below. It 2x + 1 passes the horizontal-line test, so it is one-to-one.

52. a) The graph of f (x) =

278

Chapter 4: Exponential and Logarithmic Functions b) Replace f (x) with y: y = (x + 5)3

y 10 8

f(x) =

6 4 2

Interchange x and y: x = (y + 5)3 √ Solve for y: 3 x − 5 = y

5x – 3 2x +1

√ Replace y with f −1 (x): f −1 (x) = 3 x − 5 √ 55. a) The graph of f (x) = x 4 − x2 is shown below. Since there are many horizontal lines that cross the graph more than once, the function is not one-to-one and thus does not have an inverse that is a function.

x

-10 -8 -6 -4 -2 0 -2

2 4 6 8 10

-4 -6 -8 -10

y

5x − 3 2x + 1 5y − 3 Interchange x and y: x = 2y + 1 x+3 Solve for y: =y 5 − 2x

3

b) Replace f (x) with y: y =

Replace y with f −1 (x): f −1 (x) =

2

x -3

-2

-1

0

y 10 8

1

2

3

-1

x+3 5 − 2x

53. a) The graph of f (x) = x −1 is shown below. It passes the horizontal-line test, so the function is one-toone.

-2 -3

56. a) The graph of f (x) = 2x2 − x − 1 is shown below. Since there are many horizontal lines that cross the graph more than once, the function is not one-to-one and thus does not have an inverse that is a function.

f(x) = x 3 – 1

y 3

x

-10 -8 -6 -4 -2 0 -2

4–x2

1

3

6 4 2

f(x) = x

2 4 6 8 10

2

-4 -6 -8

1 x

-10

-3

-2

-1

0

1

2

3

-1

b) Replace f (x) with y: y = x3 − 1

-2

Interchange x and y: x = y 3 − 1

Solve for y:

√ 3

-3

x + 1 = y3 x+1 = y

Replace y with f −1 (x): f −1 (x) =

√ 3

x+1

57. a) The graph of f (x) = 5x2 − 2, x ≥ 0 is shown below. It passes the horizontal-line test, so it is one-to-one.

54. a) The graph of f (x) = (x + 5)3 is shown below. It passes the horizontal-line test, so it is one-to-one.

6 4 2 -10 -8 -6 -4 -2 0 -2 -4 -6 -8 -10

y 5 4

y 10 8

f(x) = 2x 2 – x – 1

3 2 1

f(x) = (x + 5 )3

-5 -4 -3 -2 -1 0 -1 x

2 4 6 8 10

-2 -3 -4 -5

x 1 2 3 4 5 f(x) = 5x 2 – 2, x > 0

Exercise Set 4.1

279

b) Replace f (x) with y: y = 5x2 − 2 Interchange x and y: x = 5y 2 − 2

Solve for y:

x + 2 = 5y 2 x+2 = y2 5

√ 59. a) The graph of f (x) = x + 1 is shown below. It passes the horizontal-line test, so the function is oneto-one. y 5 4

x+2 =y 5 (We take the principal square root, because x ≥ 0 in the original equation.) % x+2 −1 −1 for all x Replace y with f (x): f (x) = % 5 x+2 , x ≥ −2 in the range of f (x), or f −1 (x) = 5 %

58. a) The graph of f (x) = 4x2 + 3, x ≥ 0 is shown below. It passes the horizontal-line test, so the function is one-to-one.

-10 -8 -6 -4 -2 0 -2

-2 -3 -4 -5

b) Replace f (x) with y: y =

x+1 √ Interchange x and y: x = y + 1

f(x) = 4x 2 + 3, x > 0

x2 = y + 1

x 2 4 6 8 10

Replace y with f −1 (x): f −1 (x) = x2 − 1 for all x in the range of f (x), or f −1 (x) = x2 − 1, x ≥ 0. √ 60. a) The graph of f (x) = 3 x − 8 is shown below. It passes the horizontal-line test, so the function is oneto-one. y

-10

10 8 6 4 2

b) Replace f (x) with y: y = 4x2 + 3 Interchange x and y: x = 4y 2 + 3

√

√

x −1 = y

-4 -6 -8

Solve for y:

1 2 3 4 5

2

10 8 6 4 2

x +1

x

-5 -4 -3 -2 -1 0 -1

Solve for y:

y

f(x) =

3 2 1

x − 3 = 4y x−3 = y2 4

2

x−3 =y 2 (We take the principal square root since x ≥ 0 in the original function.) √ x−3 for all x Replace y with f −1 (x): f −1 (x) = √ 2 x−3 in the range of f (x), or f −1 (x) = ,x≥3 2

-10 -8 -6 -4 -2 0 -2

3

f(x) =

x –8 x

2 4 6 8 10

-4 -6 -8 -10

b) Replace f (x) with y: y =

√ 3

x−8 √ 3 Interchange x and y: x = y − 8 Solve for y: x3 + 8 = y

Replace y with f −1 (x): f −1 (x) = x3 + 8 61. f (x) = 3x The function f multiplies an input by 3. Then to reverse this procedure, f −1 would divide each of its inputs by 3. x 1 Thus, f −1 (x) = , or f −1 (x) = x. 3 3 62. f (x) =

1 x+7 4

1 and then adds 4 7. To reverse this procedure, f −1 would subtract 7 from each of its inputs and then multiply by 4. Thus, f −1 (x) = 4(x − 7). The function f multiplies an input by

280

Chapter 4: Exponential and Logarithmic Functions

63. f (x) = −x

The outputs of f are the opposites, or additive inverses, of the inputs. Then the outputs of f −1 are the opposites of its inputs. Thus, f −1 (x) = −x. √ 64. f (x) = 3 x − 5

The function f takes the cube root of an input and then subtracts 5. To reverse this procedure, f −1 would add 5 to each of its inputs and then raise the result to the third power. Thus, f −1 (x) = (x + 5)3 . √ 65. f (x) = 3 x − 5

The function f subtracts 5 from each input and then takes the cube root of the result. To reverse this procedure, f −1 would raise each input to the third power and then add 5 to the result. Thus, f −1 (x) = x3 + 5.

70. We reflect the graph of f across the line y = x. The reflections of the labeled points are (−3, −5), (−2, 0), (−1, 3), and (0, 4). y 4

(!1, 3)

(0, 4)

2

(!2, 0) !4

!2

2

4

x

!2 !4

(!3, !5)

71. We reflect the graph of f across the line y = x. y

66. f (x) = x−1 The outputs of f are the reciprocals of the inputs. Then the outputs of f −1 are the reciprocals of its inputs. Thus, f −1 (x) = x−1 .

4 2 !4

!2

67. We reflect the graph of f across the line y = x. The reflections of the labeled points are (−5, −5), (−3, 0), (1, 2), and (3, 5). y

!4

y

2

(1, 2)

!2

2

4

4

x

2

!4

!4

(!5, !5)

y 4 2

(5, 2) (3, 1)

!2

2

4

x

!2

(2, !4)

!4

(!3, !6)

69. We reflect the graph of f across the line y = x. The reflections of the labeled points are (−6, −2), (1, −1), (2, 0), and (5.375, 1.5). y 4

(5.375, 1.5)

2 !4

!2

2

4

x

!2

68. We reflect the graph of f across the line y = x. The reflections of the labeled points are (−3, −6), (2, −4), (3, 1), and (5, 2).

(!6, !2)

x

72. We reflect the graph of f across the line y = x.

(3, 5)

!2

!4

4

!4

4

(!3, 0)

2 !2

!2

2 !2 !4

4

x (2, 0) (1, !1)

!4

73. We find (f −1 ◦f )(x) and (f ◦f −1 )(x) and check to see that each is x. # $ −1 −1 −1 7 x = (f ◦ f )(x) = f (f (x)) = f 8 # $ 8 7 x =x 7 8 # $ # $ 8 7 8 −1 −1 x = x =x (f ◦ f )(x) = f (f (x)) = f 7 8 7 # $ x+5 74. (f −1 ◦ f )(x) = 4 −5=x+5−5=x 4 4x − 5 + 5 4x = =x (f ◦ f −1 )(x) = 4 4 75. We find (f −1 ◦f )(x) and (f ◦f −1 )(x) and check to see that each is x. # $ 1−x = (f −1 ◦ f )(x) = f −1 (f (x)) = f −1 x 1 1 1 = = =x 1−x 1−x+x 1 +1 x x x

Exercise Set 4.1

281

(f ◦ f −1 )(x) = f (f −1 (x)) = f

#

1 x+1

$

=

1 x+1−1 x x+1 = x+1 = x+1 =x 1 1 1 x+1 x+1 x+1 √ 76. (f −1 ◦ f )(x) = ( 3 x + 4)3 − 4 = x + 4 − 4 = x √ √ 3 (f ◦ f −1 )(x) = 3 x3 − 4 + 4 = x3 = x # $ 2 5 x+1 −5 5 = 77. (f −1 ◦ f )(x) = f −1 (f (x)) = 2 2x 2x + 5 − 5 = =x 2 2 2 & 5x − 5 ' +1= (f ◦ f −1 )(x) = f (f −1 (x) = 5 2 x−1+1=x & x+6 ' 4x + 24 +6 4 +6 3x − 4 −1 = 3x − 4 = 78. (f ◦ f )(x) = & ' x+6 3x + 18 −1 3 −1 3x − 4 3x − 4 4x + 24 + 18x − 24 22x 3x − 4 22x 3x − 4 = · = =x 3x + 18 − 3x + 4 3x − 4 22 22 3x − 4 4x + 6 + 18x − 6 4x + 6 +6 3x − 1 3x − 1 −1 = = (f ◦ f )(x) = & 4x + 6 ' 12x + 18 3 −4 −4 3x − 1 3x − 1 22x 22x 3x − 1 22x 3x − 1 = · = =x 12x + 18 − 12x + 4 3x − 1 22 22 3x − 1

The domain and range of f are (−∞, ∞), so the domain and range of f −1 are also (−∞, ∞).

1−

79. Replace f (x) with y: y = 5x − 3 Interchange x and y: x = 5y − 3

Solve for y: x + 3 = 5y x+3 =y 5

1 3 x+3 , or x + 5 5 5 The domain and range of f are (−∞, ∞), so the domain and range of f −1 are also (−∞, ∞). Replace y with f −1 (x): f −1 (x) =

y

f

4

f !1

2 !4

!2

y 4 2 !4

!2

2

4

x

!2 !4

f " f !1

2 x 2 Interchange x and y: x = y

81. Replace f (x) with y: y =

Solve for y: xy = 2 2 y= x

2 x The domain and range of f are (−∞, 0) ∪ (0, ∞), so the domain and range of f −1 are also (−∞, 0) ∪ (0, ∞). Replace y with f −1 (x): f −1 (x) =

y 4 2 !4

!2

2 !2 !4

4

x

f " f !1

3 x+1 3 Interchange x and y: x = − y+1

82. Replace f (x) with y: y = −

Solve for y: xy + x = −3

xy = −3 − x 3 −3 − x y= , or − − 1 x x 3 Replace y with f −1 (x): f −1 (x) = − − 1 x The domain of f is (−∞, −1) ∪ (−1, ∞) and the range of f is (−∞, 0) ∪ (0, ∞). Thus the domain of f −1 is (−∞, 0) ∪ (0, ∞) and the range of f −1 is (−∞, −1) ∪ (−1, ∞). y

2

4

x

4

!2

2

!4 !4

80. Replace f (x) with y: y = 2 − x Interchange x and y: x = 2 − y Solve for y: y = 2 − x

Replace y with f −1 (x): f −1 (x) = 2 − x

!2

2 !2 !4

f

f !1

4

x

282

Chapter 4: Exponential and Logarithmic Functions 1 3 x −2 3 1 Interchange x and y: x = y 3 − 2 3 1 3 Solve for y: x+2 = y 3 3x + 6 = y 3 √ 3 3x + 6 = y

The domain of f is (−∞, 3) ∪ (3, ∞) and the range of f is (−∞, 1) ∪ (1, ∞). Thus the domain of f −1 is (−∞, 1) ∪ (1, ∞) and the range of f −1 is (−∞, 3) ∪ (3, ∞).

83. Replace f (x) with y: y =

Replace y with f −1 (x): f −1 (x) =

y 4

f !1

2

f

√ 3

3x + 6

!4

!2

The domain and range of f are (−∞, ∞), so the domain and range of f −1 are also (−∞, ∞). y

2

4

!4

Solve for y: xy + 2x = y − 1

x

!2

2x + 1 = y − xy

!4

2x + 1 = y(1 − x) 2x + 1 =y 1−x

√ 3

x−1 √ 3 Interchange x and y: x = y − 1 √ Solve for y: x+1 = 3y

84. Replace f (x) with y: y =

2x + 1 1−x The domain of f is (−∞, −2) ∪ (−2, ∞) and the range of f is (−∞, 1) ∪ (1, ∞). Thus the domain of f −1 is (−∞, 1) ∪ (1, ∞) and the range of f −1 is (−∞, −2) ∪ (−2, ∞). Replace y with f −1 (x): f −1 (x) =

(x + 1)3 = y Replace y with f −1 (x): f −1 (x) = (x + 1)3 The domain and range of f are (−∞, ∞), so the domain and range of f −1 are also (−∞, ∞). y

y 4

f !1

4

2

2 !4

!2

x

x−1 x+2 y−1 Interchange x and y: x = y+2

f !1

2 !2

4

86. Replace f (x) with y: y =

f

4

!4

2 !2

f 2

4

!4

!2

!2

x

f !1

87. Since f (f −1 (x)) = f −1 (f (x)) = x, then f (f −1 (5)) = 5 and f −1 (f (a)) = a.

x+1 85. Replace f (x) with y: y = x−3 y+1 Interchange x and y: x = y−3

88. Since f −1 (f (x)) = f (f −1 (x)) = x, then f −1 (f (p)) = p and f (f −1 (1253)) = 1253.

Solve for y: xy − 3x = y + 1

2x − 3 2 2·5−3 10 − 3 7 1 s(5) = = = =3 2 2 2 2 # $ 1 2 · 7.5 − 3 15 − 3 12 = s(7.5) = = = =6 s 7 2 2 2 2 16 − 3 13 1 2·8−3 = = =6 s(8) = 2 2 2 2

89. a) s(x) =

xy − y = 3x + 1

Replace y with f −1 (x): f −1 (x) =

4

!4

f

!4

y(x − 1) = 3x + 1 3x + 1 y= x−1

2 !2

x

3x + 1 x−1

Exercise Set 4.1

283

b) The graph of s(x) passes the horizontal-line test and thus has an inverse that is a function. 2x − 3 Replace f (x) with y: y = 2 2y − 3 Interchange x and y: x = 2 Solve for y: 2x = 2y − 3

10

2x + 3 Replace y with s (x): s (x) = 2 2·3+3 6+3 9 1 −1 = = =4 c) s (3) = 2 2 2 2 # $ 2 · 5.5 + 3 1 11 + 3 14 s−1 5 = s−1 (5.5) = = = =7 2 2 2 2 2 · 7 + 3 14 + 3 17 1 = = =8 s−1 (7) = 2 2 2 2 100 + 5x 90. C(x) = x 100 + 5x Replace C(x) with y: y = x 100 + 5y Interchange x and y: x = y 100 Solve for y: y = x−5 100 Replace y with C −1 (x): C −1 (x) = x−5 C −1 (x) gives the number of people in the group, where x is the cost per person, in dollars. 91.

y1 y2 !10 Both the domain and the range of f are the set of all real numbers. Then both the domain and the range of f −1 are also the set of all real numbers.

−1

94.

y2

!8 Xscl " 4, Yscl " 4

Both the domain and the range of f are the set of all real numbers. Then both the domain and the range of f −1 are also the set of all real numbers.

y1 " "x ! 3, y2 " x 2 # 3, x $ 0 8 y2

95.

y1

The domain of f is [3, ∞) and the range of f is [0, ∞). Then the domain of f −1 is [0, ∞) and the range of f −1 is [3, ∞). 96.

Both the domain and the range of f are the set of all real numbers. Then both the domain and the range of f −1 are also the set of all real numbers.

y1 " 2.7 ! 1.08x, 2.7 ! x y2 " 1.08

y1 " !

2 2 , y "! x 2 x 4

6

!6

!4

4

Both the domain and the range of f are (−∞, 0) ∪ (0, ∞). Since f −1 = f , f −1 has the same domain and range. 5

!2

10 !4

25

!3

12

!6

!5

92.

y1 " x 3 ! 1, 3 y2 " "x# 1 8 y1

!12

y1 " 0.8x # 1.7, x ! 1.7 y2 " 0.8 y2 y1 15

!5

15

!15

2x + 3 = 2y 2x + 3 =y 2 −1

y1 " q x ! 4, y2 " 2x # 8

93.

y1 y2

Both the domain and the range of f are the set of all real numbers. Then both the domain and the range of f −1 are also the set of all real numbers.

284

Chapter 4: Exponential and Logarithmic Functions

y1 " x 2 ! 4, x $ 0; y2 " "4 # x y1 4 y2

97.

6

!6

!4

Since it is specified that x ≥ 0, the domain of f is [0, ∞). The range of f is [−4, ∞). Then the domain of f −1 is [−4, ∞) and the range of f −1 is [0, ∞). 98.

11 · 20 + 5 225 = = 22.5 10 10 555 11 · 50 + 5 = = 55.5 D(50) = 10 10 11 · 65 + 5 720 D(65) = = = 72 10 10 11r + 5 b) Replace D(r) with y: y = 10 11y + 5 Interchange r and y: r = 10 Solve for y: 10r = 11y + 5 D(20) =

10r − 5 = 11y 10r − 5 =y 11

y1 " 3 ! x 2, x $ 0; y2 " "3 ! x 4

10r − 5 11 D−1 (r) represents the speed, in miles per hour, that the car is traveling when the reaction distance is r feet. Replace y with D−1 (r): D−1 (r) =

y2 6

!6

y1

!4

Since it is specified that x ≥ 0, the domain of f is [0, ∞). The range of f is (−∞, 3]. Then the domain of f −1 is (−∞, 3] and the range of f −1 is [0, ∞).

y1 " (3x ! 9)3, y2 "

99.

4

3 "x #9 3

6

!10

102. a) In 2007, x = 2007 − 1990 = 17.

N (17) = 0.4737(17) + 24.7702 = 32.8231 lb N (20) = 0.4737(20) + 24.7702 = 34.2442 lb

b)

y1 " 0.4737x # 24.7702, x ! 24.7702 y2 " 0.4737 y2 60 y1

# 3

0

y1 8

!3

15

In 2010, x = 2010 − 1990 = 20.

x ! 3.2 , 1.4 3 y2 " 1.4x # 3.2 5 y2

y1 "

10x ! 5 11x # 5 , y2 " 11 10 y1 y2 10

!15

y1

Both the domain and the range of f are the set of all real numbers. Then both the domain and the range of f −1 are also the set of all real numbers. 100.

y1 "

y2

!2 !2

c)

!3

Both the domain and the range of f are the set of all real numbers. Then both the domain and the range of f −1 are also the set of all real numbers. 11r + 5 101. a) D(r) = 10 11 · 0 + 5 5 D(0) = = = 0.5 10 10 115 11 · 10 + 5 = = 11.5 D(10) = 10 10

0

60

c) N −1 (x) represents the number of years after 1990 when x pounds of cheese are consumed per person per year. 103. For an even function f , f (x) = f (−x) so we have f (x) = f (−x) but x #= −x (for x #= 0). Thus f is not one-to-one and hence it does not have an inverse. 104. C and F are inverses. 105. The functions for which the coefficient of x2 is negative have a maximum value. These are (b), (d), (f), and (h). 106. The graphs of the functions for which the coefficient of x2 is positive open up. These are (a), (c), (e), and (g).

Exercise Set 4.2

285

107. Since |2| > 1 the graph of f (x) = 2x2 can be obtained by stretching the graph of f (x) = x2 vertically. Since (1( 1 ( ( 0 < ( ( < 1, the graph of f (x) = x2 can be obtained by 4 4 shrinking the graph of y = x2 vertically. Thus the graph of f (x) = 2x2 , or (a) is narrower. (2( ( ( 108. Since | − 5| > 0 and 0 < ( ( < 1, the graph of (d) is nar3 rower.

109. We can write (f) as f (x) = −2[x − (−3)]2 + 1. Thus the graph of (f) has vertex (−3, 1). 110. For the functions that can be written in the form f (x) = a(x − 0)2 + k, or f (x) = ax2 + k, the line of symmetry is x = 0. These are (a), (b), (c), and (d).

111. The graph of f (x) = x2 − 3 is a parabola with vertex (0, −3). If we consider x-values such that x ≥ 0, then the graph is the right-hand side of the parabola and it passes the horizontal line test. We find the inverse of f (x) = x2 − 3, x ≥ 0. Replace f (x) with y: y = x2 − 3

Interchange x and y: x = y 2 − 3 Solve for y:

√

x + 3 = y2

Exercise Set 4.2 1. e4 ≈ 54.5982

2. e10 ≈ 22, 026.4658 3. e−2.458 ≈ 0.0856 # $2 1 4. ≈ 0.0025 e3 5.

f (0) = −20 − 1 = −1 − 1 = −2

The only graph with y-intercept (0, −2) is (f). # $x 1 6. f (x) = − 2 # $0 1 f (0) = − = −1 2 Since the y-intercept is (0, −1), the correct graph is (a) or (c). Check another point on the graph. # $−1 1 = −2, so the point (−1, −2) is on the f (−1) = − 2 graph. Thus (c) is the correct choice. 7. f (x) = ex + 3

x+3 = y

(We take the principal square root, because x ≥ 0 in the original equation.) √ Replace y with f −1 (x): f −1√(x) = x + 3 for all x in the −1 range of f (x), or f (x) = x + 3, x ≥ −3.

Answers may vary. There are other restrictions that also make f (x) one-to-one.

This is the graph of f (x) = ex shifted up 3 units. Then (e) is the correct choice. 8. f (x) = ex+1 This is the graph of f (x) = ex shifted left 1 unit. Then (b) is the correct choice. 9.

Since the y-intercept is (0, −1), the correct graph is (a) or (c). Check another point on the graph. f (−1) = 3−(−1) − 2 = 3 − 2 = 1, so (−1, 1) is on the graph. Thus (a) is the correct choice.

3 , f (x) = 1 − x, f (x) = x. x

114. First find f −1 (x). Replace f −1 (x) with y: y = ax + b Interchange x and y: x = ay + b Solve for y: x − b = ay x−b =y a

x−b 1 b = x− a a a Now we find the values of a and b for which 1 b 1 ax + b = x − . We see that a = for a = ±1. If a = 1, a a a we have x + b = x − b, so b = 0. If a = −1, we have −x + b = −x + b, so b can be any real number. Replace y with f −1 (x): f −1 (x) =

115. If the graph y = f (x) (or the graph of y = f −1 (x)) is symmetric with respect to the line y = x, then f −1 (x) = f (x).

f (x) = 3−x − 2

f (0) = 3−0 − 2 = 1 − 2 = −1

112. No; the graph of f does not pass the horizontal-line test. 113. Answers may vary. f (x) =

f (x) = −2x − 1

10.

f (x) = 1 − ex

f (0) = 1 − e0 = 1 − 1 = 0

The only graph with y-intercept (0, 0) is (d). 11. Graph f (x) = 3x . Compute some function values, plot the corresponding points, and connect them with a smooth curve. x

y = f (x)

(x, y)

−3

1 27

−2

1 9

−1 0

1 3 1

$ 1 − 3, 27 $ # 1 − 2, 9 $ # 1 − 1, 3 (0, 1)

1

3

(1, 3)

2

9

(2, 9)

3

27

(3, 27)

#

y 8 6 4 2 !3!2!1 !1

f(x) " 3x 1 2 3

x

286

Chapter 4: Exponential and Logarithmic Functions

12. Graph f (x) = 5x . x

y = f (x)

−3

1 125

−2

1 25

−1

1 5 1

0

(x, y) 1 − 3, 125 # $ 1 − 2, 25 # $ 1 − 1, 5 (0, 1)

#

$

1

5

(1, 5)

2

25

(2, 25)

3

125

(3, 125)

y

x

y = f (x)

(x, y)

−3

64

(−3, 64)

−2

16

(−2, 16)

−1

4

(−1, 4)

0

1 1 4

(0, 1) # $ 1 1, 4 $ # 1 2, 16 $ # 1 3, 64

1

4 2 2

!4 !2 !2

4

2

1 16

3

1 64

x

f(x) " 5 x

!4

16. Graph f (x) =

13. Graph f (x) = 6x . Compute some function values, plot the corresponding points, and connect them with a smooth curve. x

y = f (x)

(x, y)

−3

1 216

−2

1 36

−1 0

1 6 1

1 − 3, 216 $ # 1 − 2, 36 $ # 1 − 1, 6 (0, 1)

1

6

(1, 6)

$

#

2

36

(2, 36)

3

216

(3, 216)

6 5 4 3 2 !3 !2 !1 !1

y = f (x)

−3

27 8

−2

9 4

1 f(x) " 6 x 1

2

3

2

4 9

3

8 27

x

!3 !2 !1 !1

x

f(x) " $~%

1

2

3

x

(x, y) $ 27 − 3, 8 $ # 9 − 2, 4 $ # 3 − 1, 2 (0, 1) $ # 2 1, 3 $ # 4 2, 9 # $ 8 3, 27

#

3 2 1 2 3

0

5 4 3 2

# $x 2 . 3

x

−1

y

y

14. Graph f (x) = 3−x . x

y = f (x)

(x, y)

−3

27

(−3, 27)

9

(−2, 9)

−1

3

(−1, 3)

−2 0

1 1 3

y (0, 1) $ # 8 1 1 1, 6 3 f (x) " 3 !x $ # 4 1 1 2 2, 9 9 # $ !4 !2 2 4 1 1 x 3 3, !2 27 27 # $x 1 15. Graph f (x) = . 4 Compute some function values, plot the corresponding points, and connect them with a smooth curve.

y 4 2

x

f(x) " $s% 2

!4 !2

4

x

!2 !4

17. Graph y = −2x . x

(x, y)

y

−3 −

1 8

#

−2 −

1 4

#

− 2, −

1 4

$

1 −1 − 2 0 −1

#

1 − 1, − 2 (0, −1)

$

1 2 3

−2 −4 −8

− 3, −

1 8

$

(1, −2) (2, −4) (3, −8)

y 4 2 !4

!2

2 !2 !4

4

x

y " !2x

Exercise Set 4.2

287

18. Graph y = 3 − 3x .

22. Graph f (x) = 2 − e−x .

(x, y)

x

y

−3

80 27

#

−2

26 9

#

−1 0

8 3 2

26 − 2, 9 $ # 8 − 1, 3 (0, 2)

1

0

(1, 0)

2

−6

(2, −6)

3

−24

− 3,

80 27

$

4

−3

−60

(−3, −60)

−2 −1

−12 0

y " 3 ! 3x

2 !4

!2

2

4

x

!2

2

1

3.75

(1, 3.75)

2

3.94

(2, 3.94)

3

3.98

(3, 3.98)

y = f (x) 1.63

(−3, 1.63)

−0.22

(−2, −0.22)

0

−2

(0, −2)

−1 1 2 3

−1.33 −2.4

!4

f(x) " !0.25x # 4

−2.64 −2.78

1

(1, 1.6)

1.9

(2, 1.9)

3

2.0

(3, 2.0)

!4

4

x

(x, y)

y

y

−1 0.0920 (−1, 0.0920)

!2

1

0.6796

(1, 0.6796)

!4

2

1.8473

(2, 1.8473)

3

5.0214

(3, 5.0214)

4

x

x

y " ~e x

4

f(x) " 0.6x ! 3

2 !2

(x, y)

y

2

4

x

!2 !4

−2 14.7781 (−2, 14.7781) −1

5.4366

(−1, 5.4366)

0

2

(0, 2)

1

0.7358

(1, 0.7358)

2

0.2707

(2, 0.2707)

3

0.0996

(3, 0.0996)

21. Graph f (x) = 1 + e−x .

y

x

y = f (x)

(x, y)

4

−3

21.1

(−3, 21.1)

2

−2

8.4

(−2, 8.4)

y

−1

3.7

(−1, 3.7)

4

0

2

(0, 2)

2

1

1.4

(1, 1.4)

!2

f(x) " 1 # e!x

2

1.1

(2, 1.1)

3

1.0

(3, 1.0)

!4

!2 !4

2 !2

2

!4 !2

4

x

y"

4

2e!x

6 5 4 3 2 1

!3 !2 !1 !1

−3 40.1711 (−3, 40.1711)

y

!4

2

2

(0, 0.25)

(3, −2.78)

!2 !2

0.25

2

(2, −2.64)

f(x) " 2 ! e!x

2 !4

24. Graph y = 2e−x .

!4

4

(0, 1)

1.6

0

!2

(−1, −1.33) (1, −2.4)

(−1, −0.8)

−2 0.0338 (−2, 0.0338)

(x, y)

−2

−3

−0.8

−3 0.0124 (−3, 0.0124)

20. Graph f (x) = 0.6x − 3. x

−1

y

(−2, −5.4)

−5.4

1 x e . 4 Choose values for x and compute the corresponding yvalues. Plot the points (x, y) and connect them with a smooth curve.

y

(−1, 0) (0, 3)

(−3, −18.1)

x

4

3

−18.1

23. Graph y =

!4

(−2, −12)

0

−3

1

19. Graph f (x) = −0.25x + 4. (x, y)

(x, y)

0

y

(3, −24)

y = f (x)

y = f (x)

−2

$

x

x

x

1

2

3

x

288

Chapter 4: Exponential and Logarithmic Functions 28. Shift the graph of y = 2x right 1 unit.

25. Graph f (x) = 1 − e−x .

Compute some function values, plot the corresponding points, and connect them with a smooth curve. x

y 8

(x, y)

y

6

−3 −19.0855 (−3, −19.0855) −2

−6.3891

(−2, −6.3891)

0

0

(0, 0)

−1

(−1, −1.7183)

−1.7183

1

0.6321

(1, 0.6321)

2

0.8647

(2, 0.8647)

3

0.9502

(3, 0.9502)

f(x) " 2 x ! 1

4 2 !4

!2

2

4

x

29. Shift the graph of y = 2x down 3 units. y 6 4

y

2

2 1

!4

1

!3 !2 !1 !2 !3 !4 !5 !6

2

3

!2

x

2 !2

4

x

f(x) " 2 x ! 3

!4

f (x) " 1 ! e!x

30. Shift the graph of y = 2x up 1 unit. y

26. Graph f (x) = ex − 2. x

6 4

(x, y)

y

2

−3 −1.9502 (−3, −1.9502) −2 −1.8647 (−2, −1.8647)

!4

!2

(0, −1)

−1

1

0.7183

(1, 0.7183)

2

5.3891

(2, 5.3891)

3

18.0855

(3, 18.0855)

2

4

x

!2

−1 −1.6321 (−1, −1.6321) 0

f(x) " 2 x # 1

!4

31. Reflect the graph of y = 3x across the y-axis, then across the x-axis, and then shift it up 4 units. y 6

f (x)" 4 ! 3!x

4

y

2

4 !4

2

!2

2

4

x

!2

2

!4 !2 !2

4

x

!4

f(x) " e x ! 2

!4

32. Shift the graph of y = 2x right 1 unit and down 3 units. y

27. Shift the graph of y = 2x left 1 unit.

4

y

2

8

!2

2 !2

4

!4

2 !4

!4

6

!2

f (x)" 2 x # 1 2

!2

4

x

4

x

f(x)" 2 x ! 1 ! 3

Exercise Set 4.2

289

33. Shift the graph of y = y

f (x) "

8

# $x 3 right 1 unit. 2

38. Stretch the graph of y = ex horizontally and reflect it across the y-axis. y

( 32 )x ! 1

4

f(x) " e !0.2x

2

6 !4

4

!2

2

4

x

!2

2

!4 !4

!2

2

4

x

34. Reflect the graph of y = 3x across the y-axis and then shift it right 4 units.

39. Shift the graph of y = ex left 1 unit and reflect it across the y-axis. y

y

8

8 6

6

f (x) " 34 ! x

4

4

2

2 2

4

6

8

!2

x

35. Shift the graph of y = 2x left 3 units and down 5 units.

2

x

8

f (x)" 2 x # 3 ! 5 2 !4

6

y

4

!6

4

40. Shrink the graph of y = ex horizontally and shift it up 1 unit.

y

!8

y " e !x # 1

!2

6

x

4

!2

2

y " e 2x # 1

!4 !4

36. Shift the graph of y = 3x right 2 units and reflect it across the x-axis. y 2 !4

!2

4

4

!2

4

!4

2 !2

2

4

!2

!8

!4

37. Shrink the graph of y = e horizontally. x

y 8 6

f(x) " e 2x

4 2 !4

!2

2

4

x

x

y

x

!6

2

41. Reflect the graph of y = ex across the y-axis and then across the x-axis; shift it up 1 unit and then stretch it vertically.

f(x)" !3x ! 2 2

!2

f(x)" 2(1 ! e!x)

6

x

290

Chapter 4: Exponential and Logarithmic Functions

42. Stretch the graph of y = ex horizontally and reflect it across the y-axis; then reflect it down across the x-axis and shift it up 1 unit. y 2 1 !200 !100

100 200

x

!1 !2

f (x) " 1 !e !0.01x

$nt # 43. a) We use the formula A = P 1 + r and substin tute 82,000 for P , 0.045 for r, and 4 for n. $4t # 0.045 = 82, 000(1.01125)4t A(t) = 82, 000 1 + 4 b) A(0) = 82, 000(1.01125)4·0 = $82, 000 A(2) = 82, 000(1.01125)4·2 ≈ $89, 677.22

A(5) = 82, 000(1.01125)4·5 ≈ $102, 561.54

A(10) = 82, 000(1.01125)4·10 ≈ $128, 278.90 $2t # 0.07 = 750(1.035)2t 44. a) A(t) = 750 1 + 2 b) A(1) = 750(1.035)2·1 ≈ $803.42

A(6) = 750(1.035)2·6 ≈ $1133.30

A(10) = 750(1.035)2·10 ≈ $1492.34

A(15) = 750(1.035)2·15 ≈ $2105.10

A(25) = 750(1.035)2·25 ≈ $4188.70 # $nt r and substitute 3000 45. We use the formula A = P 1 + n for P , 0.05 for r, and 4 for n. $4t # 0.05 = 3000(1.0125)4t A(t) = 3000 1 + 4 On Jacob’s sixteenth birthday, t = 16 − 6 = 10. A(10) = 3000(1.0125)

= 4930.86

4·10

When the CD matures $4930.86 will be available. $2t # 0.064 46. a) A(t) = 10, 000 1 + = 10, 000(1.032)2t 2 b) A(0) = 10, 000(1.032)

2·0

A(4) = 10, 000(1.032)

2·4

A(8) = 10, 000(1.032)

2·8

A(10) = 10, 000(1.032)

= $10, 000 ≈ $12, 865.82 ≈ $16, 552.94

2·10

≈ $18, 775.61

A(18) = 10, 000(1.032)2·18 ≈ $31, 079.15 $nt # r and substitute 3000 47. We use the formula A = P 1 + n for P , 0.04 for r, 2 for n, and 2 for t. # $2·2 0.04 ≈ $3247.30 A = 3000 1 + 2

# $4·3 0.03 48. A = 12, 500 1 + ≈ $13, 672.59 4 $nt # r 49. We use the formula A = P 1 + and substitute n 120,000 for P , 0.025 for r, 1 for n, and 10 for t. # $1·10 0.025 ≈ $153, 610.15 A = 120, 000 1 + 1 $4·10 # 0.025 50. A = 120, 000 1 + ≈ $153, 963.22 4 $nt # r and substitute 51. We use the formula A = P 1 + n 53,500 for P , 0.055 for r, 4 for n, and 6.5 for t. # $4(6.5) 0.055 ≈ $76, 305.59 A = 53, 500 1 + 4 $2(4.5) # 0.0675 52. A = 6250 1 + ≈ $8425.97 2 $nt # r and substitute 53. We use the formula A = P 1 + n 17,400 for P , 0.081 for r, 365 for n, and 5 for t. # $365·5 0.081 ≈ $26, 086.69 A = 17, 400 1 + 365 $365(7.25) # 0.073 ≈ $1527.81 54. A = 900 1 + 365 55. C(t) = 1.0283(1.1483)t In 2008, t = 2008 − 1996 = 12.

C(12) = 1.0283(1.1483)12 ≈ 5.4 billion gallons

In 2010, t = 2010 − 1996 = 14.

C(14) = 1.0283(1.1483)14 ≈ 7.1 billion gallons

56. N (10) = 3000(2)10/20 ≈ 4243;

N (20) = 3000(2)20/20 = 6000; N (30) = 3000(2)30/20 ≈ 8485;

N (40) = 3000(2)40/20 = 12, 000; N (60) = 3000(2)60/20 = 24, 000 57. G(x) = 433.6(1.5)x In 2009, x = 2009 − 2004 = 5.

G(5) = 433.6(1.5)5 ≈ 3293 GB

In 2014, x = 2014 − 2004 = 10.

G(10) = 433.6(1.5)10 ≈ 25, 004 GB

58. D(20) ≈ 166, 052 D(45) ≈ 458, 909 D(60) ≈ 844, 531

D(70) ≈ 1, 268, 260 59. In 2007, x = 2007 − 1996 = 11.

E(11) = 139.76(1.194)11 ≈ $982.7 billion

I(11) = 130.67(1.187)11 ≈ $861.3 billion

Exercise Set 4.2

291 75. Graph (i) is the graph of g(x) = e|x| .

In 2012, x = 2012 − 1996 = 16.

E(16) = 139.76(1.194)16 ≈ $2384.8 billion

I(16) = 130.67(1.187)16 ≈ $2029.5 billion

In 2020, x = 2020 − 1996 = 24.

E(24) = 139.76(1.194)24 ≈ $9851.1 billion I(24) = 130.67(1.187)24 ≈ $7998.2 billion

60. In 2010, t = 2010 − 1964 = 46. K(46) = 35.37(0.99)46 ≈ 22%

2

78. Graph (e) is the graph of y = |2x − 8|. 79. Graph (m) is the graph of g(x) =

61. V (0) = 1800(0.8)0 = $1800 V (1) = 1800(0.8)1 = $1440 V (5) = 1800(0.8)5 ≈ $589.82

83. They are inverses.

V (10) = 1800(0.8)10 ≈ $193.27 62. f (16) = 5728.98(1.1214)16 ≈ 35, 830 visas ≈ 50, 527 visas

84. They are inverses. 85.

= 31 − 22i

In 2000, t = 2000 − 1999 = 1.

S(1) = 29.0626(1.3438) ≈ $39.1 billion 1

In 2005, t = 2005 − 1999 = 6.

86.

S(6) = 29.0626(1.3438)6 ≈ $171.1 billion

In 2007, t = 2007 − 1999 = 8.

S(8) = 29.0626(1.3438)8 ≈ $309.0 billion 64. S(10) = 200[1 − (0.86)10 ] ≈ 155.7 words per minute S(20) = 200[1 − (0.86)20 ] ≈ 190.2 words per minute S(40) = 200[1 − (0.86)40 ] ≈ 199.5 words per minute

S(100) ≈ 200[1 − (0.86)100 ] ≈ 199.9999 words per minute ) ≈ 63%

66. V (1) = $58(1 − e−1.1(1) ) + $20 ≈ $58.69 V (2) = $58(1 − e−1.1(2) ) + $20 ≈ $71.57 V (4) = $58(1 − e−1.1(4) ) + $20 ≈ $77.29 V (6) = $58(1 − e−1.1(6) ) + $20 ≈ $77.92

V (12) = $58(1 − e−1.1(12) ) + $20 ≈ $78.00 67. Graph (c) is the graph of y = 3x − 3−x . 2

68. Graph (j) is the graph of y = 3−(x+1) . 69. Graph (a) is the graph of f (x) = −2.3x . 70. Graph (d) is the graph of f (x) = 30, 000(1.4)x . 71. Graph (l) is the graph of y = 2−|x| . 72. Graph (n) is the graph of y = 2−(x−1) . 73. Graph (g) is the graph of f (x) = (0.58)x − 1. 74. Graph (b) is the graph of y = 2x + 2−x .

(1 − 4i)(7 + 6i) = 7 + 6i − 28i − 24i2 = 7 + 6i − 28i + 24

63. S(t) = 29.0626(1.3438)t

65. f (25) = 100(1 − e

ex + e−x . 2

82. The most interest will be earned the eighth year, because the principle is greatest during that year.

V (2) = 1800(0.8)2 = $1152

−0.04(25)

ex − e−x . 2

81. Some differences are as follows: The range of f is (−∞, ∞) whereas the range of g is (0, ∞); f has no asymptotes but g has a horizontal asymptote, the x-axis; the y-intercept of f is (0, 0) and the y-intercept of g is (0, 1).

K(51) = 35.37(0.99)51 ≈ 21%

f (19) = 5728.98(1.1214)

2

77. Graph (k) is the graph of y = 2−x .

80. Graph (f) is the graph of f (x) =

In 2015, t = 2015 − 1964 = 51.

19

76. Graph (h) is the graph of f (x) = |2x − 1|.

87.

2−i 3−i 2−i = · 3+1 3+i 3−i 6 − 5i + i2 = 9 − i2 6 − 5i − 1 = 9+1 5 − 5i = 10 1 1 = − i 2 2 2x2 − 13x − 7 = 0

Setting f (x) = 0

(2x + 1)(x − 7) = 0 2x + 1 = 0

or x − 7 = 0

2x = −1 or 1 x = − or 2

x=7 x=7

1 The zeros of the function are − and 7, and the x2 # $ 1 intercepts are − , 0 and (7, 0). 2 88. h(x) = x3 − 3x2 + 3x − 1

The possible real-number solutions are of the form p/q where p = ±1 and q = ±1. Then the possibilities for p/q are 1 and −1. We try 1. ( 1 ( 1 −3 3 −1 1 −2 1 1 −2 1 0 h(x) = (x − 1)(x2 − 2x + 1)

292

Chapter 4: Exponential and Logarithmic Functions Now find the zeros of h(x).

Exercise Set 4.3

(x − 1)(x2 − 2x + 1) = 0

(x − 1)(x − 1)(x − 1) = 0

x − 1 = 0 or x − 1 = 0 or x − 1 = 0 x = 1 or

x = 1 or

x=1

The zero of the function is 1 and the x-intercept is (1, 0). 89.

x4 − x2 = 0

Setting h(x) = 0

x (x − 1) = 0 2

2

x2 (x + 1)(x − 1) = 0

x2 = 0 or x + 1 = 0 x = 0 or

or x − 1 = 0

x = −1 or

x=1

x3 + x2 − 12x = 0

x(x2 + x − 12) = 0

x(x + 4)(x − 3) = 0

x = 0 or x + 4 = 0 x = 0 or

or x − 3 = 0

x = −4 or

x=3

The zeros of the function are 0, −4, and 3, and the xintercepts are (0, 0), (−4, 0), and (3, 0). 91.

x + 6x − 16x = 0 3

2

x(x2 + 6x − 16) = 0 x(x + 8)(x − 2) = 0

x = 0 or x + 8 = 0 x = 0 or

or x − 2 = 0

x = −8 or

x=2

The solutions are 0, −8, and 2. 92.

Choose values for y and compute the corresponding xvalues. Plot the points (x, y) and connect them with a smooth curve. x

3x2 − 6 = 5x

3x2 − 5x − 6 = 0 ) −(−5) ± (−5)2 − 4 · 3 · (−6) x= 2·3 √ 5 ± 97 = 6 93. 7π ≈ 451.8078726 and π 7 ≈ 3020.293228, so π 7 is larger.

7080 ≈ 4.054 × 10147 and 8070 ≈ 1.646 × 10133 , so 7080 is larger.

(x, y)

y

1 −3 27

The zeros of the function are 0, −1, and 1, and the xintercepts are (0, 0), (−1, 0), and (1, 0). 90.

1. Graph x = 3y .

1 9

−2

1 3 1

−1 0

$ 1 , −3 27 # $ 1 , −2 9 $ # 1 , −1 3 (1, 0)

3

1

(3, 1)

9

2

(9, 2)

27

3

(27, 3)

#

y 3 2 1

x " 3y

2

!1 !2

4

6

8

x

!3

2. Graph x = 4y . x

(x, y)

y

1 −3 64

#

1 −2 16

#

1 , −3 64

$

1 4 1

−1 0

$ 1 , −2 16 $ # 1 , −1 4 (1, 0)

4

1

(4, 1)

16

2

(16, 2)

64

3

y 4

x " 4y

2 2

!2

4

6

8 x

!2 !4

(64, 3) # $y 1 3. Graph x = . 2 Choose values for y and compute the corresponding xvalues. Plot the points (x, y) and connect them with a smooth curve. x

y

(x, y)

8 −3 (8, −3) 4 −2 (4, −2) 2 −1 (2, −1) 1 1 2

0 1

1 4

2

1 8

3

(1, 0) $ # 1 ,1 2 $ # 1 ,2 4 $ # 1 ,3 8

y 3 2 1 !1 !2 !3

x " $q%

y

2 3 4 5 6 7 8

x

Exercise Set 4.3

4. Graph x = x

# $y 4 . 3

27 −3 64

#

9 −2 16

#

−1 0 1

16 9

2

64 27

3

x

0.5 −1.4 1

27 , −3 64

$

$ 9 , −2 16 # $ 3 , −1 4 (1, 0) # $ 4 ,1 3 $ # 16 ,2 9 $ # 64 ,3 27

2 2

!4 !2 !2

4

x

x " $d%

!4

y

y

−2

3 2 1

−1

y " log 3 x

2

!1 !1 !2 !3

0 2

1 16 1 4 1

4

6

8

10 x

4.2

2

4

6

8

10

x

!2

10. log3 9 = 2, because the exponent to which we raise 3 to get 9 is 2. 11. log5 125 = 3, because the exponent to which we raise 5 to get 125 is 3.

13. log 0.001 = −3, because the exponent to which we raise 10 to get 0.001 is −3. 14. log 100 = 2, because the exponent to which we raise 10 to get 100 is 2. 1 = −2, because the exponent to which we raise 2 to 4 1 get is −2. 4

15. log2

20. log 1 = 0, because the exponent to which we raise 10 to get 1 is 0.

y 4

y " log 4 x

2

−1

2

0

!2

4

1

!4

16

2

18. ln e = 1, because the exponent to which we raise e to get e is 1. 19. log 10 = 1, because the exponent to which we raise 10 to get 10 is 1.

y −2

8

2

17. ln 1 = 0, because the exponent to which we raise e to get 1 is 0.

6. y = log4 x is equivalent to x = 4y . x, or 4y

2.8

16. log8 2 =

y

9

1.4

4

f(x) " 2 ln x

1 , because the exponent to which we raise 8 to 3 1 get 2 is . 3

For y = 2, x = 32 = 9.

1

2

4

12. log2 64 = 6, because the exponent to which we raise 2 to get 64 is 6.

For y = 1, x = 31 = 3.

3

6

9. log2 16 = 4 because the exponent to which we raise 2 to get 16 is 4.

4

The equation y = log3 x is equivalent to x = 3y . We can find ordered pairs that are solutions by choosing values for y and computing the corresponding x-values. 1 For y = −2, x = 3−2 = . 9 1 −1 For y = −1, x = 3 = . 3 For y = 0, x = 30 = 1.

1 9 1 3 1

y

0

8. See Example 10. y

5. Graph y = log3 x.

x, or 3y

f (x)

0.1 −4.6

(x, y)

y

3 4 1 4 3

293

4

6

8

x

7. Graph f (x) = 2 ln x. Choose values for x and use a calculator to find the corresponding values of 2 ln x. Then plot the points and connect them with a smooth curve.

21. log5 54 = 4, because the exponent to which we raise 5 to get 54 is 4. √ 1 22. log 10 = log 101/2 = , because the exponent to which 2 1 1/2 we raise 10 to get 10 is . 2 √ 1 4 23. log3 3 = log3 31/4 = , because the exponent to which 4 1 1/4 we raise 3 to get 3 is . 4 8 , because the exponent to which we raise 10 5 8 to get 108/5 is . 5

24. log 108/5 =

294

Chapter 4: Exponential and Logarithmic Functions

25. log 10−7 = −7, because the exponent to which we raise 10 to get 10−7 is −7. 26. log5 1 = 0, because the exponent to which we raise 5 to get 1 is 0. 1 27. log49 7 = , because the exponent to which we raise 49 to 2 √ 1 get 7 is . (491/2 = 49 = 7) 2 28. log3 3−2 = −2, because the exponent to which we raise 3 to get 3−2 is −2. 3 , because the exponent to which we raise e to 4 3 is . 4

29. ln e3/4 = get e3/4 √

1 2 = log2 21/2 = , because the exponent to which we 2 1 raise 2 to get 21/2 is . 2

30. log2

45. ↓ ↓ log 5 5 = 1⇒51 = 5 ↑ ↑

The logarithm is the exponent. The base remains the same.

46. t = log4 7 ⇒ 7 = 4t 47. log 0.01 = −2 is equivalent to log10 0.01 = −2.

The logarithm is the exponent.

↓ ↓ log10 0.01 = −2⇒10−2 = 0.01 ↑ ↑ The base remains the same. 48. log 7 = 0.845 ⇒ 100.845 = 7 49. ln 30 = 3.4012 ⇒ e3.4012 = 30 50. ln 0.38 = −0.9676 ⇒ e−0.9676 = 0.38

31. log4 1 = 0, because the exponent to which we raise 4 to get 1 is 0.

51. loga M = −x ⇒ a−x = M

32. ln e−5 = −5, because the exponent to which we raise e to get e−5 is −5.

53. loga T 3 = x ⇒ ax = T 3

1 = , because the exponent to which we 33. ln e = ln e 2 1 1/2 is . raise e to get e 2 √

1/2

1 34. log64 4 = , because the exponent to which we raise 64 to 3 √ 1 get 4 is . (641/3 = 3 64 = 4) 3 35.

The exponent is the logarithm.

↓

↓ 10 = 1000⇒3 = log10 1000 ↑ ↑ The base remains the same. 3

36. 5−3 =

1 1 ⇒ log5 = −3 125 125

37. ↓ 81/3 ↑ 38. 10

↓ 1 = 2 ⇒ log 8 2 = 3 ↑

0.3010

The exponent is the logarithm.

52. logt Q = k ⇒ tk = Q 54. ln W 5 = t ⇒ et = W 5 55. log 3 ≈ 0.4771 56. log 8 ≈ 0.9031 57. log 532 ≈ 2.7259 58. log 93, 100 ≈ 4.9689 59. log 0.57 ≈ −0.2441 60. log 0.082 ≈ −1.0862 61. log(−2) does not exist. (The calculator gives an error message.) 62. ln 50 ≈ 3.9120 63. ln 2 ≈ 0.6931 64. ln(−4) does not exist. (The calculator gives an error message.) 65. ln 809.3 ≈ 6.6962

The base remains the same.

= 2 ⇒ log10 2 = 0.0310

39. e = t ⇒ loge t = 3, or ln t = 3 3

40. Qt = x ⇒ logQ x = t 41. e2 = 7.3891 ⇒ loge 7.3891 = 2, or ln 7.3891 = 2 42. e−1 = 0.3679 ⇒ loge 0.3679 = −1, or ln 0.3679 = −1 43. pk = 3 ⇒ logp 3 = k 44. e−t = 4000 ⇒ loge 4000 = −t, or ln 4000 = −t

66. ln 0.00037 ≈ −7.9020 67. ln(−1.32) does not exist. (The calculator gives an error message.) 68. ln 0 does not exist. (The calculator gives an error message.) 69. Let a = 10, b = 4, and M = 100 and substitute in the change-of-base formula. log10 100 ≈ 3.3219 log4 100 = log10 4 70. log3 20 =

log 20 ≈ 2.7268 log 3

Exercise Set 4.3

295

71. Let a = 10, b = 100, and M = 0.3 and substitute in the change-of-base formula. log10 0.3 log100 0.3 = ≈ −0.2614 log10 100

y

f !1(x) " 10 x

2 !4

log 100 ≈ 4.0229 72. logπ 100 = log π 73. Let a = 10, b = 200, and M = 50 and substitute in the change-of-base formula. log10 50 ≈ 0.7384 log200 50 = log10 200 log 1700 ≈ 4.4602 74. log5.3 1700 = log 5.3 75. Let a = e, b = 3, and M = 12 and substitute in the change-of-base formula. ln 12 ≈ 2.2619 log3 12 = ln 3 76. log4 25 =

ln 25 ≈ 2.3219 ln 4

4

x

f (x) " log x

!4

82. Graph y = ex and then reflect this graph across the line y = x to get the graph of y = ln x. y

f (x) " e x

4 2

!4

!2

2 !2

4

x

f !1(x) " ln x

!4

y 4 2 !2

2 !2

79. Graph y = 3x and then reflect this graph across the line y = x to get the graph of y = log3 x.

4

6

x

f(x) " log 2 (x # 3)

!4

Domain: (−3, ∞)

y

Vertical asymptote: x = −3

4

84. Shift the graph of y = log3 x right 2 units.

2 !4

2

83. Shift the graph of y = log2 x left 3 units.

ln 100 ≈ 2.0959 ln 9

f (x) " 3x

!2 !2

77. Let a = e, b = 100, and M = 15 and substitute in the change-of-base formula. ln 15 ≈ 0.5880 log100 15 = ln 100 78. log9 100 =

4

!2

2 !2

4

f !1(x)

y

x

" log3 x

4

!4

f(x) " log 3 (x ! 2)

2

80. Graph y = log4 x and then reflect this graph across the line y = x to get the graph of y = 4x .

2

4

6

8

x

!2 !4

y

f !1(x) " 4x

Domain: (2, ∞)

4

Vertical asymptote: x = 2

2 !4

!2

2 !2 !4

4

x

f (x) " log4 x

85. Shift the graph of y = log3 x down 1 unit. y 4

81. Graph y = log x and then reflect this graph across the line y = x to get the graph of y = 10x .

y " log 3 x ! 1

2 2

4

6

8

x

!2 !4

Domain: (0, ∞)

Vertical asymptote: x = 0

296

Chapter 4: Exponential and Logarithmic Functions

86. Shift the graph of y = log2 x up 3 units.

90. Shift the graph of y = ln x left 1 unit. y

y 8

4

6

2

4

2

y " 3 # log 2 x

2 2

4

6

8

6

8

x

!4

y " ln (x # 1)

x

Domain: (−1, ∞)

Domain: (0, ∞)

Vertical asymptote: x = 0 87. Stretch the graph of y = ln x vertically.

Vertical asymptote: x = −1

91. a) We substitute 784.242 for P , since P is in thousands. w(784.242) = 0.37 ln 784.242 + 0.05 ≈ 2.5 ft/sec b) We substitute 484.246 for P , since P is in thousands. w(484.246) = 0.37 ln 484.246 + 0.05

y 8 6 4

f (x) " 4 ln x

2 2

4

6

8

x

Domain: (0, ∞)

Vertical asymptote: x = 0 88. Shrink the graph of y = ln x vertically. y 4 2 2

4

6

8

x

!2

1

f(x) " 2 ln x

!4

Domain: (0, ∞)

Vertical asymptote: x = 0 89. Reflect the graph of y = ln x across the x-axis and then shift it up 2 units.

≈ 2.3 ft/sec c) We substitute 276.963 for P , since P is in thousands. w(276.963) = 0.37 ln 276.963 + 0.05 ≈ 2.1 ft/sec d) We substitute 2862.244 for P , since P is in thousands. w(2862.244) = 0.37 ln 2862.244 + 0.05 ≈ 3.0 ft/sec e) We substitute 533.492 for P , since P is in thousands. w(533.492) = 0.37 ln 533.492 + 0.05 ≈ 2.4 ft/sec f) We substitute 304.973 for P , since P is in thousands. w(304.973) = 0.37 ln 304.973 + 0.05 ≈ 2.2 ft/sec g) We substitute 8104.079 for P , since P is in thousands. w(8104.079) = 0.37 ln 8104.079 + 0.05 ≈ 3.4 ft/sec We substitute 122.206 for P , since P is in thousands. h) w(122.206) = 0.37 ln 122.206 + 0.05 ≈ 1.8 ft/sec

y

92. a) S(0) = 78 − 15 log(0 + 1)

4

= 78 − 15 log 1

2 2

4

6

8

x

!2 !4

4

!2

y " 2 ! ln x

Domain: (0, ∞)

Vertical asymptote: x = 0

= 78 − 15 · 0 = 78%

b) S(4) = 78 − 15 log(4 + 1) = 78 − 15 log 5

≈ 78 − 15(0.698970) ≈ 67.5%

S(24) = 78 − 15 log(24 + 1) = 78 − 15 log 25

≈ 78 − 15(1.397940) ≈ 57%

Exercise Set 4.3

297

93. a) R = log

107.85 · I0 = log 107.85 = 7.85 I0

b) R = log

108.25 · I0 = log 108.25 = 8.25 I0

c) R = log

109.6 · I0 = log 109.6 = 9.6 I0

107.9 · I0 d) R = log = log 107.9 = 7.9 I0 e) R = log

106.9 · I0 = log 106.9 = 6.9 I0

94. a) pH = − log[1.6 × 10−4 ] ≈ −(−3.8) ≈ 3.8 b) pH = − log[0.0013] ≈ −(−2.9) ≈ 2.9

c) pH = − log[6.3 × 10−7 ] ≈ −(−6.2) ≈ 6.2

9 d) L = 10 log 10 · I0 I0 = 10 log 109

= 90 decibels 98. Left to the student 99. If log b < 0, then b < 1. 100. Reflect the graph of f (x) = ln x across the line y = x to obtain the graph of h(x) = ex . Then shift this graph 2 units right to obtain the graph of g(x) = ex−2 . 101. y = 6 = 0 · x + 6

Slope: 0; y-intercept (0, 6)

102.

3x − 14 = 10y 7 3 x− = y 10 5 # $ 3 7 ; y-intercept: 0, − Slope: 10 5

d) pH = − log[1.6 × 10−8 ] ≈ −(−7.8) ≈ 7.8 e) pH = − log[6.3 × 10−5 ] ≈ −(−4.2) ≈ 4.2

95. a)

7 = − log[H + ]

−7 = log[H + ]

H + = 10−7 b)

Using the definition of logarithm

H + = 10−5.4

Using the definition of

H + ≈ 4.0 × 10−6 3.2 = − log[H + ]

−3.2 = log[H + ] H + = 10−3.2

H d)

+

Using the definition of logarithm

≈ 6.3 × 10

−4

4.8 = − log[H + ]

−4.8 = log[H + ]

H + = 10−4.8 Using the definition of logarithm

96. a) N (1) = 1000 + 200 ln 1 = 1000 units b) N (5) = 1000 + 200 ln 5 ≈ 1332 units 2510 · I0 I0 = 10 log 2510

97. a) L = 10 log

≈ 34 decibels 2, 500, 000 · I0 b) L = 10 log I0 = 10 log 2, 500, 000 6 c) L = 10 log 10 · I0 I0 = 10 log 106

= 60 decibels

#

0, −

3 13

$

104. x = −4

Slope: not defined; y-intercept: none ( 105. −5 ( 1 −6 3 10 −5 55 −290 1 −11 58 −280 The remainder ( 106. −1 ( 1 −2 −1 1 −3 f (−1) = −4

107.

f (x) = (x −

is −280, so g(−5) = −280.

√

0 1 −6 3 −3 2 3 −2 −4 7)(x +

= (x2 − 7)(x)

H + ≈ 1.6 × 10−5

≈ 64 decibels

3 13

Slope: 2, y-intercept:

logarithm c)

103. y = 2x −

5.4 = − log[H + ]

−5.4 = log[H + ]

3x − 10y = 14

108.

√

7)(x − 0)

= x3 − 7x

f (x) = (x − 4i)(x + 4i)(x − 1) = (x2 + 16)(x − 1)

= x3 − x2 + 16x − 16 109. Using the change-of-base formula, we get log5 8 = log2 8 = 3. log5 2 110. Using the change-of-base formula, we get log3 64 = log16 64. log3 16 Let log16 64 = x. Then we have

298

Chapter 4: Exponential and Logarithmic Functions 16x = 64

Using the definition of logarithm

(24 )x = 26

4. log4 (64 · 4) = log4 64 + log4 4 = 3 + 1 = 4 5. Use the product rule.

24x = 26 , so

logt 8Y = logt 8 + logt Y

4x = 6 6 3 x= = 4 2 log3 64 3 Thus, = . log3 16 2

6. log 0.2x = log 0.2 + log x 7. Use the product rule. ln xy = ln x + ln y 8. ln ab = ln a + ln b

111. f (x) = log5 x3 x3 must be positive. Since x3 > 0 for x > 0, the domain is (0, ∞). 112. f (x) = log4 x2 x must be positive, so the domain is (−∞, 0) ∪ (0, ∞). 2

9. Use the power rule. logb t3 = 3 logb t 10. loga x4 = 4 loga x 11. Use the power rule.

113. f (x) = ln |x|

log y 8 = 8 log y

|x| must be positive. Since |x| > 0 for x += 0, the domain is (−∞, 0) ∪ (0, ∞).

114. f (x) = log(3x − 4)

3x − 4 must be positive. We have 3x − 4 > 0 4 x> 3

The domain is

!

4 ,∞ . 3 "

log(2x + 5) . Observe log 2 5 that outputs are negative for inputs between − and −2. 2 " ! 5 Thus, the solution set is − , −2 . 2

115. Graph y = log2 (2x + 5) =

log(x − 3) and y2 = 4. log 2 Observe that the graph of y1 lies on or above the graph of y2 for all inputs greater than or equal to 19. Thus, the solution set is [19, ∞).

116. Graph y1 = log2 (x − 3) =

117. Graph (d) is the graph of f (x) = ln |x|. 118. Graph (c) is the graph of f (x) = | ln x|. 119. Graph (b) is the graph of f (x) = ln x2 . 120. Graph (a) is the graph of g(x) = | ln(x − 1)|.

12. ln y 5 = 5 ln y 13. Use the power rule. logc K −6 = −6 logc K 14. logb Q−8 = −8 logb Q 15. Use the power rule. √ 1 3 ln 4 = ln 41/3 = ln 4 3 √ 1 16. ln a = ln a1/2 = ln a 2 17. Use the quotient rule. M logt = logt M − logt 8 8 18. loga

19. Use the quotient rule. x log = log x − log y y 20. ln

22. logb

3 = logb 3 − logb w w

23.

loga 6xy 5 z 4 = loga 6 + loga x + loga y 5 + loga z 4 Product rule

1. Use the product rule.

= loga 6 + loga x + 5 loga y + 4 loga z

log3 (81 · 27) = log3 81 + log3 27 = 4 + 3 = 7 3. Use the product rule. log5 (5 · 125) = log5 5 + log5 125 = 1 + 3 = 4

a = ln a − ln b b

21. Use the quotient rule. r ln = ln r − ln s s

Exercise Set 4.4

2. log2 (8 · 64) = log2 8 + log2 64 = 3 + 6 = 9

76 = loga 76 − loga 13 13

Power rule 24.

loga x3 y 2 z = loga x3 + loga y 2 + loga z = 3 loga x + 2 loga y + loga z

Exercise Set 4.4 25.

299

p2 q 5 m4 b 9 = logb p2 q 5 − logb m4 b9

31.

logb

Quotient rule

=

= logb p2 + logb q 5 − (logb m4 + logb b9 )

Product rule

=

= logb p2 + logb q 5 − logb m4 − logb b9 = logb p2 + logb q 5 − logb m4 − 9

(logb b9 = 9)

= 2 logb p + 5 logb q − 4 logb m − 9 26.

logb

=

Power rule

x2 y = logb x2 y − logb b3 b3 = logb x2 + logb y − logb b3 = logb x2 + logb y − 3

= = 32.

= 2 logb x + logb y − 3

27.

2 3x3 y = ln 2 − ln 3x3 y

=

ln

=

Quotient rule

= ln 2 − (ln 3 + ln x3 + ln y)

Product rule

=

= ln 2 − ln 3 − 3 ln x − ln y

Power rule

=

= ln 2 − ln 3 − ln x − ln y 3

28.

log

5a = log 5a − log 4b2 4b2 = log 5 + log a − (log 4 + log b2 ) = log 5 + log a − log 4 − log b2

29.

30.

= 33.

= log 5 + log a − log 4 − 2 log b √ log r3 t = log(r3 t)1/2 1 = log r3 t Power rule 2 1 = (log r3 + log t) Product rule 2 1 = (3 log r + log t) Power rule 2 1 3 = log r + log t 2 2 √ 3 ln 5x5 = ln(5x5 )1/3 1 = ln 5x5 3 1 = (ln 5 + ln x5 ) 3 1 = (ln 5 + 5 ln x) 3 5 1 = ln 5 + ln x 3 3

= = = = = =

34.

35.

1 x6 loga 5 8 2 p q 1 [loga x6 − loga (p5 q 8 )] Quotient rule 2 1 [loga x6 − (loga p5 + loga q 8 )] Product rule 2 1 (loga x6 − loga p5 − loga q 8 ) 2 1 (6 loga x − 5 loga p − 8 loga q) Power rule 2 5 3 loga x − loga p − 4 loga q 2 & 3 2 3 y z logc x4 1 y3 z2 logc 4 3 x 1 (logc y 3 z 2 − logc x4 ) 3 1 (logc y 3 + logc z 2 − logc x4 ) 3 1 (3 logc y + 2 logc z − 4 logc x) 3 2 4 logc y + logc z − logc x 3 3 & 8 12 4 m n loga a3 b5 1 m8 n12 loga 3 5 Power rule 4 a b 1 (loga m8 n12 − loga a3 b5 ) Quotient rule 4 1 [loga m8 + loga n12 − (loga a3 + loga b5 )] 4 Product rule 1 8 12 (loga m + loga n − loga a3 − loga b5 ) 4 1 (loga m8 +loga n12 −3−loga b5 ) 4 (loga a3 = 3) 1 (8 loga m + 12 loga n − 3 − 5 loga b) 4 Power rule

3 5 = 2 loga m + 3 loga n − − loga b 4 4 & √ a6 b8 = loga a4 b3 loga a2 b5 1 = (loga a4 + loga b3 ) 2 1 = (4 + 3 loga b) 2 3 = 2 + loga b 2 loga 75 + loga 2 = loga (75 · 2) = loga 150

Product rule

300

Chapter 4: Exponential and Logarithmic Functions

36. log 0.01 + log 1000 = log(0.01 · 1000) = log 10 = 1 37.

46.

x3 − 8 x−2 (x − 2)(x2 + 2x + 4) = log x−2 = log(x2 + 2x + 4)

log 10, 000 − log 100 10, 000 Quotient rule = log 100 = log 100

= log

=2 47.

54 = ln 9 38. ln 54 − ln 6 = ln 6 39.

40.

41.

Power rule

= log n1/2 m3 , or √ log m3 n

Product rule √ n1/2 = n

1 log a − log 2 = log a1/2 − log 2 2 √ a1/2 a = log , or log 2 2

48.

1 loga x + 4 loga y − 3 loga x 2 = loga x1/2 + loga y 4 − loga x3 Power rule

Product rule

x1/2 y 4 = loga x3

y4 x5/2

= ln x − 3 ln(x2 − 25)

Simplifying

2 1 loga x − loga y = loga x2/5 − loga y 1/3 = 5 3 2/5 x loga 1/3 y √ 43. ln x2 − 2 ln x √ = ln x2 − ln ( x)2 Power rule √ = ln x2 − ln x [( x)2 = x] x2 x = ln x

44.

= ln x − ln(x2 − 25)3 x = ln 2 (x − 25)3

! "3 x y ! "3 x = ln 2x y 2x4 = ln 3 y

45.

ln(x2 − 4) − ln(x + 2) = ln

x2 − 4 x+2

Quotient rule

(x + 2)(x − 2) Factoring x+2 = ln(x − 2) Removing a factor of 1 = ln

=

Quotient rule

2 x2 − 9 ln + ln(x + y) 3 x+3

2 (x + 3)(x − 3) ln + ln(x + y) 3 x+3 2 = ln(x − 3) + ln(x + y) 3 = ln(x − 3)2/3 + ln(x + y) =

x y

= ln 2x + ln

Power rule

2 [ln(x2 − 9) − ln(x + 3)] + ln(x + y) 3

50.

Quotient rule

ln 2x + 3(ln x − ln y) = ln 2x + 3 ln

ln x − 3[ln(x − 5) + ln(x + 5)]

= ln x − 3 ln[(x − 5)(x + 5)] Product rule

42.

= ln

x2 − 5x − 14 Quotient rule x2 − 4 (x + 2)(x − 7) = log Factoring (x + 2)(x − 2) x−7 Removing a factor of 1 = log x−2 √ a a loga √ − loga ax = loga √ √ x x ax √ a = loga x √ = loga a − loga x 1 = loga a − loga x 2 1 = − loga x 2

49.

Quotient rule

= loga x−5/2 y 4 , or loga

log(x2 − 5x − 14) − log(x2 − 4)

= log

1 log n + 3 log m 2 = log n1/2 + log m3

= loga x1/2 y 4 − loga x3

log(x3 − 8) − log(x − 2)

= ln[(x − 3)2/3 (x + y)] 51.

4 3 ln 4x6 − ln 2y 10 2 5 3 4 = ln 22 x6 − ln 2y 10 2 5 = ln(22 x6 )3/2 − ln(2y 10 )4/5

= ln(2 x ) − ln(2 3 9

23 x9 = ln 4/5 8 2 y = ln

211/5 x9 y8

Writing 4 as 22 Power rule

y )

4/5 8

Quotient rule

Exercise Set 4.4 52.

# √ √ 5 4 120(ln x3 + ln 3 y 2 − ln 16z 5 ) # √ ! " 5 x3 3 y 2 = 120 ln √ 4 16z 5 " ! 3/5 2/3 x y = 120 ln 2z 5/4 ! 3/5 2/3 "120 x y = ln 2z 5/4 = ln

53.

54.

301

loga

x72 y 80 2120 z 150

≈ −0.74

≈ 1.146

loga 98 = loga (72 · 2)

Product rule

= 2 loga 7 + loga 2

Power rule

logb 125 = logb 53 Power rule

≈ 3(1.609) ≈ 4.827

5 = logb 5 − logb 3 3 ≈ 1.609 − 1.099 ≈ 0.51

logb

logb 15b = logb (3 · 5 · b)

= logb 3 + logb 5 + logb b

1 = logb 1 − logb 6 Quotient rule 6 = logb 1 − logb (2 · 3) = logb 1 − (logb 2 + logb 3)

= logb 1 − logb 2 − logb 3

≈ 0 − 0.693 − 1.099 ≈ −1.792

≈ 1.099 + 1.609 + 1 ≈ 3.708

(loga ax = x)

(loga ax = x)

(aloga x = x)

70. 5log5 (4x−3) = 4x − 3 71. 10log w = w

73. ln e8t = 8t

58. loga 9 cannot be found using the given information.

61.

≈ 0.099

(aloga x = x)

3

loga 2 0.301 ≈ ≈ 0.356 loga 7 0.845

logb

Quotient rule

72. eln x = x3

1 = loga 1 − loga 7 7 ≈ 0 − 0.845

= 3 logb 5

3 = logb 3 − logb b b ≈ 1.099 − 1

69. 3log3 4x = 4x

= loga 72 + loga 2

≈ −0.845

60.

logb

67. loge e|x−4| = |x − 4| √ √ 68. logq q 3 = 3

≈ 1.991

59.

63.

66. logt t2713 = 2713

≈ 2(0.845) + 0.301

57.

≈ 3.401

65. logp p3 = 3

loga 14 = loga (2 · 7)

loga

= logb 2 + logb 3 + logb 5

2 = loga 2 − loga 11 Quotient rule 11 ≈ 0.301 − 1.041

≈ 0.301 + 0.845

56.

logb 30 = logb (2 · 3 · 5)

≈ 0.693 + 1.099 + 1.609

64.

= loga 2 + loga 7

55.

62.

Product rule

(loga ax = x)

74. log 10−k = −k √ 75. logb b = logb b1/2 1 = logb b Power rule 2 1 = ·1 (logb b = 1) 2 1 = 2 √ 76. logb b3 = logb b3/2 3 = logb b 2 3 = ·1 2 3 = 2 77. f (x) = ax , g(x) = loga x Since f and g are inverses, we know that (f ◦g)(x) = x and (g ◦ f )(x) = x. Now (f ◦ g)(x) = f (g(x)) = f (loga x) = aloga x , so we know that aloga x = x. Also (g ◦ f )(x) = g(f (x)) = g(ax ) = loga ax , so we know that loga ax = x. These results are alternate proofs of the Logarithm of a Base to a Power property and the Base to a Logarithmic Power property.

302

Chapter 4: Exponential and Logarithmic Functions

78. loga ab3 += (loga a)(loga b3 ). If the first step had been correct, then so would the second step. The correct procedure follows.

93. =

loga ab3 = loga a + loga b3 = 1 + 3 loga b

=

79. The degree of f (x) = 5 − x2 + x4 is 4, so the function is quartic.

=

80. The variable in f (x) = 2x is in the exponent, so f (x) is an exponential function. ' 3 81. f (x) = − is of the form f (x) = mx + b with m = 0 and 4 3( b = − , so it is a linear function. In fact, it is a constant 4 function. 82. The variable in f (x) = 4x − 8 is in the exponent, so f (x) is an exponential function. p(x) 3 is of the form f (x) = where p(x) and q(x) x q(x) are polynomials and q(x) is not the zero polynomial, so f (x) is a rational function.

= = = = 94.

83. f (x) = −

84. f (x) = log x + 6 is a logarithmic function. 1 85. The degree of f (x) = − x3 − 4x2 + 6x + 42 is 3, so the 3 function is cubic. p(x) x2 − 1 is of the form f (x) = where p(x) x2 + x − 6 q(x) and q(x) are polynomials and q(x) is not the zero polynomial, so f (x) is a rational function.

86. f (x) =

1 87. f (x) = x + 3 is of the form f (x) = mx + b, so it is a 2 linear function. 88. The degree of f (x) = 2x2 − 6x + 3 is 2, so the function is quadratic. 89.

5log5 8 = 2x 8 = 2x (aloga x = x) 4=x The solution is 4.

90.

ln e3x−5 = −8 3x − 5 = −8 3x = −3

x = −1

loga (x2 + xy + y 2 ) + loga (x − y)

= loga [(x2 + xy + y 2 )(x − y)] Product rule 92.

= loga (x3 − y 3 )

loga (a10 − b10 ) − loga (a + b)

Multiplying

a10 − b10 , or a+b 9 8 loga (a − a b + a7 b2 − a6 b3 + a5 b4 − a4 b5 +

= loga

a3 b6 − a2 b7 + ab8 − b9 )

= loga (9 − x2 )1/2 1 = loga (9 − x2 ) 2 1 = loga [(3 + x)(3 − x)] 2 1 = [loga (3 + x) + loga (3 − x)] 2 1 1 = loga (3 + x) + loga (3 − x) 2 2 # 4 2 5 y z loga √ 4 x3 z −2 & y2 z5 4 = loga x3 z −2 & 2 7 4 y z = loga x3 ! 2 7 "1/4 y z = loga x3 ! 2 7" 1 y z = loga Power rule 4 x3 1 = (loga y 2 z 7 − loga x3 ) Quotient rule 4 1 = (loga y 2 + loga z 7 − loga x3 ) Product rule 4 1 (2 loga y + 7 loga z − 3 loga x) Power rule 4 1 = (2 · 3 + 7 · 4 − 3 · 2) 4 1 = · 28 4 =7

=

The solution is −1. 91.

95.

x−y loga # x2 − y 2 x−y loga 2 (x − y 2 )1/2 loga (x − y) − loga (x2 − y 2 )1/2 Quotient rule 1 loga (x − y) − loga (x2 − y 2 ) Power rule 2 1 loga (x − y) − loga [(x + y)(x − y)] 2 1 loga (x − y) − [loga (x + y) + loga (x − y)] 2 Product rule 1 1 loga (x − y) − loga (x + y) − loga (x − y) 2 2 1 1 loga (x − y) − loga (x + y) 2 2 √ loga 9 − x2

96. loga M + loga N = loga (M + N ) Let a = 10, M = 1, and N = 10. Then log10 1 + log10 10 = 0 + 1 = 1, but log10 (1 + 10) = log10 11 ≈ 1.0414. Thus, the statement is false.

Exercise Set 4.5

303

M N This is the quotient rule, so it is true.

97. loga M − loga N = loga 98.

107.

ln a − ln b = −xy a ln = −xy b Then, using the definition of a logarithm, we have a b e−xy = , or exy = . b a

loga M = loga M − loga N loga N loga a2 2 = = 2, but loga a 1 loga a2 − loga a = 2 − 1 = 1. Thus, the statement is false. Let M = a2 and N = a. Then

99.

loga M 1 = loga M = loga M 1/x . The statement is true x x by the power rule.

100. loga x3 = 3 loga x is true by the power rule. 101. loga 8x = loga 8 + loga x = loga x + loga 8. The statement is true by the product rule and the commutative property of addition. 102.

logN (M · N )x = x logN (M · N )

= x(logN M + logN N )

= x(logN M + 1) = x logN M + x The statement is true. ! " 1 = loga x−1 = −1 · loga x = −1 · 2 = −2 103. loga x 104.

loga x = 2 a2 = x log1/a a = n ! "n 1 = a2 a 2

1 = loga 1 − loga x = − loga x. x Let − loga x = y. Then loga x = −y and x = a−y = ! "y ! " 1 1 , so log1/a x = y. Thus, loga a−1·y = = a x − loga x = log1/a x. √ " ! x + x2 − 5 109. loga 5 √ √ " ! x + x2 − 5 x − x2 − 5 √ = loga · 5 x − x2 − 5 ! " ! " 5 1 √ √ ) = loga = loga 5(x − x2 − 5 x − x2 − 5 √ = loga 1 − loga (x − x2 − 5) √ = − loga (x − x2 − 5)

108. loga

Exercise Set 4.5 1.

Let log1/a x = n and solve for n.

3x = 81 3x = 34

Substituting a for x 2

x=4

The exponents are the same.

The solution is 4.

(a−1 )n = a2

2.

a−n = a2

2x = 32 2x = 25

−n = 2

x=5

n = −2

Thus, log1/a x = −2 when loga x = 2. 105. We use the change-of-base formula.

The solution is 5. 3.

log998 999·log999 1000 log 12 log 13 ··· = log10 11· 10 · 10 log10 11 log10 12 log10 999 log10 1000 · log10 998 log10 999 log10 11 log10 12 log10 999 = · ··· · log10 1000 log10 11 log10 12 log10 999 = log10 1000 =3 loga x + loga y − mz = 0

loga x + loga y = mz loga xy = mz amz = xy

22x = 8 22x = 23

log10 11·log11 12·log12 13 · · ·

106.

ln a − ln b + xy = 0

2x = 3 3 x= 2

The exponents are the same.

The solution is 4.

3 . 2

37x = 27 37x = 33 7x = 3 3 x= 7 The solution is

3 . 7

304

Chapter 4: Exponential and Logarithmic Functions

5.

log 2x = log 33 x log 2 = log 33 log 33 x= log 2 1.5185 x≈ 0.3010 x ≈ 5.044

2

3x

11.

12.

= 125

x log 28 = −3x

x log 28 + 3x = 0

x(log 28 + 3) = 0

4x = 10 10 5 x= = 4 2 5 The solution is . 2

x=0 The solution is 0. 13.

−c = 2c ln 5

0 = c + 2c ln 5

2

0 = c(1 + 2 ln 5)

3x − 5 = 2

0=c Dividing by 1 + 2 ln 5 The solution is 0.

3x = 7 7 x= 3

14.

7 The solution is . 3 9.

x log 15 = log 30 log 30 x= log 15 x ≈ 1.256 The solution is 1.256.

2

33 = 35x · (32 )x 2

33 = 35x · 32x 2

33 = 35x+2x

3 = 5x + 2x2

15.

0 = 2x + 5x − 3

0 = (2x − 1)(x + 3)

The solutions are −3 and

et = 1000 ln et = ln 1000

2

1 or x = −3 2

15x = 30 log 15x = log 30

2

27 = 35x · 9x

x=

e−c = 52c ln e−c = ln 52c

43x−5 = 16 =4

28x = 10−3x log 28x = log 10−3x

4x − 7 = 3

4

84x = 70

x log 84 = log 70 log 70 x= log 84 1.8451 x≈ 1.9243 x ≈ 0.959 The solution is 0.959.

54x−7 = 53

3x−5

The solutions are −3 and −1. log 84x = log 70

The solution is 5.322.

8.

x + 4x = −3

(x + 3)(x + 1) = 0 x = −3 or x = −1

x log 2 = log 40 log 40 x= log 2 1.6021 x≈ 0.3010 x ≈ 5.322 5

+4x

1 27 = 3−3 =

x + 4x + 3 = 0 2

2x = 40

4x−7

+4x

2

log 2x = log 40

7.

3x

Taking the common logarithm on both sides Power rule

The solution is 5.044. 6.

2

10.

2x = 33

t = ln 1000

1 . 2

Using loga ax = x

t ≈ 6.908 The solution is 6.908. 16.

e−t = 0.04 ln e−t = ln 0.04 −t = ln 0.04

t = − ln 0.04 ≈ 3.219 The solution is 3.219.

Exercise Set 4.5 17.

305 22.

e−0.03t = 0.08 ln e−0.03t = ln 0.08

250 = (1.87)x

log 250 = log(1.87)x

−0.03t = ln 0.08 ln 0.08 t= −0.03 −2.5257 t≈ −0.03 t ≈ 84.191

log 250 = x log 1.87 log 250 =x log 1.87 8.821 ≈ x

The solution is 8.821.

The solution is 84.191. 18.

23.

1000e0.09t = 5000

ex + e−x = 5 e2x + 1 = 5ex

e0.09t = 5

This equation is quadratic in ex . √ 5 ± 21 ex = 2 √ " ! 5 ± 21 ≈ ±1.567 x = ln 2 The solutions are −1.567 and 1.567.

0.09t = ln 5 ln 5 t= 0.09 t ≈ 17.883

The solution is 17.883. 3x = 2x−1 ln 3x = ln 2x−1

24.

ex − 6e−x = 1

e2x − 6 = ex

x ln 3 = (x − 1) ln 2

x ln 3 = x ln 2 − ln 2

ex = 3

= x(ln 2 − ln 3)

log 5

The solution is 1.099.

≈x

25.

= log 4

−x

ln 2 = 2x ln 2 =x 2 0.347 ≈ x

The solution is −0.612.

x log 3.9 = log 48 log 48 x= log 3.9 1.6812 x≈ 0.5911 x ≈ 2.844

The solution is 2.844.

Multiplying by ex

ln 2 = ln e2x

x(log 5 + log 4) = log 4 − 2 log 5 log 4 − 2 log 5 x= log 5 + log 4 x ≈ −0.612

log(3.9) = log 48

Multiplying by ex − e−x

Subtracting ex and adding e−x

2 = e2x

x log 5 + x log 4 = log 4 − 2 log 5

x

= 2e

x

2e−x = ex

x log 5 + 2 log 5 = log 4 − x log 4

(3.9) = 48

ex + e−x =3 ex − e−x ex + e−x = 3ex − 3e−x 4e

1−x

(x + 2) log 5 = (1 − x) log 4

21.

No solution

x ≈ 1.099

≈x

x

ex = −2

x = ln 3

5x+2 = 41−x x+2

or

ln ex = ln 3

=x

The solution is −1.710. 20.

e2x − ex − 6 = 0

(e − 3)(ex + 2) = 0 x

ln 2 = x ln 2 − x ln 3

ln 2 ln 2 ln 2 − ln 3 0.6931 0.6931 − 1.0986 −1.710

Multiplying by ex

e2x − 5ex + 1 = 0

ln e0.09t = ln 5

19.

250 − (1.87)x = 0

The solution is 0.347. 26.

5x − 5−x =8 5x + 5−x 5x − 5−x = 8 · 5x + 8 · 5−x −9 · 5−x = 7 · 5x

−9 = 7 · 52x Multiplying by 5x 9 − = 52x 7 The number 5 raised to any power is non-negative. Thus, the equation has no solution.

306 27.

Chapter 4: Exponential and Logarithmic Functions 35.

log5 x = 4 x=5

4

x = 625

25 = 10 + 3x

Writing an equivalent exponential equation

32 = 10 + 3x 22 = 3x 22 =x 3

The solution is 625. 28.

log2 x = −3

x = 2−3 1 x= 8

The solution is 29.

The answer checks. The solution is 36.

log x = −4

117 = −7x 117 =x − 7

The base is 10.

37.

x(x − 9) = 101

1 log64 = x 4 1 = 64x 4 1 = (43 )x 4 4−1 = 43x

x2 − 9x = 10

x2 − 9x − 10 = 0

(x − 10)(x + 1) = 0 x = 10 or x = −1

Check: For 10: log x + log(x − 9) = 1

−1 = 3x 1 − =x 3

log 10 + log(10 − 9) ?) 1 log 10 + log 1 )) 1 + 0 )) 1 ) 1 TRUE For −1: log x + log(x − 9) = 1

1 The solution is − . 3 1 25 1 25 1 52 5−2

=x

log(−1) + log(−1 − 9) ? 1 | The number −1 does not check, because negative numbers do not have logarithms. The solution is 10.

= 125x = (53 )x = 53x

38.

−2 = 3x 2 − =x 3

ln x = 1

(x + 1)(x − 1) = 23 x2 − 1 = 8

x2 = 9

The base is e.

x = e1 = e 39.

The solution is e. 34.

ln x = −2 x = e−2 , or

log2 (x + 1) + log2 (x − 1) = 3

log2 [(x + 1)(x − 1)] = 3

2 The solution is − . 3 33.

log x + log(x − 9) = 1

log10 [x(x − 9)] = 1

The solution is 10.

log125

117 . 7 The base is 10.

The answer checks. The solution is −

log x = 1 x = 101 = 10

32.

log5 (8 − 7x) = 3

125 = 8 − 7x

The solution is 0.0001.

31.

22 . 3

53 = 8 − 7x

1 . 8

x = 10−4 , or 0.0001

30.

log2 (10 + 3x) = 5

1 e2

The solution is e−2 , or

1 . e2

x = ±3 The number 3 checks, but −3 does not. The solution is 3. log2 (x + 20) − log2 (x + 2) = log2 x x + 20 = log2 x log2 x+2 x + 20 = x Using the property of x+2 logarithmic equality x + 20 = x2 + 2x Multiplying by x+2 0 = x2 + x − 20 0 = (x + 5)(x − 4)

Exercise Set 4.5 x+5 = 0

307 or x − 4 = 0

x = −5 or

43.

x=4

Check: For −5: log2 (x + 20) − log2 (x + 2) = log2 x log2 (−5 + 20) − log2 (−5 + 2) ? log2 (−5) | The number −5 does not check, because negative numbers do not have logarithms.

log8 (x + 1) − log8 x " ! x+1 log8 x x+1 x x+1 x x+1

log3 (x + 14) − log3 (x + 6) x + 14 log3 x+6 x + 14 x+6 x + 14

44. TRUE

= log3 x = log3 x

0 = (x + 7)(x − 2)

log(x+8)−log(x+1) = log x+8 = log log x+1 x+8 =6 x+1

Using the property of logarithmic equality x + 8 = 6x + 6 Multiplying by x+1

The answer checks. The solution is 42.

log(x + 5) − log(x − 3) x+5 log x−3 x+5 x−3 x+5

= 64x

= −1 = 10−1

1 10 = x+3 =

The answer checks. The solution is 45.

1 . 63

= −1

1 . 3

log x + log(x + 4) = log 12 x(x + 4) = 12

Quotient rule

2 = 5x 2 =x 5

= 64

log x(x + 4) = log 12

6 6

= 82

9x = 3 1 x= 3

= x2 + 6x

Only 2 checks. It is the solution. 41.

log x − log(x + 3) x log10 x+3 x x+3 x x+3 10x

=x

0 = x2 + 5x − 14 x = −7 or x = 2

Quotient rule

The answer checks. The solution is

The solution is 4. 40.

=2

1 = 63x 1 =x 63

For 4: log2 (x + 20) − log2 (x + 2) = log2 x log2 (4 + 20) − log2 (4 + 2) ?) log2 4 ) log2 24 − log2 6 ) ) 24 )) log2 ) 6 ) ) log2 4 ) log2 4

=2

x2 + 4x = 12

Using the property of logarithmic equality

x2 + 4x − 12 = 0

(x + 6)(x − 2) = 0

x+6 = 0

or x − 2 = 0

x = −6 or

x=2

Check: For −6: log x + log(x + 4) = log 12 2 . 5

= log 2 = log 2 =2 = 2x − 6

11 = x

The answer checks. The solution is 11.

log(−6) + log(−6 + 4) ? log 12 | The number −6 does not check, because negative numbers do not have logarithms. For 2: log x + log(x + 4) = log 12 log 2 + log(2 + 4) ?) log 12 ) log 2 + log 6 ) ) log(2 · 6) )) ) log 12 ) log 12

The solution is 2.

TRUE

308 46.

Chapter 4: Exponential and Logarithmic Functions ln x − ln(x − 4) x ln x−4 x x−4 x

51.

= ln 3

ln(x + 8) + ln(x − 1) = 2 ln x ln(x + 8)(x − 1) = ln x2

= ln 3

(x + 8)(x − 1) = x2

=3

x2 + 7x − 8 = x2

= 3x − 12

7x − 8 = 0

12 = 2x

7x = 8 8 x= 7

6=x The answer checks. The solution is 6. 47.

log4 (x + 3) + log4 (x − 3) = 2

log4 [(x + 3)(x − 3)] = 2 (x + 3)(x − 3) = 4

Product rule 2

The answer checks. The solution is 52.

log3 x + log3 (x + 1) = log3 2 + log3 (x + 3) x(x + 1) = 2(x + 3)

x2 = 25

x2 + x = 2x + 6

x = ±5

x −x−6 = 0 2

The number 5 checks, but −5 does not. The solution is 5. ln(x + 1) − ln x x+1 ln x x+1 x x+1

x = 3 or x = −2

= ln 4

The number 3 checks, but −2 does not. The solution is 3. 53.

=4

log6 x = 1 − log6 (x − 5)

log6 x + log6 (x − 5) = 1

= 4x

log6 x(x − 5) = 1

61 = x(x − 5)

The answer checks. The solution is log(2x + 1) − log(x − 2) " ! 2x + 1 log x−2 2x + 1 x−2 2x + 1

6 = x2 − 5x

0 = x2 − 5x − 6

1 . 3

=1 =1

x+1 = 0

21 = 8x 21 =x 8

The answer checks. The solution is

x=6

54.

1 256

2

−9x

=

2

−9x

= 2−8

2x 2x

x − 9x = −8 2

21 . 8

log5 (x + 4) + log5 (x − 4) = 2

log5 [(x + 4)(x − 4)] = 2

x2 − 16 = 25

or x − 6 = 0

The number −1 does not check, but 6 does. The answer is 6.

= 101 = 10 = 10x − 20

0 = (x + 1)(x − 6)

x = −1 or

Quotient rule

Multiplying by x − 2

50.

(x − 3)(x + 2) = 0

= ln 4

1 = 3x 1 =x 3

49.

8 . 7

log3 x(x + 1) = log3 2(x + 3)

x2 − 9 = 16

48.

Using the property of logarithmic equality

x2 = 41 √ x = ± 41 √ √ Only 41 checks. The solution is 41.

x2 − 9x + 8 = 0

(x − 1)(x − 8) = 0

x = 1 or x = 8

The solutions are 1 and 8.

Exercise Set 4.5 55.

309 59. e7.2x = 14.009

9x−1 = 100(3x )

Graph y1 = e7.2x and y2 = 14.009 and find the first coordinate of the point of intersection using the Intersect feature. The solution is 0.367.

(32 )x−1 = 100(3x ) 3 3

2x−2

= 100(3 ) x

2x−2

= 100 3x 3x−2 = 100

log 3

x−2

60. 0.082e0.05x = 0.034 Graph y1 = 0.082e0.05x and y2 = 0.034 and find the first coordinate of the point of intersection using the Intersect feature. The solution is −17.607.

= log 100

(x − 2) log 3 = 2 x−2 =

2 log 3

61. xe3x − 1 = 3

Graph y1 = xe3x − 1 and y2 = 3 and find the first coordinate of the point of intersection using the Intersect feature. The solution is 0.621.

2 x = 2+ log 3 x ≈ 6.192

62. 5e5x + 10 = 3x + 40

The solution is 6.192. 56.

Graph y1 = 5e5x + 10 and y2 = 3x + 40 and find the first coordinates of the points of intersection using the Intersect feature. The solutions are −10 and 0.366.

2 ln x − ln 5 = ln (x + 10)

ln x2 − ln 5 = ln (x + 10) ln

x2 = ln (x + 10) 5 x2 = x + 10 5 2 x = 5x + 50

63. 4 ln(x + 3.4) = 2.5 Graph y1 = 4 ln(x + 3.4) and y2 = 2.5 and find the first coordinate of the point of intersection using the Intersect feature. The solution is −1.532. 64. ln x2 = −x2

x2 − 5x − 50 = 0

Graph y1 = ln x2 and y2 = −x2 and find the first coordinates of the points of intersection using the Intersect feature. The solutions are −0.753 and 0.753.

(x − 10)(x + 5) = 0

x = 10 or x = −5

Only 10 checks. It is the solution. 57.

ex − 2 = −e−x 1 ex − 2 = − x e e2x − 2ex = −1 Multiplying by ex x

e2x − 2e + 1 = 0

Let u = ex . u2 − 2u + 1 = 0

(u − 1)(u − 1) = 0

u − 1 = 0 or u − 1 = 0 u = 1 or

u=1

ex = 1 or

ex = 1

x = 0 or

x=0

Replacing u with ex

The solution is 0. 58.

2 log 50 = 3 log 25 + log (x − 2) log 50 = log 25 + log (x − 2) 2

3

log 2500 = log 15, 625 + log (x − 2) log 2500 = log [15, 625(x − 2)] 2500 = 15, 625(x − 2)

2500 = 15, 625x − 31, 250

33, 750 = 15, 625x 54 =x 25

The answer checks. The solution is

65. log8 x + log8 (x + 2) = 2 log x log(x + 2) + and y2 = 2 and find the Graph y1 = log 8 log 8 first coordinate of the point of intersection using the intersect feature. The solution is 7.062. 66. log3 x + 7 = 4 − log5 x log x log x Graph y1 = + 7 and y2 = 4 − and find the first log 3 log 5 coordinate of the point of intersection using the Intersect feature. The solution is 0.141. 67. log5 (x + 7) − log5 (2x − 3) = 1 log(x + 7) log(2x − 3) − and y2 = 1 and find Graph y1 = log 5 log 5 the first coordinate of the point of intersection using the Intersect feature. The solution is 2.444. 68. Graph y1 = ln 3x and y2 = 3x − 8 and use the Intersect feature to find the points of intersection. They are (0.0001, −7.9997) and (3.445, 2.336). 69. Solving the first equation for y, we get 12.4 − 2.3x 12.4 − 2.3x . Graph y1 = and y= 3.8 3.8 y2 = 1.1 ln(x − 2.05) and use the Intersect feature to find the point of intersection. It is (4.093, 0.786). 2

54 . 25

70. Graph y1 = 2.3 ln(x + 10.7) and y2 = 10e−0.07x and use the Intersect feature to find the points of intersection. They are (−9.694, 0.014), (−3.334, 4.593), and (2.714, 5.971).

310

Chapter 4: Exponential and Logarithmic Functions 2

71. Graph y1 = 2.3 ln(x + 10.7) and y2 = 10e−0.007x and use the Intersect feature to find the point of intersection. It is (7.586, 6.684). 72. The final result would have been the same, but to find t log 2500 . we would have computed 0.08 log e It seems best to take the natural logarithm on both sides since the final computation for t is simpler. 73. Use the graph of y = ln x to estimate the x-value that corresponds to the value on the right-hand side of the equation. 74. f (x) = −x2 + 6x − 8 6 b =− =3 a) − 2a 2(−1) f (3) = −32 + 6 · 3 − 8 = 1 The vertex is (3, 1).

b) x = 3 c) Maximum: 1 at x = 3 75. g(x) = x2 − 6 0 b =− =0 a) − 2a 2·1 g(0) = 02 − 6 = −6

The vertex is (0, −6).

b) The axis of symmetry is x = 0. c) Since the coefficient of the x2 -term is positive, the function has a minimum value. It is the second coordinate of the vertex, −6, and it occurs when x = 0. 76. H(x) = 3x2 − 12x + 16 b −12 =− =2 a) − 2a 2·3 H(2) = 3 · 22 − 12 · 2 + 16 = 4 The vertex is (2, 4).

b) x = 2 c) Minimum: 4 at x = 2 77. G(x) = −2x2 − 4x − 7 b −4 =− = −1 a) − 2a 2(−2) G(−1) = −2(−1)2 − 4(−1) − 7 = −5

The vertex is (−1, −5).

b) The axis of symmetry is x = −1.

c) Since the coefficient of the x2 -term is negative, the function has a maximum value. It is the second coordinate of the vertex, −5, and it occurs when x = −1.

78.

ln(ln x) = 2 ln x = e2 2

x = ee ≈ 1618.178

2

The answer checks. The solution is ee , or 1618.178. 79.

ln(log x) = 0 log x = e0 log x = 1 x = 101 = 10

The answer checks. The solution is 10. √ √ 80. ln 4 x = ln x √ 1 ln x = ln x 4 1 (ln x)2 = ln x Squaring both sides 16 1 (ln x)2 − ln x = 0 16 Let u = ln x and substitute. 1 2 u −u = 0 16 " ! 1 u−1 = 0 u 16 1 u = 0 or u−1 = 0 16 u = 0 or u = 16 ln x = 0

or

ln x = 16

x=e

or

x = e16

x=1

or

x = e16 ≈ 8, 886, 110.521

0

Both answers check. The solutions are 1 and e16 , or 1 and 8,886,110.521. √ √ 81. ln x = ln x √ 1 ln x = ln x Power rule 2 1 ln x = (ln x)2 Squaring both sides 4 1 0 = (ln x)2 − ln x 4 Let u = ln x and substitute. 1 2 u −u = 0 4 " ! 1 u u−1 = 0 4 1 u=0 or u − 1 = 0 4 1 u=1 u=0 or 4 u=0 or u=4 ln x = 0

or

x = e0 = 1 or

ln x = 4 x = e4 ≈ 54.598

Both answers check. The solutions are 1 and e4 , or 1 and 54.598.

Chapter 5

The Trigonometric Functions Exercise Set 5.1 1. We use the definitions. 15 opp = sin φ = hyp 17 adj 8 cos φ = = hyp 17 15 opp = tan φ = adj 8 17 hyp = csc φ = opp 15 17 hyp sec φ = = adj 8 adj 8 cot φ = = opp 15 0.3 3 opp = = 2. sin β = hyp 0.5 5 adj 0.4 4 cos β = = = hyp 0.5 5 0.3 3 opp = = tan β = adj 0.4 4 0.5 5 hyp csc β = = = opp 0.3 3 hyp 0.5 5 sec β = = = adj 0.4 4 0.4 4 adj = = cot β = opp 0.3 3 3. We use the definitions. √ √ 3 3 3 opp = = sin α = hyp 6 2 3 1 adj = = cos α = hyp 6 2 √ opp 3 3 √ = = 3 tan α = adj 3

√ 2 2 3 hyp 6 = √ = √ , or opp 3 3 3 3 6 hyp = =2 sec α = adj 3 √ 1 adj 3 3 = √ = √ , or cot α = opp 3 3 3 3 csc α =

6 12 opp = = 13 hyp 13 2 5 5 adj 2 = = cos θ = 13 hyp 13 2

4. sin θ =

opp 6 12 = = 5 adj 5 2 13 13 hyp = 2 = csc θ = opp 6 12 13 hyp 13 sec θ = = 2 = 5 adj 5 2 5 adj 5 cot θ = = 2 = opp 6 12 tan θ =

5. First we use the Pythagorean theorem to find the length of the hypotenuse, c. a2 + b2 = c2 42 + 72 = c2 √

65 = c2 65 = c

Then we use the definitions to find the trigonometric function values of φ. √ 7 7 65 opp = √ , or sin φ = hyp 65 65 √ adj 4 4 65 cos φ = = √ , or hyp 65 65 7 opp = tan φ = adj 4 √ 65 hyp csc φ = = opp 7 √ 65 hyp sec φ = = adj 4 4 adj = cot φ = opp 7 6. First we use the Pythagorean theorem to find the length of the side opposite θ, a. a2 + b2 = c2 a2 + (8.2)2 = 92 a2 = 13.76 √ √ a = 13.76 = 4 0.86 ≈ 3.7

Now we find the trigonometric function values of θ. √ 13.76 3.7 opp = ≈ ≈ 0.4111 sin θ = hyp 9 9 8.2 adj = ≈ 0.9111 cos θ = hyp 9 √ opp 3.7 13.76 = ≈ ≈ 0.4512 tan θ = adj 8.2 8.2

330

Chapter 5: The Trigonometric Functions

hyp 9 9 =√ ≈ ≈ 2.4324 opp 3.7 13.76 9 hyp = ≈ 1.0976 sec θ = adj 8.2 8.2 8.2 adj =√ ≈ ≈ 2.2162 cot θ = opp 3.7 13.76 √ √ 1 3 5 1 3 3 5 = √ = √ , or √ · √ = 7. csc α = sin α 5 5 5 5 5 3 1 1 3 sec α = = = 2 cos α 2 3 √ √ 1 2 2 5 2 5 1 = √ = √ , or √ · √ = cot α = tan α 5 5 5 5 5 2 √ √ 1 3 3 2 3 2 1 8. csc β = = √ = √ , or √ · √ = sin β 4 2 2 2 2 2 2 2 3 1 1 =3 sec β = = 1 cos β 3 √ √ 1 1 2 2 1 = √ , or √ · √ = cot β = tan β 4 2 2 2 2 2 csc θ =

9. We know from the definition of the sine function that the 24 opp ratio is . Let’s consider a right triangle in which 25 hyp the hypotenuse has length 25 and the side opposite θ has length 24.

10. cos σ = 0.7 =

7 10

10

a

σ 7 a2 = 102 − 72 = 51 √ a = 51 √ √ 51 51 ; tan σ = ; sin σ = 10 7 √ 10 10 51 csc σ = √ , or ; 51 51 sec σ =

√ 7 51 10 7 ; cot σ = √ , or 7 51 51

11. We know from the definition of the tangent function that opp 2 is . Let’s consider a right triangle 2, or the ratio 1 adj in which the side opposite φ has length 2 and the side adjacent to φ has length 1.

c

2

φ 25

1

24

Use the Pythagorean theorem to find the length of the hypotenuse.

θ a Use the Pythagorean theorem to find the length of the side adjacent to θ. a2 + b2 = c2 a2 + 242 = 252 a2 = 625 − 576 = 49 a=7

Use the lengths of the three sides to find the other five ratios. 24 25 25 7 , tan θ = , csc θ = , sec θ = , cos θ = 25 7 24 7 7 cot θ = 24

a2 + b2 = c2 12 + 22 = c2 1 + 4 = c2 √

5 = c2 5=c

Use the lengths of the three sides to find the other five ratios. √ √ 2 5 1 5 2 sin φ = √ , or ; cos φ = √ , or ; 5 5 5 5 √ √ 1 5 ; sec φ = 5; cot φ = csc φ = 2 2

Exercise Set 5.1

331 14.

12.

c

17

3

θ

β

1

1 √ 2 a = ( 17) − 12 = 16

12 + 32 = c2 √

a

2

10 = c2

a=4

10 = c

√

√

3 10 3 1 10 sin θ = √ , or ; cos θ = √ , or ; 10 10 10 10 √ √ 3 10 10 √ ; sec θ = = 10 tan θ = = 3; csc θ = 1 3 1 15 3 1.5 13. csc θ = 1.5 = = = 1 10 2 We know from the definition of the cosecant function that 3 hyp . Let’s consider a right triangle in which the ratio is 2 opp the hypotenuse has length 3 and the side opposite θ has length 2.

√ √ 4 17 1 17 4 ; cos β = √ , or ; sin β = √ , or 17 17 17 17 √ 17 1 tan β = 4; csc β = ; cot β = 4 4 15. We know√from the definition of the cosine function that the 5 adj is . Let’s consider a right triangle in which ratio of 5 hyp √ the side adjacent to β has length 5 and the hypotenuse has length 5.

5

3

2

a

β 5

θ a Use the Pythagorean theorem to find the length of the side adjacent to θ. a2 + b2 = c2 a2 + 22 = 32 a = 9−4=5 √ a= 5 2

Use the lengths of the three sides to find the other five ratios. √ √ 5 2 2 5 2 ; tan θ = √ , or ; sin θ = ; cos θ = 3 3 5 5 √ √ 3 3 5 5 ; cot θ = sec θ = √ , or 5 2 5

Use the Pythagorean theorem to find the length of the side opposite β. a2 + b2 = c2 √ a2 + ( 5)2 = 52 a2 + 5 = 25 a2 = 25 − 5 = 20 √ √ a = 20 = 2 5 Use the lengths of the three sides to find the other five ratios. √ √ √ 5 5 2 5 2 5 sin β = ; tan β = √ = 2; csc β = √ , or ; 5 2 5 √ 2 5 √ 1 5 5 sec β = √ , or 5; cot β = √ = 2 5 2 5

332

Chapter 5: The Trigonometric Functions 24. See the triangle in Exercise 19. 1 cos 60◦ = 2

16.

11

25. See the triangle in Exercise 17.

10

tan 45◦ = 1

σ a a2 = 112 − 102 = 21 √ a = 21 √ √ 10 21 21 10 √ , or cos σ = ; tan σ = ; 11 21 21 √ 11 21 11 11 csc σ = ; sec σ = √ , or ; 10 21 21 √ 21 cot σ = 10 17.

2

1

√

1 2 cos 45◦ = √ , or 2 2 18. See the triangle in Exercise 19. √ 1 3 tan 30◦ = √ , or 3 3 19.

60ο

2

1

ο

30

3 sec 60◦ =

27. See the triangle in Exercise 19. csc 30◦ = 2 28. See the triangle in Exercise 19. √ 3 √ ◦ = 3 tan 60 = 1 29. We know the measure of an acute angle of a right triangle and the length of the side opposite the angle, and we want to find the length of the adjacent side. We can use the tangent ratio or the cotangent ratio. Here we use the cotangent: a cot 30◦ = 36 36 cot 30◦ = a √ 36 3 = a 62.4 m ≈ a

45ο 1

26. See the triangle in Exercise 19. √ 2 2 3 sec 30◦ = √ , or 3 3

2 =2 1

20. See the triangle in Exercise 17. √ 1 2 ◦ √ , or sin 45 = 2 2 21. See the triangle in Exercise 19. √ 1 3 cot 60◦ = √ , or 3 3 22. See the triangle in Exercise 17. √ 2 √ = 2 csc 45◦ = 1 23. See the triangle in Exercise 19. 1 sin 30◦ = 2

30. Since we know the measure of an acute angle of a right triangle and the lengths of the sides opposite and adjacent to the angle, we can use the sine, cosine, cosecant, or secant ratio to determine the length of the hypotenuse. We will use the cosecant function. hyp h = csc 45◦ = opp 90 90 csc 45◦ = h √ 90 2 = h 127.3 ≈ h

The distance from third base to first base is about 127.3 ft. 31. Using a calculator, enter 9◦ 43" . The result is 9◦ 43" ≈ 9.72◦ . 32. 52◦ 15" = 52.25◦ 33. Using a calculator, enter 35◦ 50"" . The result is 35◦ 50"" ≈ 35.01◦ . 34. 64◦ 53" ≈ 64.88◦ 35. Using a calculator, enter 3◦ 2" . The result is 3◦ 2" ≈ 3.03◦ . 36. 19◦ 47" 23"" ≈ 19.79◦ 37. Using a calculator, enter 49◦ 38" 46"" . The result is 49◦ 38" 46"" ≈ 49.65◦ . 38. 76◦ 11" 34"" ≈ 76.19◦ 39. Using a calculator, enter 0◦ 15" 5"" . The result is 15" 5"" ≈ 0.25◦ .

Exercise Set 5.1

333

40. Using a calculator, enter 68◦ 0" 2"" . The result is 68◦ 2"" ≈ 68.00◦ .

63. Use a calculator set in degree mode.

41. Using a calculator, enter 5 0 53 . The result is 5 53 ≈ 5.01◦ .

64. tan 63◦ 48" ≈ 2.0323

42. Using a calculator, enter 0◦ 44" 10"" . The result is 44" 10"" ≈ 0.74◦ .

tan 85.4◦ ≈ 12.4288

◦

"

""

◦

""

43. Enter 17.6◦ on a calculator and use the DMS feature: 17.6◦ = 17◦ 36" 44. 20.14◦ = 20◦ 8" 24"" 45. Enter 83.025◦ on a calculator and use the DMS feature: 83.025◦ = 83◦ 1" 30"" 46. 67.84◦ = 67◦ 50" 24"" 47. Enter 11.75◦ on a calculator and use the DMS feature: 11.75◦ = 11◦ 45" 48. 29.8◦ = 29◦ 48" 49. Enter 47.8268◦ on a calculator and use the DMS feature: 47.8268◦ ≈ 47◦ 49" 36"" 50. 0.253◦ ≈ 0◦ 15" 11"" 51. Enter 0.9◦ on a calculator and use the DMS feature: 0.9◦ = 0◦ 54" 52. 30.2505◦ ≈ 30◦ 15" 2"" 53. Enter 39.45◦ on a calculator and use the DMS feature: 39.45◦ = 39◦ 27" 54. 2.4◦ = 2◦ 24" 55. Use a calculator set in degree mode. cos 51◦ ≈ 0.6293 56. cot 17 ≈ 3.2709 ◦

57. Use a calculator set in degree mode. tan 4◦ 13" ≈ 0.0737 58. sin 26.1◦ ≈ 0.4399 59. Use a calculator set in degree mode. We find the reciprocal 1 or (cos 38.43)−1 . of cos 38.43◦ by entering cos 38.43 sec 38.43◦ ≈ 1.2765 60. cos 74◦ 10" 40"" ≈ 0.2727 61. Use a calculator set in degree mode. cos 40.35◦ ≈ 0.7621 62. csc 45.2◦ =

1 ≈ 1.4093 sin 45.2◦

sin 69◦ ≈ 0.9336 65. Use a calculator set in degree mode. 66. cos 4◦ ≈ 0.9976 67. Use a calculator set in degree mode. We find the reciprocal 1 or (sin 89.5)−1 . of sin 89.5◦ by entering sin 89.5 csc 89.5◦ ≈ 1.0000 68. sec 35.28◦ =

1 ≈ 1.2250 cos 35.28◦

69. Use a calculator set in degree mode. reciprocal of tan 30 25 6 ◦

"

""

(tan 30◦ 25" 6"" )−1 .

We find the 1 by entering or tan 30◦ 25" 6""

cot 30◦ 25" 6"" ≈ 1.7032 70. sin 59.2◦ ≈ 0.8590 71. On a graphing calculator, press 2nd ENTER .

SIN .5125

θ = 30.8◦ 72. tan θ = 2.032, so θ ≈ 63.8◦ . 73. tan θ = 0.2226 On a graphing calculator, press 2nd ENTER .

TAN .2226

θ = 12.5◦ 74. cos θ = 0.3842, so θ ≈ 67.4◦ . 75. sin θ = 0.9022 On a graphing calculator, press 2nd ENTER .

SIN .9022

θ = 64.4◦ 76. tan θ = 3.056, so θ ≈ 71.9◦ . 77. cos θ = 0.6879 On a graphing calculator, press 2nd ENTER .

COS .6879

θ = 46.5◦ 78. sin θ = 0.4005, so θ ≈ 23.6◦ . 1 = 2.127 tan θ 1 . Thus, tan θ = 2.127

79. cot θ =

On a graphing calculator, press 2nd ENTER . θ ≈ 25.2◦

TAN (1 ÷ 2.127)

334

Chapter 5: The Trigonometric Functions 1 = 1.147 sin θ 1 , so θ ≈ 60.7◦ . Thus, sin θ = 1.147

91. See the triangle in Exercise 85. √ √ 3 cot θ = 3, or , so θ = 30◦ . 1

80. csc θ =

92. See the triangle in Exercise 83. √ √ 2 sec θ = 2, or , so θ = 45◦ . 1

1 = 1.279 cos θ 1 . Thus, cos θ = 1.279

81. sec θ =

On a graphing calculator, press 2nd ENTER . θ ≈ 38.6◦

1 = 1.351 tan θ 1 Thus, tan θ = , so θ ≈ 36.5◦ . 1.351

82. cot θ =

83.

COS (1 ÷ 1.279)

93. The cosine and sine functions are cofunctions. The cosine and secant functions are reciprocals. 1 cos 20◦ = sin 70◦ = sec 20◦ 94. sin 64◦ = cos 26◦ =

95. The tangent and cotangent functions are cofunctions and reciprocals. 1 tan 52◦ = cot 38◦ = cot 52◦ 96. sec 13◦ = csc 77◦ =

2

45 sin θ =

1

1 csc 64◦

1 cos 13◦

97. Since 25◦ and 65◦ are complementary angles, we have sin 25◦ = cos 65◦ ≈ 0.4226, cos 25◦ = sin 65◦ ≈ 0.9063,

ο

tan 25◦ = cot 65◦ ≈ 0.4663,

1

√

csc 25◦ = sec 65◦ ≈ 2.3662,

2 , so θ = 45◦ . 2

84. See the triangle in Exercise 85. √ 3 , so θ = 60◦ . cot θ = 3 85.

60ο

2

sec 25◦ = csc 65◦ ≈ 1.1034,

cot 25◦ = tan 65◦ ≈ 2.1445. 98. Since 82◦ and 8◦ are complementary angles, we have sin 82◦ = cos 8◦ ≈ 0.9903, cos 82◦ = sin 8◦ ≈ 0.1392,

1

30ο 3 1 cos θ = , so θ = 60◦ . 2 86. See the triangle in Exercise 85. 1 sin θ = , so θ = 30◦ . 2 87. See the triangle in Exercise 83. tan θ = 1, so θ = 45◦ . 88. See the triangle in Exercise 85. √ 3 , so θ = 30◦ . cos θ = 2 89. See the triangle in Exercise 85. √ 2 2 3 , or √ , so θ = 60◦ . csc θ = 3 3 90. See the triangle in Exercise 85. √ √ 3 , so θ = 60◦ . tan θ = 3, or 1

tan 82◦ = cot 8◦ ≈ 7.1154, csc 82◦ = sec 8◦ ≈ 1.0098, sec 82◦ = csc 8◦ ≈ 7.1853,

cot 82◦ = tan 8◦ ≈ 0.1405. 99. Since 18◦ 49" 55"" and 71◦ 10" 5"" are complementary angles, we have sin 18◦ 49" 55"" = cos 71◦ 10" 5"" ≈ 0.3228, cos 18◦ 49" 55"" = sin 71◦ 10" 5"" ≈ 0.9465, 1 tan 18◦ 49" 55"" = cot 71◦ 10" 5"" = ≈ tan 71◦ 10" 5"" 1 ≈ 0.3411, 2.9321 1 csc 18◦ 49" 55"" = sec 71◦ 10" 5"" = ≈ cos 71◦ 10" 5"" 1 ≈ 3.0979, 0.3228 1 ≈ sec 18◦ 49" 55"" = csc 71◦ 10" 5"" = sin 71◦ 10" 5"" 1 ≈ 1.0565, 0.9465 cot 18◦ 49" 55"" = tan 71◦ 10" 5"" ≈ 2.9321.

Exercise Set 5.1

335

100. Since 38.7◦ and 51.3◦ are complementary angles, we have sin 51.3

◦

107.

= cos 38.7 ≈ 0.7804, ◦

108.

= cos 82 = q, ◦

cos 8 = sin 82◦ = p, ◦

tan 8 = ◦

csc 8 = ◦

sec 8◦ = cot 8 = ◦

109.

111.

and g are cofunctions, then f (θ) = g(90◦ − θ). f (x)

−4 0.1353 −2 0.3679

106.

2.7183 7.3891

f (x) ! e x/2

"3 "2 "1 "1

5 4 3 2

"3 "2 "1 "1

1.6094

1

"2 "3

2

3

4

5

x

h(x) ! ln x

y

1

2

3

4

5

x

g(x) ! log 2 x

5x = 625 Equating exponents

1

2

3

x

et = 10, 000 t = ln 10, 000 ≈ 9.21 log7 x = 3 x = 73 = 343 The solution is 343. log(3x + 1) − log(x − 1) 3x + 1 log x−1 3x + 1 x−1 3x + 1 x−1 3x + 1

=2 =2 = 102 = 100 = 100x − 100

101 = 97x 101 =x 97

y 5 4 3 2

5

"1 "1

ln et = ln 10, 000

112.

y

4

1.3863

x=4

1 . If f g(θ)

2

0.6932

4

The solution is 4.

104. If f and g are reciprocal functions, then f (θ) =

1

2

5x = 54

110.

0

0

"2 "3

103. Since 30◦ and 60◦ are complementary angles, once the function values for one angle are memorized the cofunction identities can be used to find the function values for the other angle.

x

1

"1 "1

1 1 cot 82 = = , tan 82◦ r 1 1 ◦ sec 82 = = , cos 82◦ q 1 1 csc 82◦ = = , sin 82◦ p ◦ tan 82 = r. ◦

102. Left to the student.

105.

3 2 1

3 2 1

101. Since 82◦ and 8◦ are complementary angles, we have sin 8

y

h(x)

0.5 −0.6931

cos 51.3◦ = sin 38.7◦ ≈ 0.6252, 1 ≈ 1.2481, tan 51.3◦ = cot 38.7◦ ≈ 0.8012 1 ≈ 1.2814, csc 51.3◦ = sec 38.7◦ ≈ 0.7804 1 sec 51.3◦ = csc 38.7◦ ≈ ≈ 1.5995, 0.6252 ◦ ◦ cot 51.3 = tan 38.7 ≈ 0.8012. ◦

x

f(x) !

1

2

3

2"x

x

113. Since 49.2◦ and 40.8◦ are complementary angles, we have 1 1 sin 40.8◦ = cos 49.2◦ = = ≈ 0.6534. sec 49.2◦ 1.5304

336

Chapter 5: The Trigonometric Functions

114. First find the length of the third side, a.

Exercise Set 5.2

a2 + b2 = c2 ! "2 1 a2 + = q2 q 1 a2 + 2 = q 2 q

1.

F

6

1 a =q − 2 q 2

30°

q4 − 1 a2 = q2 # q4 − 1 a= q sin α =

cos α =

tan α =

csc α =

1 1 q = 2 q q # q4 − 1 # q4 − 1 q = q q2 1 1 q # =# q4 − 1 q4 − 1 q q = q2 1 q

sec α = #

q

d

2

=#

1 116. Let h = the height of the triangle. Then Area = bh where 2 1 h sin θ = , or h = a sin θ, so Area = ab sin θ. a 2

E

To solve this triangle find F , d, and f . F = 90◦ − 30◦ = 60◦ d = sin 30◦ 6 d = 6 sin 30◦ d=3

2.

f = cos 30◦ 6 f = 6 cos 30◦ √ f = 3 3 ≈ 5.2

A = 90◦ − 45◦ = 45◦

a = cos 45◦ 10 √ a = 5 2 ≈ 7.1

q2

q4 − 1 q4 − 1 q # q4 − 1 # q = q4 − 1 cot α = 1 q 1 115. Area = ×base×height 2 1 = ab 2 1 = c sin A · b 2 ! " a sin A = , so c sin A = a c 1 = bc sin A 2

f

D

3.

b = sin 45◦ 10 √ b = 5 2 ≈ 7.1

126

A

c

C 67.3°

a B

To solve this triangle, find A, a, and c. A = 90◦ − 67.3◦ = 22.7◦ a cot 67.3◦ = 126 a = 126 cot 67.3◦ a ≈ 52.7 c 126 c = 126 csc 67.3◦

csc 67.3◦ =

c ≈ 136.6

4. T = 90◦ − 26.7◦ = 63.3◦ s csc 26.7◦ = 0.17 s ≈ 0.38 t 0.17 t ≈ 0.34

cot 26.7◦ =

Exercise Set 5.2

337

5.

9.

P

n

M

450

23.2

42° 22' p

N

A

To solve this triangle, find P , p, and n. ◦

"

◦

"

p = cot 42◦ 22" 23.2 p = 23.2 cot 42◦ 22" 6. H = 90◦ − 28◦ 34" = 61◦ 26" f csc 28◦ 34" = 17.3 f ≈ 36.2 h 17.3 h ≈ 31.8

10. A = 90◦ − 56.5◦ = 33.5◦ a cos 56.5◦ = 0.0447 a ≈ 0.0247

11.

B

B

48.3

c

A

9.73

87° 43'

C

To solve this triangle, find B, b, and c. B = 90◦ − 87◦ 43" = 2◦ 17" b cot 87◦ 43" = 9.73 b = 9.73 cot 87◦ 43" b ≈ 0.39 c 9.73 c = 9.73 csc 87◦ 43"

csc 87◦ 43" =

8.

b 0.0447 b ≈ 0.0373

sin 56.5◦ =

cot 28◦ 34" =

7.

C

B = 90◦ − A ≈ 90◦ − 77.2◦ ≈ 12.8◦ a sin A = 450 a sin 77.2◦ ≈ 450 a ≈ 450 sin 77.2◦ ≈ 439

n ≈ 34.4

p ≈ 25.4

100

a

To solve this triangle, find A, B, and a. 100 cos A = 450 A ≈ 77.2◦

P = 90 − 42 22 = 47 38 n = csc 42◦ 22" 23.2 n = 23.2 csc 42◦ 22" ◦

B

c ≈ 9.74 12.5 tan A = 18.3 A ≈ 34.3◦

B ≈ 90◦ − 34.3◦ ≈ 55.7◦ c csc 34.3◦ = 12.5 c ≈ 22.2

A

47.58° b

a

C

To solve this triangle, find B, a, and b. B = 90◦ − 47.58◦ = 42.42◦ a = sin 47.58◦ 48.3 a = 48.3 sin 47.58◦ a ≈ 35.7 b = cos 47.58◦ 48.3 b = 48.3 cos 47.58◦ b ≈ 32.6 12. A = 90◦ − 20.6◦ = 69.4◦ b cot 69.4◦ = 7.5 b ≈ 2.8 c 7.5 c ≈ 8.0

csc 69.4◦ =

338

Chapter 5: The Trigonometric Functions

13.

B c

a

35°

A

C

40

To solve this triangle find B, a, and c. B = 90◦ − A = 90◦ − 35◦ = 55◦ a tan A = 40 a tan 35◦ = 40 a = 40 tan 35◦ a ≈ 28.0

c 40 c sec 35◦ = 40 c = 40 sec 35◦ sec A =

c ≈ 48.8

14. A = 90◦ − 69.3◦ = 20.7◦ a cot 69.3◦ = 93.4 a ≈ 35.3 c 93.4 c ≈ 99.8

csc 69.3◦ = 15.

B

4.02

A

1.86

a

C

To solve this triangle find a, A, and B. 1.86 cos A = 4.02 A ≈ 62.4◦ B ≈ 90◦ − 62.4◦ ≈ 27.6◦ a sin A = 4.02 a ◦ sin 62.4 ≈ 4.02 a ≈ 4.02 sin 62.4◦ a ≈ 3.56

16.

1 10.2 = 20.4 2 A = 30◦ B = 90◦ − 30◦ = 60◦ b cos 30◦ = 20.4 b ≈ 17.7 sin A =

17. From geometry we know that when parallel lines are cut by a transversal, alternate interior angles are equal. Thus the angle of depression from the plane to the house and the angle of elevation from the house to the plane have the same measure. We can use the right triangle in the drawing in the text. We use the sine ratio to find θ. 475 sin θ = ≈ 0.5588 850 θ ≈ 34◦ 18. Since the case is an isosceles right triangle, each acute angle measures 45◦ and a line from the 90◦ angle dropped 1 perpendicular to the 21 -in. base bisects the base. 2 10.75 21.5/2 ◦ = cos 45 = x x 10.75 x= cos 45 x ≈ 15.2 in. 19. First we find the distance from point B to the raft. We know the length of the side opposite the 50◦ angle and want to find the length of the hypotenuse, c. We use the sine ratio. 40 ft sin 50◦ = c 40 ft c= ≈ 52.2 ft sin 50◦ Allowing 5 ft of rope at each end, Bryan needs about 52.2 ft + 2 · 5 ft, or about 62.2 ft of rope. 20. Let s = the length of the new side. 14.5 cos 53◦ = s 14.5 ≈ 24.1 ft s= cos 53◦ 21. We know the length of the side adjacent to the 70◦ angle and want to find the length of the opposite side. We use the tangent ratio. Let t = the height of the tree. t tan 70◦ = 40 ft t = 40 ft · tan 70◦ ≈ 110 ft The tree is about 110 ft tall.

22. We know the length of the side adjacent to the 23◦ angle and want to find the length of the opposite side. We use the tangent ratio. Let d = the distance the front legs of the easel should be from the wall. d tan 23◦ = 6 ft d = 6 ft · tan 23◦ ≈ 2.5 ft 23. Referring to the drawing in the text, consider the right triangle containing the 50◦ angle and let d = the length of the leg adjacent to the 50◦ angle. We first find d, using the tangent function with 50◦ and 30◦ . x x and tan 30◦ = tan 50◦ = d d + 670 d tan 50◦ = x and (d + 670) tan 30◦ = x

Exercise Set 5.2

339 the hypotenuse of ∆ABC and want to find the length of the side opposite the 36◦ angle. We use the sine ratio. b sin 36◦ = 15.8 cm b = 15.8 cm · sin 36◦ ≈ 9.29 cm

Substitute d tan 50◦ for x in the second equation and solve for d. (d + 670) tan 30◦ = d tan 50◦ d tan 30◦ + 670 tan 30◦ = d tan 50◦ 670 tan 30◦ = d tan 50◦ − d tan 30◦

670 tan 30◦ = d (tan 50◦ − tan 30◦ ) 670 tan 30◦ =d tan 50◦ − tan 30◦ 629.6 ≈ d

Then the length of each side of the pentagon is about 2(9.29 cm), or 18.58 cm, and the perimeter of the pentagon is about 5(18.58 cm), or 92.9 cm. 26.

Then we find x using the tangent function. x tan 50◦ ≈ 629.6 750 ≈ x

h

The height of the tower is about 750 ft.

78.2°

17.6° 10 mi

d

24.

First find d. Then solve for h. h h = tan 78.2◦ and = tan 17.6◦ d d + 10 h = d tan 78.2◦ and h = (d + 10) tan 17.6◦

d

20o 800 ft

35o x

d tan 78.2◦ = (d + 10) tan 17.6◦

We first find the distance x, using the tangent function with 35◦ and 20◦ . d d tan 35◦ = and tan 20◦ = x x + 800 d = x tan 35◦ and d = (x+800) tan 20◦

d tan 78.2◦ = d tan 17.6◦ +10 tan 17.6◦ d(tan 78.2◦ −tan 17.6◦ ) = 10 tan 17.6◦ 10 tan 17.6◦ d= tan 78.2◦ − tan 17.6◦ d ≈ 0.7097

Substitute x tan 35◦ for d in the second equation and solve for x. x tan 35◦ = (x + 800) tan 20◦ x tan 35◦ = x tan 20◦ + 800 tan 20◦ x(tan 35◦ − tan 20◦ ) = 800 tan 20◦ 800 tan 20◦ x= tan 35◦ − tan 20◦ x ≈ 866

h 0.7097 h ≈ 0.7097(tan 78.2◦ ) ≈ 3.4 mi

tan 78.2◦ ≈ 27.

670 ft

Then we find d using the tangent function. d tan 35◦ ≈ 866 d ≈ 866 tan 35◦ ≈ 606

h

63.4°

We know the length of the hypotenuse and want to find the length of the side opposite the 63.4◦ angle. We use the sine ratio. h sin 63.4◦ = 670 ft h ≈ 670 ft · sin 63.4◦ ≈ 599 ft

The height of the sand dune is about 606 ft. 25.

The kite was about 599 ft high.

B

28.

15.8 cm

x

b A

C

D

1 The measure of $ ABD is · 360◦ , or 72◦ . Then the mea5 1 sure of $ ABC is · 72◦ , or 36◦ . We know the length of 2

18.7° 100 ft 6.5° y

x 100 ft x = 100 ft · tan 18.7◦ ≈ 33.8 ft

tan 18.7◦ =

340

Chapter 5: The Trigonometric Functions y 100 ft y = 100 ft · tan 6.5◦ ≈ 11.4 ft

tan 6.5◦ =

31. We make a drawing. r

The height of the building is x + y:

θ

x + y ≈ 33.8 ft + 11.4 ft ≈ 45 ft

x

29. 7 in. r A

We use the tangent function to find the angle θ that corresponds to a pitch of 5/12. 5 tan θ = ≈ 0.4167 12 θ ≈ 22.6◦

C B

Since the width of the building is 33 ft, we have 1 x = · 33 ft = 16.5 ft. We use the cosine function to find 2 r, the length of the rafters. 16.5 ft cos 22.6◦ = r r cos 22.6◦ = 16.5 ft 16.5 ft r= cos 22.6◦ r ≈ 17.9ft

D

1 The measure of " ADC is · 360◦ , or 45◦ . Then the mea8 1 sure of " ABC is · 45◦ , or 22.5◦ . The length of the 2 1 segment CD is 7 in., so the length of BC is (7 in.), or 2 3.5 in. In " ABC, we know the length of the side opposite the 22.5◦ angle and we want to find the length of the hypotenuse. We use the sine ratio. 3.5 in. sin 22.5◦ = r 3.5 in. r= ≈ 9.15 in. sin 22.5◦ The radius of the circumscribed circle is about 9.15 in.

The rafters are about 17.9 ft long. 32.

35 ft

The length of the quilt consists of the lengths of 8 radii (2 radii in each of 4 circles), and the width consists of the lengths of 6 radii (2 radii in each of 3 circles). We have:

θ 20 ft

Length = 8(9.15) = 73.2 in.;

35 ft 20 ft θ ≈ 60.3◦

tan θ =

Width = 6(9.15) = 54.9 in. 30.

33 ft

33.

13.5ο

r θ x

81.2ο 46 ft

11 ≈ 0.9167 12 θ ≈ 42.5◦

tan θ =

1 Since the width of the house is 46 ft, we have x = · 46 ft, 2 or 23 ft. We use the cosine function to find r, the length of the rafters. 23 ft cos 42.5◦ = r 23 ft ≈ 31.2 ft r= cos 42.5◦

θ2 d2

θ1 d d1

The distance to be found is d, which is d1 − d2 .

From geometry we know that θ1 = 13.5◦ and θ2 = 81.2◦ . d2 d1 = cot θ1 = cot θ2 and 2 2 ◦ d1 = 2 cot 13.5 and d2 = 2 cot 81.2◦ d = d1 − d2 = 2 cot 13.5◦ − 2 cot 81.2◦ ≈ 8

The towns are about 8 km apart.

Exercise Set 5.2

341 37.

34.

11.3ο 55 ft

12 km 63° 20'

θ

B

b From geometry we know that θ = 11.3◦ . We know the length of the side opposite θ and want to find the length of the side adjacent to θ. We use the tangent ratio. 55 ft tan 11.3◦ = b 55 ft b= ≈ 275 ft tan 11.3◦ The boat is about 275 ft from the foot of the lighthouse. 35.

d

A

We know the length of the side adjacent to the 63◦ 20" angle and want to find the length of the opposite side. We use the tangent ratio. d tan 63◦ 20" = 12 km d = 12 km · tan 63◦ 20" ≈ 24 km

The lobster boat is about 24 km from the lighthouse. 38. 1) Use the tangent ratio for 14◦ . 2) Use the cotangent ratio for 14◦ .

15 mi

3) Find the measure of the other acute angle, 76◦ , and then use the tangent ratio for 76◦ . 4) Find the measure of the other acute angle, 76◦ , and then use the cotangent ratio for 76◦ .

37.6ο

5) Find the length of the hypotenuse. (This could be done using the cosine or secant ratio for 14◦ or the sine or cosecant ratio for 76◦ .) Then use the Pythagorean theorem to find c.

θ

S

From geometry we know that θ = 37.6◦ . We know the length of the side opposite θ and want to find the length of the side adjacent to θ. We use the tangent ratio. 15 mi tan 37.6◦ = d 15 mi ≈ 19.5 mi d= tan 37.6◦ 36.

l 10.5ο

Many would probably agree that method (1) is most efficient. 39. Sine: [0, 1); cosine: [0, 1); tangent: [0, ∞) # √ √ 40. d = (−9 − 0)2 + (3 − 0)2 = 90 = 3 10 ≈ 9.487 # 41. d = (x1 − x2 )2 + (y1 − y2 )2 # d = [8 − (−6)]2 + [−2 − (−4)]2 √ = 142 + 22 √ √ = 200 = 10 2 ≈ 14.142

50 ft φ

42. loge t = 4, or ln t = 4

θ

43. log 0.001 = −3 is equivalent to log10 0.001 = −3. Remember that the base remains the same, and the logarithm is the exponent. We have 10−3 = 0.001.

75 ft 50 75 θ ≈ 33.7◦

tan θ =

44.

φ = θ + 10.5 ≈ 33.7 + 10.5 ≈ 44.2 50 + l tan 44.2◦ ≈ 75 75 tan 44.2◦ ≈ 50 + l ◦

◦

◦

◦

5 36°

a 2

72.9 ≈ 50 + l 23 ≈ l

The length of the antenna is about 23 ft.

a sin 36 = 2 5 ◦

a 10 a = 10 sin 36◦ ≈ 5.9

sin 36◦ =

342

Chapter 5: The Trigonometric Functions

45.

48. 14, 162 ft ≈ 2.682197 mi R sin 87◦ 53" = gives R + 2.682197 ◦ " 2.682197 sin 87 53 ≈ 3298 mi R= 1 − sin 87◦ 53"

C b

h

a

36°

A

B

7

49.

A

b cos 36 = 7 b = 7 cos 36◦ ≈ 5.66

a

P

◦

c

h sin 36◦ = b

ο

V

h 5.66 h ≈ 5.66 sin 36◦ ≈ 3.3

sin 36◦ ≈ 46.

A is the actual location of the plane when heard. Plane’s speed, 200 mph ≈ 293 ft/sec. c b cot 20◦ = csc 20◦ = 3000 3000 c = 3000 csc 20◦ b = 3000 cot 20◦

V

c ≈ 8771

x

d 2 b) tan 27 = 3.93 cm d tan 27◦ = 2(3.93 cm)

8771 = 1100t 8≈t

Then the distance from P to A is given by a = 293 · 8 = 2344 ft. Then d = b − a = 8242 − 2344 = 5898 ft. 3000 3000 = tan θ = d 5898 θ ≈ 27◦

◦

Exercise Set 5.3

d = 2(3.93 cm) tan 27 ≈ 4.00 cm ◦

d ◦ c) tan 27 = 2 VP d tan 27◦ = 2V P d = 2V P tan 27◦

1.

(See part (c).)

y 187° x

The terminal side lies in quadrant III.

d = 1.02V P

47.

b ≈ 8242

The time it takes the plane to fly from P to A is approximately the same time it takes the sound to travel from P to V . We use the formula distance = rate × time to find this time. d = rt

d Let x = V P and r = . 2 2 cm a) r = = 1 cm 2 1 cm tan 27◦ = x 1 cm ≈ 1.96 cm x= tan 27◦

d = 1.02V P d VP = 1.02 V P = 0.98d 8 tan θ = 1.5 θ ≈ 79.38◦

a

P is the perceived location of the plane;

P

d)

θ d b

27° r

20

3000

2. IV 3.

y 245° 15' x

The rafters should be cut so that θ = 79.38◦ . The terminal side lies in quadrant III.

Exercise Set 5.3

343

4. III

15. We add and subtract multiples of 360◦ . Many answers are possible.

5.

y

115.3◦ + 360◦ = 475.3◦ ; 115.3◦ + 2(360◦ ) = 835.3◦ ;

800°

115.3◦ − 360◦ = −244.7◦ ;

x

115.3◦ − 2(360◦ ) = −604.7◦ 16. Answers may vary. 275◦ 10" + 360◦ = 635◦ 10" ;

The terminal side lies in quadrant I.

275◦ 10" + 2(360◦ ) = 995◦ 10" ;

6. IV

275◦ 10" − 360◦ = −84◦ 50" ;

7.

275◦ 10" − 2(360◦ ) = −444◦ 50"

y –460.5°

17. We add and subtract multiples of 360◦ . Many answers are possible.

x

−180◦ + 360◦ = 180◦ ;

−180◦ + 2(360◦ ) = 540◦ ;

−180◦ − 360◦ = −540◦ ;

The terminal side lies in quadrant III. 8. IV

−180◦ − 2(360◦ ) = −900◦

18. Answers may vary.

9.

−310◦ + 360◦ = 50◦ ;

y

−310◦ + 2(360◦ ) = 410◦ ;

–912°

−310◦ − 360◦ = −670◦ ;

x

−310◦ − 2(360◦ ) = −1030◦

19. 90◦ − 17.11◦ = 72.89◦

180◦ − 17.11◦ = 162.89◦

The complement of 17.11◦ is 72.89◦ and the supplement is 162.89◦ .

The terminal side lies in quadrant II. 10. I 20.

11.

Complement: 90◦ =

y 537°

Supplement: 180◦ = 179◦ 60" −47◦ 38" 132◦ 22"

x

21. The terminal side lies in quadrant II. 12. I 13. We add and subtract multiples of 360 . Many answers are possible. ◦

74◦ + 360◦ = 434◦ ; 74◦ + 2(360◦ ) = 794◦ ; 74◦ − 360◦ = −286◦ ;

74◦ − 2(360◦ ) = −646◦

14. Answers may vary.

−81 + 360 = 279 ; ◦

◦

◦

−81◦ + 2(360◦ ) = 639◦ ; −81 − 360 = −441 ; ◦

◦

89◦ 60" −47◦ 38" 42◦ 22"

◦

−81◦ − 2(360◦ ) = −801◦

90◦ =

89◦ 59" 60"" −12◦ 3" 14"" 77◦ 56" 46""

180◦ = 179◦ 59" 60"" −12◦ 3" 14"" 167◦ 56" 46"" The complement of 12◦ 3" 14"" is 77◦ 56" 46"" and the supplement is 167◦ 56" 46"" . 22. Complement: 90◦ − 9.038◦ = 80.962◦

Supplement: 180◦ − 9.038◦ = 170.962◦

23. 90◦ − 45.2◦ = 44.8◦

180◦ − 45.2◦ = 134.8◦

The complement of 45.2◦ is 44.8◦ and the supplement is 134.8◦ .

344

Chapter 5: The Trigonometric Functions

24. Complement: 90◦ − 67.31◦ = 22.69◦

28. r =

Supplement: 180 − 67.31 = 112.69 ◦

◦

◦

25. We first determine r. # r = x2 + y 2 # r = (−12)2 + 52 √ √ = 144 + 25 = 169 = 13

Substituting −12 for x, 5 for y, and 13 for r, the trigonometric function values of θ are 5 y sin β = = r 13 −12 12 x =− cos β = = r 13 13 y 5 5 tan β = = =− x −12 12 r 13 csc β = = y 5 13 13 r sec β = = =− x −12 12 −12 12 x =− cot β = = y 5 5 $√ √ 26. r = ( 7)2 + (−3)2 = 16 = 4 3 sin θ = − 4 √ 7 cos θ = 4 √ 3 3 7 tan θ = − √ , or − 7 7 4 csc θ = − 3 √ 4 4 7 sec θ = √ , or 7 7 √ 7 cot θ = − 3 27. We first determine r. # r = x2 + y 2 $ √ √ r = (−2 3)2 + (−4)2 = 4 · 3 + 16 √ √ √ = 12 + 16 = 28 = 2 7 √ √ Substituting −2 3 for x, −4 for y and 2 7 for r, the trigonometric function values are √ −4 2 2 7 y sin φ = = √ = − √ , or − r 7 2 7 7 √ √ √ x −2 3 3 21 cos φ = = √ = − √ , or − r 7 2 7 7 √ 2 2 3 y −4 √ = √ , or tan φ = = x 3 −2 3 3 √ √ 2 7 7 r =− csc φ = = y −4 2 √ √ √ 2 7 r 7 21 √ = − √ , or − sec φ = = x 3 −2 3 3 √ √ −2 3 3 x = cot φ = = y −4 2

√

92 + 12 =

√

82 √ 1 82 sin α = √ , or 82 82 √ 9 9 82 cos α = √ , or 82 82 1 tan α = 9 √ 82 √ = 82 csc α = 1 √ 82 sec α = 9 9 cot α = = 9 1

29. First we draw the graph of 2x + 3y = 0 and determine a quadrant IV solution of the equation. We let x = 3 and find the corresponding y-value. 2x + 3y = 0 2 · 3 + 3y = 0

Substituting

3y = −6 y = −2

Thus, (3, −2) is a point on the terminal side of the angle θ. y 2x + 3y = 0 3

θ

x –2

r (3, –2)

Using (3, −2), we determine r: # √ r = 32 + (−2)2 = 13

Then using x = 3, y = −2, and r = √ 2 13 −2 √ , sin θ = , or − 13 13 √ 3 3 13 , cos θ = √ , or 13 13 −2 2 =− . tan θ = 3 3

√

13, we find

30. A quadrant II solution of 4x + y = 0 is (−1, 4). Then # √ r = (−1)2 + 42 = 17. √ 4 17 4 sin θ = √ , or 17 17 √ 17 −1 cos θ = √ , or − 17 17 4 = −4 tan θ = −1

Exercise Set 5.3

345

31. First we draw the graph of 5x − 4y = 0 and determine a quadrant I solution of the equation. We let x = 4 and find the corresponding y-value. 5x − 4y = 0

5 · 4 − 4y = 0

Substituting

−4y = −20 y=5

3 = −3 −1 √ 3 3 2 3 √ = − √ , or − sec θ = 4 −2 2 2 2 √ √ −2 2 =2 2 cot θ = −1 csc θ =

34. tan β = 5 =

Thus, (4, 5) is a point on the terminal side of the angle θ.

5 1

y y (1, 5)

(4, 5)

26 5

5

θ

x

β

4 5x – 4y = 0

√ √ 5 26 1 26 5 ; cos β = √ , or ; sin β = √ , or 26 26 26 26 √ √ 1 26 26 √ ; sec β = = 26; cot β = csc β = 5 1 5

Using (4, 5), we determine r: √ √ r = 42 + 52 = 41 Then using x = 4, y = 5, and r = √ 5 5 41 sin θ = √ , or , 41 41 √ 4 41 4 , cos θ = √ , or 41 41 5 tan θ = . 4

√

41, we find

32. A quadrant III solution of y = 0.8x is (−10, −8). Then # √ √ r = (−10)2 + (−8)2 = 164 = 2 41. √ −8 4 41 sin θ = √ , or − 41 2 41 √ 5 41 −10 cos θ = √ , or − 41 2 41 −8 4 tan θ = = −10 5

33. First we sketch a third-quadrant angle and a reference tri−1 1 the length of the vertical angle. Since sin θ = − = 3 3 leg is 1 and the length of the is 3. The other √ hypotenuse √ leg must then have length 8, or 2 2.

35. Since θ is in quadrant IV, we have cot θ = −2 = y θ

–1 (2 , –1)

√ 5 −1 sin θ = √ , or − 5 5 √ 2 2 5 cos θ = √ , or 5 5 1 −1 tan θ = =− 2 2 √ √ 5 =− 5 csc θ = −1 √ 5 sec θ = 2 36. cos α = −

−4 4 = 5 5 y

(-4, 3) θ

–1

x

2 5

y

–2 2

x

1

x

x

3 3 5 5 , tan α = − , csc α = , sec α = − , 5 4 3 4 4 cot α = − 3 sin α =

Now we can read off the appropriate ratios: √ √ 2 2 −2 2 , or − cos θ = 3 3 √ 1 −1 2 √ = √ , or tan θ = 4 −2 2 2 2

θ -4

3

(–2 2 , –1)

5

3

2 . −1

346

Chapter 5: The Trigonometric Functions

37. cos φ =

3 5

43. Since 7560◦ = 21 · 360◦ , or 0◦ + 21 · 360◦ , then 7560◦ and 0◦ are coterminal. A point on the terminal side of 0◦ is (1, 0). Thus, 0 sin 7560◦ = sin 0◦ = = 0. 1

y θ

x

3 5

44. A point on the terminal side of 270◦ is (0, −1). Thus, −1 , which is not defined. tan 270◦ = 0

-4 (3, -4)

45. 495◦ −360◦ = 135◦ , so 495◦ and 135◦ are coterminal. Since 180◦ − 135◦ = 45◦ , the reference angle is 45◦ . Note that 495◦ is a second-quadrant√angle, so the cosine is negative. Recalling that cos 45◦ = 2/2 we have √ 2 cos 495◦ = − . 2

4 5 4 tan φ = − 3 5 csc φ = − 4 5 sec φ = 3 3 cot φ = − 4 sin φ = −

38. sin θ = −

46. 675◦ − 360◦ = 315◦ , so 675◦ and 315◦ are coterminal. The reference angle is 45◦ . Note that 675◦ is a fourthquadrant angle, so the tangent is negative. Recalling that tan 45◦ = 1, we have tan 675◦ = −1.

5 13

47. −210◦ + 360◦ = 150◦ , so −210◦ and 150◦ are coterminal. Since 180◦ − 150◦ = 30◦ , the reference angle is 30◦ . Note that −210◦ is a second-quadrant angle, so the cosecant is positive. Recalling that csc 30◦ = 2, we have

y

–12

θ

x

–5

13 (–12, –5)

12 −5 5 −12 = − , tan θ = = , 13 13 −12 12 13 13 13 13 csc θ = = − , sec θ = =− , −5 5 −12 12 12 −12 = cot θ = −5 5 cos θ =

39. Since 180◦ − 150◦ = 30◦ , the reference angle is 30◦ . Note so the cosine is negthat 150◦ is a second-quadrant angle, √ ative. Recalling that cos 30◦ = 3/2, we have √ 3 . cos 150◦ = − 2 40. −225◦ + 360◦ = 135◦ , so −225◦ and 135◦ are coterminal. The reference angle is then 45◦ . Note that −225◦ is a second-quadrant √ angle, so the secant is negative. Recalling that sec 45◦ = 2, we have √ sec(−225◦ ) = − 2. 41. −135◦ + 360◦ = 225◦ , so −135◦ and 225◦ are coterminal. Since 180◦ + 45◦ = 225◦ , the reference angle is 45◦ . Note that −135◦ is a third-quadrant angle, so the tangent is positive. Recalling that tan 45◦ = 1, we have tan(−135◦ ) = 1. 42. The reference angle is 45◦ . Note that −45◦ is a fourthquadrant √ angle, so the sine is negative. Recalling that sin 45◦ = 2/2, we have √ 2 . sin(−45◦ ) = − 2

csc(−210◦ ) = 2. 48. 360◦ − 300◦ = 60◦ , so the reference angle is 60◦ . Note that 300◦ is a fourth-quadrant √ angle, so the sine is negative. Recalling that sin 60◦ = 3/2, we have √ 3 ◦ sin 300 = − . 2 49. 570◦ −360◦ = 210◦ , so 570◦ and 210◦ are coterminal. Since 180◦ + 30◦ = 210◦ , the reference angle is 30◦ . Note that 570◦ is a third-quadrant angle, so the cotangent is positive. √ Recalling that cot 30◦ = 3, we have √ cot 570◦ = 3. 50. −120◦ + 360◦ = 240◦ , so −120◦ and 240◦ are coterminal. Since 180◦ + 60◦ = 240◦ , the reference angle is 60◦ . Note that −120◦ is a third-quadrant angle, so the cosine is negative. Recalling that cos 60◦ = 1/2, we have 1 cos(−120◦ ) = − . 2 51. 360◦ − 330◦ = 30◦ , so the reference angle is 30◦ . Note that angle, so the tangent is negative. 330◦ is a fourth-quadrant√ Recalling that tan 30◦ = 3/3, we have √ 3 tan 330◦ = − . 3 52. 855◦ − 2 · 360◦ = 135◦ , so 855◦ and 135◦ are coterminal. Since 180◦ − 135◦ = 45◦ , the reference angle is 45◦ . Note that 855◦ is a second-quadrant angle, so the cotangent is negative. Recalling that cot 45◦ = 1, we have cot 855◦ = −1. 53. A point on the terminal side of −90◦ is (0, −1). Thus, 1 sec(−90◦ ) = , which is not defined. 0

Exercise Set 5.3

347

54. A point on the terminal side of 90◦ is (0, 1). Thus, 1 sin 90◦ = = 1. 1

67. A point on the terminal side of 90◦ is (0, 1). Thus, 0 cos 90◦ = = 0. 1

55. A point on the terminal side of −180◦ is (−1, 0). Thus, −1 cos(−180◦ ) = = −1. 1

68. −135◦ + 360◦ = 225◦ , so −135◦ and 225◦ are coterminal. Since 180◦ + 45◦ = 225◦ , the reference angle is 45◦ . Note that −135◦ is a third-quadrant √ angle, so the sine is negative. Recalling that sin 45◦ = 2/2, we have √ 2 ◦ sin(−135 ) = − . 2

56. A point on the terminal side of 90◦ is (0, 1). Thus, 1 csc 90◦ = = 1. 1 57. 240◦ − 180◦ = 60◦ , so the reference angle is 60◦ . Note that angle, so the tangent is positive. 240◦ is a third-quadrant √ Recalling that tan 60◦ = 3, we have √ tan 240◦ = 3. 58. A point on the terminal side of −180◦ is (−1, 0). Thus, −1 , which is not defined. cot(−180◦ ) = 0 59. 495◦ −360◦ = 135◦ , so 495◦ and 135◦ are coterminal. Since 180◦ − 135◦ = 45◦ , the reference angle is 45◦ . Note that 495◦ is a second-quadrant √ angle, so the sine is positive. Recalling that sin 45◦ = 2/2 we have √ 2 . sin 495◦ = 2 60. 1050◦ − 2 · 360◦ = 330◦ , so 1050◦ and 330◦ are coterminal. Since 360◦ − 330◦ = 30◦ , the reference angle is 30◦ . Note that 1050◦ is a fourth-quadrant angle, so the sine is negative. Recalling that sin 30◦ = 1/2, we have 1 sin 1050◦ = − . 2 61. 225 − 180 = 45 , so the reference angle is 45 . Note that so the cosecant is negative. 225◦ is a third-quadrant angle, √ Recalling that csc 45◦ = 2, we have √ csc 225◦ = − 2. ◦

◦

◦

◦

62. −450◦ + 360◦ = −90◦ ; a point on the terminal side of −90◦ is (0, −1). Thus, −1 sin(−450◦ ) = = −1. 1 63. A point on the terminal side of 0 is (1, 0). Thus, 1 cos 0◦ = = 1. 1 ◦

64. 480 −360 = 120 , so 480 and 120 are coterminal. Since 180◦ − 120◦ = 60◦ , the reference angle is 60◦ . Note that 480◦ is a second-quadrant√angle, so the tangent is negative. Recalling that tan 60◦ = 3, we have √ tan 480◦ = − 3. ◦

◦

◦

◦

◦

65. A point on the terminal side of −90 is (0, −1). Thus, 0 cot(−90◦ ) = = 0. −1 ◦

66. 360◦ − 315◦ = 45◦ , so the reference angle is 45◦ . Note that 315◦ is a fourth-quadrant√angle, so the secant is positive. Recalling that sec 45◦ = 2, we have √ sec 315◦ = 2.

69. A point on the terminal side of 270◦ is (0, −1). Thus, 0 cos 270◦ = = 0. 1 70. A point on the terminal side of 0◦ is (1, 0). Thus, 0 tan 0◦ = = 0. 1 71. 319◦ is a fourth-quadrant angle, so the cosine and secant function values are positive and the sine, cosecant, tangent, and cotangent are negative. 72. −57◦ is a fourth-quadrant angle, so the cosine and secant function values are positive and the sine, cosecant, tangent, and cotangent are negative. 73. 194◦ is a third-quadrant angle, so the tangent and cotangent function values are positive and the sine, cosecant, cosine, and secant are negative. 74. −620◦ is a second-quadrant angle, so the sine and cosecant function values are positive and the cosine, secant, tangent, and cotangent are negative. 75. −215◦ is a second-quadrant angle, so the sine and cosecant function values are positive and the cosine, secant, tangent, and cotangent are negative. 76. 290◦ is a fourth-quadrant angle, so the cosine and secant function values are positive and the sine, cosecant, tangent, and cotangent are negative. 77. −272◦ is a first-quadrant angle, so all of the trigonometric function values are positive. 78. 91◦ is a second-quadrant angle, so the sine and cosecant function values are positive and the cosine, secant, tangent, and cotangent are negative. 79. 360◦ − 319◦ = 41◦ , so 41◦ is the reference angle for 319◦ . Note that 319◦ is a fourth-quadrant angle. Use the given trigonometric function values for 41◦ to find the trigonometric function values for 319◦ . In the fourth quadrant, the cosine and secant functions are positive and the other four are negative. sin 319◦ = − sin 41◦ = −0.6561 cos 319◦ = cos 41◦ = 0.7547

tan 319◦ = − tan 41◦ = −0.8693 1 1 csc 319◦ = = ≈ −1.5242 sin 319◦ −0.6561 1 1 ≈ 1.3250 = sec 319◦ = cos 319◦ 0.7547 1 1 = ≈ −1.1504 cot 319◦ = tan 319◦ −0.8693

348

Chapter 5: The Trigonometric Functions The airplane is about 130 km east of the airport.

80. Reference angle: 27◦

Now find s, the airplane’s distance south of the airport. s sin 30◦ = 150 s = 150 sin 30◦

Terminal side is in quadrant IV. sin 333◦ = − sin 27◦ = −0.4540

cos 333◦ = cos 27◦ = 0.8910

tan 333◦ = − tan 27◦ = −0.5095 1 1 csc 333◦ = = ≈ −2.2026 sin 333◦ −0.4540 1 1 = sec 333◦ = ≈ 1.1223 cos 333◦ 0.8910 1 1 ≈ −1.9627 = cot 333◦ = tan 333◦ −0.5095

s = 75 The airplane is 75 km south of the airport. 84.

N 100 W

81. 180◦ − 115◦ = 65◦ , so 65◦ is the reference angle for 115◦ . Note that 115◦ is a second-quadrant angle. Use the given trigonometric function values for 65◦ to find the trigonometric function values for 115◦ . In the second quadrant, the sine and cosecant functions are positive and the other four are negative.

The reference angle is 30◦ . First find n, the airplane’s distance north of the airport. n sin 30◦ = 100 n = 100 sin 30◦

tan 115◦ = − tan 65◦ = −2.1445 1 1 csc 115◦ = = ≈ 1.1034 sin 115◦ 0.9063 1 1 = ≈ −2.3663 sec 115◦ = cos 115◦ −0.4226 1 1 ≈ −0.4663 cot 115◦ = = tan 115◦ −2.1445

n = 50 mi Now find w, the airplane’s distance west of the airport. w cos 30◦ = 100 w = 100 cos 30◦ w ≈ 87 mi

82. Reference angle: 35◦

85.

N

sin 215◦ = − sin 35◦ = −0.5736

cos 215◦ = − cos 35◦ = −0.8192

83.

N

W

120° d 150

E s

138° W d

E 48° s

S

The reference angle is 48◦ . Use the formula d = rt to find the airplane’s distance d from Omaha at the end of 2 hr: km · 2 hr = 300 km d = 150 h Then find s, the airplane’s distance south of Omaha. s sin 48◦ = 300 s = 300 sin 48◦ s ≈ 223

S

The plane is about 223 km south of Omaha.

The reference angle is 30 . First find d, the airplane’s distance east of the airport. d cos 30◦ = 150 d = 150 cos 30◦ ◦

d ≈ 130

300° S

cos 115◦ = − cos 65◦ = −0.4226

tan 215◦ = tan 35◦ = 0.7002 1 1 = ≈ −1.7434 csc 215◦ = sin 215◦ −0.5736 1 1 sec 215◦ = = ≈ −1.2207 cos 215◦ −0.8192 1 1 cot 215◦ = = ≈ 1.4282 tan 215◦ 0.7002

E w

sin 115◦ = sin 65◦ = 0.9063

Terminal side is in quadrant III.

n

Exercise Set 5.3 86.

349 101. cos θ = −0.9388, 180◦ < θ < 270◦

N

W

n

d

319°

49°

E

We ignore the fact that cos θ is negative and use a calculator to find that the reference angle is approximately 20.1◦ . Since θ is a third-quadrant angle, we find θ by adding 180◦ and 20.1◦ : 180◦ + 20.1◦ = 200.1◦ . Thus, θ ≈ 200.1◦ .

S

The reference angle is 49◦ . Use the formula d = rt to find the airplane’s distance d from Chicago at the end of 2 hr: km · 2 hr = 240 km d = 120 h Then find n, the airplane’s distance north of Chicago. n sin 49◦ = 240 n = 240 sin 49◦ n ≈ 181 km

87. Use a calculator set in degree mode. tan 310.8◦ ≈ −1.1585

88. cos 205.5◦ ≈ −0.9026

89. Use a calculator set in degree mode. 1 = −1.4910 cot 146.15◦ = tan 146.15◦

90. sin(−16.4◦ ) ≈ −0.2823

91. Use a calculator set in degree mode. sin 118◦ 42" ≈ 0.8771

92. cos 273◦ 45" ≈ 0.0654

93. Use a calculator set in degree mode. cos(−295.8◦ ) ≈ 0.4352

94. tan 1086.2◦ ≈ 0.1086

95. Use a calculator set in degree mode. cos 5417◦ ≈ 0.9563 96. sec 240◦ 55" =

1 ≈ −2.0573 cos 240◦ 55"

97. Use a calculator set in degree mode. 1 ≈ 2.9238 csc 520◦ = sin 520◦ 98. sin 3824 ≈ −0.6947 ◦

99. sin θ = −0.9956, 270◦ < θ < 360◦

We ignore the fact that sin θ is negative and use a calculator to find that the reference angle is approximately 84.6◦ . Since θ is a fourth-quadrant angle, we find θ by subtracting 84.6◦ from 360◦ : 360◦ − 84.6◦ = 275.4◦ .

Thus, θ ≈ 275.4◦ .

100. tan θ = 0.2460, 180◦ < θ < 270◦ Use a calculator to find that the reference angle is approximately 13.8◦ . Then θ ≈ 180◦ + 13.8◦ ≈ 193.8◦ .

102. sec θ = −1.0485, 90◦ < θ < 180◦ 1 1 cos θ = = ≈ −0.9537 sec θ −1.0485 Ignore the fact that cos θ is negative and use a calculator to find that the reference angle is approximately 17.5◦ . Then θ ≈ 180◦ − 17.5◦ ≈ 162.5◦ . 103. tan θ = −3.054, 270◦ < θ < 360◦

We ignore the fact that tan θ is negative and use a calculator to find that the reference angle is approximately 71.9◦ . Since θ is a fourth-quadrant angle, we find θ by subtracting 71.9◦ from 360◦ :

360◦ − 71.9◦ = 288.1◦ .

Thus, θ ≈ 288.1◦ .

104. sin θ = −0.4313, 180◦ < θ < 270◦

Ignore the fact that sin θ is negative and use a calculator to find that the reference angle is approximately 25.6◦ . Then θ ≈ 180◦ + 25.6◦ ≈ 205.6◦ .

105. csc θ = 1.0480, 0 < θ < 90◦ 1 1 = ≈ 0.9542 sin θ = csc θ 1.0480 Since θ is a first-quadrant angle, use a calculator to find that θ ≈ 72.6◦ . 106. cos θ = −0.0990, 90◦ < θ < 180◦

Ignore the fact that cos θ is negative and use a calculator to find that the reference angle is approximately 84.3◦ . Then θ ≈ 180◦ − 84.3◦ ≈ 95.7◦ .

107. Given points P and Q on the terminal side of an angle θ, the reference triangles determined by them are similar. Thus, corresponding sides are proportional and the trigonometric ratios are the same. See the specific example that begins at the bottom of page 363 of the text. 108. Since sin θ = y/r and cos θ = x/r and r > 0 for all angles θ, the domain of the sine and of the cosine functions is the set of all angles θ. However, tan θ = y/x and x = 0 for all angles which are odd multiples of 90◦ . Thus, the domain of the tangent function must be restricted to avoid division by 0. 1 x2 − 25 1. The zeros of the denominator are −5 and 5, so x = −5 and x = 5 are vertical asymptotes.

109. f (x) =

2. Because the degree of the numerator is less than the degree of the denominator, the x-axis is the horizontal asymptote. There is no oblique asymptote.

350

Chapter 5: The Trigonometric Functions 3. The numerator has no zeros, so there is no xintercept. ! " 1 1 1 4. f (0) = 2 = − , so 0, − is the 0 − 25 25 25 y-intercept.

115. The first coordinate of the x-intercept is the zero of the function. Thus from Exercise 113 we know that the xintercept is (12, 0).

5. Find other function values as needed to determine the general shape and then draw the graph.

117.

116. From Exercise 114 we know that the x-intercepts are (−2, 0) and (3, 0).

12.5

y

d

1 390°

6

"6

f (x) !

x

The reference angle is 30◦ .

x!5

x ! "5

Let d be the vertical distance of the valve cap above the center of the wheel. d sin 30◦ = 12.5 d = 12.5 sin 30◦

1 x 2 " 25

110. y 25

d = 6.25

20 15

The distance above the ground is 6.25 in.+

10

13.375 in. = 19.625 in.

5 "4"3

13.375

"1 "5

1 2 3 4 x

"10

g(x) ! x3 " 2x # 1

118.

"15 "20 "25 a

x−4 x+2 The denominator is zero for x = −2, so the domain is {x|x &= −2}.

35

45°

111. f (x) =

35 – a 5

Examining the graph of the function, we see that the range is {x|x &= 1}.

x2 − 9 − 7x − 15 2x2 − 7x − 15 = (2x + 3)(x − 5); the denominator is 3 zero for x = − or x = 5, so the domain is 2 ' % & & 3 x&&x &= − and x &= 5 . 2 Examining the graph of the function, we see that the range is the set of all real numbers.

112. g(x) =

765◦ is in the first quadrant, and the reference angle is 45◦ . a cos 45◦ = 35 a = 35 cos 45◦

2x2

113. f (x) = 12 − x

Solve: 0 = 12 − x

a ≈ 24.7

The height above the ground is (35−a)+5 ≈ 35−24.7+5 ≈ 15.3 ft.

Exercise Set 5.4 1. See the answer section in the text. 2.

y

x = 12

(a) q

The zero of the function is 12. 114.

(d) k

x −x−6 = 0 2

(x − 3)(x + 2) = 0

x = 3 or x = −2

The zeros are −2 and 3.

(b) h; (e)

13p 4

(c) 2p (f)

23p 4

x

Exercise Set 5.4

351

3. See the answer section in the text. 4.

c)

y

d) 17p

(c) "S ; (e) " 6

(b) "f

x (f) "k (a) "q ; (d) "r

32 ≈ 5.09, or 32 ≈ 5.09(2π), or 5(2π) + 0.09(2π), so 2π we move completely around the circle counterclockwise 5 times and then continue about another 0.09 of the way around. 320 ≈ 50.93, or 320 ≈ 50.93(2π), or 50(2π)+0.93(2π), 2π so we move completely around the circle counterclockwise 50 times and then continue about another 0.93 of the way around. y

2π 2 . We could also say that 5. Clockwise M is at · π, or 3 3 " 2 4π 1 M is at · 2π. Counterclockwise M is at · 2π, or 3 3 3 ! 4π determines M . We could also say so the number − 3 8 that M is at · π moving counterclockwise, so again the 6" 4π result is − . 3 3π 3 . Clockwise N is at · 2π, or 4 2 1 π π Counterclockwise N is at · 2π, or , so the number − 4 2 2 determines N . ! 5π 5 . We could also say that Clockwise P is at · 2π, or 8 4 " 5 P is at · π. 4 3π 3π 3 , so the number − Counterclockwise P is at · π, or 4 4 4 ! 3 determines P . We could also say that P is at · 2π 8" 3π moving counterclockwise, so again the result is − . 4 ! 11 11π · π, or . We could also say that Clockwise Q is at 6 " 6 11 · 2π. Q is at 12 1 π π Counterclockwise Q is at · π, or so the number − 6 6 6 ! 1 determines Q. We could also say that Q is at · 2π 12 " π moving counterclockwise, so again the result is − . 6 !

b)

(c) 32 x (d) 320

8. a) 0.25 ≈ 0.04(2π), so we move counterclockwise about 0.04 of the way around the circle. b) 1.8 ≈ 0.29(2π), so we move counterclockwise about 0.29 of the way around the circle. c) 47 ≈ 7.48(2π), or 7(2π)+0.48(2π), so we move completely around the circle counterclockwise 7 times and then continue about another 0.48 of the way around. d) 500 ≈ 79.58(2π), or 79(2π) + 0.58(2π), so we move completely around the circle counterclockwise 79 times and then continue about another 0.58 of the way around. y (b) 1.8 (c) 47

(a) 0.25 x

(d) 500

9. We add and subtract multiples of 2π. Answers may vary. π 8π 9π π + 2π = + = Positive angle: 4 4 4 4 π π 8π 7π Negative angle: − 2π = − =− 4 4 4 4

2.4 ≈ 0.38, or 2.4 ≈ 0.38(2π), so we move counter2π clockwise about 0.38 of the way around the circle.

10. Answers may vary. 11π 5π + 2π = Positive angle: 3 3 π 5π − 2π = − Negative angle: 3 3

7.5 ≈ 1.19, or 7.5 ≈ 1.19(2π), or 1(2π) + 0.19(2π), so 2π we move completely around the circle counterclockwise once and then continue about another 0.19 of the way around.

11. We add and subtract multiples of 2π. Answers may vary. 7π 7π 12π 19π Positive angle: + 2π = + = 6 6 6 6 7π 7π 12π 5π Negative angle: − 2π = − =− 6 6 6 6

6. M : π, −π; N : 7. a)

(b) 7.5

(a) 2.4

4π 2π 7π 5π π 7π ,− ; P: , − ; Q: ,− 4 4 3 3 6 6

352

Chapter 5: The Trigonometric Functions

12. Answers may vary. Positive angle: π + 2π = 3π

27. −180◦ = −180◦ ·

Negative angle: π − 2π = −π

28. 90◦ = 90◦ ·

13. We add and subtract multiples of 2π. Answers may vary. 2π 2π 6π 4π Positive angle: − + 2π = − + = 3 3 3 3 2π 2π 6π 8π Negative angle: − − 2π = − − =− 3 3 3 3 14. Answers may vary. 5π 3π + 2π = Positive angle: − 4 4 3π 11π Negative angle: − − 2π = − 4 4 3π 2π π π π − = − = 15. Complement: 2 3 6 6 6 π 3π π 2π Supplement: π − = − = 3 3 3 3 π π 5π − = 2 12 12 5π 7π Supplement: π − = 12 12

16. Complement:

4π 3π π π 3π − = − = 2 8 8 8 8 3π 8π 3π 5π = − = Supplement: π − 8 8 8 8 π π π − = 18. Complement: 2 4 4 π 3π Supplement: π − = 4 4 17. Complement:

π 6π π 5π π − = − = 2 12 12 12 12 π 12π π 11π Supplement: π − = − = 12 12 12 12

3π π 2π π π π − = − = = 2 6 6 6 6 3 π 6π π 5π − = Supplement: π − = 6 6 6 6

21. 75◦ = 75◦ ·

π radians 5π radians = 180◦ 12

22. 30◦ = 30◦ ·

π radians π = radians 180◦ 6

23. 200◦ = 200◦ ·

π radians 10π = radians 180◦ 9

24. −135◦ = −135◦ ·

π radians 3π =− radians 180◦ 4

π radians 214.6π radians =− 180◦ 180 If we multiply by 5/5 to clear the decimal, we have 1073π − radians. 900

25. −214.6◦ = −214.6◦ ·

26. 37.71◦ = 37.71◦ · 419π radians 2000

π radians 37.71π = radians, or 180◦ 180

π radians π = radians 180◦ 2

π radians 12.5π = radians 180◦ 180 If we multiply by 2/2 to clear the decimal and then simplify 5π we have radians. 72

29. 12.5◦ = 12.5◦ ·

30. 6.3◦ = 6.3◦ · 7π radians 200

π radians 6.3π = radians, or 180◦ 180

31. −340◦ = −340◦ · 32. −60◦ = −60◦ · 33. 240◦ = 240◦ · 4.19 radians

34. 15◦ = 15◦ ·

π radians 17π =− radians 180◦ 9

π radians π = − radians 180◦ 3

π radians 240π = radians ≈ 180◦ 180

π radians ≈ 0.26 radians 180◦

35. −60◦ = −60◦ · −1.05 radians

36. 145◦ = 145◦ ·

π radians 60π =− radians ≈ 180◦ 180

π radians ≈ 2.53 radians 180◦

37. 117.8◦ = 117.8◦ · 2.06 radians

19. Complement:

20. Complement:

π radians = −π radians 180◦

π radians 117.8π = radians ≈ 180◦ 180

38. −231.2◦ = −231.2◦ · 39. 1.354◦ = 1.354◦ · 0.02 radians

π radians ≈ −4.04 radians 180◦

π radians 1.354π radians ≈ = 180◦ 180

40. 584◦ = 584◦ ·

π radians ≈ 10.19 radians 180◦

41. 345◦ = 345◦ ·

π radians 345π radians ≈ = 180◦ 180

6.02 radians

42. −75◦ = −75◦ ·

π radians ≈ −1.31 radians 180◦

π radians 95π = radians ≈ 180◦ 180 1.66 radians

43. 95◦ = 95◦ ·

44. 24.8◦ = 24.8◦ ·

π radians ≈ 0.43 radians 180◦

3π 180◦ 3 3π =− radians · = − · 180◦ = 4 4 π radians 4 −135◦

45. − 46.

7π 180◦ 7π = radians · = 210◦ 6 6 π radians

Exercise Set 5.4

353

47. 8π = 8π radians · 48. −

180◦ = 8 · 180◦ = 1440◦ π radians

π π 180◦ = − radians · = −60◦ 3 3 π radians

49. 1 = 1 radian ·

180◦ 180◦ ≈ ≈ 57.30◦ π radians π

50. −17.6 = −17.6 radians ·

180◦ = π radians

17.6(180◦ ) ≈ −1008.41◦ − π 51. 2.347 = 2.347 radians ·

180◦ = π radians

π π 60. (See Exercise 59.) 0◦ = 0; −30◦ = − ; −45◦ = − ; 6 4 π π 3π ◦ ◦ ◦ −60 = − ; −90 = − ; −135 = − ; 3 2 4 5π 3π −180◦ = −π; −225◦ = − ; −270◦ = − ; 4 2 7π ◦ ◦ −315 = − ; −360 = −2π 4

62.

2.347(180◦ ) ≈ 134.47◦ π 52. 25 = 25 radians · 53.

180◦ ≈ 1432.39◦ π radians

5π 5π 180◦ 5 = radians · = · 180◦ = 225◦ 4 4 π radians 4

54. −6π = −6π radians ·

180◦ = −1080◦ π radians

55. −90 = −90 radians ·

−90(180◦ ) 180◦ = ≈ π radians π

−5156.62◦ 56. 37.12 = 37.12 radians ·

2π 180◦ 2 2π = radians · = · 180◦ ≈ 51.43◦ 7 7 π radians 7

58.

π π 180◦ = radians · = 20◦ 9 9 π radians π radians = 0; 180◦ π radians π = ; 30◦ = 30◦ · 180◦ 6 π radians π = ; 45◦ = 45◦ · 180◦ 4 π ◦ ◦ π radians = ; 60 = 60 · 180◦ 3 π ◦ ◦ π radians = ; 90 = 90 · 180◦ 2 π radians 3π ; 135◦ = 135◦ · = 180◦ 4 π radians 180◦ = 180◦ · = π; 180◦ π radians 5π = 225◦ = 225◦ · ; 180◦ 4 π radians 3π = ; 270◦ = 270◦ · 180◦ 2 π radians 7π 315◦ = 315◦ · ; = 180◦ 4 π radians = 2π 360◦ = 360◦ · 180◦

59. 0◦ = 0◦ ·

63.

64.

180◦ ≈ 2126.82◦ π radians

57.

s 8 = ≈ 2.29 r 3.5 π Note that 45◦ = radians. 4 s θ= r 200 cm π = 4 r 4 r = · 200 cm ≈ 254.65 cm π s θ = , or s = r θ r 5π s = 4.2 in. · ≈ 5.50 in. 12 s θ= r 16 yd 5= r 16 yd r= = 3.2 yd 5 s θ= r θ is the radian measure of the central angle, s is arc length, and r is radius length. 132 cm Substituting 132 cm for s θ= 120 cm and 120 cm for r 11 θ= , or 1.1 The unit is understood 10 to be radians.

61. θ =

65.

1.1 = 1.1 radians ·

180◦ ≈ 63◦ π radians

66. Radius = 10 ft/2 = 5 ft s 20 ft θ= = = 4 radians r 5 ft 180◦ ≈ 229◦ 4 radians · π radians s 67. We use the formula θ = , or s = rθ. r s = rθ = (2 yd)(1.6) = 3.2 yd 68. s = rθ = (5 m)(2.1) = 10.5 m 69. Let r = the length of the minute hand. In 60 minutes the minute hand travels the circumference of a circle with radius r, or 2πr. Then in 50 minutes the minute hand 5πr 50 · 2πr, or . travels 60 3 Now find the angle through which the minute hand rotates in 50 minutes.

354

Chapter 5: The Trigonometric Functions s r 5πr θ= 3 r 5πr 1 · θ= 3 r 5π ≈ 5.24 θ= 3 The minute hand rotates through about 5.24 radians. θ=

5280 ft 12 in. · = 63, 360 in. 1 mi 1 ft 31.125 in. d r= = = 15.5625 in. 2 2 63, 360 in. s ≈ 4071 θ= = r 15.5625 in.

70. 1 mi = 1 mi ·

71. Since the linear speed must be in cm/min, the given angular speed, 7 radians/sec, must be changed to radians/min. 420 radians 7 radians 60 sec · = ω= 1 sec 1 min 1 min 15 cm d = 7.5 cm r= = 2 2 Using v = rω, we have: 420 v = 7.5 cm · Substituting and omitting 1 min the word radians cm = 3150 min The linear speed of a point on the rim is 3150 cm/min. 1m = 0.3 m 100 cm 3 radians 60 sec 180 radians ω= · = 1 sec 1 min 1 min 180 = 54 m/min v = rω = 0.3 m · 1 min

72. r = 30 cm ·

73. First convert 18.33 ft/sec to in./hr. 18.33 ft 12 in. 60 sec 60 min 18.33 ft/sec = · · · = 1 sec 1 ft 1 min 1 hr 791,856 in./hr 21 in. = 10.5 in. r= 2 Using v = rω, we have 791, 856 in./hr = 10.5 in. · ω

75, 415 radians/hr ≈ ω

Now convert 75,415 radians/hr to revolutions/hr. 75, 415 radians 1 revolution · ≈ 75, 415 radians/hr = 1 hr 2π radians 12,003 revolutions/hr The angular speed of the cylinder is about 12,003 revolutions/hr (or about 12,003 IPH). 19.25 1 mi = mi 5280 ft 5280 167 167 1 mi 11 ft = ft = ft · = 13 ft 11 in. = 13 12 12 12 5280 ft 167 mi 63, 360

74. 19 ft 3 in. = 19.25 ft ·

2.4(2π) 2.4(2π) 60 min 288π = · = 1 min 1 min 1 hr 1 hr Find Alicia’s linear speed: 19.25 288π v = rω = mi · ≈ 3.30 mph 5280 1 hr Find Zoe’s linear speed: 288π 167 mi · ≈ 2.38 mph v = rω = 63, 360 1 hr The difference in the linear speeds is about 3.30 − 2.38, or 0.92 mph. (Answers may vary slightly due to rounding differences.) 75. First find ω in radians per hour. π 2π = ω= 24 hr 12 hr Using v = rω, we have π ≈ 1047 mph. v = 4000 mi · 12 hr The linear speed of a point on the equator is about 1047 mph. 1 day π 2π · = 365.25 days 24 hr 4383 hr π ≈ 66, 659 mph v = rω = 93, 000, 000 mi · 4383 hr 1 km 1m 77. 67 cm = 67 cm · · = 0.00067 km 100 cm 1000 m 0.00067 km r= = 0.000335 km 2 v 40.423 km/h 120, 665.6716 ω= = ≈ r 0.000335 km 1 hr Convert ω from radians/hr to revolutions/hr: 120, 665.6716 radians 1 revolution · ≈ 1 hr 2π radians 19,205 revolutions/hr 76. ω =

78. First find ω in radians per hour. 1680π 14 · 2π 60 min · = ω= 1 min 1 hr hr Next find r in miles. 1 1 mi = mi r = 10 ft · 5280 ft 528 Using v = rω, we have 1680π 1 v= mi · ≈ 10 mph. 528 1 hr The speed of the river is about 10 mph. 79. First convert 22 mph to inches/second. 1 hr 1 min 22 mi 5280 ft 12 in. · · · · = v= 1 hr 1 mi 1 ft 60 min 60 sec 387.2 in./sec Using v = rω, we have in. = 23 in. · ω 387.2 sec 387.2 in./sec 16.835 so ω = ≈ . 23 in. 1 sec Then in 12 sec, 16.835 · 12 sec ≈ 202. θ = ωt = 1 sec The wheel rotates through an angle of 202 radians.

Exercise Set 5.4 80. Left to the student

355 95. a)

81. Left to the student 82. Consider Example 7. The division 4 m/2 m simplifies to the real number 2 since m/m = 1. From this point of view it would seem preferable to omit ”radians.” Also consider Example 8. If ”radians” were used throughout the solution, the answer would have been 5.24 cm - radians. Since we are finding a distance the correct unit is centimeters. Thus we omit ”radians.” Since a measure in radians is simply a real number, it is usually preferable to omit the word ”radians.” 83. For a point at a distance r from the center of rotation with a fixed angular speed k, the linear speed is given by 1 v = r · k, or r = v. Thus, the length of the radius is k directly proportional to the linear speed. s 84. We see from the formula θ = that the tire with the 15 in. r diameter will rotate through a larger angle than the tire with the 16 in. diameter. Thus the car with the 15 in. tires will probably need new tires first. 85. one-to-one 86. cosine of θ 87. exponential function 88. horizontal asymptote 89. odd function 90. natural 91. horizontal line; inverse 92. logarithm 1 of the circumference of the 93. One degree of latitude is 360 earth. c = πd, or 2πr When r = 6400 km, C = 2π · 6400 = 12, 800π km. 1 · 12, 800π km = 111.7 km. Thus 1◦ of latitude is 360 When r = 4000 mi, C = 2π · 4000 mi = 8000π mi. 1 Thus 1◦ of latitude is · 8000π mi ≈ 69.8 mi. 360 ! √ "2 21 =1 94. x2 + − 5 4 x2 = 25 2 x=± 5

100 mils = 100 mils ·

90◦ 1600 mils

= 5.625◦

= 5◦ 37" 30"" b)

Using the DMS feature on a calculator

350 mils = 350 mils ·

90◦ 1600 mils

= 19.6875◦

= 19◦ 41" 15"" Using the DMS feature on a calculator 100 grads 96. a) 48◦ = 48◦ · 90◦ 48 = · 100 grads ≈ 53.33 grads 90 5π 5π 100 grads 1000 b) = = radians · π grads ≈ 7 7 7 radians 2 142.86 grads 97. Let ω1 = the angular speed of the smaller wheel and ω2 = the angular speed of the larger wheel. The wheels have the same linear speed, so we have v = 40ω1 = 50ω2 . Convert the angular speed of the smaller wheel, ω1 , to radians per second. 20 · 2π 1 min 2 · = π/ sec ω1 = 20 rpm = 1 min 60 sec 3 Then 40ω1 = 50ω2 2 40 · π/sec = 50ω2 3 8π /sec = ω2 15 1.676/sec ≈ ω2 . The angular speed of the larger wheel is about 1.676 radians/sec.

98. The radii of the pulleys are 25 cm and 15 cm. Let ω1 = the angular speed of the larger pulley and ω2 = the angular speed of the smaller pulley. Since they are connected by the same belt, their linear speed v, will be the same. v = 25ω1 = 15ω2 . Find ω1 , the angular speed of the larger pulley, in radians/sec. The larger pulley makes 12 revolutions per minute. 12 · 2π 1 min · = 0.4π/sec ω1 = 1 min 60 sec Then 25 · 0.4π/sec = 15ω2 2 ω2 = π/sec ≈ 2.094/sec. 3

356

Chapter 5: The Trigonometric Functions

99.

3.

B

c

A

( - –25 ,√21 —) 5

y

( –25 ,√21 —) 5 x

a

b

( - –25 , -√21 —) 5

C

a = 38◦ 28" 45"" − 38◦ 27" 30"" = 1" 15"" = 1.25" = 1.25 nautical miles.

b = 82◦ 57" 15"" − 82◦ 56" 30"" = 45"" = 0.75" =

0.75 nautical miles. # √ c = a2 + b2 = (1.25)2 + (0.75)2 ≈ 1.46 nautical miles

100. Let x = the number of minutes after noon when the hands are first perpendicular. The minute hand moves through π 1 · 2π, or radians in 1 minute; the hour hand moves 60 30 1 π π 1 of this rotation, or · , or radians in through 12 12 30 360 1 minute. Then in x minutes the minute hand and hour πx πx and radians, respectively. We hand move through 30 360 want to find x such that the difference in the rotations is π . 2 πx πx π − = 30 360 2 π 11πx = 360 2 180 x= ≈ 16.3636 11 Thus the hands are perpendicular about 16.3636 minutes after 12:00 noon, or at about 12:16:22 P.M.

( –25 , -√21 —) 5

√ " √ " √ " ! ! 2 21 2 21 2 21 , , b) − , − , c) − , 5 5 5 5 5 5 ! √ " !√ " !√ " 3 1 3 1 3 1 , , b) , − , c) , 4. a) − 2 2 2 2 2 2 a)

!

5. The point determined by −π/4 is a reflection across the !√ √ " 2 2 , . Thus, x-axis of the point determined by π/4, 2 2 the coordinates of the point determined by −π/4 are √ " !√ 2 2 ,− . 2 2 6. The point determined by −β is a reflection across the xaxis determined by β. Its coordinates are √ point " ! of the 5 2 . − ,− 3 3 7. The coordinates of the point determined by π are (−1, 0). Thus, sin π = y = 0. 8. The point determined by −π/3 is a reflection across the ! √ " 3 1 , . Its cox-axis of the point determined by π/3, 2 2 √ " ! 1 3 ,− . Thus, ordinates are 2 2 " ! 1 π cos − =x= . 3 2 7π is a reflection across the origin 6 " !√ 3 1 , . Its coordinates of the point determined by π/6, 2 2 ! √ " 3 1 are − , − . Thus, 2 2 √ 3 − √ 7π x 2 = 3. = = cot 1 6 y − 2

9. The point determined by

Exercise Set 5.5 1.

y

( - –34 , √7 —) 4

( –34 , √7 —) 4 x

( - –34 , - √7 —) 4

( –34 , - √7 —) 4

√ " √ " ! √ " ! 7 3 7 3 7 3 , b) , , c) ,− − ,− 4 4 4 4 4 4 √ " √ " √ " ! ! ! 2 2 2 5 5 5 , b) − , , c) − , − ,− 2. a) 3 3 3 3 3 3 a)

!

10. The point determined by 11π/4 is a reflection across the !√ √ " 2 2 , . Its y-axis of the point determined by π/4, 2 2 ! √ √ " 2 2 . Thus, , coordinates are − 2 2 √ 2 y 11π 2 √ = −1. = = tan 4 x 2 − 2

Exercise Set 5.5 11. The coordinates of the point determined by −3π are (−1, 0). Thus, sin(−3π) = y = 0. 12. The point determined by 3π/4 is a reflection across the !√ √ " 2 2 , . Its y-axis of the point determined by π/4, 2 2 ! √ √ " 2 2 . Thus, , coordinates are − 2 2 √ 3π 1 1 2 csc = = √ = √ = 2. 4 y 2 2 2 13. The point determined by 5π/6 is a reflection across the !√ " 3 1 y-axis of the point determined by π/6, , . Its co2 2 " ! √ 3 1 . Thus, , ordinates are − 2 2 √ 5π 3 cos =x=− . 6 2 14. The point determined by −π/4 is a reflection across the !√ √ " 2 2 , x-axis of the point determined by π/4, . Its 2 2 √ " !√ 2 2 ,− . coordinates are 2 2 Thus, √ 2 " ! − y π 2 = = √ = −1. tan − 4 x 2 2 15. The coordinates of the point determined by π/2 are (0, 1). Thus, 1 1 π sec = = , which is not defined. 2 x 0 16. The coordinates of the point determined by 10π are (1, 0). Thus, cos 10π = x = 1. 17. The coordinates of the point determined by π/6 are " !√ 3 1 , . Thus, 2 2 √ π 3 cos = x = . 6 2 18. The point determined by 2π/3 is a reflection across the ! √ " 3 1 , . Its coy-axis of the point determined by π/3, 2 2 √ " ! 1 3 ordinates are − , . Thus, 2 2 √ 3 2π =y= . sin 3 2 19. The point determined by 5π/4 is a reflection across the !√ √ " 2 2 , . Its origin of the point determined by π/4, 2 2 √ " ! √ 2 2 ,− . Thus, coordinates are − 2 2

357

sin

√ 5π 2 =y=− . 4 2

20. The point determined by 11π/6 is a reflection across the " !√ 3 1 , . Its cox-axis of the point determined by π/6, 2 2 " !√ 3 1 , − . Thus, ordinates are 2 2 √ 11π 3 cos =x= . 6 2 21. The coordinates of the point determined by −5π are (−1, 0). Thus, sin(−5π) = y = 0. 22. The coordinates of the point determined by 3π/2 are (0, −1). Thus, y −1 3π = = , which is not defined. tan 2 x 0 23. The coordinates of the point determined by 5π/2 are (0, 1). Thus, x 0 5π = = = 0. cot 2 y 1 24. The point determined by 5π/3 is a reflection across the ! √ " 3 1 , . Its cox-axis of the point determined by π/3, 2 2 √ " ! 3 1 ordinates are ,− . Thus, 2 2 √ 3 − y 5π 2 = −√3. = = tan 1 3 x 2 25. Use a calculator set in radian mode. π tan ≈ 0.4816 7 26. Use a calculator set in radian mode. " ! 2π = 0.3090 cos − 5 27. Use a calculator set in radian mode. 1 ≈ 1.3065 sec 37 = cos 37 28. Use a calculator set in radian mode. sin 11.7 ≈ −0.7620 29. Use a calculator set in radian mode. 1 ≈ −2.1599 cot 342 = tan 342 30. Use a calculator set in radian mode. tan 1.3 ≈ 3.6021 31. Use a calculator set in radian mode. cos 6π = 1 32. Use a calculator set in radian mode. π ≈ 0.3090 sin 10

358

Chapter 5: The Trigonometric Functions 44. a)

33. Use a calculator set in radian mode. 1 csc 4.16 = ≈ −1.1747 sin 4.16

y 1 "f "q

34. Use a calculator set in radian mode. 1 10π = ≈ −4.4940 sec 10π 7 cos 7

"p"S"i

f

"d Adu

"u "A

b) Reflect the graph of y = cos x across the y-axis. The graph is the same as the graph in part (a). c) Reflect the graph of y = cos x across the x-axis.

36. Use a calculator set in radian mode. cos 2000 ≈ −0.3675

y

y ! "cos x

1

37. Use a calculator set in radian mode. " ! π ≈ −0.7071 sin − 4

"p "f "q

38. Use a calculator set in radian mode. 1 is not defined. cot 7π = tan 7π

d

"d

q

f

p x

"1

d) They are the same.

39. Use a calculator set in radian mode.

45. a) See Exercise 43(a).

sin 0 = 0

b) Shift the graph of y = sin x left π units.

40. Use a calculator set in radian mode. cos(−29) ≈ −0.7481

y 1

41. Use a calculator set in radian mode. 2π ≈ 0.8391 tan 9

"p "f "q

42. Use a calculator set in radian mode. 8π sin ≈ 0.8660 3 43. a)

d

"d

f Adu

p x

46. a) See Exercise 43(a).

q i S p x

b) Shift the graph of y = sin x right π units.

y ! sin x

"1

y 1

b) Reflect the graph of y = sin x across the y-axis. y

"p "f "q

1

d

q

f

p x

"1

c) Reflect the graph of y = sin x across the x-axis. The graph is the same as the graph in part (b).

d

"d

y ! sin (x " p)

y ! sin ("x)

f

d) They are the same.

"d

"d

q

c) Reflect the graph of y = sin x across the x-axis. The graph is the same as the graph in part (b).

1

"p"S"i"q "u "A

y ! sin (x # p)

"1

y

"p "f "q

q i S p x

"1

35. Use a calculator set in radian mode. 7π = −1 tan 4

"f

y ! cos x

q

f

p x

"1

c) Reflect the graph of y = sin x across the x-axis. The graph is the same as the graph in part (b). d) They are the same. 47. a)

y 1

d) They are the same. "f "q "p"S"i

y ! cos x f

"d "u "A "1

Adu

q i S p x

Exercise Set 5.5

359

b) Shift the graph of y = cos x left π units.

y

y ! "sec x

2

y

y ! cos (x # p)

1

1

"2π

3π "_ 2

"π

π "_ 2

π _ 2

π

3π _ 2

2π

x

"1 "2

"p "f "q

d

"d

q

p x

f

"1

c) Reflect the graph of y = cos x across the x-axis. The graph is the same as the graph in part (b). d) They are the same. 48. a) See Exercise 47(a). b) Shift the graph of y = cos x right π units. y

y ! cos (x " p)

1

"p "f "q

d

"d

q

f

p x

"1

c) Reflect the graph of y = cos x across the x-axis. The graph is the same as the graph in part (b). d) They are the same. 49. a)

y

y ! tan x

1 3π "_ 2

"π

π "_ 2

π _ 2

π

3π _ 2

x

2π

"1 "2

b) Reflect the graph of y = tan x across the y-axis. y ! tan ("x) 2 1 3π "_ 2

π "_ 2

"π

π _ 2

π

3π _ 2

x

2π

"1 "2

c) Reflect the graph of y = tan x across the x-axis. The graph is the same as the graph in part (b). d) They are the same. 50. a)

y

y ! sec x

2 1 "2π

3π "_ 2

"π

π "_ 2

The sine function is an odd function. cos(−x) = cos x

The cosine function is an even function. − sin x sin(−x) = tan(−x) = cos(−x) cos x sin x =− = − tan x cos x The tangent function is an odd function. 1 1 1 = =− = − csc x csc(−x) = sin(−x) − sin x sin x The cosecant function is an odd function. 1 1 sec(−x) = = = sec x cos(−x) cos x The secant function is an even function. cos x cos(−x) cot(−x) = = sin(−x) − sin x cos x =− = − cot x sin x The cotangent function is an odd function.

π _ 2

π

3π _ 2

2π

52. From the graphs and lists of properties in the text we see that the tangent and cotangent functions have period π and the sine, cosine, cosecant, and secant functions have period 2π. 53. Consider a point (x, y) on the unit circle determined by an y angle s. Then tan s = . The coordinates x and y have the x same sign in quadrants I and III, so the tangent function is positive in these quadrants; x and y have opposite signs in quadrants II and IV, so the tangent function is negative in these quadrants.

y

"2π

51. sin(−x) = − sin x

Thus the cosine and secant functions are even; the sine, tangent, cosecant, and cotangent functions are odd.

2

"2π

d) They are the same.

x

"1 "2

b) Reflect the graph of y = sec x across the y-axis. The graph is the same as the graph in part (a). c) Reflect the graph of y = sec x across the x-axis.

54. Consider a point (x, y) on the unit circle determined by an angle s. Then sin s = y. Since y is positive in quadrants I and II, the sine function is positive in these quadrants; y is negative in quadrants III and IV, so the sine function is negative in these quadrants. 55. Consider a point (x, y) on the unit circle determined by an angle s. Then cos s = x. Since x is positive in quadrants I and IV, the cosine function is positive in these quadrants; x is negative in quadrants II and III, so the cosine function is negative in these quadrants. 56. Consider a point (x, y) on the unit circle determined by an angle s. Then csc s = 1/ sin s = 1/y. From Exercise 54 we see that the cosecant function is positive in quadrants I and II and negative in quadrants III and IV.

360

Chapter 5: The Trigonometric Functions 63. See the answer section in the text.

57. Graph y = (sin x)2 .

64.

y

f(x) ! x 2

g(x) ! (x " 2)2

4 2

"4

"2

2

4

x

"2 "4

From the graph we find the following information. Domain: (−∞, ∞)

65. See the answer section in the text.

Range: [0, 1] Period: π Amplitude:

Shift the graph of f right 2 units.

1 1 (1 − 0), or 2 2

66.

y

f (x) ! x 3

4

58. Graph y = | cos x| + 1.

2 "4

"2

2

4

x

"2

g(x) ! "x 3

"4

Reflect the graph of f across the x-axis. 67. Start with y = x3 . Reflect it across the x-axis: y = −x3

Domain: (−∞, ∞)

Shift it right 2 units: y = −(x − 2)3

Range: [1, 2] Period: π Amplitude:

Shift it down 1 unit: y = −(x − 2)3 − 1 1 1 (2 − 1), or 2 2

sin x π , when 0 < x < x 2 sin π/2 ≈ 0.6369 π/2

59. f (x) =

sin 3π/8 ≈ 0.7846 3π/8 sin π/4 ≈ 0.9008 π/4

1 . x

Shrink it vertically by a factor of Shift it up 3 units: y =

1 +3 4x

1 1 1 1 : y = · , or y = 4 4 x 4x

69. Any real numbers x and −x determine points on the unit circle that are symmetric with respect to the x-axis. Their first coordinates are the same, so cos(−x) = cos x. 70. Any real numbers x and −x determine points on the unit circle that are symmetric with respect to the x-axis. Their second coordinates are opposites, so sin(−x) = − sin x.

sin π/8 ≈ 0.9750 π/8 The limiting value of

68. Start with y =

sin x as x approaches 0 is 1. x

60. ! A graph shows"that ! intervals " on which ! sin x"< cos x are 3π π 5π 9π 11π 7π ,− , − , , and , . − 4 4 4 4 4 4 In ! general sin x < cos x on " the intervals 3π π − + 2kπ, + 2kπ , k an integer. 4 4 61. The graph of the cosine function is the graph of the sine function shifted left π/2 units. (Or we can say that the graph of the sine function is the graph of the cosine function shifted right π/2 units.) 62. The numbers for which the value of the cosine function is 0 are not in the domain of the tangent function.

71. Any real numbers x and x + 2kπ determine the same point on the unit circle, so sin(x + 2kπ) = sin x. 72. Any real numbers x and x + 2kπ determine the same point on the unit circle, so cos(x + 2kπ) = cos x. 73. Any real numbers x and π −x determine points on the unit circle that are symmetric with respect to the y-axis. Their second coordinates are the same, so sin(π − x) = sin x. 74. Any real numbers x and π −x determine points on the unit circle that are symmetric with respect to the y-axis. Their first coordinates are opposites, so cos(π − x) = − cos x. 75. Any real numbers x and x−π determine points on the unit circle that are symmetric with respect to the origin. Their first coordinates are opposites, so cos(x − π) = − cos x.

Exercise Set 5.5

361 sin x cos x The domain consists of the values of x for which cos x &= 0. From the graph of the cosine function we see ' % &that the domain is & π x&&x &= + kπ, k an integer . 2

76. Any real number x and x + π determine points on the unit circle that are symmetric with respect to the origin. Their first coordinates are opposites, so cos(x + π) = − cos x.

83. f (x) =

78. Any real numbers x and x−π determine points on the unit circle that are symmetric with respect to the origin. Their second coordinates are opposites, so sin(x − π) = − sin x.

84. g(x) = log(sin x)

77. Any real number x and x + π determine points on the unit circle that are symmetric with respect to the origin. Their second coordinates are opposites, so sin(x + π) = − sin x.

π =1 2 " ! " ! π π sin + 2π = 1 sin − 2π = 1 2 2 " ! " ! π π + 2 · 2π = 1 sin − 2 · 2π = 1 sin 2 2 " ! " ! π π sin + 3 · 2π = 1 sin − 3 · 2π = 1 2 2 " ! π + k · 2π = 1, k an integer sin 2 π Thus x = + 2kπ, k an integer. 2 b) cos π = −1 cos(π + 2π) = −1 cos(π − 2π) = −1

79. a) sin

cos(π + 2 · 2π) = −1 cos(π + 3 · 2π) = −1

The domain consists of the values of x for which sin x > 0. From the graph of the sine function we see that the domain consists of the intervals (2kπ, (2k + 1)π), k an integer. 85. See the answer section in the text. 86.

y

"p "2p

p

w

2p x

87. See the answer section in the text. 88.

y

y ! !cos x!

1

cos(π − 3 · 2π) = −1

sin 2π = 0

sin(−2π) = 0

sin 3π = 0

sin(−3π) = 0

sin kπ = 0, k an integer Thus x = kπ, k an integer. 80. f (x) = x2 + 2x, g(x) = cos x f ◦ g(x) = f (g(x)) = f (cos x)

= (cos x)2 + 2 cos x

= cos2 x + 2 cos x g ◦ f (x) = g(f (x)) = g(x2 + 2x)

= cos(x + 2x) 2

cos x

The domain consists of the values of x for which cos x ≥ 0. From the graph of the cosine function we see that the domain consists ( ) of the intervals π π − + 2kπ, + 2kπ , k an integer. 2 2 1 sin x The domain consists of the values of x for which sin x &= 0. From the graph of the sine function we see that the domain consists of {x|x &= kπ, k an integer}.

82. f (x) =

q

"q

cos(π − 2 · 2π) = −1

Thus x = π + 2kπ, or x = (2k + 1)π, k an integer. c) sin 0 = 0 sin π = 0 sin(−π) = 0

81. f (x) =

"w

"1

"2p "w "p "q

cos(π + k · 2π) = 1, k an integer

√

y ! sin !x!

1

89. a)

∆OP A AP Thus, OA sin θ cos θ tan θ

∼ ∆ODB BD = OB BD = 1 = BD

b) ∆OP A OD OP OD 1 OD

∼ ∆ODB OB = OA 1 = cos θ = sec θ

c) ∆OAP OE PO OE 1 OE

∼ ∆ECO CO = AP 1 = sin θ = csc θ

d) ∆OAP ∼ ∆ECO CO CE = AO AP CE 1 = cos θ sin θ cos θ CE = sin θ CE = cot θ

q

p

w

2p

x

362

Chapter 5: The Trigonometric Functions 4. y = sin(−2x)

Exercise Set 5.6

A = 1, B = −2, C = 0, D = 0 Amplitude: |A| = |1| = 1 ! ! ! ! ! 2π ! ! 2π ! !=π Period: !! !! = !! B −2 !

1. y = sin x + 1 A = 1, B = 1, C = 0, D = 1 Amplitude: |A| = |1| = 1 ! ! ! ! ! 2π ! ! 2π ! Period: !! !! = !! !! = 2π B 1

0 C = =0 B −2 Shrink the graph of y = sin x horizontally by a factor of 2.

Phase shift:

0 C = =0 B 1 Translate the graph of y = sin x up 1 unit. Phase shift:

y

y

"2p

2

p

"p

1 "2p "w "p

y ! sin ("2x)

1

q

"q "1

w

p

2p

x

"1

x

y ! sin x $ 1

"2

2p

1 cos x 2 1 A = , B = 1, C = 0, D = 0 2 ! ! !1! 1 Amplitude: |A| = !! !! = 2 2 ! ! ! ! ! 2π ! ! 2π ! Period: !! !! = !! !! = 2π B 1 0 C = =0 Phase shift: B 1 Shrink the graph of y = cos x vertically by a factor of 2.

5. y =

1 cos x 4 1 A = , B = 1, C = 0, D = 0 4 ! ! !1! 1 Amplitude: |A| = !! !! = 4 4 ! ! ! ! ! 2π ! ! 2π ! Period: !! !! = !! !! = 2π B 1 0 C = =0 Phase shift: B 1 Shrink the graph of y = cos x vertically by a factor of 4.

2. y =

y 2

y 1

1

y ! ~ cos x "2# "#

p

"p

# "1

x

2p

"2p

"2

2#

x

1 y!" 2 cos x

"1

3. y = −3 cos x

A = −3, B = 1, C = 0, D = 0 Amplitude: |A| = | − 3| = 3 ! ! ! ! ! 2π ! ! 2π ! Period: !! !! = !! !! = 2π B 1

0 C = =0 B 1 Stretch the graph of y = cos x vertically by a factor of 3 and reflect the graph across the x-axis. Phase shift:

y 3 2 1

"w "2p

"p "q

# 1 x 2 1 A = 1, B = , C = 0, D = 0 2 Amplitude: |A| = |1| = 1 ! ! !! !! ! 2π ! ! 2π ! Period: !! !! = ! 1 ! = 4π ! 2 ! B

6. y = sin

"1 "2

w q

p

y ! "3 cos x

2p

x

"

0 C =0 = 1 B 2 Stretch the graph of y = sin x horizontally by a factor of 2. Phase shift:

Exercise Set 5.6

363

y

y 2

2

1

1 "2$ "$

$ "1 "2

2$

x

"2p "w "p

"1

( )

1 y ! sin " 2x

π 10. y = cos x − 2

A = 1, B = 2, C = 0, D = 0 Amplitude: |A| = |1| = 1 & & & & & 2π & & 2π & Period: && && = && && = π B 2 C 0 Phase shift: = =0 B 2 Shrink the graph of y = sin x horizontally by a factor of 2.

p

w

2p

x

y ! 2 sin "q x#

"2

!

7. y = sin(2x)

q

"q

"

π ,D=0 2 Amplitude: |A| = |1| = 1 & & & & & 2π & & 2π & Period: && && = && && = 2π B 1 π C π Phase shift: = 2 = B 1 2 A = 1, B = 1, C =

Translate the graph of y = cos x to the right

y 2

y

π units. 2

y ! cos "x " q#

1

1 "2$ "$

$

2$

x

"1 "2

2p

y ! sin (2x)

"2p

"1

8. y = cos x − 1

A = 1, B = 1, C = 0, D = −1

Amplitude: |A| = |1| = 1 & & & & & 2π & & 2π & Period: && && = && && = 2π B 1 C 0 Phase shift: = =0 B 1 Translate the graph of y = cos x down 1 unit. y 2 1 "2$ "$

x

p

"p

y ! cos x " 1 $

2$

x

"1

! " ( ! ") π 1 π 1 sin x + = sin x − − 2 2 2 2 1 π A = , B = 1, C = − , D = 0 2 & &2 &1& 1 Amplitude: |A| = && && = 2 2 & & & & & 2π & & 2π & Period: && && = && && = 2π B 1 π − π C = 2 =− Phase shift: B 1 2 Shrink the graph of y = sin x vertically by a factor of 2 π and translate it to the left units. 2

11. y =

"2

y 1

" 1 x 9. y = 2 sin 2 1 A = 2, B = , C = 0, D = 0 2 Amplitude: |A| = |2| = 2 & & && && & 2π & & 2π & Period: && && = & 1 & = 4π & 2 & B !

C 0 = =0 Phase shift: 1 B 2 Stretch the graph of y = sin x horizontally by a factor of 2 and stretch it vertically, also by a factor of 2.

"w "2p

y ! q sin "x # q#

"q q

"p

p

"1

12. y = cos x −

1 2

A = 1, B = 1, C = 0, D = − Amplitude: |A| = |1| = 1 & & & & & 2π & & 2π & Period: && && = && && = 2π B 1

1 2

w

2p

x

364

Chapter 5: The Trigonometric Functions

Phase shift:

C 0 = =0 B 1

Translate the graph of y = cos x down

1 unit. 2

y y ! cos x " q

1

"2p

1 sin x − 4 3 1 A = , B = 1, C = 0, D = −4 3 & & &1& 1 Amplitude: |A| = && && = 3 3 & & & & & 2π & & 2π & Period: && && = && && = 2π B 1 C 0 Phase shift: = =0 B 1 Shrink the graph of y = sin x vertically by a factor of 3 and shift it down 4 units.

15. y =

p

"p

x

2p

"1

y 1

"2 "2p "w "p

13. y = 3 cos(x − π)

A = 3, B = 1, C = π, D = 0 Amplitude: |A| = |3| = 3 & & & & & 2π & & 2π & Period: && && = && && = 2π B 1

π C = =π B 1 Stretch the graph of y = cos x vertically by a factor of 3 and shift it π units to the right.

"2p

"p "q

y

y ! 3 cos (x " p)

3 2 1

w

y ! a sin x " 4

" ( ) 1 π 1 x+ = cos (x − (−π)) 2 2 2 π 1 1 A = 1, B = , C = − , D = 2 2 2 Amplitude: |A| = |1| = 1 & & && && & 2π & & 2π & Period: && && = & 1 & = 4π & & B

16. y = cos

!

q

"1 "2

p

2p

y 1

p

"p

x

"1

Stretch the graph of y = sin x horizontally by a factor of 4, reflect it across the x-axis, and then translate it up 1 unit. y 2

y ! cos "q x # q# 2p

"2p

C 0 = 1 =0 B 4

y ! "sin "~ x# # 1

17. y = − cos(−x) + 2

A = −1, B = −1, C = 0, D = 2

Amplitude: |A| = | − 1| = 1 & & & & & 2π & & 2π & & = 2π Period: && && = && B −1 &

0 C = =0 B −1 Reflect the graph of y = cos x across the y-axis and across the x-axis and translate it up 2 units. Phase shift:

1

"2p

x

π − C = 2 = −π Phase shift: 1 B 2 Stretch the graph of y = cos x horizontally by a factor of 2 and shift it left π units.

x

" 1 14. y = − sin x +1 4 1 A = −1, B = , C = 0, D = 1 4 Amplitude: |A| = | − 1| = 1 & & && && & 2π & & 2π & Period: && && = & 1 & = 8π & 4 & B

"4p

2p

2

!

Phase shift:

w

p

"5

Phase shift:

"w

q

"q "1 "2 "3

2p

4p

x

Exercise Set 5.6

365 y 3 2 1

"2p "w "p

q

"q "1

w

p

2p

x

y ! "cos ("x ) # 2

( ! ") 1 π 1 π 18. y = sin 2x − = sin 2 x − 2 4 2 8 π 1 A = , B = 2, C = , D = 0 2 4 & & &1& 1 Amplitude: |A| = && && = 2 2 & & & & & 2π & & 2π & Period: && && = && && = π B 2 π π C = 4 = Phase shift: B 2 8 Shrink the graph of y = sin x horizontally by a factor of π 2; shrink it vertically, also by a factor of 2; and shift it 8 units to the right. !

"

y 1 "2p

y ! q sin "2x " d # p

"p

2p

x

! " ( ! ! "") 1 π 1 π 21. y = − sin 2x + = − sin 2 x − − 2 2 2 4 1 π A = − , B = 2, C = − , D = 0 2 &2 & & 1& 1 Amplitude: |A| = && − && = 2 2 & & & & & 2π & & 2π & Period: && && = && && = π B 2 π − π C = 2 =− Phase shift: B 2 4 ( ! ") π 22. y = −3 cos(4x − π) + 2 = −3 cos 4 x − +2 4 A = −3, B = 4, C = π, D = 2 Amplitude: |A| = | − 3| = 3 & & & & & 2π & & 2π & π Period: && && = && && = B 4 2 Phase shift:

π C = B 4

") ( ! 3 +2 23. y = 2 + 3 cos(πx − 3) = 3 cos π x − π A = 3, B = π, C = 3, D = 2 Amplitude: |A| = |3| = 3 & & & & & 2π & & 2π & Period: && && = && && = 2 B π

3 C = B π ! ) " ( π π π 24. y = 5−2 cos x+ = −2 cos (x−(−1)) +5 2 2 2 π π A = −2, B = , C = − , D = 5 2 2 Amplitude: |A| = | − 2| = 2 & & && && & 2π & & 2π & Period: && && = & π & = 4 & & B Phase shift:

"1

" ( ) π 1 1 x− = 2 cos (x − π) 2 2 2 π 1 A = 2, B = , C = , D = 0 2 2 Amplitude: |A| = |2| = 2 & & && && & 2π & & 2π & Period: && && = & 1 & = 4π & 2 & B

19. y = 2 cos

!

π C Phase shift: = 2 =π 1 B 2 " ( ! ! "") ! π 1 π 1 x+ = 4 sin x− − 20. y = 4 sin 4 8 4 2 π 1 A = 4, B = , C = − , D = 0 4 8 Amplitude: |A| = |4| = 4 & & && && & 2π & & 2π & Period: && && = & 1 & = 8π & & B 4

π − π C = 8 =− Phase shift: 1 B 2 4

2

π − C = π2 = −1 Phase shift: B 2

1 25. y = − cos(2πx) + 2 2 1 A = − , B = 2π, C = 0, D = 2 2 & & & 1& 1 Amplitude: |A| = && − && = 2 2 & & & & & 2π & & 2π & Period: && && = && && = 1 B 2π Phase shift:

C 0 = =0 B 2π

") ( ! π −2 26. y = −2 sin(−2x+π) − 2 = −2 sin − 2 x− 2 A = −2, B = −2, C = −π, D = −2 Amplitude: |A| = | − 2| = 2

366

Chapter 5: The Trigonometric Functions & & & & & 2π & & 2π & &=π Period: && && = && B −2 &

C −π π = = B −2 2 " ! ( ) 1 1 π 1 1 + = − sin (x − π) + 27. y = − sin x− 2 2 2 2 2 π 1 1 A = −1, B = , C = , D = 2 2 2 Amplitude: |A| = | − 1| = 1 & & && && & 2π & & 2π & Period: && && = & 1 & = 4π & 2 & B Phase shift:

π C = 2 =π Phase shift: 1 B 2

1 cos(−3x) + 1 3 1 A = , B = −3, C = 0, D = 1 3 & & &1& 1 Amplitude: |A| = && && = 3 3 & & & & & 2π & & 2π & 2π &= Period: && && = && B −3 & 3 0 C = =0 Phase shift: B −3

28. y =

29. y = cos(−2πx) + 2

A = 1, B = −2π, C = 0, D = 2 Amplitude: |A| = |1| = 1 & & & & & 2π & & 2π & &=1 Period: && && = && B −2π &

C 0 = =0 B −2π ( ! ! "") 1 1 1 30. y = sin(2πx + π) = sin 2π x − − 2 2 2 1 A = , B = 2π, C = −π, D = 0 2 & & &1& 1 Amplitude: |A| = && && = 2 2 & & & & & 2π & & 2π & Period: && && = && && = 1 B 2π Phase shift:

Phase shift:

−π 1 C = =− B 2π 2

( ! ") 1 4 1 31. y = − cos(πx − 4) = − cos π x − 4 4 π 1 A = − , B = π, C = 4, D = 0 4 & & & 1& 1 & Amplitude: |A| = & − && = 4 4 & & & & & 2π & & 2π & Period: && && = && && = 2 B π Phase shift:

4 C = B π

( ! ! "") 1 32. y = 2 sin(2πx + 1) = 2 sin 2π x − − 2π A = 2, B = 2π, C = −1, D = 0 Amplitude: |A| = |2| = 2 & & & & & 2π & & 2π & Period: && && = && && = 1 B 2π Phase shift:

−1 1 C = =− B 2π 2π

33. y = − cos 2x

Shrink the graph of y = cos x horizontally by a factor of 2 and reflect it across the x-axis. Graph (b) is the correct choice.

1 sin x − 2 2 Shrink the graph of y = sin x vertically by a factor of 2 and translate it down 2 units. Graph (e) is the correct choice. " ( ! ") ! π π = 2 cos x − − 35. y = 2 cos x + 2 2 Stretch the graph of y = cos x vertically by a factor of 2 and translate it left π/2 units. Graph (h) is the correct choice.

34. y =

1 36. y = −3 sin x − 1 2 Stretch the graph of y = sin x horizontally by a factor of 2, stretch it vertically by a factor of 3, reflect it across the x-axis, and translate it down 1 unit. Graph (d) is the correct choice. 37. y = sin(x − π) − 2

Translate the graph of y = sin x right π units and down 2 units. Graph (a) is the correct choice. ! " π 1 38. y = − cos x − 2 4 Shrink the graph of y = cos x vertically by a factor of 2, reflect it across the x-axis, and translate it to the right π/4 units. Graph (g) is the correct choice. 1 sin 3x 3 Shrink the graph of y = sin x horizontally and vertically, both by a factor of 3. Graph (f) is the correct choice. " ! π 40. y = cos x − 2

39. y =

Translate the graph of y = cos x to the right π/2 units. Graph (c) is the correct choice. 41. This graph has the same shape as y = cos x but with an 1 amplitude of and shifted up one unit. The equation is 2 1 y = cos x + 1. 2 42. This graph has the same shape as y = sin x but with a π and amplitude of 2. The equation is phase shift of 2" ! π . y = 2 sin x − 2

Exercise Set 5.6

367 y

43. This graph has the same shape as y = cos x but with a π phase shift of − and also a shift of 2 units down. The 2 ! " π − 2. equation is y = cos x + 2 44. This graph has the same shape as y = sin x but it is shrunk horizontally by a factor of 2 and reflected across the x-axis. The equation is y = − sin 2x.

2 1 "2p

p

"p "2

50.

x

y ! sin x " cos x

y 4

45. y = 2 cos x + cos 2x Graph y = 2 cos x and y = cos 2x on the same set of axes. Then graphically add some ordinates to obtain points on the graph of y = 2 cos x + cos 2x.

2p

y ! 3 cos x " sin x

2 1 "2p

"p

y 4

"1 "2 "3 "4

p

2p

x

51. y = 3 cos x + sin 2x "2p

"p

"1 "2 "3

p

2p

x

Graph y = 3 cos x and y = sin 2x on the same set of axes. Then graphically add some ordinates to obtain points on the graph of y = 3 cos x + sin 2x.

y ! 2 cos x # cos 2x

y

46.

4

y

y ! 3 cos x # sin 2x

4 1 2 1 "2p

"p

"1 "2 "3 "4

"2p p

2p

p

"p

y ! 3 cos x # cos 3x

52.

Graph y = sin x and y = cos 2x on the same set of axes. Then graphically add some ordinates to obtain points on the graph of y = sin x + cos 2x. 2 "2p

"p

48.

"1 "2

y

"2p

p

"p

2p

p

"1

2p

x

"3 "4

x

This function is the product of the functions g(x) = e−x/2 and h(x) = cos x. To find function values we can multiply ordinates of these functions. Also, note that for any real number x,

y ! 2 sin x # cos 2x

−1 ≤ cos x ≤ 1.

2p

x

"2 "3 "4

49. y = sin x − cos x

"2p

y ! 3 sin x " cos 2x

53. f (x) = e−x/2 cos x

p

"p

4 3 2 1

y ! sin x # cos 2x

y 4 3 2 1

x

"2 "3 "4

x

47. y = sin x + cos 2x

y

2p

Graph y = sin x and y = − cos x on the same set of axes. Then graphically add some ordinates to obtain points on the graph of y = sin x − cos x.

Since e−x/2 > 0 for all values of x, we can multiply to obtain −e−x/2 ≤ e−x/2 cos x ≤ e−x/2 .

This inequality tells us that f is constrained between the graphs of y = −e−x/2 and y = e−x/2 . We graph these functions using dashed lines. We see that the function intersects the x-axis only for values of x for which cos x = π 0. These values are + kπ, k an integer. We mark these 2 points on the graph. Then we can use a calculator to compute other function values.

368

Chapter 5: The Trigonometric Functions y y" f (x) " e−x/2 cos x

!9 !8 !7 !6 !5 !4 !3 !2 !1 !10 !20 y " !e−x/2 !30 !40 !50 !60

54.

1 2 3

This function is the product of the functions g(x) = x and h(x) = sin x. To find function values we can multiply ordinates of these functions. Also, note that for any real number x,

x

−1 ≤ sin x ≤ 1.

If we multiply this inequality by a negative value of x, we have −x ≥ x sin x ≥ x, or x ≤ x sin x ≤ −x.

y

y " e−0.4x f (x) "

57. f (x) = x sin x

60 50 40 30 20 10

e−x/2

For nonnegative values of x, we have

10 8 6 4 2

e−0.4x sin x

!6 !5 !4 !3 !2 !1 !2 !4 !6 !8 !10 y " !e−0.4x

−x ≤ x sin x ≤ x.

1 2 3 4

In either case we see that f is constrained between the graphs of y = −x and y = x. We start by graphing these functions using dashed lines. Also observe that f intersects the x-axis only for x = 0 and for those values of x for which sin x = 0. These values are kπ, k an integer, and they include 0. Mark these points on the graph. Then use a calculator to compute other function values.

x

55. f (x) = 0.6x2 cos x This function can be thought of as the product of the functions g(x) = 0.6x2 and h(x) = cos x. To find function values we can multiply ordinates of these functions. Also, note that for any real number x,

y " !x

−1 ≤ cos x ≤ 1.

!10!8 !6 !4 !2!2

Since 0.6x2 ≥ 0 for all values of x, we can multiply to obtain

y"x

2 4 6 8 10

x

!4 !6 !8 !10

−0.6x2 ≤ 0.6x2 cos x ≤ 0.6x2 .

This inequality tells us that f is constrained between the graphs of y = −0.6x2 and y = 0.6x2 . We graph these functions using dashed lines. We see that the function intersects the x-axis only for x = 0 and those values of x π for which cos x = 0. The latter values are + kπ, k an 2 integer. We mark these points on the graph. Then we can use a calculator to compute other function values.

y 10 8 6 4 2

f(x) " x sin x

58.

y " !x

y

y"x

10 8 6 4 2 !10!8 !6 !4 !2!2

2 4 6 8 10

x

!4

y

y " 0.6x2

!6 !8 !10

f(x) " 0.6x2 cos x

50 40 30 20 10

!9 !8 !7 !6 !5 !4 !3 !2 !1 !10

f(x) " #x| cos x 1 2 3 4 5 6 7 8 9

x

!20

The function is the product of the functions g(x) = 2−x and h(x) = sin x. To find function values we can multiply ordinates of these functions. Also, note that for any real number x,

!30 !40 !50

y " !0.6x2

56.

y " e −x/4

f (x) " e −x/4 sin x

y

−1 ≤ sin x ≤ 1.

5 4 3 2 1

!5 !4 !3 !2 !1 !1 !2 !3 !4 !5 y " −e −x/4

59. f (x) = 2−x sin x

Since 2−x > 0 for all values of x, we can multiply to obtain 1 2 3 4 5

x

−2−x ≤ 2−x sin x ≤ 2−x .

We see that f is constrained between the graphs of y = −2−x and y = 2−x . We start by graphing these functions using dashed lines. Also observe that f intersects the xaxis only for those values of x for which sin x = 0. These values are kπ, k an integer. We mark those points on the graph. Then use a calculator to compute other function values.

Exercise Set 5.6

369 sin x and use the Zero feature to find the zeros x in the interval [−12, 12]. They are approximately −9.42, −6.28, −3.14, 3.14, 6.28, and 9.42.

y

y ! 2"x

69. Graph y =

50 40 30 20 10

"9 "8 "7 "6 "5 "4 "3 "2 "1 "10

1

cos x − 1 and use the Zero feature to find the x zeros in the interval [−12, 12]. They are approximately −6.28 and 6.28. (Note that although from the graph it might appear that there is also a zero at x = 0, the function is not defined for this value of x.)

x

70. Graph y =

"20 "30 "40 "50 "x y ! "2

f(x) ! 2"x sin x

60.

71. Graph y = x3 sin x and use the Zero feature to find the zeros in the interval [−12, 12]. They are approximately −3.14, 0, and 3.14.

y ! 2"x y 10 8 6 4 2

f(x) ! 2"x cos x

"5 "4 "3 "2 "1 "2 "4 "6 "8 "10

1 2 3 4 5

x

y ! "2"x

sin2 x and use the Zero feature to find the zeros x in the interval [−12, 12]. They are approximately −3.14 and 3.14.

72. Graph y =

73. a)

p

105

61. See the answer section in the text. 62.

y ! 101.6 # 3 sin " 8 x#

y ! "x " sin x 10

"10

63. See the answer section in the text. 64.

0 97

2p

"2p

b) The maximum value occurs on day 4 when the sine function takes its maximum value, 1. It is 101.6◦ + 3◦ · 1, or 104.6◦ . The minimum value occurs on day 12 when the sine function takes its minimum value, −1. It is 101.6◦ + 3◦ (−1), or 98.6◦ .

y ! "(cos x " x) 10

74. a)

y ! 10 "1 " cos A x#

2p

"2p

12

25

"10

65. See the answer section in the text. 0

66.

y ! cos 3x # sin 3x 4

2p

"2p

67. See the answer section in the text. y ! 7.5 cos x # sin 2x 10

2p

"2p

"10

12

2π b) π = 12 6 c) $0 in July (at t = 0 and t = 12) d) $20,000 in January (at t = 6)

"4

68.

0

75. The constants B, C, and D translate the graphs and the constants A and B stretch or shrink the graphs. See the chart on page 516 of the text for a complete description of the effect of each constant. 76. The denominator B in the phase shift C/B serves to shrink or stretch the translation of C units by the same factor as the horizontal shrinking or stretching of the period. Thus, the translation must be done after the horizontal shrinking or stretching. For example, consider y = sin(2x − π). The phase shift of this function is π/2. First translate the graph of y = sin x to the right π/2 units and then shrink it

370

Chapter 5: The Trigonometric Functions

horizontally by a factor of 2. Compare this graph with the one formed by first shrinking the graph of y = sin x horizontally by a factor of 2 and then translating it to the right π/2 units. The graphs differ; the second one is correct. x+4 77. f (x) = x The function is a quotient of two polynomials, so it is rational. 78. The variable in the function is in a logarithm, so the function is logarithmic. 79. This is a polynomial function with degree 4, so it is quartic. 1 3 x + y = −5, can be written in the equivalent form 4 2 3 y = − x − 10, so it is linear. 2 81. The variable in f (x) = sin x − 3 is in a trigonometric function, so this is a trigonometric function.

90. y = tan(−x) Reflect the graph of y = tan x across the y-axis. y 3 2 1 "2p "w "p

84. The variable is in trigonometric functions, so the function is trigonometric. 85. This is a polynomial function with degree 3, so it is cubic. 86. The variable is in an exponent, so the function is exponential. ") ( ! π +6 87. y = 2 cos 3 x − 2 The maximum value of the cosine function is 1, so the maximum value of the given function is 2 · 1 + 6, or 8. The minimum value of the cosine function is −1, so the minimum value of the given function is 2(−1) + 6, or 4.

1 sin(2x − 6π) − 4 2 The maximum value of the sine function is 1, so the max7 1 imum value of the given function is · 1 − 4, or − , or 2 2 1 −3 . 2 The minimum value of the sine function is −1, so the min9 1 imum value of the given function is (−1) − 4, or − , or 2 2 1 −4 . 2 89. y = − tan x

88. y =

Reflect the graph of y = tan x across the x-axis.

"2p "w "p

Translate the graph of y = cot x down 2 units. y 3 2 1 "2p "w "p

y ! "2 # cot x

q

"q "1 "2 "3

y y ! "w csc x

3 2 1

"2p "w "p

2p

x

y ! "tan x

q

"q "1 "2 "3

p

w

2p

x

1 93. y = 2 tan x 2 Stretch the graph of y = tan x horizontally and vertically, both by a factor of 2. y 3

y ! 2 tan qx

2 1 "2p "p "q

q p w 2p x

"2 "3

94. y = cot 2x Shrink the graph of y = cot x horizontally by a factor of 2. 3 2 1

w

x

2p

3 Stretch the graph of y = csc x by a factor of and reflect 2 it across the x-axis.

y

p

w

p

3 92. y = − csc x 2

3 2 1 q

x

2p

y ! tan ("x)

y

"q "1 "2 "3

w

p

91. y = −2 + cot x

80.

82. The variable is in an exponent, so the function is exponential. 2 2 83. y = is equivalent to y = 0x + . Since the function can 5 5 be written in the form y = mx + b, it is linear.

q

"q "1 "2 "3

"2p "w "p

"q "1

y ! cot 2x

q

p

w

2p

x

Chapter 5 Review Exercises

371

95. y = 2 sec(x − π)

Phase shift:

Stretch the graph of y = sec x vertically by a factor of 2 and translate it to the right π units. y 3 2 1 "2p "w "p

q

p

w

2p

x

Phase shift:

"3

" ( ! ! "") 1 π 1 π x+ = 4 tan x− − 4 8 4 2 Stretch the graph of y = tan x horizontally and vertically, both by a factor of 4. Then translate it π/2 units to the left.

96. y = 4 tan

!

y ! 4 tan "~ x # 8 # p

y 5 4 3 2 1 p

2p

3p

4p

5p

x

" ( ! ") 3π 1 3π 1 x− = 2 csc x− 97. y = 2 csc 2 4 2 2 Stretch the graph of y = csc x horizontally and vertically, both by a factor of 2. Then translate it to the right 3π/2 units. !

y 10 8 6 4 2

) π (x − (−1)) . 4

C = −1 B

101. As t increases, e−0.8t decreases, and hence 6e−0.8t cos(6πt) decreases in the long run and approaches 0. Then the spring would be 4 inches from the ceiling when it stops bobbing. 102. The values of t for which the function is not defined are those values for which the beam is parallel to the wall.

w x

") ( ! π 98. y = 4 sec(2x − π) = 4 sec 2 x − 2 Shrink the graph of y = sec x horizontally by a factor of 2; stretch is vertically by a factor of 4; translate it to the right π/2 units. y 3 2 1 "q "1

1. Given that (−a, b) is a point on the unit circle and θ is in the second quadrant, cos θ = −a. Thus, the given statement is false. 2. It is true that the lengths of corresponding sides in similar triangles are in the same ratio. 3. 300◦ = 300◦ · is true.

π radians ≈ 5.24 radians, so the statement 180◦

4. The statement is true. See the figure on page 464 of the text. 1 1 and the amplitude of 5. The amplitude of y = sin x is 2 2 1 y = sin x is 1. Thus, the given statement is false. 2

y ! 2 csc "qx " f#

"w "p

(

Chapter 5 Review Exercises

"3p "2p "p "1 "2 "3

"r "w "q

100. Write the function as y = 3 sin Amplitude: |A| = |3| = 3 & & && && & 2π & & 2π & Period: && && = & π & = 8 & 4 & B

y ! 2 sec (x " p)

"q "1

C = 10 B

q

y ! 4 sec (2x " p)

99. Amplitude: |A| = |3000| = 3000 & & && && & 2π & & 2π & Period: && && = & π & = 90 & 45 & B

p

w

2p

x

9 9 4 π is π − π, or π. The comple6. The supplement of 13 13 13 π π π 4 π π ment of is − , or . Since π < , the statement 6 2 6 3 13 3 is false. 7. We use the definitions. √ 3 3 73 opp =√ = sin θ = hyp 73 73 √ 8 8 73 adj =√ = cos θ = hyp 73 73 opp 3 tan θ = = adj 8 √ 73 hyp = csc θ = opp 3 √ hyp 73 sec θ = = adj 8 8 adj = cot θ = opp 3

372

Chapter 5: The Trigonometric Functions √

91 10 √ a2 = 102 − ( 91)2 = 100 − 91 = 9

8. sin β =

17. Enter 22.27◦ on a calculator and use the DMS feature: 22.27◦ = 22◦ 16" 12""

a=3

18. 47◦ 33" 27"" ≈ 47.56◦

cos β =

19. tan 2184◦ ≈ 0.4452

3 10 √ 91 tan β = 3 √ 10 91 10 csc β = √ = 91 91 10 sec β = 3 √ 3 3 91 √ cot β = = 91 91 √ 1 2 (See the triangle on page 438 in the 9. cos 45◦ = √ = 2 2 text.) √ 3 1 (See the triangle on page 438 in the 10. cot 60◦ = √ = 3 3 text.) 11. 495◦ − 360◦ = 135◦ , so 495◦ and 135◦ are coterminal angles. Since 180◦ − 135◦ = 45◦ , the reference angle is 45◦ . Note that 495◦ is a second-quadrant √ angle, so the cosine 2 is negative. Recalling that cos 45◦ = , we have 2 √ 2 . cos 495◦ = − 2 12. Since 180◦ − 150◦ = 30◦ , the reference angle is 30◦ . Note that 150◦ is a second-quadrant angle, so the sine is positive. 1 Recalling that sin 30◦ = , we have 2 1 ◦ sin 150 = . 2 13. A point on the terminal side of −270◦ is (0, 1). Thus 1 sec (−270◦ ) = , which is not defined. 0

20. sec 27.9◦ ≈ 1.1315 21. cos 18◦ 13" 42"" ≈ 0.9498 22. sin 245◦ 24" ≈ −0.9092 23. cot (−33.2◦ ) ≈ −1.5282 24. sin 556.13◦ ≈ −0.2778 25. cos θ = −0.9041, 180◦ < θ < 270◦

Ignore the fact that cos θ is negative and use a calculator to find that the reference angle is approximately 25.3◦ . Since θ is a third-quadrant angle, we find θ by adding 180◦ and 25.3◦ : 180◦ + 25.3◦ = 205.3◦ . Thus, θ ≈ 205.3◦

26. tan θ = 1.0799, 0◦ < θ < 90◦

27. 28.

29.

30.

31. Since 59.1◦ and 30.9◦ are complementary angles, we have sin 30.9◦ = cos 59.1◦ ≈ 0.5135,

14. −600◦ + 2 · 360◦ = 120◦ , so −600◦ and 120◦ are coterminal angles. Since 180◦ − 120◦ = 60◦ , the reference angle is 60◦ . Note that −600◦ is a second-quadrant angle, so the √ tangent is negative. Recalling that tan 60◦ = 3, we have √ tan (−600◦ ) = − 3. √ 2 2 3 ◦ √ 15. csc 60 = = (See the triangle on page 438 in the 3 3 text.) 16. −45◦ + 360◦ = 315◦ , so −45◦ and 315◦ are coterminal angles. Since 360◦ − 315◦ = 45◦ , the reference angle is 45◦ . Note that −45◦ is a fourth-quadrant angle, so the cotangent is negative. Recalling that cot 45◦ = 1, we have cot (−45 ) = −1. ◦

Since θ is a first-quadrant angle, use a calculator to find that θ ≈ 47.2◦ . √ 3 , so θ = 60◦ . (See the triangle on page 438 in sin θ = 2 the text.) √ tan θ = 3, so θ = 60◦ . (See the triangle on page 438 in the text.) √ 2 cos θ = , so θ = 45◦ . (See the triangle on page 438 in 2 the text.) √ 2 2 3 sec θ = , or √ , so θ = 30◦ . (See the triangle on page 3 3 438 in the text.)

cos 30.9◦ = sin 59.1◦ ≈ 0.8581, 1 ≈ 0.5985, tan 30.9◦ = cot 59.1◦ ≈ 1.6709 1 csc 30.9◦ = sec 59.1◦ ≈ ≈ 1.9474, 0.5135 1 sec 30.9◦ = csc 59.1◦ ≈ ≈ 1.1654, 0.8581 ◦ ◦ cot 30.9 = tan 59.1 ≈ 1.6709. 32.

7.3 8.6 A ≈ 58.1◦

sin A =

B ≈ 90◦ − 58.1◦ = 31.9◦ b cos 58.1◦ = 8.6 b ≈ 4.5

Chapter 5 Review Exercises 33.

B c

A

51.17o

30.5

C

b

To solve this triangle, find A, b, and c. A ≈ 90◦ − 51.17◦ = 38.83◦ 30.5 cos 51.17◦ = c c ≈ 48.6 b 30.5 b ≈ 37.9

tan 51.17◦ =

34.

N N 57o 23' E 734 m

a b

b 734 b ≈ 618 a cos 57◦ 23" = 734 a ≈ 396 sin 57◦ 23" =

373 41. Complement: 90◦ − 13.4◦ = 76.6◦

Supplement: 180◦ − 13.4◦ = 166.6◦

π π π − = 2 6 3 π 5π Supplement: π − = 6 6 # √ 2 2 43. r = (−2) + 3 = 13 √ 3 3 13 y sin θ = = √ = r 13 13 √ x 2 2 13 cos θ = = − √ = − r 13 13 3 y tan θ = = − x 2 √ r 13 csc θ = = y 3 √ 13 r sec θ = = − x 2 2 x cot θ = = − y 3 42. Complement:

2 44. tan θ = √ , θ is in quadrant III. 5 $√ √ √ r = ( 5)2 + 22 = 5 + 4 = 9 = 3 y

P ≈ 396 + 618 + 734 = 1748 m 35. Let d be the distance from the observer to the mural. 6 tan 13◦ = d 6 d≈ ≈ 26 tan 13◦ Let h be the distance from eye-level to the top of the mural. h h sin 17◦ = ≈ d 26 h≈8 The mural is approximately 6 + 8, or14 ft tall.

36. 90◦ < 142◦ 11" 5"" < 180◦ , so the angle is in quadrant II. 37. −635.2◦ + 2 · 360◦ = 84.8◦ , so −635.2◦ and 84.8◦ are coterminal angles. Thus −635.2◦ is in quadrant I. 38. −392◦ + 2 · 360◦ = 328◦ , so −392◦ is in quadrant IV. 39.

65◦ + 360◦ = 425◦ 65◦ − 360◦ = −295◦

40.

7π π − 2π = 3 3 5π 7π − 4π = − 3 3

- 5 -2

θ

x

3 (- 5, -2)

sin θ = −

2 3 √

5 cos θ = − 3 √ 5 cot θ = 2 √ 3 5 3 sec θ = − √ = − 5 5 3 csc θ = − 2 45. Use the formula d = rt to find the airplane’s distance d from Minneapolis. d = (530 mph)(3.5 hr) = 1855 mi We make a drawing. Let D = the distance of the plane south of Minneapolis.

374

Chapter 5: The Trigonometric Functions N

57.

y 3 4 (- –, –) 5 5

Minneapolis

3 4 (–, –) 5 5

160o W

x

E D

θ 1855 mi 3 4 (- –, - –) 5 5

3 4 (–, - –) 5 5

S

The reference angle θ is 20◦ . D cos 20◦ = 1855 1855 cos 20◦ = D 1743 ≈ D

The plane is about 1743 mi south of Minneapolis.

Reflection across x-axis:

!

3 4 , 5 5

"

" 3 4 − ,− 5 5 " ! 3 4 Reflection across the origin: − , 5 5 Reflection across y-axis:

!

58. The coordinates of the point on the unit circle determined by π are (−1, 0). Thus, cos π = −1.

46.

59. The coordinates of the point on the unit circle √ determined √ √ − 2 5π 5π are (− 2, − 2). Thus, tan = √ = 1. by 4 4 − 2

π radians 121 π ≈ 2.53 = ◦ 180 150 π radians π −30◦ = −30◦ · = − ≈ −0.52 180◦ 6 3π 180◦ 3π = · = 270◦ 2 2 π radians 180◦ 3=3· ≈ 171.89◦ π radians 180◦ −4.5 = −4.5 · ≈ −257.83◦ π radians 180◦ = 1980◦ 11π = 11π · π radians ! " π 7π s = rθ = (7 cm) cm ≈ 5.5 cm = 4 4

47. 145.2◦ = 145.2◦ · 48. 49. 50. 51. 52. 53.

54. θ =

18 s = = 2.25 radians ≈ 129◦ r 8

55. One revolution: 2π(7) = 14π ft. "! " ! 60 sec 14π ft ≈ 37.7 ft/min Linear speed: V = 70 sec 1 min (Answers may vary slightly due to rounding differences.) 56. The radius of the wheel is 7 in. 55 mi 5280 ft · v 1 mi w = = 1 hr 1 ft r 7 in. · 12 in. =

1 12 in. 55 mi 5280 ft · · · 1 hr 1 mi 7 in. 1 ft

≈ 497, 829 radians/hr

60. The coordinates of √the √ determined ! " point on the unit circle 1 3 5π 3 5π are ,− . Thus sin =− . by 3 2 2 3 2 61. The coordinates on the unit √the point ! of " ! circle " determined 3 1 7π 1 7π by − are − , . Thus, sin − = . 6 2 2 6 2 62. The coordinates √ ! √ of"the point on the unit circle determined 1 3 1 π 1/2 3 π =√ = are , . Thus, tan = √ . by 6 2 2 6 3 3/2 3 63. The coordinates of the point on the unit circle determined by −13π are (−1, 0). Thus, cos(−13π) = −1. 64. Use a calculator set in radian mode. sin 24 ≈ −0.9056 65. Use a calculator set in radian mode. cos(−75) ≈ 0.9218 66. Use a calculator set in radian mode. cot 16π is not defined. 67. Use a calculator set in radian mode. 3π ≈ 4.3813 tan 7 68. Use a calculator set in radian mode. sec 14.3 ≈ −6.1685 69. Use a calculator set in radian mode. " ! π ≈ 0.8090 cos − 5

Chapter 5 Review Exercises

375

70.

73.

Function

I

II

III

IV

sine

+

+

cosine

+

−

−

−

−

+

tangent + − + − ! " ( ! ") π π 74. y = sin x + = sin x − − 2 2 π A = 1, B = 1, C = − , D = 0 2 Amplitude: |A| = |1| = 1 & & & & & 2π & & 2π & Period: && && = && && = 2π B 1 π − C π Phase shift: = 2 =− B 1 2

! " ( ! ") π 1 π 1 = cos 2 x − +3 75. y = 3 + cos 2x − 2 2 2 4 π 1 A = , B = 2, C = , D = 3 2 & 2& &1& 1 Amplitude: |A| = && && = 2 2 & & & & & 2π & & 2π & Period: && && = && && = π B 2 π C π Phase shift: = 2 = B 2 4

76. y = cos 2x Shrink the graph of y = cos x horizontally by a factor of 2. Graph (d) is the correct choice.

71. Period of sin, cos, sec, csc: 2π Period of tan, cot: π 72.

Function

Domain

Range

sine

(−∞, ∞)

[−1, 1]

cosine tangent

(−∞, ∞) ' π x|x &= + kπ, k an integer 2

%

1 sin x + 1 2 Shrink the graph of y = sin x vertically by a factor of 2 and translate it up 1 unit. Graph (a) is the correct choice.

77. y =

[−1, 1] (−∞, ∞)

1 x−3 2 Stretch the graph of y = sin x horizontally by a factor of 2, stretch it vertically by a factor of 2, reflect it across the x-axis, and translate it down 3 units. Graph (c) is the correct choice.

78. y = −2 sin

376

Chapter 5: The Trigonometric Functions " π 2 Reflect the graph of y = cos x across the x-axis, and transπ late it to the right units. Graph (b) is the correct choice. 2

79. y = − cos

!

x−

80.

85. The graph of the cosine function is shaped like a continuous wave, with “high” points at y = 1 and “low” points at y = −1. The maximum value of the cosine function is 1, and it occurs at all points where x = 2kπ, k an integer. 86. No; sin x is never greater than 1. 87.

81. f (x) = e

−0.7x

Domain: (−∞, ∞); range: [−3, 3], period:

cos x

This function is the product of the functions g(x) = e−0.7x and h(x) = cos x. To find function values we can multiply ordinates of these functions. Also, note that for any real number x, −1 ≤ cos x ≤ 1.

Since e−0.7x > 0 for all values of x, we can multiply to obtain −e−0.7x ≤ e−0.7x cos x ≤ e−0.7x .

This inequality tells us that f is constrained between the graphs of y = −e−0.7x and y = e−0.7x . We graph these functions using dashed lines. We see that the function intersects the x-axis only for values of x for which cos x = π 0. These values are + kπ, k an integer. We mark these 2 points on the graph. Then use a calculator to compute other function values. f(x) ! e"0.7x cos x

"9 "8 "7 "6 "5 "4 "3 "2 "1 "20

88. To generate y2 from y1 , stretch the graph of y1 vertically π by a factor of 2, translate it left units, and translate it 2 down 2 units. Thus " ! π y2 = 2 sin x + − 2. 2 89. y = log (cos x) cos x must be " domain consists of ! greater than 0, so the π π the intervals − + 2kπ, + 2kπ , k an integer. 2 2 90. Use a calculator to find the acute angle θ for which sin θ = 0.6144. Store this value. This is the reference angle for x. Keep in mind that x is a second-quadrant angle. cos x = − cos θ ≈ −0.7890

tan x = − tan θ ≈ −0.7787 cot x = − cot θ ≈ −1.2842

y

sec x = − sec θ ≈ −1.2674

100 80 60 40 20

csc x = csc θ ≈ 1.6276

1

x

"40

Chapter 5 Test

"60 "80 "100

82. The correct answer is C. (See the figure in Example 1(c) on page 493 of the text.) 83. The cosine function is defined for all real numbers, so answer B is correct. 84. Both degrees and radians are units of angle measure. A 1 of one complete positive revodegree is defined to be 360 lution. Degree notation has been in use since Babylonian times. Radians are defined in terms of intercepted arc length on a circle, with one radian being the measure of the angle for which the arc length equals the radius. There are 2π radians in one complete revolution.

2π = 4π 1/2

1. We use the definitions. √ opp 4 4 65 = √ , or sin θ = hyp 65 65 √ 7 65 adj 7 = √ , or cos θ = hyp 65 65 opp 4 = tan θ = adj 7 √ 65 hyp = csc θ = opp 4 √ hyp 65 = sec θ = adj 7 7 adj = cot θ = opp 4

Chapter 5 Test

377

2. Since 180◦ − 120◦ = 60◦ , the reference angle is 60◦ . Note that 120◦ is a second-quadrant√angle, so the sine is posi3 tive. Recalling that sin 60◦ = , we have 2 √ 3 sin 120◦ = . 2 3. −45◦ + 360◦ = 315◦ , so −45◦ and 315◦ are coterminal angles. Since 360◦ − 315◦ = 45◦ , the reference angle is 45◦ . Note that −45◦ is a fourth-quadrant angle, so the tangent is negative. Recalling that tan 45◦ = 1, we have tan (−45◦ ) = −1.

4. A point on the terminal side of 3π is (−1, 0). Thus −1 cos 3π = = −1. 1 5π π π 5π 5. − π = , so the reference angle is . Note that is a 4 4 4 4 third-quadrant angle, so the secant is negative. Recalling π √ that sec = 2, we have 4 √ 5π = − 2. sec 4

45.1 c c ≈ 55.7

sin 54.1◦ =

14. Answers may vary. 112◦ + 360◦ = 472◦ 112◦ − 360◦ = −248◦ 15. π −

5π π = 6 6

4 16. sin θ = − √ , θ is in quadrant IV. 41 √ 2 2 x + (−4) = ( 41)2 x2 + 16 = 41 x2 = 25 x=5 y θ

-4 41

7. tan 526.4◦ ≈ −0.2419

(5, -4)

8. sin (−12 ) = −0.2079 ◦

cos

5π ≈ −5.7588 9 10. cos 76.07 ≈ 0.7827 1 11. sin θ = , so θ = 30◦ . (See the triangle on page 438 in the 2 text.) 9. sec

tan csc

12. Since 28.4◦ and 61.6◦ are complementary angles, we have sin 61.6◦ = cos 28.4◦ ≈ 0.8796, cos 61.6◦ = sin 28.4◦ ≈ 0.4756, 1 tan 61.6◦ = cot 28.4◦ ≈ ≈ 1.8495, 0.5407 1 ≈ 1.1369, csc 61.6◦ = sec 28.4◦ ≈ 0.8796 1 sec 61.6◦ = csc 28.4◦ ≈ ≈ 2.1026, 0.4756 ◦ ◦ cot 61.6 = tan 28.4 ≈ 0.5407.

13.

B c

A

35.9o 45.1

x

5

6. Enter 38◦ 27" 56"" on a calculator. The result is 38◦ 27" 56"" ≈ 38.47◦ .

a C

To solve this triangle, find B, a, and c. B = 90◦ − 35.9◦ = 54.1◦ a tan 35.9◦ = 45.1 a ≈ 32.6

sec cot

√ 5 5 41 x θ = = √ , or r 41 41 y 4 θ = =− x 5 √ 41 r θ = =− y 4 √ r 41 θ= = x 5 5 x θ = =− y 4

17. 210◦ = 210◦ ·

π radians 7π = 180◦ 6

3π 3π 180◦ = · = 135◦ 4 4 π radians ! " π 16π 19. s = rθ = (16 cm) = cm ≈ 16.755 cm 3 3

18.

20. Amplitude: |A| = | − 1| = 1 & & & & & 2π & & 2π & 21. Period: && && = && && = 2π B 1

π π C = 2 = B 1 2 " ! π +1 23. y = − sin x − 2 Reflect the graph of y = sin x across the x-axis, translate π units and up 1 unit. Graph (c) is the correct it left 2 choice. 22. Phase shift:

378

Chapter 5: The Trigonometric Functions

24. Let h be the height of the kite. h sin 65◦ = 490 h ≈ 444 ft 25. Use the formula d = rt to find the distance d from Flagstaff.

1 2 f(x) ! _ 2 x sin x

"10

N Flagstaff 115o

W

25o

E

300 mi 25o

D S

The reference angle θ is 115◦ − 90◦ , or 25◦ . This is also the measure of one of the angles in the triangle shown in the drawing above. D cos 25◦ = 300 300 cos 25◦ = D 272 ≈ D

The camper is about 272 mi east of Flagstaff. 26. One revolution: 2π(6 m) = 12π m "! " ! 12π m 1.5 revolutions Linear speed: V = 1 min 1 revolution = 18π m/min ≈ 56.55 m/min

1 2 x sin x 2 This function can be thought of as the product of the func1 tions g(x) = x2 and h(x) = sin x. To find function values 2 we can multiply ordinates of these functions. Also, note that for any real number x,

27. f (x) =

−1 ≤ sin x ≤ 1. 1 Since x2 ≥ 0 for all values of x, we can multiply to obtain 2 1 1 1 − x2 ≤ x2 sin x ≤ x2 . 2 2 2 This inequality tells us that f is constrained between the 1 1 graphs of y = − x2 and y = x2 . We graph these func2 2 tions using dashed lines. We see that the function intersects the x-axis only for x = 0 and for values of x for which sin x = 0. The latter values are kπ, k an integer. These values include 0. We mark these points on the graph of the function. Then use a calculator to compute other function values.

1 x2 y! _ 2

10 x

"10 "20 "30 "40 "50

d = (50 mph)(6 hr) = 300 mi We make a drawing. Let D = the distance of the camper east of Flagstaff.

y 50 40 30 20 10

1 2 y ! "_ 2x

−3 cos x cos ! x must be > 0, so the " domain consists of the intervals π π − + 2kπ, + 2kπ , k an integer. 2 2

28. f (x) = √

Chapter 6

Trigonometric Identities, Inverse Functions, and Equations 8.

Exercise Set 6.1 1. 2.

3.

4.

(sin x − cos x)(sin x + cos x)

= sin2 x − cos2 x

tan x(cos x − csc x) ! " 1 sin x cos x − = cos x sin x 1 = sin x − cos x = sin x − sec x cos y sin y(sec y + csc y) ! " 1 1 = cos y sin y + cos y sin y = sin y + cos y

cos x csc x 1 1 1 + sin x · + cos x · + = sin x · cos x sin x cos x 1 cos x · sin x = tan x + 1 + 1 + cot x = 2 + tan x + cot x

= cos x(sin x + cos x) 10. 11.

tan2 θ − cot2 θ

= (tan θ − cot θ)(tan θ + cot θ) sin4 x − cos4 x

= (sin2 x + cos2 x)(sin2 x − cos2 x)

= sin2 x − cos2 x 12.

= (sin x + cos x)(sin x − cos x) 4 sin2 y + 8 sin y + 4 = 4(sin2 y + 2 sin y + 1) = 4(sin y + 1)2

13. 14.

2 cos2 x + cos x − 3

= (2 cos x + 3)(cos x − 1) 3 cot2 β + 6 cot β + 3 = 3(cot2 β + 2 cot β + 1) = 3(cot β + 1)2

15.

(sin φ − cos φ)

= 1 − 2 sin φ cos φ

(sin2 φ + cos2 φ = 1)

16.

sin3 x + 27

1 − 125 tan3 s

17.

sin2 x cos x cos2 x sin x sin x sin x cos x = · cos x sin x cos x sin x = cos x = tan x

18.

30 sin3 x cos x 6 cos2 x sin x

(1 + tan2 x = sec2 x)

(sin x + csc x)(sin2 x + csc2 x − 1)

= sin3 x + sin x csc2 x − sin x + csc x sin2 x+

csc3 x − csc x 1 − sin x+ = sin3 x + sin x · sin2 x 1 1 · sin2 x + csc3 x − sin x sin x 1 1 = sin3 x + − sin x + sin x + csc3 x − sin x sin x = sin3 x + csc3 x

= (sin x + 3)(sin2 x − 3 sin x + 9) = (1 − 5 tan s)(1 + 5 tan s + 25 tan2 s)

(1 + tan x)2 = sec2 x + 2 tan x

sin x cos x + cos2 x

= (sin x)3 + 33

= 1 + 2 tan x + tan2 x 7.

9.

2

= sin2 φ − 2 sin φ cos φ + cos2 φ

6.

= cos2 t

(sin x + cos x)(sec x + csc x) = sin x sec x + sin x csc x + cos x sec x+

5.

(1 − sin t)(1 + sin t)

= 1 − sin2 t

=

5 sin2 x , or 5 tan x sin x cos x

380

Chapter 6: Trigonometric Identities, Inverse Functions, and Equations sin2 x + 2 sin x + 1 sin x + 1

19.

(sin x + 1)2 sin x + 1 = sin x + 1

=

=

(cos α + 1)(cos α − 1) cos α + 1 = cos α − 1 =

sec y tan2 y 1 · 3 tan y sec y tan2 y 1 = cot y 3 27. =

4 tan t sec t + 2 sec t 6 tan t sec t + 2 sec t

21.

2 sec t(2 tan t + 1) 2 sec t(3 tan t + 1) 2 tan t + 1 = 3 tan t + 1

=

csc(−x) − csc x = cot(−x) − cot x 1 sin = cos x x sin x

=

=

22.

=

28. =

sin x 1 · sin x cos x 1 = cos x = sec x =

sin4 x − cos4 x sin2 x − cos2 x

23.

(sin2 x + cos2 x)(sin2 x − cos2 x) = sin2 x − cos2 x =1 (sin2 x + cos2 x = 1) 4 cos3 x · sin2 x

24.

!

sin x 4 cos x

4 cos3 x sin2 x 16 sin2 x cos2 x cos x = 4

"2

=

25.

sec y tan2 y · sec y 3 tan3 y

=

cos2 α − 1 cos α + 1

20.

tan2 y 3 tan3 y ÷ sec y sec y

26.

5 cos φ sin2 φ − sin φ cos φ · sin2 φ sin2 φ − cos2 φ 5 cos φ sin φ(sin φ − cos φ) = sin2 φ(sin φ + cos φ)(sin φ − cos φ) 5 cos φ = sin φ(sin φ + cos φ) 5 cot φ = sin φ + cos φ

2 1 − sin2 s − cos2 s cos s − sin s 2 1 − sin2 s − cos2 s −(− cos s + sin s) 2 1 + 2 2 sin s − cos s sin s − cos x sin s + cos s 2 1 · + sin2 s − cos2 s sin s − cos s sin s + cos s 1 + 2 sin s + 2 cos s sin2 s − cos2 s "2 ! sin x 1 − cos x cos2 x tan2 x − sec2 x

= −1

(1 + tan2 x = sec2 x)

sin2 θ − 9 10 cos θ + 5 · 2 cos θ + 1 3 sin θ + 9

29. =

(sin θ + 3)(sin θ − 3) 5(2 cos θ + 1) · 2 cos θ + 1 3(sin θ + 3)

=

5(sin θ − 3) 3 9 cos2 α − 25 cos2 α − 1 · 2 cos α − 2 6 cos α − 10

30. =

(3 cos α + 5)(3 cos α − 5)(cos α + 1)(cos α − 1) 2(cos α − 1)(2)(3 cos α − 5)

31.

(3 cos α + 5)(cos α + 1) 4 √ √ 2 sin x cos x · cos x √ = sin2 x cos2 x

32.

= sin x cos x √ √ cos2 x sin x · sin x √ = cos2 x sin2 x

=

33.

34.

= cos x sin x √ √ cos α sin2 α − cos3 α √ √ = sin α cos α − cos α cos α √ = cos α(sin α − cos α) # tan2 x − 2 tan x sin x + sin2 x # = (tan x − sin x)2 = tan x − sin x

Exercise Set 6.1 35.

36.

37.

38.

39.

40.

41.

42.

√ sin y)( sin y + 1) √ √ = (1 − sin y)(1 + sin y) √ = 1 − sin y [( sin y)2 = sin y] √ √ √ cos θ( 2 cos θ + sin θ cos θ) √ √ = 2 cos2 θ + sin θ cos2 θ √ √ = cos θ 2 + cos θ sin θ √ √ = cos θ( 2 + sin θ) $ sin x cos x $ sin x cos x = · cos x cos x $ sin x cos x = cos2 x √ sin x cos x = cos $ x sin x (Note that could also be expressed cos x √ as tan x.) $ $ cos x cos x tan x = tan x tan2 x $ sin x cos x · cos x = tan x √ sin x = tan % $x cos2 y cot2 y 2 = · 2 2 2 2 sin y √ 2 cot y = 2 % % 1 − cos β 1 − cos2 β = 1 + cos β (1 + cos β)2 # sin2 β = 1 + cos β sin β = 1 + cos β $ $ cos x cos x cos x = · sin x sin x cos x $ cos2 x = sin x cos x cos x = √ sin x cos x $ $ sin x sin x sin x = · cot x cot x sin x % sin2 x (cot x · sin x = = cos x cos x · sin x = cos x) sin x sin x = √ cos x (1 −

√

381 43.

%

1 + sin y = 1 − sin y

% %

1 + sin y 1 + sin y · 1 − sin y 1 + sin y

(1 + sin y)2 1 − sin2 y 1 + sin y = # cos2 y 1 + sin y = cos y $ $ cos2 x cot2 x cot x = = √ 44. 2 2 2 sin x 2 # √ 45. a2 − x2 = a2 − (a sin θ)2 Substituting # 2 = a2 − a2 sin θ & = a2 (1 − sin2 θ) √ = a2 cos2 θ ! = a cos θ a > 0 and cos θ > 0 " π for 0 < θ < 2 √ 2 2 a −x . Then cos θ = a x Also x = a sin θ, so sin θ = . Then a x x a sin θ = √ a = ·√ tan θ = cos θ a a2 − x2 a2 − x2 a x = √ . a2 − x2 # √ √ 46. 4 + x2 = 4 + 4 tan2 θ = 4(1 + tan2 θ) = √ 2 sec2 θ = 2 sec θ =

x

4 + x2 θ

2 x sin θ = √ 4 + x2 2 cos θ = √ 4 + x2 47.

x = 3 sec θ ! " 1 3 sec θ = x= cos θ cos θ 3 cos θ = x # √ x2 − 9 = (3 sec θ)2 − 9 Substituting √ 2 = 9 sec θ − 9 # = 9(sec2 θ − 1) √ = 9 tan2 θ " ! π = 3 tan θ tan θ > 0 for 0 < θ < 2

382

Chapter 6: Trigonometric Identities, Inverse Functions, and Equations

Then tan θ = sin θ = cos θ sin θ = 3 x

48.

√ √

x2 − 9 3 x2

√

√

x2 − a2

sin θ = cos θ =

50.

51.

√

! ! ! !

53.

54. x ! ! ! !

!!

a x2 − a2 x

a x

sin2 θ x2 = # Substituting 1 − x2 1 − sin2 θ sin2 θ = √ cos2 θ sin θ sin2 θ = sin θ · = cos θ cos θ = sin θ tan θ # √ √ 16(sec2 θ − 1) x2 − 16 16 sec2 θ − 16 = = = 2 2 x 16 sec θ 16 sec2 θ √ 4 tan2 θ tan θ 1 sin θ = = · · cos2 θ = 2 2 16 sec θ 4 sec θ 4 cos θ sin θ cos θ 4 √

3π 2π π π π = − = − 12 12 12 4 6 ! " π π π sin = sin − 12 4 6 π π π π = sin cos − cos sin 4 6 4 6 √ √ √ 2 3 2 1 · − · = 2 2 2 2 √ √ 6− 2 = 4

cos 75◦ = cos(30◦ + 45◦ ) = cos 30◦ cos 45◦ − sin 30◦ sin 45◦ √ √ √ 3 2 1 2 = · − · 2 2 2 2 √ √ 6− 2 = 4

−9

! " x2 − 9 3 cos θ = 3 x √ 3 x2 − 9 sin θ = · x 3 √ 2 x −9 . sin θ = x # √ √ x2 − a2 = a2 sec2 θ − a2 = a2 (sec2 θ − 1) = √ a tan2 θ = a tan θ ! !

49.

3

52.

55.

tan 105◦ = tan(45◦ + 60◦ ) tan 45◦ + tan 60◦ = 1 − tan 45◦ tan 60◦ √ 1+ 3 √ = 1−1· 3 √ 1+ 3 √ = 1− 3 √ √ √ 6+ 2 √ and −2 − 3.) (This is equivalent to √ 2− 6

9π 4π 3π π 5π = − = − 12 12 12 4 3 ! " 3π π 5π = tan − tan 12 4 3 3π π tan − tan 4 3 = 3π π 1 + tan tan 4 3 √ −1 − 3 √ = 1 + (−1) · 3 √ √ −1 − 3 3+1 √ , or √ = 1− 3 3−1 √ 3+ 3 √ .) (This is equivalent to 3− 3 cos 15◦ = cos(45◦ − 30◦ )

= cos 45◦ cos 30◦ + sin 45◦ sin 30◦ √ √ √ 2 3 2 1 = · + · 2 2 2 2 √ √ 6+ 2 = 4

56.

57.

9π 2π 3π π 7π = − = − 12 12 12 4 6 ! " 7π 3π π sin = sin − 12 4 6 π 3π π 3π cos − cos sin = sin 4 6 4 6 √ √ ! √ "! " 2 3 2 1 · − − = 2 2 2 2 √ √ 6+ 2 = 4 sin 37◦ cos 22◦ + cos 37◦ sin 22◦ = sin(37◦ + 22◦ ) = sin 59◦ ≈ 0.8572

Exercise Set 6.1 58.

383

cos 83◦ cos 53◦ + sin 83◦ sin 53◦

66. See the figures before Exercise 65. tan u − tan v tan(u − v) = 1 + tan u tan v 3 4 − = 4 3 3 4 1+ · 4 3 7 =− 24

= cos(83 − 53 ) ◦

59.

◦

= cos 30◦ ≈ 0.8660

cos 19◦ cos 5◦ − sin 19◦ sin 5◦

= cos(19◦ + 5◦ ) 60.

= cos 24◦ ≈ 0.9135

sin 40◦ cos 15◦ − cos 40◦ sin 15◦

= sin(40◦ − 15◦ )

67. See the figures before Exercise 65.

= sin 25◦ ≈ 0.4226

61.

62.

sin(u − v) = sin u cos v − cos u sin v 3 3 4 4 = · − · 5 5 5 5 16 7 9 − =− = 25 25 25

tan 20◦ + tan 32◦ = tan(20◦ + 32◦ ) 1 − tan 20◦ tan 32◦ = tan 52◦ ≈ 1.2799 tan 35◦ − tan 12◦ = tan(35◦ − 12◦ ) 1 + tan 35◦ tan 12◦ = tan 23◦ ≈ 0.4245

68. See the figures before Exercise 65. cos(u − v) = cos u cos v + sin u sin v 4 3 3 4 = · + · 5 5 5 5 24 12 12 + = = 25 25 25

63. See the answer section in the text. 64.

tan(u − v) sin(u − v) = cos(u − v) sin u cos v − cos u sin v = cos u cos v + sin u sin v

Use the figures and function values below for Exercises 69-72.

1 sin u cos v − cos u sin v cos u cos v · = 1 cos u cos v + sin u sin v cos u cos v sin u sin u − cos u cos v = sin u sin v 1+ cos u cos v tan u − tan u = 1 + tan u tan v

y

1 1 θ 0.7807

Use the figures and function values below for Exercises 65-68.

5

5

4

3

u 4

v 3

4 3 sin v = 5 5 3 4 cos u = cos v = 5 5 4 3 tan v = tan u = 4 3 65. See the figures above. cos(u + v) = cos u cos v − sin u sin v 4 3 3 4 = · − · 5 5 5 5 12 12 − =0 = 25 25 sin u =

0.9939

0.6249

φ

x

0.1102

sin θ = 0.6249

sin φ = 0.9939

cos θ = 0.7807

cos φ = 0.1102

tan θ ≈ 0.8004

tan φ ≈ 9.0191

69. See the figures above. tan θ + tan φ tan(θ + φ) = 1 − tan θ tan φ 0.8004 + 9.0191 = 1 − 0.8004(9.0191) ≈ −1.5790

(Answer may vary slightly due to rounding differences.)

70. See the figures before Exercise 69. sin(θ − φ) = sin θ cos φ − cos θ sin φ

= 0.6249(0.1102) − 0.7807(0.9939) ≈ −0.7071

71. See the figures before Exercise 69. cos(θ − φ) = cos θ cos φ + sin θ sin φ

= 0.7807(0.1102) + 0.6249(0.9939) ≈ 0.7071

384

Chapter 6: Trigonometric Identities, Inverse Functions, and Equations

72. See the figures before Exercise 69. cos(θ + φ) = cos θ cos φ − sin θ sin φ

= 0.7807(0.1102) − 0.6249(0.9939)

73.

≈ −0.5351

sin(α + β) + sin(α − β)

= (sin α cos β + cos α sin β)+ = 2 sin α cos β 74.

(sin α cos β − cos α sin β)

cos(α + β) − cos(α − β)

= (cos α cos β − sin α sin β)−

(cos α cos β + sin α sin β)

75.

= −2 sin α sin β cos(u + v) cos v + sin(u + v) sin v = (cos u cos v − sin u sin v) cos v+

(sin u cos v + cos u sin v) sin v

= cos u cos2 v − sin u sin v cos v+

sin u cos v sin v + cos u sin2 v

= cos u(cos2 v + sin2 v) = cos u 76.

sin(u − v) cos v + cos(u − v) sin v

= (sin u cos v − cos u sin v) cos v+

(cos u cos v + sin u sin v) sin v

= sin u cos2 v − cos u sin v cos v+

(cos u cos v sin v + sin u sin2 v

= sin u cos2 v + sin u sin2 v = sin u(cos2 v + sin2 v) = sin u 77. Left to the student 78. Left to the student 79. A trigonometric equation that is an identity is true for all possible replacements of the variables. A trigonometric equation that is not true for all possible replacements is not an identity. The equation sin2 x + cos2 x = 1 is an identity while sin2 x = 1 is not. 80. It is not possible to see the entire graph, so the graphs of the two sides of the identity could differ at a point that hasn’t been observed. However, if the two graphs are observed to be different, then we know that the equation is not an identity. " ! 81. 2x − 3 = 2 x − 3 2 2x − 3 = 2x − 3 −3 = −3

Subtracting 2x on both sides

The equation is true for all real numbers, so the solution set is the set of all real numbers.

82.

x − 7 = x + 3.4 −7 = 3.4

Subtracting x on both sides

The equation is false for all values of x. There is no solution. 83. 59◦ and 31◦ are complementary angles. Then cos 59◦ = sin 31◦ = 0.5150, so 1 1 = ≈ 1.9417. sec 59◦ = cos 59◦ 0.5150 84. tan 59◦ = cot 31◦ =

cos 31◦ 0.8572 = ≈ 1.6645 sin 31◦ 0.5150

85. Solve each equation for y to determine the slope of each line. l2 : x + y = 5 l1 : 2x = 3 − 2y 3 y = −x + y = −x + 5 2 Thus m2 = −1 and Thus m1 = −1 and 3 the y-intercept is . the y-intercept is 5. 2 The lines are parallel, so they do not form an angle. When the formula is used, the result is 0◦ . √ √ 3 86. l1 has m1 = ; l2 has m2 = 3. 3 √ √ 3 √ 3− 3 3 √ = , Then tan φ = √ 3 3 1+ 3· 3 π and φ = , or 30◦ 6 87. Find the slope of each line. l1 : y = 3

(y = 0x + 3)

Thus m = 0. l2 : x + y = 5 y = −x + 5

Thus m = −1.

Let φ be the smallest angle from l1 to l2 . ! " m2 − m1 −1 − 0 tan φ = tan φ = 1 + (−1)(0) 1 + m2 m1 −1 = = −1 1 Since tan φ is negative, we know that φ is obtuse. Thus 3π , or 135◦ . φ= 4 88. l1 has m1 = −2; l2 has m2 = 2. 4 2 − (−2) Then tan φ = =− , 1 + 2(−2) 3 and φ ≈ 126.87◦ .

Exercise Set 6.1

385

89. We add some labels to the drawing in the text.

√ # 3π 94. Let θ = . Then sin θ = −1, but sin2 θ = (−1)2 = 1. 2 Answers may vary.

R

95. See the answer section in the text. ! " π π 96. Let x = . Then sin(−x) = sin − = −1, but 2 2 π sin x = sin = 1. 2 Answers may vary.

S

47 ft 40 ft

φ

α θ

50 ft First we find the slope m1 of the line containing guy wire S. 40 m1 = tan θ = = 0.8 50 Next find the slope m2 of the line containing guy wire R. 47 = 0.94 m2 = tan φ = 50 Then α is the smallest positive angle between the guy wires. 0.94 − 0.8 tan α = ≈ 0.0799 1 + 0.94(0.8) α = 4.57◦ 90.

93. See the answer section in the text.

97. See the answer section in the text. π 98. Let θ = . Then tan2 θ + cot2 θ = 12 + 12 = 2 $= 1. 4 Answers may vary. 99.

y A

20 ft φ α

β B

10 ft

30 ft x

40 ft

Position a set of coordinate axes as shown above. Let l1 be the line containing the points (0, 0) and (40, 10). Find its slope. 10 1 = m1 = tan α = 40 4 Let l2 be the line containing the points (0, 0) and (40, 30). Find its slope. 30 3 = m2 = tan β = 40 4 Now φ is the smallest positive angle from l1 to l2 . 1 3 1 − 8 4 4 2 = = ≈ 0.4211 tan φ = 3 1 19 19 1+ · 4 4 16 φ ≈ 22.83◦ 91. See the answer section in the text. 92.

sin(x + h) − sin x h sin x cos h + cos x sin h − sin x = h sin x cos h − sin x cos x sin h + = h h " ! " ! sin h cos h − 1 + cos x = sin x h h

100.

m 2 − m1 1 + m2 m1 2 − m1 ◦ tan 30 = 3 2 1 + m1 3 2 √ − m1 3 = 3 2 3 1 + m1 3 √ √ 2 3 2 3 + m1 = − m1 3 9 3 √ √ 2 3 2 3 m1 + m1 = − 9 3 3 √ ! √ " 2 3 2− 3 +1 = m1 9 3 √ " ! √ 2− 3 2 3+9 = m1 9 3 √ 2− 3 9 m1 = · √ 3 2 3+9 √ √ 3(2 − 3) 6−3 3 = √ m1 = √ 2 3+9 2 3+9 m1 ≈ 0.0645 tan φ =

4 − m1 tan 45◦ = 3 4 1 + · m1 3 4 − 3m1 1= 3 + 4m1

3 + 4m1 = 4 − 3m1 7m1 = 1 1 m1 = 7

101. Find the slope of l1 . −2 − 7 −9 = , which is undefined so l1 is vertim1 = −3 − (−3) 0 cal. Find the slope of l2 .

386

Chapter 6: Trigonometric Identities, Inverse Functions, and Equations 6 − (−4) 10 m2 = = =5 2−0 2

108.

sin α sin β + cos α cos β sin α sin β 1+ · cos α cos β sin α cos β + cos α sin β cos α cos β = cos α cos β + sin α sin β cos α cos β sin α cos β + cos α sin β = cos α cos β + sin α sin β

tan α + tan β = 1 + tan α tan β

l1 l2

θ

φ

l3

90ο

From the drawing we see that φ = 90 + θ, where θ is the smallest positive angle from the horizontal line l3 to l2 . We find θ, recalling that the slope of a horizontal line is 0, so m3 = 0. m2 − m3 tan θ = 1 + m2 m3 5−0 tan θ = =5 1+5·0 θ ≈ 78.7◦

=

◦

Then φ ≈ 90◦ + 78.7◦ ≈ 168.7◦ .

102. Find the slope of l1 . −1 − 4 −5 5 m1 = = =− 5 − (−2) 7 7 Use the following formula to determine the slope of l2 . m2 − m1 tan φ = 1 + m2 m1 ! " 5 m2 − − 7 " ! tan 45◦ = 5 1 + m2 − 7 7m2 + 5 1= 7 − 5m2 7 − 5m2 = 7m2 + 5 2 = 12m2 1 = m2 6

103.

cos 2θ = cos(θ + θ) = cos θ cos θ − sin θ sin θ

= cos2 − sin2 θ,

or 1 − 2 sin2 θ,

or 2 cos θ − 1 2

104.

(cos2 θ = 1 − sin2 θ)

(sin2 θ = 1 − cos2 θ)

sin 2θ = sin(θ + θ) = sin θ cos θ + cos θ sin θ = 2 sin θ cos θ

105. See the answer section in the text. " ! 106. sin x − 3π = sin x cos 3π − cos x sin 3π 2 2 2 = (sin x)(0) − (cos x)(−1) = cos x

107. See the answer section in the text.

sin(α + β) cos(α − β)

Exercise Set 6.2 1. sin(3π/10) ≈ 0.8090, cos(3π/10) ≈ 0.5878

sin(3π/10) 0.8090 3π = ≈ ≈ 1.3763, 10 cos(3π/10) 0.5878 1 1 3π = ≈ ≈ 1.2361, csc 10 sin(3π/10) 0.8090 3π 1 1 sec = ≈ ≈ 1.7013, 10 cos(3π/10) 0.5878 1 1 3π cot = ≈ ≈ 0.7266 10 tan(3π/10) 1.3763 2π π 3π π π 3π = = , so and are compleb) − 2 10 10 5 10 5 ments. π 3π ≈ 0.5878, sin = cos 5 10 π 3π cos = sin ≈ 0.8090, 5 10 π 3π tan = cot ≈ 0.7266, 5 10 3π π ≈ 1.7013, csc = sec 5 10 π 3π sec = csc ≈ 1.2361, 5 10 π 3π ≈ 1.3763 cot = tan 5 10 # √ sin(π/12) π 2− 3 = =# 2. a) tan √ , 12 cos(π/12) 2+ 3 1 2 =# sec π/12 = √ , cos(π/12) 2+ 3 a) tan

π 1 2 = =# √ , 12 sin(π/12) 2− 3 # √ π 1 2+ 3 = =# cot √ , 12 tan(π/12) 2− 3 5π π and are complements, we have b) Since 12 12 # # √ √ 5π 2+ 3 2− 3 5π sin = , cos = , 12 2 12 2 # √ 2+ 3 5π 2 5π =# =# tan √ , sec √ , 12 12 2− 3 2− 3 # √ 5π 2 2− 3 5π =# =# csc √ , cot √ . 12 12 2+ 3 2+ 3 csc

Exercise Set 6.2

387

3. We sketch a second quadrant triangle.

! " π cot θ − = 2

y 3

1

θ x

4.

–2 2

√ √ 2 2 −2 2 =− , 3 3 √ 1 2 √ =− tan θ = , 4 −2 2 3 csc θ = = 3, 1 √ 3 2 3 √ =− , sec θ = 4 −2 2 √ √ −2 2 = −2 2 cot θ = 1 π b) Since θ and − θ are complements, we have 2 √ " ! π 2 2 sin − θ = cos θ = − , 2 3 " ! π 1 cos − θ = sin θ = , 2 3 " ! √ π tan − θ = cot θ = −2 2, 2 √ ! " π 3 2 − θ = sec θ = − , csc 2 4 " ! π sec − θ = csc θ = 3, 2 √ " ! 2 π − θ = tan θ = − cot 2 4 " ' ! "( ! π π c) sin θ − = sin − −θ = 2 2 √ " √ " ! ! 2 2 2 2 π −θ =− − = , − sin 2 3 3 " ' ! "( ! " ! π π π −θ = cos −θ cos θ− = cos − 2 2 2 1 = , 3 " ' ! "( ! π π = tan − −θ = tan θ− 2 2 " ! √ √ − tan π2 −θ = −(−2 2) = 2 2, ! " 1 3 π 1 "= √ = √ = ! = csc θ − π 2 2 2 2 2 sin θ − 2 3 √ 3 2 , 4 " ! 1 1 π "= ! = = 3, sec θ − 1 π 2 cos θ − 3 2

y φ

a) cos θ =

4 5

5.

6.

7.

8.

√ 1 2 "= √ = π 4 2 2 tan θ − 2 1

!

x -3

3 5 3 a) sin φ = − , tan φ = − , csc φ = − , 5 4 3 5 4 sec φ = , cot φ = − 4 3 π b) Since φ and − φ are complements we have 2 " ! " ! 4 π 3 π − φ = , cos −φ =− , sin 2 5 2 5 " ! " ! 4 π 5 π − φ = − , csc −φ = , tan 2 3 2 4 " ! " ! π 5 π 3 sec − φ = − , cot −φ =− . 2 3 2 4 " ! " ! π π c) Since sin + φ = cos φ and cos +φ = 2 2 − sin φ, we have " ! " ! 4 π 3 π + φ = , cos +φ = , sin 2 5 2 5 " ! " ! π 4 π 5 tan + φ = , csc +φ = , 2 3 2 4 " ! " ! π 5 π 3 sec + φ = , cot +φ = . 2 3 2 4 " ! 1 1 π ! "= = − csc x = sec x + π 2 − sin x cos x + 2 ! " ' ! "( π π " cos x− ! cos − −x π 2 2 " = "( = ! ' ! = cot x− π π 2 sin x− sin − −x 2 2 " ! π −x cos sin x 2 "= ! = − tan x π − cos x −x − sin 2 " ' ! "( ! π π = tan − −x = tan x − 2 2 ! "( " ' ! π π − sin −x −x sin − 2 2 "( = " = ' ! ! π π −x −x cos − cos 2 2 " ! π − tan − x = − cot x 2 " ! 1 π 1 ! "= = = sec x csc x + π 2 cos x sin x + 2

388

Chapter 6: Trigonometric Identities, Inverse Functions, and Equations

9. Make a drawing.

cos 2θ = cos θ − sin θ =

y

"2

−

!

4 − 5

"2

=−

7 25

sin 2θ 24 =− cos 2θ 7 (We could have found tan 2θ using a double-angle identity.)

4 θ 3

Since sin 2θ is positive and cos 2θ is negative, 2θ is in quadrant II.

x

3 From this drawing we find that cos θ = and 5 4 tan θ = . 3 4 3 24 sin 2θ = 2 sin θ cos θ = 2 · · = 5 5 25 ! "2 ! "2 3 4 − cos 2θ = cos2 θ − sin2 θ = 5 5 9 16 7 = − =− 25 25 25 4 8 2· 24 2 tan θ 3 3 tan 2θ = = ! "2 = 7 = − 7 1 − tan2 θ 4 − 1− 9 3

12.

y

y

17 15

θ

x

–8

sin 2θ = 2 sin θ cos θ = 2 · cos 2θ = cos2 θ − sin2 θ = −

161 289

15 17 !

" 8 240 =− 17 289 "2 ! "2 8 15 − − = 17 17 !

−

240 − 240 sin 2θ 289 tan 2θ = = = 161 cos 2θ 161 − 289 Since sin 2θ and cos 2θ are both negative, 2θ is in quadrant III.

(We could have found tan 2θ by dividing: sin 2θ .) tan 2θ = cos 2θ Since sin 2θ is positive and cos 2θ is negative, 2θ is in quadrant II.

13. Make a drawing. y

13

12

θ 5

x

y θ x

4 4 From this drawing we find that sin θ = − and tan θ = . 5 3 ! "! " 4 3 24 sin 2θ = 2 sin θ cos θ = 2 − − = 5 5 25

x

5 and cos θ = From this drawing we find that sin θ = 13 12 − . 13 ! " 12 120 5 − =− sin 2θ = 2 sin θ cos θ = 2 · 13 13 169 "2 ! "2 ! 12 5 119 − = cos 2θ = cos2 θ−sin2 θ = − 13 13 169 120 − sin 2θ 120 169 = tan 2θ = =− 119 cos 2θ 119 169 (We could have found tan 2θ using a double-angle identity.)

11. Make a drawing.

5

θ

-12

120 12 5 · = 13 13 169 ! "2 ! "2 5 12 119 cos 2θ = cos2 θ − sin2 θ = − =− 13 13 169 120 sin 2θ 120 = 169 = − tan 2θ = 119 cos 2θ 119 − 169 Since sin 2θ is positive and cos 2θ is negative, 2θ is in quadrant II.

-3

13

5

sin 2θ = 2 sin θ cos θ = 2 ·

-4

3 − 5

!

tan 2θ =

5

10.

2

2

Since sin 2θ is negative and cos 2θ is positive, 2θ is in quadrant IV. 14.

y

90 10

x

- 10

! √ "! √ " 10 90 = sin 2θ = 2 sin θ cos θ = 2 − 10 10

Exercise Set 6.2

−

389

√ 2 900 2 · 30 3 =− =− 100 100 5

cos 2θ = cos θ − sin θ = 2

2

!√

90 10

"2

−

!

−

√

10 10

"2

=

10 80 4 90 − = = 100 100 100 5 3 − 3 sin 2θ = 5 =− tan 2θ = 4 cos 2θ 4 5 Since sin 2θ is negative and cos 2θ is positive, 2θ is in quadrant IV. 15.

cos 4x = cos[2(2x)] = 1 − 2 sin2 2x

= 1 − 2(2 sin x cos x)2

= 1 − 8 sin2 x cos2 x or cos 4x = cos[2(2x)]

135◦ 1 − cos 135◦ 18. tan 67.5◦ = tan = = 2 sin 135◦ ! √ " 2 √ √ 1− − 2+ 2 2 2+2 √ 2 √ = √ = = 2+1 2 2 2 2 $ 225◦ 1 − cos 225◦ = = 19. sin 112.5◦ = sin 2 2 % % # √ √ √ 1 − (− 2/2) 2+ 2 2+ 2 = = 2 4 2 (We choose the positive square root since 112.5◦ is in quadrant II where the sine function is positive.) π $ π 1 + cos(π/4) 4 = = 20. cos = cos 8 2 2 % % # √ √ √ 1 + ( 2/2) 2+ 2 2+ 2 = = 2 4 2 (We choose the positive square root since π/8 is in quadrant I where the cosine function is positive.) 21. tan 75◦ = tan

= cos2 2x − sin2 2x

1 √ 1 2√ √ , or 2 + 3 = 1 + (− 3/2) 2− 3

= (cos2 x − sin2 x)2 − (2 sin x cos x)2

= cos4 x − 6 sin x cos2 x + sin x or cos 4x 2

4

= cos[2(2x)] = 2 cos2 2x − 1

= 2(2 cos2 x − 1)2 − 1

16.

= 8 cos4 x − 8 cos2 x + 1 sin4 θ = sin2 θ · sin2 θ 1 − cos 2θ 1 − cos 2θ · = 2 2

5π $ 1 − cos(5π/6) 5π = sin 6 = = 22. sin 2 2 2 % % # √ √ √ 1 − (− 3/2) 2+ 3 2+ 3 = = 2 4 2 (We choose the positive square root since 5π/12 is in quadrant I where the sine function is positive.) 23. First find cos θ. sin2 θ + cos2 θ = 1

1 − 2 cos 2θ + cos2 2θ 4 1 + cos 4θ 1 − 2 cos 2θ + 2 = 4

(0.3416)2 + cos2 θ = 1

=

2 − 4 cos 2θ + 1 + cos 4θ 8 3 − 4 cos 2θ + cos 4θ = 8 % $ √ ◦ ◦ 30 1 + cos 30 1 + 3/2 = = = 17. cos 15◦ = cos 2 2 2 % # √ √ 2+ 3 2+ 3 = 4 2

cos2 θ = 1 − (0.3416)2 cos θ ≈ 0.9398

(θ is in quadrant I.)

sin 2θ = 2 sin θ cos θ

≈ 2(0.3416)(0.9398)

=

(We choose the positive square root since 15◦ is in quadrant I where the cosine function is positive.)

150◦ sin 150◦ = = 2 1 + cos 150◦

24.

25.

≈ 0.6421 $ 1 + cos θ θ cos = 2 2 $ 1 + 0.9398 ≈ 2 ≈ 0.9848 $ 1 − cos θ θ sin = 2 2

≈

$

1 − 0.9398 2

≈ 0.1735

(θ is in quadrant I.) We found cos θ in Exercise 23. !

" θ is in quadrant I. 2

We found cos θ in Exercise 23.

390

Chapter 6: Trigonometric Identities, Inverse Functions, and Equations

26. sin 4θ = 2 sin 2θ cos 2θ

34.

We found sin 2θ ≈ 0.6421 in Exercise 23. Now we find cos 2θ:

cos 2θ = 1 − 2 sin θ = 1 − 2(0.3416) ≈ 0.7666 2

2

Then sin 4θ ≈ 2(0.6421)(0.7666) ≈ 0.9845. ! " x x = cos x 27. 2 cos2 − 1 = cos 2 · 2 2 28.

cos4 x − sin4 x

= (cos2 x + sin2 x)(cos2 x − sin2 x)

= (1) · (cos 2x) = cos 2x 29.

(sin x − cos x)2 + (sin 2x)

= (sin2 x − 2 sin x cos x + cos2 x)+ (2 sin x cos x)

= sin x + cos2 x 2

=1 30.

(sin x + cos x)2 = sin2 x + 2 sin x cos x + cos2 x = 1 + sin 2x

31. 32.

2 2 − sec2 x = − 1 = 2 cos2 x − 1 = cos 2x sec2 x sec2 x 1 + sin 2x + cos 2x 1 + sin 2x − cos 2x

1 + (2 sin x cos x) + (2 cos2 x − 1) = 1 + (2 sin x cos x) − (1 − 2 sin2 x) 2 cos x(sin x + cos x) = 2 sin x(cos x + sin x) = cot x 33.

(−4 cos x sin x + 2 cos 2x)2 + (2 cos 2x + 4 sin x cos x)2 = (16 cos2 x sin2 x − 16 sin x cos x cos 2x+

4 cos2 2x) + (4 cos2 2x + 16 sin x cos x cos 2x+

16 sin x cos2 x) 2

= 8(cos2 x − sin2 x)2 + 32 cos2 x sin2 x

= 8(cos4 x − 2 cos2 x sin2 x + sin4 x)+ 32 cos2 x sin2 x

= 8 cos x − 16 cos x sin x + 8 sin x+ 2

2

4

32 cos2 x sin2 x

= 8 cos4 x + 16 cos2 x sin2 x + 8 sin4 x = 8(cos4 x + 2 cos2 x sin2 x + sin4 x) = 8(cos2 x + sin2 x)2 =8

= (2 sin x cos x)(cos2 x − sin2 x) = (sin 2x)(cos 2x) 1 = sin 4x 2

35. Using a graphing calculator we see that the graphs of cos 2x and y2 = cos x + sin x are the same y1 = cos x − sin x on [−2π, 2π]. The correct choice is (d). See the answer section in the text for the algebraic proof. 36. Using a graphing calculator we see that the graphs of x y1 = 2 cos2 and y2 = 1 + cos x are the same on 2 [−2π, 2π]. The correct choice is (d). ! " 1 + cos x x = 1 + cos x 2 cos2 = 2 2 2 37. Using a graphing calculator we see that the graphs of sin 2x and y2 = sin x are the same on y1 = 2 cos x [−2π, 2π]. The correct choice is (d). See the answer section in the text for the algebraic proof. 38. Using a graphing calculator we see that the graphs of θ θ y1 = 2 sin cos and y2 = sin θ are the same on 2 2 [−2π, 2π]. The correct choice is (c). ! " θ θ θ = sin θ 2 sin cos = sin 2 · 2 2 2 39. Each has amplitude 1 and is periodic. The period of y1 = x sin x is 2π, of y2 = sin 2x is π, and of y3 = sin is 4π. 2 40. In the first line, cos 4x $= 2 cos 2x. In the second line, cos 2x $= cos2 x+sin2 x. If the second line had been correct, the third line would have been correct also. 41. 1 − cos2 x = sin2 x

(sin2 x + cos2 x = 1)

42. sec2 x − tan2 x = 1

(1 + tan2 x = sec2 x)

43. sin2 x − 1 = − cos2 x

(sin2 x + cos2 x = 1)

44. 1 + cot2 x = csc2 x 45. csc2 x − cot2 x = 1

= 8 cos2 2x + 32 cos2 x sin2 x

4

2 sin x cos3 x − 2 sin3 x cos x

(1 + cot2 x = csc2 x)

46. 1 + tan2 x = sec2 x 47. 1 − sin2 x = cos2 x

(sin2 x + cos2 x = 1)

48. sec2 x − 1 = tan2 x

(1 + tan2 x = sec2 x)

49. The functions f (x) = A sin(Bx − C) + D, or f (x) = A cos(Bx − C) + D for which |A| = 2 are (a) and (e). 50. The functions f (x) = A sin(Bx − C))+ D, ) or ) 2π ) ) f (x) = A cos(Bx − C) + D for which ) )) = π are (b), (c), B and (f).

Exercise Set 6.2

391

51. The function f (x) = A sin(Bx − C) +)D, )or ) 2π ) f (x) = A cos(Bx − C) + D for which )) )) = 2π is (d). B

56.

52. The function f (x) = A sin(Bx − C) + D, or C π f (x) = A cos(Bx − C) + D for which = is (e). B 4 53. Observe that 141◦ = 51◦ + 90◦ . Find sin 51◦ : sin2 51◦ + cos2 51◦ = 1 sin2 51◦ + (0.6293)2 = 1 sin2 51◦ = 1 − (0.6293)2 sin 51◦ ≈ 0.7772

in quadrant I.)

cos 141◦ = − sin 51◦ ≈ −0.7772 sin 141◦ 0.6293 tan 141◦ = ≈ ≈ −0.8097 cos 141◦ −0.7772 1 1 ≈ ≈ 1.5891 csc 141◦ = sin 141◦ 0.6293 1 1 ≈ −1.2867 sec 141◦ = ≈ cos 141◦ −0.7772 1 1 ≈ −1.2350 cot 141◦ = ≈ tan 141◦ −0.8097 " ! π 54. sin − x [sec x − cos x] 2 "' ( ! 1 π − cos x = − sin x − 2 cos x " ' ! "( ! ! π π − x = sin − x − = sin 2 2 "" ! π − sin x − 2 ! " ' ! " 2 π 1 − cos x sin x − = = −(− cos x) cos x 2 ( − cos x = 1 − cos2 x = sin2 x

55.

" ! π cos(π − x) + cot x sin x − 2 = cos π cos x + sin π sin x+ " ( ' ! π = − cos x cot x(− cos x) sin x − 2 = − cos x + 0 − cot x cos x = − cos x(1 + cot x)

" π cos x − sin − x sin x 2 cos x − cos(π − x) tan x "( ' ! π sin x cos x − sin − x − 2 = cos x − (cos π cos x + sin π sin x) tan x " ! π cos x + sin x − sin x 2 = cos x − (−1 · cos x) tan x cos x − cos x sin x = cos x + cos x tan x cos x(1 − sin x) cos x(1 + tan x) 1 − sin x = 1 + tan x ! " π cos2 y sin y + cos2 y cos y 2 " = ! π sin2 y cos y sin2 y sin −y 2 cos2 y = sin2 y = cot2 y =

(51◦ is

sin 141◦ = cos 51◦ ≈ 0.6293

!

57.

58.

3π 3π ≤ 2θ ≤ 2π, so ≤ θ ≤ π; sin θ is positive; cos θ and 2 4 tan θ are negative. cos 2θ = 1 − 2 sin2 θ 7 = 1 − 2 sin2 θ 12 5 2 sin2 θ = 12 5 sin2 θ = 24 $ √ 5 30 = sin θ = 24 12 cos 2θ = 2 cos2 θ − 1 7 = 2 cos2 θ − 1 12 19 = 2 cos2 θ 12 19 = cos2 θ 24 $ 19 = cos θ, or − 24 √ 114 − = cos θ 12 √ 30 $ √ sin θ 5 95 12 tan θ = =− = √ =− cos θ 19 19 114 − 12

392

Chapter 6: Trigonometric Identities, Inverse Functions, and Equations

3π 59. Since π < θ ≤ , tan θ is positive and sin θ and cos θ are 2 negative. $ 1 − cos θ θ tan = − 2 1 + cos θ $ 5 1 − cos θ − =− 3 1 + cos θ 1 − cos θ 25 = 9 1 + cos θ 25 + 25 cos θ = 9 − 9 cos θ 34 cos θ = −16 8 16 =− cos θ = − 34 17 % ! "2 # 8 15 sin θ = − 1 − cos2 θ = − 1 − − =− 17 17

g = 9.78049[1 + 0.005288 sin2 φ− 0.000024 sin2 φ(1 − sin2 φ)]

g = 9.78049(1 + 0.005264 sin2 φ +

0.000024 sin4 φ)

Exercise Set 6.3 Note: Answers for the odd-numbered exercises 1-29 are in the answer section in the text. 2.

15 − 15 sin θ 17 = tan θ = = 8 cos θ 8 − 17

cos θ + cos2 θ + sin2 θ sin θ cos θ cos θ + 1 = sin θ cos θ We started with the left side and deduced the right side, so the proof is complete. =

60. a) We substitute 42◦ for φ. N (42◦ ) = 6066 − 31 cos(2 · 42◦ ) = 6066 − 31 cos 84◦ ≈ 6062.76 ft

4.

b) We substitute 90◦ for φ. N (90◦ ) = 6066 − 31 cos(2 · 90◦ ) = 6066 − 31 cos 180◦

= 6066 − 31(−1) = 6066 + 31 = 6097 ft

c) We substitute 2 cos2 φ − 1 for cos 2φ. N (φ) = 6066 − 31(2 cos2 φ − 1), or

N (φ) = 6097 − 62 cos2 φ

g = 9.78049[1 + 0.005288 sin 42 − 2

≈ 9.80359 m/sec

◦

0.000006 sin2 (2 · 42◦ )]

g = 9.78049[1 + 0.005288 sin 40◦ − 2

2

= =

b) Substitute 40◦ for φ.

≈ 9.80180 m/sec

1 + tan y sec y = 1 + cot y csc y We start with the left side. 1 + tan y 1 + cot y sin y 1+ cos y = cos y 1+ sin y =

61. a) Substitute 42◦ for φ.

2

1 + cos θ sin θ cos θ + 1 + = sin θ cos θ sin θ cos θ We start with the left side. sin θ 1 + cos θ + sin θ cos θ sin θ sin θ 1 + cos θ cos θ · + · = sin θ cos θ cos θ sin θ

0.000006 sin2 (2 · 40◦ )]

c) g = 9.78049[1 + 0.005288 sin2 φ− 0.000006(2 sin φ cos φ)2 ] g = 9.78049(1 + 0.005288 sin2 φ− 0.000024 sin2 φ cos2 φ)

=

cos y + sin y cos y sin y + cos y sin y cos y + sin y sin y · cos y sin y + cos y sin y cos y tan y

Now we stop and work with the right side. sin y sec y = csc y cos y = tan y We have deduced the same expression from each side, so the proof is complete.

Exercise Set 6.3 6.

393

sin x + cos x sin x = sec x + csc x sec x We start with the left side. sin x + cos x sec x + csc x sin x + cos x = 1 1 + cos x sin x sin x + cos x = sin x + cos x sin x cos x sin x cos x = (sin x + cos x) · sin x + cos x = sin x cos x 1 = sin x · sec x sin x = sec x We started with the left side and deduced the right side, so the proof is complete. sec θ 2 − sec2 θ We begin with the left side. 1 sec 2θ = cos 2θ 1 = cos2 θ − sin2 θ Now we stop and work with the right side.

10.

= tan u + cot v We started with the left side and deduced the right side, so the proof is complete. 12. cos4 x − sin4 x = cos 2x

We start with the left side. cos4 x − sin4 x

= (cos2 x − sin2 x)(cos2 x + sin2 x)

2

8. sec 2θ =

= cos2 x − sin2 x = cos 2x

We started with the left side and deduced the right side, so the proof is complete. 14.

sec2 θ 1 + tan2 θ = 2 − sec2 θ 2 − 1 − tan2 θ =

1 + tan2 θ 1 − tan2 θ

sin2 θ cos2 θ = sin2 θ 1− cos2 θ cos2 θ + sin2 θ = cos2 θ − sin2 θ 1 = 2 cos θ − sin2 θ We have deduced the same expression from each side, so the proof is complete.

cos(u − v) = tan u + cot v sin u sin v We start with the left side. cos(u − v) cos u sin v cos u cos v + sin u sin v = cos u sin v sin u sin v cos u cos v + = cos u sin v cos u sin v sin u cos v + = sin v cos u = cot v + tan u

1+

16.

2 tan t tan 3t − tan t = 1 + tan 3t tan t 1 − tan2 t We start with the left side. tan 3t − tan t = tan(3t − t) 1 + tan 3t tan t = tan 2t 2 tan t = 1 − tan2 t We started with the left side and deduced the right side, so the proof is complete. 2 + sin 2β cos3 β − sin3 β = cos β − sin β 2 We start with the left side. cos3 β − sin3 β cos β − sin β

(cos β − sin β)(cos2 β + cos β sin β + sin2 β) cos β − sin β = 1 + cos β sin β =

Now we stop and work with the right side. 2 + 2 sin β cos β 2 + sin 2β = 2 2 = 1 + sin β cos β We have deduced the same expression from each side, so the proof is complete.

394

Chapter 6: Trigonometric Identities, Inverse Functions, and Equations

18. cos2 x(1 − sec2 x) = − sin2 x

26.

We start with the left side. cos2 x(1 − sec2 x) = cos2 x(− tan2 x) " ! sin2 x = cos2 x − cos2 x = − sin2 x

sin2 x + cos2 x cos x sin x = sin x + cos x cos x sin x sin2 x + cos2 x = sin x + cos x 1 = cos x + sin x We started with the left side and deduced the right side, so the proof is complete.

We started with the left side and deduced the right side, so the proof is complete. 20.

cos θ + sin θ = 1 + tan θ cos θ We start with the left side. cos θ sin θ cos θ + sin θ = + cos θ cos θ cos θ = 1 + tan θ We started with the left side and deduced the right side, so the proof is complete.

22.

tan y + cot y = sec y csc y We start with the left side. sin y cos y + tan y + cot y cos y sin y = 1 csc y sin y =

sin2 y + cos2 y sin y · cos y sin y 1

sin2 y + cos2 y cos y 1 = cos y = sec y

=

We started with the left side and deduced the right side, so the proof is complete. 24. tan θ − cot θ = (sec θ − csc θ)(sin θ + cos θ) We start with the left side. cos θ sin θ − tan θ − cot θ = cos θ sin θ

sin2 θ − cos2 θ cos θ sin θ Now we stop and work with the right side. =

(sec θ − csc θ)(sin θ + cos θ) " ! 1 1 − (sin θ + cos θ) = cos θ sin θ "! " ! sin θ + cos θ sin θ − cos θ = sin θ cos θ 1

sin2 θ − cos2 θ sin θ cos θ We have deduced the same expression from each side, so the proof is complete. =

tan x + cot x 1 = sec x + csc x cos x + sin x We start with the left side. sin x cos x + tan x + cot x cos x sin x = 1 1 sec x + csc x + cos x sin x

28.

csc θ + 1 cot θ = csc θ − 1 cot θ We start with the left side. cot θ csc θ + 1 cot θ = · csc θ − 1 csc θ − 1 csc θ + 1 =

cot θ(csc θ + 1) csc2 θ − 1

cot θ(csc θ + 1) cot2 θ csc θ + 1 = cot θ We started with the left side and deduced the right side, so the proof is complete. =

30. sec4 s − tan2 s = tan4 s + sec2 s

We start with the left side. sec4 s − tan2 s = sec4 s − (sec2 s − 1) = sec4 s − sec2 s + 1

Now we stop and work with the right side. tan4 s + sec2 s = (tan2 s)2 + sec2 s = (sec2 s − 1)2 + sec2 s

= sec4 s − 2 sec2 s + 1 + sec2 s

= sec4 s − sec2 s + 1

We have deduced the same expression from each side, so the proof is complete. 31. See the answer section in the text. 32. Prove identity (5). x−y x+y cos 2 sin 2 2 ' ! " ! "( x+y x−y x+y x−y 1 = 2· sin + + sin − 2 2 2 2 2

Using identity (3)

2x 2y = sin + sin 2 2 = sin x + sin y

Exercise Set 6.3

395

Prove identity (6). x−y x+y sin 2 cos 2 2 ' ! " ! "( 1 x+y x−y x+y x−y = 2· sin − sin + − 2 2 2 2 2

38.

39.

Using identity (4)

2y 2x − sin = sin 2 2 = sin x − sin y

Prove identity (7). x−y x+y cos 2 cos 2 2 ' ! " ! "( 1 x+y x−y x+y x−y = 2· cos − + cos + 2 2 2 2 2

40. 41.

Using identity (2)

2x 2y + cos = cos 2 2 = cos y + cos x Prove identity (8). x+y x−y 2 sin sin 2 2 ' ! " ! "( x+y x−y x+y x−y 1 cos − − cos + = 2· 2 2 2 2 2

Using identity (1)

2x 2y − cos 2 2 = cos y − cos x sin 3θ − sin 5θ

3θ − 5θ 3θ + 5θ sin 2 2 8θ −2θ = 2 cos sin 2 2 = 2 cos 4θ sin(−θ) = 2 cos

= −2 cos 4θ sin θ 34. sin 7x − sin 4x = 2 cos 35.

43.

7 cos θ sin 7θ 1 = 7 · [sin(7θ + θ) + sin(7θ − θ)] 2 7 = (sin 8θ + sin 6θ) 2 1 cos 2t sin t = (sin 3t − sin t) 2 cos 55◦ sin 25◦ 1 = [sin(55◦ + 25◦ ) − sin(55◦ − 25◦ )] 2 1 = (sin 80◦ − sin 30◦ ) 2 ! " 1 1 sin 80◦ − = 2 2 1 1 ◦ = sin 80 − 2 4 7 cos 5θ cos 7θ 1 = 7 · [cos (−2θ) + cos 12θ] 2 7 = (cos 2θ + cos 12 θ) 2 See the answer section in the text.

44. tan 2x(cos x + cos 3x) = sin x + sin 3x

= cos

33.

42.

2 sin 7θ cos 3θ 1 = 2 · (sin 10θ + sin 4θ) 2 = sin 10θ + sin 4θ

(sin(−θ) = − sin θ)

3x 11x sin 2 2

sin 8θ + sin 5θ 8θ + 5θ 8θ − 5θ = 2 sin cos 2 2 3θ 13θ cos = 2 sin 2 2

36.

cos θ − cos 7θ 8θ 6θ sin = 2 sin 2 2 = 2 sin 4θ sin 3θ

37.

sin 7u sin 5u 1 = [cos(7u − 5u) − cos(7u + 5u)] 2 1 = (cos 2u − cos 12u) 2

We start with the left side. ! " 4x 2x sin 2x 2 cos cos tan 2x(cos x + cos 3x) = cos 2x 2 2 sin 2x (2 cos 2x cos x) = cos 2x = 2 sin 2x cos x Now we stop and work with the right side. 4x −2x sin x + sin 3x = 2 sin cos 2 2 = 2 sin 2x cos(−x) = 2 sin 2x cos x We have deduced the same expression from each side, so the proof is complete. 45. See the answer section in the text. sin x + sin y x+y = 46. tan 2 cos x + cos y We start with the right side. x+y x−y cos 2 sin sin x + sin y 2 2 = x+y x−y cos x + cos y 2 cos cos 2 2 x+y sin 2 = x+y cos 2 x+y = tan 2 We started with the right side and deduced the left side,

396

Chapter 6: Trigonometric Identities, Inverse Functions, and Equations so the proof is complete.

47. See the answer section in the text. φ−θ cos θ − cos φ θ+φ tan = 2 2 cos θ + cos φ We start with the right side. φ+θ φ−θ 2 sin sin cos θ − cos φ 2 2 = φ+θ φ−θ cos θ + cos φ 2 cos cos 2 2 φ−θ θ+φ sin sin 2 · 2 = θ+φ φ−θ cos cos 2 2 θ+φ φ−θ = tan tan 2 2 We started with the right side and deduced the left side, so the proof is complete.

48. tan

49. See the answer section in the text. 50. sin 2θ + sin 4θ + sin 6θ = 4 cos θ cos 2θ sin 3θ We start with the left side. sin 2θ + sin 4θ + sin 6θ −2θ 6θ cos + sin 6θ = 2 sin 2 2 = 2 sin 3θ cos(−θ) + sin 6θ = 2 sin 3θ cos θ + 2 sin 3θ cos 3θ = 2 sin 3θ(cos θ + cos 3θ) " ! 2θ 4θ = 2 sin 3θ 2 cos cos 2 2 = 2 sin 3θ(2 cos 2θ cos θ) = 4 cos θ cos 2 θ sin 3θ We started with the left side and deduced the right side, so the proof is complete. cos x + cot x = 1 + csc x cos x. The proof is in the answer section in the text.

51. Expression B completes the identity

52. Expression E completes the identity cot x + csc x = sin x . To prove it we start with the left side. 1 − cos x cos x 1 cot x + csc x = + sin x sin x cos x + 1 = sin x Now we stop and work with the right side. sin x 1 + cos x sin x = · 1 − cos x 1 − cos x 1 + cos x =

53. Expression A completes the identity sin x cos x + 1 = sin3 x − cos3 x . The proof is in the answer section in the sin x − cos x text. 54. Expression F completes the identity 2 cos2 x − 1 = cos4 x − sin4 x. To prove it we start with the left side. 2 cos2 x − 1 = 2 cos2 x − (sin2 x + cos2 x)

= cos2 x − sin2 x

Now we stop and work with the right side. cos4 x − sin4 x

= (cos2 x + sin2 x)(cos2 x − sin2 x)

= cos2 x − sin2 x

We have deduced the same expression from each side, so the proof is complete. 1 = cot x sin2 x tan x + cot x. The proof is in the answer section in the text.

55. Expression C completes the identity

56. Expression D completes the identity (cos x + sin x)(1 − sin x cos x) = cos3 x + sin3 x. To prove it we start with the right side. cos3 x + sin3 x = (cos x + sin x)(cos2 x − cos x sin x + sin2 x)

= (cos x + sin x)(1 − sin x cos x)

We started with the right side and deduced the left side, so the proof is complete. 57. a) x $= kπ, k an integer; the tangent function is not defined for these values of x. b) sin x = 0 for x = kπ, k an integer; cos x = −1 for x = kπ, k an odd integer; thus the restriction x $= kπ, k an integer applies. c) The sine and cosine functions are defined for all real numbers so there are no restrictions.

58. The expression tan(x + 450◦ ) can be simplified using the sine and cosine sum formulas but cannot be simplified using the tangent sum formula because while sin 450◦ and cos 450◦ are both defined, tan 450◦ is undefined. 59. f (x) = 3x − 2

a) Find some ordered pairs. When x = 0, f (0) = 3 · 0 − 2 = −2. When x = 2, f (2) = 3 · 2 − 2 = 4.

Plot these points and draw the graph.

sin x(1 + cos x) 1 − cos2 x

sin x(1 + cos x) sin2 x 1 + cos x = sin x We have deduced the same expression from each side, so the proof is complete.

y

f(x) " 3x ! 2

4

=

2 2

!2

x#2 f !1(x) " 3

!2

4

x

Exercise Set 6.3

397

b) Since there is no horizontal line that intersects the graph more than once, the function is oneto-one.

c) Replace f (x) with y: y = x2 − 4

Interchange x and y: x = y 2 − 4 √ Solve for y: y = x + 4 (We choose the positive square root since y ≥ 0.) √ Replace y with f −1 (x): f −1 (x) = x + 4

c) Replace f (x) with y: y = 3x − 2 Interchange x and y: x = 3y − 2 x+2 Solve for y: y = 3

x+2 3 d) Find some ordered pairs or reflect the graph of f (x) across the line y = x. The graph is shown in part (a) above. Replace y with f −1 (x): f −1 (x) =

d) Find some ordered pairs or reflect the graph of f (x) across the line y = x. The graph is shown in part (a) above. √ 62. f (x) = x + 2 a)

y

60. f (x) = x3 + 1

6

a) y

f (x) "

4

x3

f(x) " !x # 2

#1

2

4

x

b) Since there is no horizontal line that intersects the graph more than once, the function is oneto-one.

Solve for y: y = x2 − 2, x ≥ 0

c) Replace f (x) with y: y = x3 + 1

d) See the graph in part (a). √ 3

x−1

63.

x(2x − 5) = 0

61. f (x) = x − 4, x ≥ 0

x = 0 or 2x − 5 = 0 5 x = 0 or x= 2

a) Find some ordered pairs. When x = 0, f (0) = 02 − 4 = −4.

The solutions are 0 and

When x = 1, f (1) = 12 − 4 = −3. When x = 2, f (2) = 22 − 4 = 0. When x = 3, f (3) = 3 − 4 = 5. 2

Plot these points and draw the graph.

f !1(x) " !x # 4

f (x) " x 2 ! 4, x $ 0

4

64.

5 . 2

3x2 + 5x − 10 = 18 3x2 + 5x − 28 = 0

(3x − 7)(x + 4) = 0 7 or x = −4 x= 3 The solutions are −4 and

4

!4 !2

2x2 = 5x 2x2 − 5x = 0

2

y

x

Replace y with f −1 (x): f −1 (x) = x2 − 2, x ≥ 0

3

d) See the graph in part (a).

6

b) Since there is no horizontal line that intersects the graph more than once, the function is oneto-one. √ c) Replace f (x) with y: y = x + 2 √ Interchange x and y: x = y + 2

!4

Replace y with f −1 (x): f −1 (x) =

4 !2

!2

Interchange x and y: x = y + 1 √ Solve for y: y = 3 x − 1

4

!2

3 f !1(x) " !x !1

2 !4 !2

f !1(x) " x 2 ! 2, x $ 0

x

!2 !4

b) Since there is no horizontal line that intersects the graph more than once, the function is oneto-one.

7 . 3

65. x4 + 5x2 − 36 = 0

Let u = x2 and substitute. u2 + 5u − 36 = 0 (u + 9)(u − 4) = 0 u+9 = 0

u = −9

x = −9 2

or u − 4 = 0 or

u=4

or

x2 = 4

x = ±3i or

x = ±2

The solutions are ±3i and ±2.

398

Chapter 6: Trigonometric Identities, Inverse Functions, and Equations

66. x2 − 10x + 1 = 0 # −(−10) ± (−10)2 − 4 · 1 · 1 x= 2·1 √ √ 10 ± 96 10 ± 4 6 = = 2 2 √ = 5±2 6 √ The solutions are 5 ± 2 6. √ 67. x−2 = 5 x − 2 = 25

74.

= =

Squaring both sides

x = 27

=

This answer checks. The solution is 27. √ 68. x = x+7+5 √ x−5 = x+7

=

x2 − 10x + 25 = x + 7 x2 − 11x + 18 = 0

(x − 2)(x − 9) = 0

=

x = 2 or x = 9

=

Only 9 checks. The solution is 9. 69. See the answer section in the text. 70. ln | sec θ + tan θ| = − ln | sec θ − tan θ| We start with the left side. ln | sec θ + tan θ| ) ) ) sec θ − tan θ )) = ln )) sec θ + tan θ · sec θ − tan θ ) ) 2 ) ) sec θ − tan2 θ ) ) = ln )) sec θ − tan θ ) ) ) ) ) 1 ) = ln )) sec θ − tan θ ) = ln |1| − ln | sec θ − tan θ|

=

1. The only number√in the restricted range [−π/2, π/2] √ with a sine of − 3/2 is −π/3. Thus, sin−1 (− 3/2) = −π/3, or −60◦ . 2. cos−1

= − ln | sec θ − tan θ|

We started with the left side and deduced the right side, so the proof is complete. 71. See the answer section in the text. I1 cos θ sin θ = # (I1 cos φ)2 + (I2 sin φ)2 I1 cos φ sin θ = # (I1 cos φ)2 + (I1 sin φ)2 I1 cos φ sin θ = & 2 I1 (cos2 φ + sin2 φ) I1 cos φ I1 sin θ = cos φ.

sin θ =

73. See the answer section in the text.

E1 − E2 2 ! " ! " √ √ π π 2Et cos θ + − 2Et cos θ − P P 2 " '! √ π π 2Et cos θ cos − sin θ sin − P P "( ! π π /2 cos θ cos + sin θ sin P P ! " √ π 2Et − 2 sin θ sin P 2 √ π − 2Et sin θ sin P

Exercise Set 6.4

= 0 − ln | sec θ − tan θ|

72.

=

E1 + E 2 2 ! " ! " √ √ π π 2Et cos θ + + 2Et cos θ − P P 2 " '! √ π π 2Et cos θ cos − sin θ sin + P P ! "( π π /2 cos θ cos + sin θ sin P P ! " √ π 2Et 2 cos θ cos P 2 √ π 2Et cos θ cos P

π 1 = , or 60◦ 2 3

3. The only number in the restricted range (−π/2, π/2) with a tangent of 1 is π/4. Thus, tan−1 1 = π/4, or 45◦ . 4. sin−1 0 = 0, or 0◦

(I1 = I2 )

5. The √only number in the restricted √ range [0, π] with a cosine of 2/2 is π/4. Thus, cos−1 ( 2/2) = π/4, or 45◦ . √ 6. sec−1 2 = π/4, or 45◦ 7. The only number in the restricted range (−π/2, π/2) with a tangent of 0 is 0. Thus, tan−1 0 = 0, or 0◦ . √ 3 π −1 8. tan = , or 30◦ 3 6 9. The √ range [0, π] with a cosine √only number in the restricted of 3/2 is π/6. Thus, cos−1 ( 3/2) = π/6, or 30◦ . √ 10. cot−1 (− 3/3) = −π/3, or −60◦

Exercise Set 6.4

399

11. The only number in the restricted range [−π/2, 0)∪(0, π/2] with a cosecant of 2 is π/6. Thus, csc−1 2 = π/6, or 30◦ . π 1 = , or 30◦ 2 6

12. sin−1

36. tan β =

π 14. tan−1 (−1) = − , or −45◦ 4

15. The only number in the restricted range![−π/2, " π/2] 1 π π 1 −1 with a sine of − is − . Thus, sin = − , or − 2 6 2 6 −30◦ . ! √ " 3π 2 = , or 135◦ . 16. cos−1 − 2 4 17. The only number in the restricted range [0, π] with a cosine of 0 is π/2. Thus, cos−1 0 = π/2, or 90◦ . √ 3 π −1 18. sin = , or 60◦ 2 3 19. The only number in the restricted range [0, π/2) ∪ (π/2, π] with a secant of 2 is π/3. Thus, sec−1 2 = π/3, or 60◦ . 20. csc−1 (−1) = −π/2, or −90◦ 21. tan−1 0.3673 ≈ 0.3520, or 20.2◦ 22. cos−1 (−0.2935) ≈ 1.8687, or 107.1◦ 23. sin−1 0.9613 ≈ 1.2917, or 74.0◦ 24. sin−1 (−0.6199) ≈ −0.6686, or −38.3◦ 26. tan−1 158 ≈ 1.5645, or 89.6◦ " ! 1 27. csc−1 (−6.2774) = sin−1 ≈ −6.2774 −0.1600, or −9.2◦ " ! 1 28. sec−1 1.1677 = cos−1 ≈ 0.5426, or 31.1◦ 1.1677

32. cos

!

51. Find tan(arcsin 0.1) We wish to find the tangent of an angle whose sine is 0.1, 1 . or 10

10

(−0.8192) ≈ −0.9600, or −55.0

3 11 √ The length of the other leg is 3 11. √ 11 1 . Thus, tan(arcsin 0.1) = √ , or 33 3 11

(−0.2716) ≈ 1.8459, or 105.8◦

33. sin−1 : [−1, 1]; cos−1 : [−1, 1]; tan−1 : (−∞, ∞) 34. sin−1 : [−π/2, π/2]; cos−1 : [0, π]; tan−1 : (−π/2, π/2) 52.

θ 2000 ft

y

19

d

θ

θ sin θ =

π π 2000 , − ≤ θ ≤ , so θ = sin−1 d 2 2

1

θ

◦

35.

π 2

√ " 3 π π −1 = = sin 49. sin cos 6 2 3 "( ' ! π π 50. sin−1 tan − = sin−1 (−1) = − 4 2 −1

29. tan−1 1.091 ≈ 0.8289, or 47.5◦ ! " 1 30. cot−1 1.265 = tan−1 ≈ 0.6689, or 38.3◦ 1.265 −1

" 50 . d

38. tan[tan−1 (−4.2)] = −4.2 ' ! "( !√ " π 2 π −1 −1 cos − = cos = 39. cos 4 2 4 √ ! " 2π 3 π 40. sin−1 sin = sin−1 = 3 2 3 " ! π π π = because is in the range of the arcsine 41. sin−1 sin 5 5 5 function. " ! ! √ " π 3 2π = cot−1 − =− 42. cot−1 cot 3 3 3 ! " √ π 2π 43. tan−1 tan = tan−1 (− 3) = − 3 3 " ! π π π = because is in the range of the arc44. cos−1 cos 7 7 7 cosine function. √ " ! 3 π 1 −1 45. sin tan = sin = 3 6 2 √ " ! 3 π 1 = cos = 46. cos sin−1 2 3 2 √ " ! 2 π 47. tan cos−1 = tan = 1 2 4 48. cos−1 (sin π) = cos−1 (0) =

25. cos−1 (−0.9810) ≈ 2.9463, or 168.8◦

31. sin

!

37. Since 0.3 is in the interval [−1, 1], sin(sin−1 0.3) = 0.3.

13. The only number in the√restricted range [−π/2, 0)∪(0, √ π/2] with a cotangent of − 3 is −π/6. Thus, cot−1 (− 3) = −π/6, or −30◦ .

−1

50 π π , − < β < , so β = tan−1 d 2 2

!

" 2000 . d

3 x

4 √ " √ ! 4 3 4 19 = √ , or . cos tan−1 4 19 19

400

Chapter 6: Trigonometric Identities, Inverse Functions, and Equations

53. sin−1

!

54. tan−1

!

7π 6

sin

tan −

55. Find sin

!

"

= sin−1

3π 4

"

!

−

√ " 3 π =− 2 6

= tan−1 1 =

π 4

" a . 3

tan−1

We draw right triangles whose legs have lengths |a| and 3 a so that tan θ = . 3 y

a2 + 9 θ θ

56.

a, positive x

3

58.

y

y

p2 + 9 θ θ

a2 + 9 " ! a −1 a =√ sin tan 3 a2 + 9

tan

!

cos

57. Find cot

x2 – 9

−1

x

!

3 x

"

=

sin−1

√

x2 − 9 3

" p . q

We draw right triangles with one leg of length |p| and hyp potenuse q so that sin θ = . q y

q

p, positive q2

θ θ

–

p2

cot

!

sin−1

p q

x

p, negative

q

"

=

#

q 2 − p2 p

x

p, negative

" p p = sin−1 # 2 3 p +9 ! " 1 1 60. tan sin−1 2 2 " ! −1 1 sin sin x sin x 2 " Using tan = ! = 1 2 1 + cos x 1 + cos sin−1 2 1 2 = π 1 + cos 6 1 2√ = 3 1+ 2 √ 1 √ , or 2 − 3 = 2+ 3 √ " ! " √ ! 3 1 3 1 sin−1 = cos · 60◦ = 61. cos 2 2 2 2 We could also have used a half-angle identity: √ " % √ ! 1 3 1 + cos[sin−1 ( 3/2)] −1 cos sin = 2 2 2 * + + 1 + cos π , 3 = 2 * + +1 + 1 , 2 = 2 √ 3 = 2 tan

3

3

p, positive

p2 + 9

y

θ

1

1 – x2 x θ θ x, negative x, positive √ √ 1 − x2 sin(cos−1 x) = , or 1 − x2 . 1 " ! p −1 . 59. Find tan sin # p2 + 9 We draw the#right triangle with one leg of length |p| and hypotenuse p2 + 9.

a, negative

x

1

1 – x2

!

Exercise Set 6.4

401

62. We use the identity sin 2θ = 2 sin θ cos θ. " ! 3 sin 2 cos−1 5 ! " ! " 3 3 = 2 sin cos−1 cos cos−1 5 5 4 3 See the triangle in = 2· · 5 5 Exercise 61. 24 = 25 √ ! " 2 3 −1 63. Evaluate cos sin + cos−1 . 2 5 This is the cosine of a sum so we use the identity cos(u + v) = cos u cos v − sin u sin v. √ " ! 2 3 + cos−1 cos sin−1 2 5 √ " ! " ! 2 3 = cos sin−1 cos cos−1 − 2 5 √ " ! " ! 2 3 sin sin−1 sin cos−1 2 5 "! " √ ! " ! 3 2 3 π = cos − · sin cos−1 4 5 2 5 √ √ ! " 2 3 2 3 · − · sin cos−1 = 2 5 2 5 " ! 3 We draw a triangle in order to find sin cos−1 . 5

We will use the identity sin(u+v) = sin u cos v +cos u sin v. We draw a triangle with an angle whose sine is x and another with an angle whose cosine is y.

1

x, positive 1 – x2

u u

x, negative 1

1

1 – y2

1

1 – y2

v v y, negative y, positive sin(sin−1 x + cos−1 y) = sin(sin−1 x) cos(cos−1 y)+cos(sin−1 x) sin(cos−1 y) # √ 1 − x2 1 − y2 · = x·y+ 1 1 # = xy + (1 − x2 )(1 − y 2 )

66. We use the identity cos(u − v) = cos u cos v + sin u sin v. We will also use the triangles in Exercise 63. cos(sin−1 x − cos−1 y)

5 θ

65. Evaluate sin(sin−1 x + cos−1 y).

4

3

Our expression simplifies to √ √ √ √ √ 2 3 2 4 3 2−4 2 2 · − · = =− . 2 5 2 5 10 10 64. We use the identity sin(u + v) = sin u cos v + cos u sin v. " ! −1 1 −1 3 + cos sin sin 2 5 " ! " ! 1 3 cos cos−1 + = sin sin−1 2 5 " ! " ! −1 1 −1 3 sin cos cos sin 2 5 1 3 π 4 = · + cos · See the triangle in 2 5 6 5 Exercise 61. √ 3 4 1 3 · = · + 2 5 2 5 √ 3+4 3 = 10

= cos(sin−1 x) cos(cos−1 y)+sin(sin−1 x) sin(cos−1 y) # √ 1 − x2 1 − y2 = ·y+x· 1 1 # √ = y 1 − x2 + x 1 − y 2

67. Evaluate sin(sin−1 0.6032 + cos−1 0.4621). We will use the identity sin(u + v) = sin u cos v + cos u sin v. We draw a triangle with an angle whose sine is 0.6032 and another with an angle whose cosine is 0.4621. 1 1 u 0.7976

0.8868

0.6032 v 0.4621

sin(sin−1 0.6032 + cos−1 (0.4621) = sin(sin−1 0.6032) cos(cos−1 0.4621)+ cos(sin−1 0.6032) sin(cos−1 0.4621) = 0.6032(0.4621) + 0.7976(0.8868) ≈ 0.9861

402

Chapter 6: Trigonometric Identities, Inverse Functions, and Equations

68. We will use the identity cos(u−v) = cos u cos v +sin u sin v and the triangles below.

83. See the answer section in the text. 84.

) ) x tan x )) sin−1 √ 2+1 x ) . ) x sin(tan−1 x) )) sin sin−1 √ 2 x +1 ) ) x ) √ x √ x2 + 1 ) x2 + 1 −1

1

1 0.8752

0.7325

u

v

0.6808

0.4838

cos(sin−1 0.7325 − cos−1 0.4838)

85. See the answer section in the text.

= cos(sin−1 0.7325) cos(cos−1 0.4838)+

86.

sin(sin−1 0.7325) sin(cos−1 0.4838) = 0.6808(0.4838) + 0.7325(0.8752) ≈ 0.9705

Answers will vary slightly depending on when rounding is done. 69. The ranges are restricted so that the inverses of the trigonometric functions are also functions. 70. The graphs have different domains and ranges. The graph of y = sin−1 x is the reflection of the portion of the graph π π of y = sin x for − ≤ x ≤ , across the line y = x. 2 2 5π . It 71. The range of the arcsine function does not include 6 ' ( π π is − , . 2 2

87.

) √ ) 1 − x2 ) cos−1 x ) tan−1 ) x ) √ ) 1 − x2 . ) cos(cos−1 x) ) cos tan−1 ) x ) x)x

h y

α

x Let θ = α − β h+y , tan α = x y tan β = , x

y

β

x

α = tan−1 β = tan−1

h+y x y x

h+y y − tan−1 . x x When x = 20 ft, y = 7 ft, and h = 25 ft we have 25 + 7 7 − tan−1 θ = tan−1 20 20 ≈ 57.99◦ − 19.29◦

72. periodic

Thus, θ = tan−1

73. radian measure 74. similar 75. angle of depression 76. angular speed

= 38.7◦ .

77. supplementary

1 1 − 4 tan−1 ≈ 3.141592654 5 239 This expression seems to approximate π.

88. 16 tan−1

78. amplitude 79. acute

Exercise Set 6.5

80. circular 81. See the answer section in the text. 82.

3 2 Since cos x is positive the solutions are in quadrants I and 11π π + 2kπ, where k is any inteIV. They are + 2kπ or 6 6 ger. The solutions can also be expressed as 30◦ + k · 360◦ or 330◦ + k · 360◦ , where k is any integer. √ 2 2. sin x = − 2 Since sin x is negative the solutions are in quadrants III 5π 7π + 2kπ or + 2kπ, where k is any and IV. They are 4 4 integer. The solutions can also be expressed as 225◦ + k · 360◦ or 315◦ + k · 360◦ , where k is any integer. 1. cos x =

π 2 π sin(tan−1 x + cot−1 x) sin 2 sin(tan−1 x) cos(cot−1 x)+ 1 cos(tan−1 x) sin(cot−1 x) x 1 1 x √ ·√ +√ ·√ 1 + x2 1 + x2 1 + x2 1 + x2 tan−1 x + cot−1 x

x2 + 1 x2 + 1 1

) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) )

√

Exercise Set 6.5 √ 3. tan x = − 3

Since tan x is negative the solutions are in quadrants II 5π 2π + 2kπ or + 2kπ. This can be and IV. They are 3 3 2π condensed as + kπ, where k is any integer. The solu3 tions can also be expressed as 120◦ + k · 180◦ , where k is any integer.

1 2 Since cos x is negative the solutions are in quadrants II 2π 4π and III. They are + 2kπ or + 2kπ, where k is any 3 3 integer. The solutions can also be expressed as 120◦ + k · 360◦ or 240◦ + k · 360◦ , where k is any integer.

4. cos x = −

1 2 Since sin x is positive the solutions are in quadrants I and 5π π II. They are + 2kπ or + 2kπ, where k is any integer. 6 6 The solutions can also be expressed as 30◦ + k · 360◦ or 150◦ + k · 360◦ , where k is any integer.

5. sin x =

6. tan x = −1

Since tan x is negative the solutions are in quadrants II 7π 3π + 2kπ or + 2kπ, where k is any and IV. They are 4 4 3π + kπ, where k is any integer. This can be condensed as 4 integer. The solutions can also be expressed as 135◦ + k · 180◦ , where k is any integer. √ 2 7. cos x = − 2 Since cos x is negative the solutions are in quadrants II 5π 3π + 2kπ or + 2kπ, where k is any and III. They are 4 4 integer. The solutions can also be expressed as 135◦ + k · 360◦ or 225◦ + k · 360◦ , where k is any integer. √ 3 8. sin x = 2 Since sin x is positive the solutions are in quadrants I and π 2π + 2kπ, where k is any integer. II. They are + 2kπ or 3 3 The solutions can also be expressed as 60◦ + k · 360◦ or 120◦ + k · 360◦ , where k is any integer. 9.

403 10.

sin x = −0.9184

Now x = arcsin(−0.9184) ≈ −66.69◦ . Since sin x is negative the solutions are in quadrants III and IV. The solutions in [0, 360◦ ) are 180◦ + 66.69◦ , or 246.69◦ , and 360◦ − 66.69◦ , or 293.31◦ . √ 11. 2 sin x + 3 = 0 √ 2 sin x = − 3 √ 3 sin x = − 2 5π 4π and . The solutions in [0, 2π) are 3 3 12. 2 tan x − 4 = 1 2 tan x = 5

tan x = 2.5 Now x = arctan 2.5 ≈ 68.20◦ . Since tan x is positive, the solutions are in quadrants I and III. The solutions in [0, 360◦ ) are 68.20◦ and 180◦ + 68.20◦ , or 248.20◦ . 13.

2 cos2 x = 1 1 cos2 x = 2

√ 2 1 cos x = ± √ , or ± 2 2

The solutions in [0, 2π) are 14.

7π π 3π 5π , , , and . 4 4 4 4

csc2 x − 4 = 0 1 −4 = 0 sin2 x 1 sin2 x = 4 sin x = ±

1 2

π 5π 7π 11π , , , 6 6 6 6 2 sin2 x + sin x = 1 x=

15.

2 sin2 x + sin x − 1 = 0

(2 sin x − 1)(sin x + 1) = 0

2 sin x − 1 = 0 or sin x + 1 = 0

2 cos x − 1 = −1.2814

2 sin x = 1 or 1 or sin x = 2

2 cos x = −0.2814 cos x = −0.1407

Using a calculator we find that the reference angle, arccos(−0.1407) is x ≈ 98.09◦ . Since cos x is negative, the solutions are in quadrants II and III. Thus, one solution is 98.09◦ . The reference angle for 98.09◦ is 180◦ − 98.09◦ , or 81.91◦ , so the other solution in [0◦ , 360◦ ) is 180◦ + 81.91◦ , or 261.91◦ .

sin x + 3 = 2.0816

sin x = −1 sin x = −1

The solutions in [0, 2π) are 16.

3π π 5π , , and . 6 6 2

sin2 x + 2 cos x = 3 cos x + 2 cos x − 3 = 0 2

(cos x + 3)(cos x − 1) = 0 cos x + 3 = 0

or cos x − 1 = 0

cos x = −3 or

cos x = 1

Since cosine values are never less than −1, cos x = −3 has no solution. Using cos x = 1, we find that the solution in [0, 2π) is 0.

404 17.

Chapter 6: Trigonometric Identities, Inverse Functions, and Equations √

22.

3 cos x = 0 √ cos x(2 cos x − 3) = 0 √ cos x = 0 or 2 cos − 3 = 0 √ 2 cos2 x −

5 sin2 x − 8 sin x − 3 = 0 √ 8 ± 64 + 60 sin x = 10 √ 8 ± 124 sin x = 10

3 2 π 3π π 11π The solutions in [0, 2π) are , , , and . 2 2 6 6 cos x = 0 or

18.

cos x =

sin x ≈ −0.3136 or sin x ≈ 1.9136

2 sin2 θ + 7 sin θ = 4

Since sine values are never greater than 1, sin x ≈ 1.9136 has no solution. Using sin x ≈ −0.3136, we find that x = arcsin(−0.3136) ≈ −18.28◦ . Thus, the solutions in [0, 360◦ ) are 180◦ + 18.28◦ , or 198.28◦ , and 360◦ − 18.28◦ , or 341.72◦ . (Answers may vary slightly due to rounding differences.)

2 sin2 θ + 7 sin θ − 4 = 0

(2 sin θ − 1)(sin θ + 4) = 0

2 sin θ − 1 = 0 or sin θ + 4 = 0 1 sin θ = or sin θ = −4 2 Since sine values are never less than −1, sin θ = −4 has 1 no solution. Using sin θ = , we find that the solutions in 2 π 5π [0, 2π) are and . 6 6 19.

23. cos2 x + 6 cos x + 4 = 0 √ √ −6 ± 36 − 16 −6 ± 20 cos x = = 2 2 cos x ≈ −0.7639 or cos x ≈ −5.2361

Since cosine values are never less than −1, cos x ≈ −5.2361 has no solution. Using cos x = −0.7639, we find that x = arccos(−0.7639) ≈ 139.81◦ . Thus, one solution in [0, 360◦ ) is 139.81◦ . The reference angle for this angle is 180◦ − 139.81◦ = 40.19◦ . Then the other solution in [0, 360◦ ) is 180◦ + 40.19◦ = 220.19◦ .

6 cos2 φ + 5 cos φ + 1 = 0 (3 cos φ + 1)(2 cos φ + 1) = 0 or 2 cos φ + 1 = 0 1 1 cos φ = − cos φ = − or 3 2 ! " 1 1 Using cos φ = − , we find that φ = arccos − ≈ 3 3 ◦ ◦ ◦ 109.47 , so one solution in [0, 360 ) is 109.47 . The reference angle for this angle is 180◦ −109.47◦ , or 70.53◦ . Thus, another solution is 180◦ + 70.53◦ , or 250.53◦ . 1 Using cos φ = − , we find that the other solutions in 2 [0, 360◦ ) are 120◦ and 240◦ .

3 cos φ + 1 = 0

20.

24.

Using tan x ≈ −1.2656, we find that x = arctan(−1.2656) ≈ −51.70◦ . Thus, two solutions in [0, 360◦ ) are 180◦ − 51.70◦ , or 128.30◦ , and 360◦ − 51.70◦ , or 308.30◦ .

2 sin t cos t + 2 sin t − cos t − 1 = 0

(2 sin t − 1)(cos t + 1) = 0

2 sin t − 1 = 0 or cos t + 1 = 0 1 or cos t = −1 sin t = 2 π 5π or t=π t= , 6 6

Using tan x ≈ 2.7656, we find that x = arctan 2.7656 ≈ 70.12◦ . Then the other solutions in [0, 360◦ ) are 70.12◦ and 180◦ +70.12◦ , or 250.12◦ . (Answers may vary slightly due to rounding differences.) 25.

sin 2x cos x − sin x = 0

2 sin x cos2 x − sin x = 0 sin x(2 cos2 x − 1) = 0

√

2 1 cos x = ± √ , or ± 2 2 π 3π 5π 7π The solutions in [0, 2π) are 0, π, , , , and . 4 4 4 4 sin x = 0 or

7 = cot2 x + 4 cot x 0 = cot2 x + 4 cot x − 7 √ −4 ± 16 + 28 Using the cot x = 2 quadratic formula √ −4 ± 44 cot x = 2 cot x ≈ 1.3166 or cot x ≈ −5.3166

(2 sin x cos x) cos x − sin x = 0

sin x = 0 or 2 cos2 x − 1 = 0 1 sin x = 0 or cos2 x = 2

2 tan2 x = 3 tan x + 7 2 tan2 x − 3 tan x − 7 = 0 √ 3 ± 9 + 56 tan x = 4 √ 3 ± 65 tan x = 4 tan x ≈ −1.2656 or tan x ≈ 2.7656

2 sin t(cos t + 1) − (cot t + 1) = 0

21.

5 sin2 x − 8 sin x = 3

Using cot x ≈ 1.3166, we find that x = arccot 1.3166 ≈ 37.22◦ . Thus, two solutions in [0, 360◦ ) are 37.22◦ and 180◦ + 37.22◦ , or 217.22◦ . Using cot x ≈ −5.3166 we find that x = arccot (−5.3166) ≈ −10.65◦ . Then the other solutions in [0, 360◦ ) are 180◦ − 10.65◦ , or 169.35◦ , and 360◦ − 10.65◦ , or 349.35◦ .

Exercise Set 6.5

405 3 sin2 x = 3 sin x + 2

26.

31.

27.

sin x = 0 sin x = 0

2 sin x cos2 x + sin x = 0 sin x(2 cos2 x + 1) = 0

x = 0, π or

0 = sin x(2 sin x + 1)

sin x = 0

or

or

1 2 2π 4π , x= 3 3

All values check.

All values check. 33.

(2 sec x + 1)(tan x + 1) = 0 2 sec x + 1 = 0

or tan x + 1 = 0 1 tan x = −1 sec x = − or 2 3π 7π No solution x= , 4 4 Both values check. The solutions in [0, 2π) are 34.

4 sin2 x cos x = 0

sin2 x = 0

2 sin 2x(cos 2x − 1) = 0

or cos x = 0 π 3π x = 0, π or x= , 2 2 All values check.

cos 2x = 1 2x = 0, 2π x = 0, π

π All values check. The solutions in [0, 2π) are 0, , π, and 2 3π . 2 30. tan x sin x − tan x = 0 tan x(sin x − 1) = 0

or sin x − 1 = 0

sin x = 1 π x= 2

π does not check, but the other values do. Thus 2 the solutions in [0, 2π) are 0 and π.

or cos x = 0

sin x = 0

or cos 2x − 1 = 0

2x = 0, π, 2π, 3π or 3π π or x = 0, , π, 2 2

The value

sin 2x sin x − cos 2x cos x = − cos x

2 sin2 x cos x − cos x + 2 sin2 x cos x = − cos x

2 sin 2x cos 2x − 2 sin 2x = 0

x = 0, π

3π 7π and . 4 4

(2 sin x cos x) sin x − (1 − 2 sin2 x) cos x = − cos x

sin[2(2x)] − 2 sin 2x = 0

or

2 sec x tan x + 2 sec x + tan x + 1 = 0 2 sec x(tan x + 1) + (tan x + 1) = 0

sin 4x − 2 sin 2x = 0

or

2x = π, 3π π 3π x= , 2 2

x = 0, π or

cos x = −

x = 0, π or

cos 2x = −1

x = 0, π or

1 sin x = − 2 7π 11π , x= 6 6

sin x = 0

tan x = 0

cos 2x sin x + sin x = 0 or cos 2x + 1 = 0

or 2 cos x + 1 = 0

tan x = 0

1 2 No solution

cos2 x = −

sin x = 0

sin x = 0

sin 2x = 0

or

0 = 2 sin x + sin x

sin x(2 cos x + 1) = 0

sin 2x = 0

sin x = 0

sin x(cos 2x + 1) = 0 2

7π , and All values check. The solutions in [0, 2π) are 0, π, 6 11π . 6 28. 2 sin x cos x + sin x = 0

29.

or 2 cos2 x + 1 = 0

x = 0, π or

or 2 sin x + 1 = 0 or

sin x = 0

Both values check. The solutions are 0 and π. 32.

cos 2x − sin x = 1

1 − 2 sin2 x − sin x = 1

sin 2x cos x + sin x = 0 (2 sin x cos x) cos x + sin x = 0

3 sin x − 3 sin x − 2 = 0 √ 3 ± 33 sin x = 6 sin x ≈ 1.4574 or sin x ≈ −0.4574 Since sine values are never greater than 1, sin x ≈ 1.4574 has no solution. Using sin x ≈ −0.4574, we find that x = arcsin(−0.4574) ≈ −27.22◦ . Thus, the solutions in [0, 360◦ ) are 180◦ + 27.22◦ , or 207.22◦ , and 360◦ − 27.22◦ , or 332.78◦ . 2

35.

sin 2x + sin x + 2 cos x + 1 = 0 2 sin x cos x + sin x + 2 cos x + 1 = 0 sin x(2 cos x + 1) + 2 cos x + 1 = 0 (sin x + 1)(2 cos x + 1) = 0 sin x + 1 = 0

or 2 cos x + 1 = 0

sin x = −1 or x=

3π or 2

1 2 2π 4π x= , 3 3

cos x = −

2π 4π , , and All values check. The solutions in [0, 2π) are 3 3 3π . 2

406 36.

Chapter 6: Trigonometric Identities, Inverse Functions, and Equations 39.

tan2 x + 4 = 2 sec2 x + tan x tan x + 4 = 2(1 + tan x) + tan x 2

2

tan2 x + 4 = 2 + 2 tan2 x + tan x 0 = tan2 x + tan x − 2

0 = (tan x + 2)(tan x − 1)

tan x + 2 = 0

or

x ≈ 2.034, 5.176 or

All values check.

sin 2x = 2x =

sec x − 2 tan x = 0

1 + tan2 x − 2 tan2 x = 0 1 − tan2 x = 0 tan2 x = 1 tan x = ±1 π 3π 5π 7π x= , , , 4 4 4 4

π 3π 5π All values check. The solutions in [0, 2π) are , , , 4 4 4 7π . 4 38.

sin 2x + 1 =

tan x = 1 π 5π x= , 4 4

2

2

x=

1 = 3 tan x 1 = tan2 x 3 2

√

3 = tan x 3 π 5π 7π 11π , , , =x 6 6 6 6 3π π and check. The solutions in These values and also 2 2 π π 5π 7π 3π 11π [0, 2π) are , , , , , and . 6 2 6 6 2 6 ±

Squaring

both sides 3 2 1 2 π 5π 13π 17π , , , 6 6 6 6 π 5π 13π 17π , , , 12 12 12 12

13π 17π and do not check, but the other values 12 12 π 5π do. The solutions in [0, 2π) are and . 12 12 √ 40. 3 cos x − sin x = 1 √ 3 cos x = 1 + sin x The values

3 cos2 x = 1 + 2 sin x + sin2 x 3(1 − sin2 x) = 1 + 2 sin x + sin2 x 3 − 3 sin2 x = 1 + 2 sin x + sin2 x

cot x = tan(2x − 3π) tan 2x − tan 3π cot x = 1 + tan 2x tan 3π tan 2x − 0 cot x = 1 + tan 2x · 0 cot x = tan 2x 2 tan x 1 = tan x 1 − tan2 x In this form of the equation, x cannot be π/2 or 3π/2. Therefore π/2 and 3π/2 must also be checked in the original equation. 1 − tan2 x = 2 tan2 x

√

6 √ 6 cos x + sin x = 2 6 cos2 x + 2 sin x cos x + sin2 x = 4

or tan x − 1 = 0

tan x = −2

37.

2 cos x + 2 sin x =

0 = 4 sin2 x + 2 sin x − 2 0 = 2 sin2 x + sin x − 1

0 = (2 sin x − 1)(sin x + 1)

2 sin x − 1 = 0 or sin x + 1 = 1 or sin x = sin x = 2 π 5π x= , or x= 6 6 5π The value does not check, but 6 π check. The solutions in [0, 2π) are 6 41.

0

−1 3π 2 the other values do 3π and . 2

sec2 x + 2 tan x = 6 1 + tan2 x + 2 tan x = 6 tan2 x + 2 tan x − 5 = 0 √ √ −2 ± 24 −2 ± 4 + 20 = tan x = 2 2 tan x ≈ 1.4495 or tan x ≈ −3.4495

Using tan x ≈ 1.4495, we find that x = arctan 1.4495 ≈ 0.967. Then two possible solutions in [0, 2π) are 0.967 and π + 0.967, or 4.109. Using tan x ≈ −3.4495, we find that x = arctan(−3.4495) ≈ −1.289. Thus, the other two possible solutions in [0, 2π) are π − 1.289, or 1.853, and 2π − 1.289, or 4.994. All values check. The solutions in [0, 2π) are 0.967, 1.853, 4.109, and 4.994. (Answers may vary slightly due to rounding differences.)

Exercise Set 6.5 42.

407 50. Find the zero of y1 = x cos x−2 The only solution in [0, 2π) is 5.114.

5 cos 2x + sin x = 4 5(1 − 2 sin x) + sin x = 4 2

51. Find the points of intersection of y1 = cos x − 2 and y2 = x2 − 3x or find the zeros of y1 = cos x − 2 − x2 + 3x. The solutions in [0, 2π) are 0.422 and 1.756.

0 = 10 sin2 x − sin x − 1

√

1 ± 41 20 sin x ≈ 0.3702 or sin x ≈ −0.2702

sin x =

Using sin x ≈ 0.3702, we find that x = arcsin 0.3702 ≈ 0.379. Then two possible solutions in [0, 2π) are 0.379 and π − 0.379, or 2.763.

Using sin x ≈ −0.2702, we find that x = arcsin(−0.2702) ≈ −0.274. Then the other two possible solutions in [0, 2π) are π + 0.274, or 3.416, and 2π − 0.274, or 6.009.

(Answers may vary slightly due to rounding differences.) " ! 43. cos(π − x) + sin x − π = 1 2 (− cos x) + (− cos x) = 1 −2 cos x = 1

1 2 2π 4π , x= 3 3

x 52. Find the points of intersection of y1 = sin x and y2 = tan 2 x or find the zeros of y1 = sin x − tan . The solutions in 2 π 3π [0, 2π) are 0, 1.571, and 4.712 or 0, , and . 2 2 53. a) Using the sine regression feature on a graphing calculator, we get y = 7 sin(−2.6180x + 0.5236) + 7 b) In December, x = 12; when x = 12, y ≈ 10.5. Thus, total sales in December are about $10,500. In July, x = 7; when x = 7, y ≈ 13.062. Thus, total sales in July are about $13,062. 54. a) Using the sine regression feature on a graphing calculator, we get y = 7.8787 sin(0.0166x − 1.2723) + 12.1840.

cos x = −

Both values check. The solutions in [0, 2π) are 44.

√ 2 sin2 x − 1 " ! = −1 π 2 cos −x +1 2 √ sin2 x − 1 2 = −1 sin x + 1 2 √ (sin x + 1)(sin x − 1) 2 = −1 sin x + 1 2 √ 2 −1 sin x − 1 = 2 √ 2 sin x = 2 π 3π x= , 4 4 Both values check.

b) On April 22, x = 112; when x = 112, y ≈ 16.5. Thus, there are approximately 16.5 hours of daylight on April 22. On July 4, x = 185; when x = 185, y ≈ 19.9. Thus, there are approximately 19.9 hours of daylight on July 4.

2π 4π and . 3 3

45. Left to the student

On December 15, x = 349; when x = 349, y ≈ 4.5. Thus, there are approximately 4.5 hours of daylight on December 15. 55. Yes; first note that 7π/6 = π/6+π. Since π/6+kπ includes both odd and even multiples of π it is equivalent to π/6 + 2kπ and 7π/6 + 2kπ. 56. “Possible” replacements for the variables are those for which the expressions in the identity are defined. 57. B = 90◦ − 55◦ = 35◦ 201 tan 55◦ = b 201 b= tan 55◦ b ≈ 140.7 201 c 201 c= sin 55◦ c ≈ 245.4

46. Left to the student

sin 55◦ =

47. Find the points of intersection of y1 = x sin x and y2 = 1 or find the zeros of y1 = x sin x−1. The solutions in [0, 2π) are 1.114 and 2.773. 48. Find the points of intersection of y1 = x2 +2 and y2 = sin x or find the zeros of y1 = x2 + 2 − sin x. There are no solutions in [0, 2π). 49. Find the point of intersection of y1 = 2 cos2 x and y2 = x + 1 or find the zero of y1 = 2 cos2 x − x − 1. The only solution in [0, 2π) is 0.515.

58.

3.8 14.2 R ≈ 15.5◦

sin R =

T ≈ 90◦ − 15.5◦ ≈ 74.5◦

408

Chapter 6: Trigonometric Identities, Inverse Functions, and Equations t2 + (3.8)2 = (14.2)2

65.

t2 = (14.2)2 − (3.8)2 = 187.2 59.

60.

x 4 = 27 3 x = 36

cos x = 1

t ≈ 13.7

x=0 This value checks.

Multiplying by 27

66.

sin x = 0

0.2(0.7) = 14 0.01 √ 3 61. | sin x| = 2 √ √ 3 3 sin x = or sin x = − 2 2 4π 5π π 2π or x= , x= , 3 3 3 3

x = 0, π Both values check.

h=

π 2π 4π All values check. The solutions in [0, 2π) are , , , 3 3 3 5π and . 3

1 2 1 1 cos x = or cos x = − 2 2 2π 4π π 5π or x= , x= , 3 3 3 3 All values check. √ √ 63. tan x = 4 3 √ √ ( tan x)4 = ( 4 3)4

67.

68.

69.

" π t , 0 ≤ t ≤ 12 8 ! " π 101.6◦ + 3◦ sin t Substituting 8 ! " π t 3◦ sin 8 ! " π sin t 8 ! 0.4855, 2.6561 0 ≤ t ≤ 12, so " 3π π 0≤ t≤ 8 2 1.24, 6.76

T (t) = 101.6◦ + 3◦ sin 103◦ = 1.4◦ = 0.4667 ≈

√ tan x = ± 3 π 2π 4π 5π x= , , , 3 3 3 3 4π π and check. They are the solutions in [0, 2π). Only 3 3 √ 64. 12 sin x − 7 sin x + 1 = 0 √ Let u = sin x. 12u2 − 7u + 1 = 0

π t≈ 8

t≈

!

Both values check. The patient’s temperature was 103◦ at t ≈ 1.24 days and t ≈ 6.76 days. 70.

All values check. (Answers may vary slightly due to rounding differences.)

eln sin x = 1 sin x = 1 π x= 2 This value checks.

tan2 x = 3

or 3u − 1 = 0 1 or u= 3 √ 1 sin x = or 3 1 or sin x = 9 or x ≈ 0.111, 3.031

sin(ln x) = −1 3π + 2kπ, k an integer ln x = 2 x = e3π/2+2kπ , k an integer x is in the interval [0, 2π) when k ≤ −1. Thus, the possible solutions are e3π/2+2kπ , k an integer, k ≤ −1. These values check and are the solutions.

62. | cos x| =

4u − 1 = 0 1 u= 4 √ 1 sin x = 4 1 sin x = 16 x ≈ 0.063, 3.079

esin x = 1 ln esin x = ln 1

0.2 0.01 = 0.7 h

(4u − 1)(3u − 1) = 0

ln(cos x) = 0

Note that

0 ≤ t ≤ 240, so

−10 ≤ t − 10 ≤ 230 and π π π (−10) ≤ (t − 10) ≤ (230), or 45 45 45 π 46π 2π ≤ (t − 10) ≤ . − 9 45 9 ( ' π 3000 = 5000 cos (t − 10) 45 π 0.6 = cos (t − 10) 45 We will consider only first quadrant solutions. Solutions in the fourth quadrant represent positions south of the equator.

Chapter 6 Review Exercises

409

π (t − 10) ≈ 0.9273, 7.2105, 13.4937 45 t ≈ 23.28 min, 113.28 min, 203.28 min 71. N (φ) = 6066 − 31 cos 2φ

We consider φ in the interval [0◦ , 90◦ ] since we want latitude north. 6040 = 6066 − 31 cos 2φ Substituting −26 = −31 cos 2φ

(0◦ ≤ 2φ ≤ 180◦ )

0.8387 ≈ cos 2φ 2φ ≈ 33.0

◦

φ ≈ 16.5◦

The value checks. At about 16.5◦ N the length of a British nautical mile is found to be 6040 ft. 72. Using the result of Exercise 53, Exercise Set 6.2, we have 9.8 = 9.78049(1 + 0.0005264 sin2 φ+ 0.000024 sin4 φ). We solve for 0◦ ≤ φ ≤ 90◦ . φ ≈ 37.95615◦ N

3 . 5 (See the triangle in Exercise 63, Exercise Set 6.4) Then 4 3 4 sin θ = . Thus, arccos = arcsin . 5 5 5 4 3 arccos x = arccos − arcsin 5 5 arccos x = 0

73. Sketch a triangle having an angle θ whose cosine is

x=1 74. sin−1 x = tan−1

2. If α and β are acute angles and sin(α + β) = 1, then π α + β = . Also, if sin(α − β) = 0, then α − β = 0, or 2 α = β. Then we have π α+β = 2 π α+α = Since α = β 2 π 2α = 2 π α= . 4 The statement is true. 3. If the terminal side of θ is in quadrant IV, then tan θ < 0 and cos θ > 0, so tan θ < cos θ. The statement is true. 7π 5π = − cos , so the statement is false. 12 12 2 5. If sin θ = − , then θ is a third or fourth quadrant angle. 5 If θ is in quadrant III, then tan θ > 0 and cos θ < 0, so tan θ $< cos θ and the statement is false. 4. cos

7. sin2 x + cos2 x = 1

1 sin−1 x = α + β where α = tan−1 , and 3 1 β = tan−1 . 2

8. (tan y − cot y)(tan y + cot y) = tan2 y − cot2 y 9.

y

10 α

1

5 x

1

β

3

x

= sin α cos β + cos α sin β 2 3 1 1 = √ ·√ +√ ·√ 10 5 10 5 √ 2 . = 2 sin x = 5 cos x sin x =5 cos x tan x = 5 x ≈ 1.3734, 4.5150

Then sin x cos x = sin(1.3734) cos(1.3734) ≈ 0.1923.

(The result is the same if x ≈ 4.5150 is used.)

(cos x + sec x)2 "2 ! 1 = cos x + cos x ! 2 "2 cos x + 1 = cos x =

2

Then x = sin(α + β)

75.

π 1. The statement is true. For example, let s = . Then ! 2"2 π 2 2 π 2 2 = 1 = 1, but sin s = sin = sin s = sin 2 2 2 π ≈ 0.6243 $= 1. sin 4

6. 1 + cot2 x = csc2 x

1 1 + tan−1 , or 3 2

y

Chapter 6 Review Exercises

(cos2 x + 1)2 cos2 x

10. sec x csc x − csc2 x = csc x(sec x − csc x)

11. 3 sin2 y − 7 sin y − 20 = (3 sin y + 5)(sin y − 4) 1000 − cos3 u

12.

= 103 − (cos u)3

= (10 − cos u)(100 + 10 cos u + cos2 u) sec4 x − tan4 x sec2 x + tan2 x

13.

(sec2 x + tan2 x)(sec2 x − tan2 x) sec2 x + tan2 x 2 = sec x − tan2 x =

=1

410 14.

Chapter 6: Trigonometric Identities, Inverse Functions, and Equations 2 sin2 x · cos3 x

!

cos x 2 sin x

2 sin2 x cos2 x = · cos3 x 4 sin2 x 1 1 = · 2 cos x 1 = sec x 2 15.

16.

17.

"2

2 3 − cos y − sin y sin2 y − cos2 y 2 −1 3 − · = cos y − sin y sin2 y − cos2 y −1 3 2 = + cos y − sin y cos2 y − sin2 y cos y + sin y 2 3 = · + cos y − sin y cos y + sin y cos2 y − sin2 y 3 cos y + 3 sin y + 2 = cos2 y − sin2 y ! "2 1 cot2 x 1 cot x + = + 2 csc x csc x csc2 x csc2 x cot2 x + 1 csc2 x

csc2 x csc2 x =1

19.

20.

cos x = tan x

√

22.

=

18.

$

4 sin x cos2 x 1 cos x = · 4 sin x 16 sin2 x cos x 1 = cot x 4 # sin2 x + 2 cos x sin x + cos2 x # = (sin x + cos x)2

= sin x + cos x $ $ 1 + sin x 1 + sin x 1 − sin x = · 1 − sin x 1 − sin x 1 − sin x % 1 − sin2 x = (1 − sin x)2 % cos2 x = (1 − sin x)2 cos x = 1 − sin x

= = = =

$

cos x cos x · tan x cos x

$

cos2 x sin x cos x = √ sin x =

3 sin x cos2 x + cos x sin x · cos2 x sin2 x − cos2 x 3 sin x cos x(cos x + sin x) = cos2 x(sin x + cos x)(sin x − cos x) 3 sin x 3 sin x = · = cos x(sin x − cos x) sin x − cos x cos x 3 tan x = sin x − cos x

=

21.

9 + x2 # 9 + (3 tan θ)2 √ 9 + 9 tan2 θ # 9(1 + tan2 θ) √ 9 sec2 θ

= 3 sec θ " ! 3π 3π 3π = cos x cos − sin x sin 23. cos x + 2 2 2 24. tan(45◦ − 30◦ ) =

tan 45◦ − tan 30◦ 1 + tan 45◦ tan 30◦

25. cos 27◦ cos 16◦ +sin 27◦ sin 16◦ = cos(27◦ −16◦ ) = cos 11◦ 26.

cos 165◦ = cos(45◦ + 120◦ )

= cos 45◦ cos 120◦ − sin 45◦ sin 120◦ ! √ "! " ! √ "! √ " 2 1 2 3 = − − 2 2 2 2 √ √ √ √ 2 6 − 2− 6 =− − =− 4 4 4 √ 2 π π and 0 ≤ β ≤ , then β = and thus 27. If sin β = 2 2 4 tan β = 1. tan α − tan β tan(α − β) = 1 + tan α tan β √ 3−1 √ = 1+ 3·1 √ √ 3−1 1− 3 √ · √ = 1+ 3 1− 3 √ −4 + 2 3 = −2 √ = 2− 3 # # 28. cos θ = 1 − sin2 θ sin φ = 1 − cos2 φ # # ≈ 1 − (0.2341)2 ≈ 1 − (0.5812)2 ≈ 0.8138

≈ 0.9722

cos(θ + φ) = cos θ cos φ − sin θ sin φ

≈ (0.8138)(0.2341) − (0.5812)(0.9722)

29.

≈ −0.3745 " π π π = cos x cos − sin x sin cos x + 2 2 2 = cos x · 0 − sin x · 1 !

= − sin x

Chapter 6 Review Exercises 30.

cos

!

411

" π π π − x = cos cos x + sin sin x 2 2 2 = 0 · cos x + 1 · sin x = sin x

31.

!

sin x −

π 2

"

π π − cos x sin 2 2 = sin x · 0 − cos x · 1 = sin x cos

= − cos x

32. a) Recall the drawing from Exercise 11 in Section 6.2. y θ x

-3 -4

33.

! " π cos α 3 tan α + = = − cot α = − 2 − sin α 4 " ! π − sin α 4 cot α + = = − tan α = − 2 cos α 3 " ! " ! 1 5 π 5 = = − csc α = − − = sec α + 2 − sin α 4 4 ! " π 1 5 csc α + = = sec α = − 2 cos α 3 " ! 1 π ! " csc x − = π 2 sin x − 2 =−

5

Replacing θ with α, we have: 4 −4 =− sin α = 5 5 4 −4 = tan α = −3 3 3 −3 = cot α = −4 4 5 5 sec α = =− −3 3 5 5 =− csc α = −4 4 π b) Since α and − α are complements, we have 2 ! " π 3 sin − α = cos α = − 2 5 " ! π 4 cos − α = sin α = − 2 5 " ! 3 π − α = cot α = tan 2 4 " ! π 4 cot − α = tan α = 2 3 " ! 5 π sec − α = csc α = − 2 4 " ! π 5 csc − α = sec α = − 2 3 " ! π c) Since sin + α = cos α and 2" ! π cos + α = − sin α, we have 2 " ! 3 π = cos α = − sin α + 2 5 " ! " ! π 4 4 = − sin α = − − = cos α + 2 5 5

sin

!

1 cos x = − sec x

1 " π −x 2

=−

34. First find sin θ. sin2 θ + cos2 θ = 1 ! "2 4 =1 sin2 θ + − 5 16 sin2 θ + =1 25 9 sin2 θ = 25 3 sin θ = − 5

Choosing the negative

square root since θ is in quadrant III "! " ! 4 24 3 sin 2θ = 2 sin θ cos θ = 2 − − = 5 5 25 "2 ! "2 ! 4 3 cos 2θ = cos2 θ − sin2 θ = − − − = 5 5 9 7 16 − = 25 25 25 24 24 25 sin 2θ 24 25 = = · = tan 2θ = 7 cos 2θ 25 7 7 25 Since sin 2θ and cos 2θ are both positive, 2 θ is in quadrant I. * + + 1 − cos π , 4

1 π π = sin · = = 8 2 4 2 * √ + 2 # $ + √ √ ,1 − 2 = 2− 2 = 2− 2 2 4 2

35. sin

!

"

412

Chapter 6: Trigonometric Identities, Inverse Functions, and Equations 1 − sin x cos x = cos x 1 + sin x We start with the left side.

36. First find cos β.

41. Prove:

sin2 β + cos2 β = 1 (0.2183)2 +cos2 β = 1 cos2 β = 1 − (0.2183)2 # cos β = 1−(0.2183)2

β is in quadrant I

cos β ≈ 0.9759

sin 2β = 2 sin β cos β

1 − sin x 1 + sin x 1 − sin x = · cos x cos x 1 + sin x =

cos2 x cos x(1 + sin x) cos x = 1 + sin x We started with the left side and deduced the right side, so the proof is complete. =

≈ 2(0.2183)(0.9759)

≈ 0.4261 $ 1 + cos β β cos = 2 2 $ 1 + 0.9759 ≈ 2 ≈ 0.9940

1 + cos 2θ = cot θ sin 2θ We start with the left side.

42. Prove:

To find cos 4β, we first find cos 2β. # cos 2β = 1 − sin2 2β # ≈ 1 − (0.4261)2

1 + cos2 θ − sin2 θ 1 + cos 2θ = sin 2θ 2 sin θ cos θ

≈ 0.9047

cos 4β = cos(2(2 β)) = cos2 (2 β) − sin2 (2 β)

! " x x = cos 2 · = cos x 2 2

(sin x + cos x)2 − sin 2x

= (sin x + 2 sin x cos x + cos x) − 2 sin x cos x 2

2

= sin2 x + cos2 x =1 39.

2 sin x cos3 x + 2 sin3 x cos x = 2 sin x cos x(cos2 x + sin2 x) = 2 sin x cos x · 1 = sin 2x

40.

1 − sin2 θ + cos2 θ 2 sin θ cos θ

=

cos2 θ + cos2 θ 2 sin θ cos θ

2 cos2 θ 2 sin θ cos θ cos θ = sin θ = cot θ

≈ 0.6369

38.

=

=

≈ (0.9047)2 − (0.4261)2 37. 1 − 2 sin2

1 − sin2 x cos x(1 + sin x)

2 cos x 2 cot x sin2 x sin x = · 2 2 cos x cot x − 1 sin2 x −1 2 sin x 2 sin x cos x = cos2 x − sin2 x sin 2x = cos 2x = tan 2x

We started with the left side and deduced the right side, so the proof is complete. y tan y + sin y = cos2 2 tan y 2 We start with the right side. ! $ "2 1 + cos y 2 y = ± cos 2 2 1 + cos y = 2 1 + cos y tan y = · 2 tan y tan y + cos y tan y = 2 tan y sin y tan y + cos y · cos y = 2 tan y tan y + sin y = 2 tan y We started with the right side and deduced the left side, so the proof is complete.

43. Prove:

Chapter 6 Review Exercises sin x − cos x tan2 x − 1 44. Prove: = 2 cos x sin x + cos x We start with the right side.

413 58.

!

cos 2 sin

−1

4 5

sin2 x −1 2 cos2 x tan2 x − 1 = cos x · sin x + cos x sin x + cos x cos2 x =

sin2 x − cos2 x x(sin x + cos x)

cos2

(sin x + cos x)(sin x − cos x) cos2 x(sin x + cos x) sin x − cos x = cos2 x We started with the right side and deduced the left side, so the proof is complete. =

45.

46.

47. 48. 49.

1 3 cos 2θ sin θ = 3 · [sin(2θ + θ) − sin(2θ − θ)] 2 3 = (sin 3θ − sin θ) 2 ! " 5θ −3θ sin θ − sin 4θ = 2 cos sin 2 2 3θ 5θ sin = −2 cos 2 2 ! " 1 π sin−1 − = − , or −30◦ 2 6 √ 3 π cos−1 = , or 30◦ 2 6 π tan−1 1 = , or 45◦ 4

b 57. The angle θ = tan−1 is an acute angle of a right trian3 gle where √ b is the side √ opposite θ, 3 is the side adjacent 2 + 32 , or b b2 + 9, is the hypotenuse. Then to θ, and " ! adj 3 b = cos θ = =√ . cos tan−1 3 hyp b2 + 9

= 1 − 2 sin

2

!

sin

−1

! "2 4 = 1−2 5 32 = 1− 25 7 =− 25

4 5

"

2 2 Since cos x is negative, the solutions are in quadrants II 3π 5π and IV. They are + 2kπ or + 2kπ, where k is any 4 4 integer. The solutions can also be expressed as 135◦ + k · 360◦ or 225◦ + k · 360◦ , where k is any integer. √ 60. tan x = 3 59. cos x = −

Since tan x is positive, the solutions are in quadrants I and 4π π + 2kπ, where k is any integer. III. They are + 2kπ or 3 3 π This can be condensed as + kπ, where k is any integer. 3 The solutions can also be expressed as 60◦ +k ·180◦ , where k is any integer. 61.

4 sin2 x = 1 1 sin2 x = 4 sin x = ±

1 2

1 1 or sin x = 2 2 7π 11π π 5π x= , or x= , 6 6 6 6

sin x = −

50. sin−1 0 = 0, or 0◦ 51. cos−1 (−0.2194) ≈ 1.7920, or 102.7◦ " ! 1 52. cot−1 (2.381) = tan−1 ≈ 0.3976, or 22.8◦ 2.381 " ! 1 1 53. cos cos−1 = 2 2 √ " √ ! 3 3 54. tan−1 tan = 3 3 ! " π π = 55. sin−1 sin 7 7 √ " √ ! 2 π 2 −1 = cos = 56. cos sin 2 4 2

√

"

π 5π 7π All values check. The solutions in [0, 2π) are , , , 6 6 6 11π and . 6 62.

sin 2x sin x − cos x = 0

(2 sin x cos x) sin x − cos x = 0

2 sin2 x cos x − cos x = 0 cos x(2 sin2 x − 1) = 0

cos x = 0 or 2 sin2 x − 1 = 0 1 cos x = 0 or sin2 x = 2

√ 1 2 sin x = ± √ , or ± 2 2 π 3π π 3π 5π 7π x= , , or x = , , , 2 2 4 4 4 4 cos x = 0 or

π π All values check. The solutions in [0, 2π) are , , 4 2 3π 5π 3π 7π , , , and . 4 4 2 4

63.

2 cos2 x + 3 cos x = −1

2 cos2 x + 3 cos x + 1 = 0

(2 cos x + 1)(cos x + 1) = 0

414

Chapter 6: Trigonometric Identities, Inverse Functions, and Equations 2 cos x + 1 = 0

1 or 2 2π 4π , or x= 3 3

cos x = −

x=π

sin2 x − 7 sin x = 0

sin x(sin x − 7) = 0

sin x = 0 or sin x − 7 = 0 sin x = 0 or x = 0, π

sin x = 7

No solution

0 and π check. They are the solutions in [0, 2π). 65.

csc2 x − 2 cot2 x = 0

1 + cot2 x − 2 cot2 x = 0 (1 + cot x)(1 − cot x) = 0 cot x = −1

or 1 − cot x = 0 or

tan x = −1 or 3π 7π x= , or 4 4

cot x = 1

tan x = 1 π 5π x= , 4 4

π 3π 5π , , All values check. The solutions in [0, 2π) are , 4 4 4 7π . and 4 66.

sin 4x + 2 sin 2x = 0 2 sin 2x cos 2x + 2 sin 2x = 0

sin 2x = 0

or cos 2x + 1 = 0 or

2x = 0, π, 2π, 3π or 3π π or x = 0, , π, 2 2

6 tan2 x = 5 tan x + sec2 x 6 tan2 x = 5 tan x + 1 + tan2 x 5 tan2 x − 5 tan x − 1 = 0

We solve for tan x using the quadratic formula with a = 5, b = −5, and c = −1. √ 5±3 5 tan x = 10 tan x ≈ 1.1708 or tan x ≈ −0.1708

Using tan x = 1.1708, we find that x = arctan 1.1708 ≈ 0.864. Then the two possible solutions in [0, 2π) are 0.864 and 0.864 + π, or 4.006. Using tan x ≈ −0.1708, we find that x = arctan(−0.1708) ≈ −0.169. Then the other two possible solutions in [0, 2π) are −0.169 + π, or 2.972, and −0.169 + 2π, or 6.114.

69. The domain of the function cos−1 x is [−1, 1], so answer B is correct. ! " ! " 7π 1 π 70. sin−1 sin = sin−1 − = − , so answer A is 6 2 6 correct. 71. Expression B completes the identity. cos x 1 − cos x · csc x − cos x cot x = sin x sin x =

1 − cos2 x sin x

sin2 x sin x = sin x =

2 sin 2x(cos 2x + 1) = 0 2 sin 2x = 0

68.

The solutions in [0, 2π) are 0.864, 2.972, 4.006, 6.114.

1 − cot2 x = 0

1 + cot x = 0

The values

cos x = −1

2π All values check. The solutions in [0, 2π) are , π, and 3 4π . 3 64.

11π 19π and do not check, but the other values 12 12 23π 7π and . do. The solutions in [0, 2π) are 12 12

or cos x + 1 = 0

cos 2x = −1

2x = π, 3π π 3π x= , 2 2

π All values check. The solutions in [0, 2π) are 0, , π, and 2 3π . 2 √ 67. 2 cos x + 2 sin x = 2 √ 2 cos x + sin x = 2 2 Squaring both sides cos2 x+2 cos x sin x+sin2 x = 4 1 sin 2x + 1 = 2 1 sin 2x = − 2 7π 11π 19π 23π 2x = , , , 6 6 6 6 7π 11π 19π 23π , , , x= 12 12 12 12

72. Expression D completes the identity. cos x 1 cos x cos x 1 − = − · sin x cos x sin x sin x cos x sin x cos x =

cos2 x 1 − sin x cos x sin x cos x

=

1 − cos2 x sin x cos x

= = = = =

sin2 x sin x cos x sin x cos x sin x cos x · cos x cos x sin x cos x cos2 x sin x cos x 1 − sin2 x

Chapter 6 Review Exercises

415 Answers may vary. Method 2 may be the most efficient because it involves straightforward factorization and simplification. Method 1(a) requires a manipulation that is not obvious, multiplying by a particular expression equal to 1.

73. Expression A completes the identity. cos x −1 cot x − 1 = sin x sin x 1 − tan x 1− cos x cos x sin x − sin x = sin x cos x sin x − cos x cos x = =

=

=

cos x cos x − sin x · sin x cos x − sin x cos x sin x 1 sin x 1 cos x csc x sec x

74. Expression C completes the identity. sin x (cos x + 1)2 + sin2 x cos x + 1 + = sin x cos x + 1 sin x(cos x + 1) cos2 x + 2 cos x + 1 + sin2 x sin x(cos x + 1) 2 cos x + 2 = sin x(cos x + 1) =

2(cos x + 1) sin x(cos x + 1) 2 = sin x =

75. Find the point of intersection of y1 = x cos x and y2 = 1 or find the zero of y1 = x cos x − 1. The only solution in [0, 2π) is 4.917. 76. Find the points of intersection of y1 = 2 sin2 x and y2 = x + 1 or find the zeros of y1 = 2 sin2 x − x − 1. There are no solutions in [0, 2π). 77. a) 2 cos2 x − 1 = cos 2x

= cos2 x − sin2 x

= 1 · (cos2 x − sin2 x)

= (cos2 x + sin2 x)(cos2 x − sin2 x)

= cos4 x − sin4 x

b) cos4 x − sin4 x = (cos2 x + sin2 x)(cos2 x − sin2 x) = 1 · (cos2 x − sin2 x) = cos2 x − sin2 x = cos 2x

c)

= 2 cos2 x − 1 2 cos2 x − 1 cos4 x − sin4 x

cos 2x

(cos2 x + sin2 x)(cos2 x − sin2 x) 1 · (cos2 x − sin2 x) cos2 x − sin2 x cos 2x

78. The ranges of the inverse trigonometric functions are restricted in order that they might be functions. 79.

l1 : x + y = 3 y = −x + 3

l2 : 2x − y = 5 2x − 5 = y

l1 has m1 = −1; l2 has m2 = 2. 3 2 − (−1) = = −3 Then tan φ = 1 + 2(−1) −1 and φ = −71.6◦ + 180◦ = 108.4◦ 80.

81.

cos(u + v) = cos u cos v − sin u sin v " ! " ! π π = cos u cos v − cos − u cos −v 2 2 " ! π cos − x [csc x − sin x] 2 ( ' 1 = sin x − sin x sin x = 1 − sin2 x = cos2 x

1 π , ≤ 2θ < π 5 2 Find cos 2θ. Since 2θ is in quadrant II, the value of the cosine function is negative. # cos 2θ = − 1 − sin2 2θ % ! "2 1 = − 1− 5 $ 24 =− 25 √ 2 6 =− 5 π π π Since ≤ 2θ < π, we have ≤ θ < , so all the function 2 4 2 values of θ are positive. ! " $ 1 − cos 2θ 2θ = sin θ = sin 2 2 * √ " * ! + √ + + 2 6 2 6 + +1 − − , 1 + , 5 5 = = 2 2 % √ 1 6 + = 2 5

82. sin 2θ =

416

Chapter 6: Trigonometric Identities, Inverse Functions, and Equations

cos θ = cos

!

2θ 2

"

1 + cos 2θ = 2 * √ + 2 6 + ,1 − 5 = 2 % √ 1 6 − = 2 5 $

3.

sin θ cos θ $ $ √ √ 1 6 5+2 6 + = $ 2 √5 = $ 10√ 1 6 5−2 6 − 2 5 10 % √ % 5+2 6 10 √ = · 10 5−2 6 % √ 5+2 6 √ = 5−2 6

4.

=

%

1 − sin2 θ (1 + sin θ)2

√

4 − x2 =

= 5.

√ √ 2 2 . Then tan−1 ≈ 0.6155 but 85. Let x = 2 2 √ 2 π sin−1 2 4 = 1. √ = π 2 −1 cos 4 2 cos x e =1 86.

3π π and . 2 2

Chapter 6 Test 2 cos2 x − cos x − 1 (2 cos x + 1)(cos x − 1) = cos x − 1 cos x − 1 = 2 cos x + 1 "2 ! sec2 x 1 1 sec x = − − 2 tan x tan x tan2 x tan2 x

sin 75 = sin(45◦ + 30◦ ) ◦

= sin 45◦ cos 30◦ + cos 45◦ sin 30◦ ! √ "! √ " ! √ "! " 2 3 2 1 = + 2 2 2 2 √ √ √ √ 6 2 6+ 2 + = = 4 4 4 ! " π π π = tan − tan 12 3 4 π π tan − tan 3 4 = π π 1 + tan tan 3 4 √ 3−1 √ = 1+ 3·1 √ √ 3−1 3 √ ·√ = 1+ 3 3 √ 3− 3 = √ 3+3

cos(u − v) = cos u cos v + sin u sin v ! "! " ! "! " 5 12 12 5 = + 13 13 13 13 60 60 + = 169 169 120 = 169 8. θ is an acute angle in a reference triangle where −2 is the # 2 − (−2)2 = 3 side adjacent to θ, 3 is the hypotenuse and √ 5 is the side opposite θ. √ ! " π opp 5 cos − θ = sin θ = = 2 hyp 3

9. We make a drawing. y

sec2 x − 1 = tan2 x tan2 x tan2 x =1

cos2 θ (1 + sin θ)2 cos θ = 1 + sin θ # 4 − (2 sin θ)2 # 4 − 4 sin2 θ & 4(1 − sin2 θ) √ 4 cos2 θ

7.

ln ecos x = ln 1

=

%

= 2 cos θ

6.

2.

1 − sin θ 1 + sin θ · 1 + sin θ 1 + sin θ

=

84.

1.

$

=

83. ln esin t = loge esin t = sin t

Both values check. The solutions in [0, 2π) are

1 − sin θ = 1 + sin θ

=

tan θ =

cos x = 0 π 3π x= , 2 2

$

θ x

-3 -4

5

Chapter 6 Test

417

3 From the drawing we see that cos θ = − . 5 "! " ! 4 3 24 sin 2θ = 2 sin θ cos θ = 2 − − = 5 5 25 Since sin 2θ is positive, we know that 2 θ is in quadrant I or II. To determine which we also find cos 2θ. ! "2 4 32 7 =1− =− cos 2θ = 1 − 2 sin2 θ = 1 − 2 − 5 25 25 Since sin 2θ is positive and cos 2θ is negative, 2 θ is in quadrant II. * /π0 + + 1 + cos π , π 6 6 = = = cos 10. cos 12 2 2 * √ + % 3 + # √ √ ,1 + 2 = 2+ 3 = 2+ 3 2 4 2 11. First, find cos θ. # cos θ = 1 − sin2 θ # ≈ 1 − (0.6820)2

12.

1 + cos β 1 − cos β We start with the right side. 1 + cos β 1 + cos β 1 + cos β = · 1 − cos β 1 − cos β 1 + cos β

15. Prove: (csc β + cot β)2 =

=

1 + 2 cos β + cos2 β sin2 β 1 1 cos β cos2 β = +2· · + 2 sin β sin β sin β sin2 β =

= csc2 β + 2 csc β cot β + cot2 β = (csc β + cot β)2 We started with the right side and deduced the left side, so the proof is complete. tan α 1 + sin α = 1 + csc α sec α We start with the left side.

16. Prove:

1 + sin α 1 + sin α = 1 1 + csc α 1+ sin α

≈ 0.7314 (θ is in quadrant I) $ θ 1 + cos θ cos = 2 2 $ 1 + 0.7314 ≈ 2 ≈ 0.9304

1 + sin α sin α + 1 sin α = (1 + sin α) ·

=

Now we stop and work with the right side.

= sin2 x + 2 sin x cos x + cos2 x − 1 + 4 sin x cos x

sin α tan α cos α = 1 sec α cos α

= 6 sin x cos x

= 3(2 sin x cos x) = 3 sin 2x 13. Prove: csc x − cos x cot x = sin x

=

We start with the left side. cos x 1 − cos θ · csc x − cos x cot x = sin x sin x 1 − cos2 x sin x

sin x sin x = sin x =

2

We start with the left side. (sin x + cos x)2 = sin2 x + 2 sin x cos x + cos2 x = 1 + 2 sin x cos x = 1 + sin 2x We started with the left side and deduced the right side, so the proof is complete.

sin α · cos α cos α

= sin α We have obtained the same expression from each side, so the proof is complete. 17.

α − 8α 8α + α sin 2 2 ! " 7α 9α sin − = 2 sin 2 2 9α 7α = −2 sin sin 2 2

cos 8α − cos α = 2 sin

We started with the left side and deduced the right side, so the proof is complete. 14. Prove: (sin x + cos x)2 = 1 + sin 2x

sin α 1 + sin α

= sin α

(sin x + cos x)2 − 1 + 2 sin 2x

=

1 + 2 cos β + cos2 β 1 − cos2 β

18.

1 4 sin β cos 3β = 4 · [sin (β + 3β) + sin(β − 3β)] 2 = 2[sin 4β + sin (−2β)]

= 2(sin 4β − sin 2β) √ " 2 = −45◦ 19. sin−1 − 2 √ π 20. tan−1 3 = 3 !

418

Chapter 6: Trigonometric Identities, Inverse Functions, and Equations 7π does not check but the other two values do. The solu6 11π π and . tions in [0, 2π) are 2 6

21. cos−1 (−0.6716) ≈ 2.3072 " ! " √ ! π 1 3 = cos = 22. cos sin−1 2 6 2

5 23. The angle θ = sin−1 is an acute angle of a right triangle x where 5 is the √ √ side opposite θ, x is the hypotenuse, and x2 − 52 , or x2 − 25, is the side adjacent to θ. Then we have " ! 5 opp 5 . = tan θ = =√ tan sin−1 x adj x2 − 25 " ! " ! 1 1 π π 24. cos sin−1 + cos−1 = cos + = 2 2 6 3 " ! π π 2π + = cos = 0 cos 6 6 2 25.

4 cos2 x = 3 3 cos2 x = 4

√

3 2 π 5π 7π 11π x= , , , 6 6 6 6

cos x = ±

π 5π 7π All values check. The solutions in [0, 2π) are , , , 6 6 6 11π . and 6 √ 26. 2 sin2 x = 2 sin x √ 2 sin2 x − 2 sin x = 0 √ sin x(2 sin x − 2) = 0 √ sin x = 0 or 2 sin x − 2 = 0 √ 2 sin x = 0 or sin x = 2 π 3π x = 0, π or x= , 4 4 π 3π All values check. The solutions in [0, 2π) are 0, , , 4 4 and π. √ 27. 3 cos x + sin x = 1 √ 3 cos x = 1 − sin x 3 cos2 x = 1 − 2 sin x + sin2 x Squaring both sides 3(1 − sin2 x) = 1 − 2 sin x + sin2 x 3 − 3 sin2 x = 1 − 2 sin x + sin2 x

0 = 4 sin2 x − 2 sin x − 2

0 = 2 sin2 x − sin x − 1 Dividing by 2 0 = (2 sin x + 1)(sin x − 1)

2 sin x + 1 = 0

or sin x − 1 = 0 1 or sin x = 1 sin x = − 2 7π 11π π x= , or x= 6 6 2

28.

3π < θ < 2π, so cos θ is positive. 2 ! " 2θ cos θ = cos 2 $ 1 + cos 2θ = 2 * * + 11 + + +1 + 5 , , 6 = 6 = 2 2 $ $ 11 1 11 = · = 6 2 12

Chapter 7

Applications of Trigonometry There are two angles less than 180◦ having a sine of 0.8427. They are 57.4◦ and 122.6◦ . This gives us two possible solutions.

Exercise Set 7.1 1.

Solution I

B

If B ≈ 57.4◦ , then

a

38°

C ≈ 180◦ − (36.5◦ + 57.4◦ ) ≈ 86.1◦ .

c 21°

A

b

C

To solve this triangle find A, a, and c. A = 180◦ − (38◦ + 21◦ ) = 121◦

Use the law of sines to find a and c. Find a: b a = sin A sin B 24 a = sin 121◦ sin 38◦ 24 sin 121◦ a= ≈ 33 sin 38◦ Find c: b c = sin C sin B 24 c = sin 21◦ sin 38◦ 24 sin 21◦ c= ≈ 14 sin 38◦ 2. B = 180◦ − (131◦ + 23◦ ) = 26◦ a b a 10 = , or ◦ = sin A sin B sin 131 sin 26◦ a ≈ 17

3.

b c 10 c = , or = sin C sin B sin 23◦ sin 26◦ c≈9 B 24 A

36.5° 34

Find c: a c = sin C sin A 24 c = sin 86.1◦ sin 36.5◦ 24 sin 86.1◦ c= ≈ 40 sin 36.5◦ Solution II If B ≈ 122.6◦ , then

C ≈ 180◦ − (36.5◦ + 122.6◦ ) ≈ 20.9◦ .

Find c: a c = sin C sin A 24 c = sin 20.9◦ sin 36.5◦ 24 sin 20.9◦ c= ≈ 14 sin 36.5◦

4. A = 180◦ − (118.3◦ + 45.6◦ ) = 16.1◦ 42.1 a b a = = , or sin A sin B sin 16.1◦ sin 118.3◦ a ≈ 13.3

42.1 c b c = , or = sin C sin B sin 45.6◦ sin 118.3◦ c ≈ 34.2

5. Find B: b c = sin B sin C 30.3 24.2 = sin B sin 61◦ 10" 24.2 sin 61◦ 10" ≈ 0.6996 30.3 Then B ≈ 44◦ 24" or B ≈ 135◦ 36" . An angle of 135◦ 36" cannot be an angle of this triangle because it already has an angle of 61◦ 10" and the two would total more than 180◦ . Thus B ≈ 44◦ 24" . sin B =

C

To solve this triangle find B, C, and c. Find B: a b = sin B sin A 24 34 = sin B sin 36.5◦ 34 sin 36.5◦ sin B = ≈ 0.8427 24

Find A:

A ≈ 180◦ − (61◦ 10" + 44◦ 24" ) ≈ 74◦ 26"

420

Chapter 7: Applications of Trigonometry Find A:

Find a:

a c = sin A sin C a 30.3 = sin 74◦ 26" sin 61◦ 10" a=

6.

A ≈ 180◦ − (83.78◦ + 83.78◦ ) ≈ 12.44◦

Find a:

c a = sin A sin C 56.78 a = sin 12.44◦ sin 83.78◦ 56.78 sin 12.44◦ a= ≈ 12.30 yd sin 83.78

30.3 sin 74◦ 26" ≈ 33.3 sin 61◦ 10"

c a 13.5 17.2 = , or = sin C sin A sin C sin 126.5◦ sin C ≈ 0.6309

Then C ≈ 39.1 or C ≈ 140.9 . An angle of 140.9 cannot be an angle of this triangle because it already has an angle of 126.5◦ and these two would total more than 180◦ . Thus C ≈ 39.1◦ . ◦

◦

◦

B ≈ 180◦ − (126.5◦ + 39.1◦ ) ≈ 14.4◦ b a b 17.2 = = , or sin B sin A sin 14.4◦ sin 126.5◦ b ≈ 5.3

7. Find A: A = 180◦ − (37.48◦ + 32.16◦ ) = 110.36◦ Find a:

c a = sin A sin C 3 a = sin 110.36◦ sin 32.16◦ 3 sin 110.36◦ a= ≈ 5 mi sin 32.16◦ Find b: c b = sin B sin C b 3 = sin 37.48◦ sin 32.16◦ 3 sin 37.48◦ b= ≈ 3 mi sin 32.16◦ 8.

a 2345 2345 b = , or = sin B sin A sin B sin 124.67◦ sin B ≈ 0.8224

Then B ≈ 55.33◦ or B ≈ 124.67◦ . An angle of 55.33◦ cannot be an angle of this triangle because it already has an angle of 124.67◦ and the two would total 180◦ . An angle of 124.67◦ cannot be an angle of this triangle either because the two 124.67◦ angles would total more than 180◦ . There is no solution. 9. Find B: b c = sin B sin C 56.78 56.78 = sin B sin 83.78◦ 56.78 sin 83.78◦ ≈ 0.9941 sin B = 56.78 ◦ Then B ≈ 83.78 or B ≈ 96.22◦ . An angle of 96.22◦ cannot be an angle of this triangle because it already has an angle of 83.78◦ and the two would total 180◦ . Thus, B ≈ 83.78◦ .

10. B = 180◦ − (129◦ 32" + 18◦ 28" ) = 32◦ 1204 a b a = = , or sin A sin B sin 129◦ 32" sin 32◦ a ≈ 1752 in.

b c 1204 c = , or = sin C sin B sin 18◦ 28" sin 32◦ c ≈ 720 in.

11. Find B: a b = sin B sin A 10.07 20.01 = sin B sin 30.3◦ 10.07 sin 30.3◦ ≈ 0.2539 sin B = 20.01 ◦ Then B ≈ 14.7 or B ≈ 165.3◦ . An angle of 165.3◦ cannot be an angle of this triangle because it already has an angle of 30.3◦ and the two would total more than 180◦ . Thus, B ≈ 14.7◦ . Find C:

C ≈ 180◦ − (30.3◦ + 14.7◦ ) ≈ 135.0◦ Find c:

12.

c a = sin C sin A 20.01 c = sin 135.0◦ sin 30.3◦ 20.01 sin 135.0◦ c= ≈ 28.04 cm sin 30.3◦ c 4.157 3.446 b = , or = sin B sin C sin B sin 51◦ 48" sin B ≈ 0.9480

Then B ≈ 71◦ 26" or B ≈ 108◦ 34" Solution I

If B ≈ 71◦ 26" , then

A ≈ 180◦ − (71◦ 26" + 51◦ 48" ) ≈ 56◦ 46" . 3.446 a c a = = , or sin A sin C sin 56◦ 46" sin 51◦ 48" a ≈ 3.668 km Solution II

If B ≈ 108◦ 34" , then

A ≈ 180◦ − (108◦ 34" + 51◦ 48" ) ≈ 19◦ 38" . c a 3.446 a = , or = sin A sin C sin 19◦ 38" sin 51◦ 48" a ≈ 1.473 km

Exercise Set 7.1

421

13. Find B: b a = sin B sin A 15.6 18.4 = sin B sin 89◦ 18.4 sin 89◦ sin B = ≈ 1.1793 15.6 Since there is no angle having a sine greater than 1, there is no solution. 14.

c 56.2 22.1 a = , or = sin A sin C sin A sin 46◦ 32" sin A ≈ 1.8456

Since there is no angle having a sine greater than 1, there is no solution.

20.

K = K =

21.

K = K = K ≈

22.

1 bc sin A 2 1 K = (18.2)(23.7) sin 113◦ ≈ 198.5 cm2 2 K =

23.

C 42°

15. Find B:

50 m

B = 180◦ − (32.76◦ + 21.97◦ ) = 125.27◦

112° A

Find b:

a b = sin B sin A b 200 = sin 125.27◦ sin 32.76◦ 200 sin 125.27◦ b= ≈ 302 m sin 32.76◦ Find c: a c = sin C sin A 200 c ◦ = sin 21.97 sin 32.76◦ 200 sin 21.97◦ c= ≈ 138 m sin 32.76◦ 16.

a 45.6 23.8 c = , or = sin C sin A sin C sin 115◦ sin C ≈ 1.7365

18.

19.

1 ac sin B 2 1 K = (7.2)(3.4) sin 42◦ 2 K ≈ 8.2 ft2

First find B: B = 180◦ − (112◦ + 42◦ ) = 26◦

Now we find c: c b = sin C sin B 50 c = sin 42◦ sin 26◦ 50 sin 42◦ c= ≈ 76.3 m sin 26◦ The crater is about 76.3 m wide. 24.

A

B

c 50°

50°

10 ft

10 ft

C

K =

First we find C: C = 180◦ − (50◦ + 50◦ ) = 80◦

Substituting

Now we use the law of sines to find c: a c = sin C sin A c 10 = sin 80◦ sin 50◦ 10 sin 80◦ c= ≈ 12.86 sin 50◦ The speakers should be placed about 12.86 ft apart, or about 12 ft 10 in. apart.

1 bc sin A 2 1 K = (10)(13) sin 17◦ 12" ≈ 19 in2 2 K =

1 ab sin C 2 1 K = · 4 · 6 · sin 82◦ 54" 2 K ≈ 12 yd2

B

c

Let c = the width of the crater.

Since there is no angle having a sine greater than 1, there is no solution. 17.

1 ab sin C 2 1 (1.5)(2.1) sin 75.16◦ ≈ 1.5 m2 2 1 ac sin B 2 1 (46.12)(36.74) sin 135.2◦ 2 596.98 ft2

K =

Substituting

25.

1 bc sin A 2 1 K = · 42 · 53 sin 135◦ 2 K ≈ 787 ft2

K =

422

Chapter 7: Applications of Trigonometry

26. The amount of rope for side s of triangle ST Q ! available " 1 is 38 ft − 21 ft − 2 4 ft , or 8 ft. To determine if it is 2 possible to have a triangle ST Q with S = 35◦ , s = 8 ft, and q = 21 ft, we use the law of sines. s 21 8 q = , or = sin Q sin S sin Q sin 35◦

29.

C (Fire)

C = 180◦ − (110◦ + 30◦ ) = 40◦

The distance from Tower A to the fire is b: b c = sin B sin C 45 b = sin 30◦ sin 40◦ 45 sin 30◦ b= ≈ 35 mi sin 40◦ The distance from Tower B to the fire is a: c a = sin A sin C a 45 ◦ = sin 110 sin 40◦ 45 sin 110◦ a= ≈ 66 mi sin 40◦

7° p 51° R

Find R: R = 90◦ − 7◦ = 83◦

Find Q: Q = 180◦ − (51◦ + 83◦ ) = 46◦

C 230 km A

B c

30. C

65° 10'

B

a

7.2 km

5.1 km A

Find B: B = 90◦ − 65◦ 10" = 24◦ 50"

a 85°

A

B = 255◦ − 180◦ − 45◦ = 30◦

Q

28.

45 mi

b

A = 45◦ + (360◦ − 295◦ ) = 110◦

27.

Find p: (p is the length of the pole.) p Q = sin P sin Q p 47 = sin 51◦ sin 46◦ 47 sin 51◦ p= ≈ 51 sin 46◦ The pole is about 51 ft long.

45°

295°

Since there is no angle whose sine is greater than 1, the triangle described above does not exist. Thus, the rancher does not have enough rope.

47 ft

255°

a

sin Q ≈ 1.5056

P

B

283°

Find B: B = 283◦ − 180◦ − 85◦ = 18◦

Find C: C = 180◦ − (85◦ + 18◦ ) = 77◦ Find a: 230 b a a = , or = sin A sin B sin 85◦ sin 18◦ a ≈ 742

Find c: 230 c b c = = , or sin C sin B sin 77◦ sin 18◦ c ≈ 725

The plane flew a total of 742 + 725, or 1467 km.

Find C: b 7.2 5.1 c = , or = sin C sin B sin C sin 24◦ 50" sin C ≈ 0.5929

Then C ≈ 36◦ 20" or C ≈ 143◦ 40" .

If C ≈ 143◦ 40" , then

A ≈ 180◦ − (24◦ 50" + 143◦ 40" ) ≈ 11◦ 30" .

Find a: b a 5.1 a = , or = sin A sin B sin 11◦ 30" sin 24◦ 50" a ≈ 2.4 If C ≈ 36◦ 20" , then

A ≈ 180◦ − (24◦ 50" + 36◦ 20" ) ≈ 118◦ 50" .

Find a: b a 5.1 a = , or = sin A sin B sin 118◦ 50" sin 24◦ 50" a ≈ 10.6

The boat is either 2.4 km or 10.6 km from lighthouse B.

Exercise Set 7.1

423 36. Since there is no angle whose cosine is greater than 1, there is no angle A for which cos A = 1.5612.

31. M 31o 20'

F

18 mi

78o 40'

m

31o 20'

64o 10'

C

M = 78◦ 40" − 31◦ 20" = 47◦ 20"

C = 31◦ 20" + 64◦ 10" = 95◦ 30"

F = 180◦ − (47◦ 20" + 95◦ 30" ) = 37◦ 10" m f = sin M sin F 18 m = sin 47◦ 20" sin 37◦ 10" 18 sin 47◦ 20" ≈ 22 sin 37◦ 10" The freighter is about 22 mi from Cheboygan. m=

32.

C

38. With a calculator set in Degree mode, enter 125◦ 3" 42"" as 125" 3" 42" . We find that 125◦ 3" 42"" ≈ 125.06◦ . 39. The distance of −5 from 0 is 5, so | − 5| = 5. √ 3 π 40. cos = 6 2 √ 1 2 41. sin 45◦ = √ , or 2 2 √ 3 ◦ ◦ 42. sin 300 = − sin 60 = − 2 ! " 2π π 1 43. cos − = − cos = − 3 3 2 44. (1 − i)(1 + i) = 1 − i2 = 1 − (−1) = 2

50 ft B

41°

37. With a calculator set in Degree mode, enter 18◦ 14" 20"" as 18" 14" 20" . We find that 18◦ 14" 20"" ≈ 18.24◦ .

45. See the answer section in the text. 46. s1

58 ft

A

a = 28 ft + 22 ft = 50 ft b = 22 ft + 36 ft = 58 ft a 58 50 c = , or = sin C sin A sin C sin 41◦ sin C ≈ 0.7610 Then C ≈ 50◦ or C ≈ 130◦ .

S

If C ≈ 130◦ , then φ ≈ 180◦ − (41◦ + 130◦ ) ≈ 9◦ .

s2

A = s2 h, h = s1 sin S, so A = s1 s2 sin S. 47. See the answer section in the text. 48. First find the length x of the diagonal of the 11 in. by 12 in. base: x2 = 112 + 122 x2 = 265

Note that angle B above is labeled φ in the text. If C ≈ 50◦ , then φ ≈ 180◦ − (41◦ + 50◦ ) ≈ 89◦ .

x ≈ 16.2788

Now consider the triangle formed by x, d, and the 15 in. side of the figure:

Using the law of sines, we find that b ≈ 76 ft when φ ≈ 89◦ and b ≈ 12 ft when φ ≈ 9◦ . Since b must be at least the sum of the two radii 36 ft and 28 ft, only φ ≈ 89◦ can be a solution. 33. The law of sines involves two angles of a triangle and the sides opposite them. Three of these four values must be known in order to find the fourth. Thus, we must know the measure of one angle in order to use the law of sines. 34. In Cases 1 and 6, b < h; in Case 2, b = h, in Cases 3, 4, 5, 7, and 8, b > h. 35. cos A = 0.2213 Using a calculator set in Radian mode, press .2213 SHIFT COS , or 2nd COS .2213 ENTER . The calculator returns 1.347649005, so A ≈ 1.348 radians. Set the calculator in Degree mode and repeat the keystrokes above. The calculator returns 77.21460028, so A ≈ 77.2◦ .

h

F d E

50°

15 in.

16.2788 in. D

Find F : 15 16.2788 = sin F sin 50◦ sin F ≈ 0.8314

Then F ≈ 56.24◦ or F ≈ 123.76◦ .

The drawing in the text indicates that F is an acute angle, so we consider F ≈ 56.24◦ .

424

Chapter 7: Applications of Trigonometry 7 "" = 59.875" 8 d sin 29◦ = 59.875 d = 59.875 sin 29◦

Find D:

4" 11

D = 180◦ − (50◦ + 56.24◦ ) = 73.76◦ Find d: 15 d = sin 73.76◦ sin 50◦ d ≈ 18.8 in.

d ≈ 29.0""

We subtract to find the desired distance. Note that the length of wall 1, 29" 9"" , can be expressed as 357"" . Also, 1 "" the distance from C to B, 15" 10 , can be expressed as 8 190.125"" . Then we have 357"" − (9"" + d + 190.125"" + a) = 357"" − (9"" + 29.0"" + 190.125"" + 24.6"" ) = 104.275 ≈ 104.3"" .

49. First we determine the distance from point A to wall 1.

20o 160o

A

13 3' 6 — " 16 125o

b

Exercise Set 7.2 1.

B

a

9"

B 24

a

wall 1 B = 180◦ − 125◦ = 55◦ 13 "" 13 "" 13 "" 3" 6 = 36"" + 6 = 42 16 16 16 b ◦ sin 55 = 13 42 16 13 b = 42 · sin 55◦ 16 b ≈ 35.1""

30° A

C

To solve this triangle find a, B, and C. From the law of cosines, a2 = b2 + c2 − 2bc cos A

a2 = 122 + 242 − 2 · 12 · 24 cos 30◦ a2 ≈ 221 a ≈ 15

The distance from point A to wall 1 is 9"" + 35.1"" , or 44.1"" .

Next we use the law of cosines again to find a second angle.

Now find the distance from point A to wall 4. First we find the length a in the drawing above. Since B = 55◦ , we have A = 90◦ − 55◦ = 35◦ . a sin 35◦ = 13 42 16 13 a = 42 · sin 35◦ 16 a ≈ 24.6""

b2 = a2 + c2 − 2ac cos B

122 = 152 + 242 − 2(15)(24) cos B

144 = 225 + 576 − 720 cos B

−657 = −720 cos B

cos B ≈ 0.9125 B ≈ 24◦

Now we find the third angle.

Now we consider a right triangle with hypotenuse the distance from D to C.

C ≈ 180◦ − (30◦ + 24◦ ) ≈ 126◦ 2.

D

12

b2 = a2 + c2 − 2ac cos B, or

b2 = 122 + 152 − 2 · 12 · 15 cos 133◦ b ≈ 25

4' 11 –7 " 8

119o

a2 = b2 + c2 − 2bc cos A, or

122 = 252 + 152 − 2 · 25 · 15 cos A

−0.9413 ≈ cos A

d

C

C = 180◦ − 119◦ = 61◦ D = 90◦ − 61◦ = 29◦

20◦ ≈ A

C ≈ 180◦ − (20◦ + 133◦ ) ≈ 27◦

Exercise Set 7.2 3.

425 6.

C

c2 = (25.4)2 + (73.8)2 − 2(25.4)(73.8) cos 22.28◦

14

12

B

c ≈ 51.2 cm

a2 = b2 + c2 − 2bc cos A

A

20

(25.4)2 = (73.8)2 + (51.2)2 − 2(73.8)(51.2) cos A cos A ≈ 0.9822

To solve this triangle find A, B, and C.

A ≈ 10.82◦

Find A: a2 = b2 + c2 − 2bc cos A

B ≈ 180◦ − (10.82◦ + 22.28◦ ) ≈ 146.90◦ .

122 = 142 + 202 − 2 · 14 · 20 cos A

7. Find A: a2 = b2 + c2 − 2bc cos A

144 = 196 + 400 − 560 cos A

162 = 202 + 322 − 2 · 20 · 32 cos A

−452 = −560 cos A

256 = 400 + 1024 − 1280 cos A

cos A ≈ 0.8071

−1168 = −1280 cos A

Thus A ≈ 36.18◦ .

cos A ≈ 0.9125

Find B: b2 = a2 + c2 − 2ac cos B

A ≈ 24.15◦ .

142 = 122 + 202 − 2 · 12 · 20 cos B

Find B: b2 = a2 + c2 − 2ac cos B

196 = 144 + 400 − 480 cos B

202 = 162 + 322 − 2 · 16 · 32 cos B

−348 = −480 cos B

400 = 256 + 1024 − 1024 cos B

cos B ≈ 0.7250

−880 = −1024 cos B

Thus B ≈ 43.53 . ◦

cos B ≈ 0.8594

Then C ≈ 180◦ − (36.18◦ + 43.53◦ ) ≈ 100.29◦ . 4.

B ≈ 30.75◦ .

a2 = b2 + c2 − 2bc cos A, or

(22.3)2 = (22.3)2 + (36.1)2 − 2(22.3)(36.1) cos A 0.8094 ≈ cos A

Then C ≈ 180◦ − (24.15◦ + 30.75◦ ) ≈ 125.10◦ . 8.

2(23.78)(25.74) cos 72.66◦

b2 = a2 + c2 − 2ac cos B, or

b ≈ 29.38 km

(22.3)2 = (22.3)2 + (36.1)2 − 2(22.3)(36.1) cos B

a2 = b2 + c2 − 2bc cos A, or

0.8094 ≈ cos B

(23.78)2 = (29.38)2 + (25.74)2 −

35.96◦ ≈ B

2(29.38)(25.74) cos A

C ≈ 180◦ − (35.96◦ + 35.96◦ ) ≈ 108.08◦

0.6349 ≈ cos A

5. To solve this triangle find b, A, and C.

50.59◦ ≈ A

Find b: b2 = a2 + c2 − 2ac cos B

C ≈ 180◦ − (50.59◦ − 72.66◦ ) ≈ 56.75◦ 9. Find A: a2 = b2 + c2 − 2bc cos A

b2 = 782 + 162 − 2 · 78 · 16 cos 72◦ 40" b2 ≈ 5596

22 = 32 + 82 − 2 · 3 · 8 cos A

b ≈ 75 m

4 = 9 + 64 − 48 cos A

Find A: a2 = b2 + c2 − 2bc cos A

−69 = −48 cos A

cos A ≈ 1.4375

782 = 752 + 162 − 2 · 75 · 16 cos A

Since there is no angle whose cosine is greater than 1, there is no solution.

6084 = 5625 + 256 − 2400 cos A 203 = −2400 cos A

cos A ≈ −0.0846 a ≈ 94 51

10.

"

Then C = 180 − (94 51 + 72 40 ) = 12 29 ◦

◦

"

b2 = a2 + c2 − 2ac cos B, or b2 = (23.78)2 + (25.74)2 −

35.96◦ ≈ A

◦

c2 = a2 + b2 − 2ab cos C, or

◦

"

◦

"

a2 = b2 + c2 − 2bc cos A, or a2 = (15.8)2 + (18.4)2 − a ≈ 25.5 yd

2(15.8)(18.4) cos 96◦ 13"

426

Chapter 7: Applications of Trigonometry b2 = a2 + c2 − 2ac cos B

14.

(15.8) = (25.5) + (18.4) − 2(25.5)(18.4) cos B 2

2

2

(11.2)2 = (5.4)2 + 72 − 2(5.4)(7) cos A

cos B ≈ 0.7877 B ≈ 38◦ 2"

cos A ≈ −0.6254 A ≈ 128.71◦

C ≈ 180 − (96 13 + 38 2 ) ≈ 45 45 ◦

◦

"

◦ "

◦

"

b2 = a2 + c2 − 2ac cos B, or

(5.4)2 = (11.2)2 + 72 − 2(11.2)(7) cos B

11. Find A: a2 = b2 + c2 − 2bc cos A

cos B ≈ 0.9265

(26.12) = (21.34) + (19.25) − 2

2

B ≈ 22.10◦

2

2(21.34)(19.25) cos A

682.2544 = 455.3956 + 370.5625 − 821.59 cos A

−143.7037 = −821.59 cos A cos A ≈ 0.1749

Find B:

C ≈ 180◦ − (128.71◦ + 22.10◦ ) ≈ 29.19◦ 15. Find a: a2 = b2 + c2 − 2bc cos A

a2 = (10.2)2 + (17.3)2 − 2(10.2)(17.3) cos 53.456◦

A ≈ 79.93◦

a2 ≈ 193.19

a ≈ 13.9 in.

b2 = a2 + c2 − 2ac cos B

Find B: b2 = a2 + c2 − 2ac cos B

(21.34)2 = (26.12)2 + (19.25)2 −

2(26.12)(19.25) cos B

(10.2)2 = (13.9)2 + (17.3)2 − 2(13.9)(17.3) cos B

455.3956 = 682.2544 + 370.5625 − 1005.62 cos B

104.04 = 193.21 + 299.29 − 480.94 cos B

−597.4213 = −1005.62 cos B

−388.46 = −480.94 cos B

B ≈ 53.55◦

B ≈ 36.127◦

cos B ≈ 0.5941

cos B ≈ 0.8077

C ≈ 180◦ − (79.93◦ + 53.55◦ ) ≈ 46.52◦ 12.

a2 = b2 + c2 − 2bc cos A, or

c = a + b − 2ab cos C, or 2

2

Find C:

C ≈ 180◦ − (53.456◦ + 36.127◦ ) ≈ 90.417◦

2

c2 = 62 + 92 − 2 · 6 · 9(cos 28◦ 43" ) c ≈ 5 mm

a = b + c = 2bc cos A, or 2

2

2

62 = 92 + 52 − 2 · 9 · 5 cos A

0.7778 ≈ cos A 38◦ 57" ≈ A

B ≈ 180◦ − (38◦ 57" + 28◦ 43" ) = 112◦ 20" 13. Find c: c2 = a2 + b2 − 2ab cos C

c2 = (60.12)2 + (40.23)2 − c2 ≈ 2040

2(60.12)(40.23) cos 48.7◦

c ≈ 45.17 mi

Find A:

a2 = b2 + c2 − 2bc cos A

(60.12)2 = (40.23)2 + (45.17)2 −

2(40.23)(45.17) cos A 3614.4144 = 1618.4529 + 2040.3289− 3634.3782 cos A −44.3674 = −3634.3782 cos A cos A ≈ 0.0122 A ≈ 89.3◦

B ≈ 180◦ − (89.3◦ + 48.7◦ ) ≈ 42.0◦

16.

a2 = b2 + c2 − 2bc cos A, or

172 = (15.4)2 + (1.5)2 − 2(15.4)(1.5) cos A

cos A ≈ −1.0734

Since there is no angle whose cosine is less than −1, there is no solution. 17. We are given two angles and the side opposite one of them. The law of sines applies. C = 180◦ − (70◦ + 12◦ ) = 98◦ a b = sin A sin B 21.4 a = sin 70◦ sin 12◦ 21.4 sin 70◦ a= ≈ 96.7 sin 12◦ c b = sin C sin B 21.4 c = sin 98◦ sin 12◦ 21.4 sin 98◦ c= ≈ 101.9 sin 12◦ 18. We are given two sides and the included angle. The law of cosines applies. b2 = a2 + c2 − 2ac cos B

b2 = 152 + 72 − 2 · 15 · 7 cos 62◦ b ≈ 13

Exercise Set 7.2

427

a2 = b2 + c2 − 2bc cos A

b2 = a2 + c2 − 2ac cos B

15 = 13 + 7 − 2 · 13 · 7 cos A 2

2

(6.2)2 = (3.6)2 + (4.1)2 − 2(3.6)(4.1) cos B

2

A ≈ 92◦

38.44 = 12.96 + 16.81 − 29.52 cos B 8.67 = −29.52 cos B

C ≈ 180◦ − (92◦ + 62◦ ) ≈ 26◦

cos B ≈ −0.2937

19. We are given all three sides of the triangle. The law of cosines applies. a2 = b2 + c2 − 2bc cos A

(3.3)2 = (2.7)2 + (2.8)2 − 2(2.7)(2.8) cos A 10.89 = 7.29 + 7.84 − 15.12 cos A

B ≈ 107.08◦

C ≈ 180◦ − (33.71◦ + 107.08◦ ) ≈ 39.21◦ 24. We are given two angles and the side opposite one of them. The law of sines applies. A = 180◦ − (110◦ 30" + 8◦ 10" ) = 61◦ 20" a 0.912 c a = = , or sin A sin C sin 61◦ 10" sin 8◦ 10" a ≈ 5.633

−4.24 = −15.12 cos A cos A ≈ 0.2804 A ≈ 73.71◦

b2 = a2 + c2 − 2ac cos B

(2.7)2 = (3.3)2 + (2.8)2 − 2(3.3)(2.8) cos B 7.29 = 10.89 + 7.84 − 18.48 cos B

−11.44 = −18.48 cos B

25.

cos B ≈ 0.6190 B ≈ 51.75

b c b 0.912 = , or = sin B sin C sin 110◦ 30" sin 8◦ 10" b ≈ 6.014 N P

◦

X

C ≈ 180 − (73.71 + 51.75 ) ≈ 54.54 ◦

◦

◦

◦

20. We are given two sides and the angle opposite one of them. The law of sines applies. a b 1.5 2.5 = = , or sin A sin B sin 58◦ sin B sin B ≈ 1.4134, so there is no solution.

500 ft

355°

r2 ≈ 134, 876 r ≈ 367

The distance between the poachers and the game is about 367 ft.

c = 60 + 40 − 2(60)(40) cos 47 2

2

◦

c ≈ 44

a2 = b2 + c2 = 2bc cos A

R

r2 = 5002 + 3752 − 2 · 500 · 375 cos 47◦

22. We are given two sides of the triangle and the included angle. The law of cosines applies. 2

26.

C

60 = 40 + 44 − 2(40)(44) cos A 2

2

2

cos A ≈ −0.0182 A ≈ 91◦

B ≈ 180◦ − (47◦ + 91◦ ) ≈ 42◦ 23. We are given all three sides of the triangle. The law of cosines applies. a2 = b2 + c2 − 2bc cos A

(3.6)2 = (6.2)2 + (4.1)2 − 2(6.2)(4.1) cos A 12.96 = 38.44 + 16.81 − 50.84 cos A

−42.29 = −50.84 cos A cos A ≈ 0.8318 A ≈ 33.71◦

375 ft

Angle XRP = 360◦ − 355◦ = 5◦ , so R = 42◦ + 5◦ = 47◦ . Use the law of cosines to find r, the distance from the poachers to the game.

21. We are given the three angles of a triangle. Neither law applies. This triangle cannot be solved using the given information.

c2 = a2 + b2 − 2ab cos C

42°

w 122 ft 35° A

30 ft

B

D

Angle ABC = 180◦ − 35◦ = 145◦ . Then w2 = 302 + 1222 − 2 · 30 · 122 cos 145◦ w ≈ 148 ft

148 ft − 122 ft = 26 ft, so the aerialists will have to walk about 26 ft farther. Find the new approach angle A: 148 122 = sin A sin 145◦ sin A ≈ 0.4728 A ≈ 28◦

The new approach angle is about 28◦ .

428

Chapter 7: Applications of Trigonometry

27. Use the law of cosines to find the distance d from A to B. d2 = (0.5)2 + (1.3)2 − 2(0.5)(1.3) cos 110◦

Find a: a2 = b2 + c2 − 2bc cos A

d2 ≈ 2.38

a2 = 502 + 402 − 2 · 50 · 40 cos 47◦ a2 ≈ 1372

d ≈ 1.5

He skated about 1.5 mi. 28.

a ≈ 37

The ships are about 37 nautical mi apart.

P d F 60.5 ft

31. 150 km/h × 3 hr = 450 km 200 km/h × 3 hr = 600 km

T

34 ft

N

25°

450 km

A

200°

B

320°

d

Since the baseball diamond is a square, angle F BP = 1 · 90◦ , or 45◦ , and angle T BP = 45◦ − 25◦ , or 20◦ . Use 2 the law of cosines to find the distance d that the pitcher travels. d2 = (60.5)2 + 342 − 2(60.5)(34) cos 20◦

600 km

The angle opposite d is 320◦ − 200◦ , or 120◦ . Use the law of cosines to find d: d2 = 4502 + 6002 − 2 · 450 · 600 cos 120◦

d ≈ 30.8 ft

29. Find S: s2 = u2 + t2 − 2ut cos S

(45.2)2 = (31.6)2 + (22.4)2 − 2(31.6)(22.4) cos S

−0.3834 ≈ cos S

S

B

d ≈ 912

The planes are about 912 km apart. 32.

112.5 ≈ S ◦

Find T : s t = sin T sin S 45.2 22.4 = sin T sin 112.5◦ sin T ≈ 0.4579

d 46 ft

45°

65 ft

d2 = 652 + 462 − 2 · 65 · 46 cos 45◦

T ≈ 27.2◦

Find U :

33.

d ≈ 46 ft

B

30. 25 knots × 2 hr = 50 nautical mi

100° 10 ft

20 knots × 2 hr = 40 nautical mi N

d

10 ft

h

80°

80° 14 ft

A

a

C

b

U ≈ 180◦ − (112.5◦ + 27.2◦ ) ≈ 40.3◦

D

Let b represent the other base, d represent the diagonal, and h the height. a) Find d:

50 15°

32°

40

A

Find A: A = 15◦ + 32◦ = 47◦

Using the law of cosines, d2 = 102 + 142 − 2 · 10 · 14 cos 80◦ d2 ≈ 247 d ≈ 16

The length of the diagonal is about 16 ft.

Exercise Set 7.2

429

b) Find h: h = sin 80◦ 10 h = 10 sin 80◦

35. Place the figure on a coordinate system as shown.

h ≈ 9.85

Q(x, y)

Find # CBD: Using the law of sines, 16 10 = # sin CBD sin 100◦ 10 sin 100◦ sin # CBD = ≈ 0.6155 16 Thus # CBD ≈ 38◦ .

d 109° R

P

r = 1.4 cm

Then # CDB ≈ 180◦ − (100◦ + 38◦ ) ≈ 42◦ . Find b: Using the law of sines, 16 b = sin 42◦ sin 100◦ 16 sin 42◦ b= ≈ 10.87 sin 100◦ 1 Area = h(b1 + b2 ) 2 1 Area ≈ (9.85)(10.87 + 14) 2 ≈ 122

r = 1.1 cm

r = 1.8 cm

Let (x, y) be the coordinates of point Q. The coordinates of point P are (1.4 + 1.1, 0), or (2.5, 0). The length of QR is 1.8 + 1.4, or 3.2. We use the law of cosines to find d. d2 = (2.5)2 + (3.2)2 − 2(2.5)(3.2) cos 109◦ d2 ≈ 21.70 d ≈ 4.7

The length P Q is about 4.7 cm. 36.

C 32.8 ft

The area is about 122 ft . (Answers may vary due to rounding differences.)

a

2

40° 50' A

34.

B

44 ft

B

a2 = b2 + c2 − 2bc cos A

a2 = (32.8)2 + 442 − 2(32.8)(44) cos 40.8◦ 20 ft 19° 19° 20 ft h

b = a + c − 2ac cos B, or 2

2

b2 = 202 + 202 − 2 · 20 · 20 cos 38◦ b ≈ 13

h , so h ≈ 19 20 1 1 Area = bh ≈ · 13 · 19 ≈ 124 ft2 2 2 (Answers may vary due to rounding differences.) cos 19◦ =

37. Using the law of cosines it is necessary to solve the quadratic equation (11.1)2 = a2 + (28.5)2 − 2a(2.85) cos 19◦ , or

b 2

a ≈ 28.8 ft

0 = a2 −[2(2.85) cos 19◦ ]a+[(28.5)2 −(11.1)2 ].

The law of sines requires less complicated computations. 38. The law of sines involves two angles of a triangle and the sides opposite them. Three of these four values must be known in order to find the fourth. Given SAS, only two of these four values are known. 39. This is a polynomial function with degree 4, so it is a quartic function. 40. y − 3 = 17x, or y = 17x + 3

This function can be written in the form y = mx + b, so it is linear.

41. The variable is in a trigonometric function, so this is trigonometric. 42. The variable is in an exponent, so this is an exponential function. 43. This is the quotient of two polynomials, so it is a rational function.

430

Chapter 7: Applications of Trigonometry We substitute in the expression for K 2 : ! "2 1 1 −a2 + b2 + c2 K 2 = b2 c2 − 4 4 2 1 2 2 1 1 = b c − · (−a2 + b2 + c2 )2 4 4 4 1 = [4b2 c2 − (−a2 + b2 + c2 )2 ] 16 1 [2bc+(−a2 +b2 +c2 )][2bc−(−a2 +b2 +c2 )] = 16 1 2 = [(b +2bc+c2 )−a2 ][a2 −(b2 −2bc+c2 )] 16 1 [(b + c)2 − a2 ][a2 − (b − c)2 ] = 16 1 = (b+c+a)(b + c − a)(a + b − c)(a − b + c) 16 # $ 1 1 = (a + b + c) · (a + b + c) − a · 2 2 $ # $ # 1 1 (a + b + c) − b · (a + b + c) − c 2 2

44. This is a polynomial function with degree 3, so it is a cubic function. 45. The variable is in an exponent, so this is an exponential function. 46. The variable is in a logarithmic function, so it is logarithmic. 47. The variable is in a trigonometric function, so this is trigonometric. 48. This is a polynomial function with degree 2, so it is a quadratic function. 49.

5045 5045 – x

x

h α

78

o

β

72o

α = 90◦ − 78◦ = 12◦

Thus, K 2 = s(s − a)(s − b)(s − c), where 1 s = (a + b + c). 2 % Then K = s(s − a)(s − b)(s − c), where 1 s = (a + b + c). 2 In Exercise 36, a = 44 ft, b = 32.8 ft, and c ≈ 28.8 ft. 1 1 s = (a + b + c) ≈ (44 + 32.8 + 28.8) ≈ 52.8 2 2 % K ≈ 52.8(52.8 − 44)(52.8 − 32.8)(52.8 − 28.8)

β = 90 − 72◦ = 18◦

x , so x = h tan 12◦ h 5045 − x tan 18◦ = , or h 5045 − h tan 12◦ tan 18◦ = h Substituting for x tan 12◦ =

h tan 18◦ = 5045 − h tan 12◦

h tan 18 + h tan 12◦ = 5045 ◦

h(tan 18◦ + tan 12◦ ) = 5045 5045 tan 18 + tan 12◦ h ≈ 9386 h=

51.

K ≈ 472.3 ft2

θ

◦

50.

1 bc sin A 2 1 K 2 = b2 c2 sin2 A Squaring both sides 4 1 K 2 = b2 c2 (1 − cos2 A) 4 1 1 K 2 = b2 c2 − b2 c2 cos2 A 4 4 Now, the law of cosines gives us

h

K =

a2 = b2 + c2 − 2bc cos A so −a2 + b2 + c2 bc cos A = and 2 "2 ! 2 −a + b2 + c2 . b2 c2 cos2 A = 2

θ — 2

a

The canyon is about 9386 ft deep.

a

b Let b represent the base, h the height, and θ the included angle. Find h: θ h = cos a 2 h = a cos

θ 2

Find b: b2 = a2 + a2 − 2 · a · a cos θ b2 = 2a2 − 2a2 cos θ

b2 = 2a2 (1 − cos θ) √ √ b = 2a 1 − cos θ

Exercise Set 7.3

431

Find A (area): 1 A = bh 2 " ! ' 1 &√ √ θ 2a 1 − cos θ a cos A= 2 2 "! ( " ! ( 1 − cos θ 1 + cos θ a A= a 2 2 % 1 2% 1 2 A = a 1 − cos2 θ = a sin2 θ 2 2 1 2 A = a sin θ 2 The maximum area occurs when θ is 90◦ , because the sine function takes its maximum value, 1, at θ = 90◦ . 52. We add labels to the drawing in the text.

Exercise Set 7.3 1.

Imaginary 4

2

!4 !2

|4 + 3i| = 2.

√

42 + 32 =

16 + 9 =

√

25 = 5

Imaginary

2

25°

!2 ! 3i 5000 ft

35° y

70°

3.

25°

z

4

Real

!2 !4

| − 2 − 3i| =

O 105°

2

!4 !2 R

C (Carrier)

√

4

D

x

Real

!4

35°

60°

4

!2

P (Plane)

60°

4 " 3i

2

%

(−2)2 + (−3)2 =

√

13

√

29

Imaginary

4 S (Submarine)

i

2

We find x using ∆P CO: x cot 60◦ = 5000 x = 5000 cot 60◦

2

!4 !2

4

Real

!2 !4

x = 2886.75 We find y using ∆P SO: y cot 25◦ = 5000 y = 5000 cot 25◦

|i| = |0 + 1 · i| = 4.

12 = 1

Imaginary 4

y ≈ 10, 722.53

2

Finally, we use ∆OCS and the law of cosines to find z, the distance from the submarine to the carrier. z 2 = x2 + y 2 − 2xy cos 70◦

2

!4 !2

!5 ! 2i

z 2 = (2886.75)2 + (10, 722.53)2 −

2(2886.75)(10, 722.53) cos 70◦

z ≈ 10, 106 ft

√

Real

!4

| − 5 − 2i| = 5.

4

!2

%

(−5)2 + (−2)2 =

Imaginary

4 2 2

!4 !2 !2

4!i

Real

!4

|4 − i| =

% √ √ 42 + (−1)2 = 16 + 1 = 17

432 6.

Chapter 7: Applications of Trigonometry 11. From the graph we see that standard notation for the number is 0 + 4i, or 4i.

Imaginary 6 " 3i

4 2 2

!2

4

6

!2

Real

!4

|6 + 3i| = 7.

√

62 + 32 =

√

√ 45 = 3 5

12. Standard notation: −2 + 2i

Imaginary 4 2

3 2

!4 !2

4

Real

!2 !4

|3| = |3 + 0i| = 8.

√

32 = 3

Imaginary 4 2 2

!4 !2

4

!2i

Real

!4

| − 2i| =

%

Find trigonometric notation: √ r = 42 = 4 4 0 π sin θ = = 1, cos θ = = 0. Thus, θ = , or 90◦ , and we 4 4 2 have ! " π π , or 4i = 4 cos +i sin 2 2 ◦ ◦ 4i = 4(cos 90 +i sin 90 ).

(−2)2 = 2

9. From the graph we see that standard notation for the number is 3 − 3i. Find trigonometric notation: % √ √ r = 32 + (−3)2 = 18 = 3 2 √ 1 2 −3 sin θ = √ = − √ , or − 2 3 2 2 √ 1 2 3 cos θ = √ = √ , or 2 3 2 2 7π Thus, θ = , or 315◦ , and we have 4! " √ 7π 7π + i sin , or 3 − 3i = 3 2 cos 4 4 √ 3 − 3i = 3 2(cos 315◦ + i sin 315◦ ).

10. Standard notation is −1.

Find trigonometric notation: % r = (−1)2 = 1 0 −1 sin θ = = 0, cos θ = = −1. Thus, θ = π, or 180◦ , 1 1 and we have −1 = 1(cos π + i sin π), or

−1 = 1(cos 180◦ + i sin 180◦ ).

Find trigonometric notation: % √ √ r = (−2)2 + 22 = 8 = 2 2 √ 1 2 2 sin θ = √ = √ , or 2 2 2 2 √ 1 −2 2 . cos θ = √ = − √ , or − 2 2 2 2 3π , or 135◦ , and we have Thus, θ = 4 " ! √ 3π 3π −2 + 2i = 2 2 cos + i sin , or 4 4 √ ◦ ◦ −2 + 2i = 2 2(cos 135 + i sin 135 ).

13. Find trigonometric notation for 1 − i. % √ r = 12 + (−1)2 = 2 √ √ 2 1 2 −1 sin θ = √ , or − , cos θ = √ , or 2 2 2 2 7π Thus, θ = , or 315◦ , and we have !4 " √ 7π 7π 1 − i = 2 cos + i sin , or 4 4 √ 1 − i = 2(cos 315◦ + i sin 315◦ ). √ 14. Find trigonometric notation for −10 3 + 10i. ) √ √ r = (−10 3)2 + 102 = 400 = 20 √ √ 1 −10 3 10 3 = , cos θ = =− sin θ = 20 2 20 2 5π ◦ , or 150 , so we have Thus, θ = 6 " ! √ 5π 5π + i sin , or −10 3 + 10i = 20 cos 6 6 √ −10 3 + 10i = 20(cos 150◦ + i sin 150◦ ). 15. Find trigonometric notation for −3i. % r = (−3)2 = 3 −3 0 = −1, cos θ = = 0 sin θ = 3 3 3π , or 270◦ , and we have Thus, θ = 2 " ! 3π 3π , or + i sin −3i = 3 cos 2 2 −3i = 3(cos 270◦ + i sin 270◦ ).

Exercise Set 7.3 16. Find trigonometric notation for −5 + 5i. % √ √ r = (−5)2 + 52 = 50 = 5 2 √ 5 1 2 sin θ = √ = √ , or 2 5 2 2 √ 1 −5 2 cos θ = √ = − √ , or − 2 5 2 2 3π , or 135◦ , and we have Thus, θ = 4 ! " √ 3π 3π , or + i sin −5 + 5i = 5 2 cos 4 4 √ −5 + 5i = 5 2(cos 135◦ + i sin 135◦ ). √ 17. Find trigonometric notation for 3 + i. )√ √ r = ( 3)2 + 12 = 4 = 2 √ 1 3 π sin θ = , cos θ = . Thus, θ = , or 30◦ , 2 2 6 and we have ! " √ π π 3 + i = 2 cos + i sin , or 6 6 √ 3 + i = 2(cos 30◦ + i sin 30◦ ). 18. Find trigonometric notation for 4. √ r = 42 = 4 0 4 sin θ = = 0, cos θ = = 1. Thus, 0 = 0, or 0◦ and we 4 4 have 4 = 4(cos 0 + i sin 0), or 4 = 4(cos 0◦ + i sin 0◦ ). 2 19. Find trigonometric notation for . 5 *! " 2 2 2 r= = 5 5

2 0 = 0, cos θ = 5 = 1. Thus, θ = 0, or 0◦ , sin θ = 2 2 5 5 and we have 2 r = (cos 0 + i sin 0), or 5 2 r = (cos 0◦ + i sin 0◦ ). 5

20. Find trigonometric notation for 7.5i. % r = (7.5)2 = 7.5 0 π 7.5 = 1, cos θ = = 0. Thus, θ = , or 90◦ , sin θ = 7.5 7.5 2 and we have " ! π π , or 7.5i = 7.5 cos + i sin 2 2 7.5i = 7.5(cos 90◦ + i sin 90◦ ).

433 √ √ 21. Find trigonometric notation for −3 2 − 3 2i. ) √ √ √ r = (−3 2)2 + (−3 2)2 = 36 = 6 √ √ √ √ −3 2 2 2 −3 2 =− , cos θ = =− . Thus sin θ = 6 2 6 2 √ √ 5π θ= , or 225◦ , and we have −3 2 − 3 2i = " ! 4 5π 5π + i sin , or 6(cos 225◦ + i sin 225◦ ). 6 cos 4 4 √ 9 9 3 22. Find trigonometric notation for − − i. 2 2 *! ( ( √ "2 "2 ! 9 3 9 81 243 324 + − = + = = r= − 2 2 4 4 4 √ 81 = 9 √ 9 9 3 √ − − 3 1 2 2 =− , cos θ = = − . Thus sin θ = 9 2 9 2 4π ◦ θ= , or 240 , and we have 3 √ ! " 9 9 3 4π 4π i = 9 cos + i sin − − , or 2 2 3 3 ◦ ◦ 9(cos 240 + i sin 240 ). 23. 3(cos 30◦ + i sin 30◦ ) = 3 cos 30◦ + (3 sin 30◦ )i √ √ 3 3 3 ◦ = a = 3 cos 30 = 3 · 2 2 1 3 ◦ b = 3 sin 30 = 3 · = 2 2 √ 3 3 3 Thus 3(cos 30◦ + i sin 30◦ ) = + i. 2 2 24. 6(cos 120◦ + i sin 120◦ ) = 6 cos 120◦ + (6 sin 120◦ )i ! " 1 ◦ = −3 a = 6 cos 120 = 6 − 2 √ √ 3 =3 3 b = 6 sin 120◦ = 6 · 2 √ Thus 6(cos 120◦ + i sin 120◦ ) = −3 + 3 3i. 25. 10(cos 270◦ + i sin 270◦ ) = 10 cos 270◦ + (10 sin 270◦ )i a = 10 cos 270◦ = 10 · 0 = 0

b = 10 sin 270◦ = 10(−1) = −10

Thus 10(cos 270◦ + i sin 270◦ ) = 0 + (−10)i = −10i. 26. 3(cos 0◦ + i sin 0◦ ) = 3 cos 0◦ + (3 sin 0◦ )i a = 3 cos 0◦ = 3 · 1 = 3 b = 3 sin 0◦ = 3 · 0 = 0

Thus 3(cos 0◦ + i sin 0◦ ) = 3 + 0i = 3. " ! " ! √ √ √ π π π π = 8 cos + i 27. 8 cos + i sin 8 cos 4 4 4 4 √ √ π √ 2 =2 a = 8 cos = 8 · 4 2 √ √ 2 π √ =2 b = 8 sin = 8 · 4 2 " ! √ π π = 2 + 2i. Thus 8 cos + i sin 4 4

434

28.

29.

30.

31.

Chapter 7: Applications of Trigonometry " ! " π π π π 5 cos + i sin = 5 cos + 5 sin i 3 3 3 3 π 1 5 a = 5 cos = 5 · = 3 2 2 √ √ 5 3 π 3 = b = 5 sin = 5 · 3 2 2 √ ! " π π 5 5 3 Thus 5 cos + i sin = + i. 3 3 2 2 " " ! ! π π π π = 2 cos + 2 sin i 2 cos + i sin 2 2 2 2 π a = 2 cos = 2 · 0 = 0 2 π b = 2 sin = 2 · 1 = 2 2 " ! π π = 0 + 2i = 2i. Thus 2 cos + i sin 2 2 # ! " ! "$ 3π 3π 3 cos − + i sin − = 4 4 " # ! "$ ! 3π 3π + 3 sin − i 3 cos − 4 4 √ " ! √ " ! 2 3 2 3π a = 3 cos − =3 − =− 4 2 2 √ " ! √ " ! 3π 2 3 2 b = 3 sin − =3 − =− 4 2 2 " ! "$ # ! 3π 3π Thus 3 cos − + i sin − = 4 4 √ √ 3 2 3 2 − i. − 2 2 √ 2[cos(−60◦ ) + i sin(−60◦ )] = √ √ 2 cos(−60◦ ) + [ 2 sin(−60◦ )]i √ √ 1 √ 2 a = 2 cos(−60◦ ) = 2 · = 2 2 √ ! √ " √ √ 3 6 =− b = 2 sin(−60◦ ) = 2 − 2 2 √ √ √ 2 6 Thus 2[cos(−60◦ ) + i sin(−60◦ )] = − i. 2 2 !

32. 4(cos 135◦ + i sin 135◦ ) = 4 cos 135◦ + (4 sin 135◦ )i ! √ " √ 2 ◦ a = 4 cos 135 = 4 − = −2 2 2 √ √ 2 =2 2 b = 4 sin 135◦ = 4 · 2 √ √ Thus 4(cos 135◦ + i sin 135◦ ) = −2 2 + 2 2i. 33.

12(cos 48◦ + i sin 48◦ ) 3(cos 6◦ + i sin 6◦ ) 12 [cos(48◦ − 6◦ ) + i sin(48◦ − 6◦ )] = 3 = 4(cos 42◦ + i sin 42◦ )

34.

35.

!

" ! " π π π π 5 cos + i sin · 2 cos + i sin 3 3 4 4 # ! " ! "$ π π π π = 5 · 2 cos + + i sin + 3 4 3 4 " ! 7π 7π = 10 cos + i sin 12 12 2.5(cos 35◦ + i sin 35◦ ) · 4.5(cos 21◦ + i sin 21◦ )

= 2.5(4.5)[cos(35◦ + 21◦ ) + i sin(35◦ + 21◦ )]

36.

= 11.25(cos 56◦ + i sin 56◦ ) ! " 2π 2π 1 cos + i sin 2 3 3 ! " π π 3 cos + i sin 8 6 6 1# ! " ! "$ 2π π 2π π = 2 cos − + i sin − 3 3 6 3 6 8 ! " 4 π π = cos + i sin 3 2 2

37. (1 − i)(2 + 2i)

Find trigonometric notation for 1 − i. % √ r = 12 + (−1)2 = 2 √ √ 1 −1 2 2 , cos θ = √ = sin θ = √ = − 2 2 2 2 √ 7π Thus θ = , or 315◦ , and 1−i = 2(cos 315◦ +i sin 315◦ ). 4 Find trigonometric notation for 2 + 2i. √ √ √ r = 22 + 22 = 8 = 2 2 √ √ 2 2 2 2 , cos θ = √ = sin θ = √ = 2 2 2 2 2 2 √ π Thus θ = , or 45◦ , and 2 + 2i = 2 2(cos 45◦ + i sin 45◦ ). 4 (1 − i)(2 + 2i) √ √ = 2(cos 315◦ +i sin 315◦ ) · 2 2(cos 45◦ +sin 45◦ ) √ √ = 2 · 2 2[cos(315◦ + 45◦ ) + i sin(315◦ + 45◦ )] = 4(cos 360◦ + i sin 360◦ )

= 4(cos 0◦ + i sin 0◦ ), or 4 ) √ √ 38. For 1 + i 3: r = 12 + ( 3)2 = 2 √ 3 1 sin θ = and cos θ = , so θ = 60◦ . 2 2 √ 1 + i 3 = 2(cos 60◦ + i sin 60◦ ) From Example 3(a) we know that 1 + i = √ 2(cos 45◦ + i sin 45◦ ). √ 1 + i = 2(cos 45◦ + i sin 45◦ ) √ 2(cos 60◦ + i sin 60◦ ) · 2(cos 45◦ + i sin 45◦ ) = √ 2 2(cos 105◦ + i sin 105◦ )

Exercise Set 7.3 Using identities for the √ and difference of angles we √ sum 2− 6 ◦ find that cos 105 = and 4 √ √ 6+ 2 . Thus, sin 105◦ = 4 √ ◦ 2 2(cos 105 + i sin 105◦ ) √ √ √ " !√ √ 2− 6 6+ 2 + i =2 2 4 4 √ √ = 1 − 3 + ( 3 + 1)i

39.

1−i 1+i Find trigonometric notation for 1 − i. % √ r = 12 + (−1)2 = 2 √ √ −1 2 1 2 sin θ = √ = − , cos θ = √ = 2 2 2 2 √ 7π , or 315◦ , and 1−i = 2(cos 315◦ +i sin 315◦ ) Thus θ = 4 From Example 3(a) we know that 1 + i = √ 2(cos 45◦ + i sin 45◦ ). 1−i 1+i √ 2(cos 315◦ + i sin 315◦ ) = √ 2(cos 45◦ + i sin 45◦ ) √ 2 = √ [cos(315◦ − 45◦ ) + i sin(315◦ − 45◦ )] 2 = 1(cos 270◦ + i sin 270◦ ) = 1[0 + i(−1)] = −i

40. From Exercise 35 we know that 1 − i = √ 2(cos 315◦ + i sin 315◦ ). √ From Example 3(b) we know that 3 − i =

2(cos 330◦ + i sin 330◦ ). 1−i √ 3−i √ 2(cos 315◦ + i sin 315◦ ) = 2(cos 330◦ + i sin 330◦ ) √ 2 [cos(−15◦ ) + i sin(−15◦ )] = 2 Using identities for the√ sum√and difference of angles we 2+ 6 find that cos(−15◦ ) = and 4 √ √ 2− 6 sin(−15◦ ) = . Thus, 4 √ 2 [cos(−15◦ ) + i sin(−15◦ )] 2 √ !√ √ √ √ " 2 2+ 6 2− 6 = + i 2 4 4 √ √ 1+ 3 1− 3 + i = 4 4

435 √ 41. (3 3 − 3i)(2i)

√ Find trigonometric notation for 3 3 − 3i. ) √ √ r = (3 3)2 + (−3)2 = 36 = 6 √ √ 1 3 3 −3 3 = − , cos θ = = sin θ = 6 2 6 2 √ Thus θ = 330◦ , and 3 3 − 3i = 6(cos 330◦ + i sin 330◦ ).

Find trigonometric notation for 2i. √ r = 22 = 2 2 0 sin θ = = 1, cos θ = = 0 2 2 Thus θ = 90◦ , and 2i = 2(cos 90◦ + i sin 90◦ ). √ (3 3 − 3i)(2i)

= 6(cos 330◦ + i sin 330◦ ) · 2(cos 90◦ + i sin 90◦ )

= 6 · 2[cos(330◦ + 90◦ ) + i sin(330◦ + 90◦ )] = 12(cos 420◦ + i sin 420◦ )

= 12(cos 60◦ + i sin 60◦ ) √ " ! 3 1 +i· = 12 2 2 √ = 6 + 6 3i ) √ √ 42. For 2 3 + 2i: r = (2 3)2 + 22 = 4 √ √ 1 2 3 3 2 = , so θ = 30◦ . sin θ = = and cos θ = 4 2 4 2 √ 2 3 + 2i = 4(cos 30◦ + i sin 30◦ ) From Exercise 41 we know that 2i = 2(cos 90◦ + i sin 90◦ ). √ (2 3 + 2i)(2i) = 4(cos 30◦ + i sin 30◦ ) · 2(cos 90◦ + i sin 90◦ )

= 8(cos 120◦ + i sin 120◦ ) √ " ! 1 3 = 8 − +i· 2 2 √ = −4 + 4 3i √ 2 3 − 2i √ 43. 1 + 3i

√ Find trigonometric notation for 2 3 − 2i. ) √ √ r = (2 3)2 + (−2)2 = 16 = 4 √ √ 1 2 3 3 −2 = − , cos θ = = sin θ = 4 2 4 2 √ Thus θ = 330◦ , and 2 3 − 2i = 4(cos 330◦ + i sin 330◦ ) √ Find trigonometric notation for 1 + 3i. ) √ √ r = 12 + ( 3)2 = 4 = 2 √ 3 1 sin θ = , cos θ = 2 2 √ Thus θ = 60◦ , and 1 + 3i = 2(cos 60◦ + i sin 60◦ )

436

Chapter 7: Applications of Trigonometry √ 2 3 − 2i √ 1 + 3i 4(cos 330◦ + i sin 330◦ ) = 2(cos 60◦ + i sin 60◦ ) 4 = [cos(330◦ − 60◦ ) + i sin(330◦ − 60◦ )] 2 = 2(cos 270◦ + i sin 270◦ ) = 2[0 + i · (−1)]

49.

= 27(cos 60◦ + i sin 60◦ ) √ " ! 1 3 = 27 +i· 2 2 √ 27 27 3 = + i 2 2 50.

) √ √ 44. For 3 − 3 3i: r = 32 + (−3 3)2 = 6 √ √ 3 3 1 −3 3 =− and cos θ = = , so θ = 300◦ . sin θ = 6 2 6 2 √ 3 − 3 3i = 6(cos 300◦ + i sin 300◦ ) From Example 3(b) we know that √ 3 − i = 2(cos 330◦ + i sin 330◦ ). √ 3 − 3 3i √ 3−i 6(cos 300◦ + i sin 300◦ ) = 2(cos 330◦ + i sin 330◦ ) = 3[cos(−30◦ ) + i sin(−30◦ )] ! "$ #√ 3 1 +i· − =3 2 2 √ 3 3 3 − i = 2 2 "$3 # ! π π 45. 2 cos + i sin 3 3 " ! "$ # ! π π = 23 cos 3 · + i sin 3 · 3 3 = 8(cos π + i sin π) [2(cos 120◦ + i sin 120◦ )]4 = 24 [cos(4 · 120◦ ) + i sin(4 · 120◦ )]

= 16(cos 480◦ + i sin 480◦ ) = 16(cos 120◦ + i sin 120◦ )

47. From Exercise 39 we know that 1 + i = √ 2(cos 45◦ + i sin 45◦ ). √ (1 + i)6 = [ 2(cos 45◦ + i sin 45◦ )]6 √ = ( 2)6 [cos(6 · 45◦ ) + i sin(6 · 45◦ )]

48.

[2(cos 10◦ + i sin 10◦ )]9 = 512(cos 90◦ + i sin 90◦ )

= −2i

46.

[3(cos 20◦ + i sin 20◦ )]3

= 8(cos 270◦ + i sin 270◦ ), or " ! 3π 3π 8 cos + i sin 2 2

= 512(0 + i · 1) = 512i

51. From Exercise 13 we know that 1 − i = √ 2(cos 315◦ + i sin 315◦ ). √ (1 − i)5 = [ 2(cos 315◦ + i sin 315◦ )]5 √ = ( 2)5 (cos 1575◦ + i sin 1575◦ )

52.

= 64(cos 180◦ + i sin 180◦ ) = 64(−1 + i · 0) = −64

1 1 53. Find trigonometric notation for √ − √ i: 2 2 *! "2 ! "2 √ 1 1 √ + −√ = 1=1 r= 2 2 1 √ −√ 1 2 2 = − √ , or − sin θ = 1 2 2 1 √ √ 1 2 2 = √ , or cos θ = 1 2 2 1 1 ◦ Thus θ = 315 , so √ − √ i = 2 2 1(cos 315◦ + i sin 315◦ ). "12 ! 1 1 √ −√ i = [1(cos 315◦ + i sin 315◦ )]12 2 2 = 1(cos 3780◦ + i sin 3780◦ ) = 1(cos 180◦ + i sin 180◦ ) = 1(−1 + i · 0)

√ (− 3 + i)5 = [2(cos 150◦ + i sin 150◦ )]5 = 32(cos 750◦ + i sin 750◦ ) = 32(cos 30 + i sin 30 ), or " ! π π 32 cos + i sin 6 6 ◦

◦

= 25/2 (cos 135◦ + i sin 135◦ ) √ " ! √ √ 2 2 +i· =4 2 − 2 2 = −4 + 4i √ (2 + 2i)4 = [2 2(cos 45◦ + i sin 45◦ )]4

54.

= −1 "10 3 1 + i = [1(cos 30◦ + i sin 30◦ )]10 2 2 = 1(cos 300◦ + i sin 300◦ ) ! √ "$ # 3 1 = 1 +i· − 2 2 √ 3 1 i = − 2 2

!√

Exercise Set 7.3

437 59. i = 1(cos 90◦ + i sin 90◦ )

55. −i = 1(cos 270◦ + i sin 270◦ ) (−i)1/2

i1/3

= [1(cos 270◦ + i sin 270)]1/2 " # ! 270◦ 360◦ = 11/2 cos +k· + 2 2 ! "$ 270◦ 360◦ i sin +k· , k = 0, 1 2 2

= [1(cos 90◦ + i sin 90◦ )]1/3 " ! ◦ "$ # ! ◦ 90 360◦ 90 360◦ = 1 cos +k · +i sin +k · , 3 3 3 3 k = 0, 1, 2 ◦ ◦ ), 1(cos 150√ + i sin 150◦ ), The roots are 1(cos 30◦ + i sin 30√ 3 1 3 1 and 1(cos 270◦ + i sin 270◦ ), or + i, − + i, and 2 2 2 2 −i.

= 1[cos(135 + k · 180 ) + i sin(135 + k · 180 )], ◦

◦

◦

◦

k = 0, 1

The roots are 1(cos 135◦ + i sin 135◦ ), for k = 0

60.

= [64(cos 270◦ + i sin 270◦ )]1/3 # ! " 270◦ 360◦ = 4 cos +k· + 3 3 "$1/3 ! 270◦ 360◦ , k = 0, 1, 2 i sin +k· 3 3

and 1(cos 315◦ + i sin 315◦ ), for k = 1, √ √ √ √ 2 2 2 2 or − + i and − i. 2 2 2 2 56.

(1 + i)1/2 √ = [ 2(cos 45◦ + i sin 45◦ )]1/2 # ! ◦ " 45 360◦ 1/2 1/2 = (2 ) cos +k· + 2 2 "$ ! ◦ 360◦ 45 +k· , k = 0, 1 i sin 2 2 The roots are √ 4 2(cos 22.5◦ + i sin 22.5◦ ) and

√ 4

2(cos 202.5◦ + i sin 202.5◦ ). √ 57. 2 2 − 2 2i = 4(cos 315◦ + i sin 315◦ ) √ √ (2 2 − 2 2i)1/2 √

= [4(cos 315◦ + i sin 315◦ )]1/2 " # ! 315◦ 360◦ = 41/2 cos +k· + 2 2 "$ ! 360◦ 315◦ +k· , k = 0, 1 i sin 2 2 The roots are

◦ ◦ The roots are 4(cos 90◦ + i sin 90◦ ), 4(cos √ 210 + i sin √210 ), ◦ ◦ and 4(cos 330 +i sin 330 ), or 4i, −2 3−2i, and 2 3−2i. √ 61. (2 3 − 2i)1/3

= 4(cos 330◦ + i sin 330◦ )1/3 " # ! 360◦ 330◦ +k· = 41/3 cos + 3 3 ! "$ 330◦ 360◦ i sin +k· , k = 0, 1, 2 3 3 √ The roots are 3 4(cos 110◦ + i sin 110◦ ), √ 3 4(cos 230◦ + i sin 230◦ ), and √ 3 4(cos 350◦ + i sin 350◦ ). √ 62. (1 − 3i)1/3 = [2(cos 300◦ + i sin 300◦ )]1/3 " # ! 300◦ 360◦ = 21/3 cos +k· + 3 3 "$ ! 300◦ 360◦ i sin +k· , k = 0, 1, 2 3 3 √ The roots are 3 2(cos 100◦ + i sin 100◦ ), √ 3 2(cos 220◦ + i sin 220◦ ), and √ 3 2(cos 340◦ + i sin 340◦ ).

2(cos 157.5◦ + i sin 157.5◦ ), for k = 0 and 2(cos 337.5◦ + i sin 337.5◦ ), for k = 1. √ (− 3 − i)1/2

58.

= [2(cos 210◦ + i sin 210◦ )]1/2 " # ! 210◦ 360◦ = 21/2 cos +k· + 2 2 "$ ! 210◦ 360◦ i sin +k· , k = 0, 1 2 2 The roots are √ 2(cos 105◦ + i sin 105◦ ) and

√

2(cos 285◦ + i sin 285◦ ).

(−64i)1/3

63.

161/4 = [16(cos 0◦ + i sin 0◦ )]1/4 " ! "$ # ! ◦ 360◦ 0 360◦ +k · +i sin 0◦ +k · , = 2 cos 4 4 4 k = 0, 1, 2, 3 The roots are 2(cos 0◦ + i sin 0◦ ), 2(cos 90◦ + i sin 90◦ ), 2(cos 180◦ + i sin 180◦ ), and 2(cos 270◦ + i sin 270◦ ), or 2, 2i, −2, and −2i.

438

Chapter 7: Applications of Trigonometry

i

2i !1

!2

1

2 !i

!2i

67. 64.

= [8(cos 0◦ + i sin 0◦ )]1/10 # ! " ! ◦ "$ 360◦ 0 360◦ 0◦ = 81/10 cos +k· +i sin +k· , 10 10 10 10 k = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 √ √ The roots are 10 8(cos 0◦ + i sin 0◦ ), or 10 8; √ √ 10 8(cos 36◦ + i sin 36◦ ); 10 8(cos 72◦ + i sin 72◦ ), √ √ 10 8(cos 108◦ + i sin 108◦ ); 10 8(cos 144◦ + i sin 144◦ ); √ √ 10 8(cos 180◦ + i sin 180◦ ), or − 10 8; √ √ 10 8(cos 216◦ + i sin 216◦ ); 10 8(cos 252◦ + i sin 252◦ ); √ 10 8(cos 288◦ + i sin 288◦ ); and √ 10 8(cos 324◦ + i sin 324◦ ).

i1/4 = [1(cos 90◦ + i sin 90◦ )]1/4 # ! ◦ " ! ◦ "$ 90 360◦ 90 360◦ = 1 cos +k · +i sin +k · , 4 4 4 4 k = 0, 1, 2, 3 The roots are cos 22.5◦ +i sin 22.5◦ , cos 112.5◦ +i sin 112.5◦ , cos 202.5◦ + i sin 202.5◦ , and cos 292.5◦ + i sin 292.5◦ .

i

!1

1

68.

!i

65.

= [1(cos 180◦ + i sin 180◦ )]1/5 " ! ◦ "$ # ! ◦ 180 360◦ 180 360◦ = 1 cos +k· +i sin +k· , 5 5 5 5 k = 0, 1, 2, 3, 4 The roots are cos 36◦ + i sin 36◦ ; cos 108◦ + i sin 108◦ ; cos 180◦ + i sin 180◦ , or −1; cos 252◦ + i sin 252◦ ; and cos 324◦ + i sin 324◦ .

i

1

69.

!i

66.

11/6 = [1(cos 0◦ + i sin 0◦ )]1/6 " ! ◦ "$ # ! ◦ 0 360◦ 0 360◦ +k · +i sin +k · , = 1 cos 6 6 6 6 k = 0, 1, 2, 3, 4, 5 The roots are cos 0 + i sin 0 , cos 60 + i sin 60 , ◦

◦

◦

◦

cos 120◦ + i sin 120◦ , cos 180◦ + i sin 180◦ , cos 240◦ + i sin 240◦ , and cos 300◦ + i sin 300◦ , √ √ √ √ 1 1 1 3 3 3 3 1 i, − + i, −1, − − i, and − i. or 1, + 2 2 2 2 2 2 2 2

(−4)1/9 = [4(cos 180◦ + i sin 180◦ )]1/9 # ! " 180◦ 360◦ 1/9 =4 cos +k· + 9 9 ! "$ 360◦ 180◦ +k· , i sin 9 9 k = 0, 1, 2, 3, 4, 5, 6, 7, 8 √ 9 The roots are 4(cos 20◦ + i sin 20◦ ); √ √ √ 9 9 √ 4 4 3 9 4(cos 60◦ + i sin 60◦ ), or + i; 2 2 √ √ 9 4(cos 100◦ + i sin 100◦ ); 9 4(cos 140◦ + i sin 140◦ ); √ √ 9 4(cos 180◦ + i sin 180◦ ), or − 9 4; √ √ 9 4(cos 220◦ + i sin 220◦ ); 9 4(cos 260◦ + i sin 260◦ ); √ √ √ 9 9 √ 4 4 3 9 4(cos 300◦ + i sin 300◦ ), or − i; 2 2 √ and 9 4(cos 340◦ + i sin 340◦ ).

(−1)1/5

!1

81/10

(−1)1/6 = [1(cos 180◦ + i sin 180◦ )]1/6 " # ! 360◦ 180◦ +k· + = 1 cos 6 6 "$ ! 180◦ 360◦ i sin +k· , k = 0, 1, 2, 3, 4, 5 6 6 The roots are cos 30◦ + i sin 30◦ , cos 90◦ + i sin 90◦ , cos 150◦ + i sin 150◦ , cos 210◦ + i sin 210◦ , cos 270◦ + i sin 270◦ , √ 3 1 + i, and cos 330◦ + i sin 330◦ , or 2 2 √ √ √ 3 1 3 1 3 1 i, − + i, − − i, −i, and − i. 2 2 2 2 2 2

Exercise Set 7.3 70.

439

= [12(cos 0◦ + i sin 0◦ )]1/4 # ! ◦ " ! ◦ "$ 0 360◦ 0 360◦ = 121/4 cos +i sin , +k · +k · 4 4 4 4 k = 0, 1, 2, 3 √ The roots are 4 12(cos 0◦ + i sin 0◦ ), √ √ 4 12(cos 90◦ + i sin 90◦ ), 4 12(cos 180◦ + i sin 180◦ ), √ √ √ and 4 12(cos 270◦ + i sin 270◦ ), or 4 12, 4 12i, √ √ − 4 12, and − 4 12i. 71. x3 = 1 The solutions of this equation are the cube roots of 1. These √ were found in Example 11 in the text. They are 1, √ 3 1 3 1 i, and − − i. − + 2 2 2 2 72.

x5 − 1 = 0

x5 = 1

Find the fifth roots of 1. 11/5 = [1(cos 0◦ + i sin 0◦ )]1/5 " ! ◦ "$ # ! ◦ 0 360◦ 0 360◦ = 1 cos +k· +i sin +k · , 5 5 5 5 k = 0, 1, 2, 3, 4 The solutions are cos 0◦ + i sin 0◦ , or 1; cos 72◦ + i sin 72◦ ; cos 144◦ + i sin 144◦ ; cos 216◦ + i sin 216◦ ; and cos 288◦ + i sin 288◦ . 73.

x4 + i = 0 x4 = −i

Find the fourth roots of −i. (−i)1/4 = [1(cos 270◦ + i sin 270◦ )]1/4 " # ! 360◦ 270◦ +k· + = 1 cos 4 4 "$ ! 270◦ 360◦ +k· i sin , k = 0, 1, 2, 3 4 4 The solutions are cos 67.5◦ + i sin 67.5◦ , ◦

◦

75.

x6 + 64 = 0 x6 = −64

Find the sixth roots of −64. (−64)1/6 = [64(cos 180◦ +i sin 180◦ )]1/6 # ! " 180◦ 360◦ = 2 cos +k· + 6 6 "$ ! 360◦ 180◦ +k· , k = 0, 1, 2, 3, 4, 5 i sin 6 6

The solutions are 2(cos 30◦ + i sin 30◦ ), 2(cos 90◦ + i sin 90◦ ), 2(cos 150◦ + i sin 150◦ ), ◦ 270◦ ), and√ 2(cos 210◦ + i sin 210◦ ), 2(cos √ 270 + i sin√ ◦ ◦ 2(cos 330 √+ i sin 330 ), or 3 + i, 2i, − 3 + i, − 3 − i, −2i, and 3 − i. √ 76. x5 + 3 + i = 0 √ x5 = − 3 − i √ Find the fifth roots of − 3 − i. √ (− 3 − i)1/5 = [2(cos 210◦ + i sin 210◦ )]1/5 # ! " √ 210◦ 360◦ 5 = 2 cos +k· + 5 5 "$ ! 360◦ 210◦ +k· , k = 0, 1, 2, 3, 4 i sin 5 5 √ The solutions are 5 2(cos 42◦ + i sin 42◦ ); √ √ 5 2(cos 114◦ + i sin 114◦ ); 5 2(cos 186◦ + i sin 186◦ ); √ 5 2(cos 258◦ + i sin 258◦ ); and √ √ √ 5 5 √ 2 3 2 5 2(cos 330◦ + i sin 330◦ ), or − i. 2 2

77. Left to the student 78. Left to the student

cos 157.5 + i sin 157.5 , cos 247.5 + i sin 247.5 , and ◦

◦

cos 337.5◦ + i sin 337.5◦ . 74.

√ √ 3 2 3 2 and 3(cos 315◦ + i sin 315◦ ), or + i, 2 2 √ √ √ √ √ √ 3 2 3 2 3 2 3 2 3 2 3 2 + i, − − i, and − i. − 2 2 2 2 2 2

121/4

x4 + 81 = 0 x = −81 4

Find the fourth roots of −81. (−81)1/4 = [81(cos 180◦ + i sin 180◦ )]1/4 " # ! 180◦ 360◦ +k· + = 3 cos 4 4 "$ ! 180◦ 360◦ +k· , k = 0, 1, 2, 3 i sin 4 4

The solutions are 3(cos 45◦ + i sin 45◦ ),

3(cos 135◦ + i sin 135◦ ), 3(cos 225◦ + i sin 225◦ ),

79. The square roots of 1 − i are √ 4 2(cos 157.5◦ + i sin 157.5◦ ) and √ 4 2(cos 337.5◦ + i sin 337.5◦ ), or approximately 1.0987 + 0.4551i and −1.0987 − 0.4551i. Imaginary i

Real 1

The roots are opposites because they are reflections of each other across the origin.

440

Chapter 7: Applications of Trigonometry

80. Trigonometric notation is not unique because there are infinitely many angles coterminal with a given angle. Standard notation is unique because any point has a unique ordered pair (a, b) associated with it. 81.

π 180◦ π radians = radians · 12 12 π radians π(180◦ ) 12π = 15◦

=

180◦ 82. 3π radians = 3π radians · = 540◦ π radians 83.

330◦ = 330◦ ·

π radians 180◦

330π radians 180 11π radians = 6 =

84. −225◦ = −225◦ ·

π radians 5π =− radians 180◦ 4

85. Use the Pythagorean theorem. r2 = 9 + 36 = 45 √ √ r = 45 = 3 5 y 4 2

(0, 3) 4

!4 !2 !2

91. x2 + (1 − i)x + i = 0

Use the quadratic formula. a = 1, b = 1 − i, c = i # −(1 − i) ± (1 − i)2 − 4 · 1 · i x= 2·1 √ −1 + i ± 1 − 2i + i2 − 4i = 2 √ −1 + i ± −6i = 2 Now we find the square roots of −6i. (−6i)1/2

= [6(cos 270◦ + i sin 270◦ )]1/2 " ! ◦ "% $ ! ◦ √ 270 360◦ 270 360◦ = 6 cos +k· +i sin +k · , 2 2 2 2 k = 0, 1 √ The roots are 6(cos 135◦ + i sin 135◦ ) and √ √ √ 6(cos 315◦ + i sin 315◦ ), or − 3 + 3i and √ √ 3 − 3i.

r2 = 32 + 62

86.

5π is a reflection across the y6 π axis of the point determined by . The coordinates of the ! √6 " π 3 1 point determined by are , , so the coordinates 6 2 2 ! √ " 5π 3 1 of the point determined by are − , . Thus, 6 2 2 1 5π =y= . sin 6 2

90. The point determined by

(2, !1)

x

!!q, !4" 2π is a reflection across the y3 π axis of the point determined by . The coordinates of the ! 3√ " 1 π 3 point determined by are , , so the coordinates 3 2 2 √ " ! 2π 1 3 of the point determined by are − , . Thus, 3 2 2 √ 3 2π sin =y= . 3 2 π 88. The coordinates of the point determined by are 6 " !√ 3 1 , . Thus, 2 2 √ π 3 cos = x = . 6 2 π 89. The coordinates of the point determined by are 4 !√ √ " 2 2 , . Thus, 2 2 √ 2 π . cos = x = 4 2 87. The point determined by

Then the solutions of the original equation are √ √ √ √ −1 − 3 1 + 3 −1 + i − 3 + 3i = + i, or x= 2 2 2 √ √ 1+ 3 1+ 3 − + i 2 2 and √ √ √ √ −1 + 3 1 − 3 −1 + i + 3 − 3i = + i, or x= 2 2 2 √ √ 1− 3 1− 3 + i. − 2 2 92. 3x2 + (1 + 2i)x + (1 − i) = 0 # −(1 + 2i) ± (1 + 2i)2 − 4 · 3 · (1 − i) x= 2·3 √ −1 − 2i ± −15 + 16i = 6 Find the square roots of −15 + 16i.

(−15 + 16i)1/2 &√ '1/2 ≈ 481(cos 133.15◦ + i sin 133.15◦ ) $ ! " √ 133.15◦ 360◦ 4 ≈ 481 cos +k· + 2 2 "% ! 360◦ 133.15◦ +k· , k = 0, 1 i sin 2 2 The square roots are approximately √ 4 481(cos 66.575◦ + i sin 66.575◦ ) and

Exercise Set 7.4

441

√ 4

481(cos 246.575◦ + i sin 246.575◦ ), or 1.8618 + 4.2972i and −1.8618 − 4.2972i.

93.

(cos θ + i sin θ)−1 1 cos θ − i sin θ = · cos θ + i sin θ cos θ − i sin θ cos θ − i sin θ = cos2 θ + sin2 θ = cos θ − i sin θ √ 94. z = a + bi, |z| = a2 + b2 ; % √ −z = −a − bi, | − z| = (−a)2 + (−b)2 = a2 + b2 ;

!z! # 1

i

Then the solutions of the original equation are −1 − 2i + 1.8618 + 4.2972i ≈ 0.1436 + 0.3829i x≈ 6 and −1−2i−1.8618−4.2972i ≈ −0.4770−1.0495i. x≈ 6

!1

1 !i

101.

z+z = 3 (a + bi) + (a − bi) = 3

2a = 3 3 a= 2

z"z#3

so |z| = | − z|.

95. See the answer section in the text.

w

96. |zz| = |(a + bi)(a − bi)| = |a2 + b2 | = a2 + b2 ;

|z 2 | = |(a + bi)2 | = |a2 + 2abi − b2 | = % √ (a2 − b2 )2 + (2ab)2 = a4 − 2a2 b2 + b4 + 4a2 b2 = % √ a4 + 2a2 b2 + b4 = (a2 + b2 )2 = a2 + b2 ; so |zz| = |z |. 2

97. See the answer section in the text. 98. |z · w| = |r1 (cos θ1 + i sin θ1 ) · r2 (cos θ2 + i sin θ2 )| = |r1 r2 [cos(θ1 + θ2 ) + i sin(θ1 + θ2 )]| = ) r12 r22 cos2 (θ1 + θ2 ) + r12 r22 sin2 (θ1 + θ2 ) = ) r12 r22 [cos2 (θ1 + θ2 ) + sin2 (θ1 + θ2 )] = % r12 r22 = r1 r2 ;

|z| · |w| = ) ) r12 cos2 θ1 + r12 sin2 θ1 r22 cos2 θ2 + r22 sin2 θ2 = ) ) r12 (cos2 θ1 + sin2 θ1 ) r22 (cos2 θ2 + sin2 θ2 ) = % % r12 r22 = r1 r2 ; so |z · w| = |z| · |w|.

99. See the answer section in the text. 100. Graph: |z| = 1

Let z = a + bi Then |a + bi| = 1 √ a2 + b2 = 1 a2 + b2 = 1

The graph of a2 + b2 = 1 is a circle whose radius is 1 and whose center is (0, 0).

Exercise Set 7.4 See the text answer section for odd exercise answers 1-11. Even exercises 2-12 are shown below. 90$

120$ 150$ 180$

2

6 8 10

210$ 240$

60$

12

270$

30$ 0 4 360$

2

4

330$

300$

13. Answers may vary. A: (4, 30◦ ), (4, 390◦ ), (−4, 210◦ ); B: (5, 300◦ ), (5, −60◦ ), (−5, 120◦ );

C: (2, 150◦ ), (2, −210◦ ), (−2, 330◦ );

D: (3, 225◦ ), (3, −135◦ ), (−3, 45◦ )

14. Answers may vary. " ! " ! " ! 9π 5π π A: 2, , 2, , − 2, ; 4 4 4 ! " ! " ! " 3π π π B: 5, , 5, − , − 5, ; 2 2 2 " ! " ! " ! 13π π 11π , 4, − , − 4, − ; C: 4, 12 12 12 " ! " ! " ! 5π π 2π , 3, − , − 3, D: 3, 3 3 3

442 15. (0, −3) % √ r = 02 + (−3)2 = 9 = 3 −3 tan θ = , which is not defined; therefore, since (0, −3) 0 3π lies on the negative y-axis, θ = 270◦ , or . 2 " ! 3π . Thus, (0, −3) = (3, 270◦ ), or 3, 2 16. (−4, 4) % √ √ r = (−4)2 + 42 = 32 = 4 2 3π 4 = −1, so θ = 135◦ , or . (The point is in tan θ = −4 4 quadrant II.) " ! √ 3π √ . Thus, (−4, 4) = (4 2, 135◦ ), or 4 2, 4 √ 17. (3, −3 3) ) √ √ √ r = 32 + (−3 3)2 = 9 + 27 = 36 = 6 √ √ 5π −3 3 = − 3, so θ = 300◦ , or . (The point is tan θ = 3 3 in quadrant IV.) ! " √ 5π Thus, (3, −3 3) = (6, 300◦ ), or 6, . 3 √ 18. (− 3, 1) ) √ √ r = (− 3)2 + 12 = 4 = 2 √ 5π 1 3 tan θ = √ , or − , so θ = 150◦ , or . (The point is 3 6 − 3 in quadrant II.) " ! √ 5π . Thus, (− 3, 1) = (2, 150◦ ), or 2, 6 √ 19. (4 3, −4) ) √ √ √ r = (4 3)2 + (−4)2 = 48 + 16 = 64 = 8 √ 1 3 11π −4 , so θ = 330◦ , or . (The tan θ = √ = − √ , or − 3 6 4 3 3 point is in quadrant IV.) ! " √ 11π . Thus, (4 3, −4) = (8, 330◦ ), or 8, 6 √ 20. (2 3, 2) ) √ √ r = (2 3)2 + 22 = 16 = 4 √ 1 3 π 2 tan θ = √ = √ , or , so θ = 30◦ , or . (The point 3 6 2 3 3 is in quadrant I.) " ! √ π . Thus, (2 3, 2) = (4, 30◦ ), or 4, 6 √ √ 21. (− 2, − 2) ) √ √ √ √ r = (− 2)2 + (− 2)2 = 2 + 2 = 4 = 2 √ 5π − 2 . (The point is in tan θ = √ = 1, so θ = 225◦ , or 4 − 2 quadrant III.)

Chapter 7: Applications of Trigonometry √ √ Thus, (− 2, − 2) = (2, 225◦ ), or

!

2,

" 5π . 4

√ 22. (−3, 3 3) ) √ √ r = (−3)2 + (3 3)2 = 36 = 6 √ √ 3 3 2π tan θ = = − 3, so θ = 120◦ , or . (The point is in −3 3 quadrant II.) ! " √ 2π . Thus, (−3, 3 3) = (6, 120◦ ), or 6, 3 √ 23. (1, 3) ) √ √ r = 12 + ( 3)2 = 4 = 2 √ 3 √ π tan θ = = 3, so θ = 60◦ , or . (The point is in 1 3 quadrant I.) ! " √ π Thus, (1, 3) = (2, 60◦ ), or 2, . 3 24. (0, −1) % r = 02 + (−1)2 = 1 −1 tan θ = , which is not defined; therefore, since (0, −1) 0 3π lies on the negative y-axis, θ = 270◦ , or . Thus, 2 ! " 3π (0, −1) = (1, 270◦ ), or 1, . 2 √ " ! √ 5 2 5 2 ,− 25. 2 2 *! √ " √ "2 ( ! 2 5 2 25 25 √ 5 2 r= + − = + = 25 = 5 2 2 2 2 √ 5 2 − 7π . (The point is tan θ = √2 = −1, so θ = 315◦ , or 4 5 2 2 in quadrant IV.) √ " " ! ! √ 7π 5 2 5 2 ,− = (5, 315◦ ), or 5, . Thus, 2 2 4 √ " ! 3 3 3 26. − , − 2 2 *! √ "2 ( "2 ! 3 3 3 9 27 √ r= − + − = + = 9=3 2 2 4 4 √ 3 3 − 2 = √3, so θ = 240◦ , or 4π . (The point is tan θ = 3 3 − 2 in quadrant III.) √ " ! ! " 4π 3 3 3 Thus, − , − = (3, 240◦ ), or 3, . 2 2 3 27. (5, 60◦ ) x = r cos θ = 5 cos 60◦ = 5 ·

1 5 = 2 2

Exercise Set 7.4

443

√ √ 3 5 3 y = r sin θ = 5 sin 60◦ = 5 · = 2 2 √ " ! 5 5 3 , (5, 60◦ ) = 2 2

28. (0, −23◦ )

x = 0 cos(−23◦ ) = 0

y = 0 cos(−23◦ ) = 0 (0, −23◦ ) = (0, 0)

29. (−3, 45◦ )

√

√ 2 3 2 x = r cos θ = −3 cos 45 = −3 · =− 2 2 √ √ 2 2 3 =− y = r sin θ = −3 sin 45◦ = −3 · 2 2 √ √ " ! 3 2 3 2 ,− (−3, 45◦ ) = − 2 2 ◦

30. (6, 30◦ )

x = 6 cos 30 = 6 · ◦

√

√ 3 =3 3 2

1 y = 6 sin 30◦ = 6 · = 3 2 √ (6, 30◦ ) = (3 3, 3) 31. (3, −120◦ )

" ! 1 3 =− x = 3 cos(−120 ) = 3 − 2 2 √ ! √ " 3 3 3 y = 3 sin(−120◦ ) = 3 − =− 2 2 √ " ! 3 3 3 (3, −120◦ ) = − , − 2 2 ! " π 32. 7, 6 √ √ 3 7 3 π = x = 7 cos = 7 · 6 2 2 1 7 π y = 7 sin = 7 · = 6 2 2 ! " ! √ " π 7 3 7 7, = , 6 2 2 " ! 5π 33. − 2, 3 5π 1 x = −2 cos = −2 · = −1 3 2 ! √ " √ 3 5π = −2 − = 3 y = −2 sin 3 2 " ! √ 5π = (−1, 3) − 2, 3 ◦

35. (2, 210◦ )

! √ " √ 3 =− 3 x = 2 cos 210◦ = 2 − 2 ! " 1 = −1 y = 2 sin 210◦ = 2 − 2 √ (2, 210◦ ) = (− 3, −1) ! " 7π 36. 1, 4 √ 7π 2 x = 1 · cos = 4 2 √ 7π 2 =− y = 1 · sin 4 2 √ " ! " !√ 7π 2 2 1, = ,− 4 2 2 " ! 5π 37. − 6, 6 ! √ " √ 3 5π = −6 − =3 3 x = −6 cos 6 2 5π 1 x = −6 sin = −6 · = −3 6 2 ! " √ 5π − 6, = (3 3, −3) 6 38. (4, 180◦ ) x = 4 cos 180◦ = 4(−1) = −4 y = 4 sin 180◦ = 4 · 0 = 0 (4, 180◦ ) = (−4, 0) 39.

3x + 4y = 5 3r cos θ + 4r sin θ = 5 (x = r cos θ, y = r sin θ) r(3 cos θ + 4 sin θ) = 5

40.

5x + 3y = 4 5r cos θ + 3r sin θ = 4 r(5 cos θ + 3 sin θ) = 4

41.

x=5 r cos θ = 5

42.

(x = r cos θ)

y=4 r sin θ = 4

43.

x2 + y 2 = 36 (r cos θ)2 +(r sin θ)2 = 36 (x = r cos θ, y = r sin θ) r2 cos2 θ + r2 sin2 θ = 36 r2 (cos2 θ + sin2 θ) = 36 r2 = 36 (cos2 θ + sin2 θ = 1)

34. (1.4, 225 ) ◦

√ " √ 2 = −0.7 2 x = 1.4 cos 225 = 1.4 − 2 ! √ " √ 2 = −0.7 2 y = 1.4 sin 225◦ = 1.4 − 2 √ √ (1.4, 225◦ ) = (−0.7 2, −0.7 2) ◦

!

r=6 44.

x − 4y 2 = 4 2

(r cos θ)2 − 4(r sin θ)2 = 4 r2 cos2 θ − 4r2 sin2 θ = 4 r2 (cos2 θ − 4 sin2 θ) = 4

444

Chapter 7: Applications of Trigonometry x2 = 25y

45.

56.

(r cos θ)2 = 25r sin θ

Substituting for x and y

r cos θ = 25r sin θ 2

2

2 1 − sin θ r − r sin θ = 2 r=

r−y = 2

2x − 9y + 3 = 0

46.

r = 2+y

2r cos θ − 9r sin θ + 3 = 0

47.

r2 = 4 + 4y + y 2

r(2 cos θ − 9 sin θ) + 3 = 0

x2 + y 2 = 4 + 4y + y 2

y − 5x − 25 = 0 2

(r sin θ)2 − 5r cos θ − 25 = 0 48.

Substituting for x and y

x2 = 4y + 4 57.

r − 9 cos θ = 7 sin θ

r2 − 9r cos θ = 7r sin θ

r2 sin2 θ − 5r cos θ − 25 = 0 x + y = 8y 2

2

(r cos θ) + (r sin θ)2 = 8r sin θ 2

r2 cos2 θ + r2 sin2 θ = 8r sin θ r2 (cos2 θ + sin2 θ) = 8r sin θ

x2 + y 2 − 9x = 7y

x − 9x + y 2 − 7y = 0 2

58.

r + 5 sin θ = 7 cos θ r2 + 5(r sin θ) = 7(r cos θ)

r2 = 8r sin θ

(x2 + y 2 ) + 5y = 7x

x2 − 2x + y 2 = 0

49.

(r cos θ)2 − 2r cos θ + (r sin θ)2 = 0 r2 cos2 θ − 2r cos θ + r2 sin2 θ = 0

59.

r2 − 2r cos θ = 0, or

r2 = 2r cos θ

3x2 y = 81

tan θ = 5π = 3 √ − 3= √ − 3x = y=

62.

y x y 5π Substituting for θ x 3 y x y, or √ − 3x

r = cos θ − sin θ

r2 = r cos θ − r sin θ

x2 + y 2 = x − y 63. r = sin θ

Make a table of values. Note that the points begin to repeat at θ = 360◦ . Plot the points and draw the graph.

(y = r sin θ)

r = −3 sin θ

r2 = −3(r sin θ)

120$

x + y = −3y 2

2

90$

60$ 30$

150$

r + r cos θ = 3

% ± x2 + y 2 + x = 3 % ± x2 + y 2 = 3 − x

180$

x + y = (3 − x) 2

2

"

r = 3 cos θ

tan

Substituting for r

r sin θ = 2

54.

55.

61.

x2 + y 2 = 25 Squaring y tan θ = x ! " 3π y 3π tan = θ= 4 x 4 y −1 = x y = −x y=2

1 cos θ

x2 + y 2 = 3x

r3 cos2 θ sin θ = 27

53.

sec θ =

r2 = 3r cos θ

3r3 cos2 θ sin θ = 81

52.

!

x=5

3(r2 cos2 θ)(r sin θ) = 81

r=5 % 2 ± x + y2 = 5

r = 5 sec θ 1 r = 5· cos θ r cos θ = 5

60.

3(r cos θ)2 (r sin θ) = 81

51.

x − 7x + y 2 + 5y = 0 2

r2 (cos2 θ + sin2 θ) − 2r cos θ = 0

50.

Multiplying both sides by r

2

Squaring both sides

x2 + y 2 = 9 − 6x + x2 y 2 = −6x + 9

1

0 2 360$ 330$

210$ 240$

270$

r # sin u

300$

Exercise Set 7.4 64.

120$

445 90$

69. r = 2 − cos 3θ

60$

Make a table of values. Note that the points begin to repeat at θ = 120◦ . Plot the points and draw the graph.

30$

150$ 180$

0 2 360$

1

90o 60o

120o

330$

210$ 240$

150o

300$

270$

r # 1 ! cos u

30o

0 360o

180o

65. r = 4 cos 2θ

210o

Make a table of values. Note that the points begin to repeat at θ = 180◦ . Plot the points and draw the graph.

330o

240o

300o 270o

r = 2 – cos 3θ

90$

120$

60$

70.

180$

0 360$

2

330$

210$ 240$

300$

270$

r # 4 cos 2u

66.

90o

30$

150$

0 4 360$

2

330$

210$ 300$

270$

r # 1 ! 2 sin u

Make a table of values. Note that the points begin to repeat at θ = 3600 . Plot the points and draw the graph. 90°

60°

150°

30°

180°

1

2

210°

0 360° 330°

240°

120°

90°

30° 2

4

210°

0 360° 330°

240°

270°

300o

300°

r # 2 sec θ

71. Use the ANGLE feature on a graphing calculator. (3, 7) = (7.616, 66.8◦ ), or (7.616, 1.166) √ 72. (−2, − 5) = (3, 228.2◦ ), or (3, 3.983)

(The calculator might return a negative angle. Convert to the smallest positive angle by adding 360◦ or 2π.)

74. (0.9, −6) = (6.067, 278.5◦ ), or (6.067, 4.861)

(The calculator might return a negative angle. Convert to the smallest positive angle by adding 360◦ or 2π.)

75. Use the ANGLE feature on a graphing calculator. (3, −43◦ ) = (2.19, −2.05) ! " π 76. − 5, = (−4.50, −2.17) 7

78. (2.8, 166◦ ) = (−2.72, 0.68)

60°

150° 180°

330o

77. Use the ANGLE feature on a graphing calculator. " ! 3π = (1.30, −3.99) − 4.2, 5

300°

270°

r # cos θ

68.

210o

73. Use the ANGLE feature on a graphing calculator. √ (− 10, 3.4) = (4.643, 132.9◦ ), or (4.643, 2.230)

67. r = cos θ

120°

0 360o

180o

270o

30$

240$

30o

240o

60$

150$ 180$

150o

1 r = –––––––– 1 + cos θ

90$

120$

60o

120o

79. Graph (d) is the graph of r = 3 sin 2θ. 80. Graph (i) is the graph of r = 4 cos θ. 81. Graph (g) is the graph of r = θ. 82. Graph (c) is the graph of r2 = sin 2θ. 83. Graph (j) is the graph of r =

5 . 1 + cos θ

446

Chapter 7: Applications of Trigonometry

84. Graph (f) is the graph of r = 1 + 2 sin θ.

101.

x = 12

85. Graph (b) is the graph of r = 3 cos 2θ. 86. Graph (a) is the graph of r = 3 sec θ. 87. Graph (e) is the graph of r = 3 sin θ.

102.

4 − 5y = 3

−5y = −1 1 y= 5

89. Graph (k) is the graph of r = 2 sin 3θ.

The solution is

90. Graph (l) is the graph of r sin θ = 6.

92.

Adding 4 and subtracting x

The solution is 12.

88. Graph (h) is the graph of r = 4 cos 5θ.

91. See the answer section in the text.

2x − 4 = x + 8

1 . 5

103. y = 2x − 5

Make a table of values by choosing values for x and finding the corresponding y-values. Plot points and draw the graph.

r # 3u 30

y 20

!10

4 2

!20

93. See the answer section in the text.

4

!4 !2

x

!2

94.

r # 10 2u

!4

8

104.

y 4

8

!1 !1

2

95. See the answer section in the text.

4

!4 !2 !2

96.

y # 2x ! 5

r # cos 2u ! 2

4x ! y # 6

!4

5

x

!6 5

!5

105. x = −3

Note that any point on the graph has −3 for its first coordinate. Thus, the graph is a vertical line 3 units left of the y-axis.

!5

97. See the answer section in the text.

y

98.

r # sin 2u " cos u 2

4 2 2

!1

2

!4 !2 !2

!2

99. Rectangular coordinates are unique because any point has a unique ordered pair (x, y) associated with it. Polar coordinates are not unique because there are infinitely many angles coterminal with a given angle and also because r can be positive or negative depending on the angle used. 100. One example is the equation of a circle not centered at the origin. Often, in rectangular coordinates, we must complete the square in order to graph the circle.

!4

106.

4

x # !3

y 4 2

y#0 4

!4 !2 !2 !4

x

x

Exercise Set 7.5 107.

447 θ 2

r = sec2

2.

y

1

r=

θ 2 1 r= 1 + cos θ 2 2 r= 1 + cos θ r + r cos θ = 2 cos2

x2 + y 2 = (2 − x)2

x2 + y 2 = 4 − 4x + x2 y 2 = −4x + 4

(4, 120°)

3.

(4, 60°)

x

DJ

(–1, –1)

% ± x2 + y 2 + x = 2

108.

(2, 1) (3, 1)

(0, 0) OF

% √ √ −→ |DJ| = [3 − (−1)]2 + [1 − (−1)]2 = 16 + 4 = 20 % √ √ −→ |OF | = (2 − 0)2 + (1 − 0)2 = 4 + 1 = 5 −→ −→ Since |DJ| = & |OF |, the vectors are not equivalent. y (3, 4) AB

(4, 180°)

(–2, 2)

(4, 0°)

(4, 240°)

(3, 1)

(–1, –1)

(4, 300°)

x

DJ

Exercise Set 7.5 1.

y (–4, 4) GE (–4, 1)

(3, 4) BJ (3, 1)

x

4.

First we find the length of each vector using the distance formula. % √ √ −→ |DJ| = [3 − (−1)]2 + [1 − (−1)]2 = 16 + 4 = 20 % √ √ −→ |AB| = [3 − (−2)]2 + (4 − 2)2 = 25 + 4 = 29 −→ −→ Since |DJ| = & |AB|, the vectors are not equivalent. (–2, 5) (–4, 4)

First we find the length of each vector using the distance formula. % √ −→ |GE| = [−4 − (−4)]2 + (4 − 1)2 = 9 = 3 % √ −→ |BJ| = (3 − 3)2 + (4 − 1)2 = 9 = 3 −→ −→ Thus, |GE| = |BJ|.

Both vectors point down. To verify that they have the same direction we calculate the slopes of the lines that they lie on. 3 4−1 −→ = (undefined) Slope of GE = −4 − (−4) 0 3 −→ 4 − 1 Slope of BJ = = (undefined) 3−3 0 Both vectors have undefined slope, so they lie on vertical lines and hence have the same direction. −→ −→ Since GE and BJ have the same magnitude and same direction, they are equivalent.

y

CG

(0, 0)

FO

(2, 1)

x

% √ √ −→ |CG| = [−2 − (−4)]2 + (5 − 4)2 = 4 + 1 = 5 % √ √ −→ |F O| = (2 − 0)2 + (1 − 0)2 = 4 + 1 = 5 −→ −→ Thus, |CG| = |F O|. 1 5−4 −→ = Slope of CG = −2 − (−4) 2 1 −→ 1 − 0 Slope of F O = = 2−0 2 The slopes are the same. −→ −→ Since CG and F O have the same magnitude and same direction, they are equivalent.

448

Chapter 7: Applications of Trigonometry

5.

7.

y

y (–4, 4)

(3, 4) BH

(3, 4) BJ

EG

(1, 2)

(–4, 1)

x

(3, 1)

x

(–1, –1) DK (–3, –3)

First we find the length of each vector using the distance formula. % √ √ −−→ |DK| = [−3 − (−1)]2 + [−3 − (−1)]2 = 4 + 4 = 8 % √ √ −−→ |BH| = (3 − 1)2 + (4 − 2)2 = 4 + 4 = 8 −−→ −−→ Thus, |DK| = |BH|. Both vectors point down and to the left. We calculate the slopes of the lines that they lie on. −2 −−→ −3 − (−1) = =1 Slope of DK = −3 − (−1) −2 2 −−→ 4 − 2 Slope of BH = = =1 3−1 2 The slopes are the same. −−→ −−→ Since DK and BH have the same magnitude and same direction, they are equivalent.

6.

The vectors clearly have different directions, so they are not equivalent. 8.

(–2, 5)

y

GC

(–4, 4)

FO

(0, 0)

(2, 1)

x

The vectors clearly have different directions, so they are not equivalent. 9.

y

y (–4, 4)

(3, 4)

(3, 4) GA (–2, 2)

BA

BH (–2, 2)

(1, 2)

x

x (–1, –1) DI (–6, –3)

% √ √ −→ |BA| = (−2 − 3)2 + (2 − 4)2 = 25 + 4 = 29 % √ √ −→ |DI| = [−6 − (−1)]2 + [−3 − (−1)]2 = 25 + 4 = 29 −→ −→ Thus, |BA| = |DI|. −2 2 2−4 −→ = = Slope of BA = −2 − 3 −5 5 −2 2 −→ −3 − (−1) = = Slope of DI = −6 − (−1) −5 5 The slopes are the same. −→ −→ Since BA and DI have the same magnitude and same direction, they are equivalent.

The vectors clearly have different directions, so they are not equivalent. 10.

(–2, 5) (–4, 4)

y

CG (3, 1)

(–1, –1)

x

JD

% √ √ −→ |JD| = (−1 − 3)2 + (−1 − 1)2 = 16 + 4 = 20 % √ √ −→ |CG| = [−4 − (−2)]2 + (4 − 5)2 = 4 + 1 = 5 −→ −→ Since |JD| = & |CG|, the vectors are not equivalent.

Exercise Set 7.5

449

11.

To find θ use the fact that triangle OAB is a right triangle. 45 tan θ = 32 θ ≈ 55◦

y (3, 4) AB

(–2, 2)

x

14.

C

B

(–1, –1) ID

Both vectors point up and to the right. We calculate the slopes of the lines that they lie on. −2 2 2−4 −→ = = Slope of AB = −2 − 3 −5 5 −2 2 −→ −3 − (−1) Slope of ID = = = −6 − (−1) −5 5 The slopes are the same. −→ −→ Since AB and ID have the same magnitude and same direction, they are equivalent.

12.

|v| = 60 + 50 √ |v| = 602 + 502 ≈ 78 N

C

A = 180 − 47 = 133

#

◦

◦

Use the law of cosines to find the magnitude of the resultant v. |v|2 = 6002 + 4102 − 2 · 600 · 410 cos 133◦ √ |v| ≈ 863, 643 ≈ 929 N

Use the law of sines to find θ.

929 410 = sin θ sin 133◦ 410 sin 133◦ sin θ = ≈ 0.3228 929 θ ≈ 19◦

x

OF

C

45 N

A

Use the Pythagorean theorem to find the magnitude of the resultant v. |v|2 = 322 + 452 √ |v| = 322 + 452 ≈ 55 N

B

v

325 N

325 N

o

64 O #

32 N

A

600 N ◦

B

θ

θ

O

% √ √ −→ |OF | = (2 − 0)2 + (1 − 0)2 = 4 + 1 = 5 % √ √ −−→ |HB| = (3 − 1)2 + (4 − 2)2 = 4 + 4 = 8 −→ −−→ Since |OF | = & |HB|, the vectors are not equivalent.

O

410 N

v

47o

16.

v

B

410 N

(1, 2)

45 N

2

50 60 θ ≈ 40◦

15.

(3, 4)

C

2

tan θ =

HB

(0, 0)

A

60 N

2

y

(2, 1)

50 N

θ

O

First we find the length of each vector using the distance formula. % √ √ −→ |AB| = [3 − (−2)]2 + (4 − 2)2 = 25 + 4 = 29 % √ √ −→ |ID| = [−1 − (−6)]2 + [−1 − (−3)]2 = 25 + 4 = 29 −→ −→ Thus, |AB| = |ID|.

13.

v

50 N

(–6, –3)

θ

255 N

A

A = 180◦ − 64◦ = 116◦ |v|2 = 2552 + 3252 − 2 · 255 · 325 cos 116◦ √ |v| ≈ 243, 310 ≈ 493 N 325 493 = sin θ sin 116◦ 325 sin 116◦ ≈ 0.5925 sin θ = 493 θ ≈ 36◦

450

Chapter 7: Applications of Trigonometry

17.

C

20.

B u+v

35

C

35

B

u+v

30

30

o

75

θ

O

A

45

O

We use the Pythagorean theorem to find the magnitude of u + v. |u + v|2 = 352 + 452 √ |u + v| = 352 + 452 To find the direction of u + v we note that since OAB is a right triangle 35 tan θ = 45 θ ≈ 38◦ . C

B 43

150o O

A = 180 − 75◦ = 105◦ |u + v|2 = 252 + 302 − 2 · 25 · 30 cos 105◦ √ |u + v| ≈ 1913.23 ≈ 43.7 43.7 30 = sin θ sin 105◦ 30 sin 105◦ sin θ = ≈ 0.6631 43.7 θ ≈ 42◦

19.

(Answers may vary slightly due to rounding differences.) 21.

u+v

o

θ

30 54

C

O

θ

o

63

θ

A

20

A = 180 − 117 = 63 ◦

◦

Use the law of cosines to find |u + v|.

|u + v|2 = 202 + 202 − 2 · 20 · 20 cos 63◦ √ |u + v| ≈ 436.81 ≈ 20.9

Triangle OAB is isosceles so θ and angle OBA have the same measure. 1 1 Thus, θ = (180◦ − 63◦ ) = (117◦ ) = 58.5◦ ≈ 59◦ . 2 2 (If we use the law of sines and |u + v| ≈ 20.9 to find θ, we get θ ≈ 58◦ .) 22.

67

20

117o O

12

o

u+v

◦

B

u+v

B 20

A

43 27.3 = sin θ sin 30◦ 43 sin 30◦ sin θ = ≈ 0.7875 27.3 ◦ θ ≈ 52

12

C

43

A = 180◦ − 150◦ = 30◦ |u + v|2 = 542 + 432 − 2 · 54 · 43 cos 30◦ √ |u + v| ≈ 743.18 ≈ 27.3 #

25

105o A

◦

|u + v| ≈ 57.0

18.

θ

C

B

o

113 10

A

A = 180◦ − 67◦ = 113◦

Use the law of cosines to find |u + v|.

|u + v|2 = 102 + 122 − 2 · 10 · 12 cos 113◦ √ |u + v| ≈ 337.78 ≈ 18.4

Use the law of sines to find θ. 18.4 12 = sin θ sin 113◦ 12 sin 113◦ sin θ = ≈ 0.6003 18.4 θ ≈ 37◦

30

u+v

30

123o O

o

57

θ

30

A

A = 180◦ − 123◦ = 57◦ |u + v|2 = 302 + 302 − 2 · 30 · 30 cos 57◦ √ |u + v| ≈ 819.65 ≈ 28.6

Triangle OAB is isosceles so θ and angle OBA have the same measure. 1 1 Thus, θ = (180◦ − 57◦ ) = (123◦ ) = 61.5◦ ≈ 62◦ . 2 2

Exercise Set 7.5

451

23. 47 θ

O

23

C u+v

B

26.

N 120

o

27

153

80

W

47

o

o

10

60

o

110

o

o

20 70

o

o

E

α

A

200

A = 180◦ − 27◦ = 153◦

d

Use the law of cosines to find |u + v|.

|u + v|2 = 232 + 472 − 2 · 23 · 47 cos 153◦ √ |u + v| ≈ 4664.36 ≈ 68.3

24.

Use the law of sines to find θ. 47 68.3 = sin θ sin 153◦ 47 sin 153◦ sin θ = ≈ 0.3124 68.3 θ ≈ 18◦ C

74

u+v

S

Note:

20°

B

180◦ − (20◦ + 90◦ ) = 70◦

74 70°

180◦ − 70◦ = 110◦

o

72

108o A 32

θ

O

A = 180◦ − 72◦ = 108◦ |u + v|2 = 322 + 742 − 2 · 32 · 74 cos 108◦ √ |u + v| ≈ 7963.50 ≈ 89.2

25.

89.2 74 = sin θ sin 108◦ 74 sin 108◦ sin θ = ≈ 0.7890 89.2 ◦ θ ≈ 52 C

B

v

80°

90◦ − 80◦ = 10◦ 10°

180◦ − (10◦ + 110◦ ) = 60◦

Use the law of cosines to find |d|.

|d|2 = 1202 + 2002 − 2 · 120 · 200 cos 60◦ √ |d| = 30, 400 ≈ 174

Use the law of sines to find θ in this triangle. 120

10 ft/sec

θ

O

θ

5 ft/sec

A

110°

174

60o

200

Use the Pythagorean theorem to find the speed of the balloon, |v|. |v|2 = 52 + 102 √ |v| = 52 + 102 ≈ 11 ft/sec

Use the fact that triangle OAB is a right triangle to find θ. 10 =2 tan θ = 5 θ ≈ 63◦

174 200 = sin θ sin 60◦ 200 sin 60◦ ≈ 0.9954 sin θ = 174 θ ≈ 85◦ .

Now we can find α.

α = 180◦ − (80◦ + 85◦ ) = 15◦ .

452

Chapter 7: Applications of Trigonometry The ship is 174 nautical miles from the starting point in the direction S 15◦ E. (Answers may vary due to rounding differences.)

27.

30.

10 km/h

y

75

o

16 km/h

w

150

θ

750 F

35o θ

320o

55o o 50 150

|w| =

x

|F|2 = 7502 + 1502 − 2 · 750 · 150 cos 75◦ √ |F| = 526, 766 ≈ 726 lb

10 16 θ ≈ 32◦

180◦ − θ = 180◦ − 32◦ = 148◦ . 31.

β α

θ

32o

|e| = cos 45◦ 100 |e| = 100 cos 45◦ ≈ 70.7 |s| = sin 45◦ 100 |s| = 100 sin 45◦ ≈ 70.7

x

58o

The easterly and southerly components are both 70.7.

210

32. 150 lb

|d|2 = 1702 + 2102 − 2 · 170 · 210 cos 68◦ √ |d| = 46, 253 ≈ 215 km 170 215 = sin β sin 68◦ 170 sin 68◦ sin β = ≈ 0.7331 215 β ≈ 47◦

Then α = 47 − 32 = 15 , so the direction of the plane is 360◦ − 15◦ , or 345◦ . ◦

29.

s

100

y

d

e 45°

The boat is moving in the direction 35◦ + 12◦ , or 47◦ .

170

+ 162 ≈ 18.9 km/h

The wind is blowing from the direction

Use the law of sines to find θ. 726 150 = sin θ sin 75◦ 150 sin 75◦ sin θ = ≈ 0.1996 726 ◦ θ ≈ 12

10o

θ 102

tan θ =

Use the law of cosines to find |F|.

28.

√

52° a

|a| = 150 cos 52◦ ≈ 92

|b| = 150 sin 52◦ ≈ 118

The horizontal component is about 92 lb, and the vertical component is about 118 lb.

◦

y

b

33.

225 mph 17°

20o 150 θ

70o

20o

25 150 θ ≈ 10◦

25

a |a| = 225 cos 17◦ ≈ 215.17

70o 70o x

sin θ =

Then the airplane’s actual heading will be about 90◦ − (10◦ + 20◦ ), or 60◦ .

|b| = 225 sin 17◦ ≈ 65.78

The horizontal component is about 215.17 mph forward, and the vertical component is about 65.78 mph up.

Exercise Set 7.5

453 38.

34. 97 lb

72 mph

b

b 45°

38° a

a

|a| = 72 cos 45◦ ≈ 50.91

|a| = 97 cos 38◦ ≈ 76.44

|b| = 97 sin 38 ≈ 59.72

|b| = 72 sin 45◦ ≈ 50.91

◦

The horizontal component is about 76.44 lb forward, and the vertical component is about 59.72 lb down. 35.

The horizontal component is about 50.91 mph forward, and the vertical component is about 50.91 mph up. 39. b 25°

780 lb

a 100 25°

|a| = 100 cos 25◦ ≈ 90.6

60° a

|b| = 100 sin 25◦ ≈ 42.3

|a| = 780 cos 60 = 390 ◦

The magnitude of the component perpendicular to the incline is about 90.6 lb; the magnitude of the component parallel to the incline is about 42.3 lb.

|b| = 780 sin 60◦ ≈ 675.5

The horizontal component is about 390 lb forward, and the vertical component is about 675.5 lb up.

40.

36.

b 30°

1200 lb

b

a 450 30°

45° a

|a| = 450 cos 30◦ ≈ 389.7 kg

|a| = 1200 cos 45◦ ≈ 848.5

|b| = 1200 sin 45 ≈ 848.5 ◦

The horizontal component is about 848.5 lb toward the balloon, and the vertical component is about 848.5 lb up.

41.

|b| = 450 sin 30◦ = 225 kg f 37°

37. 80 37°

n

200 35° w

305°

|f | = 80 sin 37◦ ≈ 48.1 lb 42.

f 9°

|w| = 200 cos 35 ≈ 164 ◦

|n| = 200 sin 35◦ ≈ 115

The westerly component is about 164 km/h, and the northerly component is about 115 km/h.

3500

9°

|f | = 3500 sin 9◦ ≈ 547.5 lb 43. A vector is a quantity that has both magnitude and direction. −→ −→ 44. Vectors QR and RQ have opposite directions, so they are not equivalent.

454

Chapter 7: Applications of Trigonometry −→ 2. CD = '5 − 1, 7 − 5( = '4, 2( √ √ √ √ −→ |CD| = 42 + 22 = 16 + 4 = 20, or 2 5

45. natural 46. half-angle 47. linear speed

−→ 3. F E = '8 − 11, 4 − (−2)( = '−3, 6( % √ √ √ −→ |F E| = (−3)2 + 62 = 9 + 36 = 45, or 3 5

48. cosine 49. identity

−→ 4. BA = '9 − 9, 0 − 7)( = '0, −7( % √ −→ |BA| = 02 + (−7)2 = 49 = 7

50. cotangent of θ 51. coterminal 52. sines 53. horizontal line; inverse 54. reference angle; acute 55.

y

7

30° x

45° 8

−→ 5. KL = '8 − 4, −3 − (−3)( = '4, 0( √ √ −→ |KL| = 42 + 02 = 16 = 4

−−→ 6. GH = '−3 − (−6), 2 − 10( = '3, −8( % √ √ −−→ |GH| = 32 + (−8)2 = 9 + 64 = 73 % √ √ 7. |u| = (−1)2 + 62 = 1 + 36 = 37 % √ −→ 8. |ST | = (−12)2 + 52 = 169 = 13

9. u + w = '5 + (−1), −2 + (−3)( = '4, −5(

10. w + u = '−1 + 5, −3 + (−2)( = '4, −5( 11.

a)

|3w − v| = |3'−1, −3( − '−4, 7(| = |'−3, −9( − '−4, 7(|

(x, y)

= |'−3, −(−4), −9 − 7(| 7

= |'1, −16(| % = 12 + (−16)2 √ = 257

y

45° x

12.

x = 7 cos 45◦ ≈ 4.950

= '−24, 42( + '25, −10(

y = 7 sin 45◦ ≈ 4.950

= '−24 + 25, 42 − 10(

The cliff is located at (4.950, 4.950). b)

(4.950, 4.950)

= '1, 32(

13. v − u = '−4, 7( − '5, −2( = '−4 − 5, 7 − (−2)( = '−9, 9( 14.

30° 8

T

15.

a = 8 sin 30◦ = 4

Exercise Set 7.6 −−→ 1. M N = '−3 − 6, −2 − (−7)( = '−9, 5( % √ √ −−→ |M N | = (−9)2 + 52 = 81 + 25 = 106

5u − 4v = 5'5, −2( − 4'−4, 7(

= '25, −10( − '−16, 28(

b = 8 cos 30◦ ≈ 6.928

Then the coordinates of T are (4.950 − 4, 4.950 − 6.928), or (0.950, −1.978).

|2w| = |2'−1, −3(|

= |'−2, −6(| % = (−2)2 + (−6)2 √ √ = 40, or 2 10

b

a

|6v + 5u| = 6'−4, 7( + 5'5, −2(

= '25 − (−16), −10 − 28( = '41, −38(

16. −5v = −5'−4, 7( = '20, −35( 17.

|3u| −| v| = |3'5, −2(| − |'−4, 7(|

= |'15, −6(| − |'−4, 7(| % % = 152 + (−6)2 − (−4)2 + 72 √ √ = 261 − 65

Exercise Set 7.6

455

18.

|v| + |u| = |'−4, 7(| + |'5, −2(| % % = (−4)2 + 72 + 52 + (−2)2 √ √ = 65 + 29

19.

v + u + 2w = '−4, 7( + '5, −2( + 2'−1, −3( = '−4, 7( + '5, −2( + '−2, −6(

= '−4 + 5 + (−2), 7 + (−2) + (−6)( 20.

= '−1, −1(

w − (u + 4v) = '−1, −3( − ('5, −2( + 4'−4, 7()

= '−1, −3( − ('5, −2( + '−16, 28() = '−1, −3( − '−11, 26(

= '−1 − (−11), −3 − 26( = '10, −29(

21. 2v + O = 2v = 2'−4, 7( = '−8, 14( 22.

10|7w − 3u| = 10|7'−1, −3( − 3'5, −2(| = 10|'−7, −21( − '15, −6(|

= 10|'−7 − 15, −21 − (−6)(| = 10|'−22, −15(| % = 10 (−22)2 + (−15)2 √ = 10 709

23. u · w = 5(−1) + (−2)(−3) = −5 + 6 = 1 24. w · u = −1 · 5 + (−3)(−2) = −5 + 6 = 1 25. u · v = 5(−4) + (−2)(7) = −20 − 14 = −34 26. v · w = −4(−1) + 7(−3) = 4 − 21 = −17 27. See the answer section in the text.

√ √ 34. |'3, 4(| = 32 + 42 = 25 = 5 +3 4, 1 '3, 4( = , 5 5 5 % √ 35. |'1, −10(| = 12 + (−10)2 = 101 + 1 10 , 1 √ '1, −10( = √ , −√ 101 101 101 % √ 2 2 36. |'6, −7(| = 6 + (−7) = 85 + 6 7 , 1 √ '6, −7( = √ , − √ 85 85 85 % √ √ 37. |'−2, −8(| = (−2)2 + (−8)2 = 68 = 2 17 + 2 8 , 1 √ '−2, −8( = − √ , − √ = 2 17 2 17 +2 17 1 , 4 − √ , −√ 17 17 % √ √ 38. |'−3, −3(| = (−3)2 + (−3)2 = 18 = 3 2 + 1 1 , 1 √ '−3, −3( = − √ , − √ 3 2 2 2 39. '−4, 6( = −4i + 6j 40. −15i + 9j 41. '2, 5( = 2i + 5j 42. 2i − j 43. Horizontal component: −3 − 4 = −7 Vertical component: 3 − (−2) = 5 We write the vector as −7i + 5j.

44. Horizontal component: 1 − (−3) = 4 Vertical component: 4 − (−4) = 8 We write the vector as 4i + 8j.

28. u

u ! 2v

45. (a) 4u − 5w = 4(2i + j) − 5(i − 5j) = 8i + 4j − 5i + 25j

!2v

29. See the answer section in the text.

(b) 3i + 29j = '3, 29( 46. (a) v + 3w = −3i − 10j + 3(i − 5j)

30. qu

= 3i + 29j

= −3i − 10j + 3i − 15j

qu ! w !w

31. (a) w = u + v

= −25j

(b) −25j = 0i − 25j = '0, −25( 47. (a) u − (v + w) = 2i + j − (−3i − 10j + i − 5j) = 2i + j − (−2i − 15j)

(b) v = w − u 32. Since the diagonals of a parallelogram intersect at their midpoints, P is the midpoint of u + w. Thus 1 v = (u + w). 2 % √ 33. |'−5, 12(| = (−5)2 + 122 = 169 = 13 + 5 12 , 1 '−5, 12( = − , 13 13 13

= 2i + j + 2i + 15j = 4i + 16j (b) 4i + 16j = '4, 16(

48. (a) (u − v) + w = (2i + j − (−3i − 10j)) + i − 5j = (2i + j + 3i + 10j) + i − 5j = 5i + 11j + i − 5j = 6i + 6j (b) '6, 6(

456

Chapter 7: Applications of Trigonometry

49.

54.

y

u

π/2

x (1, 0)

u= 50.

!

" ! " π π i + sin j = 0i + 1j = j, or '0, 1( cos 2 2

55.

θ = tan−1 2 ≈ 63◦ .

u π/3

56. (1, 0)

51.

!

cos

√ " ! " + 1 √3 , π 1 3 π i + sin j= i+ j, or , 3 3 2 2 2 2 y

57.

4π/3

52.

+

!

√ " ! " 4π 1 3 4π i + sin j=− i− j, or 3 3 2 2 √ , 3

cos

1 − ,− 2 2

y

3π/2

x (1, 0)

u

u= 53.

!

cos

" ! " 3π 3π i + sin j = 0i − 1 · j = −j, or '0, −1( 2 2

5 −5 = −2 2 5 θ = tan−1 2 The vector is in the third quadrant, so θ is a third-quadrant angle. The reference angle is 5 tan−1 ≈ 68◦ 2 and θ ≈ 180◦ + 68◦ , or 248◦ . tan θ =

1 −1 =− 5 5 ! " 1 θ = tan−1 − 5 θ is a fourth-quadrant angle. The reference angle is 1 tan−1 ≈ 11◦ 5 and θ ≈ 360◦ − 11◦ , or 349◦ . tan θ =

tan θ =

6 5

6 5 The vector is in the first quadrant, so θ is a first-quadrant angle. Then 6 θ = tan−1 ≈ 50◦ . 5 1 −4 58. = tan θ = −8 2 −1 1 θ = tan 2 θ is a third-quadrant angle. The reference angle is 1 tan−1 ≈ 27◦ 2 and θ ≈ 180◦ + 27◦ , or 207◦ . % 59. |u| = (3 cos 45◦ )2 + (3 sin 45◦ )2 % = 9 cos2 45◦ + 9 sin2 45◦ ) = 9(cos2 45◦ + sin2 45◦ ) √ √ = 9·1= 9=3 θ = tan−1

(1, 0)

u

u=

2 =2 1 θ = tan−1 2

tan θ =

The vector is in the first quadrant, so θ is a first-quadrant angle. Then

y

u=

−3 3 =− 4 4 ! " 3 θ = tan−1 − 4 θ is a fourth-quadrant angle. The reference angle is 3 tan−1 ≈ 37◦ 4 and θ ≈ 360◦ − 37◦ , or 323◦ . tan θ =

The vector is given in terms of the direction angle, 45◦ . % 60. |w| = (6 cos 150◦ )2 + (6 sin 150◦ )2 % = 36 cos2 150◦ + 36 sin2 150◦ ) = 36(cos2 150◦ + sin2 150◦ ) √ = 36 · 1 = 6

The vector is given in terms of the direction angle, 150◦ .

Exercise Set 7.6 -0 . "2 ! √ "2 ( . 3 1 3 √ 1 / + = 1=1 61. |v| = + = − 2 2 4 4 √

3 √ ! " √ 3 2 − =− 3 tan θ = 2 = 1 2 1 − 2 √ θ = tan−1 (− 3)

The vector is in the second quadrant, so θ is a secondquadrant angle. The reference angle is √ tan−1 3 = 60◦ and θ = 180◦ − 60◦ , or 120◦ . % √ 62. |u| = (−1)2 + (−1)2 = 2 −1 =1 −1 θ = tan−1 1

tan θ =

θ is a third-quadrant angle. The reference angle is tan−1 1 = 45◦ and θ = 180◦ + 45◦ , or 225◦ . 63. u = '2, −5(, v = '1, 4(

u · v = 2 · 1 + (−5)(4) = −18 % √ |u| = 22 + (−5)2 = 29 √ √ |v| = 12 + 42 = 17 u·v −18 cos α = =√ √ |u| |v| 29 17 −1 √ −18 √ α = cos 29 17 α ≈ 144.2◦

64. a = '−3, −3(, b = '−5, 2(

a · b = −3(−5) + (−3)(2) = 9 % √ |a| = (−3)2 + (−3)2 = 18 % √ |b| = (−5)2 + 22 = 29 9 a·b =√ √ |a| |b| 18 29 −1 √ 9√ α = cos 18 29 α ≈ 66.8◦

cos α =

457 66. v = '−4, 2(, t = '1, −4(

v · t = −4 · 1 + 2(−4) = −12 % √ |v| = (−4)2 + 22 = 20 % √ |t| = 12 + (−4)2 = 17 −12 v·t =√ √ |v| |t| 20 17 −12 α = cos−1 √ √ 20 17 α ≈ 130.6◦

cos α =

67. a = i + j, t = 2i − 3j

a · b = 1 · 2 + 1(−3) = −1 √ √ |a| = 12 + 12 = 2 % √ |b| = 22 + (−3)2 = 13

a·b −1 =√ √ |a| |b| 2 13 −1 √ −1 √ α = cos 2 13 α ≈ 101.3◦

cos α =

68. u = 3i + 2j, t = −i + 4j

u · v = 3(−1) + 2 · 4 = 5 √ √ |u| = 32 + 22 = 13 % √ |v| = (−1)2 + 42 = 17

u·v 5 =√ √ |u| |v| 13 17 5 α = cos−1 √ √ 13 17 α ≈ 70.3◦ √ ! " ! " π π π 3 1 69. For θ = , u = cos i + sin j= i + j. 6 6 6 2 2 3π For θ = , 4 √ √ " ! " ! 3π 3π 2 2 u = cos i + sin j=− i+ j. 4 4 2 2 cos α =

y u#f

u#A x

65. w = '3, 5(, r = '5, 5(

w · r = 3 · 5 + 5 · 5 = 40 √ √ |w| = 32 + 52 = 34 √ √ |r| = 52 + 52 = 50 40 w·r =√ √ cos α = |w| |r| 34 50 40 α = cos−1 √ √ 34 50 α ≈ 14.0◦

u # "3 i " qj 2

u # ! "2 i" 2

"2 j 2

π 70. For θ = − , # !4 "$ # ! "$ π π u = cos − i + sin − j= 4 4 √ √ 2 2 i− j. 2 2

458

Chapter 7: Applications of Trigonometry 3π For θ = − , "$ # ! "$ # !4 3π 3π i + sin − j= u = cos − 4 4 √ √ 2 2 − i− j. 2 2 y

√

√ " ! " ! √ √ 2 3 2 13 3 13 13 √ i − √ j , or 13 i− j . 13 13 13 13

76. Find the magnitude of 5i + 12j. √ √ 52 + 132 = 169 = 13 ! " 5i + 12j Then the desired vector is 13 = 13 ! " 12 5 i+ j . 13 13 13 77. See the answer section in the text.

x u # ! "2 i! 2

"2 j 2

78.

y

j u # "2 i ! "2 2 2

4 2

5π π 3π , or . Then 71. u = (cos θ)i + (sin θ)j where θ = + 2 4√ 4 √ " ! " ! 5π 5π 2 2 u = cos i + sin j=− i− j. 4 4 2 2 y

!2

4

x

v

!4 !6

79. Consider the drawing representing the situation in Exercise 26, Exercise Set 7.5.

j

3π/4

u 2

!4 !2

N

θ

120

x

u

W

80

o

10

60

o

110

20

o

70

o

o

o

E

α

200

11π π . Then 72. u = (cos θ)i + (sin θ)j where θ = 2π − , or 6 √ 6 " ! " ! 11π 11π 1 3 u = cos i + sin j= i − j. 6 6 2 2 y

120(cos 10◦ )i + 120(sin 10◦ )j.

u u

S

The vector representing the first part of the ship’s trip can be given by

i

j

d

The vector representing the second part of the trip can be given by

x

200(cos 250◦ )i + 200(sin 250◦ )j. Then the resultant is 120(cos 10◦ )i + 120(sin 10◦ )j + 200(cos 250◦ )i+

73. Find the magnitude of −i + 3j. % √ (−1)2 + 32 = 10

200(sin 250◦ )j

Then the desired unit vector is √ √ 1 3 10 3 10 −i + 3j √ = − √ i + √ j, or − i+ j. 10 10 10 10 10

74. Find the magnitude of 6i − 8j. % √ 62 + (−8)2 = 100 = 10

Then the desired unit vector is

6i − 8j 3 4 = i − j. 10 5 5

75. Find the magnitude of 2i − 3j. % √ 22 + (−3)2 = 13 " ! √ 2i − 3j = Then the desired vector is 13 √ 13

= [120(cos 10 ) + 200(cos 250 )]i+ ◦

◦

[120(sin 10◦ ) + 200(sin 250◦ )]j ≈ 49.77i − 167.10j

Then the distance from the starting point is % (49.77)2 + (−167.10)2 ≈ 174 nautical miles.

Now we find the direction angle of the resultant. −167.10 tan θ = 49.77 −167.10 θ = tan−1 49.77 θ is a fourth-quadrant angle. The reference angle is

Exercise Set 7.6

459

167.10 ≈ 73◦ 49.77 and θ ≈ 360◦ − 73◦ , or 287◦ .

82. Refer to the drawing in this manual accompanying the solution for Exercise 28, Exercise Set 7.5. The vector representing the first part of the flight can be given by

tan−1

Thus, the ship’s bearing is about S17◦ E. (This answer differs from the answer found in Section 7.5 due to rounding differences.) 80. Refer to the drawing in this manual accompanying the solution for Exercise 27, Exercise Set 7.5. The vector representing the boat’s velocity can be given by 750(cos 55◦ )i + 750(sin 55◦ )j. The vector representing the wind can be given by 150(cos 310◦ )i + 150(sin 310◦ )j. Then the resultant is the sum of these vectors: 526.60i + 499.46j. Then magnitude of the resultant force is % (526.60)2 + (499.46)2 ≈ 726 lb.

Now we find the direction angle of the resultant. 499.46 tan θ = 526.60 499.46 ≈ 43◦ θ = tan−1 526.60 Then the direction from north is 90◦ − 43◦ , or 47◦ .

81. Refer to the drawing in this manual accompanying the solution for Exercise 29, Exercise Set 7.5. We can use the Pythagorean theorem to find the magnitude of the vector representing the desired velocity of the airplane: √ 1502 − 252 ≈ 148 Then the vector representing the desired velocity can be given by 148(cos 20◦ )i + 148(sin 20◦ )j. The vector representing the wind can be given by

210(cos 58◦ )i + 210(sin 58◦ )j. The vector representing the second part of the flight can be given by 170(cos 170◦ )i + 170(sin 170◦ )j. The resultant is the sum of these vectors: −56.13i + 207.61j

The magnitude of the resultant is % (−56.13)2 + (207.61)2 ≈ 215 mi.

Now we find the direction angle of the resultant. 207.61 tan φ = −56.13 207.61 φ = tan−1 −56.13 φ is a second-quadrant angle. The reference angle is 207.61 ≈ 75◦ tan−1 56.13 and φ ≈ 180◦ − 75◦ = 105◦ . Then the direction of the airplane is 345◦ .

83. We draw a force diagram with the initial point of each vector at the origin. y

S

R 150°

60°

30°

x

W

25(cos 290◦ )i + 25(sin 290◦ )j. The vector representing the desired velocity is the resultant of the vectors representing the airplane’s actual velocity and the wind, so we subtract to find the vector representing the airplane’s actual velocity. [148(cos 20◦ )i + 148(sin 20◦ )j]− [25(cos 290◦ )i + 25(sin 290◦ )j] = [148(cos 20 ) − 25(cos 290 )]i+ ◦

≈ 130.52i + 74.11j

R = |R|[(cos 150◦ )i + (sin 150◦ )j] S = |S|[(cos 60◦ )i + (sin 60◦ )j]

W = 1000(cos 270◦ )i + 1000(sin 270◦ )j = −1000j

Substituting for R, S, and W in R + S + W = O, we have

◦

[148(sin 20 ) − 25(sin 290 )]j ◦

We express each vector in terms of its magnitude and direction angle.

◦

Now we find the direction angle of this vector. Note that α is a first-quadrant angle. 74.11 tan α = 130.52 74.11 ≈ 30◦ α = tan−1 130.52 Thus, the airplane’s actual bearing is 60◦ .

|R|[(cos 150◦ )i + (sin 150◦ )j]+ |S|[(cos 60◦ )i + (sin 60◦ )j] − 1000j = 0i + 0j. This gives us a system of equations.

|R|(cos 150◦ ) + |S|(cos 60◦ ) = 0, |R|(sin 150◦ ) + |S|(sin 60◦ ) = 1000

Solving this system, we get |R| = 500, |S| ≈ 866.

The tension in the cable on the left is 500 lb and in the cable on the right is about 866 lb.

460

Chapter 7: Applications of Trigonometry

84.

86.

y

R

y R

S 135° 30°

45°

S

x

50°

x

50° 40°

W

R = |R|[(cos 50◦ )i + (sin 50◦ )j]

R = |R|[(cos 135◦ )i + (sin 135◦ )j]

S = |S|[(cos 30 )i + (sin 30 )j] ◦

S = |S|[(cos 180◦ )i + (sin 180◦ )j] = −|S|i

◦

W = 2500(cos 270◦ )i + 2500(sin 270◦ )j = −2500j

W = 200(cos 270◦ )i + 200(sin 270◦ )j = −200j

Then we have

Then we have

|R|[(cos 135 )i + (sin 135 )j]+ |S|[(cos 30◦ )i + (sin 30◦ )j] − 2500j = 0i + 0j. ◦

|R|[(cos 50◦ )i + (sin 50◦ )j] − |S|i − 200j = 0i + 0j.

◦

This gives us a system of equations.

This gives us a system of equations.

|R|(cos 50◦ ) − |S| = 0,

|R|(cos 135◦ ) + |S|(cos 30◦ ) = 0,

|R|(sin 50◦ )

|R|(sin 135◦ ) + |S|(sin 30◦ ) = 2500

85. We draw a force diagram with the initial point of each vector at the origin.

= 200

Solving this system, we get

Solving this system, we get

|R| ≈ 2241 kg, |S| ≈ 1830 kg.

W

|R| ≈ 261 lb, |S| ≈ 168 lb. 87.

u + v = 'u1 , u2 ( + 'v1 , v2 ( = 'u1 + v1 , u2 + v2 ( = 'v1 + u1 , v2 + u2 ( = 'v1 , v2 ( + 'u1 , u2 ( = v+u

S

88. 42°

138°

We express each vector in terms of its magnitude and direction angle. R = |R|[(cos 0◦ )i + (sin 0◦ )j] = |R|i S = |S|[(cos 138◦ )i + (sin 138◦ )j]

W = 150(cos 270◦ )i + 150(sin 270◦ )j = −150j

Substituting for R, S, and W in R + S + W = O, we have |R|i + |S|[(cos 138◦ )i + (sin 138◦ )j] − 150j = 0i + 0j. This gives us a system of equations. |R| + |S|(cos 138◦ ) = 0, |S|(sin 138◦ ) = 150

Solving this system, we get |R| ≈ 167, |S| ≈ 224.

= v1 u1 + v2 u2

R

W

The tension in the cable is about 224 lb, and the compression in the boom is about 167 lb. (Answers may vary slightly due to rounding differences.)

u · v = u1 v1 + u2 v2 = v·u

89. The terminal point of a unit vector in standard position is a point on the unit circle. 90. Answers may vary. For u = 3i − 4j and w = 2i − 4j find v where v = u + w 1 91. − x − y = 15 5 1 −y = x + 15 5 1 Multiplying by −1 y = − x − 15 5 With the equation written in the form y = mx + b we see 1 that the slope is − and the y-intercept is (0, −15). 5 92. y = 7 = 0x + 7 Slope: 0; y-intercept: (0, 7) 93.

x3 − 4x2 = 0

x2 (x − 4) = 0

x2 = 0 or x − 4 = 0 x = 0 or

x=4

The zeros are 0 and 4.

Chapter 7 Review Exercises 94.

461

6x2 + 7x = 55 6x2 + 7x − 55 = 0

(2x − 5)(3x + 11) = 0

2x − 5 = 0 or 3x + 11 = 0

2x = 5 or 3x = −11 11 5 or x=− x= 2 3 5 11 The zeros are and − . 2 3 95. (a) Assume neither |u| nor |v| is zero.

If u · v = |u||v| cos θ = 0, then cos θ = 0, or θ = 90◦ and the vectors are perpendicular.

(b) Answers may vary. Let u = i and v = j. Then u · v = 1 · 0 + 0 · 1 = 0.

−→ −→ 96. Let P Q = 'x, y(. Then QP = '−x, −y( and P Q + QP = 'x, y( + '−x, −y( = '0, 0( = 0 97. Find the magnitude of u = '3, −4(: % √ |u| = 32 + (−4)2 = 25 = 5

Chapter 7 Review Exercises 1. For any point (x, y) on the unit circle, so the statement is true.

2. The statement is false. We must know the measure of an angle in order to use the law of sines to solve a triangle. 3. Two vectors may lie on lines with the same slope, but they can have opposite directions. Thus, the statement is false. 4. The vectors '8, −2( and '−8, 2( have opposite directions, so the statement is false. 5. The statement is false. If we know only the three angle measures, we cannot solve a triangle. 6. We must know two sides of a triangle and the included angle in order to use the law of cosines to solve a triangle. Thus, the statement is true. 7. a2 = b2 + c2 − 2bc cos A, or

(23.4)2 = (15.7)2 + (8.3)2 − 2(15.7)(8.3) cos A cos A ≈ −0.8909

The unit vector in the same direction as u is 3 4 3i − 4j = i − j. 5 5 5 The unit vector in the opposite direction to u is " ! 3i − 4j 3 4 − , or − i + j. 5 5 5

These are the only unit vectors parallel to '3, −4(. % √ 98. |v| = (−1)2 + 22 = 5

Then to find a vector of length 2 whose direction is the 2 opposite of the direction of v, we multiply v by − √ , or 5 √ 2 5 − . We have 5 ! √ √ √ " 2 5 4 5 2 5 i− j. − − i + 2j = 5 5 5 This is the only vector that satisfies the given criteria.

99. B has coordinates (x, y) such that and

x − 2 = 3, or x = 5 y − 9 = −1, or y = 8.

A ≈ 153◦

b2 = a2 + c2 − 2ac cos B, or

(15.7)2 = (23.4)2 + (8.3)2 − 2(23.4)(8.3) cos B cos B ≈ 0.9524 B ≈ 18◦

C ≈ 180◦ − (153◦ + 18◦ ) ≈ 9◦ 8. A = 180◦ − (27◦ + 35◦ ) = 118◦ 19 a b a = , or ◦ = sin A sin B sin 118 sin 27◦ a ≈ 37 in. b c 19 c = , or = sin C sin B sin 35◦ sin 27◦ c ≈ 24 in.

9. B = 180◦ − (133◦ 28" + 31◦ 42" ) = 14◦ 50" b a 890 a = , or = sin A sin B sin 133◦ 28" sin 14◦ 50" a ≈ 2523 m 890 c b c = , or = sin C sin B sin 31◦ 42" sin 14◦ 50" c ≈ 1827 m

Thus point B is (5,8).

100. A has coordinates (x, y) such that and

−2 − x = 4, or x = −6

% √ x2 + y 2 = 1 = 1,

10.

5 − y = −2, or y = 7.

Thus point A is (−6, 7). Then the vector from A to the origin is '0 − (−6), 0 − 7(, or '6, −7(, or 6i − 7j. 11.

b 8 4 c = , or = sin C sin B sin C sin 37◦ 8 sin 37◦ ≈ 1.2036 sin C = 4 Since there is no angle with a sine greater than 1, there is no solution. 1 bc sin A 2 1 K = (9.8)(7.3) sin 67.3◦ ≈ 33 m2 2 K =

462

Chapter 7: Applications of Trigonometry

12. From Section 7.1, Exercise 46, we know that the area of a parallelogram is the product of two adjacent sides and the sine of the included angle.

16. 175 km/h · 2 hr = 350 km 220 km/h · 2 hr = 440 km

Area = (3.21)(7.85) sin 147◦ ≈ 13.72 ft2 13.

N

1 bc sin A 2 1 K = (15)(12.5) sin 42◦ ≈ 63 ft2 2 The area of the floor of the sandbox is 63 ft2 . K =

350 km 305.6o W

14. Let a = 11 m, b = 9 m, and c = 6 m.

195.5o

440 km

a2 = b2 + c2 − 2bc cos A, or

(11)2 = (9)2 + (6)2 − 2(9)(6) cos A

S

cos A ≈ −0.0370

θ = 305.6 − 195.5 = 110.1◦ d2 = (350)2 + (440)2 − 2(350)(440) cos 110.1◦ ◦

A ≈ 92◦

b2 = a2 + c2 − 2ac cos B, or

(9)2 = (11)2 + (6)2 = 2(11)(6) cos B 17.

cos B ≈ 0.5758 B ≈ 55

E

θ d

◦

d ≈ 650 km

◦

C ≈ 180◦ − (92◦ + 55◦ ) ≈ 33◦

The angles of the garden are 92◦ , 55◦ , and 33◦ . 15.

B

c

|2 − 5i| =

a 18.

A

52.3o

52.3o 513 ft

% √ 22 + (−5)2 = 29

C

B = 180◦ − (52.3◦ + 52.3◦ ) = 75.4◦ Use the law of sines to find a. b a = sin A sin B 513 a = sin 52.3◦ sin 75.4◦ 513 sin 52.3◦ a= sin 75.4◦ a ≈ 419

|4| = |4 + 0i| =

√

42 + 02 = 4

19.

Sides a and c are the same length, so the length of each of the other two sides is about 419 ft.

|2i| = |0 + 2i| =

√

02 + 22 = 2

Chapter 7 Review Exercises

463 26. 7(cos 0◦ + i sin 0◦ ) = 7 cos 0◦ + (7 sin 0◦ )i

20.

a = 7 cos 0◦ = 7 · 1 = 7 b = 7 sin 0◦ = 7 · 0 = 0

| − 3 + i| =

% √ (−3)2 + 12 = 10

21. Find trigonometric notation for 1 + i. % √ r = (1)2 + (1)2 = 2 √ √ 2 1 2 1 , cos θ = √ = sin θ = √ = 2 2 2 2 π θ = , or 45◦ , and we have 4 ! " √ √ π π , or 2(cos 45◦ + i sin 45◦ ). 1 + i = 2 cos + i sin 4 4 22. Find trigonometric notation for −4i. % r = (−4)2 = 4 0 −4 = −1, cos θ = = 0 sin θ = 4 4 3π ◦ θ= , or 270 , and we have 2 " ! 3π 3π −4i = 4 cos + i sin , or 4(cos 270◦ + i sin 270◦ ). 2 2 √ 23. Find trigonometric notation for −5 3 + 5i. ) √ √ r = (−5 3)2 + (5)2 = 100 = 10 √ √ 5 1 −5 3 3 sin θ = = , cos θ = =− 10 2 10 2 5π θ= , or 150◦ , and we have 6 ! " √ 5π 5π + i sin , or −5 3 + 5i = 10 cos 6 6 ◦ ◦ 10(cos 150 + i sin 150 ).

7(cos 0◦ + i sin 0◦ ) = 7 + 0i = 7 ! " ! " 2π 2π 2π 2π 27. 5 cos + i sin = 5 cos + 5 sin i 3 3 3 3 ! " 2π 1 5 a = 5 cos =5 − =− 3 2 2 √ !√ " 3 5 3 2π b = 5 sin =5 = 3 2 2 √ " ! 2π 5 5 3 2π + i sin =− + i 5 cos 3 3 2 2 # ! " ! "$ ! " ! ! "" π π π π 28. 2 cos − + i sin − = 2 cos − + 2 sin − i 3 3 3 3 " ! π 1 a = 2 cos − =2· =1 3 2 " ! √ " ! √ π 3 =2 − =− 3 b = 2 sin − 3 2 " ! "$ # ! √ π π + i sin − = 1 − 3i 2 cos − 3 3 √ 29. (1 + i 3)(1 − i) ) √ √ √ For 1 + i 3, r = (1)2 + ( 3)2 = 4 = 2. √ 3 1 , cos θ = , so θ = 60◦ . sin θ = 2 2 % √ For 1 − i, r = (1)2 + (1)2 = 2. √ √ 1 2 2 1 , cos θ = √ = , so θ = −45◦ . sin θ = − √ = − 2 2 2 2 √ 2(cos 60◦ + i sin 60◦ ) · 2(cos(−45◦ ) + i sin(−45◦ )) = √ 2 2(cos 15◦ + i sin 15◦ ) Using identities√for sum of angles, √ and difference √ √ we find 2 + 6 6 − 2 that cos 15◦ = and sin 15◦ = . Thus 4 4 √ 2 2(cos 15◦ + i sin 15◦ ) √ √ " √ !√ √ 2+ 6 6− 2 + i =2 2 4 4 √ √ = 1 + 3 + (−1 + 3)i.

3 24. Find trigonometric notation for . 4 *! " 2 3 3 = r= 4 4

3 0 4 sin θ = = 0, cos θ = =1 3 3 4 4 θ = 0, or 0◦ , and we have 3 3 3 = (cos 0 + i sin 0), or (cos 0◦ + i sin 0◦ ) 4 4 4

25. 4(cos 60◦ + i sin 60◦ ) = 4 cos 60◦ + (4 sin 60◦ )i 1 a = 4 cos 60◦ = 4 · = 2 2 √ √ 3 =2 3 b = 4 sin 60◦ = 4 · 2 √ 4(cos 60◦ + i sin 60◦ ) = 2 + 2 3i

30.

2 − 2i 2 + 2i

% √ √ (2)2 + (−2)2 = 8 = 2 2. √ √ −2 2 2 2 sin θ = √ = − , cos θ = √ = , so θ = 315◦ . 2 2 2 2 2 2 % √ √ For 2 + 2i, r = (2)2 + (2)2 = 8 = 2 2. √ √ 2 2 2 2 sin θ = √ = , cos θ = √ = , so θ = 45◦ . 2 2 2 2 2 2 √ 2 2[cos(315◦ ) + i sin(315◦ )] √ = cos(270◦ ) + i sin(270◦ ) 2 2[cos 45◦ + i sin 45◦ ] = 0 + (−1)i = −i For 2 − 2i, r =

464

Chapter 7: Applications of Trigonometry

√ 2 + 2 3i √ 3−i ) √ √ √ For 2 + 2 3i, r = (2)2 + (2 3)2 = 16 = 4. √ √ 3 2 1 2 3 = , cos θ = = , so θ = 60◦ . sin θ = 4 2 4 2 )√ √ √ 2 2 For 3 − i, r = ( 3) + (−1) = 4 = 2. √ 3 1 , so θ = 330◦ . sin θ = − , cos θ = 2 2 4(cos 60◦ + i sin 60◦ ) = 2(cos(−270◦ ) + i sin(−270◦ ) 2[cos(330◦ ) + i sin(330◦ )] = 2(0 + i · 1) = 2i √ 32. i(3 − 3 3i) √ For i, r = 02 + 12 = 1. 1 0 sin θ = = 1, cos θ = = 0, so θ = 90◦ . 1 1 ) √ √ √ For 3 − 3 3i, r = (3)2 + (−3 3)2 = 36 = 6. √ √ 3 3 3 3 1 sin θ = − =− , cos θ = = , so θ = 300◦ . 6 2 6 2 1(cos 90◦ + i sin 90◦ ) · 6[cos(300◦ ) + i sin(300◦ )] " !√ √ 3 1 = 6(cos 390◦ + i sin 390◦ ) = 6 + i = 3 3 + 3i 2 2 31.

33.

" ! √ 135◦ 135◦ 4 The roots are 2 cos + i sin and 2" ! !2 " ◦ ◦ √ √ 495 495 3π 3π 4 4 2 cos + i sin , or 2 cos + i sin 2 8 8 " ! 2 √ 11π 11π 4 . and 2 cos + i sin 8 8 √ 38. (3 3 − 3i)1/3 = [6(cos 330◦ + i sin 330◦ )]1/3 = # ! " ! "$ √ 330◦ 360◦ 330◦ 360◦ 3 ( 6 cos +k· +i sin +k· , 3 3 3 3 k = 0, 1, 2 √ ◦ The roots are 3 6(cos 110◦ + i sin √ √ 110 ), 3 6(cos 230◦ + i sin 230◦ ), and 3 6(cos 350◦ + i sin 350◦ ). 39. (81)1/4 = [81(cos 0◦ + i sin 0◦ )]1/4 = # ! ◦ " ! ◦ "$ 0 360◦ 0 360◦ 3 cos +k· +i sin +k· , 4 4 4 4 k = 0, 1, 2, 3 The roots are 3(cos 0◦ + i sin 0◦ ) = 3, 3(cos 90◦ + i sin 90◦ ) = 3i, 3(cos 180◦ + i sin 180◦ ) = −3, and 3(cos 270◦ + i sin 270◦ ) = −3i.

[2(cos 60◦ + i sin 60◦ )]3 = 23 [cos(3 · 60◦ ) + i sin(3 · 60◦ )]

= 8(cos 180◦ + i sin 180◦ ) 34.

(1 − i)4 = = =

35.

(1 + i)6 = =

# ! "$4 √ 7π 7π 2 cos + i sin 4 4 " ! "$ # ! √ 4 7π 7π ( 2) cos 4 · + i sin 4 · 4 4 4(cos 7π + i sin 7π) √ [ 2(cos 45◦ + i sin 45◦ )]6 √ ( 2)6 [cos(6 · 45◦ ) + i sin(6 · 45◦ )]

= 8(cos 270◦ + i sin 270◦ ) = 8[0 + i(−1)]

36.

= −8i √ "10 ! 3 1 + i = (cos 60◦ + i sin 60◦ )10 2 2 = cos(10 · 60◦ ) + i sin(10 · 60◦ )

= cos 600◦ + i sin 600◦ √ 1 3 =− − i 2 2 √ 37. (−1 + i)1/2 = [ 2(cos 135◦ + i sin 135◦ )]1/2 = " ! "$ # ! √ 360◦ 135◦ 360◦ 135◦ +k· +i sin +k· , ( 2)1/2 cos 2 2 2 2 k = 0, 1

40. (1)1/5 = [1(cos 0◦ + i sin 0◦ )]1/5 = " ! ◦ "$ # ! ◦ 360◦ 0 360◦ 0 +k· +i sin +k· , 1 cos 5 5 5 5 k = 0, 1, 2, 3, 4 The roots are cos 0◦ + i sin 0◦ = 1, cos 72◦ + i sin 72◦ , cos 144◦ + i sin 144◦ , cos 216◦ + i sin 216◦ , and cos 288◦ + i sin 288◦ .

Chapter 7 Review Exercises 41.

465 47. (−6, −120◦ )

x4 − i = 0 x4 = i

! " 1 x = −6 cos(−120◦ ) = −6 − =3 2 √ ! " √ 3 y = −6 sin(−120◦ ) = −6 − =3 3 2 √ (−6, −120◦ ) = (3, 3 3)

Find the fourth roots of i. (1)1/4 = [1(cos 90◦ + i sin 90◦ )]1/4 = # ! ◦ " ! ◦ "$ 90 360◦ 90 360◦ 1 cos +k· +i sin +k· , 4 4 4 4 k = 0, 1, 2, 3 The solutions are cos 22.5◦ + i sin 22.5◦ , cos 112.5◦ + i sin 112.5◦ , cos 202.5◦ + i sin 202.5◦ , and cos 292.5◦ + i sin 292.5◦ . 42.

48.

49.

x3 + 1 = 0 Find the third roots of −1.

y=3

50.

= [1(cos 180 + i sin 180 )] = (−1) # ! " ! "$ ◦ ◦ ◦ 180 360 180 360◦ 1 cos +k· +i sin +k· , 3 3 3 3 k = 0, 1, 2 √ 3 1 ◦ ◦ i, The solutions are cos 60 + i sin 60 = + 2 2 cos 180◦ + i sin 180◦ = −1, and √ 1 3 cos 300◦ + i sin 300◦ = − i. 2 2 ◦

◦

1/3

r2 (cos2 θ + sin2 θ) = 9 r2 = 9 r=3 51. 52.

D: (1, 300 ), (1, −60 ), (−1, 120 ) √ √ 44. (−4 2, 4 2) ) √ √ √ r = (−4 2)2 + (4 2)2 = 64 = 8 √ 4 2 3π √ = −1, so θ = 135◦ , or . tan θ = 4 −4 2 √ √ (−4 2, 4 2) = 8(cos 135◦ + i sin 135◦ ), or (8, 135◦ ) ! " ! " 3π 3π 3π = 8 cos + i sin , or 8, 4 4 4

r=6 % 2 x + y2 = 6

53.

r + r sin θ = 1 %

◦

45. (0, −5) % r = 02 + (−5)2 = 5 3π −5 tan θ = undefined, so θ = 270◦ , or . 0 2 (0, −5) = 5(cos 270◦ + i sin 270◦ ), or (5, 270◦ ) ! " ! " 3π 3π 3π = 5 cos + i sin , or 5, 2 2 2 " ! π 46. 3, 4 √ √ 2 3 2 π = x = 3 cos = 3 · 4 2 2 √ √ 3 2 2 π = y = 3 sin = 3 · 4 2 2 √ " " ! √ ! 3 2 3 2 π = , 3, 4 2 2

r sin θ − 4r cos θ − 16 = 0

x2 + y 2 = 36

C: (4, 60◦ ), (4, 420◦ ), (−4, 240◦ ); ◦

y 2 − 4x − 16 = 0

2

2

A: (5, 120◦ ), (5, 480◦ ), (−5, 300◦ ); B: (3, 210◦ ), (−3, 30◦ ), (−3, 390◦ );

x2 + y 2 = 9 r2 cos2 θ + r2 sin2 θ = 9

43. Answers may vary.

◦

r(5 cos θ − 2 sin θ) = 6 r sin θ = 3

x = −1 3

1/3

5x − 2y = 6

5r cos θ − 2r sin θ = 6

x2 + y 2 + y = 1 % x2 + y 2 = 1 − y

x2 + y 2 = 1 − 2y + y 2

x2 + 2y = 1 54.

3 1 − cos θ r − r cos θ = 3 r=

r−x = 3

r = 3+x

r2 = 9 + 6x + x2 x2 + y 2 = 9 + 6x + x2 55.

y 2 − 6x = 9 r − 2 cos θ = 3 sin θ

r2 − 2r cos θ = 3r sin θ

Multiplying by r

x + y − 2x = 3y 2

2

x2 − 2x + y 2 − 3y = 0

56. Graph (b) is the graph of r = 2 sin θ. 57. Graph (d) is the graph of r2 = cos 2 θ. 58. Graph (a) is the graph of r = 1 + 3 cos θ. 59. Graph (c) is the graph of r sin θ = 4.

466

Chapter 7: Applications of Trigonometry

60. A = 180◦ − 120◦ = 60◦ |u + v|2 = 122 + 152 − 2 · 12 · 15 cos 60◦ √ |u + v| = 189 ≈ 13.7 13.7 15 = sin θ sin 60◦ 15 sin 60◦ sin θ = ≈ 0.9482 13.7 ◦ θ ≈ 71

15 25 θ ≈ 31◦

tan θ =

The wind is blowing from the direction 180◦ − θ = 180◦ − 31◦ = 149◦ . 66. We make a drawing and use the given information and facts from geometry to label some of the angles.

61. A = 180◦ − 25◦ = 155◦ |u + v|2 = 412 + 602 − 2 · 41 · 60 cos 155◦ √ |u + v| ≈ 9740.03 ≈ 98.7 98.7 60 = sin θ sin 155◦ 60 sin 155◦ sin θ = ≈ 0.2569 98.7 ◦ θ ≈ 15

75o 90 φ

θ 75o

10o

α 100

d

θ = 75◦ − 10◦ = 65◦ |d|2 = 902 + 1002 − 2 · 90 · 100 cos 65◦ √ |d| ≈ 10, 492.87 ≈ 102.4 nautical miles

62.

102.4 100 = sin φ sin 65◦ 100 sin 65◦ sin φ = ≈ 0.8851 102.4 φ ≈ 62◦

63.

Now we can find α.

α ≈ 180◦ − (75◦ + 62◦ ) ≈ 43◦

64.

C 230

B

52o

v

θ O

65.

A

500

A = 180◦ − 52◦ = 128◦ |v|2 = 2302 + 5002 − 2 · 230 · 500 cos 128◦ √ |v| ≈ 444, 502.14 ≈ 666.7 N 500 666.7 = sin θ sin 128◦ 500 sin 128◦ sin θ = ≈ 0.5910 666.7 θ ≈ 36◦

15 km/h

25 km/h θ

φ θ

√

−→ 68. T R = '0 − (−2), 7 − 13( = '2, −6( % √ √ 69. |u| = 52 + (−6)2 = 25 + 36 = 61

70.

152 + 252 ≈ 29 km/h

4u + w = 4'3, −4( + '−2, −5(

= '12, −16( + '−2, −5(

= '12 + (−2), −16 + (−5)( 71.

= '10, −21(

2w − 6v = 2'−2, −5( − 6'−3, 9(

= '−4, −10( − '−18, 54(

= '−4 − (−18), −10 − 54(

72.

w

|w| =

The ship is 102.4 nautical miles from the starting point in the direction S43◦ E. −→ 67. AB = '−2 − 2, −5 − (−8)( = '−4, 3(

= '14, −64(

|u| + |2w| = |'3, −4(| + |2'−2, −5(|

= |'3, −4(| + |'−4, −10(| % % = (3)2 + (−4)2 + (−4)2 + (−10)2 √ √ = 25 + 116 √ = 5 + 116

73. u · v = 3(−2) + (−4)(−5) = −6 + 20 = 14

Chapter 7 Review Exercises

467

% √ √ √ (−6)2 + (−2)2 = 36 + 4 = 40 = 2 10 1 2 3 1 1 √ '−6, −2( = − √ , − √ 2 10 10 10

We use triangle OAB and the law of cosines to find |v|.

74. |'−6, −2(| =

|v|2 = 1602 + 202 − 2 · 160 · 20 · cos 130◦

|v| ≈ 173.5

Now we use the law of sines to find α. 173.5 20 = sin α sin 130◦ 20 sin 130◦ sin α = 173.5 α = 5.1◦

75. |'−9, 4(| = −9i + 4j

1 −1 = −4 4 ! " 1 θ = tan−1 4 θ is a! third-quadrant angle. The reference angle is " 1 −1 ◦ tan ≈ 14.0 and θ = 180◦ + 14.0◦ = 194.0◦ . 4 % √ 77. |u| = (−5)2 + (−3)2 = 34 3 −3 = tan θ = −5 5 3 θ = tan−1 5 θ is a! third-quadrant angle. The reference angle is " 3 −1 ◦ tan ≈ 31.0 and θ = 180◦ + 31.0◦ = 211.0◦ . 5 76. tan θ =

78. u = '3, −7(, v = '2, 2(

79.

C 160 80o

α

O

B

v

E 160

20

50o

5u − 8v = 5(2i + 5j) − 8(−3i + 10j) = 10i + 25j + 24i − 80j

81.

= 34i − 55j

u − (v + w) = 2i + 5j − (−3i + 10j + 4i + 7j) = 2i + 5j − (i + 17j) = 2i + 5j − i − 17j

82.

= i − 12j

|u − v| = |2i + 5j − (−3i + 10j)| = |5i − 5j| % = 52 + (−5)2 √ = 50 √ =5 2

N

W

80.

= |2i + 5j + 3i − 10j|

u · v = 3 · 2 + (−7)(2) = −8 % √ |u| = 32 + (−7)2 = 58 √ √ |v| = 22 + 22 = 8 −8 u·v =√ √ cos α = |u||v| 58 8 −8 α = cos−1 √ √ 58 8 α ≈ 111.8◦

310o

Then the airplane’s actual heading is 80◦ +α = 80◦ +5.1◦ = 85.1◦ .

A S

The measure of # ADC is 180◦ − (80◦ + 50◦ ) = 50◦ .

Since the measures of opposite angles of a parallelogram are equal, # ABC is also 50◦ . The sum of the measures of the angles of a parallelogram is 360◦ and # OAB = 360◦ − (50◦ + 50◦ ) # OCB, so the measure of each is = 2 260◦ = 130◦ . 2

83.

3|w| + |v| = 3|4i + 7j| + | − 3i + 10j| % √ = 3 42 + 72 + (−3)2 + 102 √ √ = 3 65 + 109

84.

−→ P Q = (−4 − 1)i + (2 − (−3))j

= −5i + 5j √ √ ! " ! " π π 2 2 π i + sin j= i+ j 85. For θ = , u = cos 4 4 4 2 2 √ √ ! " ! " 2 2 5π 5π 5π For θ = , u = cos i + sin j=− i− j 4 4 4 2 2

468

Chapter 7: Applications of Trigonometry

π 2π 86. u = (cos θ)i + (sin θ)j where θ = + 2 3 √ " ! " ! 7π 7π 3 i + sin j=− i− u = cos 6 6 2

=

7π . 6

97. |u| =

√

√

169 = 13 1 New vector: 3 · (12i + 5j) 13 15 36 i+ j = 13 13

1 j 2

122 + 52 =

98. From Section 7.1, Exercise 46, we know that the area of a parallelogram is the product of two adjacent sides and the sine of the included angle. 18.4 = (3.42)(6.97) sin θ 18.4 ≈ 0.7719 sin θ = (3.42)(6.97) θ ≈ 50.52◦ 87. Find the magnitude of 3i − j. % √ 32 + (−1)2 = 10 ! " √ 3i − j √ , or Then the desired vector is 10 10 √ " ! √ √ 3 10 10 10 i− j . 10 10 88. Answer D is correct. See the solution for Exercise 13 in Exercise Set 7.3. 89.

r = 100 x2 + y 2 = 100 % 2 ( x2 + y 2 ) = 1002 %

180◦ − 50.52◦ ≈ 129.48◦

Chapter 7 Test 1. A = 180◦ − (54◦ + 43◦ ) = 83◦ a b 18 b = , or = sin A sin B sin 83◦ sin 54◦ b ≈ 14.7 ft c a c 18 = = , or sin A sin C sin 83◦ sin 43◦ c ≈ 12.4 ft 2.

x2 + y 2 = 10, 000

Answer A is correct. 90. (−2, 5) = (5.385, 111.8◦ ), or (5.385, 1.951)

For B ≈ 70.1◦

91. Use the ANGLE feature on a graphing calculator. √ (−4.2, 7) = (4.964, 147.8◦ ), or (4.964, 2.579)

A = 180◦ − (36◦ + 70.1◦ ) = 73.9◦ a b a 8 = , or = sin A sin B sin 73.9◦ sin 70.1◦ a ≈ 8.2 m

92. (2, −15◦ ) = (1.93, −0.52) 93. Use the ANGLE feature on a graphing calculator. " ! π = (−1.86, −1.35) − 2.3, 5 94. A nonzero complex number has n different complex nth roots. Thus, 1 has three different complex cube roots, one of which is the real number 1. The other two roots are complex conjugates. Since the set of real numbers is a subset of the set of complex numbers, the real cube root of 1 is also a complex root of 1.

For B ≈ 109.9◦

A = 180◦ − (36◦ + 109.9◦ ) = 34.1◦ a b a 8 = , or = sin A sin B sin 34.1◦ sin 109.9◦ a ≈ 4.8 m

3. a2 = b2 + c2 − 2bc cos A (16.1)2 = (9.8)2 + (11.2)2 − 2(9.8)(11.2) cos A cos A ≈ −0.1719

95. A triangle has no solution when a sine or a cosine value is less than −1 or greater than 1. A triangle also has no solution if the sum of the angle measures calculated is greater than 180◦ . A triangle has only one solution if only one possible answer is found, or if one of the possible answers has an angle sum greater than 180◦ . A triangle has two solutions when two possible answers are found and neither results in an angle sum greater than 180◦ . 96. With 5◦ deviation, a 150-yd shot will actually land about 13.1 yd to the side and a 300-yd shot will land about 26.2 yd to the side. Peck’s numbers were high.

b c 8 5 = , or = sin B sin C sin B sin 36◦ 8 sin 36◦ sin B = ≈ 0.9405 5 ◦ B ≈ 70.1 or B ≈ 109.9◦

A ≈ 99.9◦

b2 = a2 + c2 − 2ac cos B (9.8)2 = (16.1)2 + (11.2)2 − 2(16.1)(11.2) cos B cos B ≈ 0.8003 B ≈ 36.8◦

C ≈ 180◦ − (99.9◦ + 36.8◦ ) ≈ 43.3◦ 4.

1 ab sin C 2 1 K = (7)(13) sin 106.4◦ ≈ 43.6 cm2 2 K =

Chapter 7 Test

469

5.

52 m A

C 44o

108o

c

B #

ABC = 180◦ − (108◦ + 44◦ ) = 28◦

We use the law of sines to find c. c b = sin # ABC sin # ACB c 52 = sin 28◦ sin 44◦ 52 sin 44◦ =c sin 28◦ 77 ≈ c

The distance from A to B is about 77 m.

9. Find trigonometric notation for 3 − 3i. % √ √ r = 32 + (−3)2 = 18 = 3 2 √ √ −3 2 2 3 sin θ = √ = − , cos θ = √ = 2 2 3 2 3 2 7π , or 315◦ , and we have θ= 4 ! " √ 7π 7π + i sin , or 3 − 3i = 3 2 cos 4 4 √ 3 2(cos 315◦ + i sin 315◦ ) " ! 2π 2π 10. 2 cos + i sin 3 3 " ! π π 8 cos + i sin # !6 "6 ! "$ 2 2π π 2π π = cos − + i sin − 8 3 6 3 6 ! " π π 1 1 1 = cos + i sin = (0 + i · 1) = i 4 2 2 4 4 √ 11. (1 − i)8 = [ 2(cos 315◦ + i sin 315◦ )]8 √ = ( 2)8 [cos(8 · 315◦ ) + i sin(8 · 315◦ )] = 16(cos 0◦ + i sin 0◦ )

6. 210 km/h · 3 hr = 630 km

= 16(1 + i · 0)

180 km/h · 3 hr = 540 km

= 16 √

N B 630 km 290o C

W c

E 185o

540 km A

S

In triangle ABC, #

A = 290◦ − 185◦ = 105◦ .

We use the law of cosines to find c. c2 = a2 + b2 − 2ab cos C

c2 = 6302 + 540◦ − 2 · 630 · 540 cos 105◦

7.

c ≈ 930 km

12. (−1, 3) ) √ √ r = (−1)2 + ( 3)2 = 4 = 2 √ √ 3 tan θ = = − 3, so θ = 120◦ . −1 √ (−1, 3) = 2(cos 120◦ + i sin 120◦ ) " ! 2π 13. − 1, 3 ! " 2π 1 1 = −1 − x = −1 · cos = 3 2 2 √ !√ " 3 3 2π y = −1 · sin = −1 =− 3 2 2 √ " " ! ! 1 3 2π = ,− − 1, 3 2 2 14.

x2 + y 2 = 10 r2 = 10 √ r = 10

15.

8. |2 − 3i| =

% √ 22 + (−3)2 = 13

16. Graph (a) is the graph of r = 3 cos θ.

470

Chapter 7: Applications of Trigonometry

17.

C

B

5

u+v 63o θ

O

8

A

A = 180◦ − 63◦ = 117◦ |u + v|2 = 82 + 52 − 2 · 8 · 5 cos 117◦ √ |u + v| = 125.32 ≈ 11.2

18.

5 11.2 = sin θ sin 117◦ 5 sin 117◦ sin θ = ≈ 0.3978 11.2 ◦ θ ≈ 23.4

2u − 3v = 2(2i − 7j) − 3(5i + j) = 4i − 14j − 15i − 3j = −11i − 17j

19. Find the magnitude of −4i + 3j. % √ (−4)2 + 32 = 25 = 5 4 3 1 Then the desired vector is (−4i + 3j) = − i + j. 5 5 5 20. The area of a parallelogram is the product of two adjacent sides and the sine of the included angle. 72.9 = (15.4)(9.8) sin θ 72.9 ≈ 0.4830 sin θ = (15.4)(9.8) θ = 28.9◦ 180◦ − 28.9◦ ≈ 151.1◦

The measures of the angles are 28.9◦ and 151.1◦ .

Chapter 8

Systems of Equations and Matrices 9. Graph x + 2y = 1 and x + 4y = 3 and find the coordinates of the point of intersection.

Exercise Set 8.1 1. Graph (c) is the graph of this system.

y 5 4

2. Graph (e) is the graph of this system. 3. Graph (f) is the graph of this system.

(–1, 1)

4. Graph (a) is the graph of this system. 5. Graph (b) is the graph of this system.

-5

10. Graph 3x+4y = 5 and x−2y = 5 and find the coordinates of the point of intersection.

5 4 3 2 1

y

x+y=2 x

-5 -4 -3 -2 -1 0 -1

5 4

1 2 3 4 5

-2 -3 -4

3x + 4y = 5 3x + y = 0

-5

3 2 1

x

-5 -4 -3 -2 -1 0 -1

The solution is (−1, 3).

x – 2y = 5

1 2 3 4 5

-2 -3 -4

(3, –1)

-5

8. Graph x + y = 1 and 3x + y = 7 and find the coordinates of the point of intersection.

The solution is (3, −1).

y

11. Graph y + 1 = 2x and y − 1 = 2x and find the coordinates of the point of intersection.

5 4

-5 -4 -3 -2 -1 0 -1

x + 2y = 1

The solution is (−1, 1).

y

3 2 1

x

1 2 3 4 5

-2 -3 -4

7. Graph x + y = 2 and 3x + y = 0 and find the coordinates of the point of intersection.

x+y=1

x + 4y = 3

-5 -4 -3 -2 -1 0 -1

6. Graph (d) is the graph of this system.

(–1, 3)

3 2 1

3x + y = 7 y x

5 4

1 2 3 4 5

-2 -3 -4 -5

(3, –2)

3 2 1 -5 -4 -3 -2 -1 0 -1 y – 1 = 2x

The solution is (3, −2).

y + 1 = 2x

x

1 2 3 4 5

-2 -3 -4 -5

The graphs do not intersect, so there is no solution.

472

Chapter 8: Systems of Equations and Matrices

12. Graph 2x−y = 1 and 3y = 6x−3 and find the coordinates of the point of intersection.

15. Graph 2y = x−1 and 3x = 6y +3 and find the coordinates of the point of intersection. y

y 5

5 4

4

2x – y = 1

3 2 1

3 1

x

-5 -4 -3 -2 -1 0 -1

1

2

3

4

5

–2 –3

3x = 6y + 3

–4 –5

-5

The graphs coincide so there are infinitely many solutions. Solving either equation for y, we have y = 2x − 1, so the solutions can be expressed as (x, 2x−1). Similarly, solving y+1 , so the solutions either equation for x, we get x = ! "2 y+1 can also be expressed as ,y . 2 13. Graph x − y = −6 and y = −2x and find the coordinates of the point of intersection.

The graphs coincide so there are infinitely many solutions. x−1 Solving either equation for y, we get y = , so the so2 " ! x−1 lutions can be expressed as x, . Similarly, solving 2 either equation for x, we get x = 2y + 1, so the solutions can also be expressed as (2y + 1, y). 16. Graph y = 3x+2 and 3x−y = −3 and find the coordinates of the point of intersection. y

y

5

5

4

4

3

3

2

y = 3x + 2

3x – y = – 3 1

2

x–y =–6

x

–5 –4 –3 –2 –1 0 –1

1 2 3 4 5

-2 -3 -4

(– 2, 4)

2y = x – 1

2

3y = 6x – 3

1

–5 –4 –3 –2 –1 0 –1

x

–5 –4 –3 –2 –1 0 –1

1

2

3

4

5

2

3

4

5

–2 –3

y = – 2x

–2

x 1

–3

–4

–4

–5

–5

The graphs do not intersect, so there is no solution. The solution is (−2, 4).

17.

14. Graph 2x + y = 5 and x = −3y and find the coordinates of the point of intersection. y

4 2 1

–2

Solve equation (1) for either x or y. We choose to solve for y. Then substitute 9 − x for y in equation (2) and solve the resulting equation.

2x + y = 5

3

–5 –4 –3 –2 –1 0 –1

(1) (2)

y =9−x

5

x = – 3y

x + y = 9, 2x − 3y = −2

x 1

2

3

(3, – 1)

–3 –4

4

5

2x − 3(9 − x) = −2 2x − 27 + 3x = −2 5x − 27 = −2 5x = 25

–5

The solution is (3, −1).

x=5 Now substitute 5 for x in either equation (1) or (2) and solve for y. 5+y = 9

Using equation (1)

y=4 The solution is (5, 4).

Exercise Set 8.1 18.

473

3x − y = 5, (1) 1 x+y= (2) 2 Solve equation (2) for y. 1 y = −x 2 Substitute in equation (1) and solve for x. " ! 1 −x = 5 3x − 2 1 3x − + x = 5 2 11 4x = 2 11 x= 8 Back-substitute to find y. 1 11 +y = Using equation (2) 8 2 7 y=− 8 ! " 11 7 The solution is ,− . 8 8

Substitute 2 for x in equation (1) to find y. y = 2 · 2 − 6 = 4 − 6 = −2

The solution is (2, −2). 22.

y = 2x + 3 3x + 5(2x + 3) = 2 3x + 10x + 15 = 2 13x = −13 x = −1

Back-substitute to find y. 2(−1) − y = −3 1=y

The solution is (−1, 1). 23. x − 5y = 4, (1) y = 7 − 2x (2)

Use equation (2) and substitute 7 − 2x for y in equation (1). Then solve for x. x − 5(7 − 2x) = 4

x − 35 + 10x = 4

y + 4 − 2y = 7

11x − 35 = 4

−y + 4 = 7

11x = 39 39 x= 11 39 Substitute for x in equation (2) to find y. 11 78 1 39 y =7−2· =7− =− 11 11 11 ! " 39 1 The solution is ,− . 11 11

−y = 3

y = −3

Substitute −3 for y in equation (2) to find x. x = −3 + 4 = 1

The solution is (1, −3). 20. x + 4y = 6, (1) x = −3y + 3 (2)

Back-substitute to find x. x = −3 · 3 + 3 = −6

The solution is (−6, 3). 21. y = 2x − 6, (1)

5x − 3y = 16 (2)

Use equation (1) and substitute 2x − 6 for y in equation (2). Then solve for x. 5x − 3(2x − 6) = 16 5x − 6x + 18 = 16 −x + 18 = 16

−x = −2 x=2

Using equation (2)

−2 − y = −3

Use equation (2) and substitute y + 4 for x in equation (1). Then solve for y.

y=3

(2)

Substitute 2x + 3 for y in equation (1) and solve for x.

x = y + 4 (2)

−3y + 3 + 4y = 6

(1)

2x − y = −3

Solve equation (2) for y.

19. x − 2y = 7, (1)

Substitute −3y + 3 for x in equation (1) and solve for y.

3x + 5y = 2,

24.

5x + 3y = −1, (1) x+ y=1

(2)

Solve equation (2) for either x or y. We choose to solve for x. x=1−y

Substitute 1 − y for x in equation (1) and solve for y. 5(1 − y) + 3y = −1 5 − 5y + 3y = −1 −2y = −6 y=3

Back-substitute to find x. x+3 = 1 Using equation (2) x = −2

The solution is (−2, 3).

474 25.

Chapter 8: Systems of Equations and Matrices x + 2y = 2

2x − 3y = 5, (1) 5x + 4y = 1

x = −2y + 2

(2)

Substitute −2y + 2 for x in equation (2) and solve for y.

Solve one equation for either x or y. We choose to solve equation (1) for x.

4(−2y + 2) + 4y = 5

2x − 3y = 5

−8y + 8 + 4y = 5

2x = 3y + 5 5 3 x= y+ 2 2 5 3 Substitute y + for x in equation (2) and solve for y. 2 2 ! " 3 5 5 y+ + 4y = 1 2 2 25 15 y+ + 4y = 1 2 2 25 23 y+ =1 2 2 23 23 y=− 2 2 y = −1

Substitute −1 for y in either equation (1) or (2) and solve for x. 2x − 3(−1) = 5 Using equation (1)

−4y + 8 = 5

−4y = −3 3 y= 4

3 Substitute for y in either equation (1) or equation (2) 4 and solve for x. x + 2y = 2 Using equation (1) 3 x+2· = 2 4 3 x+ = 2 2 1 x= 2 ! " 1 3 The solution is , . 2 4 28.

4x + y = 3

2x + 3 = 5

x=1 The solution is (1, −1).

y = −4x + 3

Substitute −4x + 3 for y in equation (1) and solve for x.

3x + 4y = 6, (1) 2x + 3y = 5

2x − (−4x + 3) = 2

(2)

2x + 4x − 3 = 2

Solve one equation for either x or y. We choose to solve equation (2) for x. 5 3 x=− y+ 2 2 3 5 Substitute − y + for x in equation (1) and solve for y. 2 2 " ! 5 3 3 − y+ + 4y = 6 2 2 9 15 + 4y = 6 − y+ 2 2 3 1 − y=− 2 2 y=3 Back-substitute to find x. 3x + 4 · 3 = 6 Using equation (1) 3x = −6

The solution is (−2, 3). x + 2y = 2, (1) 4x + 4y = 5

6x = 5 5 x= 6 Back-substitute to find y. 5 4· +y = 3 6 10 +y = 3 3 1 y=− 3 ! " 5 1 The solution is ,− . 6 3 29.

x + 2y = 7,

(1)

x − 2y = −5

(2)

We add the equations to eliminate y.

x = −2

27.

(2)

Solve one equation for either x or y. We choose to solve equation (2) for y since y has a coefficient of 1 in that equation.

2x = 2

26.

2x − y = 2, (1)

(2)

Solve one equation for either x or y. We choose to solve equation (1) for x since x has a coefficient of 1 in that equation.

x + 2y = 7 x − 2y = −5 2x =2

Adding

x=1 Back-substitute in either equation and solve for y.

Exercise Set 8.1 1 + 2y = 7

475 Using equation (1)

The equation 0 = 0 is true for all values of x and y. Thus, the system of equations has infinitely many solutions. 1 1 Solving either equation for y, we can write y = x − so 2" ! 4 1 1 . the solutions are ordered pairs of the form x, x − 4 2 Equivalently, if we solve either equation for x we get x = 4y + 2 so the solutions can also be expressed as (4y + 2, y). Since there are infinitely many solutions, the system of equations is consistent and dependent.

2y = 6 y=3 The solution is (1, 3). Since the system of equations has exactly one solution it is consistent and independent. 30.

3x + 4y = −2

(1)

−3x − 5y = 1 −y = −1

(2) Adding

y=1

34.

Back-substitute to find x. 3x + 4 · 1 = −2 Using equation (1) 3x = −6 x = −2

The solution is (−2, 1). Since the system of equations has exactly one solution it is consistent and independent. 31.

x − 3y = 2,

(1)

6x + 5y = −34

(2)

Multiply equation (1) by −6 and add it to equation (2) to eliminate x. −6x + 18y = −12 6x + 5y = −34 23y = −46 y = −2

Back-substitute to find x. x − 3(−2) = 2 Using equation (1) x+6 = 2

x = −4

The solution is (−4, −2). Since the system of equations has exactly one solution it is consistent and independent. 32.

x + 3y = 0,

(1)

20x − 15y = 75

(2)

Multiply equation (1) by 5 and add. 5x + 15y = 0 20x − 15y = 75 25x = 75

(2)

Multiply equation (1) by 3 and equation (2) by −2 and add. 6x + 18y = 21 −6x − 18y = −20 0=1 We get a false equation so there is no solution. Since there is no solution the system of equations is inconsistent and independent. 35. 2x = 5 − 3y,

(1)

4x = 11 − 7y

(2)

We rewrite the equations. 2x + 3y = 5,

(1a)

4x + 7y = 11

(2a)

Multiply equation (2a) by −2 and add to eliminate x. −4x − 6y = −10 4x + 7y = 11 y=1

Back-substitute to find x. 2x = 5 − 3 · 1 Using equation (1) 2x = 2 x=1 The solution is (1, 1). Since the system of equations has exactly one solution it is consistent and independent. 2x = y + 5

Back-substitute to find y. 3 + 3y = 0 Using equation (1) 3y = −3

(1)

(2)

x − y = 2, (1a) 2x − y = 5

(2a)

Dividing equation (1) by 7 Rewriting equation (2)

Multiply equation (1a) by −1 and add.

y = −1

The solution is (3, −1). Since the system of equations has exactly one solution it is consistent and independent. 3x − 12y = 6, (1) 2x − 8y = 4

(1)

3x + 9y = 10

36. 7(x − y) = 14,

x=3

33.

2x + 6y = 7,

(2)

Multiply equation (1) by 2 and equation (2) by −3 and add. 6x − 24y = 12 −6x + 24y = −12 0=0

−x + y = −2 2x − y = 5 x =3

Back-substitute to find y. 3−y = 2

Using equation (1a)

1=y

The solution is (3, 1). Since the system of equations has exactly one solution it is consistent and independent.

476 37.

Chapter 8: Systems of Equations and Matrices 1 1 x+ y= 5 2 1 3 x− y= 5 2 4 x = 5 x=

0.3x − 0.2y = −0.9,

0.2x − 0.3y = −0.6

First, multiply each equation by 10 to clear the decimals. 3x − 2y = −9 2x − 3y = −6

(1) (2)

Now multiply equation (1) by 3 and equation (2) by −2 and add to eliminate y. 9x − 6y = −27

x = −3

Back-substitute to find y. Using equation (1)

−9 − 2y = −9

The solution is (−3, 0). Since the system of equations has exactly one solution it is consistent and independent. 38.

0.2x − 0.3y = 0.3,

0.4x + 0.6y = −0.2

First, multiply each equation by 10 to clear the decimals. 2x − 3y = 3

4x + 6y = −2

(1) (2)

Now multiply equation (1) by 2 and add. 4x − 6y = 6

4x + 6y = −2 8x =4 1 x= 2 Back-substitute to find y. 1 4 · + 6y = −2 Using equation (2) 2 2 + 6y = −2

6y = −4 2 y=− 3 " ! 1 2 The solution is , − . Since the system of equations 2 3 has exactly one solution it is consistent and independent. 39.

1 1 x + y = 6, (1) 5 2 1 3 x − y = 2 (2) 5 2 We could multiply both equations by 10 to clear fractions, but since the y-coefficients differ only by sign we will just add to eliminate y.

8 10

The solution is (10, 8). Since the system of equations has exactly one solution it is consistent and independent.

−2y = 0 y=0

2

Back-substitute to find y. 1 1 · 10 + y = 6 Using equation (1) 5 2 1 2+ y = 6 2 1 y=4 2 y=8

−4x + 6y = 12 5x = −15

3(−3) − 2y = −9

6

40.

3 2 x + y = −17, 3 5 1 1 x − y = −1 2 3 Multiply the first equation by 15 and the second by 6 to clear fractions. 10x + 9y = −255 (1) 3x − 2y = −6

(2)

Now multiply equation (1) by 2 and equation (2) by 9 and add. 20x + 18y = −510 27x − 18y = −54 47x = −564 x = −12

Back-substitute to find y. 3(−12) − 2y = −6

Using equation (2)

−36 − 2y = −6 −2y = 30

y = −15

The solution is (−12, −15). Since the system of equations has exactly one solution it is consistent and independent. 41. Familiarize. Let x = the number of injuries that occur in snowboarding each winter and y = the number of injuries that occur in skiing. Translate. The total number of snowboarding and skiing injuries is 288,400 so we have one equation: x + y = 288, 400 Skiing accounts for 400 more injuries than snowboarding, so we have a second equation: y = x + 400 Carry out. We solve the system of equations x + y = 288, 400, (1) y = x + 400. (2)

Exercise Set 8.1

477

Substitute x + 400 for y in equation (1) and solve for x. x + (x + 400) = 288, 400 2x + 400 = 288, 400 2x = 288, 000 x = 144, 000 Back-substitute in equation (2) to find y. y = 144, 000 + 400 = 144, 400 Check. 144, 400+144, 000 = 288, 400 injuries and 144,400 is 400 more than 144,000. The answer checks. State. Snowboarding accounts for 144,000 injuries each winter and skiing accounts for about 144,400 injuries. 42. Let x = the amount spent on textbooks and course materials and y = the amount spent on computer equipment. Solve: x = y + 183, x + y = 819. x = $501, y = $318 43. Familiarize. Let m = the number of streets named Main Street and s = the number of streets named Second Street. Translate. The total number of streets bearing one of these names is 15,684, so we have one equation: m + s = 15, 684 A second equation comes from the fact that there are 260 more streets named Second Street than Main Street: s = m + 260 Carry out. We solve the system of equations m + s = 15, 684, (1) s = m + 260. (2) Substitute m + 260 for s in equation (1) and solve for m. m + (m + 260) = 15, 684 2m + 260 = 15, 684 2m = 15, 424 m = 7712 Back-substitute in equation (2) to find s. s = 7712 + 260 = 7972 Check. The total number of streets is 7712 + 7972, or 15,684 and 7972 is 260 more than 7712. The answer checks. State. There are 7712 streets named Main Street and 7972 streets named Second Street. 44. Let a = the cost of each adult’s admission and c = the cost of each child’s admission. Solve: a = c + 5, 2a + 5c = 59. a = $12, c = $7 45. Familiarize. Let w = the number of visitors hosted by the website and m = the number hosted by the museums, in millions. Translate. The total number of visitors was 142 million, so we have one equation:

m + w = 142 A second equation comes from the fact that the museums had 94 million fewer visitors than the website: m = w − 94

Carry out. We solve the system of equations m + w = 142, (1) m = w − 94. (2)

Substitute w − 94 for m in equation (1) and solve for w. (w − 94) + w = 142 2w − 94 = 142

2w = 236 w = 118

Back-substitute in equation (2) to find m. m = 118 − 94 = 24

Check. The total number of visitors is 24 + 118, or 142 million, and 24 million is 94 million less than 118 million. The answer checks State. The Smithsonian Institution hosted 24 million visitors and the Smithsonian website hosted 118 million visitors. 46. Let x = the number of standard-delivery packages and y = the number of express-delivery packages. Solve:

x + y = 120, 3.5x + 7.5y = 596.

x = 76 packages, y = 44 packages 47. Familiarize. Let x = the number of video rentals and y = the number of boxes of popcorn given away. Then the total cost of the rentals is 1 · x, or x, and the total cost of the popcorn is 2y. Translate. The number of new members is the same as the number of incentives so we have one equation: x + y = 48. The total cost of the incentive was $86. This gives us another equation: x + 2y = 86. Carry out. We solve the system of equations x + y = 48, (1) x + 2y = 86. (2) Multiply equation (1) by −1 and add. −x − y = −48 x + 2y = 86 y = 38

Back-substitute to find x. x + 38 = 48 Using equation (1) x = 10 Check. When 10 rentals and 38 boxes of popcorn are given away, a total of 48 new members have signed up. Ten rentals cost the store $10 and 38 boxes of popcorn cost 2 · 38, or $76, so the total cost of the incentives was $10 + $76, or $86. The solution checks.

478

Chapter 8: Systems of Equations and Matrices State. Ten rentals and 38 boxes of popcorn were given away.

48. Let x = the number of tickets sold for pavilion seats and y = the number sold for lawn seats. Solve:

x + y = 1500, 25x + 15y = 28, 500.

x = 600, y = 900 49. Familiarize and Translate. We use the system of equations given in the problem. y = 70 + 2x

(1)

y = 175 − 5x,

(2)

Carry out. Substitute 175 − 5x for y in equation (1) and solve for x. 175 − 5x = 70 + 2x 105 = 7x

Adding 5x and subtracting 70

15 = x Back-substitute in either equation to find y. We choose equation (1). y = 70 + 2 · 15 = 70 + 30 = 100

Check. original checks. point of

Substituting 15 for x and 100 for y in both of the equations yields true equations, so the solution We could also check graphically by finding the intersection of equations (1) and (2).

State. The equilibrium point is (15, $100). 50.

Solve: y = 240 + 40x, y = 500 − 25x.

x = 4, y = 400, so the equilibrium point is (4, $400). 51. Familiarize and Translate. We find the value of x for which C = R, where C = 14x + 350, R = 16.5x. Carry out. When C = R we have: 14x + 350 = 16.5x 350 = 2.5x 140 = x Check. When x = 140, C = 14 · 140 + 350, or 2310 and R = 16.5(140), or 2310. Since C = R, the solution checks. State. 140 units must be produced and sold in order to break even. 52. Solve C = R, where C = 8.5x + 75, R = 10x. x = 50 53. Familiarize and Translate. We find the value of x for which C = R, where C = 15x + 12, 000, R = 18x − 6000.

Carry out. When C = R we have: 15x + 12, 000 = 18x − 6000 18, 000 = 3x

Subtracting 15x and adding 6000

6000 = x

Check. When x = 6000, C = 15·6000+12, 000, or 102,000 and R = 18 · 6000 − 6000, or 102,000. Since C = R, the solution checks. State. 6000 units must be produced and sold in order to break even. 54. Solve C = R, where C = 3x + 400, R = 7x − 600.

x = 250

55. Familiarize. Let x = the number of servings of spaghetti and meatballs required and y = the number of servings of iceberg lettuce required. Then x servings of spaghetti contain 260x Cal and 32x g of carbohydrates; y servings of lettuce contain 5y Cal and 1 · y or y, g of carbohydrates. Translate. One equation comes from the fact that 400 Cal are desired: 260x + 5y = 400. A second equation comes from the fact that 50g of carbohydrates are required: 32x + y = 50. Carry out. We solve the system 260x + 5y = 400, (1) 32x + y = 50.

(2)

Multiply equation (2) by −5 and add. 260x + 5y = 400

−160x − 5y = −250 100x = 150 x = 1.5 Back-substitute to find y. 32(1.5) + y = 50

Using equation (2)

48 + y = 50 y=2 Check. 1.5 servings of spaghetti contain 260(1.5), or 390 Cal and 32(1.5), or 48 g of carbohydrates; 2 servings of lettuce contain 5 · 2, or 10 Cal and 1 · 2, or 2 g of carbohydrates. Together they contain 390 + 10, or 400 Cal and 48 + 2, or 50 g of carbohydrates. The solution checks. State. 1.5 servings of spaghetti and meatballs and 2 servings of iceberg lettuce are required. 56. Let x = the number of servings of tomato soup and y = the number of slices of whole wheat bread required. Solve: 100x + 70y = 230, 18x + 13y = 42. x = 1.25, y = 1.5

Exercise Set 8.1

479

57. Familiarize. It helps to make a drawing. Then organize the information in a table. Let x = the speed of the boat and y = the speed of the stream. The speed upstream is x − y. The speed downstream is x + y. 46 km 2 hr ! Downstream 51 km "

(x + y) km/h !

3 hr

(x − y) km/h

Upstream

Distance

Speed

Time

Downstream

46

x+y

2

Upstream

51

x−y

3

51 = (x − y)3

or

x + y = 23,

(1)

x − y = 17

(2)

!

Carry out. We begin by adding equations (1) and (2). x + y = 23

y = 15, 000,

Multiplying the second equation by 100 to clear the decimals, we have: x + y = 15, 000,

(1)

Carry out. We begin by multiplying equation (1) by −7 and adding. −7x − 7y = −105, 000 7x + 9y = 123, 000 2y = 18, 000 y = 9000 Back-substitute to find x. x + 9000 = 15, 000 Using equation (1) x = 6000 Check. The total investment is $6000+$9000, or $15,000. The total interest is 0.07($6000) + 0.09($9000), or $420 + $810, or $1230. The solution checks. State. $6000 was invested at 7% and $9000 was invested at 9%.

x − y = 17 2x = 40

60. Let x = the number of short-sleeved shirts sold and y = the number of long-sleeved shirts sold.

x = 20 Back-substitute to find y. 20 + y = 23

x+

0.07x + 0.09y = 1230

7x + 9y = 123, 000. (2)

Translate. Using d = rt in each row of the table, we get a system of equations. 46 = (x + y)2

We have a system of equations:

Solve:

Using equation (1)

x+

y = 36,

12x + 18y = 522. x = 21, y = 15

y=3 Check. The speed downstream is 20+3, or 23 km/h. The distance traveled downstream in 2 hr is 23 · 2, or 46 km. The speed upstream is 20 − 3, or 17 km/h. The distance traveled upstream in 3 hr is 17 · 3, or 51 km. The solution checks. State. The speed of the boat is 20 km/h. The speed of the stream is 3 km/h. 58. Let x = the speed of the plane and y = the speed of the wind. Speed

Time

Distance

Downwind

x+y

3

3000

Upwind

x−y

4

3000

Solve: (x + y)3 = 3000, (x − y)4 = 3000.

x = 875 km/h, y = 125 km/h 59. Familiarize. Let x = the amount invested at 7% and y = the amount invested at 9%. Then the interest from the investments is 7%x and 9%y, or 0.07x and 0.09y. Translate. The total investment is $15,000. x + y = 15, 000 The total interest is $1230. 0.07x + 0.09y = 1230

61. Familiarize. Let x = the number of pounds of French roast coffee used and y = the number of pounds of Kenyan coffee. We organize the information in a table.

Amount Price per pound Total cost

French roast x

Kenyan

Mixture

y

10 lb

$9.00

$7.50

$8.40

$9x

$7.50y

$8.40(10), or $84

Translate. The first and third rows of the table give us a system of equations. x+

y = 10,

9x + 7.5y = 84 Multiply the second equation by 10 to clear the decimals. x+

y = 10,

(1)

90x + 75y = 840

(2)

Carry out. Begin by multiplying equation (1) by −75 and adding. −75x − 75y = −750 90x + 75y = 840 15x = 90 x=6

480

Chapter 8: Systems of Equations and Matrices 65. Left to the student

Back-substitute to find y. 6 + y = 10 Using equation (1)

66. Left to the student

y=4 Check. The total amount of coffee in the mixture is 6 + 4, or 10 lb. The total value of the mixture is 6($9)+4($7.50), or $54 + $30, or $84. The solution checks. State. 6 lb of French roast coffee and 4 lb of Kenyan coffee should be used. 62. Let x = the monthly sales, C = the earnings with the straight-commission plan, and S = the earnings with the salary-plus-commission plan. Then 8% of sales is represented by 8%x, or 0.08x, and 1% of sales is represented by 1%x, or 0.01x. Solve: C = 0.08x, S = 1500 + 0.01x. x ≈ 21, 428.57, so the two plans pay the same amount for monthly sales of about $21,428.57. 63. Familiarize. Let x = the speed of the plane and y = the speed of the wind. We organize the information in a table. Distance

Speed

Time

LA to NY

3000

x+y

5

NY to LA

3000

x−y

6

Translate. Using d = rt in each row of the table, we get a system of equations. 3000 = (x + y)5 x + y = 600, (1) or 3000 = (x − y)6 x − y = 500 (2) Carry out. We begin by adding equations (1) and (2). x + y = 600

67. a) b(x) = −0.0657894737x + 63.95394737; c(x) = 1.186842105x + 48.24078947

b) Graph y1 = b(x) and y2 = c(x) and find the first coordinate of the point of intersection of the graphs. It is approximately 12.5, so we estimate that chicken consumption will equal beef consumption about 12.5 yr after 1995. 68. a) b(x) = −0.7240223464x + 60.60111732; g(x) = 0.7240223464x + 39.39888268

b) Graph y1 = b(x) and y2 = g(x) and find the first coordinate or the point of intersection of the graphs. It is approximately 15, so the percentages of brandname and generic drugs sold will be equal about 15 years after 1995. 69. When a variable is not alone on one side of an equation or when solving for a variable is difficult or produces an expression containing fractions, the elimination method is preferable to the substitution method. 70. The solution of the equation 2x + 5 = 3x − 7 is the first coordinate of the point of intersection of the graphs of y1 = 2x + 5 and y2 = 3x − 7. The solution of the system of equations y = 2x + 5, y = 3x − 7 is the ordered pair that is the point of intersection of y1 and y2 . 71. Familiarize. Let h = the number of hardback books sold, in millions. Then h + 40 = the number of paperback books sold. Translate. paperback Hardback plus sales sales $% &  # $% & #   ( ( ( h + (h + 40)

x − y = 500 2x = 1100 x = 550 Back-substitute to find y. 550 + y = 600

Carry out. We solve the equation.

total sales.  # $%  & ( ( = 110 is

h + (h + 40) = 110

y = 50 Check. The speed with the wind is 550 + 50, or 600 mph. Then the distance traveled with a tailwind is 600 · 5, or 3000 mi. The speed against the wind is 550−50, or 500 mi. The distance traveled against the wind is 500·6, or 3000 mi. The distances are both 3000 mi, so the answer checks.

If h = 35, then h + 40 = 35 + 40 = 75.

State. The speed of the plane is 550 mph, and the speed of the wind is 50 mph.

Check. Total sales were 35+75, or 110 million, and 75 million is 40 million more than 35 million. The answer checks.

64. Let d = the distance traveled by the slower plane and t = the time the planes travel.

State. 35 million hardback books and 75 million paperback books were sold.

Distance

Speed

Time

Slower plane

d

190

t

Faster plane

780 − d

200

t

Solve:

d = 190t, 780 − d = 200t.

t = 2 hr

2h + 40 = 110 2h = 70 h = 35

72. Let s = the amount of soyfood sales in 2000, in billions of dollars. Solve: s + 0.46s = 4.1 s ≈ $2.8 billion

Exercise Set 8.1

481

73. Substituting 15 for f (x), we solve the following equation. 15 = x − 4x + 3 2

0 = x2 − 4x − 12

0 = (x − 6)(x + 2)

x − 6 = 0 or x + 2 = 0 x = 6 or

x = −2

If the output is 15, the input is 6 or −2.

74. Substituting 8 for f (x), we solve the following equation. 8 = x2 − 4x + 3 0 = x2 − 4x − 5

0 = (x − 5)(x + 1)

x − 5 = 0 or x + 1 = 0 x = 5 or

x = −1

Given an output of 8, the corresponding inputs are 5 and −1.

75. f (−2) = (−2)2 − 4(−2) + 3 = 15 76.

x2 − 4x + 3 = 0

(x − 1)(x − 3) = 0 x = 1 or x = 3

State. Nancy jogged 4 km on each trip. 78. Let x = the number of one-turtleneck orders and y = the number of two-turtleneck orders. Solve: x + 2y = 1250, 15x + 25y = 16, 750. y = 400 customers 79. Familiarize and Translate. We let x and y represent the speeds of the trains. Organize the information in a table. Using d = rt, we let 3x, 2y, 1.5x, and 3y represent the distances the trains travel. First situation: 3 hours

x km/h

The total time is 1 hr. x+y =1 The total distance is 6 km. 8x + 4y = 6 Carry out. Solve the system x + y = 1, (1) 8x + 4y = 6. (2) Multiply equation (1) by −4 and add. −4x − 4y = −4

8x + 4y = 6 4x =2 1 x= 2 This is the time we need to find the distance spent jogging, so we could stop here. However, we will not be able to check the solution unless we find y also so we continue. We back-substitute. 1 + y = 1 Using equation (1) 2 1 y= 2 1 Then the distance jogged is 8· , or 4 km, and the distance 2 1 walked is 4 · , or 2 km. 2 1 1 Check. The total time is hr + hr, or 1 hr. The total 2 2 distance is 4 km + 2 km, or 6 km. The solution checks.

2 hours Central

y km/h

3 hours Central

216 km Second situation: 1.5 hours x km/h !

Union

216 km

77. Familiarize. Let x = the time spent jogging and y = the time spent walking. Then Nancy jogs 8x km and walks 4y km. Translate.

y km/h !

Union

Distance traveled in first situation

Distance traveled in second situation

3x

1.5x

2y

3y

216

216

Train1 (from Union to Central) Train2 (from Central to Union) Total

The total distance in each situation is 216 km. Thus, we have a system of equations. 3x + 2y = 216, (1) 1.5x + 3y = 216

(2)

Carry out. Multiply equation (2) by −2 and add. 3x +

−3x −

2y = 216

6y = −432 −4y = −216 y = 54

Back-substitute to find x. 3x + 2 · 54 = 216 Using equation (1) 3x + 108 = 216

3x = 108 x = 36 Check. If x = 36 and y = 54, the total distance the trains travel in the first situation is 3 · 36 + 2 · 54, or 216 km. The total distance they travel in the second situation is 1.5 · 36 + 3 · 54, or 216 km. The solution checks.

State. The speed of the first train is 36 km/h. The speed of the second train is 54 km/h.

482

Chapter 8: Systems of Equations and Matrices Carry out. We solve the system of equations

80. Let x = the amount of mixture replaced by 100% antifreeze and y = the amount of 30% mixture retained. Amount Percent of antifreeze Amount of antifreeze Solve:

Replaced x

Retained y

Total 16 L

100%

30%

50%

100%x

30%y

50%×16, or 8 L

x+

−49x − 49y = −441 49x + 51y = 447 2y = 6 y=3

Back-substitute to find x. x+3 = 9

x + y = 16,

x=6

4 L 7

Then in the city the car is driven 49(6), or 294 mi; on the highway it is driven 51(3), or 153 mi. Check. The number of gallons of gasoline used is 6 + 3, or 9. The number of miles driven is 294 + 153 = 447. The answer checks.

81. Substitute the given solutions in the equation Ax + By = 1 to get a system of equations. 3A − B = 1, (1)

−4A − 2B = 1

(1)

Multiply equation (1) by −49 and add.

x + 0.3y = 8. x=4

y = 9,

49x + 51y = 447. (2)

State. The car was driven 294 mi in the city and 153 mi on the highway.

(2)

Multiply equation (1) by −2 and add. −6A + 2B = −2

−4A − 2B = 1 −10A = −1 1 A = 10 Back-substitute to find B. ! " 1 −B = 1 Using equation (1) 3 10 3 −B = 1 10 7 −B = 10 7 B=− 10 7 1 We have A = and B = − . 10 10

84. First we convert the given distances to miles: 5 300 mi = mi, 300 ft = 5280 88 25 500 mi = mi 500 ft = 5280 264

5 88 , or Then at 10 mph, Heather can run to point P in 10 25 5 1 hr, and she can run to point Q in 264 , or hr 176 10 528 d (using d = rt, or = t). r Let d = the distance, in miles, from the train to point P in the drawing in the text, and let r the speed of the train, in miles per hour.

x + 1 + y = 3y. x = 5 people

x + y = 9. A second equation comes from the fact that the car is driven 447 mile: 49x + 51y = 447.

Time

d

r

1 176

r " ! 1 Solve: d=r , 176 ! " 25 5 5 d+ + =r . 88 264 528 r = 40 mph

5 528

Going to Q

Solve: x = 2 + y,

Translate. The fact that 9 gal of gasoline were used gives us one equation:

Speed

Going to P

82. Let x = the number of people ahead of you and y = the number of people behind you.

83. Familiarize. Let x and y represent the number of gallons of gasoline used in city driving and in highway driving, respectively. Then 49x and 51y represent the number of miles driven in the city and on the highway, respectively.

Distance d+

5 88

+

25 264

Exercise Set 8.2 1.

x + y + z = 2,

(1)

6x − 4y + 5z = 31, (2) 5x + 2y + 2z = 13

(3)

Multiply equation (1) by −6 and add it to equation (2). We also multiply equation (1) by −5 and add it to equation (3).

Exercise Set 8.2 x+

483

y+ z=2

(1)

− 10y − z = 19

x + 6(−1) + 3 · 4 = 4

(4)

− 3y − 3z = 3

x = −2

(5)

Multiply the last equation by 10 to make the y-coefficient a multiple of the y-coefficient in equation (4). x+

y+

z=2

(1)

z = 19

(4)

− 30y − 30z = 30

(6)

− 10y −

The solution is (−2, −1, 4). 3.

y+

− 10y −

z=2

(1)

z = 19

(4)

− 27z = −27

(7)

(2) (3)

(1)

3y + z = 7

(4)

3y − 3z = 3

(5)

Multiply equation (4) by −1 and add it to equation (5).

−27z = −27 z=1

x − y + 2z = −3

(1)

− 4z = −4

(6)

3y + z = 7

Back-substitute 1 for z in equation (4) and solve for y. −10y − 1 = 19

(4)

Solve equation (6) for z.

−10y = 20

−4z = −4

y = −2

z=1

Back-substitute 1 for z for −2 and y in equation (1) and solve for x. x + (−2) + 1 = 2

Back-substitute 1 for z in equation (4) and solve for y. 3y + 1 = 7 3y = 6

x−1 = 2

y=2

x=3

Back-substitute 1 for z and 2 for y in equation (1) and solve for x. x − 2 + 2 · 1 = −3

The solution is (3, −2, 1). x + 6y + 3z = 4, (1) 2x + y + 2z = 3, (2) 3x − 2y + z = 0

x = −3

(3)

The solution is (−3, 2, 1).

Multiply equation (1) by −2 and add it to equation (2). We also multiply equation (1) by −3 and add it to equation (3). x + 6y + 3z = 4, − 11y − 4z = −5,

− 20y − 8z = −12

(1) (4) (5)

Multiply equation (5) by 11. x+

x + 2y + 3z = 4

2x + y + z = −3

x − y + 2z = −3

Solve equation (7) for z.

2.

(1)

Multiply equation (1) by −1 and add it to equation (2). We also multiply equation (1) by −2 and add it to equation (3).

Multiply equation (4) by −3 and add it to equation (6). x+

x − y + 2z = −3

6y + 3z = 4,

(1)

− 11y − 4z = −5,

(4)

− 220y − 88z = −132

(6)

Multiply equation (4) by −20 and add it to equation (6). x + 6y + 3z = 4,

− 11y − 4z = −5,

− 8z = −32

Complete the solution. −8z = −32 z=4

−11y − 4 · 4 = −5 −11y = 11

y = −1

(1) (4) (7)

4.

x + y + z = 6,

(1)

2x − y − z = −3, (2) x − 2y + 3z = 6

(3)

x + y + z = 6,

(1)

Multiply equation (1) by −2 and add it to equation (2). Also, multiply equation (1) by −1 and add it to equation (3). − 3y − 3z = −15, (4)

− 3y + 2z = 0

(5)

x + y + z = 6,

(1)

Multiply equation (4) by −1 and add it to equation (5). − 3y − 3z = −15, (4) 5z = 15

Complete the solution. 5z = 15 z=3 −3y − 3 · 3 = −15 −3y = −6 y=2

(6)

484

Chapter 8: Systems of Equations and Matrices x + 2y + 5z = 4,

x+2+3 = 6

− 7y − 23z = −19

The solution is (1, 2, 3). 5.

x + 2y + 5z = 4,

2x − 4y + z = 0, (2)

(3)

54z = 30

54z = 30 5 z= 9

(1)

5 = −7 9 8 −y = − 9 8 y= 9 5 8 x+2· +5· = 4 9 9

(5)

−y − 11 ·

Multiply equation (5) by 2 to make the y-coefficient a multiple of the y-coefficient of equation (4). x + 2y −

z = 5,

(1)

− 8y + 3z = −10, (4) − 8y + 10z = −24

(6)

Multiply equation (4) by −1 and add it to equation (6). x + 2y − z = 5,

5 9 ! " 5 8 5 The solution is − , , . 9 9 9

(1)

x=−

− 8y + 3z = −10, (4) 7z = −14

(7)

Solve equation (7) for z. 7z = −14

7.

z = −2

8x + y + z = 2

Back-substitute −2 for z in equation (4) and solve for y. −8y − 6 = −10

6.

2x + 3y − z = 1,

(1)

x + 2y + 5z = 4,

(2)

3x − y − 8z = −7

(3)

x + 2y + 5z = 4,

(2)

2x + 3y − z = 1,

(1)

Interchange equations (1) and (2).

3x − y − 8z = −7

(3)

Multiply equation (2) by −2 and add it to equation (1). Multiply equation (2) by −3 and add it to equation (3).

(3)

x + 2y − z = −8, (1)

−8y = −4 1 y= 2 1 Back-substitute for y and −2 for z in equation (1) and 2 solve for x. 1 x + 2 · − (−2) = 5 2 x+1+2 = 5 x=2 " 1 2, , −2 . 2

(2)

Multiply equation (1) by −2 and add it to equation (2). Also, multiply equation (1) by −8 and add it to equation (3).

−8y + 3(−2) = −10

The solution is

x + 2y − z = −8, (1)

2x − y + z = 4,

!

(4)

Complete the solution.

− 8y + 3z = −10, (4) − 4y + 5z = −12

(5) (2)

− y − 11z = −7

Multiply equation (1) by −2 and add it to equation (2). Also, multiply equation (1) by −3 and add it to equation (3). x + 2y − z = 5,

(4)

Multiply equation (4) by −7 and add it to equation (5).

x + 2y − z = 5, (1)

3x + 2y + 2z = 3

(2)

− y − 11z = −7,

x=1

− 5y + 3z = 20,

− 15y + 9z = 66

(4) (5)

Multiply equation (4) by −3 and add it to equation (5). x + 2y − z = −8, (1) − 5y + 3z = 20, 0=6

(4) (6)

Equation (6) is false, so the system of equations has no solution. 8.

x + 2y − z = 4,

4x − 3y + z = 8, 5x − y

= 12

(1) (2) (3)

Multiply equation (1) by −4 and add it to equation (2). Also, multiply equation (1) by −5 and add it to equation (3). x + 2y − z = 4,

(1)

− 11y + 5z = −8, (4) − 11y + 5z = −8

(5)

Multiply equation (4) by −1 and add it to equation (5).

Exercise Set 8.2

485

x + 2y − z = 4,

(1)

0=0

(6)

9y − 5z = −11

−5z = −9y − 11 9y + 11 z= 5 Back-substitute in equation (2) to find an expression for x in terms of y. 9y + 11 =6 x − 4y + 5 11 9 =6 x − 4y + y + 5 5 11 19 11y + 19 x= y+ = 5 5 5 ! " 11y + 19 9y + 11 The solutions are given by , y, , where 5 5 y is any real number.

− 11y + 5z = −8, (4)

The equation 0 = 0 tells us that equation (3) of the original system is dependent on the first two equations. The system of equations has infinitely many solutions and is equivalent to x + 2y − z = 4, (1) 4x − 3y + z = 8. (2)

To find an expression for the solutions, we first solve equation (4) for either y or z. We choose to solve for y. −11y + 5z = −8

−11y = −5z − 8 5z + 8 y= 11 Back-substitute in equation (1) to find an expression for x in terms of z. " ! 5z + 8 −z = 4 x+2 11 10z 16 x+ + −z = 4 11 11 28 z + 28 z x= + = 11 11 11 ! " z + 28 5z + 8 , , z , where z The solutions are given by 11 11 is any real number. 9.

2x + y − 3z = 1,

(1)

4x − 7y − z = 13

(3)

x − 4y + z = 6,

(2)

4x − 7y − z = 13

(3)

x − 4y + z = 6,

10.

x + 8y + 11z = 2

x + 3y + 4z = 1, (1) − 5y − 7z = 0, (4) 5y + 7z = 1

Multiply equation (2) by −2 and add it to equation (1). Also, multiply equation (2) by −4 and add it to equation (3). x − 4y + z = 6,

(2)

9y − 5z = −11, (4)

9y − 5z = −11

(5)

Multiply equation (4) by −1 and add it to equation (5). x − 4y + z = 6,

(1)

0=0

(6)

9y − 5z = −11, (4)

The equation 0 = 0 tells us that equation (3) of the original system is dependent on the first two equations. The system of equations has infinitely many solutions and is equivalent to 2x + y − 3z = 1, (1) x − 4y + z = 6. (2)

To find an expression for the solutions, we first solve equation (4) for either y or z. We choose to solve for z.

(5)

Add equation (4) to equation (5). x + 3y + 4z = 1, (1) − 5y − 7z = 0, (4)

(2)

(1)

(3)

Multiply equation (1) by −3 and add it to equation (2). Also, multiply equation (1) by −1 and add it to equation (3).

0=1

(6)

Equation (6) is false, so the system of equations has no solution.

Interchange equations (1) and (2). 2x + y − 3z = 1,

x + 3y + 4z = 1, (1) 3x + 4y + 5z = 3, (2)

11.

4a + 9b 8a

= 8,

(1)

+ 6c = −1, (2)

6b + 6c = −1

(3)

Multiply equation (1) by −2 and add it to equation (2). 4a + 9b

= 8,

(1)

− 18b + 6c = −17, (4) 6b + 6c = −1

(3)

Multiply equation (3) by 3 to make the b-coefficient a multiple of the b-coefficient in equation (4). 4a + 9b

= 8,

(1)

− 18b + 6c = −17, (4) 18b + 18c = −3

(5)

Add equation (4) to equation (5). 4a + 9b

= 8,

(1)

− 18b + 6c = −17, (4) 24c = −20

Solve equation (6) for c. 24c = −20 5 20 =− c=− 24 6

(6)

486

Chapter 8: Systems of Equations and Matrices 5 for c in equation (4) and solve for b. 6 −18b + 6c = −17 " ! 5 = −17 −18b + 6 − 6 −18b − 5 = −17

Interchange equations (1) and (3).

Back-substitute −

x − 2y 2x

The solution is 12.

3p

!

3y − 2z = 6,

4y + z = 19

x − 2y

12y + 3z = 57

3p

+ 2r = 11

11z = 33

z=3 3y − 2z = 6

3y − 2 · 3 = 6 3y − 6 = 6

3y = 12

(1)

y=4 Back-substitute 4 for y in equation (3) and solve for x.

= 1, (3)

x − 2y = −9

x − 2 · 4 = −9

(4)

x − 8 = −9

Multiply equation (2) by −18 and add it to equation (4). p − 6q

q−

= 1,

(3)

7r = 4,

(2)

128r = −64

(5)

Complete the solution.

128r = −64 1 r=− 2 ! " 1 q−7 − =4 2 1 q= 2 1 p−6· = 1 2 p=4 " ! 1 1 The solution is 4, , − . 2 2 13.

2x

+ z = 1,

(1)

3y − 2z = 6,

(2)

x − 2y

= −9

(6)

Back-substitute 3 for z in equation (2) and solve for y.

q − 7r = 4, (2)

18q + 2r = 8

(2)

11z = 33

Multiply equation (3) by −3 and add it to equation (1). p − 6q

(5)

Solve equation (6) for z.

(3)

(2)

(2)

= −9, (3)

3y − 2z = 6,

Interchange equations (1) and (3). (3)

(4)

Multiply equation (2) by −4 and add it to equation (5).

(2)

= 1,

(2)

= −9, (3)

3y − 2z = 6,

1 2 5 , ,− . 2 3 6

q − 7r = 4,

(1)

Multiply equation (4) by 3 to make the y-coefficient a multiple of the y-coefficient in equation (2).

"

=1

(2)

= −9, (3)

x − 2y

q − 7r = 4,

p − 6q

+ z=1

x − 2y

+ 2r = 11, (1)

p − 6q

3y − 2z = 6,

Multiply equation (3) by −2 and add it to equation (1).

−18b = −12 2 12 = b= 18 3 2 Back-substitute for b in equation (1) and solve for a. 3 4a + 9b = 8 2 4a + 9 · = 8 3 4a + 6 = 8 4a = 2 1 a= 2

= −9, (3)

(3)

x = −1

The solution is (−1, 4, 3). 14.

3x x − 2y

+ 4z = −11, (1) = 5,

4y − z = −10

(2) (3)

Interchange equations (1) and (2). x − 2y

3x

= 5,

(2)

+ 4z = −11, (1)

4y − z = −10

(3)

Multiply equation (2) by −3 and add it to equation (1). x − 2y

= 5,

(2)

6y + 4z = −26, (4) 4y − z = −10

(3)

Multiply equation (3) by 3 to make the y-coefficient a multiple of the y-coefficient in equation (4). x − 2y

= 5,

(2)

6y + 4z = −26, (4)

12y − 3z = −30

(5)

Exercise Set 8.2

487

Multiply equation (4) by −2 and add it to equation (5). x − 2y

= 5,

(2)

6y + 4z = −26, (4) − 11z = 22

16.

w + x − y + z = 0,

−w + 2x + 2y + z = 5,

(2)

−2w + x + y − 3z = −7

(4)

−w + 3x + y − z = −4, (3)

(6)

Solve equation (6) for z.

Add equation (1) to equation (2) and equation (3). Also, multiply equation (1) by 2 and add it to equation (4).

−11z = 22

w + x − y + z = 0,

z = −2

3x + y + 2z = 5,

Back-substitute −2 for z in equation (4) and solve for y.

4x

6y + 4z = −26

w+

6y = −18

Back-substitute −3 for y in equation (2) and solve for x. x − 2y = 5 x+6 = 5

x = −1

w+ x+ y+ z=2 w + 2x + 2y + 4z = 1

(2) (3)

−w + 3x + y − z = −2

(4)

Multiply equation (1) by −1 and add to equation (2). Add equation (1) to equation (3) and to equation (4).

=0

(1) (5)

= −12, (8)

3x − y − z = −7

3x + y + 2z = 5,

(1)

−w + x − y − z = −6

4x + 2y

12x

w + x − y + z = 0,

The solution is (−1, −3, −2).

= −4

x − y + z = 0,

(7)

Multiply equation (5) by −4 and add it to equation (8). Also, multiply equation (5) by −1 and add it to equation (7).

x − 2(−3) = 5

2x

(7)

3x + y + 2z = 5,

y = −3

x + y + 3z = −1

(5)

Multiply equation (6) by 3.

6y − 8 = −26

w+ x+ y+ z=2

(1)

= −4, (6)

3x − y − z = −7

6y + 4(−2) = −26

15.

(1)

(1) (5) (6) (7)

Solve equation (6) for x. 2x = −4

(5)

− 4y − 8z = −32, (9) − 2y − 3z = −12

(10)

Multiply equation (10) by 2. w + x − y + z = 0, 3x + y + 2z = 5,

(1) (5)

− 4y − 8z = −32, (9) − 4y − 6z = −24

(11)

Multiply equation (9) by −1 and add it to equation (11). w + x − y + z = 0, 3x + y + 2z = 5,

(1) (5)

− 4y − 8z = −32, (9) 2z = 8

x = −2

(1)

(12)

Complete the solution.

Back-substitute −2 for x in equation (7) and solve for y. 4(−2) + 2y = 0 −8 + 2y = 0 2y = 8 y=4 Back-substitute −2 for x and 4 for y in equation (5) and solve for z. −2 + 4 + 3z = −1 3z = −3 z = −1

Back-substitute −2 for x, 4 for y, and −1 for z in equation (1) and solve for w. w−2+4−1 = 2

w=1

The solution is (1, −2, 4, −1).

2z = 8 z=4 −4y − 8 · 4 = −32 −4y = 0 y=0

3x + 0 + 2 · 4 = 5

3x = −3 x = −1

w−1−0+4 = 0

w = −3

The solution is (−3, −1, 0, 4). 17. Familiarize. Let c, r, and w represent the amount spent on cross-training shoes, running shoes, and walking shoes in 2004, respectively, in billions of dollars.

488

Chapter 8: Systems of Equations and Matrices Translate. Total sales were $6.7 billion. c + r + w = 6.7 The amount spent on walking shoes was $0.3 billion more than the total amount spent on cross-training and running shoes. w = c + r + 0.3 The amount spent on running shoes was $0.4 billion more than the amount spent on cross-training shoes. We have a system of equations c + r + w = 6.7,

w = c + r + 0.3, or −c − r + w = 0.3, r = c + 0.4

−c + r

= 0.4

Carry out. Solving the system of equations, we get (1.4, 1.8, 3.5). Check. Total sales were 1.4 + 1.8 + 3.5, or $6.7 billion. Also, $3.5 billion is $0.3 billion more than the total of $1.4 billion and $1.8 billion, and $1.8 billion is $0.4 billion more than $1.4 billion. The answer checks. State. $1.4 billion was spent on cross-training shoes, $1.8 billion was spent on running shoes, and $3.5 billion was spent on walking shoes. 18. Let x, y, and z represent the amounts spent on infant/preschool toys, dolls, and games/puzzles, respectively, in billions of dollars. Solve: x + y + z = 8.2, y + z = x + 2, z = y − 0.3

x = $3.1 billion, y = $2.7 billion, z = $2.4 billion 19. Familiarize. Let x y, and z represents the number of milligrams of caffeine in an 8-oz serving of brewed coffee, Red Bull energy drink, and Mountain Dew, respectively. Translate. The total amount of caffeine in one serving of each beverage is 197 mg. x + y + z = 197 One serving of brewed coffee has 6 mg more caffeine than two servings of Mountain Dew. x = 2z + 6 One serving of Red Bull contains 37 mg less caffeine than one serving each of brewed coffee and Mountain Dew. y = x + z − 37

We have a system of equations x + y + z = 197, x = 2z + 6, y = x + z − 37

Carry out. (80, 80, 37).

x + y + z = 197, or

x

20. Let x, y, and z represent the number of adults who do spring housecleaning in March, April, and May, respectively. Solve: x + y + z = 70, y = x + z + 14, y + z = 3x + 2 x = 17, y = 42, z = 11

r = c + 0.4 c + r + w = 6.7,

State. One serving each of brewed coffee, Red Bull energy drink, and Mountain Dew contains 80 mg, 80 mg, and 37 mg of caffeine, respectively.

− 2z = 6,

−x + y − z = −37

Solving the system of equations, we get

Check. The total amount of caffeine is 80 + 80 + 37, or 197 mg. Also, 80 mg is 6 mg more than twice 37 mg, and 80 mg is 37 mg less than the total of 80 mg and 37 mg, or 117 mg. The answer checks.

21. Familiarize. Let x, y, and z represent the number of grams of carbohydrates in 1 cup of raw lettuce, 6 raw asparagus spears, and 1 cup of raw tomatoes, respectively. Translate. Together, 1 cup of raw lettuce, 6 raw asparagus spears, and 1 cup of raw tomatoes contain 12 grams of carbohydrates. x + y + z = 12 One cup of raw lettuce and 6 raw asparagus spears have one-half the carbohydrates of 1 cup of raw tomatoes. 1 x+y = z 2 One cup each of raw lettuce and raw tomatoes have 3 times the carbohydrates of 6 raw asparagus spears. x + z = 3y We have x + y + z = 12, 1 x + y = z, 2 x + z = 3y or x + y + z = 12, 1 x + y − z = 0, 2 x − 3y + z = 0. Carry out. Solving the system of equations, we get (1, 3, 8). Check. The total carbohydrate content is 1+3+8, or 12 g. One cup of raw lettuce and 6 raw asparagus spears contain 1 + 3, or 4 g, of carbohydrates. This is one-half of 8 g, the carbohydrate content of 1 cup of raw tomatoes. One cup each of raw lettuce and raw tomatoes contain 1 + 8, or 9 g, of carbohydrates. This is 3 times the 3 g of carbohydrates in 6 raw asparagus spears. The solution checks. State. One cup of raw lettuce contains 1 g of carbohydrates, 6 raw asparagus spears contain 3 g of carbohydrates, and 1 cup of raw tomatoes contains 8 g of carbohydrates. 22. Let b, s, and v represent the numbers of Americans who spent leisure time reading books, surfing the Internet, and playing video games, respectively, in 2004, in millions. Solve: b + s + v = 162.2, b = v + 49.9, s + v = b + 7.2 b = 77.5 million, s = 57.1 million, v = 27.6 million

Exercise Set 8.2

489

23. Familiarize. Let x = the number of orders under 10 lb, y = the number of orders from 10 lb up to 15 lb, and z = the number of orders of 15 lb or more. Then the total shipping charges for each category of order are $3x, $5y, and $7.50z. Translate. The total number of orders was 150. x + y + z = 150 Total shipping charges were $680. 3x + 5y + 7.5z = 680 The number of orders under 10 lb was three times the number of orders weighing 15 lb or more. x = 3z x+

y+

z = 150,

3x + 5y + 7.5z = 680, or 30x + 50y + 75z = 6800, x = 3z

x

Carry out. (60, 70, 20).

34y + 10z = 41.5, 9x +

Check. 1.25 servings of ground beef contains 306.25 Cal, no carbohydrates, and 11.25 mg of calcium; 1 baked potato contains 145 Cal, 34 g of carbohydrates, and 8 mg of calcium; 0.75 servings of strawberries contains 33.75 Cal, 7.5 g of carbohydrates, and 15.75 mg of calcium. Thus, there are a total of 306.25 + 145 + 33.75, or 485 Cal, 34 + 7.5, or 41.5 g of carbohydrates, and 11.25 + 8 + 15.75, or 35 mg of calcium. The solution checks.

− 3z = 0

26. Let x, y, and z represent the number of servings of chicken, mashed potatoes, and peas to be used, respectively. Solve: 140x + 160y + 125z = 415, 27x +

Solving the system of equations, we get

Check. The total number of orders is 60 + 70 + 20, or 150. The total shipping charges are $3 · 60 + $5 · 70 + $7.50(20), or $680. The number of orders under 10 lb, 60, is three times the number of orders weighing over 15 lb, 20. The solution checks. State. There were 60 packages under 10 lb, 70 packages from 10 lb up to 15 lb, and 20 packages weighing 15 lb or more. 24. Let x, y, and z represent the number of orders of $25 or less, from $25.01 to $75, and over $75, respectively. Solve: x + y + z = 600, 4x + 6y + 7z = 3340, x = z + 80. x = 180, y = 320, z = 100 25. Familiarize. Let x, y, and z represent the number of servings of ground beef, baked potato, and strawberries required, respectively. One serving of ground beef contains 245x Cal, 0x or 0 g of carbohydrates, and 9x mg of calcium. One baked potato contains 145y Cal, 34y g of carbohydrates, and 8y mg of calcium. One serving of strawberries contains 45z Cal, 10z g of carbohydrates, and 21z mg of calcium. Translate. The total number of calories is 485. 245x + 145y + 45z = 485 A total of 41.5 g of carbohydrates is required. 34y + 10z = 41.5 A total of 35 mg of calcium is required. 9x + 8y + 21z = 35 We have a system of equations.

8y + 21z = 35

Carry out. Solving the system of equations, we get (1.25, 1, 0.75).

State. 1.25 servings of ground beef, 1 baked potato, and 0.75 serving of strawberries are required.

We have a system of equations x + y + z = 150,

245x + 145y + 45z = 485,

4y +

8z = 50.5,

64x + 636y + 139z = 553. x = 1.5, y = 0.5, z = 1 27. Familiarize. Let x, y, and z represent the amounts invested at 3%, 4%, and 6%, respectively. Then the annual interest from the investments is 3%x, 4%y, and 6%z, or 0.03x, 0.04y, and 0.06z. Translate. A total of $5000 was invested. x + y + z = 5000 The total interest is $243. 0.03x + 0.04y + 0.06z = 243 The amount invested at 6% is $1500 more than the amount invested at 3%. z = x + 1500 We have a system of equations. x + y + z = 5000, 0.03x + 0.04y + 0.06z = 243, z = x + 1500 or x + y + z = 5000, 3x + 4y + 6z = 24, 300, −x

+ z = 1500

Carry out. Solving the system of equations, we get (1300, 900, 2800). Check. The total investment was $1300+$900+$2800, or $5000. The total interest was 0.03($1300) + 0.04($900) + 0.06($2800) = $39 + $36 + $168, or $243. The amount invested at 6%, $2800, is $1500 more than the amount invested at 3%, $1300. The solution checks. State. $1300 was invested at 3%, $900 at 4%, and $2800 at 6%.

490

Chapter 8: Systems of Equations and Matrices

28. Let x, y, and z represent the amounts invested at 2%, 3%, and 4%, respectively. Solve: 0.02x + 0.03y + 0.04z = 126,

A golfer who shoots par on every hole has a score of 72. 3x + 4y + 5z = 72

y = x + 500,

The sum of the number of par-3 holes and the number of par-5 holes is 8.

z = 3y.

x+z =8

x = $300, y = $800, z = $2400 29. Familiarize. Let x, y, and z represent the prices of orange juice, a raisin bagel, and a cup of coffee, respectively. The new price for orange juice is x + 50%x, or x + 0.5x, or 1.5x; the new price of a bagel is y + 20%y, or y + 0.2y, or 1.2y. Translate. Orange juice, a raisin bagel, and a cup of coffee cost $4.10. x + y + z = 4.1 After the price increase, orange juice, a raisin bagel, and a cup of coffee will cost $5.10. 1.5x + 1.2y + z = 5.1 After the price increases, orange juice will cost twice as much as coffee. 1.5x = 2z We have a system of equations. x + y + z = 4.1,

10x + 10y + 10z = 41,

1.5x+1.2y+z = 5.1, or 15x + 12y + 10z = 51, 1.5x = 2z Carry out. (1.2, 2, 0.9).

15x

− 20z = 0

Solving the system of equations, we get

Check. If orange juice costs $1.20, a bagel costs $2, and a cup of coffee costs $0.90, then together they cost $1.20 + $2 + $0.90, or $4.10. After the price increases orange juice will cost 1.5($1.20), or $1.80 and a bagel will cost 1.2($2) or $2.40. Then orange juice, a bagel, and coffee will cost $1.80 + $2.40 + $0.90, or $5.10. After the price increase the price of orange juice, $1.80, will be twice the price of coffee, $0.90. The solution checks. State. Before the increase orange juice cost $1.20, a raisin bagel cost $2, and a cup of coffee cost $0.90. 30. Let x, y, and z represent the prices of a carton of milk, a donut, and a cup of coffee, respectively. Solve: x + 2y + z = 5, 3y + 2z = 5.50, x + y + 2z = 5.25. x = $1.75, y = $1, z = $1.25 Then 2 cartons of milk and 2 donuts will cost 2($1.75) + 2($1), or $5.50. They will not have enough money. They need 5/c more. 31. Familiarize. Let x, y, and z represent the number of par3, par-4, and par-5 holes, respectively. A golfer who shoots par on every hole has a score of 3x from the par-3 holes, 4y from the par-4 holes, and 5z from the par-5 holes. Translate. The total number of holes is 18. x + y + z = 18

We have a system of equations. x + y + z = 18, 3x + 4y + 5z = 72, + z=8

x

Carry out. We solve the system. The solution is (4, 10, 4). Check. The total number of holes is 4 + 10 + 4, or 18. A golfer who shoots par on every hole has a score of 3 · 4 + 4 · 10 + 5 · 4, or 72. The sum of the number of par-3 holes and the number of par-5 holes is 4 + 4, or 8. The solution checks. State. There are 4 par-3 holes, 10 par-4 holes, and 4 par-5 holes. 32. Let x, y, and z represent the number of par-3, par-4, and par-5 holes, respectively. Solve: x + y + z = 18, 3x + 4y + 5z = 70, y = 2z. x = 6, y = 8, z = 4 33. a) Substitute the data points (0, 279), (2, 9741), and (3, 4459) in the function f (x) = ax2 + bx + c. 279 = a · 02 + b · 0 + c

9741 = a · 22 + b · 2 + c 4459 = a · 32 + b · 3 + c

We have a system of equations. c = 279, 4a + 2b + c = 9741, 9a + 3b + c = 4459 Solving the system of equations, we get " ! 10, 013 34, 219 , , 279 , so − 3 3 10, 013 2 34, 219 f (x) = − x + x + 279, where x is the 3 3 number of years after 2002. b) In 2003, x = 1. 10, 013 2 34, 219 ·1 + · 1 + 279 ≈ 8348 fines f (1) = − 3 3 34. a) Solve: 22 = a · 02 + b · 0 + c, 25 = a · 52 + b · 5 + c,

or

22 = a · 82 + b · 8 + c, c = 22,

25a + 5b + c = 25, 64a + 8b + c = 22. 8 1 8 1 a = − , b = , c = 22, so f (x) = − x2 + x + 22, 5 5 5 5 where x is the number of years after 1995.

Exercise Set 8.2

491

1 2 8 b) f (9) = − · 9 + · 9 + 22 ≈ 20% 5 5 35. a) Substitute the data points (0, 36), (10, 42), and (14, 36) in the function f (x) = ax2 + bx + c. 36 = a · 02 + b · 0 + c

42 = a · 102 + b · 10 + c 36 = a · 142 + b · 14 + c

We have a system of equations. c = 36, 100a + 10b + c = 42, 196a + 14b + c = 36 Solving the system of equations, we get (−0.15, 2.1, 36), so f (x) = −0.15x2 + 2.1x + 36, where f (x) is in thousands and x is the number of years after 1990. b) In 2008, x = 2008 − 1990, or 18.

f (18) = −0.15(18)2 + 2.1(18) + 36 ≈ 25

There were about 25 thousand marriages in Arkansas in 2008.

36. a) Solve: 112 = a · 02 + b · 0 + c,

114 = a · 102 + b · 10 + c,

or

112 = a · 132 + b · 13 + c, c = 112,

100a + 10b + c = 114, 169a + 13b + c = 112. 13 1 ,b= , c = 112, so 15 15 13 1 f (x) = − x2 + x + 112, where x is the number 15 15 of years after 1990. 13 1 2 · 20 + 112 ≈ 103 lb b) f (20) = − · 20 + 15 15 a=−

37. a) Using the quadratic regression feature on a graphing calculator, we have f (x) = 0.1822373078x2 − 11.23256978x + 468.7226133, where x is the number of years after 1920. b) In 2008, x = 2008 − 1920, or 88; f (88) ≈ 892, so we estimate that there will be about 892 morning newspapers in 2008. 38. a) f (x) = 0.0187903168x2 − 0.1413997864x + 7.747733001, where x is the number of years after 1995 and f (x) is in millions. b) f (10) ≈ 8.213, so we estimate that there were about 8.213 million children enrolled in preprimary school in 2005. 39. Add a non-zero multiple of one equation to a non-zero multiple of the other equation, where the multiples are not opposites. 40. Answers will vary.

41. Perpendicular 42. The leading-term test 43. A vertical line 44. A one-to-one function 45. A rational function 46. Inverse variation 47. A vertical asymptote 48. A horizontal asymptote 2 1 3 49. − − = −1, x y z 2 1 1 − + = −9, x y z 2 4 1 + − = 17 x y z 1 1 1 First substitute u for , v for , and w for and solve x y z for u, v, and w. 2u − v − 3w = −1,

2u − v + w = −9, u + 2v − 4w = 17

Solving this system we get (−1, 5, −2). 1 1 If u = −1, and u = , then −1 = , or x = −1. x x 1 1 1 If v = 5 and v = , then 5 = , or y = . y y 5 1 1 1 If w = −2 and w = , then −2 = , or z = − . z z 2 " ! 1 1 The solution of the original system is − 1, − . 5 2 2 3 2 50. + − = 3, x y z 1 2 3 − − = 9, x y z 2 9 7 − + = −39 x y z 1 1 1 Substitute u for , v for , and w for and solve for u, x y z v, and w. 2u + 2v − 3w = 3, u − 2v − 3w = 9,

7u − 2v + 9w = −39

Solving this system we get (−2, −1, −3). 1 1 1 = −2, or x = − ; = −1, or y = −1; and Then x 2 y 1 1 = −3, or z = − . The solution of the original system is z 3 " ! 1 1 − , −1, − . 2 3

492

Chapter 8: Systems of Equations and Matrices

51. Label the angle measures at the tips of the stars a, b, c, d, and e. Also label the angles of the pentagon p, q, r, s, and t. A a

53. Substituting, we get 3 A + B + 3C = 12, 4 4 A + B + 2C = 12, 3 2A + B + C = 12, or 4A + 3B + 12C = 48,

E

p

e

q

t

b

4A + 3B + 6C = 36, Clearing fractions

B

2A + B +

Solving the system of equations, we get (3, 4, 2). The equation is 3x + 4y + 2z = 12.

r s

d

54.

C

Using the geometric fact that the sum of the angle measures of a triangle is 180◦ , we get 5 equations. p + b + d = 180

55. Substituting, we get 59 = a(−2)3 + b(−2)2 + c(−2) + d,

q + c + e = 180

13 = a(−1)3 + b(−1)2 + c(−1) + d,

r + a + d = 180

−1 = a · 13 + b · 12 + c · 1 + d,

s + b + e = 180

−17 = a · 23 + b · 22 + c · 2 + d, or

t + a + c = 180

−8a + 4b − 2c + d = 59,

Adding these equations, we get

−a + b − c + d = 13,

(p + q + r + s + t) + 2a + 2b + 2c + 2d + 2e = 5(180).

a + b + c + d = −1,

The sum of the angle measures of any convex polygon with n sides is given by the formula S = (n − 2)180. Thus p + q + r + s + t = (5 − 2)180, or 540. We substitute and solve for a + b + c + d + e. 540 + 2(a + b + c + d + e) = 900 2(a + b + c + d + e) = 360 a + b + c + d + e = 180 The sum of the angle measures at the tips of the star is 180◦ . 52. Let h, t, and u represent the hundred’s, ten’s, and unit’s digit of the year, respectively. (We know the thousand’s digit is 1.) Using the given information we know the following: 1 + h + t + u = 24, u = 1 + h, t = k · 3 and u = m · 3

where k, m are positive integers. We know h > 5 (there was no transcontinental railroad before 1600); and, since u = 1 + h = m · 3, h $= 6, h $= 7, h $= 9.

Thus h = 8 and u = 9. Then 1 + 8 + t + 9 = 24, or t = 6.

Solve: 1 = B − M − 2N,

2 = B − 3M + 6N, 3 1=B− M −N 2 1 1 1 1 B = 2, M = , N = , so y = 2 − x − z. 2 4 2 4

c

D

The year is 1869.

C = 12.

8a + 4b + 2c + d = −17.

Solving the system of equations, we get (−4, 5, −3, 1), so y = −4x3 + 5x2 − 3x + 1. 56.

Solve: −39 = a(−2)3 + b(−2)2 + c(−2) + d, −12 = a(−1)3 + b(−1)2 + c(−1) + d, −6 = a · 13 + b · 12 + c · 1 + d,

16 = a · 33 + b · 32 + c · 3 + d, or

−8a + 4b − 2c + d = −39, −a + b − c + d = −12, a + b + c + d = −6,

27a + 9b + 3c + d = 16.

a = 2, b = −4, c = 1, d = −5, so y = 2x3 − 4x2 + x − 5. 57. Familiarize and Translate. Let a, s, and c represent the number of adults, students, and children in attendance, respectively. The total attendance was 100. a + s + c = 100 The total amount of money taken in was $100. 1 (Express 50 cents as dollar.) 2 1 10a + 3s + c = 100 2 The resulting system is

Exercise Set 8.3

493

a+ s+

c = 100, 1 10a + 3s + c = 100. 2 Carry out. Multiply the first equation by −3 and add it to the second equation to obtain 5 5 (c − 80) where (c − 80) is a pos7a − c = −200 or a = 2 14 itive multiple of 14 (because a must be a positive integer). That is (c − 80) = k · 14 or c = 80 + k · 14, where k is a positive integer. If k > 1, then c > 100. This is impossible since the total attendance is 100. Thus k = 1, so 5 5 (94 − 80) = · 14 = 5, c = 80 + 1 · 14 = 94. Then a = 14 14 and 5 + s + 94 = 100, or s = 1.

12.

Check. The total attendance is 5 + 1 + 94, or 100.

15.

The total amount of money taken in was 1 $10 · 5 + $3 · 1 + $ · 94 = $100. The result checks. 2 State. There were 5 adults, 1 student, and 94 children in attendance.

Exercise Set 8.3 1. The matrix has 3 rows and 2 columns, so its order is 3 × 2. 2. The matrix has 4 rows and 1 column, so its order is 4 × 1. 3. The matrix has 1 row and 4 columns, so its order is 1 × 4. 4. The matrix has 1 row and 1 column, so its order is 1 × 1. 5. The matrix has 3 rows and 3 columns, so its order is 3 × 3. 6. The matrix has 2 rows and 4 columns, so its order is 2 × 4. 7. We omit the variables and replace the equals signs with a vertical line.   7 2 −1   1 4 −5 

8. 

3

8

2

2

−3 15

 

9. We omit the variables, writing zeros for the missing terms, and replace the equals signs with a vertical line.   1 −2 3 12   0 −4 8   2 0 3 1 7 

 10. 

1 0

1 −1 7 3

−2 −5



 2 1  0 6

11. Insert variables and replace the vertical line with equals signs. 3x − 5y = 1, x + 4y = −2

x + 2y = −6,

4x + y = −3

13. Insert variables and replace the vertical line with equals signs. 2x + y − 4z = 12,

3x 14.

+ 5z = −1,

x−y+ z=2

−x − 2y + 3z = 6, 4y + z = 2,

2x − y

=9

4x + 2y = 11, 3x − y = 2

Write the augmented matrix. We will use Gaussian elimination.   4 2 11   3 −1 2 Multiply row 2 by 4 to make the first number in row 2 a multiple of 4.   4 2 11   12 −4 8 Multiply row 1 by −3 and add it to row 2.   11 4 2   0 −10 −25 Multiply row 1 by    

1

1 2

0

1

1 1 and row 2 by − . 4 10

 11 4   5  2

Write the system of equations that corresponds to the last matrix. 11 1 , (1) x+ y = 2 4 5 (2) y= 2 Back-substitute in equation (1) and solve for x. 1 5 11 x+ · = 2 2 4 5 11 x+ = 4 4 3 6 x= = 4 2 " ! 3 5 , . The solution is 2 2

494 16.

Chapter 8: Systems of Equations and Matrices Write the system of equations that corresponds to the last matrix. 2 3 x− y = − , (1) 5 5 114 y=− (2) 29 Back-substitute in equation (1) and solve for x. ! " 114 3 2 − =− x− 5 29 5 228 3 x+ =− 145 5 63 315 =− x=− 145 29 " ! 63 114 . The solution is − , − 29 29

2x + y = 1, 3x + 2y = −2

Write the augmented matrix. We will use Gauss-Jordan elimination.   1 2 1   3 2 −2 Multiply row 2 by 2.   1 2 1   6 4 −4 Multiply row 1 by −3 and add it to row 2.   1 2 1   0 1 −7 Multiply row 2 by −1 and add it to row 1.   8 2 0   0 1 −7 Multiply row 1 by  

1

0

4

0

1 −7

1 . 2

 

The solution is (4, −7). 17.

5x − 2y = −3,

2x + 5y = −24

Write the augmented matrix. We will use Gaussian elimination.   5 −2 −3   2 5 −24 Multiply row 2 by 5 to make the first number in row 2 a multiple of 5.   −3 5 −2   10 25 −120 Multiply row 1 by −2 and add it to row 2.   −3 5 −2   0 29 −114 Multiply row 1 by    

1 −

2 5

0

1 −

1 1 and row 2 by . 5 29 

3 5   114  −

29

18.

2x + y = 1, 3x − 6y = 4

Write the augmented matrix. We will use Gaussian elimination.   2 1 1   3 −6 4 Multiply row 2 by 2.   2 1 1   6 −12 8 Multiply row 1 by −3 and add it to row 2.   2 1 1   0 −15 5 Multiply row 1 by    

1 0

1 1 and row 2 by − . 2 15

1 2

1  2    1 1 − 3

We have:

1 x + y = 1, 2

(1)

1 (2) 3 Back-substitute in equation (1) and solve for x. ! " 1 1 1 − = x+ 2 3 2 2 x= 3 " ! 2 1 ,− . The solution is 3 3 y=−

Exercise Set 8.3 19.

495 

3x + 4y = 7,



−5x + 2y = 10

Write the augmented matrix. We will use Gaussian elimination.   3 4 7   −5 2 10

   

  

1

4 3

0

1

Write the system of equations that corresponds to the last matrix. 7 4 x + y = , (1) 3 3 5 y= (2) 2 Back-substitute in equation (1) and solve for x. 4 5 7 x+ · = 3 2 3 7 10 = x+ 3 3 3 x = − = −1 3 " ! 5 . The solution is − 1, 2 20.

5x − 3y = −2, 4x + 2y = 5

Write the augmented matrix. We will use Gaussian elimination.   5 −3 −2   4 2 5 Multiply row 2 by 5.   5 −3 −2   20 10 25 Multiply row 1 by −4 and add it to row 2.

33



1

−

0

3 5

−

1

1 1 and row 2 by . 5 22

2  5    3 2

2 3 x − y = − , (1) 5 5 3 (2) y= 2 Back-substitute in (1) and solve for x. 2 3 3 x− · = − 5 2 5 2 9 =− x− 10 5 1 x= 2 ! " 1 3 The solution is , . 2 2

1 1 and row 2 by . 3 26

7  3    5 2

22

We have:

Multiply row 1 by 5 and add it to row 2.   7 3 4   0 26 65



0



Multiply row 1 by

Multiply row 2 by 3 to make the first number in row 2 a multiple of 3.   3 4 7   −15 6 30

Multiply row 1 by

5 −3 −2

21.

3x + 2y = 6, 2x − 3y = −9

Write the augmented matrix. We will use Gauss-Jordan elimination.   6 3 2   2 −3 −9 Multiply row 2 by 3 to make the first number in row 2 a multiple of 3.   6 3 2   6 −9 −27 Multiply row 1 by −2 and add it to row 2.   3 2 6   0 −13 −39 Multiply row 2 by −  

3 2

6

0 1 3

1 . 13

 

Multiply row 2 by −2 and add it to row 1.   3 0 0   0 1 3

496

Chapter 8: Systems of Equations and Matrices

Multiply row 1 by  

0

1

0

0

1 3

1 . 3

Interchange the rows.   −1 2 −3   4 −8 12

 

Multiply row 1 by 4 and add it to row 2.   −1 2 −3   0 0 0

We have x = 0, y = 3. The solution is (0, 3). 22.

x − 4y = 9,

2x + 5y = 5

Write the augmented matrix. We will use Gauss-Jordan elimination.   1 −4 9   2 5 5

The last row corresponds to 0 = 0, so the system is dependent and equivalent to −x + 2y = −3. Solving this system for x gives us x = 2y + 3. Then the solutions are of the form (2y + 3, y), where y is any real number. 25.

3x − 9y = −6 Write the augmented matrix.   4 −2 6   3 −9 −6

Multiply row 1 by −2 and add it to row 2.   9 1 −4   0 13 −13 Multiply row 2 by  

1 −4 0

9

1 −1

1 . 13

1 Multiply row 1 by − . 2   1 −3 −2   3 −9 −6

 

Multiply row 1 by −3 and add it to row 2.   1 −3 −2   0 0 0

Multiply row 2 by 4 and add it to row 1.   5 1 0   0 1 −1

The last row corresponds to the equation 0 = 0 which is true for all values of x and y. Thus, the system of equations is dependent and is equivalent to the first equation −2x + 6y = 4, or x − 3y = −2. Solving for x, we get x = 3y − 2. Then the solutions are of the form (3y − 2, y), where y is any real number.

The solution is (5, −1). 23.

x − 3y = 8,

2x − 6y = 3

Write the augmented matrix.   1 −3 8   2 −6 3 Multiply row 1 by −2 and add it to row 2.   8 1 −3   0 0 −13 The last row corresponds to the false equation 0 = −13, so there is no solution. 24.

4x − 8y = 12,

−x + 2y = −3

Write the augmented matrix.   4 −8 12   −1 2 −3

−2x + 6y = 4,

26.

6x + 2y = −10,

−3x − y = 6 Write the augmented matrix.   6 2 −10   −3 −1 6 Interchange the rows.   6 −3 −1   6 2 −10 Multiply row 1 by 2 and add it to row 2.   −3 −1 6   0 0 2 The last row corresponds to the false equation 0 = 2, so there is no solution.

Exercise Set 8.3 27.

497 28.

x + 2y − 3z = 9,

x − y + 2z = 0,

x − 2y + 3z = −1,

2x − y + 2z = −8,

2x − 2y + z = −3

3x − y − 4z = 3

Write the augmented matrix. We will use Gauss-Jordan elimination.   9 1 2 −3   2 −8   2 −1 3 −1 −4 3

Write the augmented matrix. We will use Gauss-Jordan elimination.   1 −1 2 0    1 −2 3 −1  2 −2 1 −3

Multiply row 1 by −2 and add it to row 2. Also, multiply row 1 by −3 and add it to row 3.   9 1 2 −3   8 −26   0 −5 0 −7 5 −24

Multiply row 1 by −1 and add it to row 2. Also, multiply row 1 by −2 and add it to row 3.   0 1 −1 2   1 −1   0 −1 0 0 −3 −3

1 Multiply row 2 by − to get a 1 in the second row, second 5 column.   1 2 −3 9  8 26    0 1 −  5 5  0 −7 5 −24

Multiply row 2 by −1.   0 1 −1 2   1 −1 1   0 0 0 −3 −3 Add  1   0 0

Multiply row 2 by −2 and add it to row 1. Also, multiply row 2 by 7 and add it to row 3.  7  1 − 1 0  5 5    8 26     0 1 −  5 5    31 62 0 0 − 5 5

1 Multiply row 3 by − . 3   1 0 1 1    0 1 −1 1  0 0 1 1

5 Multiply row 3 by − to get a 1 in the third row, third 31 column.   7 1 − 1 0  5 5     8 26    0 1 −  5 5  0 0 1 −2 1 Multiply row 3 by − and add it to row 1. Also, multiply 5 8 row 3 by and add it to row 2. 5   1 0 0 −1   2   0 1 0 0 0 1 −2 We have x = −1, y = 2, z = −2. (−1, 2, −2).

The solution is

row 2 to row 1.  1 0 1  1 −1 1  0 −3 −3

Multiply row 3 by −1 and add it to row 1. Also, add row 3 to row 2.   1 0 0 0    0 1 0 2  0 0 1 1 The solution is (0, 2, 1). 29.

4x − y − 3z = 1, 8x + y − z = 5, 2x + y + 2z = 5

Write the augmented matrix. We will use Gauss-Jordan elimination.   4 −1 −3 1   1 −1 5   8 2 1 2 5 First interchange rows 1 and 3 so that each number below the first number in the first row is a multiple of that number.

498

Chapter 8: Systems of Equations and Matrices 

2

1

2 5



30.

 5  1

 1 −1  8 4 −1 −3

x + 2y − z = 5,

2x − 4y + z = 0

Multiply row 1 by −4 and add it to row 2. Also, multiply row 1 by −2 and add it to row 3.   5 2 1 2    0 −3 −9 −15  0 −3 −7 −9 Multiply row 2 by −1 and add it to row 3.   5 2 1 2    0 −3 −9 −15  6 0 0 2 1 Multiply row 2 by − to get a 1 in the second row, second 3 column.   2 1 2 5    0 1 3 5  0 0 2 6 Multiply row 2 by −1 and add it to row 1.   2 0 −1 0   3 5   0 1 0 0 2 6 1 Multiply row 3 by to get a 1 in the third row, third 2 column.   2 0 −1 0   3 5   0 1 0 0 1 3 Add it to  2  0  0

row 3 to row 1. Also multiply row 3 by −3 and add row 2.  3 0 0  1 0 −4  0 1 3

Finally, multiply row 1 by 

1

   0

0

0

0

1

0

0

1

3 2 −4 3

3x + 2y + 2z = 3,

1 . 2

   

3 We have x = , y = −4, z = 3. 2 ! " 3 , −4, 3 . 2

The solution is

Write the augmented matrix. We will use Gauss-Jordan elimination.   3 2 2 3   2 −1 5   1 2 −4 1 0 Interchange the first two rows.   1 2 −1 5   2 2 3   3 2 −4 1 0 Multiply row 1 by −3 and add it to row 2. Also, multiply row 1 by −2 and add it to row 3.   5 1 2 −1   5 −12   0 −4 0 −8 3 −10 1 Multiply row 2 by − . 4   1 2 −1 5   5  0 3  1 −   4 0 −8 3 −10 Multiply row 2 by −2 and add it to row 1. Also, multiply row 2 by 8 and add it to row 3.   3 −1 1 0   2     5  0 1 − 3    4 0 0 −7 14 1 Multiply row 3 by − . 7   3 1 0 −1   2     5  0 1 − 3    4 0 0 1 −2 3 Multiply row 3 by − and add it to row 1. Also, multiply 2 5 row 3 by and add it to row 2. 4   1 0 0 2  1    0 1 0  2  0

0

1 −2 ! " 1 The solution is 2, , −2 . 2

Exercise Set 8.3 31.

499 

x − 2y + 3z = 4,

 

3x + y − z = 0, 2x + 3y − 5z = 1

Write the augmented matrix. We will use Gaussian elimination.   1 −2 3 −4   0  1 −1  3 2 3 −5 1

−1

0

0

13 −8

and add it to row 2. Also, multiply to row 3.   

32

0 −36

144

1 1 and multiply row 3 by − . 11 36 

9 1 4 −1   16 4   0 1 −  11 11  0 0 1 −4 

We have x + 4y −

z = 9, (1) 16 4 z= , (2) y− 11 11 z = −4. (3)

Back-substitute 3 10 ·3 = y− 7 30 = y− 7

for z in equation (2) and solve for y. 12 7 12 7 42 y= =6 7 Back-substitute 6 for y and 3 for z in equation (1) and solve for x. x − 2 · 6 + 3 · 3 = −4

Back-substitute in (2) to find y. 16 4 y − (−4) = 11 11 16 16 = y+ 11 11 y=0

x − 3 = −4

Back-substitute in (1) to find x.

x = −1

x − 4 · 0 − (−4) = 9

The solution is (−1, 6, 3).

Write the augmented matrix. We will use Gaussian elimination.

5

Multiply row 2 by −

Now write the system of equations that corresponds to the last matrix. x − 2y + 3z = −4, (1) 12 10 z= , (2) y− 7 7 z=3 (3)

x + 4y − z = 9,

1 −5

Multiply row 2 by 13 and add it to row 3.   9 1 4 −1   4 −16   0 −11

1 and multiply row 3 by −1. 7   1 −2 3 −4  10 12    0 1 −  7 7  0 0 1 3

−3x + y − 5z = 5

1 −5

 9  5

Multiply row 3 by 11.   1 4 −1 9   4 −16   0 −11 0 143 −88 352

−3

2x − 3y + 2z = 2,

4 −1

Multiply row 1 by −2 row 1 by 3 and add it  9 1 4 −1  4 −16  0 −11

Multiply row 2 by

32.

−3

−3

Multiply row 2 by −1 and add it to row 3.   1 −2 3 −4   7 −10 12   0 0

1



2 2

Interchange the first two rows.   1 4 −1 9   2 2   2 −3

Multiply row 1 by −3 and add it to row 2. Also, multiply row 1 by −2 and add it to row 3.   1 −2 3 −4   7 −10 12   0 9 0 7 −11

0

2 −3

x=5

The solution is (5, 0, −4). 33.

2x − 4y − 3z = 3,

x + 3y + z = −1,

5x + y − 2z = 2

Write the augmented matrix.

500

Chapter 8: Systems of Equations and Matrices 

2

  1 5

−4

−3

3

3





 1 −1  1 −2 2

1

  4 2

5 3



4

 1  7 −7 1

Interchange the first two rows to get a 1 in the first row, first column.   1 3 1 −1   3   2 −4 −3 2 5 1 −2

Multiply row 1 by −4 and add it to row 2. Also, multiply row 1 by −2 and add it to row 3.   4 1 1 −3    0 1 13 −15  0 1 13 −15

Multiply row 1 by −2 and add it to row 2. Also, multiply row 1 by −5 and add it to row 3.   1 3 1 −1   5   0 −10 −5 7 0 −14 −7

Multiply row 2 by −1 and add it to row 3.   4 1 1 −3    0 1 13 −15 

1 Multiply row 2 by − to get a 1 in the second row, second 10 column.   1 3 1 −1  1 1   0 1 −   2 2  0 −14 −7 7

Solve the second equation for y.

0

1

3

  0 

1



0

0

1 1 2 0

 −1 1  −  2  0

The last row corresponds to the equation 0 = 0. This indicates that the system of equations is dependent. It is equivalent to x + 3y +

z = −1, 1 1 y+ z=− 2 2 We solve the second equation for y. 1 1 y=− z− 2 2 Substitute for y in the first equation and solve for x. ! " 1 1 x+3 − z− + z = −1 2 2 3 3 x − z − + z = −1 2 2 1 1 x= z+ 2 2 " ! 1 1 1 1 The solution is z + , − z − , z , where z is any real 2 2 2 2 number. x + y − 3z = 4,

4x + 5y + z = 1, 2x + 3y + 7z = −7

Write the augmented matrix.

0

0

0

We have a dependent system of equations that is equivalent to x + y − 3z = 4, y + 13z = −15.

y = −13z − 15

Substitute for y in the first equation and solve for x. x − 13z − 15 − 3z = 4

Multiply row 2 by 14 and add it to row 3.

34.

1 −3

x = 16z + 19

The solution is (16z + 19, −13z − 15, z), where z is any real number. 35.

p + q + r = 1, p + 2q + 3r = 4, 4p + 5q + 6r = 7 Write the augmented matrix.   1 1 1 1    1 2 3 4  4

5

6 7

Multiply row 1 by −1 and add it to row 2. Also, multiply row 1 by −4 and add it to row 3.   1 1 1 1    0 1 2 3  0 1 2 3 Multiply row 2 by −1 and add it to row 3.   1 1 1 1    0 1 2 3  0 0 0 0 The last row corresponds to the equation 0 = 0. This indicates that the system of equations is dependent. It is equivalent to p + q + r = 1, q + 2r = 3.

Exercise Set 8.3

501

We solve the second equation for q. q = −2r + 3

Substitute for y in the first equation and solve for p. p − 2r + 3 + r = 1 p−r+3 = 1

p = r−2

The solution is (r − 2, −2r + 3, r), where r is any real number. 36.

m + n + t = 9, m − n − t = −15,

3m + n + t = 2

Write the augmented matrix.   9 1 1 1    1 −1 −1 −15  3 1 1 2 Multiply row 1 by −1 and add it to row 2. Also, multiply row 1 by −3 and add it to row 3.   9 1 1 1    0 −2 −2 −24  0 −2 −2 −25 Multiply row 2 by −1 and add it to row 3.   9 1 1 1    0 −2 −2 −24  0 0 0 −1 The last row corresponds to the false equation 0 = −1. Thus, the system of equations has no solution. 37.

a + b − c = 7, a − b + c = 5,

3a + b − c = −1

Write the augmented matrix.   7 1 1 −1   1 5   1 −1 3 1 −1 −1 Multiply row 1 by −1 and add it to row 2. Also, multiply row 1 by −3 and add it to row 3.   7 1 1 −1   2 −2   0 −2 0 −2 2 −22 Multiply row 2 by −1 and add it to row 3.   7 1 1 −1   2 −2   0 −2 0 0 0 −20 The last row corresponds to the false equation 0 = −20. Thus, the system of equations has no solution.

38.

a − b + c = 3,

2a + b − 3c = 5,

4a + b − c = 11

Write the augmented matrix. We will use Gaussian elimination.   1 −1 1 3   5  1 −3  2 4 1 −1 11 Multiply row 1 by −2 and add it to row 2. Also, multiply row 1 by −4 and add it to row 3.   3 1 −1 1   3 −5 −1   0 0 5 −5 −1 Multiply row 3 by 3.   3 1 −1 1   3 −5 −1   0 0 15 −15 −3 Multiply row 2 by −5 and add it to row 3.   3 1 −1 1   3 −5 −1   0 0 0 10 2 1 1 and multiply row 3 by . 3 10   1 −1 1 3  5 1   0 1 − −    3 3    1  0 0 1 5 Multiply row 2 by

We have x−y+

z = 3, (1) 5 1 y − z = − , (2) 3 3 1 (3) z= . 5 Back-substitute in equation (2) to find y. 5 1 1 y− · =− 3 5 3 y=0 Back-substitute in equation (1) to find x. 1 x−0+ = 3 5 14 x= 5 " ! 14 1 , 0, . The solution is 5 5

502 39.

Chapter 8: Systems of Equations and Matrices 1 1 1 Multiply row 2 by − , row 3 by − , and row 6 by − . 2 4 6   −5 1 1 1 1    0 1 2 − 1 − 13   2 2     3  1   0 0 1 − −  2 2  0 0 0 1 −1

−2w + 2x + 2y − 2z = −10, w + x + y + z = −5,

3w + x − y + 4z = −2, w + 3x − 2y + 2z = −6

Write the augmented matrix. We will use Gaussian elimination.   −2 2 2 −2 −10  1 1 1 1 −5       3 1 −1 4 −2  1

3

2

−2

Write the system of equations that corresponds to the last matrix. w + x + y + z = −5, (1) 13 1 x + 2y − z = − , (2) 2 2 1 3 y − z = − , (3) 2 2 z = −1 (4)

−6

Interchange rows 1 and 2. 1



1

 −2 2    3 1 1

3

1

1

−5



2 −2 −10    −1 4 −2  2

−2

Back-substitute 1 y − (−1) 2 1 y+ 2 y

−6

Multiply row 1 by 2 and add it to row 2. Multiply row 1 by −3 and add it to row 3. Multiply row 1 by −1 and add it to row 4.   1 1 1 1 −5  0 4 4 0 −20       0 −2 −4 1 13  0

2 −3

1

Back-substitute in equation (2) and solve for x. 13 1 x + 2(−2) − (−1) = − 2 2 13 1 x−4+ = − 2 2 x = −3

−1

Interchange rows 2 and 3. 

1

1

1

1



−5

 0 −2 −4 1 13    0 4 4 0 −20 0

2 −3

1

Multiply row 2 row 4.  1 1 1  0 −2 −4    0 0 −4 0

0 −7

by 2 and add it to row 3. Add row 2 to 1 −5

2 2

1

1

 0 −2    0 0

−4

1

0 −28

8

0

−4

1 −5

2



13    6  48

Multiply row 3 by −7 and add it to row 4. 

1

1

1

 0 −2 −4    0 0 −4 0

0

1 −5

1 2

0 −6



13    6  6

The solution is (1, −3, −2, −1). 40.

−w + 2x − 3y + z = −8,

−w + x − y − z = −14

12

1

w=1

w + x + y + z = 22,

Multiply row 4 by 4. 

w − 3 − 2 − 1 = −5

−w + x + y − z = −4,



13    6 

1

Back-substitute in equation (1) and solve for w.

   

−1

in equation (3) and solve for y. 3 =− 2 3 =− 2 = −2

Write the augmented elimination.  −1 2 −3 1  −1 1 1 −1    1 1 1 1 −1

1 −1

−1

matrix. We will use Gauss-Jordan −8



−4    22 

−14

Multiply row 1 by −1. 

1 −2

 −1    1 −1

1 1

3 −1 1 −1 1

1 −1

1

8



−4    22 

−1 −14

Add row 1 to row 2 and to row 4. Multiply row 1 by −1 and add it to row 3.

Exercise Set 8.3 

1

−2

503

3 −1

 0 −1 4 −2    0 3 −2 2 0 −1

2 −2

8



4    14 

−6

Multiply row 2 by −1. 

1 −2

 0    0

3 −1

1 −4



2 −4    2 14 

3 −2

0 −1

8

2 −2

−6

Multiply row 2 by −3 and add it to row 3. Also, add row 2 to row 4.   8 1 −2 3 −1  0 1 −4 2 −4       0 0 10 −4 26  0

0 −2

0 −10

Interchange row 3 and row 4. 

1 −2

 0    0 0

1 −4



2 −4    0 −10 

0 −2

0

8

3 −1

10 −4

26

1 Multiply row 3 by − . 2  1 −2 3 −1 8  0 1 −4 2 −4    0 0 1 0 5 0

0

10 −4

1 −2

 0    0 0

3 −1

1 −4

2

0

1

0

0

0 −4

   

26

8

0

0

0

1



−4    5 

−24

1 Multiply row 4 by − . 4  1 −2 3 −1 8  0 1 −4 2 −4    0 0 1 0 5 6

0

0

1

6

1 −2 0

 0    0 0

0 −1

1

0

0

1

0

0

0

1



4    5 

0

6

Multiply row 2 by 2 and add it to row 1. 7

1

0

0

0

 0    0

1

0

0

1

0

0

0 4    0 5 



0



1 6

The solution is (7, 4, 5, 6). 41. Familiarize. Let x = the number of hours the Houlihans were out before 11 P.M. and y = the number of hours after 11 P.M. Then they pay the babysitter $5x before 11 P.M. and $7.50y after 11 P.M. Translate. The Houlihans were out for a total of 5 hr. x+y =5 They paid the sitter a total of $30. 5x + 7.5y = 30 Carry out. Use Gaussian elimination or Gauss-Jordan elimination to solve the system of equations. y = 5,

5x + 7.5y = 30. The solution is (3, 2). The coordinate y = 2 indicates that the Houlihans were out 2 hr after 11 P.M., so they came home at 1 A.M. Check. The total time is 3 + 2, or 5 hr. The total pay is $5 · 3 + $7.50(2), or $15 + $15, or $30. The solution checks. State. The Houlihans came home at 1 A.M.

42. Let x, y, and z represent the amount spent on advertising in fiscal years 2004, 2005, and 2006, respectively, in millions of dollars. Solve: x + y + z = 11, z = 3x,

    

Add row 4 to row 1. Also, multiply row 4 by −2 and add it to row 2.   1 −2 3 0 14  0 1 −4 0 −16       0 5  0 1 0 0



x+ 

Multiply row 3 by −10 and add it to row 4. 

Multiply row 3 by −3 and add it to row 1. Also, multiply row 3 by 4 and add it to row 2.

y =z−3

x = $2 million, y = $3 million, z = $6 million 43. Familiarize. Let x, y, and z represent the amounts borrowed at 8%, 10%, and 12%, respectively. Then the annual interest is 8%x, 10%y, and 12%z, or 0.08x, 0.1y, and 0.12z. Translate. The total amount borrowed was $30,000. x + y + z = 30, 000 The total annual interest was $3040. 0.08x + 0.1y + 0.12z = 3040 The total amount borrowed at 8% and 10% was twice the amount borrowed at 12%.

504

Chapter 8: Systems of Equations and Matrices x + y = 2z We have a system of equations. x + y + z = 30, 000, 0.08x + 0.1y + 0.12z = 3040, x + y = 2z, or x+

y+

z = 30, 000,

0.08x + 0.1y + 0.12z = 3040, x+

y−

2z = 0

Carry out. Using Gaussian elimination or Gauss-Jordan elimination, we find that the solution is (8000, 12, 000, 10, 000). Check. The total amount borrowed was $8000+$12, 000+ $10, 000, or $30,000. The total annual interest was 0.08($8000) + 0.1($12, 000)+ 0.12($10, 000), or $640 + $1200 + $1200, or $3040. The total amount borrowed at 8% and 10%, $8000 + $12, 000 or $20,000, was twice the amount borrowed at 12%, $10,000. The solution checks. State. The amounts borrowed at 8%, 10%, and 12% were $8000, $12,000 and $10,000, respectively. 44. Let x and y represent the number of 39/ c and 24/ c stamps purchased, respectively. Solve: x+ y = 60, 0.39x + 0.24y = 21.15 x = 45, y = 15 45. Answers will vary. 46. See Example 4 in this section of the text. 47. The function has a variable in the exponent, so it is an exponential function. 48. The function is of the form f (x) = mx + b, so it is linear. 49. The function is the quotient of two polynomials, so it is a rational function. 50. This is a polynomial function of degree 4, so it is a quartic function. 51. The function is of the form f (x) = loga x, so it is logarithmic. 52. This is a polynomial function of degree 3, so it is a cubic function. 53. The function is of the form f (x) = mx + b, so it is linear. 54. This is a polynomial function of degree 2, so it is quadratic. 55. Substitute to find three equations. 12 = a(−3)2 + b(−3) + c −7 = a(−1)2 + b(−1) + c −2 = a · 12 + b · 1 + c

We have a system of equations. 9a − 3b + c = 12, a − b + c = −7, a + b + c = −2

Write the augmented matrix. We will use Gaussian elimination.   9 −3 1 12    1 −1 1 −7  1 1 1 −2 Interchange the first two rows.   1 −1 1 −7    9 −3 1 12  1 1 1 −2 Multiply row 1 by −9 and add it to row 2. Also, multiply row 1 by −1 and add it to row 3.   1 −1 1 −7   6 −8 75   0 0 2 0 5 Interchange row 2 and row 3.   1 −1 1 −7   2 0 5   0 0 6 −8 75 Multiply row 2 by −3 and add it to row 3.   1 −1 1 −7   2 0 5   0 0

0 −8

60

1 1 and row 3 by − . 2 8   −7 1 −1 1  5    0 1 0   2    15  0 0 1 − 2 Multiply row 2 by

Write the system of equations that corresponds to the last matrix. x − y + z = −7, 5 y = , 2 15 z=− 2 15 5 for z in the first equation Back-substitute for y and − 2 2 and solve for x. 5 15 x− − = −7 2 2 x − 10 = −7 x=3 " ! 5 15 , so the equation is The solution is 3, , − 2 2 15 5 y = 3x2 + x − . 2 2

Exercise Set 8.3 56.

Solve:

505 b) If the system has exactly one solution, we have:

0 = a(−1)2 + b(−1) + c,

2k − 2 $= −3k + 3

−3 = a · 1 + b · 1 + c, 2

5k $= 5

−22 = a · 32 + b · 3 + c, or

k $= 1

a − b + c = 0,

c) If the system has infinitely many solutions, we have:

a + b + c = −3,

9a + 3b + c = −22 1 3 1 3 a = −2, b = − , c = , so y = −2x2 − x + . 2 2 2 2   1 5  57.  3 2 Multiply row 1 by −3 and add it to row 2.   1 5   0 −13 Multiply row 2 by −  

58.

1

5

0

1



1 . 13

Row-echelon form



Multiply row 2 by −5 and add it to row 1.   1 0   Reduced row-echelon form 0 1 x − y + 3z = −8,

2x + 3y −

z = 5,

3x + 2y + 2kz = −3k

Write the augmented matrix.   −8 1 −1 3   3 −1 5   2 3 2 2k −3k Multiply row 1 by −2 and add it to row 2. Also, multiply row 1 by −3 and add it to row 3.   −8 1 −1 3   5 −7 21   0 0

5

2k − 9

−3k + 24

Multiply row 2 by −1 and add it to row 3.   1 −1 3 −8   5 −7 21   0 0

0

2k − 2

−3k + 3

a) If the system has no solution we have: 2k − 2 = 0 and −3k + 3 $= 0 k = 1 and

k $= 1

This is impossible, so there is no value of k for which the system has no solution.

2k − 2 = 0 and −3k + 3 = 0 k = 1 and

k = 1, or

k=1 59.

y = x + z, 3y + 5z = 4, x + 4 = y + 3z, or x − y + z = 0, 3y + 5z = 4,

x − y − 3z = −4

Write the augmented matrix. We will use Gauss-Jordan elimination.   1 −1 1 0   4  3 5  0 1 −1 −3 −4 Multiply row 1 by −1 and add it to row 3.   0 1 −1 1   3 5 4   0 0 0 −4 −4 1 Multiply row 3 by − . 4   1 −1 1 0   3 5 4   0 0 0 1 1 Multiply row 3 by −1 and add it to row 1. Also, multiply row 3 by −5 and add it to row 2.   1 −1 0 −1   3 0 −1   0 0 0 1 1 Multiply row 2 by

1 . 3 

1 −1 0 −1  1   0 1 0 −   3  0 0 1 1 

Add   1    0  0

row 2 to row 1.  4 0 0 −  3  1  1 0 −  3  1 0 1

506

Chapter 8: Systems of Equations and Matrices Read the solution from the last matrix. It is " ! 4 1 − ,− ,1 . 3 3

60.

61.

3x + y + 3z = −5

Write the augmented matrix.   1 −4 2 7   3 1 3 −5

x + y = 2z, 2x − 5z = 4,

x − z = y + 8, or

x + y − 2z = 0,

2x

Multiply row 1 by −3 and add it to row 2.   7 1 −4 2   0 13 −3 −26

− 5z = 4,

x−y− z=8

Write the augmented matrix. We will use Gauss-Jordan elimination.   1 1 −2 0   0 −5 4   2 1 −1 −1 8

1 . 13   7 1 −4 2     3 0 1 − −2 13 Multiply row 2 by

Multiply row 1 by −2 and add it to row 2. Also, multiply row 1 by −1 and add it to row 3.   1 1 −2 0    0 −2 −1 4  0 −2 1 8

Write the system of equations that corresponds to the last matrix. x − 4y + 2z = 7, 3 y − z = −2 13 Solve the second equation for y. 3 z−2 y= 13 Substitute in the first equation and solve for x. " ! 3 z − 2 + 2z = 7 x−4 13 12 x − z + 8 + 2z = 7 13 14 x = − z−1 13 ! " 14 3 The solution is − z − 1, z − 2, z , where z is any 13 13 real number.

1 Multiply row 2 by − . 2   1 1 −2 0   1  0 1 −2    2 8 0 −2 1 Multiply row 2 by −1 and add it to row 1. Also, multiply row 2 by 2 and add it to row 3.   5 2 1 0 −   2     1  0 1 −2    2 0 0 2 4 Multiply row 5  1 0 −2   1  0 1  2 0 0 1 

Multiply row 3 by 

1 0   0 1 0 0

1 3 by . 2  2    −2   2

5 row 3 by and add it to row 1. Also, multiply 2 1 − and add it to row 2. 2  7 0  0 −3  2 1

The solution is (7, −3, 2).

x − 4y + 2z = 7,

62.

x − y − 3z = 3,

−x + 3y + z = −7

Write the augmented matrix.   3 1 −1 −3   −1 3 1 −7 Add row 1 to row 2.   3 1 −1 −3   0 2 −2 −4 1 . 2   1 −1 −3 3   0 1 −1 −2 Multiply row 2 by

Exercise Set 8.4

507

We have x − y − 3z = 3,

−x + 2y = −2,

y − z = −2.

3x − 5y = 1

Then y = z − 2. Substitute in the first equation and solve for x. x − (z − 2) − 3z = 3

Write the augmented matrix.   2 −3 −1   2 −2   −1

x − z + 2 − 3z = 3

3

x = 4z + 1

The solution is (4z+1, z−2, z), where z is any real number. 63.

4x + 5y = 3, 3x − 2y = −15

Write the augmented matrix.   3 4 5   9  1  −2

Multiply row 1 by 2 and add it to row 2. Also, multiply row 1 by 3 and add it to row 3.   −1 2 −2    0 1 −5  0 1 −5

−15

Multiply row 2 by 2 and row 3 by 4.   3 4 5   2 18   −4 12 −8 −60

Multiply row 2 by −1 and add it to row 3.   −1 2 −2    0 1 −5  0 0 0

Add row 1 to row 2. Also, multiply row 1 by −3 and add it to row 3.   3 4 5   7 21   0 0 −23 −69 Multiply 

4

5

  0 0

1 1

We have a dependent system that is equivalent to −x + 2y = −2, (1) y = −5. (2)

Back-substitute in equation (1) to find x. −x + 2(−5) = −2

1 1 row 2 by and row 3 by − . 7 23  3  3  3

Multiply row 2 by −1 and add it to row 3.   4 5 3    0 1 3  0 0 0 The last row corresponds to the equation 0 = 0. Thus we have a dependent system that is equivalent to 4x + 5y = 3, (1) y = 3. (2) Back-substitute in equation (1) to find x. 4x + 5 · 3 = 3 4x + 15 = 3 4x = −12 x = −3

The solution is (−3, 3).

1

−5

Interchange the first two rows.   −1 2 −2    2 −3 −1  3 −5 1

−2x + y = 9,

3 −2

2x − 3y = −1,

64.

−x − 10 = −2 −x = 8

x = −8

The solution is (−8, −5).

Exercise Set 8.4 1.

/

5

x

0

=

/

y

−3

0

Corresponding entries of the two matrices must be equal. Thus we have 5 = y and x = −3. 1 2 1 2 6x −9 2. = 25 5y 6x = −9 and 25 = 5y 3 and 5=y x=− 2 1 2 1 2 3 2x 3 −2 3. = y −8 1 −8

Corresponding entries of the two matrices must be equal. Thus, we have: 2x = −2 and y = 1 x = −1

and y = 1

508

4.

Chapter 8: Systems of Equations and Matrices 1

x−1 4 y + 3 −7

x−1 = 0

x=1

5.

6. 7.

8. 9.

10. 11.

12.

2

=

1

0 4 −2 −7

2

13.

and y + 3 = −2 and

y = −5 1 2 1 2 1 2 −3 5 A+B = + 4 3 2 −1 1 2 1 + (−3) 2 + 5 = 4+2 3 + (−1) 1 2 −2 7 = 6 2 1 2 −2 7 B+A=A+B= (See Exercise 5.) 6 2 1 2 1 2 1 3 0 0 E+0 = + 2 6 0 0 1 2 1+0 3+0 = 2+0 6+0 1 2 1 3 = 2 6 1 2 1 2 2·1 2·2 2 4 2A = = 2·4 2·3 8 6 1 2 3 3 3F = 3 −1 −1 1 2 3·3 3·3 = 3 · (−1) 3 · (−1) 1 2 9 9 = −3 −3 1 2 1 2 −1 · 1 −1 · 1 −1 −1 (−1)D = = −1 · 1 −1 · 1 −1 −1 1 2 1 2 3 3 9 9 3F = 3 = , −1 −1 −3 −3 1 2 1 2 1 2 2 4 2A = 2 = 4 3 8 6 1 2 1 2 9 9 2 4 3F + 2A = + −3 −3 8 6 1 2 9+2 9+4 = −3 + 8 −3 + 6 1 2 11 13 = 5 3 1 2 1 2 1 2 1 2 −3 5 4 −3 A−B= − = 4 3 2 −1 2 4

B−A =

1

−3 2

=

1

5 −1

−3 2

5 −1

15.

16. 17.

18.

19.

20.

2

−

1

1 2 4 3

2

+

1

−1 −4

−2 −3

2

[B − A = B + (−A)] 2 −3 + (−1) 5 + (−2) = 2 + (−4) −1 + (−3) 1 2 −4 3 = −2 −4 1 21 2 1 2 −3 5 AB = 4 3 2 −1 1 2 1(−3) + 2 · 2 1 · 5 + 2(−1) = 4(−3) + 3 · 2 4 · 5 + 3(−1) 2 1 1 3 = −6 17 1 21 2 −3 5 1 2 BA = 2 −1 4 3 1 2 −3 · 1 + 5 · 4 −3 · 2 + 5 · 3 = 2 · 1 + (−1)4 2 · 2 + (−1)3 1 2 17 9 = −2 1 1 2 0 0 0F = 0 = 0 0 1 21 2 1 −1 1 1 CD = −1 1 1 1 1 2 1 · 1 + (−1) · 1 1 · 1 + (−1) · 1 = −1 · 1 + 1 · 1 −1 · 1 + 1 · 1 1 2 0 0 = 0 0 1 21 2 1 3 3 3 EF = 2 6 −1 −1 1 2 1 · 3 + 3(−1) 1 · 3 + 3(−1) = 2 · 3 + 6(−1) 2 · 3 + 6(−1) 1 2 0 0 = 0 0 1 21 2 1 2 1 0 AI = 4 3 0 1 1 2 1·1+2·0 1·0+2·1 = 4·1+3·0 4·0+3·1 1 2 1 2 = 4 3 1 2 1 2 IA = A = 4 3 1

14.

2

Exercise Set 8.4 1

21.

1

−1 0 7 3 −5 2

509 2





6  −4  1

28. 2

−1 · 6 + 0(−4) + 7 · 1 3 · 6 + (−5)(−4) + 2 · 1 1 2 1 = 40   1 4 / 0 22. 6 −1 2  −2 0  5 −3 / 0 = 6·1+(−1)(−2)+2·5 6·4+(−1)·0+ 2(−3) / 0 = 18 18   2 −2 4 1 3 −6 23.  5 1  −1 4 −1 −3   −2 · 3 + 4(−1) −2(−6) + 4 · 4 5 · 3 + 1(−1) 5(−6) + 1 · 4  = −1 · 3 + (−3)(−1) −1(−6) + (−3) · 4   −10 28 =  14 −26  0 −6   1 2 −3 1 0 2 −1 0  24. 0 2 −1  0 5 4 5 0 4 1 2 −6 + 0 + 0 2−2+0 0+1+0 = 0 + 0 + 20 0 + 10 + 0 0 − 5 + 16 1 2 −6 0 1 = 20 10 11   2 1 1 −6 5 8 25.  −5  0 4 −1 3 =

This product is not defined because the number of columns of the first matrix, 1, is not equal to the number of rows of the second matrix, 2.    2 0 0 0 −4 3 26.  0 −1 0   2 1 0  0 0 3 −1 0 6   0 + 0 + 0 −8 + 0 + 0 6+0+0 0−1+0 0+0+0  = 0−2+0 0+0−3 0 + 0 + 0 0 + 0 + 18   0 −8 6 0  =  −2 −1 −3 0 18    1 −4 3 3 0 0 27.  0 8 0   0 −4 0  −2 −1 5 0 0 1   3 + 0 + 0 0 + 16 + 0 0 + 0 + 3 =  0 + 0 + 0 0 − 32 + 0 0 + 0 + 0  −6 + 0 + 0 0+4+0 0+0+5   3 16 3 =  0 −32 0  −6 4 5

1

4 −5

2



 2 0  6 −7  0 −3

This product is not defined because the number of columns of the first matrix, 1, is not equal to the number of rows of the second matrix, 3. / 0 29. a) B = 150 80 40 b) $150 + 5% · $150 = 1.05($150) = $157.50 $80 + 5% · $80 = 1.05($80) = $84 $40 + 5% · $40 = 1.05($40) = $42

We write the matrix that corresponds to these amounts. / 0 R = 157.5 84 42 / 0 / 0 c) B+R = 150 80 40 + 157.5 84 42 / 0 = 307.5 164 82

The entries represent the total budget for each type of expenditure for June and July. / 0 30. a) A = 40 20 30 b) 1.1(40) = 44, 1.1(20) = 22, 1.1(30) = 33 / 0 B = 44 22 33 / 0 / 0 c) A + B = 40 20 30 + 44 22 33 / 0 = 84 42 63

The entries represent the total amount of each type of produce ordered for both weeks. / 0 31. a) C = 140 27 3 13 64 / 0 P = 180 4 11 24 662 / 0 B = 50 5 1 82 20 b)

C + 2P + 3B / 0 140 / 27 3 13 64 + 0 360 8 22 48 1324 + / 0 150 15 3 246 60 / 0 = 650 50 28 307 1448 =

The entries represent the total nutritional value of 1 serving of chicken, 1 cup of potato salad, and 3 broccoli spears. / 0 32. a) P = 290 15 9 39 / 0 G = 70 2 0 17 / 0 M = 150 8 8 11 b)

3P + 2G + 2M / 0 870 45 27 117 + / 0 140 4 0 34 + / 0 300 16 16 22 / 0 = 1310 65 43 173 =

The entries represent the total nutritional value of 3 slices of pizza, 1 cup of gelatin, and 2 cups of whole milk.

510

Chapter 8: Systems of Equations and Matrices 

45.29  53.78 33. a) M =   47.13 51.64 / b) N = 65 48

34.

35.

36.

37.

6.63 4.95 8.47 7.12 93

10.94 9.83 12.66 11.57 0 57

7.42 6.16 8.29 9.35



8.01 12.56   9.43  10.72

c) NM = / 0 12, 851.86 1862.1 3019.81 2081.9 2611.56 d) The entries of NM represent the total cost, in cents, of each  item for the day’s meals.  1 2.5 0.75 0.5  0 0.5 0.25 0   a) M =   0.75 0.25 0.5 0.5  0.5 0 0.5 1 / 0 b) C = 15 28 54 83 / 0 c) CM = 97 65 86.75 117.5 d) The entries of CM represent the total cost, in cents, of each menu item.   8 15 a) S =  6 10  4 3 / 0 b) C = 3 1.5 2 / 0 c) CS = 41 66 d) The entries of CS represent the total cost, in dollars, of ingredients for each coffee shop.   900 500 a) M =  450 1000  600 700 / 0 b) P = 5 8 4 / 0 c) PM = 10, 500 13, 300 d) The entries of PM represent the total profit from each distributor. / 0 a) P = 6 4.5 5.2   / 0 8 15 b) PS = 6 4.5 5.2  6 10  4 3 / 0 = 95.8 150.6

The profit from Mugsey’s Coffee Shop is $95.80, and the profit from The Coffee Club is $150.60. / 0 38. a) C = 20 25 15   500 / 0 900 b) CM = 20 25 15  450 1000  600 700 / 0 = 38, 250 45, 500

The total production costs for the products shipped to Distributors 1 and 2 are $38,250 and $45,500, respectively.

39.

2x − 3y = 7,

x + 5y = −6 Write the coefficients on the left in a matrix. Then write the product of that matrix and the column matrix containing the variables, and set the result equal to the column matrix containing the constants on the right. 1 21 2 1 2 2 −3 x 7 = 1 5 y −6

40. 41.

1

−1 1 5 −4

21

x y

2

=

1

3 16

2

x + y − 2z = 6,

3x − y + z = 7, 2x + 5y − 3z = 8

Write the coefficients on the left in a matrix. Then write the product of that matrix and the column matrix containing the variables, and set the result equal to the column matrix containing the constants on the right.      1 1 −2 x 6  3 −1 1  y  =  7  2 5 −3 z 8      3 −1 1 x 1 2 −1   y  =  3  42.  1 4 3 −2 z 11 43.

3x − 2y + 4z = 17, 2x + y − 5z = 13

Write the coefficients on the left in a matrix. Then write the product of that matrix and the column matrix containing the variables, and set the result equal to the column matrix containing the constants on the right.   1 2 x 1 2 17 3 −2 4  y = 2 1 −5 13 z   1 2 x 1 2 3 2 5  9 y = 44. 4 −3 2 10 z 45.

−4w + x − y + 2z = 12, w + 2x − y − z = 0,

−w + x + 4y − 3z = 1, 2w + 3x + 5y − 7z = 9

Write the coefficients on the left in a matrix. Then write the product of that matrix and the column matrix containing the variables, and set the result equal to the column matrix containing the constants on the right.      −4 1 −1 2 w 12  1 2 −1 −1   x   0    =   −1 1 4 −3   y   1  2 3 5 −7 z 9      12 2 4 −5 w 2  −1     4 −1 12   x  =  5  46.     2 −1   y 13  4 0 0 2 10 1 z 5 47. Left to the student 48. Left to the student 49. No; see Exercise 17, for example. 50. She could make decisions regarding cost and profit, including how many of each product to prepare and distribute to each shop.

Exercise Set 8.4

511

51. f (x) = x2 − x − 6 −1 1 b =− = a) − 2a 2·1 2 ! " ! "2 1 1 1 25 f = − −6=− 2 2 2 4 ! " 1 25 The vertex is ,− . 2 4

3 b) The axis of symmetry is x = − . 2 c) Since the coefficient of x2 is negative, the function has a maximum value. It is the second coordinate 17 of the vertex, . 4 d) Plot some points and draw the graph of the function. y

1 . 2 c) Since the coefficient of x2 is positive, the function has a minimum value. It is the second coordinate 25 of the vertex, − . 4 d) Plot some points and draw the graph of the function. b) The axis of symmetry is x =

4 2 !4

!2

2

4

x

!2 !4 !6

f(x) " x 2 ! x ! 6

52. f (x) = 2x2 − 5x − 3 −5 5 b =− = a) − 2a 2·2 4 ! " ! "2 ! " 5 5 5 49 f =2 −5 −3=− 4 4 4 8 " ! 5 49 ,− . The vertex is 4 8 5 b) x = 4 49 c) Minimum: − 8 d) y 2 !2

4

x

!4

f(x) " !x 2 ! 3x # 2

2

!4

2 !2

y

!4

!2

54. f (x) = −3x2 + 4x + 4 4 2 b =− = a) − 2a 2(−3) 3 ! " ! " ! "2 2 2 2 16 f +4 = −3 +4= 3 3 3 3 " ! 2 16 , . The vertex is 3 3 2 b) x = 3 16 Maximum: c) 3 d) y 4 2 !4

2

4

x

!4 !6

f(x) " 2x 2 ! 5x ! 3

53. f (x) = −x2 − 3x + 2 b −3 3 =− =− a) − 2a 2(−1) 2 " ! "2 ! " ! 3 3 17 3 −3 − =− − +2= f − 2 2 2 4 " ! 3 17 . The vertex is − , 2 4

2

4

x

!2 !4

f(x) " !3x 2 # 4x # 4

2 1 2 1 −1 ,B= 0 2 1 21 2 0 −1 −2 1 (A + B)(A − B) = 2 3 2 −1 1 2 −2 1 = 2 −1

55. A =

!2

!2

1

−1 2

0 1

A2 − B2 1 21 2 1 −1 0 −1 0 1 = − 2 1 2 1 0 1 2 1 2 1 0 1 −3 = − 0 1 0 4 1 2 0 3 = 0 −3

−1 2

Thus (A + B)(A − B) $= A2 − B2 .

21

1 0

−1 2

2

512

Chapter 8: Systems of Equations and Matrices 2 1 2 1 −1 ,B= 0 2 1 21 2 0 −1 0 −1 (A + B)(A + B) = 2 3 2 3 1 2 −2 −3 = 6 7

56. A =

1

−1 2

0 1

We found A2 and B2 in Exercise 55. 1 21 2 1 −1 0 1 −1 −2 2AB = 2 = 2 1 0 2 4 A2 + 2AB + B2 1 2 1 1 0 −2 = + 0 1 4 1 2 0 −1 = 4 5

2 0

2

+

1

1 0

−3 4

60.

2 0

AB =

21

1 −1 0 2

2

=

1

−1 2

A2 + BA − AB − B2 1 2 1 2 1 1 0 −3 −1 −1 = + − 0 1 4 2 2

=

1

−2 1 2 −1

(a11 +b11 )+c11 (a21 +b21 )+c21 · · · (am1 +bm1 )+cm1

· · · (a1n +b1n )+c1n · · · (a2n +b2n )+c2n · · · · · · · · · (amn +bmn )+cmn

           

= (A + B) + C 61. See the answer section in the text.

and we also found A2 and B2 . 1 21 2 1 1 −1 −1 0 −3 BA = = 0 2 2 1 4 0 1

  =  

2

62.

57. In 1 Exercise 255 we found that (A + B)(A − B) = −2 1 2 −1

−1 2



2

Thus (A + B)(A + B) $= A2 + 2AB + B2 .

1

A + (B + C)  a11 +(b11 +c11 ) · · · a1n +(b1n +c1n )  a21 +(b21 +c21 ) · · · a2n +(b2n +c2n )  · · · =  · · ·  · · · am1 +(bm1 +cm1 ) · · · amn +(bmn +cmn )

−1 2 1 0

2

2

2 1 − 0 1 2 1 −3 0 4

2

Thus (A + B)(A − B) = A2 + BA − AB − B2 . 58. In Exercise 56 we found that 1 2 −2 −3 (A + B)(A + B) = . 6 7 We found A2 and B2 in Exercise 55, and we found BA and AB in Exercise 55. A2 + BA + AB + B2 1 2 1 2 1 2 1 0 −3 −1 −1 1 = + + + 0 1 4 2 2 0 1 2 1 −3 0 4 1 2 −2 −3 = 6 7 Thus (A + B)(A + B) = A2 + BA + AB + B2 . 59. See the answer section in the text.

k(A + B)  k(a11 + b11 ) · · · k(a1n + b1n )  k(a21 + b21 ) · · · k(a2n + b2n )  · · =  · ·  · · k(am1 + bm1 ) · · · k(amn + bmn ) 

  =  

ka11 + kb11 ka21 + kb21 · · · kam1 + kbm1

ka1n + kb1n ka2n + kb2n · · · · · · kamn + kbmn ··· ···

           

= kA + kB 63. See the answer section in the text.

Exercise Set 8.5 2 1 1 −3 = −2 7 1 21 2 1 1 −3 7 3 AB = = −2 7 2 1

1. BA =

1

7 2

3 1

21

1 0 0 1

2

1 0 0 1

2

Since BA = I = AB, B is the inverse of A. 2. BA = I = AB,  2 3 3. BA =  3 3 1 1   3 −2 0  4 −3 0  1 −1 1

so B is the inverse of A.   2 −1 −1 6 4  1 0 −2  = 1 1 0 −3

Since BA $= I, B is not the inverse of A.   1 0 −24 8  $= I, so B is not the inverse of A. 4. BA =  0 1 0 0 17

Exercise Set 8.5 2 3 2 5 3 Write the augmented matrix. 2 1 3 2 1 0 5 3 0 1 Multiply row 2 by 3. 2 1 3 2 1 0 15 9 0 3 Multiply row 1 by −5 and add it to row 2. 2 1 1 0 3 2 0 −1 −5 3 Multiply row 2 by 2 and add it to row 1. 2 1 3 0 −9 6 0 −1 −5 3 1 Multiply row 1 by and row 2 by −1. 23 1 1 0 −3 2 5 −3 0 1 1 2 −3 2 Then A−1 = . 5 −3 1 2 3 5 6. A = 1 2 Write the augmented matrix. 2 1 3 5 1 0 1 2 0 1 Interchange the rows. 2 1 1 2 0 1 3 5 1 0 Multiply row 1 by −3 and add it to row 2. 1 2 1 2 0 1 0 −1 1 −3 Multiply row 2 by 2 and add it to row 1. 2 1 1 0 2 −5 0 −1 1 −3 Multiply row 2 by −1. 2 1 2 −5 1 0 3 0 1 −1 1 2 2 −5 Then A−1 = . −1 3 1 2 6 9 7. A = 4 6 Write the augmented matrix. 2 1 6 9 1 0 4 6 0 1 Multiply row 2 by 3. 2 1 6 9 1 0 12 18 0 3 Multiply row 1 by −2 and add it to row 2. 2 1 6 9 1 0 0 0 −2 3 We cannot obtain the identity matrix on the left since the second row contains only zeros to the left of the vertical line. Thus, A−1 does not exist. 5. A =

1

513 8. A =

1

−4 −6 2 3

2

Write the augmented matrix. 1 2 −4 −6 1 0 2 3 0 1 Multiply row 2 by 2. 2 1 −4 −6 1 0 4 6 0 2 Add row 1 to row 2. 1 2 −4 −6 1 0 0 0 1 2 We cannot obtain the identity matrix on the left since the second row contains only zeros to the left of the vertical line. Thus, A−1 does not exist. 1 2 4 −3 9. A = 1 −2 Write the augmented matrix. 2 1 4 −3 1 0 1 −2 0 1 Interchange the rows. 2 1 1 −2 0 1 4 −3 1 0

Multiply row 1 by −4 and add it to row 2. 2 1 1 1 −2 0 0 5 1 −4

1 Multiply row 2 by . 5   1 1 −2 0  4  1 − 0 1 5 5 Multiply row 2 by 2 and add it to row 1.  3  2 1 0 −  5 5    1 4 0 1 − 5 5  2 3  −  5 5  Then A−1 =  . 4 1 − 5 5 1 2 0 −1 10. A = 1 0 Write the augmented matrix. 2 1 0 −1 1 0 1 0 0 1 Interchange the rows. 2 1 1 0 0 1 0 −1 1 0

Multiply row 2 by −1. 2 1 0 1 1 0 0 1 −1 0 1 2 0 1 −1 Then A = . −1 0

514

Chapter 8: Systems of Equations and Matrices 



 3 1 0 1 1  11. A =  1 1 −1 2 Write the  3 1  1 1 1 −1

augmented matrix.  0 1 0 0 1 0 1 0  2 0 0 1

Interchange the first  1 1 1 0 1  3 1 0 1 0 1 −1 2 0 0

two rows.  0 0  1

Multiply row 1 by −3 and add it to row 2. Also, multiply row 1 by −1 and add it to row 3.   1 0 1 1 1 0  0 −2 −3 1 −3 0  0 −2 1 0 −1 1 1 Multiply row 2 by − . 2   0 1 0 1 1 1   3 3 1  0 1 − 0    2 2 2 0 −1 1 0 −2 1

Multiply row 2 by −1 and add it to row 1. Also, multiply row 2 by 2 and add it to row 3.   1 1 1 − 0 1 0 −   2 2 2     3 1 3  0 1 − 0    2 2 2 2 1 0 0 4 −1 1 Multiply row 3 by . 4   1 1 1 1 0 − − 0   2 2 2   3 1 3    0 1 − 0    2 2 2   1 1 1 0 0 1 − 4 2 4 1 Multiply row 3 by and add it to row 1. Also, multiply 2 3 row 3 by − and add it to row 2. 2  1 1  3 − 1 0 0  8 4 8    1 3 3    0 1 0 − −   8 4 8    1 1 1 0 0 1 − 4 2 4  3 1 1  −  8 4 8    3 3   1 −1 Then A =  − − .  8 4 8    1 1 1 − 4 2 4   1 0 1 1 0  12. A =  2 1 −1 1 Write the augmented matrix.

1 0 1 1  2 1 0 0 1 −1 1 0

0 1 0

Add  1  0 0

3. 0 1 1

 0 0  1

Multiply row 1 by −2 and add it to row 2. Also, multiply row 1 by −1 and add it to row 3.   1 0 0 1 0 1  0 1 −2 −2 1 0  0 −1 0 −1 0 1 row 2 to row 1 0 1 1 −2 −2 0 −2 −3

 0 0  1

1 Multiply row 3 by − . 2   1 0 1 1 0 0  0 1 −2 −2 1 0     3 1 1  − − 0 0 1 2 2 2 Multiply row 3 by −1 and add it to row 1. Also, multiply row 3 by 2 and add it to row 2.   1 1 1 − 1 0 0  2 2 2      0 1 0 1 0 −1    3 1 1  0 0 1 − − 2 2 2   1 1 1 −  2 2 2    . Then A−1 =  1 0 −1    3 1 1  − − 2 2 2   1 −4 8 13. A =  1 −3 2  2 −7 10 Write the augmented matrix.   1 −4 8 1 0 0  1 −3 2 0 1 0  2 −7 10 0 0 1

Multiply row 1 by −1 and add it to row 2. Also, multiply row 1 by −2 and add it to row 3.   1 0 0 1 −4 8  0 1 −6 −1 1 0  0 1 −6 −2 0 1

Since the second and third rows are identical left of the vertical line, it will not be possible to obtain the identity matrix on the left side. Thus, A−1 does not exist.   −2 5 3 14. A =  4 −1 3  7 −2 5 Write the augmented  −2 5 3 1 0  4 −1 3 0 1 7 −2 5 0 0 Multiply row 3 by 2.

matrix.  0 0  1

Exercise Set 8.5

515



−2 5 3 1  4 −1 3 0 14 −4 10 0

0 1 0

 0 0  2

Multiply row 1 by 2 and add it to row 2. Also, multiply row 1 by 7 and add it to row 3.   −2 5 3 1 0 0  0 9 9 2 1 0  0 31 31 7 0 2

To the left of the vertical line, row 3 is a multiple of row 2 so it will not be possible to obtain the identity matrix on the left. Thus, A−1 does not exist.   2 3 2 3 4  15. A =  3 −1 −1 −1 Write the augmented matrix.   2 3 2 1 0 0  3 3 4 0 1 0  −1 −1 −1 0 0 1 Interchange rows 1 and  −1 −1 −1 0 0  3 3 4 0 1 2 3 2 1 0

3.



1 0  0

Multiply row 1 by 3 and add it to row 2. Also, multiply row 1 by 2 and add it to row 3.   −1 −1 −1 0 0 1  0 0 1 0 1 3  0 1 0 1 0 2 Multiply  1 1  0 0 0 1

row 1 by −1.

Multiply  1 0  0 1 0 0

row 2 by −1 and add it to row 1.  1 −1 0 −3 1 0 2  0 1 0 1 3

1 0 1 0 0 1

Interchange  1 1 1  0 1 0 0 0 1

Multiply  1 0  0 1 0 0

0 1 0

 −1 3  2

rows 2 and 3.  0 0 −1 1 0 2  0 1 3

row 3 by −1 and add it to row 1.  0 −1 −1 −6 1 0 2  0 1 0 1 3   −1 −1 −6 −1 0 2 . Then A =  1 0 1 3   1 2 3 16. A =  2 −1 −2  −1 3 3 Write the augmented matrix.   1 2 3 1 0 0  2 −1 −2 0 1 0  −1 3 3 0 0 1

Multiply row 1 by −2 and add it to row 2. Also, add row 1 to row 3.   1 2 3 1 0 0  0 −5 −8 −2 1 0  0 5 6 1 0 1 Multiply  1 2   0 1  0 5

1 row 2 by − . 5 0 0 3 1 8 2 1 − 0 5 5 5 0 1 6 1

   

Multiply row 2 by −2 and add it to row 1. Also, multiply row 2 by −5 and add it to row 3.   2 1 1 0 1 0 −   5 5 5     2 1 8  0 1 − 0    5 5 5 1 1 0 0 −2 −1

1 Multiply row 3 by − . 2   2 1 1 0 1 0 −   5 5 5     8 2 1   0 1 − 0   5 5 5    1 1  1 − − 0 0 1 2 2 2 1 Multiply row 3 by and add it to row 1. Also, multiply 5 8 row 3 by − and add it to row 2. 5   3 3 1 1 0 0 −  10 10 10     2 3 4    0 1 0 −  5 5 5     1 1 1  0 0 1 − − 2 2 2 

 3 3 1 −  10 10 10     3 4  2 , or Then A−1 =   −5 5 5     1 1 1  − − 2 2 2   0.3 0.3 −0.1  −0.4 0.6 0.8 . 0.5 −0.5 −0.5   1 2 −1 0 1  17. A =  −2 1 −1 0 Write the augmented matrix.   1 2 −1 1 0 0  −2 0 1 0 1 0  1 −1 0 0 0 1

Multiply row 1 by 2 and add it to row 2. Also, multiply row 1 by −1 and add it to row 3.

516

Chapter 8: Systems of Equations and Matrices 

 0 −1.5 −1 −1 −1.5 −2.5. 0 −1 −0.5  3 −1 2 −1  1 0

 1 2 −1 1 0 0  0 4 −1 2 1 0  0 −3 1 −1 0 1 

Add row 3  1 −1  0 1 0 −3

to row 1 and also to row 2.  0 0 1 0 0 1 1 1  1 −1 0 1

Add row 2 to row 1. Also, multiply row 2 by 3 and add it to row 3.   1 0 0 1 1 2  0 1 0 1 1 1  0 0 1 2 3 4   1 1 2 −1 Then A =  1 1 1 . 2 3 4   7 −1 −9 0 −4  18. A =  2 −4 0 6 Write the augmented matrix.   7 −1 −9 1 0 0  2 0 −4 0 1 0  −4 0 6 0 0 1

Interchange row 1 and row 2.   2 0 −4 0 1 0  7 −1 −9 1 0 0  −4 0 6 0 0 1 Multiply row 2 by  2 0 −4  14 −2 −18 −4 0 6

2. 0 2 0

1 0 0

 0 0  1

Multiply row 1 by −7 and add it to row 2. Also, multiply row 1 by 2 and add it to row 3.   1 0 2 0 −4 0  0 −2 10 2 −7 0  2 1 0 0 −2 0

Multiply row 3 by −2 and add it to row 1. Also, multiply row 3 by 5 and add it to row 2.   2 0 0 0 −3 −2  0 −2 0 2 3 5  0 0 −2 0 2 1

1 1 . Also, multiply rows 2 and 3 by − . 2 2  3 −1  0 0 − 2  3 5  0 −1 − −  2 2   1  0 −1 − 1 2   3 0 − −1   2     5 3 −1 , or Then A =  − −1 −  2 2     1  0 −1 − 2

Multiply   1 0    0 1    0 0

row 1 by



1 19. A =  0 1

Write the augmented matrix.   1 3 −1 1 0 0  0 2 −1 0 1 0  1 1 0 0 0 1

Multiply row 1 by −1 and add it to row 3.   1 3 −1 1 0 0  0 0 1 0  2 −1 0 −2 1 −1 0 1 Add row 3  1 1  0 0 0 −2

to row 1 and also to row 2.  0 0 0 0 0 −1 1 1  1 −1 0 1

Since the second row consists only of zeros to the left of the vertical line, it will not be possible to obtain the identity matrix on the left side. Thus, A−1 does not exist.   −1 0 −1 0  20. A =  −1 1 0 1 1 Write  −1  −1 0

the augmented matrix.  0 −1 1 0 0 1 0 0 1 0  1 1 0 0 1

Multiply row 1 by  1 0 1 −1  −1 1 0 0 0 0 1 1 Add  1  0 0

row 0 1 1

−1. 0 1 0

 0 0  1

1 to row 2.  1 −1 0 0 1 −1 1 0  0 0 1 1

Since the second and third rows are identical to the left of the vertical line, it will not be possible to obtain the identity matrix on the left side. Thus, A−1 does not exist.   1 2 3 4  0 1 3 −5   21. A =   0 0 1 −2  0 0 0 −1 Write the augmented matrix.  1 2 3 4 1 0 0 0  0 1 3 −5 0 1 0 0   0 0 1 −2 0 0 1 0 0 0 0 −1 0 0 0 1 Multiply  1 2  0 1   0 0 0 0

row 4 by −1.

   

 0 3 4 1 0 0 3 −5 0 1 0 0   1 −2 0 0 1 0  0 1 0 0 0 −1

Exercise Set 8.5

517

Multiply row 4 by −4 and add it to row 1. Multiply row 4 by 5 and add it to row 2. Also, multiply row 4 by 2 and add it to row 3.   4 1 2 3 0 1 0 0  0 1 3 0 0 1 0 −5     0 0 1 0 0 0 1 −2  0 0 0 1 0 0 0 −1

Multiply  1 2  0 1   0 0 0 0

Multiply  1 0  0 1   0 0 0 0

row 3 by −3 and add it to row 1 and to row 2.  0 0 1 0 −3 10 0 0 0 1 −3 1   1 0 0 0 1 −2  0 −1 0 1 0 0

row 2 by −2 and add it to row 1.  3 8 0 0 1 −2 1 −3 1  0 0 0  1 0 0 0 1 −2  0 0 −1 0 1 0   1 −2 3 8  0 1 −3 1  . Then A−1 =   0 0 1 −2  0 0 0 −1   −2 −3 4 1  0 1 1 0   22. A =   0 4 −6 1  −2 −2 5 1 Write the augmented matrix.  −2 −3 4 1 1 0 0  0 1 1 0 0 1 0   0 4 −6 1 0 0 1 −2 −2 5 1 0 0 0

Multiply row 1 by −1 and  −2 −3 4 1 1  0 0 1 1 0   0 4 −6 1 0 0 1 1 0 −1

 0 0   0  1

add it to row 4.  0 0 0 1 0 0   0 1 0  0 0 1

Since the second and fourth rows are identical to the left of the vertical line, it will not be possible to get the identity matrix on the left side. Thus, A−1 does not exist.   1 −14 7 38  −1 2 1 −2   23. A =   1 2 −1 −6  1 −2 3 6 Write the augmented matrix.  1 −14 7 38 1 0  −1 2 1 −2 0 1   1 2 −1 −6 0 0 1 −2 3 6 0 0 Add it to  1  0   0 0

row 1 row 3 −14 −12 16 12

0 0 1 0

 0 0   0  1

to row 2. Also, multiply row 1 by −1 and add and to row 4.  1 0 0 0 7 38 8 36 1 1 0 0   −8 −44 −1 0 1 0  −4 −32 −1 0 0 1

Add row 2 to row 4.  1 −14 7 38 1  0 −12 1 8 36   0 16 −8 −44 −1 0 0 4 4 0

1 Multiply row 4 by . 4  1 1 −14 7 38  0 −12 8 36 1   0 16 −8 −44 −1  0 0 0 1 1

0 1 0 1

0 0 1 0

 0 0   0  1

 0 0 0 1 0 0   0 1 0   1 1 0 4 4 Multiply row 4 by −38 and add it to row 1. Multiply row 4 by −36 and add it to row 2. Also, multiply row 4 by 44 and add it to row 3.  19  19 0 − 1 − 1 −14 −31 0  2 2      0 −12 −28 0 1 −8 0 −9     0 16 36 0 −1 11 1 11   1 1 0 0 0 1 1 0 4 4 1 . Multiply row 3 by 36   19 19 0 − 1 − 1 −14 −31 0  2 2      0 −12 −28 0 1 −8 0 −9     11 1 11 4 1   0 1 0 −  9 36 36 36 36     1 1  0 0 1 1 0 0 4 4 Multiply row 3 by 31 and add it to row 1. Multiply row 3 by 28 and add it to row 2. Also, multiply row 3 by −1 and add it to row 4.   5 1 31 1 2 0 0 − − 1 −  9 36 36 36 36     2 5 7 4  4   0 0 0 −  9 9 9 9 9     1 11 1 11  4   0 1 0 −  9 36 36 36 36     1 1 1 1  4 0 − 0 1 − − − 9 36 18 36 18 1 Multiply row 2 by and add it to row 1. Also, multiply 2 row 2 by −1 and add it to row 3. Add row 2 to row 4.   1 5 1 1 − 1 0 0 0  4 4 4 4     2 5 7 4    0 4 0 0 −  9 9 9 9 9     3    0 0 1 0 −1 −1 −3  4 4 4 4     1 1 3 1  0 0 0 1 − 4 2 4 2 9 Multiply row 2 by . 4

518

Chapter 8: Systems of Equations and Matrices 

1

0

0

0

    0 1 0 0     0 0 1 0    0 0 0 1



Then A−1



     =     

1 4 1 2 1 − 4 1 4

1 4 5 4 1 − 4 1 2

5 4 7 4 3 − 4 3 4

1 4 1 2 1 − 4 1 4

1 4 5 4 1 − 4 1 2

5 4 7 4 3 − 4 3 4

 1 4    −1    3   4   1  − 2 −

 1 4    −1   , or 3   4   1  − 2  −

0.25 0.25 1.25 −0.25  0.5 1.25 1.75 −1   .  −0.25 −0.25 −0.75 0.75  0.25 0.5 0.75 −0.5   10 20 −30 15  3 −7 14 −8   24. A =   −7 −2 −1 2  4 4 −3 1 Write the augmented matrix.  10 20 −30 15 1 0  3 −7 14 −8 0 1   −7 −2 −1 2 0 0 4 4 −3 1 0 0 Multiply rows  10 20  30 −70   −70 −20 20 20

0 0 1 0

 0 0   0  1

2 and 3 by 10. Also, multiply row 4 by 5.  −30 15 1 0 0 0 0 0  140 −80 0 10  0 10 0  −10 20 0 −15 5 0 0 0 5

Multiply row 1 by −3 and add it to row 2. Multiply row 1 by 7 and add it to row 3. Also, multiply row 1 by −2 and add it to row 4.   1 0 0 0 10 20 −30 15  0 −130 230 −125 −3 10 0 0     0 120 −220 125 7 0 10 0  0 0 5 0 −20 45 −25 −2

Interchange rows 2 and 4.  10 20 −30 15 1  0 −20 45 −25 −2   0 7 120 −220 125 0 −130 230 −125 −3 Multiply row 4 by 2.  10 20 −30 15 1  0 −2 −20 45 −25   0 120 −220 125 7 0 −260 460 −250 −6

 0 0 0 0 0 5   0 10 0  10 0 0  0 0 0 0 0 5   0 10 0  20 0 0

Add row 2 to row 1. Multiply row 2 by 6 and add it to row 3. Also, multiply row 2 by −13 and add it to row 4.



10 0 15 −10 −1 0  0 −20 45 −25 −2 0   0 0 0 50 −25 −5 0 0 −125 75 20 20

 0 5 0 5   10 30  0 −65

Multiply rows 1 and 2 by 10 and multiply row 4 by 2.   50 100 0 150 −100 −10 0 0  0 −200 50  450 −250 −20 0 0    0 0 50 −25 −5 0 10 30  40 40 0 −130 0 0 −250 150 Multiply row 3 by −3 and add it to row 1. by −9 and add it to row 2. Also, multiply add it to row 4.  5 0 −30 100 0 0 −25  0 −200 0 −25 25 0 −90   0 0 50 −25 −5 0 10 50 0 0 0 25 15 40

Multiply row 3 row 3 by 5 and  −40 −220   30  20

Add row 4 to rows 1, 2, and 3.   20 −20 100 0 0 0 20 40  0 −200 0 0 40 40 −40 −200     0 60 50  0 50 0 10 40 0 0 0 25 15 40 50 20 1 1 , multiply row 2 by − , multiply Multiply row 1 by 100 200 1 1 , and multiply row 4 by . row 3 by 50 25   2 1 1 1 1 0 0 0 −  5 5 5 5      1 1 1   0 1 0 0 − − 1   5 5 5     1 4 6  0 0 1 0 1    5 5 5    3 8 4  2 0 0 0 1 5 5 5  2 1 1 1 −  5 5 5 5     1 − − 1 1 1   5 5 5 , or =   1 4 6   1   5 5 5    3 8 4 2 5 5 5   0.2 0.4 0.2 −0.2 −0.2 −0.2 0.2 1 .   0.2 0.8 1.2 1 0.6 1.6 2 0.8 

Then A−1

25. Write an equivalent matrix equation, AX = B. 1 21 2 1 2 x −4 11 3 = 7 2 y 5 Then we have X = A−1 B. 1 2 1 21 2 1 2 x 2 −3 −4 −23 = = y −7 11 5 83

The solution is (−23, 83). 1 2 1 21 2 1 2 26. x −3 5 −6 28 = = y 5 −8 2 −46 The solution is (28, −46).

Exercise Set 8.5 27. Write an equivalent matrix equation, AX = B.      3 1 0 x 2  2 −1 2   y  =  −5  1 1 1 z 5

Then we have X = A−1 B.        3 1 −2 2 −9 x 1 1  y  =  0 −3 6   −5  =  45  = 9 9 −3 2 5 5 9 z   −1  5  1

519 32.

−x + 4y = −5

Write an equivalent matrix equation, AX = B. 1 21 2 1 2 1 −6 x 5 = −1 4 y −5   2 1 2 −2 −3 1 5 5 Then X = A−1 B =  1 1  = . −5 0 − − 2 2 The solution is (5, 0). 33.

The solution is (−1, 5, 1).        −3 2 −1 −4 5 x 1 1 28.  y  =  12 −3 4   −3  =  −35  = 5 5 7 −3 4 1 −15  z  1  −7  −3

x−y+z=4

x − 2y = 6

30.

34.

−3x + 5y − 6z = 4

Write an equivalent matrix equation, AX = B.      1 2 3 x −1  2 −3 4  y  =  2  −3 5 −6 z 4      −2 27 17 −1 124 3 2   2  =  14 . Then X = A−1 B =  0 1 −11 −7 4 −51

4x + y = −7

31.

5x + y = 2, 3x − 2y = −4

Write an equivalent matrix equation, AX = B.  2 1 1 2 1 21 2 2 5 1 x  13 13  =  −4 = 3 −2 y 3 5 − 13 13 1 2 2 −4 1 2 0 . Then X = A−1 B = 2 The solution is (0, 2).

x + 2y + 3z = −1,

2x − 3y + 4z = 2,

2x − 3y = 7,

Write an equivalent matrix equation, AX = B. 1 21 2 1 2 2 −3 x 7 = 4 1 y −7  1 3  1 2 1 2 7 −1  14 14  −1 Then X = A B =   −7 = −3 . 2 1 − 7 7 The solution is (−1, −3).

= 3,

Write an equivalent matrix equation, AX = B.      1 0 1 x 1  2 1 0  y  =  3  1 −1 1 z 4   1 1 1     −  2 2 2 1 3   −1     Then X = A B =   1 0 −1  3 = −3 .  3 1 1 4 −2 − − 2 2 2 The solution is (3, −3, −2).

4x + 3y = 2, Write an equivalent matrix equation, AX = B. 1 21 2 1 2 4 3 x 2 = 1 −2 y 6  2 3 1 2 1 2 2  11 11  2 Then X = A−1 B =   6 = −2 . 4 1 − 11 11 The solution is (2, −2).

+ z = 1,

x 2x + y

The solution is (1, −7, −3).

29.

x − 6y = 5,

The solution is (124, 14, −51).

35.

2x + 3y + 4z = 2, x − 4y + 3z = 2,

5x + y + z = −4

Write an equivalent matrix equation, AX = B.      2 3 4 x 2  1 −4 3   y  =  2  5 1 1 z −4  1 1 25  −  16 112 112        −1 2  1 9 1   2  =  0 . Then X=A−1 B =  − −  8 56 56   −4  1   11 3 13 − 16 112 112 The solution is (−1, 0, 1).

520 36.

Chapter 8: Systems of Equations and Matrices x+ y 3x

x + y = 145

= 2,

The number of hot dogs sold is 45 more than the number of sausages.

+ 2z = 5,

2x + 3y − 3z = 9

Write an equivalent matrix     1 1 0 x  3 0 2  y  =  2 3 −3 z  6 3 −  7 7   13 3 Then X=A−1 B =   7 −7   9 1 − 7 7 The solution is (3, −1, −2). 37.

x = y + 45

equation, AX = B.  2 5  9 2

We have a system of equations: x + y = 145, x = y + 45,

7      2 3 2 5  = −1 . −   7 9 −2  3 − 7

2w − 3x + 4y − 5z = 0,

−w − 3x − 6y + 4z = 6

Write  2  3   1 −1

an equivalent matrix equation, AX = B.     −3 4 −5 w 0  x   2  −2 7 −3   =  1 −1 1  y   1  −3 −6 4 z 6   0 26 11 127 9 2 1  −8 −19 39 −34  −1    Then X = A B = 203 −37 39 −48 −5  1 −55 47 −11 20 6   1 −1     0 . 1

State. Stefan sold 95 hot dogs and 50 Italian sausages. 40. Let x = the price of a lab record book and y = the price of a highlighter. Solve: 4x + 3y = 13.93, 

  = 

Translate. Four tons of topsoil, 3 tons of mulch, and 6 tons of pea gravel costs $2825.

5w − 4x + 3y − 2z = −6,

4x + 3y + 6z = 2825

w + 4x − 2y + 3z = −5,

Five tons of topsoil, 2 tons of mulch, and 5 tons of pea gravel costs $2663.

2w − 3x + 6y − 9z = 14,

Then

matrix  w  x   y z

5x + 2y + 5z = 2663

equation, AX = B.    −6   −5   =   14  −3



 18 75 1 45 −6  −5 1  −10 59 33 −25 −1   X=A B= 302  112 −87 23 −173   14 82 −61 −29 −97 −3

The solution is (−2, 1, 2, −1).

  −2   1 =   2 −1

Pea gravel costs $17 less per ton than topsoil. z = x − 17

We have a system of equations. 4x + 3y + 6z = 2825,

The total number of items sold was 145.

5x + 2y + 5z = 2663,



 . 

39. Familiarize. Let x = the number of hot dogs sold and y = the number of sausages. Translate.

3x + 2y = 10.25 1 21 2 1 2 4 3 x 13.93 Writing = , we find that x = 3 2 y 10.25 $2.89, y = $0.79. 41. Familiarize. Let x, y, and z represent the prices of one ton of topsoil, mulch, and pea gravel, respectively.

The solution is (1, −1, 0, 1).

3w − 5x + 2y − 4z = −3

x − y = 45.

Check. The total number of items is 95 + 50, or 145. The number of hot dogs, 95, is 45 more than the number of sausages. The solution checks.

w + x − y + z = 1,

Write an equivalent  5 −4 3 −2  1 4 −2 3   2 −3 6 −9 3 −5 2 −4

x + y = 145,

Carry out. Write an equivalent matrix equation, AX = B. 1 21 2 1 2 1 1 x 145 = 1 −1 y 45   1 1 1 2 1 2  2 2   145 = 95 , so the Then X = A−1 B =   1 45 50 1  − 2 2 solution is (95, 50).

3w − 2x + 7y − 3z = 2,

38.

or

z = x − 17, or 4x + 3y + 6z = 2825, 5x + 2y + 5z = 2663, x

− z = 17

Carry out. Write an equivalent matrix equation, AX = B.      4 3 6 x 2825  5 2 5   y  =  2663  1 0 −1 z 17

Exercise Set 8.5

521

1 3 3  5 10 10  Then X = A−1 B =  1  1 −1  1 3 7 − − 5 10 10   239  179 , so the solution is (239, 179, 222). 222 

−







 2825    2663  =   17

50.

51.

2x2 + x = 7 2x + x − 7 = 0 2

a = 2, b = 1, c = −7 √ −b ± b2 − 4ac x= 2a 4 √ −1 ± 12 − 4 · 2 · (−7) −1 ± 1 + 56 = = 2·2 4 √ −1 ± 57 = 4 √ √ −1 + 57 −1 − 57 The solutions are and , or 4 4 √ −1 ± 57 . 4

State. The price of topsoil is $239 per ton, of mulch is $179 per ton, and of pea gravel is $222 per ton. 42. Let x, y, and z represent the amounts invested at 2.2%, 2.65%, and 3.05%, respectively. Solve: x + y + z = 8500,

52.

0.022x + 0.0265y + 0.0305z = 230, z = x + 1500 1 0.0265 0

2 −1 5 6 −4 6 15 60 198 2 5 20 66 194

f (3) = 194

Check. Four tons of topsoil, 3 tons of mulch, and 6 tons of pea gravel costs 4 · $239 + 3 · $179 + 6 · $222, or $956 + $537 + $1332, or $2825. Five tons of topsoil, 2 tons of mulch, and 5 tons of pea gravel costs 5 · $239 + 2 · $179 + 5 · $222, or $1195 + $358 + $1110, or $2663. The price of pea gravel, $222, is $17 less than the price of topsoil, $239. The solution checks.

Writing  1  0.022 −1

3 33

    1 x 8500 0.0305   y  =  230 , 1 z 1500

1 6 − = 1, LCD is (x + 1)(x − 1) x+1 x−1 " ! 6 1 − = (x+1)(x−1)·1 (x + 1)(x − 1) x+1 x−1 x − 1 − 6(x + 1) = x2 − 1 x − 1 − 6x − 6 = x2 − 1

0 = x2 + 5x + 6 0 = (x + 3)(x + 2)

we find that x = $2500, y = $2000, z = $4000.

x = −3 or x = −2

43. Left to the student 44. Left to the student 45. Left to the student 46. Left to the student 1 2 1 0 47. No; for example, let A = B = . 0 1 1 1 2 0.5 2 0 Then A + B = and (A + B)−1 = 0 0 2 1 2 1 2 1 1 0 1 0 2 0 but A−1 + B−1 = + = 0 1 0 1 0 2 1 2 1 3 2 11 48. No; for example, let A = and B = 5 3 7 1 2 47 13 Then AB = and (AB)−1 = 1 2 76 21 −21 13 , 76 −47 but

1 21 2 −3 2 2 −3 = A−1 B−1 = −7 11 1 2 5 −3 −20 31 . 31 −48 3 49. −2 3 1 −6 4 −8 −2 16 −40 1 −8 20 −48 f (−2) = −48

0 0.5 2 . 3 2

2 ,

2 .

Both numbers check. √ √ 53. 2x + 1 − 1 = 2x − 4 √ √ ( 2x + 1 − 1)2 = ( 2x − 4)2 Squaring both sides √ 2x + 1 − 2 2x + 1 + 1 = 2x − 4 √ 2x + 2 − 2 2x + 1 = 2x − 4 √ 2 − 2 2x + 1 = −4 Subtracting 2x √ Subtracting 2 −2 2x + 1 = −6 √ 2x + 1 = 3 Dividing by −2 √ Squaring both ( 2x + 1)2 = 32 sides 2x + 1 = 9 2x = 8 x=4 The number 4 checks. It is the solution. √ 54. x − x − 6 = 0 √ Let u = x. u2 − u − 6 = 0

(u − 3)(u + 2) = 0 u = 3 or √ x = 3 or x=9

u = −2 √ x = −2

No solution

The number 9 checks. It is the solution.

522

Chapter 8: Systems of Equations and Matrices

55. f (x) = x3 − 3x2 − 6x + 8

We use synthetic division to find one factor. We first try x − 1. 3 1 3 1 −3 −6 8 1 −2 −8 1 −2 −8 0

Since f (1) = 0, x − 1 is a factor of f (x). We have f (x) = (x − 1)(x2 − 2x − 8). Factoring the trinomial we get f (x) = (x − 1)(x − 4)(x + 2). 56. f (x) = x4 + 2x3 − 16x2 − 2x + 15 We try x − 1. 3 1 3 1 2 −16 −2 15 1 3 −13 −15 1 3 −13 −15 0

We have f (x) = (x − 1)(x3 + 3x2 − 13x − 15). Now we factor the cubic polynomial. We try x + 1. 3 −1 3 1 3 −13 −15 −1 −2 15 1 2 −15 0

Now we have f (x) = (x − 1)(x + 1)(x2 + 2x − 15) = (x − 1)(x + 1)(x + 5)(x − 3). 57. A = [x] Write the augmented matrix. 0 / x 1 1 Multiply by . x 2 1 1 1 x Then A 58. A =

1

−1

exists if and only if x $= 0. A

x 0 0 y

−1

1 2 1 = . x

2

Write the augmented matrix. 2 1 x 0 1 0 0 y 0 1 Multiply row 1 by   

1 0

0 1

1 x 0

0

1 1 and row 2 by . x y



 1  y

Then A−1 exists if and only if x $= 0 and y $= 0, or if and only if xy $= 0.   1 0   x A−1 =  1  0 y



 0 0 x 59. A =  0 y 0  z 0 0

Write the augmented matrix.   0 0 x 1 0 0  0 y 0 0 1 0  z 0 0 0 0 1

Interchange  z 0 0  0 y 0 0 0 x Multiply  1 0     0 1   0 0

row 1 0 0 0 1 1 0

row 1 0

0

0

0

and row 3.  1 0  0 1 1 1 by , row 2 by , and row 3 by . z y x 1  0 z   1  0   y  0 0

1 x Then A−1 exists if and only if x $= 0 and y $= 0 and z $= 0, or if and only if xyz $= 0.  1  0 0  z    1   −1 A = 0 0    y   1 0 0 x   x 1 1 1  0 y 0 0   60. A =   0 0 z 0  0 0 0 w 1

Write the augmented matrix.   x 1 1 1 1 0 0 0  0 y 0 0 0 1 0 0     0 0 z 0 0 0 1 0  0 0 0 w 0 0 0 1 1 Multiply row 4 by − and add it to row 1. w   1 x 1 1 0 1 0 0 −  w     0 y 0 0 0 1 0 0     0 0 z 0 0 0 1 0  0 0 0 w 0 0 0 1 1 Multiply row 3 by − and add it to row 1. z   1 1 x 1 0 0 1 0 − −  z w     0 y 0 0 0 1 0 0     0 0 z 0 0 0 1 0  0 0 0 w 0 0 0 1 1 Multiply row 2 by − and add it to row 1. y   1 1 1 1 − x 0 0 0 − −  y z w     0 y 0 0 0 1 0 0     0 0 z 0 0 0 1 0  0 0 1 0 0 0 w 0

Exercise Set 8.6

523

Multiply row 1 by 1 by . w  1 0 0 0     0 1 0 0     0 0 1 0   0 0 0 1

1 x

1 1 1 , row 2 by , row 3 by , and row 4 x y z

1  xw    0 0  0   1  0 0 0   z  1 0 0 0 w Then A−1 exists if and only if w $= 0 and x $= 0 and y $= 0 and z $= 0, or if and only if wxyz $= 0.  1 1 1 1  − − −  x xy xz xw    1    0 0 0    y A−1 =   1    0 0 0    z   1 0 0 0 w −

1 xy 1 y

−

1 xz

−

Exercise Set 8.6 1. 2. 3. 4. 5. 6. 7. 8.

9.

3 3 3 5 3 33 3 3 −2 −4 3 = 5(−4) − (−2) · 3 = −20 + 6 = −14 3 3 3 −8 6 3 3 3 3 −1 2 3 = −8 · 2 − (−1) · 6 = −16 + 6 = −10 3 3 3 4 −7 3 3 3 3 −2 3 3 = 4 · 3 − (−2)(−7) = 12 − 14 = −2 3 3 3 −9 −6 3 3 3 = −9 · 4 − 5(−6) = −36 + 30 = −6 3 5 4 3 3 √ 3 3 −2 − 5 3 √ √ 3 √ 3 = −2 · 3 − (− 5)(− 5) = −6 − 5 = −11 3− 5 3 3 3 3√ 3 5 −3 3 √ √ 3 = 5(2) − 4(−3) = 2 5 + 12 3 3 4 2 3 3 3 3x 4 3 2 3 3 3 2 3 x x 3 = x · x − x · 4 = x − 4x 3 2 3 3 y −2 3 3 = y 2 · 3 − y(−2) = 3y 2 + 2y 3 3y 3 3   7 −4 −6 0 −3  A= 2 1 2 −5

M11 is the determinant of the matrix formed by deleting the first3 row and 3 first column of A: 3 0 −3 3 3 = 0(−5) − 2(−3) = 0 + 6 = 6 M11 = 33 2 −5 3 M32 is the determinant of the matrix formed by deleting the third 3 row and 3 second column of A: 3 7 −6 3 3 = 7(−3) − 2(−6) = −21 + 12 = −9 M32 = 33 2 −3 3

M22 is the the second 3 37 M22 = 33 1

determinant of the matrix formed by deleting row 3and second column of A: −6 33 = 7(−5) − 1(−6) = −35 + 6 = −29 −5 3

10. See matrix A in Exercise 9 above: 3 3 32 03 3=2 · 2 − 1 · 0 = 4 − 0 = 4 M13 = 33 1 23 3 3 3 −4 −6 3 3 3 = −4(−3) − 0(−6) = 12 − 0 = 12 M31 = 3 0 −3 3 3 3 3 7 −4 3 3 = 7 · 2 − 1(−4) = 14 + 4 = 18 M23 = 33 1 2 3 11. In Exercise 9 we found that M11 = 6. A11 = (−1)1+1 M11 = 1 · 6 = 6

In Exercise 9 we found that M32 = −9.

A32 = (−1)3+2 M32 = −1(−9) = 9

In Exercise 9 we found that M22 = −29.

A22 = (−1)2+2 (−29) = 1(−29) = −29

12. In Exercise 10 we found that M13 = 4. A13 = (−1)1+3 M13 = 1 · 4 = 4

In Exercise 10 we found that M31 = 12.

A31 = (−1)3+1 M31 = 1 · 12 = 12

In Exercise 10 we found that M23 = 18.

A23 = (−1)2+3 M23 = −1 · 18 = −18   7 −4 −6 0 −3  13. A =  2 1 2 −5 |A|

= 2A21 + 0A22 + (−3)A23 3 3 3 3 −4 −6 3 3 3 +0+(−3)(−1)2+3 3 7 = 2(−1)2+1 33 3 31 2 −5

= 2(−1)[−4(−5) − 2(−6)] + 0+

3 −4 33 2 3

(−3)(−1)[7 · 2 − 1(−4)]

= −2(32) + 0 + 3(18) = −64 + 0 + 54 = −10

14. See matrix A in Exercise 13 above. |A|

= −4A12 + 0A22 + 2A32 3 3 3 3 3 2 −3 3 3 3 3 + 0 + 2(−1)3+2 3 7 −6 3 = −4(−1)1+2 33 3 3 1 −5 2 −3 3 = −4(−1)[2(−5) − 1(−3)] + 0+ 2(−1)[7(−3) − 2(−6)]

= 4(−7) + 0 − 2(−9) = −28 + 0 + 18 = −10

524

Chapter 8: Systems of Equations and Matrices 

 7 −4 −6 0 −3  15. A =  2 1 2 −5 |A|

= −6A13 + (−3)A23 + (−5)A33 3 3 3 3 32 03 3 3 3 + (−3)(−1)2+3 3 7 −4 3 + = −6(−1)1+3 33 3 3 1 2 1 2 3 3 3 3 7 −4 3 3 (−5)(−1)3+3 33 2 0 3 = −6 · 1(2 · 2 − 1 · 0) + (−3)(−1)[7 · 2 − 1(−4)]+ −5 · 1(7 · 0 − 2(−4))

= −6(4) + 3(18) − 5(8) = −24 + 54 − 40

= −10

16. See matrix A in Exercise 15 above. |A|

= 7A11 − 4A12 − 6A13 3 3 3 3 3 0 −3 3 3 3 3 + (−4)(−1)1+2 3 2 −3 3 + = 7(−1)1+1 33 3 1 −5 3 2 −5 3 3 3 32 03 3 (−6)(−1)1+3 33 1 23 = 7·1[0(−5)−2(−3)]+(−4)(−1)[2(−5)−1(−3)]+ (−6)(1)(2 · 2 − 1 · 0)

= 7(6) + 4(−7) − 6(4) = 42 − 28 − 24 = −10 

 1 0 0 −2  4 1 0 0   17. A =   5 6 7 8  −2 −3 −1 0

M41 is the determinant of the matrix formed by deleting the fourth row and the first column of A.   0 0 −2 0  M41 =  1 0 6 7 8 We will expand M41 across the first row. 3 3 3 3 30 03 3 3 3 + 0(−1)1+2 3 1 0 3 + M41 = 0(−1)1+1 33 36 83 7 83 3 3 3 3 3+1 3 1 0 3 (−2)(−1) 36 73 = 0 + 0 + (−2)(1)(1 · 7 − 6 · 0) = 0 + 0 − 2(7) = −14

M33 is the determinant of the matrix formed by deleting the third row and the third column of A.   1 0 −2 1 0  M33 =  4 −2 −3 0 We will expand M33 down the third column.

3 3 3 3 34 3 1 33 0 33 2+3 3 1 M33 = −2(−1)1+3 33 +0(−1) 3−2 −33+ −2 −33 3 3 31 03 3 0(−1)3+3 33 4 13 = −2(1)[4(−3) − (−2)(1)] + 0 + 0 = −2(−10) + 0 + 0 = 20

18. See matrix A in Exercise 17 above.   4 0 0 7 8  M12 =  5 −2 −1 0

We will expand M12 across the first row. 3 3 3 3 3 3 7 83 3 3 +0(−1)1+2 3 5 83+ M12 = 4(−1)1+1 33 3−2 03 −1 03 3 3 3 5 7 33 0(−1)1+3 33 −2 −1 3 = 4 · 1[7 · 0 − (−1)8] + 0 + 0

M44

= 4(8) = 32   1 0 0 = 4 1 0  5 6 7

We will expand M44 across the first row. 3 3 3 3 3 1 03 3 3 3 +0(−1)1+2 34 03+ M44 = 1(−1)1+1 33 3 3 6 7 5 73 3 3 34 13 3 0(−1)1+3 33 5 63 = 1 · 1(1 · 7 − 6 · 0) + 0 + 0

= 1(7) = 7   1 0 0 −2  4 1 0 0   19. A =   5 6 7 8  −2 −3 −1 0

A24 = (−1)2+4 M24 = 1 · M24 = M24 3 3 3 1 0 0 33 3 = 33 5 6 7 33 3 −2 −3 −1 3

We will expand across the first row. 3 3 3 1 0 0 33 3 3 5 6 7 33 3 3 −2 −3 −1 3 3 3 3 36 3 7 33 1+2 3 5 + 0(−1) = 1(−1)1+1 33 3 3−2 −3 −1 3 3 3 5 6 33 0(−1)1+3 33 −2 −3 3 = 1 · 1[6(−1) − (−3)(7)] + 0 + 0 = 1(15) = 15

A43 = (−1)4+3 M43 = −1 · M43 3 3 3 1 0 −2 3 3 3 = −1 · 33 4 1 0 33 35 6 8 3

We will expand across the first row.

3 7 33 + −13

Exercise Set 8.6 3 3 3 1 0 −2 3 3 3 −1 · 33 4 1 0 33 35 6 8 3 3 1 31 = −1 1(−1)1+1 33 6

3 3 3 033 1+2 34 + 0(−1) 3 35 8 32 3 34 13 3 (−2)(−1)1+3 33 5 63

525 

 1 0 0 −2  4 1 0 0   21. A =   5 6 7 8  −2 −3 −1 0

3 033 + 83

|A|

= 1 · A11 + 0 · A12 + 0 · A13 + (−2)A14

= −1[1·1(1·8−6 · 0) + 0 + (−2)·1(4·6−5·1)]

= −1[1(8) + 0 − 2(19)]

= −1(8 − 38) = −1(−30) = 30

20. See matrix A in Exercise 19 above. A22 = (−1)2+2 M22 = M22 3 3 3 1 0 −2 33 3 = 33 5 7 8 33 3 −2 −1 0 3

We will expand across the first row. 3 3 3 1 0 −2 33 3 3 5 7 8 33 3 3 −2 −1 0 3 3 3 3 3 7 83 3 3 + 0(−1)1+2 3 5 = 1(−1)1+1 33 3−2 −1 03 3 3 3 5 7 33 (−2)(−1)1+3 33 −2 −1 3

We will expand each determinant across the first row. We have: 3 3 3 7 83 3 +0+0+ 1(−1)1+1 33 −1 03 3 3 3 3 2 1 3 3 7 33 7 33 1+1 3 6 1+2 3 5 + 1(−1) 3 +0 2 4(−1) 3 −3 −13 −2 −1 3

3 833 + 03

= 1 · 1[7 · 0 − (−1)8] + 2[4 · 1[6(−1) − (−3) · 7]+ 1(−1)[5(−1) − (−2) · 7]

= 1·1[7·0−(−1)8] + 0 + (−2)·1[5(−1)−(−2)7]

= 1(8) + 0 − 2(9) = 8 + 0 − 18

= −1[1 · 1(1(−1) − (−3) · 0) + 0 + 0]

= 8 + 102 = 110

= 7A33 − A43 3 3 1 3 = 7(−1)3+3 33 4 3 −2 3 31 3 (−1)4+3 33 4 35

A34 = (−1)3+4 M34 = −1 · M34 3 3 3 1 0 0 33 3 = −1 · 33 4 1 0 33 3 −2 −3 −1 3

= −1[1(−1)] = 1

= 1(8) + 2[4(15) − 1(9)] = 8 + 2(51) 22. See matrix A in Exercise 21 above. |A| = 0 · A13 + 0 · A23 + 7A33 + (−1)A43

= −10

We will expand across the first row. 3 3 3 1 0 0 33 3 −1 · 33 4 1 0 33 3 −2 −3 −1 3 3 3 3 1 31 3 0 33 1+2 3 4 + 0(−1) = −1 1(−1)1+1 33 3 3−2 −3 −1 3 32 3 4 1 33 0(−1)1+3 33 −2 −3 3

= A11 + (−2)A14 3 3 3 1 0 0 33 3 7 8 33 + = (−1)1+1 33 6 3 −3 −1 0 3 3 3 3 4 1 0 33 3 (−2)(−1)1+4 33 5 6 7 33 3 −2 −3 −1 3 3 3 3 3 3 4 3 1 1 0 33 0 0 33 3 3 6 7 33 7 8 33 + 2 33 5 = 33 6 3 −2 −3 −1 3 3 −3 −1 0 3

3 0 33 + −13

3 −2 33 0 33− 0 3 3 0 −2 33 1 0 33 6 8 3

0 1 −3

We will expand each determinant down the third column. We have: 3 3 2 1 3 4 13 3 +0 + 0 + 7 − 2(−1)1+3 33 −2 3 3 3 3 3 32 1 34 13 3 3 3 + 0 + 8(−1)3+3 3 1 0 3 − 2(−1)1+3 33 34 13 5 63 = 7[−2(4(−3) − (−2) · 1)] + [−2(4 · 6 − 5 · 1)+ 8(1 · 1 − 4 · 0)]

= 7[−2(−10)] + [−2(19) + 8(1)] = 7(20) + (−30) = 140 − 30 = 110

526

Chapter 8: Systems of Equations and Matrices

23. We will expand across the first row. We could have chosen any other row or column just as well. 3 3 3 3 23 3 3 1 3 3 3 3 13 3 −2 3 3 3 3 3 −6 3 3 3 4 3 3 3 3 33 1 3 3 3 3 + 1 · (−1)1+2 3 −2 1 3 + = 3(−1)1+1 33 3 3 −6 3 4 −6 3 3 3 −2 2(−1)1+3 33 3

26. We will expand across the third row. 3 3 3 3 −1 3 3 x 1 3 3 3 2 3 x x3 3x 3 3 3 3 13 3 0 x 3 3 3 3 1 −1 3 3 3 + x(−1)3+2 3 x2 = 0(−1)3+1 33 3x x x 3 3 3x 1(−1)3+3 33 2 x

3 3 33 43

= 0 + x(−1)[x·x−x2 (−1)]+ 1 · 1(x · x−x2 · 1)

= 3·1[3(−6)−4 · 1]+1(−1)[−2(−6) − 3 · 1]+

= −x(x2 + x2 ) + 1(x2 − x2 )

2 · 1(−2 · 4−3 · 3)

= 3(−22) − (9) + 2(−17) = −109

27.

3 3 −2 Dy = 33 3

1 · 1[3 · 4 − 2(−2)]

= 1 · 14 − 3 · 13 + 1 · 16

28.

x(−1)

3 3 3 x −1 3 3 3 3 2 x2 3

= 0(−1)[2·1−(−3)x2 ]+ x · 1[x · 1−(−3)(−1)]+ x(−1)[x · x2 − 2(−1)]

= 0 + x(x − 3) − x(x3 + 2)

= x2 − 3x − x4 − 2x = −x4 + x2 − 5x

3 3 33 = −2 · 1 − 3 · 3 = −2 − 9 = −11 13

Dx −25 25 = =− D 2 2 −11 11 Dy = =− y= D 2 2 " ! 25 11 . The solution is − , − 2 2 x=

= 1 · 1[2 · 5 − (−1) · 4] + 3(−1)[3 · 5 − (−1)(−2)]+

3+2

−2x + 4y = 3,

3 3 33 4 3 3 = 3(−7) − 1 · 4 = −21 − 4 = −25 Dx = 33 1 −7 3

3 3 3 3 −2 3 3 1(−1)3+3 33 2 4 3

25. We will expand down the second column. We could have chosen any other row or column just as well. 3 3 3 3 −1 3 3 x 0 3 3 3 3 x2 3 3 2 x 3 3 3 3 13 3 −3 x 3 3 3 3 3 2 x2 3 3 3 3 + x(−1)2+2 3 x −1 3 + = 0(−1)1+2 33 3 3 −3 1 −3 1 3

= −x(2x2 ) + 1(0) = −2x3

3x − 7y = 1 3 3 3 −2 4 3 3 = −2(−7) − 3 · 4 = 14 − 12 = 2 D = 33 3 −7 3

24. We will expand down the third column. 3 3 3 3 13 3 3 −2 3 3 3 3 4 33 3 2 3 3 3 3 5 13 3 −1 3 3 3 3 3 2 43 3 3 3 + 3(−1)2+3 3 3 −2 3 + = 1(−1)1+3 33 3 −1 5 3 −1 5 3

= −9

3 1 33 x3

3 −1 33 + x 3

5x − 4y = −3,

7x + 2y = 6 3 3 3 5 −4 3 3 = 5 · 2 − 7(−4) = 10 + 28 = 38 D = 33 7 2 3 3 3 −3 Dx = 33 6

3 −4 33 = −3 · 2 − 6(−4) = −6 + 24 = 18 2 3

3 3 3 5 −3 3 3 3 = 5 · 6 − 7(−3) = 30 + 21 = 51 Dy = 3 7 6 3

18 9 Dx = = D 38 19 51 Dy y= = D 38 ! " 9 51 The solution is , . 19 38 x=

Exercise Set 8.6 29.

527

2x − y = 5,

32.

3x + 6y = −0.5 3 3 32 33 3 = 2 · 6 − 3 · 3 = 12 − 9 = 3 D = 33 3 63 3 3 3 −1 3 3 3 = −1 · 6 − (−0.5)3 = Dx = 33 −0.5 6 3 −6 + 1.5 = −4.5 3 3 3 2 −1 3 3 = 2(−0.5) − 3(−1) = −1 + 3 = 2 Dy = 33 3 −0.5 3

x − 2y = 1 3 3 3 2 −1 3 3 = 2(−2) − 1(−1) = −4 + 1 = −3 D = 33 1 −2 3

3 35 Dx = 33 1 3 32 Dy = 33 1

3 −1 33 = 5(−2) − 1(−1) = −10 + 1 = −9 −2 3 3 5 33 = 2 · 1 − 1 · 5 = 2 − 5 = −3 13

Dx −9 = =3 D −3 −3 Dy = =1 y= D −3 The solution is (3, 1). x=

30.

3x + 4y = −2,

5x − 7y = 1 3 3 33 4 3 3 3 = 3(−7) − 5 · 4 = −21 − 20 = −41 D=3 5 −7 3 3 3 −2 Dx = 33 1 3 33 Dy = 33 5

31.

33.

Dx −4.5 3 = = −1.5, or − D 3 2

y=

Dy 2 = D 3

!

" 3 2 . − , 2 3

2x + 5y = 7, 3x − 2y = 1 3 3 32 5 3 3 = 2(−2) − 3 · 5 = −4 − 15 = −19 D = 33 3 −2 3

3 4 33 = −2(−7) − 1 · 4 = 14 − 4 = 10 −7 3

3 3 37 5 3 3 3 = 7(−2) − 1 · 5 = −14 − 5 = −19 Dx = 3 1 −2 3

3 −2 33 = 3 · 1 − 5(−2) = 3 + 10 = 13 1 3

2x + 9y = −2,

x=

The solution is

3 3 32 73 3 3 = 2 · 1 − 3 · 7 = 2 − 21 = −19 Dy = 3 3 13

10 10 Dx = =− D −41 41 13 13 Dy y= = =− D −41 41 " ! 10 13 . The solution is − , − 41 41 x=

2x + 3y = −1,

−19 Dx = =1 D −19 −19 Dy y= = =1 D −19 The solution is (1, 1). x=

34.

3x + 2y = 7,

4x − 3y = 3 3 3 32 9 3 3 = 2(−3) − 4 · 9 = −6 − 36 = −42 D = 33 4 −3 3

2x + 3y = −2 3 3 33 23 3=3·3−2·2=9−4=5 D = 33 2 33

3 32 Dy = 33 4

3 3 33 7 3 3 = 3(−2) − 2 · 7 = −6 − 14 = −20 Dy = 33 2 −2 3

3 3 −2 Dx = 33 3

3 9 33 = −2(−3) − 3 · 9 = 6 − 27 = −21 −3 3

3 −2 33 = 2 · 3 − 4(−2) = 6 + 8 = 14 3 3

Dx −21 1 = = D −42 2 14 1 Dy y= = =− D −42 3 " ! 1 1 ,− . The solution is 2 3 x=

3 3 3 7 23 3 = 7 · 3 − (−2) · 2 = 21 + 4 = 25 Dx = 33 −2 3 3

25 Dx = =5 D 5 Dy −20 y= = = −4 D 5 The solution is (5, −4). x=

528 35.

Chapter 8: Systems of Equations and Matrices 3x + 2y − z = 4,

37.

3x − 2y + z = 5,

x − 4y + 2z = 13,

4x − 5y − z = −1 3 3 3 3 2 −1 3 3 3 D = 33 3 −2 1 33 = 42 3 4 −5 −1 3 3 3 3 4 2 −1 33 3 Dx = 33 5 −2 1 33 = 63 3 −1 −5 −1 3 3 33 3 Dy = 33 3 34 3 33 3 Dz = 33 3 34

2x + 4y + 3z = 1 3 3 3 3 5 −1 3 3 3 D = 33 1 −4 2 33 = −67 32 4 3 3 3 3 3 −2 5 −1 3 3 3 Dx = 33 13 −4 2 33 = −201 3 1 4 3 3 3 3 3 3 −2 −1 3 3 3 Dy = 33 1 13 2 33 = 134 32 1 3 3 3 3 3 3 5 −2 3 3 3 Dz = 33 1 −4 13 33 = −67 32 4 1 3

3 −1 33 1 33 = 39 −1 3 3 2 4 33 −2 5 33 = 99 −5 −1 3 4 5 −1

x=

Dx 63 3 = = D 42 2

y=

Dy 39 13 = = D 42 14

99 33 Dz = = D 42! 14 " 3 13 33 , , The solution is . 2 14 14 (Note that we could have used Cramer’s rule to find only two of the values and then used substitution to find the remaining value.)

x= y=

3x − y + 2z = 1, x − y + 2z = 3,

−2x + 3y + z = 1 3 3 3 3 −1 2 3 3 3 D = 33 1 −1 2 33 = −14 3 −2 3 1 3 3 3 3 1 −1 2 3 3 3 Dx = 33 3 −1 2 33 = 14 31 3 13 3 3 3 3 1 23 3 3 Dy = 33 1 3 2 33 = 12 3 −2 1 1 3 3 3 3 3 −1 1 3 3 3 Dz = 33 1 −1 3 33 = −22 3 −2 3 1 3 x= y=

Dx 14 = = −1 D −14

Dy 12 6 = =− D −14 7

−22 11 Dz = = D −14 7 " ! 6 11 . The solution is − 1, − , 7 7 z=

Dx −201 = =3 D −67

134 Dy = = −2 D −67

−67 Dz = =1 D −67 The solution is (3, −2, 1).

z=

36.

3x + 5y − z = −2,

z=

(Note that we could have used Cramer’s rule to find only two of the values and then used substitution to find the remaining value.) 38.

3x + 2y + 2z = 1, 5x − y − 6z = 3,

2x + 3y + 3z = 4 3 3 33 2 2 33 3 D = 33 5 −1 −6 33 = 25 32 3 3 3 3 3 31 2 2 33 3 Dx = 33 3 −1 −6 33 = −25 34 3 3 3 3 3 33 1 2 3 3 3 Dy = 33 5 3 −6 33 = 100 32 4 3 3 3 3 33 2 13 3 3 Dz = 33 5 −1 3 33 = −50 32 3 43 x=

Dx −25 = = −1 D 25

y=

100 Dy = =4 D 25

Dz −50 = = −2 D 25 The solution is (−1, 4, −2). z=

Exercise Set 8.6 39.

529

x − 3y − 7z = 6,

2x + 3y + z = 9,

8x

4x + y 3 31 3 D = 33 2 34 3 36 3 Dx = 33 9 37 3 31 3 Dy = 33 2 34 3 31 3 Dz = 33 2 34

4x + 9y =8 3 3 30 6 63 3 3 D = 33 8 0 6 33 = 576 34 9 03 3 3 3 −1 6 6 3 3 3 Dx = 33 −1 0 6 33 = 288 3 8 9 03

=7 −3 3 1

3 −7 33 1 33 = 57 0 3

3 −7 33 1 33 = 57 0 3 3 6 −7 33 9 1 33 = 171 7 0 3 3 −3 6 33 3 9 33 = −114 1 73 −3 3 1

x=

Dx 57 = =1 D 57

y=

171 Dy = =3 D 57

− 2z = 8,

2x + y + 4z = 13 3 3 3 1 −2 −3 3 3 3 D = 33 3 0 −2 33 = 25 32 1 4 3 3 3 3 4 −2 −3 3 3 3 0 −2 33 = 100 Dx = 33 8 3 13 1 4 3 3 3 3 1 4 −3 3 3 3 Dy = 33 3 8 −2 33 = −75 3 2 13 4 3 3 3 3 1 −2 4 3 3 3 Dz = 33 3 0 8 33 = 50 3 2 1 13 3

x=

Dx 288 1 = = D 576 2

y=

384 2 Dy = = D 576 3

z=

(Note that we could have used Cramer’s rule to find only two of the values and then used substitution to find the remaining value.) 3x

3 −1 6 33 −1 6 33 = 384 8 03 3 6 −1 33 0 −1 33 = −480 9 8 3

−480 5 Dz = =− D 576 6 " ! 1 2 5 , ,− . The solution is 2 3 6 (Note that we could have used Cramer’s rule to find only two of the values and then used substitution to find the remaining value.)

−114 Dz = = −2 D 57 The solution is (1, 3, −2).

x − 2y − 3z = 4,

+ 6z = −1,

3 30 3 Dy = 33 8 34 3 30 3 Dz = 33 8 34

z=

40.

6y + 6z = −1,

41.

42.

3x + 5y 2x

D

Dx

Dy

Dz

= 2, − 3z = 7,

4y + 2z = −1 3 3 33 5 0 3 3 3 = 33 2 0 −3 33 = 16 30 4 2 3 3 3 3 2 5 0 3 3 3 = 33 7 0 −3 33 = −31 3 −1 4 2 3 3 3 33 2 0 33 3 = 33 2 7 −3 33 = 25 3 0 −1 2 3 3 3 33 5 2 3 3 3 = 33 2 0 7 33 = −58 3 0 4 −1 3

x=

100 Dx = =4 D 25

x=

y=

−75 Dy = = −3 D 25

−31 31 Dx = =− D 16 16

y=

25 Dy = D 16

Dz 50 = =2 D 25 The solution is (4, −3, 2). z=

−58 29 Dz = =− D 16 8 " ! 31 25 29 . The solution is − , , − 16 16 8 z=

530

Chapter 8: Systems of Equations and Matrices 52. The graph of f (x) = x2 − 4 is shown below. It fails the horizontal-line test, so it is not one-to-one.

43. Left to the student 44. Left to the student 45. Enter matrix A and use the “det(” option from the MATRIX MATH menu.

y 5 4

|A| = 110

3 2 1

46. Enter matrix A and use the “det(” option from the MATRIX MATH menu.

f(x) = x 2 – 4 x

-5 -4 -3 -2 -1 0 -1

|A| = −195

-2 -3 -4

47. Left to the student 48. Left to the student 3 3 3a b 3 49. If 33 1 1 33 = 0, then a1 = ka2 and b1 = kb2 for some a2 b2 number k. This means that the equations a1 x + b1 y = c, and a2 x + b2 y = c2 are either dependent, if c1 = kc2 , or inconsistent, if c1 $= kc2 .

-5

53. The graph of f (x) = |x| + 3 is shown below. It fails the horizontal-line test, so it is not one-to-one. y

50. If a1 x + b1 y = c1 and a2 x + b2 y = c2 are parallel lines, then a1 = 3 ka2 , b13 = kb32 , and c31 $= kc2 , for 3 some 3number 3 a1 b1 3 3 c1 b1 3 3a c 3 3 3 3 3 k. Then 3 = 0, 3 $= 0, and 33 1 1 33 $= 0. a2 b2 3 c2 b2 3 a2 c2

5 4

x

-5 -4 -3 -2 -1 0 -1

1 2 3 4 5

-2 -3 -4

y 5 4

-5 f(x) = 3x + 2

3 2 1

x 1 2 3 4 5

√ 54. The graph of f (x) = 3 x + 1 is show below. Since it passes the horizontal-line test, the function is one-to-one.

-2 -3 -4

y 10 8

-5

6 4 2

We find a formula for f −1 (x). Replace f (x) with y: y = 3x + 2

-10 -8 -6 -4 -2 0 -2

Interchange x and y: x = 3y + 2 x−2 Solve for y: y = 3 Replace y with f

f(x) = |x | + 3

3 2 1

51. The graph of f (x) = 3x + 2 is shown below. Since it passes the horizontal-line test, the function is one-to-one.

-5 -4 -3 -2 -1 0 -1

1 2 3 4 5

−1

(x): f

−1

f(x) =

3

x +1 x

2 4 6 8 10

-4 -6 -8 -10

x−2 (x) = 3

We find a formula for f −1 (x). √ Replace f (x) with y: y = 3 x + 1 √ Interchange x and y: x = 3 y + 1 Solve for y: y = (x − 1)3

Replace y with f −1 (x): f −1 (x) = (x − 1)3

55.

(3 − 4i) − (−2 − i) = 3 − 4i + 2 + i = (3 + 2) + (−4 + 1)i = 5 − 3i

56. (5 + 2i) + (1 − 4i) = 6 − 2i

Exercise Set 8.6 57.

531

(1 − 2i)(6 + 2i) = 6 + 2i − 12i − 4i2 =

6 + 2i − 12i + 4 = 10 − 10i 58.

59.

3 + i 4 + 3i 12 + 9i + 4i + 3i2 3+i = = · = 4 − 3i 4 − 3i 4 + 3i 16 − 9i2 9 + 13i 9 13 12 + 9i + 4i − 3 = = + i 16 + 9 25 25 25 3 3 3 x 53 3 3 3 −4 x 3 = 24

√ (−∞, − 10) : f (−4) = (−4)2 − 10 = 6 > 0 √ √ (− 10, 10) : f (0) = 02 − 10 = −10 < 0 √ ( 10, ∞) : f (4) = 42 − 10 = 6 > 0 √ √ √ √ The solution set is {y|− 10 < y < 10}, or (− 10, 10). 3 3 3x + 3 43 3 3 63. 3 x − 3 5 3 = −7 (x + 3)(5) − (x − 3)(4) = −7 5x + 15 − 4x + 12 = −7

x · x − (−4)(5) = 24 Evaluating the determinant

x + 27 = −7

x2 + 20 = 24 x2 = 4

x = ±2

The solutions are −2 and 2. 3 3 3y 23 3 3 60. 33 y3 = y

x = −34

The solution is −34. 3 3 3 m + 2 −3 3 3 3 64. 3 m + 5 −4 3 = 3m − 5 −4m − 8 + 3m + 15 = 3m − 5 −m + 7 = 3m − 5

y2 − 6 = y

−4m = −12

y2 − y − 6 = 0

(y − 3)(y + 2) = 0

y − 3 = 0 or y + 2 = 0 y = 3 or

y = −2

The solutions are 3 and −2. 3 3 3 x −3 3 3 3 61. 3 −1 x 3 ≥ 0 x · x − (−1)(−3) ≥ 0 x −3 ≥ 0 2

We solve the related equation. x2 − 3 = 0

x2 = 3

√ x=± 3

√ √ The numbers − 3 and 3 divide the x-axis into three intervals. We let f (x) = x2 − 3 and test a value in each interval. √ (−∞, − 3) : f (−2) = (−2)2 − 3 = 1 > 0 √ √ (− 3, 3) : f (0) = 02 − 3 = −3 < 0 √ ( 3, ∞) : f (2) = 22 − 3 = 1 > 0 √ √ The function is positive in (−∞, − 3), and ( 3, ∞). We also include the endpoints of the intervals since the √ inequality symbol is ≥.√The solution set is {x|x ≤ − 3 or x ≥ √ √ 3}, or (−∞, − 3] ∪ [ 3, ∞). 3 3 3 y −5 3 3 62. 3 3 −2 y 3 < 0 y 2 − 10 < 0

Solve the related equation. y 2 − 10 = 0

y 2 = 10 √ y = ± 10 √ √ The numbers − 10 and 10 divide the axis into three intervals. We let f (y) = y 2 − 10 and test a value in each interval.

65.

m=3 3 3 32 x 1 3 3 3 3 1 2 −1 3 = −6 3 3 3 3 4 −2 3 −x − 2 = −6 −x = −4 x=4

The solution is 4. 3 3 3 3 66. 3 x 2 x 3 3 3 −1 1 3 = −10 3 3 3 1 −2 2 3 −5x − 10 = −10 −5x = 0 x=0

67. Answers may vary. 3 3 3 L −W 3 3 3 32 2 3

68. Answers may vary. 3 3 3 π −π 3 3 3 3h r 3 69. Answers may vary. 3 3 3 a b3 3 3 3 −b a 3 70. Answers may vary. 3h h 33 3 − 3 3 23 32 3 3 3 3 b a

71. Answers may vary. 3 3 3 2πr 2πr 3 3 3 3 −h r 3

Evaluating the determinant

532

Chapter 8: Systems of Equations and Matrices 2. Next determine which half-plane to shade by testing a point not on the line. Here we use (2, 2) as a check.

72. Answers may vary. 3 3 3 xy Q 3 3 3 3 Q xy 3

y+x ≥ 0

2 + 2 ?3 0 3 4 3 0 TRUE

Exercise Set 8.7

Since 4 ≥ 0 is true, (2, 2) is a solution. Thus shade the half-plane containing (2, 2).

1. Graph (f ) is the graph of y > x. 2. Graph (c) is the graph of y < −2x.

y

3. Graph (h) is the graph of y ≤ x − 3.

4

4. Graph (a) is the graph of y ≥ x + 5.

2

5. Graph (g) is the graph of 2x + y < 4.

2

!4 !2

6. Graph (d) is the graph of 3x + y < −6.

y#x'0

9. Graph: y > 2x

12.

y y!x$0

1. We first graph the related equation y = 2x. We draw the line dashed since the inequality symbol is >.

x

!4

7. Graph (b) is the graph of 2x − 5y > 10. 8. Graph (e) is the graph of 3x − 9y < 9.

4

!2

4 2 2

!4 !2

4

x

!2

2. To determine which half-plane to shade, test a point not on the line. We try (1, 1) and substitute:

!4

y > 2x 13. Graph: y > x − 3

1 ?3 2 · 1 3 1 3 2 FALSE

1. We first graph the related equation y = x − 3. Draw the line dashed since the inequality symbol is >.

Since 1 > 2 is false, (1, 1) is not a solution, nor are any points in the half-plane containing (1, 1). The points in the opposite half-plane are solutions, so we shade that half-plane and obtain the graph.

2. To determine which half-plane to shade, test a point not on the line. We try (0, 0). y > x−3

0 ?3 0 − 3 3 0 3 −3 TRUE

y 4

Since 0 > −3 is true, (0, 0) is a solution. Thus we shade the half-plane containing (0, 0).

2 2

!4 !2 !2

4

y

x

y & 2x

!4

4 2

10.

4

!4 !2

y

!2 2y $ x

4

y&x!3

2 2

!4 !2

4

x

!2 !4

14.

y y%x#4 2

11. Graph: y + x ≥ 0

1. First graph the related equation y + x = 0. Draw the line solid since the inequality is ≥.

2

!4 !2 !2 !4

4

x

x

Exercise Set 8.7

533

15. Graph: x + y < 4

18.

y

1. First graph the related equation x + y = 4. Draw the line dashed since the inequality is <.

4 2

2. To determine which half-plane to shade, test a point not on the line. We try (0, 0).

2

!4 !2

x+y < 4

!4

0 + 0 ?3 4 3 0 3 4 TRUE

Since 0 < 4 is true, (0, 0) is a solution. Thus shade the half-plane containing (0, 0).

1. First graph the related equation 3y + 2x = 6. Draw the line solid since the inequality is ≥.

2. To determine which half-plane to shade, test a point not on the line. We try (0, 0).

x#y$4

3y + 2x ≥ 6

2 2

4

2x ! 5y $ 10

19. Graph: 3y + 2x ≥ 6

y

!4 !2

x

3 · 0 + 2 · 0 ?3 6 3 0 3 6 FALSE

x

!2

Since 0 ≥ 6 is false, (0, 0) is not a solution. We shade the half-plane which does not contain (0, 0).

!4

y

16.

y 4

4 x!y'5

2 2

!4 !2

4

2

!4 !2

x

!2

!2

!4

!4

17. Graph: 3x − 2y ≤ 6

20.

4

2y # x % 4

2

!4 !2

4

x

!2

3x − 2y ≤ 6

!4

3(0) − 2(0) ?3 6 3 0 3 6 TRUE

21.

Since 0 ≤ 6 is true, (0, 0) is a solution. Thus shade the half-plane containing (0, 0). y

Graph: 3x − 2 ≤ 5x + y −2 ≤ 2x + y

Adding −3x

1. First graph the related equation 2x + y = −2. Draw the line solid since the inequality is ≤.

2. To determine which half-plane to shade, test a point not on the line. We try (0, 0).

4

2x + y ≥ −2

2 4

!4 !2

3y # 2x ' 6

y

1. First graph the related equation 3x − 2y = 6. Draw the line solid since the inequality is ≤.

2. To determine which half-plane to shade, test a point not on the line. We try (0, 0).

x

x

!2

3x ! 2y % 6

2(0) + 0 ?3 −2 3 0 3 −2 TRUE

Since 0 ≥ −2 is true, (0, 0) is a solution. Thus shade the half-plane containing the origin.

534

Chapter 8: Systems of Equations and Matrices 2. To determine which half-plane to shade we test a point not on the line. We try (0, 0).

y 4

y > −3

2 2

!4 !2

4

0 ? −3 TRUE

Since 0 > −3 is true, (0, 0) is a solution. We shade the half-plane containing (0, 0).

x

!2 !4

y 4

3x ! 2 % 5x # y

2

22.

y

2

!4 !2

4 2

!4 4

!4 !2

4

x

!2

2x ! 6y ' 8 # 2y

y & !3

x

!2

26.

!4

y 4

1. We first graph the related equation x = −4. Draw the line dashed since the inequality is <. 2. To determine which half-plane to shade, test a point not on the line. We try (0, 0).

2

!4 !2

4

x

!2 !4

27. Graph: −4 < y < −1

x < −4

This is a conjunction of two inequalities

0 ? −4 FALSE

Since 0 < −4 is false, (0, 0) is not a solution. Thus, we shade the half-plane which does not contain the origin. y

−4 < y and y < −1.

We can graph −4 < y and y < −1 separately and then graph the intersection, or region in both solution sets. y 6

4

5 4

2

3 2 1

2

!2 !2

4

x

x $ !4

!4

24.

x%5

2

23. Graph: x < −4

-2 -3 -4

6

y 5 4

y'5 2

4

x

!2

3 2 1 -6 -5 -4 -3 -2 -1 -1

25. Graph: y > −3

–4 < y

6

4

!4 !2

1 2 3 4 5 6

-5 -6

y

2

x

-6 -5 -4 -3 -2 -1 -1

1. First we graph the related equation y = −3. Draw the line dashed since the inequality is >.

-2 -3 -4 -5 -6

y < –1 1 2 3 4 5 6

x

Exercise Set 8.7

535 35. Graph (b) is the correct graph.

y

36. Graph (e) is the correct graph.

4

!4 $ y $ !1

2

2

!4 !2

4

x

!2

28.

The other line goes through (0, 0) and (1, 3). We find the slope. 3−0 =3 m= 1−0 This line has slope 3 and y-intercept (0, 0), so its equation is y = 3x + 0, or y = 3x.

y 4

!3 % x % 3

2 2

!4 !2

4

37. First we find the related equations. One line goes through (0, 4) and (4, 0). We find its slope: 0−4 −4 m= = = −1 4−0 4 This line has slope −1 and y-intercept (0, 4), so its equation is y = −x + 4.

x

!2

Observing the shading on the graph and the fact that the lines are solid, we can write the system of inqualities as

!4

y ≤ −x + 4,

29. Graph: y ≥ |x|

1. Graph the related equation y = |x|. Draw the line solid since the inequality symbol is ≥.

2. To determine the region to shade, observe that the solution set consists of all ordered pairs (x, y) where the second coordinate is greater than or equal to the absolute value of the first coordinate. We see that the solutions are the points on or above the graph of y = |x|. y 4 2 2

!4 !2 !2

4

x

y ' !x!

!4

y ≤ 3x.

Answers may vary. 38. First we find the related equations. One line goes through (−1, 2) and (0, 0). We find its slope: 0−2 = −2 m= 0 − (−1) Then the equation of the line is y = −2x + 0, or y = −2x. The other line goes through (0, −3) and (3, 0). We find its slope: 3 0 − (−3) = =1 m= 3−0 3 Then the equation of this line is y = x − 3.

Observing the shading on the graph and the fact that the lines are solid, we can write the system of inequalities as y ≤ −2x,

y ≥ x − 3.

Answers may vary. 30.

y % !x # 2!

y 4 2 2

!4 !2

4

x

!2 !4

31. Graph (f ) is the correct graph. 32. Graph (c) is the correct graph. 33. Graph (a) is the correct graph. 34. Graph (d) is the correct graph.

39. The equation of the vertical line is x = 2 and the equation of the horizontal line is y = −1. The lines are dashed and the shaded area is to the left of the vertical line and above the horizontal line, so the system of inequalities can be written x < 2, y > −1. 40. The equation of the vertical line is x = −1 and the equation of the horizontal line is y = 2. The lines are dashed and the shaded area is to the right of the vertical line and below the horizontal line, so the system of inequalities can be written x > −1, y < 2.

536

Chapter 8: Systems of Equations and Matrices

41. First we find the related equations. One line goes through (0, 3) and (3, 0). We find its slope: 0−3 −3 m= = = −1 3−0 3 This line has slope −1 and y-intercept (0, 3), so its equation is y = −x + 3. The other line goes through (0, 1) and (1, 2). We find its slope: 1 2−1 = =1 m= 1−0 1 This line has slope 1 and y-intercept (0, 1), so its equation is y = x + 1. Observe that both lines are solid and that the shading lies below both lines, to the right of the y-axis, and above the x-axis. We can write this system of inequalities as

44.

y ≤ x−5 y 4 2

45.

Graph: y ≥ x,

y ≤ x−4

We graph the related equations y = x and y = x − 4 using solid lines. The arrows at the ends of the lines indicate the half-plane containing the solution set for each inequality. There is not a region common to both solution sets, so there are no vertices. y

y ≥ 0.

42. One line goes through (−2, 0) and (0, 2). We find its slope: 2 2−0 = =1 m= 0 − (−2) 2 Then the equation of this line is y = x + 2. The other line is the vertical line x = 2. Observe that both lines are solid and that the shading lies below y = x + 2, to the left of x = 2, to the right of the y-axis, and above the x-axis. We can write this system of inequalities as

4 2 2

!4 !2

x

!4

46.

Graph: y ≥ x,

y ≤ 2−x y

y ≥ 0.

4

Graph: y ≤ x,

2

y ≥ 3−x

We graph the related equations y = x and y = 3 − x using solid lines. The arrows at the ends of the lines indicate the half-plane containing the solution set for each inequality. Shade the region common to both solution sets.

4

x

!2 !4

Graph: y ≥ −3, x≥1

4

"w, w#

2 2

!4 !2

4

x

!2

We graph the related equations y = −3 and x = 1 using solid lines. The arrows at the ends of the lines indicate the half-plane containing the solution set for each inequality. Shade the region common to both solution sets. y

!4

We find the vertex

(1, 1) 2

!4 !2

47.

y

y = 3 − x.

4

!2

x ≥ 0,

y = x,

x

!4

y ≤ x + 1,

43.

4

!2

x ≥ 0,

y ≤ x + 2,

2

!4 !2

y ≤ −x + 3,

x ≤ 2,

Graph: y ≥ x,

!

3 3 , 2 2

"

4

by solving the system

2 2

!4 !2

4

!2 !4

(1, !3)

x

Exercise Set 8.7

537

We find the vertex (1, −3) by solving the system

51.

y = −3,

x−y ≤ 2

x = 1. 48.

Graph: x + y ≤ 1, We graph the related equations x + y = 1 and x − y = 2 using solid lines. The arrows at the ends of the lines indicate the half-plane containing the solution set for each inequality. Shade the region common to both solution sets.

Graph: y ≤ −2, x≥2

y

y 4

4

2

2 4

!4 !2

x

4

!4 !2

!4

49.

!4

Graph: x ≤ 3,

We find the vertex

y ≥ 2 − 3x

We graph the related equations x = 3 and y = 2−3x using solid lines. The arrows at the ends of the lines indicate the half-plane containing the solution set for each inequality. Shade the region common to both solution sets.

52.

by solving the system

y 2

2

4

1

x

1

!2 !1

2

x

!1

!4

!2

(3, !7)

53.

We find the vertex (3, −7) by solving the system x = 3,

y = 2 − 3x. Graph: x ≥ −2,

Graph: 2y − x ≤ 2,

y + 3x ≥ −1

We graph the related equations 2y − x = 2 and y + 3x = −1 using solid lines. The arrows at the ends of the lines indicate the half-plane containing the solution set for each inequality. Shade the region common to both solution sets.

y ≤ 3 − 2x

y

y (!2, 7)

"

y + 3x ≤ 2

!2

50.

3 1 ,− 2 2

Graph: y + 3x ≥ 0,

2

!8

!

x − y = 2.

4

!6

"w, !q#

x + y = 1,

y

!4 !2

x

!2

(2, !2)

4

"!¢, ∞#

8 6

2 2

!4 !2

4

x

!2 !4 !4

4

x

We find the vertex 2y − x = 2,

y + 3x = −1.

!

4 5 − , 7 7

"

by solving the system

538 54.

Chapter 8: Systems of Equations and Matrices 57.

Graph: y ≤ 2x + 1,

Graph: 4x − 3y ≥ −12, 4x + 3y ≥ −36,

y ≥ −2x + 1,

y ≤ 0,

x≤2

x≤0

y 6

Shade the intersection of the graphs of the four inequalities. y

(2, 5)

4

8

(0, 1) 4

!4 !2 !2

55.

Graph:

x

(0, 0)

(2, !3)

8

x

(0, !3)

(!12, 0)

x − y ≤ 2,

(!4, !6)

y≤4

We find the vertex (−12, 0) by solving the system

We graph the related equations x − y = 2, x + 2y = 8, and y = 4 using solid lines. The arrows at the ends of the lines indicate the half-plane containing the solution set for each inequality. Shade the region common to all three solution sets.

4y + 3x = −36, y = 0.

We find the vertex (0, 0) by solving the system y = 0,

y

x = 0.

(0, 4)

2

!2

We find the vertex (0, −3) by solving the system

(6, 4)

2 (4, 2) 4

6

4y − 3x = −12, x = 0.

x

We find the vertex (−4, −6) by solving the system 4y − 3x = −12,

!4

We find the vertex (0, 4) by solving the system

58.

x + 2y = 8,

4y + 3x = −36. Graph: 8x + 5y ≤ 40, x + 2y ≤ 8,

y = 4.

x ≥ 0,

We find the vertex (6, 4) by solving the system

y≥0

x − y = 2,

y

y = 4.

We find the vertex (4, 2) by solving the system

6

x − y = 2,

2

Graph: x + 2y ≤ 12,

(0, 0)

2x + y ≤ 12,

!2

x ≥ 0,

59.

y≥0

4

6

(5, 0)

2 2

x

Shade the intersection of the graphs of the given inequalities.

(4, 4)

4

8

Graph: 3x + 4y ≥ 12, 1≤x≤3

(0, 6)

!2

2

5x + 6y ≤ 30,

y

!2

24 "40 11 , 11#

(0, 4)

x + 2y = 8.

(0, 0)

4

!8 !4

x + 2y ≥ 8,

56.

4

(6, 0)

x

Exercise Set 8.7

539 y

"1, 256 # "3, e# "3, !#

4

"1, ##

2 2

!2

4

6

y

C B

D

A

E

x

!2

x

!4

We find the vertex

Vertex A: (0, 0)

!

1,

!

5 3, 2

"

by solving the system

!

3 3, 4

"

by solving the system

!

1,

9 4

"

by solving the system

25 6

"

by solving the system

5x + 6y = 30, x = 1. We find the vertex 5x + 6y = 30, x = 3. We find the vertex 3x + 4y = 12, x = 3. We find the vertex

x = 1. y − x ≤ 3,

2≤x≤5

y

17 · 0 − 3 · 4 + 60 = 48

8 2 − 3 · 4 + 60 = 66 3 3 3 3 17 · 7 − 3 · + 60 = 176 4 4 17 · 7 − 3 · 0 + 60 = 179 17 ·

The minimum value of P is 48 when x = 0 and y = 4.

(5, 6)

62. We graph the system of inequalities and find the vertices:

4 2

P = 17x − 3y + 60 17 · 0 − 3 · 0 + 60 = 60

The maximum value of P is 179 when x = 7 and y = 0.

(5, 8)

(2, 5) 6

Evaluate the objective function P at each vertex.

B(0, 4) " ! 8 ,4 C !3 " 3 D 7, 4 E(7, 0)

Graph: y − x ≥ 1,

8

Vertex C: We solve the system 6x + 8y ! = 48 " and y = 4. The 8 coordinates of point C are ,4 . 3 Vertex D: We solve the system 6x + 8y " and x = 7. The ! = 48 3 coordinates of point D are 7, . 4 Vertex E: We solve the system x = 7 and y = 0. The coordinates of point E are (7, 0). Vertex A(0, 0)

3x + 4y = 12, 60.

Vertex B: We solve the system x = 0 and y = 4. The coordinates of point B are (0, 4).

(2, 3)

y

4

6

8

6

x

4 5 (—, 4) 4 5 3 2 1

61. Find the maximum and minimum values of P = 17x − 3y + 60, subject to 6x + 8y ≤ 48,

-6 -5 -4 -3 -2 -1 -1

5 (3, —) 4 1 2 3 4 5 6

-2 -3 -4

0 ≤ y ≤ 4,

0 ≤ x ≤ 7.

Graph the system of inequalities and determine the vertices.

(3, 4)

-5 -6

Vertex " ! 4 ,4 5 (3, 4) " ! 5 3, 4

Q = 28x − 4y + 72 2 78 ←− Minimum 5 140 151 ←− Maximum

x

540

Chapter 8: Systems of Equations and Matrices

63. Find the maximum and minimum values of

Vertex G = 16x + 14y (0, 0)

F = 5x + 36y, subject to

0 ←− Minimum

(0, 5.8) 81.2 ←− Maximum

5x + 3y ≤ 34,

3x + 5y ≤ 30, x ≥ 0, y ≥ 0.

Graph the system of inequalities and find the vertices. y

(2, 3)

74

(4, 0)

64

65. Let x = the number of jumbo biscuits and y = the number of regular biscuits to be made per day. The income I is given by I = 0.10x + 0.08y subject to the constraints x + y ≤ 200,

B

2x + y ≤ 300,

C

x ≥ 0,

x A

y ≥ 0.

D

We graph the system of inequalities, determine the vertices, and find the value if I at each vertex.

Vertex A: (0, 0) Vertex B: We solve the system 3x + 5y = 30 and x = 0. The coordinates of point B are (0, 6). Vertex C: We solve the system 5x + 3y = 34 and 3x + 5y = 30. The coordinates of point C are (5, 3). Vertex D: We solve the system 5x + 3y ! = 34"and y = 0. The 34 ,0 . coordinates of point D are 5 Evaluate the objective function F at each vertex. Vertex A(0, 0) B(0, 6)

F = 5x + 36y 5 · 0 + 36 · 0+ = 0

y 300

C(5, 3) 5 · 5 + 39 · 3 = 133 " ! 34 34 D ,0 5· + 36 · 0 = 34 5 5 The maximum value of F is 216 when x = 0 and y = 6. The minimum value of F is 0 when x = 0 and y = 0. 64. We graph the system of inequalities and find the vertices:

(0, 0)

-5 -6

x

300

(0, 200)

0.10(0) + 0.08(200) = 16

(150, 0)

0.10(150) + 0.08(0) = 15

The company will have a maximum income of $18 when 100 of each type of biscuit is made. 66. Let x = the number of gallons the car uses and y = the number of gallons the moped uses. Find the maximum value of M = 20x + 100y subject to x+y

≤

12,

0 ≤ x ≤ 10, 0 ≤ y ≤ 3.

(2, 3)

x 1 2 3 4 5 6 (4, 0)

200

I = 0.10x + 0.08y 0.10(0) + 0.08(0) = 0

6 (0, 5.8) 5

-6 -5 -4 -3 -2 -1 (0, 0) -1 -2 -3 -4

100

Vertex (0, 0)

y

3 2 1

(100, 100) (150, 0)

100

(100, 100) 0.10(100) + 0.08(100) = 18

5 · 0 + 36 · 6 = 216

4

(0, 200)

200

y 12 11 10 9 8 7 6 5 4 3

(10, 3) (0, 3)

(9, 3)

2 1 (0, 0)

x 1 2 3 4 5 6 7 8 9

11 12 (10, 0)

Exercise Set 8.7

541

Vertex M = 20x + 100y (0, 0) 0

Vertex (0, 0)

P = 40x + 30y 0

(0, 3)

300

(0, 240)

7200

(9, 3)

480

(80, 160) 8000

(10, 2) 400

(160, 0)

(10, 0) 200 The maximum number of miles is 480 when the car uses 9 gal and the moped uses 3 gal. 67. Let x = the number of units of lumber and y = the number of units of plywood produced per week. The profit P is given by P = 20x + 30y

The maximum profit of $8000 occurs when 80 acres of corn and 160 acres of oats are planted. 69. Let x = the number of sacks of soybean meal to be used and y = the number of sacks of oats. The minimum cost is given by C = 15x + 5y subject to the constraints

subject to the constraints x + y ≤ 400,

50x + 15y ≥ 120, 8x + 5y ≥ 24,

x ≥ 100,

5x + y ≥ 10,

y ≥ 150.

We graph the system of inequalities, determine the vertices and find the value of P at each vertex. y

x ≥ 0, y ≥ 0.

Graph the system of inequalities, determine the vertices, and find the value of C at each vertex.

400

y 12 10

(100, 300)

300 200

A

8 6 4

(250, 150) (100, 150)

100

B C

2 x 100

200

300

400

Vertex P = 20x + 30y (100, 150) 20 · 100 + 30 · 150 = 6500

(100, 300) 20 · 100 + 30 · 300 = 11, 000 (250, 150) 20 · 250 + 30 · 150 = 9500

The maximum profit of $11,000 is achieved by producing 100 units of lumber and 300 units of plywood. 68. Let x = the corn acreage and y = the oats acreage. Find the maximum value of P = 40x + 30y subject to x + y ≤ 240,

1

Vertex A(0, 10) ! " 6 B ,4 5 ! " 24 24 C , 13 13 D(3, 0)

D

2

x ≥ 0,

C = 15x + 5y 15 · 0 + 5 · 10 = 50

6 + 5 · 4 = 38 5 24 24 12 15 · +5· = 36 13 13 13 15 · 3 + 5 · 0 = 45 15 ·

12 24 The minimum cost of $36 is achieved by using , or 13 13 11 24 11 1 sacks of soybean meal and , or 1 sacks of oats. 13 13 13

C = 15x + 8y

y ≥ 0.

subject to the constraints 50x + 20y ≥ 120,

y

8x + 6y ≥ 24,

300

5x + 8y ≥ 10,

(0, 240)

x ≥ 0,

200 (80, 160) 100

y ≥ 0.

(160, 0) x 100

200

x

3

70. Let x = the number of sacks of soybean meal to be used and y = the number of sacks of alfalfa. Find the minimum value of

2x + y ≤ 320,

(0, 0)

6400

300

542

Chapter 8: Systems of Equations and Matrices y 6 5

72. Let x = the amount invested in City Bank and y = the amount invested in People’s Bank. Find the maximum value of

A

4 3

I = 0.06x + 0.065y

2 1

subject to

B C 1

2

Vertex A(0, 6) " ! 12 12 , B 7 7 C(3, 0)

x + y ≤ 22, 000,

x

2000 ≤ x ≤ 14, 000,

3

0 ≤ y ≤ 15, 000.

C = 15x + 8y 15 · 0 + 8 · 6 = 48

y

12 12 3 +8· = 39 15 · 7 7 7 15 · 3 + 8 · 0 = 45

12 5 3 , or 1 The minimum cost of $39 is achieved by using 7 7 7 12 5 sacks of soybean meal and , or 1 sacks of alfalfa. 7 7

(2000, 15,000)

(14,000 , 8000)

71. Let x = the amount invested in corporate bonds and y = the amount invested in municipal bonds. The income I is given by I = 0.08x + 0.075y subject to the constraints x + y ≤ 40, 000,

6000 ≤ x ≤ 22, 000,

(2000, 0)

y (6000, 30,000)

x

Vertex (2000, 0)

I = 0.06x + 0.065y 120

(2000, 15, 000)

1095

(7000, 15, 000)

1395

(14, 000, 0)

840

The maximum interest income is $1395 when $7000 is invested in City Bank and $15,000 is invested in People’s Bank. 73. Let x = the number of P1 airplanes and y = the number of P2 airplanes to be used. The operating cost C, in thousands of dollars, is given by

(10,000, 30,000) 30,000

C = 12x + 10y

20,000

(22,000, 18,000)

subject to the constraints 40x + 80y ≥ 2000,

10,000 (22,000, 0) 10,000

20,000

30,000

x 40,000

40x + 30y ≥ 1500,

120x + 40y ≥ 2400, x ≥ 0,

(6000, 0)

Vertex (6000, 0)

I = 0.08x + 0.075y 480

(6000, 30, 000)

2730

(10, 000, 30, 000) 3050 (22, 000, 18, 000) 3110 (22, 000, 0)

(14,000 , 0)

(14, 000, 8, 000) 1360

y ≤ 30, 000.

We graph the system of inequalities, determine the vertices, and find the value of I at each vertex.

40,000

(7000, 15,000)

1760

The maximum income of $3110 occurs when $22,000 is invested in corporate bonds and $18,000 is invested in municipal bonds.

y ≥ 0.

Graph the system of inequalities, determine the vertices, and find the value of C at each vertex. y 60 50 40 30 20 10

(0, 60) (6, 42) (30, 10) (50, 0) x 10 20 30 40 50

Exercise Set 8.7 Vertex (0, 60)

543

C = 12x + 10y 12 · 0 + 10 · 60 = 600

(6, 42)

y

12 · 6 + 10 · 42 = 492

(30, 10) 12 · 30 + 10 · 10 = 460 (50, 0)

(0, 5) (2, 4)

12 · 50 + 10 · 0 = 600

The minimum cost of $460 thousand is achieved using 30 P1 ’s and 10 P2 ’s. 74. Let x = the number of P2 airplanes and y = the number of P3 airplanes to be used. Find the minimum value of C = 10x + 15y

x (4, 0)

(0, 0)

Vertex P = 34x + 31y (0, 0) 34 · 0 + 31 · 0 = 0 (0, 5)

subject to

34 · 0 + 31 · 5 = 155

(2, 4)

80x + 40y ≥ 2000,

34 · 2 + 31 · 4 = 192

(4, 0)

30x + 40y ≥ 1500,

34 · 4 + 31 · 0 = 136

The maximum profit per day is $192 when 2 knit suits and 4 worsted suits are made.

40x + 80y ≥ 2400, x ≥ 0,

76. Let x = the number of smaller gears and y = the number of larger gears produced each day. Find the maximum value of

y ≥ 0.

y

P = 25x + 10y

60

subject to

50 (0, 50)

4x + y ≤ 24,

40 (10, 30)

30

x + y ≤ 9,

(30, 15)

20 10

(60, 0)

x ≥ 0, y ≥ 0.

x

10 20 30 40 50 60 70 80

y

Vertex (0, 50)

C = 10x + 15y 10 · 0 + 15 · 50 = 750

(10, 30) 10 · 10 + 15 · 30 = 550 (30, 15) 10 · 30 + 15 · 15 = 525 (60, 0)

10 · 60 + 15 · 0 = 600

The minimum cost of $525 thousand is achieved using 30 P2 ’s and 15 P3 ’s. 75. Let x = the number of knit suits and y = the number of worsted suits made. The profit is given by P = 34x + 31y

30 20 10 (0, 9) (5, 4) (0, 0)

2x + 4y ≤ 20,

4x + 2y ≤ 16, x ≥ 0, y ≥ 0.

Graph the system of inequalities, determine the vertices, and find the value of P at each vertex.

4

6

8

x

10

Vertex P = 25x + 10y (0, 0) 25 · 0 + 10 · 0 = 0 (0, 9) (5, 4)

subject to

2

(6, 0)

(6, 0)

25 · 0 + 10 · 9 = 90

25 · 5 + 10 · 4 = 165 25 · 6 + 10 · 0 = 150

The maximum profit of $165 per day is achieved when 5 smaller gears and 4 larger gears are produced. 77. Let x = the number of pounds of meat and y = the number of pounds of cheese in the diet in a week. The cost is given by C = 3.50x + 4.60y subject to 2x + 3y ≥ 12, 2x + y ≥ 6, x ≥ 0, y ≥ 0.

544

Chapter 8: Systems of Equations and Matrices Graph the system of inequalities, determine the vertices, and find the value of C at each vertex.

Graph the system of inequalities, determine the vertices, and find the value of T at each vertex. y

y

3000

6 (0, 6) 5 4

2000 (1.5, 3)

3 2 1

(6, 0)

1000

x

(0, 525) (550, 250)

1 2 3 4 5 6

(0, 0)

Vertex C = 3.50x + 4.60y (0, 6) 3.50(0) + 4.60(6) = 27.60

(600, 0)

(1.5, 3) 3.50(1.5) + 4.60(3) = 19.05

Vertex (0, 0)

T =x+y 0+0=0

(6, 0)

(0, 525)

0 + 525 = 525

3.50(6) + 4.60(0) = 21.00

The minimum weekly cost of $19.05 is achieved when 1.5 lb of meat and 3 lb of cheese are used. 78. Let x = the number of teachers and y = the number of teacher’s aides. Find the minimum value of C = 35, 000x + 18, 000y subject to x + y ≤ 50,

x

250 500 750 1000

(550, 250) 550 + 250 = 800 (600, 0)

600 + 0 = 600

The maximum total number of 800 is achieved when there are 550 of A and 250 of B. 80. Let x = the number of animal A and y = the number of animal B. Find the maximum value of T =x+y

x + y ≥ 20,

subject to

y ≥ 12,

x + 0.2y ≤ 1080,

x ≥ 2y.

0.5x + y ≤ 810, x ≥ 0,

y

y ≥ 0.

100 50 (——, —) 3 3

y 6000

(24, 12)

5000

(38, 12) x

4000 3000 2000

Vertex (24, 12) " ! 100 50 , 3 3 (38, 12)

C = 35, 000x + 18, 000y 35, 000·24+18, 000·12 = 1, 056, 000 100 50 +18, 000· ≈ 1, 466, 667 3 3 35, 000·38+18, 000·12 = 1, 546, 000

1000 (0, 810) (0, 0)

500 1000 1500 2000 (1080, 0)

Vertex (0, 0)

T =x+y 0+0=0

(0, 810)

0 + 810 = 810

35, 000·

The minimum cost of $1,056,000 is achieved when 24 teachers and 12 teacher’s aides are hired. 79. Let x = the number of animal A and y = the number of animal B. The total number of animals is given by T =x+y subject to x + 0.2y ≤ 600, 0.5x + y ≤ 525, x ≥ 0, y ≥ 0.

(1020, 300)

x

(1020, 300) 1020 + 300 = 1320 (1080, 0)

1080 + 0 = 1080

The maximum total number of 1320 is achieved when there are 1020 of A and 300 of B. 81. Left to the student 82. Left to the student 83. Answers will vary. Exercise 59 can be used as a model, if desired.

Exercise Set 8.7

545 9 <0 2 Function values are positive on (−3, −2). This can also be x−1 determined from the graph of y = − 4. The solution x+2 set is {x| − 3 < x < 2}, or (−3, −2).

84. The graph of a linear equation consists of a set of points on a line. The graph of a linear inequality consists of the set of points in a half-plane and might also include the points on the line that is the boundary of the half-plane. 85.

−5 ≤ x + 2 < 4 −7 ≤ x < 2

Subtracting 2

(−2, ∞): f (0) = −

89.

The solution set is {x| − 7 ≤ x < 2}, or [−7, 2).

y ≤ 2 − x2

86. |x − 3| ≥ 2

First graph the related equations y = x2 −2 and y = 2−x2 using solid lines. The solution set consists of the region above the graph of y = x2 − 2 and below the graph of y = 2 − x2 .

x − 3 ≤ −2 or x − 3 ≥ 2 x≤1

or

x≥5

The solution set is {x|x ≤ 1 or x ≥ 5}, or (−∞, 1] ∪ [5, ∞). 87.

x2 − 2x ≤ 3

x − 2x − 3 ≤ 0 2

x2 − 2x − 3 = 0

(x + 1)(x − 3) = 0

Graph: y ≥ x2 − 2,

y

Polynomial inequality

4

Related equation

Using the principle of zero products or by observing the graph of y = x2 − 2x − 3, we see that the solutions of the related equation are −1 and 3. These numbers divide the x-axis into the intervals (−∞, −1), (−1, 3), and (3, ∞). We let f (x) = x2 − 2x − 3 and test a value in each interval.

2

!4 !2

Factoring

90.

y < x+1 y ≥ x2

(−1, 3): f (0) = −3 < 0

y

(3, ∞): f (4) = 5 > 0

4

Function values are negative on (−1, 3). This can also be determined from the graph of y = x2 − 2x − 3. Since the inequality symbol is ≤, the endpoints of the interval must be included in the solution set. It is {x| − 1 ≤ x ≤ 3} or [−1, 3]. x−1 > 4 Rational inequality x+2 x−1 −4 > 0 x+2 x−1 − 4 = 0 Related equation x+2 x−1 The denominator of f (x) = − 4 is 0 when x = −2, x+2 so the function is not defined for x = −2. We solve the related equation f (x) = 0. x−1 −4 = 0 x+2 x − 1 − 4(x + 2) = 0 Multiplying by x + 2 x − 1 − 4x − 8 = 0

2 2

!4 !2

4

!4

91. See the answer section in the text. 92.

y 2 1 1

!2 !1

2

x

!1 !2

!x! # !y! % 1

93. See the answer section in the text. 94.

y !x ! y! & 0

−3x = 9

Thus, the critical values are −3 and −2. They divide the x-axis into the intervals (−∞, −3), (−3, −2), and (−2, ∞). We test a value in each interval. 3 (−∞, −3): f (−4) = − < 0 2 (−3, −2): f (−2.5) = 3 > 0

x

!2

−3x − 9 = 0

x = −3

x

!4

(−∞, −1): f (−2) = 5 > 0

88.

4

x

546

Chapter 8: Systems of Equations and Matrices Vertex (0, 0)

95. Let x = the number of less expensive speaker assemblies and y = the number of more expensive assemblies. The income is given by

(0, 10)

I = 350x + 600y

80 · 0 + 300 · 10 = 3000

(25, 9.5) 80 · 25 + 300(9.5) = 4850

subject to

(70, 5)

y ≤ 44

(95, 0)

x + y ≤ 60,

80 · 70 + 300 · 5 = 7100 80 · 95 + 300 · 0 = 7600

The maximum income of $7600 is achieved by making 95 chairs and 0 sofas.

x + 2y ≤ 90, x ≥ 0, y ≥ 0.

Exercise Set 8.8

Graph the system of inequalities, determine the vertices, and find the value of I at each vertex.

1.

y

x+7 A B = + (x − 3)(x + 2) x−3 x+2

x+7 A(x + 2) + B(x − 3) = Adding (x − 3)(x + 2) (x − 3)(x + 2) Equate the numerators:

(2, 44)

x + 7 = A(x + 2) + B(x − 3)

(30, 30)

Let x + 2 = 0, or x = −2. Then we get

(0, 44)

−2 + 7 = 0 + B(−2 − 3)

x

(60, 0)

5 = −5B

(0, 0)

Vertex (0, 0)

−1 = B

I = 350x + 600y 350 · 0 + 600 · 0 = 0

(0, 44)

Next let x − 3 = 0, or x = 3. Then we get 3 + 7 = A(3 + 2) + 0

350 · 0 + 600 · 44 = 26, 400

(2, 44)

10 = 5A

350 · 2 + 600 · 44 = 27, 100

2=A

(30, 30) 350 · 30 + 600 · 30 = 28, 500 (60, 0)

The decomposition is as follows: 1 2 − x−3 x+2

350 · 60 + 600 · 0 = 21, 000

The maximum income of $28,500 is achieved when 30 less expensive and 30 more expensive assemblies are made.

2.

96. Let x = the number of chairs and y = the number of sofas produced. Find the maximum value of

2x A B = + (x + 1)(x − 1) x+1 x−1

A(x − 1) + B(x + 1) 2x = (x + 1)(x − 1) (x + 1)(x − 1) 2x = A(x − 1) + B(x + 1)

I = 80x + 300y subject to

Let x = 1: 2 · 1 = 0 + B(1 + 1)

20x + 100y ≤ 1900,

2 = 2B

x + 50y ≤ 500,

1=B

2x + 20y ≤ 240,

Let x = −1: 2(−1) = A(−1 − 1) + 0

x ≥ 0,

−2 = −2A

y ≥ 0.

1=A

y

The decomposition is

20

3.

15 10 (0, 10)

I = 80x + 300y 80 · 0 + 300 · 0 = 0

5

(0, 0)

(25, 9.5) (70, 5) (95, 0) 100 200 300 400 500

x

7x − 1 6x2 − 5x + 1 7x − 1 = (3x − 1)(2x − 1) B A = + 3x − 1 2x − 1 =

1 1 + . x+1 x−1

Factoring the denominator

A(2x − 1) + B(3x − 1) (3x − 1)(2x − 1)

Adding

Exercise Set 8.8

547 3x2 −11x−26 = A(x − 2)(x + 1)+ B(x + 2)(x + 1)+C(x + 2)(x − 2)

Equate the numerators: 7x − 1 = A(2x − 1) + B(3x − 1) 1 Let 2x − 1 = 0, or x = . Then we get 2 ! " ! " 1 1 −1 = 0+B 3· −1 7 2 2 1 5 = B 2 2 5=B

Let x + 2 = 0 or x = −2. Then we get

3(−2)2 −11(−2)−26 = A(−2−2)(−2+1)+0+0 12 + 22 − 26 = A(−4)(−1) 8 = 4A 2=A

Next let x − 2 = 0, or x = 2. Then, we get

3 · 22 − 11 · 2 − 26 = 0 + B(2 + 2)(2 + 1) + 0

1 Next let 3x − 1 = 0, or x = . We get 3 ! " ! " 1 1 −1 = A 2· −1 +0 7 3 3 ! " 2 7 −1 = A −1 3 3 4 1 =− A 3 3 −4 = A

12 − 22 − 26 = B · 4 · 3 −36 = 12B −3 = B

Finally let x + 1 = 0, or x = −1. We get

3(−1)2 −11(−1)−26 = 0 + 0 + C(−1 + 2)(−1 − 2) 3 + 11 − 26 = C(1)(−3) −12 = −3C 4=C

The decomposition is as follows: 5 4 + − 3x − 1 2x − 1 4.

13x + 46 13x + 46 = 12x2 − 11x − 15 (4x + 3)(3x − 5) B A = + 4x + 3 3x − 5

The decomposition is as follows: 3 4 2 − + x+2 x−2 x+1 6.

A(3x − 5) + B(4x + 3) (4x + 3)(3x − 5) ! " 5 5 5 Let x = : 13 · + 46 = 0 + B 4 · + 3 3 3 3 203 29 = B 3 3 7=B =

=

3x2 − 11x − 26 (x2 − 4)(x + 1)

5. =

B(x − 4)(x + 1)+C(x − 4)(x − 2)

Let x = 4: 5 · 42 + 9 · 4 − 56 = A(4 − 2)(4 + 1) + 0 + 0 60 = 10A

= =

B C A + + x+2 x−2 x+1

−18 = −6B 3=B

Let x = −1: 5(−1)2 + 9(−1) − 56 = 0 + 0 + C(−1 − 4)(−1 − 2) −60 = 15C −4 = C

Factoring the denominator

A(x−2)(x+1)+B(x+2)(x+1)+C(x+2)(x−2) (x+2)(x−2)(x+1) Adding

Equate the numerators:

6=A Let x = 2: 5 · 22 + 9 · 2 − 56 = 0 + B(2 − 4)(2 + 1) + 0

5 7 + . 4x + 3 3x − 5

3x2 − 11x − 26 (x + 2)(x − 2)(x + 1)

A(x−2)(x+1)+B(x−4)(x+1)+C(x−4)(x−2) (x−4)(x−2)(x+1)

5x2 + 9x − 56 = A(x − 2)(x + 1)+

3 Let x = − : 4 " 1 ! " 2 ! 3 3 + 46 = A 3 − −5 +0 13 − 4 4 145 29 =− A 4 4 −5 = A The decomposition is −

5x2 + 9x − 56 (x − 4)(x − 2)(x + 1) A B C = + + x−4 x−2 x+1

The decomposition is 7.

3 4 6 + − . x−4 x−2 x+1

9 (x + 2)2 (x − 1) C B A + + = x + 2 (x + 2)2 x−1 =

A(x + 2)(x − 1) + B(x − 1) + C(x + 2)2 (x + 2)2 (x − 1) Adding

548

Chapter 8: Systems of Equations and Matrices 2x2 + 3x + 1 = A(x − 1)(2x − 1)+ B(x + 1)(2x − 1)+C(x + 1)(x − 1)

Equate the numerators: 9 = A(x + 2)(x − 1) + B(x − 1) + C(x + 2)2

(1)

Let x + 1 = 0, or x = −1. Then, we get

Let x − 1 = 0, or x = 1. Then, we get

2(−1)2 + 3(−1) + 1 = A(−1 − 1)[2(−1) − 1]+0+0

9 = 0 + 0 + C(1 + 2)2

2 − 3 + 1 = A(−2)(−3)

9 = 9C

0 = 6A

1=C

0=A

Next let x + 2 = 0, or x = −2. Then, we get

Next let x − 1 = 0, or x = 1. Then, we get

9 = 0 + B(−2 − 1) + 0

2 · 12 + 3 · 1 + 1 = 0 + B(1 + 1)(2 · 1 − 1) + 0

9 = −3B

2+3+1 = B·2·1

−3 = B

6 = 2B

To find A we first simplify equation (1).

3=B

9 = A(x2 + x − 2) + B(x − 1) + C(x2 + 4x + 4)

1 Finally we let 2x − 1 = 0, or x = . We get 2 ! "2 ! " ! "! " 1 1 1 1 +3 +1 = 0+0+C +1 −1 2 2 2 2 2 ! " 1 3 3 1 + +1 = C · · − 2 2 2 2 3 3=− C 4 −4 = C

= Ax2 + Ax − 2A + Bx − B + Cx2 + 4Cx + 4C

= (A + C)x2 +(A + B + 4C)x+(−2A − B + 4C)

Then we equate the coefficients of x2 . 0 = A+C 0 = A+1

Substituting 1 for C

−1 = A

The decomposition is as follows: 3 1 1 − + − 2 x + 2 (x + 2) x−1 8.

The decomposition is as follows: 4 3 − x − 1 2x − 1

x2 − x − 4 (x − 2)3 A C B = + + x − 2 (x − 2)2 (x − 2)3 =

A(x − 2)2 + B(x − 2) + C (x − 2)3

x2 − x − 4 = A(x − 2)2 + B(x − 2) + C

(1)

Let x = 2: 2 − 2 − 4 = 0 + 0 + C −2 = C

Simplify equation (1).

x2 − x − 4 = Ax2 − 4Ax + 4A + Bx − 2B + C

= Ax2 + (−4A + B)x + (4A − 2B + C)

1 = A and

−1 = −4A + B, so B = 3. The decomposition is

2 3 1 + − . x − 2 (x − 2)2 (x − 2)3

2x + 3x + 1 (x2 − 1)(2x − 1) 2

9.

2x2 + 3x + 1 Factoring the (x + 1)(x − 1)(2x − 1) denominator A B C + + = x + 1 x − 1 2x − 1

=

=

A(x−1)(2x−1)+B(x+1)(2x−1)+C(x+1)(x−1) (x + 1)(x − 1)(2x − 1) Adding

Equate the numerators:

x2 − 10x + 13 (x − 3)(x − 2)(x − 1) A B C = + + x−3 x−2 x−1

=

2

Then

x2 − 10x + 13 (x2 − 5x + 6)(x − 1)

10.

=

A(x−2)(x−1)+B(x−3)(x−1)+C(x−3)(x−2) (x − 3)(x − 2)(x − 1)

x2 − 10x + 13 = A(x − 2)(x − 1)+ B(x − 3)(x − 1)+C(x − 3)(x − 2)

Let x = 3: 32 − 10 · 3 + 13 = A(3 − 2)(3 − 1) + 0 + 0 −8 = 2A −4 = A

Let x = 2: 22 − 10 · 2 + 13 = 0 + B(2 − 3)(2 − 1) + 0 −3 = −B 3=B

Let x = 1: 12 − 10 · 1 + 13 = 0 + 0 + C(1 − 3)(1 − 2) 4 = 2C 2=C The decomposition is −

3 2 4 + + . x−3 x−2 x−1

Exercise Set 8.8

549

x4 − 3x3 − 3x2 + 10 (x + 1)2 (x − 3)

11. =

x4 − 3x3 − 3x2 + 10 x3 − x2 − 5x − 3

12.

10x3 − 15x2 − 35x 20x − 30 = 10x − 5 + 2 x2 − x − 6 x −x−6

Dividing

Multiplying the denominator

20x − 30 20x − 30 = x2 − x − 6 (x − 3)(x + 2) A B = + x−3 x+2

Since the degree of the numerator is greater than the degree of the denominator, we divide. x− 2 x3 − x2 − 5x − 3 x4 −3x3 −3x2 + 0x+10 x4 − x3 −5x2 − 3x −2x3 +2x2 + 3x+10 −2x3 +2x2 +10x+ 6 − 7x+ 4

A(x + 2) + B(x − 3) (x − 3)(x + 2) 20x − 30 = A(x + 2) + B(x − 3) =

Let x = 3: 20 · 3 − 30 = A(3 + 2) + 0

The original expression is thus equivalent to the following: −7x + 4 x−2+ 3 x − x2 − 5x − 3 We proceed to decompose the fraction. −7x + 4 (x + 1)2 (x − 3) B C A = + + 2 x + 1 (x + 1) x−3 A(x + 1)(x − 3) + B(x − 3) + C(x + 1)2 = (x + 1)2 (x − 3) Adding

30 = 5A 6=A

Let x = −2: 20(−2) − 30 = 0 + B(−2 − 3) −70 = −5B 14 = B

The decomposition is 10x − 5 + 13.

Equate the numerators: −7x + 4 = A(x + 1)(x − 3) + B(x − 3)+ C(x + 1)

2

(Ax + B)(x − 1) + C(x2 + 2) (x2 + 2)(x − 1) Equate the numerators: =

(1)

Let x − 3 = 0, or x = 3. Then, we get −7 · 3 + 4 = 0 + 0 + C(3 + 1)

Let x − 1 = 0, or x = 1. Then we get −12 + 2 · 1 − 13 = 0 + C(12 + 2)

−17 = 16C 17 − =C 16 Let x + 1 = 0, or x = −1. Then, we get

−1 + 2 − 13 = C(1 + 2) −12 = 3C −4 = C

−7(−1) + 4 = 0 + B(−1 − 3) + 0

To find A and B we first simplify equation (1).

11 = −4B 11 =B − 4 To find A we first simplify equation (1).

−x2 + 2x − 13

= Ax2 − Ax + Bx − B + Cx2 + 2C

= (A + C)x2 + (−A + B)x + (−B + 2C) Equate the coefficients of x2 :

= A(x − 2x − 3) + B(x − 3) + C(x + 2x + 1)

−1 = A + C

= (A+C)x2 + (−2A+B −2C)x + (−3A−3B +C)

Equate the constant terms:

2

2

= Ax − 2Ax − 3A + Bx − 3B + Cx − 2Cx + C 2

Adding

−x2 + 2x − 13 = (Ax + B)(x − 1) + C(x2 + 2)

2

−7x + 4

−x2 + 2x − 13 (x2 + 2)(x − 1) Ax + B C = 2 + x +2 x−1

14 6 + . x−3 x+2

Substituting −4 for C, we get A = 3.

2

−13 = −B + 2C

Then equate the coefficients of x . 2

Substituting −4 for C, we get B = 5.

0=A+C 17 17 for C, we get A = . Substituting − 16 16 The decomposition is as follows: 17/16 11/4 17/16 − − x+1 (x + 1)2 x−3 The original expression is equivalent to the following: 17/16 11/4 17/16 − − x−2+ x+1 (x + 1)2 x−3

The decomposition is as follows: 4 3x + 5 − x2 + 2 x − 1 14.

26x2 + 208x (x2 + 1)(x + 5) C Ax + B + = 2 x +1 x+5

=

(Ax + B)(x + 5) + C(x2 + 1) (x2 + 1)(x + 5)

(1)

550

Chapter 8: Systems of Equations and Matrices 26x2 + 208x = (Ax + B)(x + 5) + C(x2 + 1)

Substituting 3 for A, we obtain B = −2.

(1)

The decomposition is as follows: 3 2 10 − + 2x − 1 x + 2 (x + 2)2

Let x = −5: 26(−5)2 + 208(−5) = 0 + C[(−5)2 + 1] −390 = 26C −15 = C

Simplify equation (1).

5x3 + 6x2 + 5x (x − 1)(x + 1)4 B C A D E = + + + + x−1 x+1 (x+1)2 (x+1)3 (x+1)4

26x2 + 208x

=

= Ax2 + 5Ax + Bx + 5B + Cx2 + C = (A + C)x2 + (5A + B)x + (5B + C) 26 = A + C

= [A(x+1)4 +B(x−1)(x+1)3 +C(x−1)(x+1)2+

26 = A − 15

D(x−1)(x+1)+E(x−1)]/[(x − 1)(x + 1)4 ]

41 = A

5x3 + 6x2 + 5x =

0 = 5B + C

A(x + 1)4 + B(x − 1)(x + 1)3 + C(x − 1)(x + 1)2 +

0 = 5B − 15

D(x − 1)(x + 1) + E(x − 1)

3=B

A=1 Let x = −1: 5(−1)3 + 6(−1)2 + 5(−1) = E(−1 − 1) −4 = −2E 2=E

A(x + 2)2 + B(2x − 1)(x + 2) + C(2x − 1) (2x − 1)(x + 2)2 Adding

Simplify equation (1). 5x3 + 6x2 + 5x = (A + B)x4 + (4A + 2B + C)x3 + (6A + C + D)x2 + (4A − 2B − C + E)x+

Equate the numerators:

(A − B − C − D − E)

6 + 26x − x2 = A(x + 2)2 + B(2x − 1)(x + 2)+ C(2x − 1) (1)

0 = A+B

1 Let 2x − 1 = 0, or x = . Then, we get 2 ! "2 ! "2 1 1 1 =A 6 + 26 · − +2 +0+0 2 2 2 ! "2 1 5 6 + 13 − = A 4 2 75 25 = A 4 4 3=A

0 = 1+B −1 = B

5 = 4A + 2B + C 5 = 4 · 1 + 2(−1) + C 3=C

0 = A−B−C −D−E

0 = 1 − (−1) − 3 − D − 2

D = −3

Let x + 2 = 0, or x = −2. We get

The decomposition is 1 3 3 2 1 − + − + . x − 1 x + 1 (x + 1)2 (x + 1)3 (x + 1)4

6 + 26(−2) − (−2)2 = 0 + 0 + C[2(−2) − 1] 6 − 52 − 4 = −5C −50 = −5C 10 = C

To find B we first simplify equation (1). 6 + 26x − x2

= A(x2 + 4x + 4) + B(2x2 + 3x − 2) + C(2x − 1)

= Ax2 +4Ax+4A+2Bx2 +3Bx−2B +2Cx−C

= (A+2B)x2 + (4A+3B +2C)x + (4A−2B −C)

Equate the coefficients of x2 : −1 = A + 2B

2

16 = 16A

6 + 26x − x2 (2x − 1)(x + 2)2 A B C = + + 2x − 1 x + 2 (x + 2)2 =

(1)

Let x = 1: 5 · 1 + 6 · 1 + 5 · 1 = A(1 + 1)4 3

15 41x + 3 − . The decomposition is 2 x +1 x+5 15.

5x3 + 6x2 + 5x (x2 − 1)(x + 1)3

16.

17.

6x3 + 5x2 + 6x − 2 2x2 + x − 1 Since the degree of the numerator is greater than the degree of the denominator, we divide. 3x+ 1 2x2 + x − 1 6x3 +5x2 +6x−2 6x3 +3x2 −3x 2x2 +9x−2 2x2 + x−1 8x−1 The original expression is equivalent to

Exercise Set 8.8

551

8x − 1 2x2 + x − 1 We proceed to decompose the fraction. 8x − 1 8x − 1 = Factoring the 2x2 + x − 1 (2x − 1)(x + 1) denominator B A + = 2x − 1 x + 1 3x + 1 +

A(x + 1) + B(2x − 1) (2x − 1)(x + 1) Equate the numerators:

19.

A(x2 +2x−5)+(Bx+C)(x−3) (x − 3)(x2 + 2x − 5) Equate the numerators: 2x2 − 11x + 5 = A(x2 + 2x − 5)+ (Bx + C)(x − 3)

(1)

Let x − 3 = 0, or x = 3. Then, we get

8x − 1 = A(x + 1) + B(2x − 1)

2 · 32 − 11 · 3 + 5 = A(32 + 2 · 3 − 5) + 0

Let x + 1 = 0, or x = −1. Then we get

18 − 33 + 5 = A(9 + 6 − 5)

8(−1) − 1 = 0 + B[2(−1) − 1]

−10 = 10A

−8 − 1 = B(−2 − 1)

−1 = A

−9 = −3B

To find B and C, we first simplify equation (1).

3=B

2x2 − 11x + 5 = Ax2 + 2Ax − 5A + Bx2 − 3Bx+

1 Next let 2x − 1 = 0, or x = . We get 2 ! " ! " 1 1 −1 = A +1 +0 8 2 2 ! " 3 4−1 = A 2 3 3= A 2 2=A The decomposition is 3 2 + . 2x − 1 x + 1 The original expression is equivalent to 3 2 + . 3x + 1 + 2x − 1 x + 1 18.

Adding

=

Adding

=

2x2 − 11x + 5 (x − 3)(x2 + 2x − 5) A Bx + C = + x − 3 x2 + 2x − 5

2x3 + 3x2 − 11x − 10 −3x − 13 = 2x − 1 + 2 x2 + 2x − 3 x + 2x − 3 Dividing A B −3x − 13 = + x2 + 2x − 3 x+3 x−1 A(x − 1) + B(x + 3) (x + 3)(x − 1) −3x − 13 = A(x − 1) + B(x + 3) =

Let x = −3: −3(−3) − 13 = A(−3 − 1) + 0

Cx − 3C

= (A + B)x2 + (2A − 3B + C)x+ (−5A − 3C)

Equate the coefficients of x2 : 2=A+B

Substituting −1 for A, we get B = 3. Equate the constant terms: 5 = −5A − 3C

Substituting −1 for A, we get C = 0. The decomposition is as follows: 3x 1 + − x − 3 x2 + 2x − 5 20.

3x2 − 3x − 8 (x − 5)(x2 + x − 4) Bx + C A + = x − 5 x2 + x − 4 =

A(x2 + x − 4) + (Bx + C)(x − 5) (x − 5)(x2 + x − 4)

3x2 − 3x − 8 = A(x2 + x − 4)+

(Bx + C)(x − 5)

Let x = 1: −3 · 1 − 13 = 0 + B(1 + 3) −16 = 4B −4 = B

4 1 − . The decomposition is 2x − 1 + x+3 x−1

2

52 = 26A

−4 = −4A 1=A

(1)

Let x = 5: 3 · 5 − 3 · 5 − 8 = A(5 + 5 − 4) + 0 2

2=A Simplify equation (1). 3x2 − 3x − 8

= Ax2 + Ax − 4A + Bx2 − 5Bx + Cx − 5C

= (A + B)x2 + (A − 5B + C)x + (−4A − 5C) 3 = A+B 3 = 2+B 1=B

552

Chapter 8: Systems of Equations and Matrices 36x + 1 = A(3x + 2) + B(4x − 5)

−8 = −4A − 5C

or 36x + 1 = (3A + 4B)x + (2A − 5B)

−8 = −4 · 2 − 5C

Then equate corresponding coefficients.

0=C

x 2 + . The decomposition is x − 5 x2 + x − 4

36 = 3A + 4B 1 = 2A − 5B

A = 8 and B = 3. The decomposition is 8 3 + . 4x − 5 3x + 2 24.

= A(x + 1)2 + B(3x + 5)(x + 1) + C(3x + 5) = A(x2 + 2x + 1) + B(3x2 + 8x + 5) + C(3x + 5) or −4x2 − 2x + 10

61 = 7A − 3B

Then A = 7 and B = −4. The decomposition is 4 7 − . 6x − 3 x + 7

Then equate corresponding coefficients. −2 = 2A + 8B + 3C 10 = A + 5B + 5C

Coefficients of x2 -terms Coefficients of x-terms Constant terms

25.

We solve this system of three equations and find A = 5, B = −3, C = 4. The decomposition is 3 4 5 − + . 3x + 5 x + 1 (x + 1)2

22.

= (Ax + B)(x − 2) + C(3x2 + 1)

= Ax2 −2Ax+Bx−2B +3Cx2 +C

or

26x2 − 36x + 22

−4x2 − 9x + 8

= A(2x − 1) + B(x − 4)(2x − 1) + C(x − 4) 2

= (A + 3C)x2 + (−2A + B)x + (−2B + C)

= A(4x − 4x + 1) + B(2x − 9x + 4) + C(x − 4) or 26x2 −36x+22 = (4A+2B)x2 +(−4A−9B +C)x+ 2

Then equate corresponding coefficients. −4 = A + 3C

−9 = −2A + B

(A + 4B − 4C)

8 = −2B + C

Solving the system of equations 26 = 4A + 2B,

Coefficients of x-terms Constant terms

A = 2, B = −5, C = −2. The decomposition is 2 2x − 5 − . 3x2 + 1 x − 2

22 = A + 4B − 4C

we get A = 6, B = 1, and C = −3.

36x + 1 36x + 1 = 23. 12x2 − 7x − 10 (4x − 5)(3x + 2) The decomposition looks like B A + . 4x − 5 3x + 2 Add and equate the numerators.

Coefficients of x2 -terms

We solve this system of equations and find

−36 = −4A − 9B + C,

Then the decomposition is 1 3 6 + − . x − 4 2x − 1 (2x − 1)2

−4x2 − 9x + 8 (3x2 + 1)(x − 2) The decomposition looks like Ax + B C + . 3x2 + 1 x−2 Add and equate the numerators. −4x2 − 9x + 8

26x2 − 36x + 22 A B C = + + (x − 4)(2x − 1)2 x − 4 2x − 1 (2x − 1)2 Add and equate numerators.

2

−17x + 61 A B = + 6x2 + 39x − 21 6x − 3 x + 7 −17x + 61 = (A + 6B)x + (7A − 3B) −17 = A + 6B,

= (A + 3B)x2 +(2A + 8B + 3C)x+(A + 5B + 5C)

−4 = A + 3B

Constant terms

We solve this system of equations and find

−4x2 − 2x + 10 21. (3x + 5)(x + 1)2 The decomposition looks like A B C . + + 3x + 5 x + 1 (x + 1)2 Add and equate the numerators. −4x2 − 2x + 10

Coefficients of x-terms

26.

Ax + B C 11x2 − 39x + 16 = 2 + (x2 + 4)(x − 8) x +4 x−8

11x2 − 39x + 16 = (A + C)x2 + (−8A + B)x+ (−8B + 4C)

11 = A + C, −39 = −8A + B,

16 = −8B + 4C

Then A = 5, B = 1, and C = 6.

Exercise Set 8.8

553 x4 − x3 − 5x2 − x − 6 = 0

The decomposition is 5x + 1 6 + . x2 + 4 x − 8

27. See the procedure on pages 753 and 754 of the text. One of the algebraic methods referred to in Step 5 involves substituting values for the variable that allow us to find the constants. The other method involves equating numerators, equating coefficients of the like terms, and then using a system of equations to find the constants. Answers will vary regarding the method preferred. 28. The denominator of the second fraction, x2 −5x+6, can be factored into linear factors with real coefficients: (x−3)(x− 2). Thus, the given expression is not a partial fraction decomposition.

(x + 2)(x3 − 3x2 + x − 3) = 0

(x + 2)[x2 (x − 3) + (x − 3)] = 0 (x + 2)(x − 3)(x2 + 1) = 0

x+2 = 0

x = −2 or

x3 + 5x2 + 5x − 3 = 0

(x + 3)(x2 + 2x − 1) = 0

x + 3 = 0 or x2 + 2x − 1 = 0

The solution of the first equation is −3. We use the quadratic formula to solve the second equation. √ −b ± b2 − 4ac x= 2a 4 √ −2 ± 22 − 4 · 1 · (−1) −2 ± 8 = = 2·1 2 √ √ −2 ± 2 2 2(−1 ± 2) = = 2√ 2 = −1 ± 2 √ √ The solutions are −3, −1 + 2, and −1 − 2.

x3 − 3x2 + x − 3 = 0

(x − 3)(x2 + 1) = 0

x − 3 = 0 or x2 + 1 = 0 x = 3 or

x = 3 or

x2 = −1 x = ±i

The solutions are 3, i, and −i. 31. f (x) = x3 + x2 − 3x − 2

We use synthetic division to factor the polynomial. Using the possibilities found by the rational zeros theorem we find that x + 2 is a factor: 3 −2 3 1 1 −3 −2 −2 2 2 1 −1 −1 0 We have x3 + x2 − 3x − 2 = (x + 2)(x2 − x − 1). x3 + x2 − 3x − 2 = 0

(x + 2)(x − x − 1) = 0 2

x + 2 = 0 or x2 − x − 1 = 0

The solution of the first equation is −2. We use the quadratic formula to solve the second equation. √ −b ± b2 − 4ac x= 2a 4 −(−1) ± (−1)2 − 4 · 1 · (−1) = 2·1 √ 1± 5 = 2 √ √ 1+ 5 1− 5 The solutions are −2, and . 2 2 32. f (x) = x4 − x3 − 5x2 − x − 6 3 −2 3 1 −1 −5 −1 −6 −2 6 −2 6 1 −3 1 −3 0

x = ±i

33. f (x) = x3 + 5x2 + 5x − 3 3 5 5 −3 −3 3 1 −3 −6 3 1 2 −1 0

x (x − 3) + (x − 3) = 0 2

x = 3 or

The solutions are −2, 3, i, and −i.

29. The degree of the numerator is equal to the degree of the denominator, so the first step should be to divide the numerator by the denominator in order to express the fraction as a quotient + remainder/denominator. 30.

or x − 3 = 0 or x2 + 1 = 0

34.

A P (x) 9x3 − 24x2 + 48x = + (x − 2)4 (x + 1) x + 1 (x − 2)4 Add and equate numerators.

9x3 − 24x2 + 48x = A(x − 2)4 + P (x)(x + 1)

Let x = −1: 9(−1)3 − 24(−1)2 + 48(−1) = A(−1 − 2)4 + 0 −81 = 81A −1 = A

Then 9x3 − 24x2 + 48x = −(x − 2)4 + P (x)(x + 1)

9x3 − 24x2 + 48x = −x4 + 8x3 − 24x2 + 32x− 16 + P (x)(x + 1)

x + x + 16x + 16 = P (x)(x + 1) 4

3

x3 + 16 = P (x) Now decompose

Dividing by x + 1

x + 16 . (x − 2)4 3

x3 + 16 C B D E + = + + (x − 2)4 x − 2 (x − 2)2 (x − 2)3 (x − 2)4 Add and equate numerators. x3 + 16 = B(x − 2)3 + C(x + 2)2 + D(x − 2) + E

Let x = 2: 2 + 16 = 0 + 0 + 0 + E 3

24 = E Simplify equation (1).

(1)

554

Chapter 8: Systems of Equations and Matrices 1 1 1 x 2 2 2 2a 4a 4a − 2 + + x + a2 x+a x−a

x3 + 16 = Bx3 +(−6B +C)x2 +(12B −4C +D)x+ (−8B + 4C − 2D + E)

1=B

36.

0 = −6B + C

Multiplying by ex /ex y Let y = ex , decompose 2 , and then substitute 2y + 3y + 1 1 1 − . ex for y. The result is x e + 1 2ex + 1

0 = −6 · 1 + C 6=C

0 = 12B − 4C + D

0 = 12 · 1 − 4 · 6 + D

12 = D

The decomposition is 12 24 1 6 1 + + . + + − x + 1 x − 2 (x − 2)2 (x − 2)3 (x − 2)4 35. = =

x x4 − a4

x (x2 + a2 )(x + a)(x − a)

C D Ax + B + + x2 + a2 x+a x−a

Factoring the denominator

=[(Ax + B)(x + a)(x − a) + C(x2 + a2 )(x − a)+ D(x + a )(x + a)]/[(x + a )(x + a)(x − a)] 2

2

2

2

Equate the numerators:

x = (Ax + B)(x + a)(x − a) + C(x2 + a2 )(x − a)+ D(x + a )(x + a) 2

2

Let x − a = 0, or x = a. Then, we get a = 0 + 0 + D(a2 + a2 )(a + a)

a = D(2a2 )(2a) a = 4a3 D 1 =D 4a2 Let x + a = 0, or x = −a. We get

−a = 0 + C[(−a)2 + a2 ](−a − a) + 0

−a = C(2a )(−2a) 2

−a = −4a3 C 1 =C 4a2 Equate the coefficients of x3 : 0=A+C +D 1 Substituting 2 for C and for D, we get 4a 1 A = − 2. 2a Equate the constant terms: 0 = −Ba − Ca + Da 1 Substitute 2 for C and for D. Then solve for B. 4a 1 1 0 = −Ba2 − 2 · a3 + 2 · a3 4a 4a 0 = −Ba2 2

3

ex 1 = e−x + 3 + 2ex 1 + 3ex + 2e2x

37.

1 + 2 ln x 1 + ln x2 = (ln x + 2)(ln x − 3)2 (ln x + 2)(ln x − 3)2 Let u = ln x. Then we have: 1 + 2u (u + 2)(u − 3)2 A B C = + + u + 2 u − 3 (u − 3)2

A(u − 3)2 + B(u + 2)(u − 3) + C(u + 2) (u + 2)(u − 3)2 Equate the numerators: =

1 + 2u = A(u − 3)2 + B(u + 2)(u − 3) + C(u + 2)

Let u − 3 = 0, or u = 3.

1 + 2 · 3 = 0 + 0 + C(5)

7 = 5C 7 =C 5 Let u + 2 = 0, or u = −2.

1 + 2(−2) = A(−2 − 3)2 + 0 + 0

−3 = 25A 3 − =A 25 To find B, we equate the coefficients of u2 : 0=A+B 3 3 for A and solving for B, we get B = . 25 25 1 + 2u The decomposition of is as follows: (u + 2)(u − 3)2 3 3 7 − + + 25(u + 2) 25(u − 3) 5(u − 3)2 Substituting ln x for u we get 3 3 7 + + . − 25(ln x + 2) 25(ln x − 3) 5(ln x − 3)2 Substituting −

Chapter 8 Review Exercises

3

0=B

The decomposition is as follows:

1. The statement is true. See page 677 in the text. 2. The statement is false. See page 677 in the text. 3. The statement is true. See page 715 in the text. 4. The statement is false. See page 717 in the text. 5. (a)

Chapter 8 Review Exercises 6. (e)

555 16.

−3x − 3y = 6

7. (h)

3x + 3y = −6

9. (b)

−3x − 3y = 6 0=0

10. (g)

The equation 0 = 0 is true for all values of x and y. Thus the system of equations has infinitely many solutions. Solving either equation for y, we get y = −x − 2, so the solutions are ordered pairs of the form (x, −x − 2). Equivalently, if we solve either equation for x we get x = −y − 2 so the solutions can also be expressed as (−y − 2, y).

11. (c) 12. (f) 5x − 3y = −4, (1) 3x − y = −4

(2)

Multiply equation (2) by −3 and add.

17.

5x − 3y = −4

(3)

Multiply equations (2) and (3) by 2.

Back-substitute to find y. 3(−2) − y = −4

2x − 4y + 3z = −3

(1)

6x + 4y − 4z = 8

(5)

−10x + 4y − 2z = 14

Using equation (2)

−6 − y = −4

(4)

Multiply equation (1) by 5 and add it to equation (4).

−y = 2 y = −2

Multiply equation (1) by −3 and add it to equation (5).

The solution is (−2, −2). 2x + 3y = 2,

(1)

5x − y = −29

(2)

(1)

16y − 13z = 17

(7)

2x − 4y + 3z = −3

(1)

0 = 16

(8)

(6)

Add equation (6) to equation (7).

2x + 3y = 2

− 16y + 13z = −1

15x − 3y = −87 17x = −85

(6)

Equation (8) is false, so the system of equations has no solution.

x = −5

Back-substitute to find y. 5(−5) − y = −29

2x − 4y + 3z = −3 − 16y + 13z = −1

Multiply equation (2) by 3 and add.

Using equation (2)

−25 − y = −29 −y = −4 y=4

18.

x + 5y + 3z = 0, (1) 3x − 2y + 4z = 0, (2) 2x + 3y − z = 0

(3)

Multiply equation (1) by −3 and add it to equation (2).

The solution is (−5, 4). x + 5y = 12, (1) 5x + 25y = 12

(2)

3x + 2y − 2z = 4

x = −2

15.

2x − 4y + 3z = −3, (1)

−5x + 2y − z = 7,

−9x + 3y = 12 −4x =8

14.

(2)

Multiply equation (1) by 3 and add.

8. (d)

13.

x + y = −2, (1)

(2)

Solve equation (1) for x. x = −5y + 12

Substitute in equation (2) and solve for y. 5(−5y + 12) + 25y = 12 −25y + 60 + 25y = 12 60 = 12

We get a false equation, so there is no solution.

Multiply equation (1) by −2 and add it to equation (3). x + 5y + 3z = 0

(1)

− 17y − 5z = 0

(4)

− 7y − 7z = 0

(5)

Multiply equation (5) by 17. x+

5y +

3z = 0

(1)

− 17y −

5z = 0

(4)

− 119y − 119z = 0

(6)

Multiply equation (4) by −7 and add it to equation (6). x + 5y + 3z = 0

(1)

− 17y − 5z = 0

(4)

− 84z = 0

(7)

556

Chapter 8: Systems of Equations and Matrices 20. Systems 13, 14, 16, 18, and 19 each have at least one solution, so they are consistent. Equations 15 and 17 have no solution, so they are inconsistent.

Complete the solution. −84z = 0 z=0

21. Systems 13, 14, 15, 17, 18, and 19 each have either no solution or exactly one solution, so the equations in those systems are independent. System 16 has infinitely many solutions, so the equations in that system are dependent.

−17y − 5 · 0 = 0

−17y = 0 y=0

x+5·0+3·0 = 0

22.

x=0

2x − 5y = −8

The solution is (0, 0, 0). 19.

= 6, (2)

y−z

x

Write the augmented matrix. We will use Gaussian elimination.   5 1 2   2 −5 −8

= 5, (1)

x−y

z − w = 7, (3) +w=8

(4)

Multiply row 1 by −2 and add it to row 2.   5 1 2   0 −9 −18

Multiply equation (1) by −1 and add it to equation (4). =5

(1)

=6

(2)

z−w=7

(3)

x−y

y−z

+w=3

y

1 Multiply row 2 by − . 9   1 2 5   0 1 2

(5)

Multiply equation (2) by −1 and add it to equation (5). x−y

y−z

=5

(1)

=6

(2)

z−w=7

(3)

z + w = −3

(6)

Multiply equation (3) by −1 and add it to equation (6). =5

(1)

=6

(2)

z− w=7

(3)

x−y

y−z

x + 2y = 5,

2w = −10

2w = −10 w = −5

Back-substitute −5 for w in equation (3) and solve for z. z−w = 7

z − (−5) = 7 z+5 = 7

z=2 Back-substitute 2 for z in equation (2) and solve for y. y−z = 6 y−2 = 6

y=8

Back-substitute 8 for y in equation (1) and solve for x. x−y = 5 x−8 = 5

x = 13

Writing the solution as (w, x, y, z), we have (−5, 13, 8, 2).

(1)

y=2

(2)

Back-substitute in equation (1) and solve for x. x + 2(2) = 5 x=1

(7)

Solve equation (7) for w.

We have: x + 2y = 5

The solution is (1, 2). 23.

3x + 4y + 2z = 3 5x − 2y − 13z = 3

4x + 3y − 3z = 6

Write the augmented matrix. We will use Gaussian elimination.   3 4 2 3    5 −2 −13 3  4 3 −3 6 Multiply row 2 and row 3 by 3.   3 4 2 3    15 −6 −39 9  12 9 −9 18 Multiply row 1 by −5 and add it to row 2. Multiply row 1 by −4  3 3 4 2   0 −26 −49 −6 0

−7

−17

6

and add it to row 3.   

Chapter 8 Review Exercises

557 Multiply row 1 by −3 and add it to row 3.   1 3 1 0    0 −10 −5 0 

Multiply row 3 by 26.   3 4 2 3   −49 −6   0 −26 0 −182 −442 156

0



1     0  0

4 3

2 1 3 3 49 26 13 1 −2

1 0

1 1 1 , row 2 by − , and row 3 by − . 3 26 99 

0

(2)

The solution is (−3, 4, −2). 3x + 5y + z = 0, 2x − 4y − 3z = 0, x + 3y + z = 0

Write the augmented matrix. We will use Gaussian elimination.   3 5 1 0    2 −4 −3 0  1 3 1 0 Interchange rows 1 and 3.   1 3 1 0    2 −4 −3 0  5

0 0

The system of equations is equivalent to x + 3y + z = 0 (1)

(1)

x = −3

3

0

The last row corresponds to the equation 0 = 0. This indicates that the equations are dependent.

Back-substitute in equation (2) and solve for y. 49 3 y + (−2) = 26 13 3 49 = y− 13 13 52 y= =4 13 Back-substitute in equation (1) and solve for x. 2 4 x + (4) + (−2) = 1 3 3 16 4 − =1 x+ 3 3

24.

Multiply row 2 by −2 and add it to row 3.   1 3 1 0    0 −10 −5 0   

    

2 4 x+ y+ z = 1 3 3 3 49 y+ z = 26 13 z = −2

0

−2

Multiply row 3 by 5.   1 3 1 0    0 −10 −5 0    0 −20 −10 0

Multiply row 2 by −7 and add it to row 3.   3 3 4 2    0 −26 −49 −6  0 0 −99 198 Multiply row 1 by

−4

1 0

Multiply row 1 by −2 and add it to row 2.

−10y − 5z = 0. (2)

Solve equation (2) for y. −10y − 5z = 0 y=−

z 2

z Substitute − for y in equation (1) and solve for x. ! 2 " z x+3 − +z = 0 2 z x= 2 ! " z z The solution is , − , z , where z is any real number. 2 2 w + x + y + z = −2,

25.

−3w − 2x + 3y + 2z = 10,

2w + 3x + 2y − z = −12, 2w + 4x − y + z = 1

Write the augmented matrix. We will use Gauss-Jordan elimination.   1 1 1 1 −2  −3 −2 3 2 10       2 3 2 −1 −12  2

4 −1

1

1

Multiply row 1 by 3 and add it to row 2. Multiply row 1 by −2 and add it to row 3. Multiply row 1 by −2 and add it to row 4.   1 1 1 1 −2  0 1 4  6 5      0 1 0 −3 −8  0

2 −3 −1

5

558

Chapter 8: Systems of Equations and Matrices Multiply row 2 by −1 and add it to row 1.

Multiply row 2 by −1 and add it to row 3. Multiply row 2 by −2 and add it to row 4. 

1

0

 0 1    0 0 0

0

−5

−4

−6

−8

6

−15

5

−11

−6



4    −12  −3

0

0 −15

−3

−11

    

Multiply row 3 by 5 and add it to row 4. 3 0 0 8 12  0 1 0 −3 −8       0 0 3 4 6  

0

0

9

27

1 Multiply row 4 by . 9  3 0 0 8 12  0 1 0 −3 −8    0 0 3 6 4 0

0

0

1

3

   

Multiply row 4 by 3 and add it to row 2.

Multiply row 4 by −4 and add it to row 3. 3

0

0

0

 0 1 0 0    0 0 3 0 0

0

Multiply 1 0  0 1    0 0 

0

0

0

1

−12



1    −6  3

1 rows 1 and 3 by . 3  0 0 −4 1  0 0   1 0 −2  0

1

The total interest is $167. 0.03x + 0.035y = 167 We have a system of equations. x+

y = 5000,

0.03x + 0.035y = 167 Multiplying the second equation by 1000 to clear the decimals, we have: x+

y = 5000,

(1)

30x + 35y = 167, 000. (2) Carry out. We begin by multiplying equation (1) by −30 and adding. −30x − 30y = −150, 000 30x + 35y = 167, 000 5y = 17, 000 y = 3400



Multiply row 4 by −8 and add it to row 1.



27. Familiarize. Let x = the amount invested at 3% and y = the amount invested at 3.5%. Then the interest from the investments is 3%x and 3.5%y, or 0.03x and 0.035y.

x + y = 5000

Multiply row 3 by −2 and add it to row 2.

0

x = 31, y = 44

The total investment is $5000.

Multiply row 3 by 5 and add it to row 1.



0.05x + 0.10y = 5.95

Translate.

Multiply row 1 by 3. 1 Multiply row 3 by − . 2  3 0 −15 −12 −18  0 1 4 6 5    0 0 3 4 6

26. Let x = the number of nickels and y = the number of dimes. Solve: x+ y = 75,

3

The solution is (−4, 1, −2, 3).

Back-substitute to find x. x + 3400 = 5000 Using equation (1) x = 1600 Check. The total investment is $1600 + $3400, or $5000. The total interest is 0.03($1600) + 0.035($3400), or $48 + $119, or $167. The solution checks. State. $1600 was invested at 3% and $3400 was invested at 3.5%. 28. Let x, y, and z represent the number of servings of bagels, cream cheese, and bananas, respectively. Solve: 200x + 100y + 105z = 460, 2x + 10y +

z = 9,

29x + 24y +

7z = 55

1 x = 1, y = , z = 2 2 29. Familiarize. Let x, y, and z represent the scores on the first, second, and third tests, respectively. Translate. The total score on the three tests is 226. x + y + z = 226 The sum of the scores on the first and second tests exceeds the score on the third test by 62. x + y = z + 62

Chapter 8 Review Exercises

559

The first score exceeds the second by 6.

34.

x=y+6 We have a system of equations. x + y + z = 226, x + y = z + 62, x=y+6 or

x + y + z = 226, x + y − z = 62, Solving the system of equations, we get

Carry out. (75, 69, 82).

Check. The sum of the scores is 75 + 69 + 82, or 226. The sum of the scores on the first two tests is 75 + 69, or 144. This exceeds the score on the third test, 82, by 62. The score on the first test, 75, exceeds the score on the second test, 69, by 6. The solution checks.

36.

State. The scores on the first, second, and third tests were 75, 69, and 82, respectively. 37.

30. a) Solve: 7.3 = a · 02 + b · 0 + c, 6.5 = a · 22 + b · 2 + c,

or

6.7 = a · 32 + b · 3 + c c = 7.3,

4a + 2b + c = 6.5, 9a + 3b + c = 6.7 a = 0.2, b = −0.8, c = 7.3, so f (x) = 0.2x2 − 0.8x + 7.3, where x is the number of years after 2000. b) f (5) = 0.2(5)2 − 0.8(5) + 7.3 = 8.3 lb 31.

32.

33.



1 −1 3 A+B =  2 −2 0  1 + (−1) =  2+1 −2 + 0  0 −1 1 = 3 −2 1 

1 −3A = −3  2 −2 

1 −A = −1  2 −2

   1 −1 0 −1 0 6 3 −2  ·  1 −2 0  AB =  2 −2 0 1 0 1 −3   −1 − 1 + 0 0 + 2 + 0 6 + 0 + 0 =  −2 + 3 + 0 0 − 6 − 2 12 + 0 + 6  2+0+0 0 + 0 + 1 −12 + 0 − 3   −2 2 6 18  =  1 −8 2 1 −15

35. A and B do not have the same order, so it is not possible to find A + B.

= 6

x−y



  0 −1 −2  +  1 1 0

 0 6 −2 0  1 −3  −1 + 0 0+6 3 + (−2) −2 + 0  0+1 1 + (−3)  6 −2  −2

   −1 0 −3 3 0 3 −2  =  −6 −9 6  0 1 6 0 −3   −1 0 −1 3 −2  =  −2 0 1 2

 1 0 −3 2  0 −1



   1 −1 0 −1 0 6 3 −2  −  1 −2 0 A−B =  2 −2 0 1 0 1 −3   2 −1 −6 5 −2  = 1 −2 −1 4 

   −1 0 6 1 −1 0 0 · 2 3 −2  BA =  1 −2 0 1 −3 −2 0 1   −1 + 0 − 12 1 + 0 + 0 0+0+6 −1 − 6 + 0 0 + 4 + 0  = 1 − 4 + 0 0+2+6 0+3+0 0−2−3   −13 1 6 4  =  −3 −7 8 3 −5 

   1 −1 0 −1 0 6 3 −2  + 3  1 −2 0  38. A + 3B =  2 −2 0 1 0 1 −3     1 −1 0 −3 0 18 3 −2  +  3 −6 0  = 2 −2 0 1 0 3 −9   −2 −1 18 =  5 −3 −2  −2 3 −8   46.1 5.9 10.1 8.5 11.4   39. a) M =  54.6 4.6 9.6 7.6 10.6   48.9 5.5 12.7 9.4 9.3  51.3 4.8 11.3 6.9 12.7 / 0 b) N = 32 19 43 38 / 0 c) NM = 6564.7 695.1 1481.1 1082.8 1448.7

d) The entries of NM represent the total cost, in cents, for each item for the day’s meal. 1 2 −2 0 40. A = 1 3 Write the augmented matrix. 2 1 −2 0 1 0 1 3 0 1 Interchange rows. 2 1 1 3 0 1 −2 0 1 0

560

Chapter 8: Systems of Equations and Matrices Interchange rows  1 0 0 0  0 2 2 0   0 4 −5 0 0 0 0 1

Multiply row 1 by 2 and add it to row 2. 1 2 1 3 0 1 0 6 1 2 1 Multiply row 2 by . 6   1 3 0 1  1 1  0 1 6 3 Multiply row 2 by −3 and add it to row 1.   1 0 1 0 −   2   1 1 0 1 6 3   1 0 −   2 A−1 =   1 1 6 3   0 0 3 41. A =  0 −2 0  4 0 0 Write the  0 0  0 −2 4 0

Multiply row 2  1 0 0  0 18 18   0 0 −9 0 0 0 Multiply row 3  1 0 0  0 18 0   0 0 −9 0 0 0

Write the augmented matrix.  1 0 0 0 1 0 0 0  0 4 −5 0 0 1 0 0   0 2 2 0 0 0 1 0 0 0 0 1 0 0 0 1

   

0 0 1 0

0 1 0 0

 0 0   0  1

by 9. 0 0 0 1

43.

 1 0 0 0 0 0 9 0   0 1 −2 0  0 0 0 1

by 2 and add it to row 2.  0 0 0 1 0 0 0 2 5 0   0 0 1 −2 0  1 0 0 0 1

Multiply row 2 by  1 0 0 0 1   0 1 0 0 0     0 0 1 0 0  0 0 0 1 0  1 0 1   0  9 A−1 =    0 −1  9 0 0

1 and 3.  0 1 1 0  0 0

1 1 1 Multiply row 1 by , row 2 by − , and row 3 by . 4 2 3   1 0   1 0 0 0 4     1   0 1 0 0 − 0   2     1 0 0 1 0 0 3   1 0  0 4      1  A−1 =  0 − 0   2     1 0 0 3   1 0 0 0  0 4 −5 0   42. A =   0 2 2 0  0 0 0 1

1 0 0 0

Multiply row 2 by −2 and add it to row 3.   0 0 1 0 0 0 1 0  0 2 2 0 0 0 1 0     0 0 −9 0 0 1 −2 0  0 0 0 1 0 0 0 1

augmented matrix.  3 1 0 0 0 0 1 0  0 0 0 1

Interchange rows  4 0 0 0  0 −2 0 0 0 0 3 1

2 and 3.

1 1 and row 3 by − . 18 9  0 0 0 1 5  0   9 18   1 2 − 0   9 9 0 0 1  0 0 5  0   18   2 0   9 0 1

3x − 2y + 4z = 13, x + 5y − 3z = 7,

2x − 3y + 7z = −8

44.

Write the coefficients on the left in a matrix. Then write the product of that matrix and the column matrix containing the variables, and set the result equal to the column matrix containing the constants on the right.      3 −2 4 x 13  1 5 −3   y  =  7  2 −3 7 z −8 2x + 3y = 5, 3x + 5y = 11 Write an equivalent matrix equation, AX = B. 1 21 2 1 2 2 3 x 5 = 3 5 y 11 Then, X = A−1 B =

1

5 −3 −3 2

The solution is (−8, 7).

21

5 11

2

=

1

−8 7

2

Chapter 8 Review Exercises 45.

561

5x − y + 2z = 17,

3x + 2y − 3z = −16,

46.

4x − 3y − z = 5 Write an equivalent matrix equation, AX = B.      5 −1 2 x 17  3 2 −3   y  =  −16  4 −3 −1 z 5 Then,   11 7 1  80     80 80    17 1   9 13 21 −1   −16  =  −2  X=A B= −  80 80 80    5 5  17 11 13  − − 80 80 80 The solution is (1, −2, 5).

50. We will expand down the third column. 3 3 3 1 −1 2 3 3 3 3 −1 2 0 3 3 3 3 −1 3 1 3 3 3 3 3 3 −1 2 3 3 3 3 + 0(−1)2+3 3 1 −1 3 + = 2(−1)1+3 33 3 −1 3 3 −1 3 3 3 3 3 1 −1 3 3 1(−1)3+3 33 −1 2 3

= 2 · 1[−1(3) − (−1)(2)] + 0 + 1 · 1[1(2) − (−1)(−1)] = 2(−1) + 0 + 1(1)

51.

2w + 3x − 2y − z = 2,

−w + 5x + 4y − 2z = 3,

The solution is (2, −1, 1, −3). 3 3 3 3 47. 33 1 −2 33 = 1 · 4 − 3(−2) = 4 + 6 = 10 3 4 3 3√ 3 √ 3 √ √ 3 = 3(− 3) − (−3)(−5) = −3 − 15 48. 33 3 −5 −3 − 3 3 = −18 49. We will expand across the first row. 3 3 3 2 −1 1 3 3 3 3 1 2 −1 3 3 3 3 3 4 −3 3 3 3 3 3 2 −1 3 3 3 + (−1)(−1)1+2 3 1 = 2(−1)1+1 33 3 33 4 −3 3 3 31 23 3 1(−1)1+3 33 3 43

3 −1 33 + −3 3

= 2 · 1[2(−3) − 4(−1)] + (−1)(−1)[1(−3) − 3(−1)]+ 1 · 1[1(4) − 3(2)]

= 2(−2) + 1(0) + 1(−2) = −6

5x − 2y = 19,

7x + 3y = 15 3 3 3 5 −2 3 3 = 5(3) − 7(−2) = 29 D = 33 7 3 3

w − x − y + z = −1,

3w − 2x + 5y + 3z = 4 Write an equivalent matrix equation, AX = B.      1 −1 −1 1 w −1  2     3 −2 −1   x   2    =  −1 5 4 −2   y   3  3 −2 5 3 z 4 Then,    −55 22 −19 13 −1   1 24 −9   67 −8  2 = X = A−1 B =     −26 1 −3 7 3 47 143 −29 40 −15 4   2  −1     1 −3

= −1

3 3 3 19 −2 3 3 = 19(3) − 15(−2) = 87 Dx = 33 15 3 3 3 35 Dy = 33 7

3 19 33 = 5(15) − 7(19) = −58 15 3

Dx 87 = =3 D 29 Dy −58 = = −2 y= D 29 The solution is (3, −2). x=

52.

x + y = 4, 4x + 3y = 11 3 3 31 13 3 = 1(3) − 4(1) = −1 D = 33 4 33 3 3 4 Dx = 33 11 3 31 Dy = 33 4

3 1 33 = 4(3) − 11(1) = 1 33

3 4 33 = 1(11) − 4(4) = −5 11 3

1 Dx = = −1 D −1 −5 Dy y= = =5 D −1 The solution is (−1, 5). x=

562 53.

Chapter 8: Systems of Equations and Matrices 3x − 2y + z = 5,

56.

4x − 5y − z = −1,

3x + 2y − z = 4 3 3 3 3 −2 1 3 3 3 D = 33 4 −5 −1 33 = 42 3 3 2 −1 3 3 3 3 5 −2 1 3 3 3 Dx = 33 −1 −5 −1 33 = 63 3 4 2 −1 3 3 3 33 5 1 33 3 Dy = 33 4 −1 −1 33 = 39 3 3 4 −1 3 3 3 3 3 −2 5 3 3 3 Dz = 33 4 −5 −1 33 = 99 33 2 4 3

Dx 63 3 = = D 42 2 39 13 Dy y= = = D 42 14 99 33 Dz z= = = D 42! 14 " 3 13 33 The solution is , , . 2 14 14

57.

x=

54.

2x − y − z = 2,

3x + 2y + 2z = 10, x − 5y − 3z = −2 3 3 3 2 −1 −1 3 3 3 D = 33 3 2 2 33 = 14 3 1 −5 −3 3 3 3 3 2 −1 −1 3 3 3 2 2 33 = 28 Dx = 33 10 3 −2 −5 −3 3 3 3 3 2 2 −1 3 3 3 Dy = 33 3 10 2 33 = −14 3 1 −2 −3 3 3 3 3 2 −1 2 3 3 3 10 33 = 42 Dz = 33 3 2 3 1 −5 −2 3

28 Dx = =2 D 14 −14 Dy = = −1 y= D 14 Dz 42 z= = =3 D 14 The solution is (2, −1, 3).

58. Graph the system of inequalities and find the vertices, (2, 4), (2, 8) and (10, 0). Evaluate T = 6x + 10y at each vertex. (2, 4) : 6(2) + 10(4) = 52 Minimum (2, 8) : 6(2) + 10(8) = 92 Maximum (10, 0) : 6(10) + 10(0) = 60 59. Let x = the number of questions answered from group A and y = the number of questions answered from group B. Find the maximum value of S = 7x + 12y subject to x + y ≥ 8,

8x + 10y ≤ 80, x ≥ 0, y≥0

Graph the system of inequalities and find the vertices, (0, 8), (8, 0) and (10, 0). Evaluate S = 7x + 12y at each vertex. (0, 8) : 7(0) + 12(8) = 96 (8, 0) : 7(8) + 12(0) = 56 (10, 0) : 7(10) + 12(0) = 70 The maximum score of 96 occurs when 0 questions from group A and 8 questions from group B are answered correctly.

x=

55.

60.

5 A B C = + + (x+2)2 (x+1) x+2 (x+2)2 x+1 A(x+2)(x+1)+B(x+1)+C(x+2)2 (x + 2)2 (x + 1) Equate the numerators. =

5 = A(x + 2)(x + 1) + B(x + 1) + C(x + 2)2 Let x = −1 : 5 = 0 + 0 + C(−1 + 2)

2

5=C

Let x = −2 :

5 = 0 + B(−2 + 1) + 0 5 = −B

−5 = B

(1)

Chapter 8 Review Exercises To find A, choose any value for x except −1 and −2 and replace B with −5 and C with 5. We let x = 0.

563 64. Answers will vary. During a holiday season, a caterer sold a total of 55 food trays. She sold 15 more seafood trays than cheese trays. How many of each were sold?

5 = A(0 + 2)(0 + 1) − 5(0 + 1) + 5(0 + 2)2

5 = 2A − 5 + 20

−10 = 2A −5 = A

The decomposition is − 61.

5 5 5 − + . x + 2 (x + 2)2 x+1

−8x + 23 −8x + 23 = 2x2 + 5x − 12 (2x − 3)(x + 4) B A + = 2x − 3 x + 4

A(x + 4) + B(2x − 3) (2x − 3)(x + 4) Equate the numerators.

65. In general, (AB)2 $= A2 B2 . (AB)2 = ABAB and A2 B2 = AABB. Since matrix multiplication is not commutative, BA $= AB, so (AB)2 $= A2 B2 . 66. Let x, y, and z represent the amounts invested at 4%, 5%, 1 and 5 %, respectively. 2 Solve: x + y + z = 40, 000, 0.04x + 0.05y + 0.055z = 1990, 0.055z = 0.04x + 590

=

−8x + 23 = A(x + 4) + B(2x − 3) ! " ! " 3 3 3 Let x = : −8 + 23 = A +4 +0 2 2 2 11 A −12 + 23 = 2 11 11 = A 2 2=A Let x = −4 : −8(−4) + 23 = 0 + B[2(−4) − 3] 32 + 23 = −11B 55 = −11B

−5 = B

The decomposition is 62.

2x + y = 7, (1) x − 2y = 6

5 2 − . 2x − 3 x + 4

(2)

Multiply equation (1) by 2 and add. 4x + 2y = 14 x − 2y = 6 5x = 20 x=4 Back-substitute to find y. 2·4+y = 7

Using equation (1)

8+y = 7 y = −1

The solution is (4, −1), so answer C is correct. 63. Interchanging columns of a matrix is not a row-equivalent operation, so answer A is correct. (See page 705 in the text.)

x = $10, 000, y = $12, 000, z = $18, 000 67.

4 2 + = 8, 3x 5y 5 3 − = −6 4x 2y 1 1 Let a = and b = . Then we have: x y 2 4 a + b = 8, 3 5 3 5 a − b = −6 4 2 Multiply the first equation by 15 and the second by 4 to clear the fractions. 10a + 12b = 120, (1) 5a − 6b = −24

(2)

Multiply equation (2) by 2 and add. 10a + 12b = 120 10a − 12b = −48 20a = 72 18 a= 5 Back-substitute to find b. 18 + 12b = 120 Using equation (1) 10 · 5 36 + 12b = 120 12b = 84 b=7 Now find x and y. 1 1 b= a= x y 1 1 18 = 7= 5 x y 1 5 y= x= 18 7 " ! 5 1 . The solution is , 18 7

564 68.

Chapter 8: Systems of Equations and Matrices 3 4 1 − + = −2, (1) x y z 1 2 5 + − = 1, (2) x y z 7 3 2 + + = 19 (3) x y z Multiply equations (2) and (3) by 3. 3 4 1 − + = −2 (1) x y z 3 6 15 + − =3 (4) x y z 9 6 21 + + = 57 (5) x y z Multiply equation (1) by −5 and add it to equation (2).

Multiply equation (1) by −7 and add it to equation (3). 4 1 3 − + = −2 (1) x y z 11 23 − = 13 (6) y z 1 37 − = 71 (7) y z Multiply equation (7) by 23. 4 1 3 − + = −2 (1) x y z 23 11 − = 13 (6) y z 851 23 − = 1633 (8) y z Multiply equation (6) by −37 and add it to equation (8). 4 1 3 − + = −2 (1) x y z 11 23 − = 13 (6) y z 384 = 1152 (9) z Complete the solution. 384 = 1152 z 1 =z 3 23 11 − y 1/3 23 − 33 y 23 y 1 2

= 13 = 13

3 4 1 − + = −2 x 1/2 1/3 3 − 8 + 3 = −2 x 3 − 5 = −2 x 3 =3 x 1=x " ! 1 1 . The solution is 1, , 2 3 (We could also have solved this system of equations by first 1 1 1 substituting a for , b for , and c for and proceeding x y z as we did in Exercise 61 above.) 69.

70.

Chapter 8 Test 1.

3x + 2y = 1,

(1)

2x − y = −11

(2)

Multiply equation (2) by 2 and add. 3x + 2y = 1

= 46

4x − 2y = −22 7x = −21

=y

Back-substitute to find y.

x = −3

2(−3) − y = −11

Using equation (2).

−6 − y = −11 −y = −5 y=5

The solution is (−3, 5). Since the system of equations has exactly one solution, it is consistent and independent.

Chapter 8 Test 2.

565

2x − y = 3,

(1)

2y = 4x − 6

Multiply equation (1) by −3 and add it to equation (4).

(2)

Multiply equation (1) by −5 and add it to equation (5). 4x +

Solve equation (1) for y. y = 2x − 3

Subsitute in equation (2) and solve for x. 2(2x − 3) = 4x − 6

(1)

3y = 3x − 8

2y +

z=4

2y − 17z = −28

−10y + 17z = 4

(1) (7) (6)

Multiply equation (7) by 5 and add it to equation (6). 4x + 2y + 2y −

z=4

(1)

17z = −28

(7)

−68z = −136

(8)

Solve equation (8) for z. −68z = −136 z=2

2y − 34 = −28 2y = 6

Subsitute in equation (2) and solve for y. 3y = 3(y + 4) − 8 3y = 3y + 12 − 8

0=4 We get a false equation so there is no solution. Since there is no solution the system of equations is inconsistent and independent. 2x − 3y = 8, (1)

(2)

−10x + 4y = −18 − 11y = 22

4x + 8 = 4

4x = −4 x = −1

6. Familiarize. Let x and y represent the number of student and nonstudent tickets sold, respectively. Then the receipts from the student tickets were 3x and the receipts from the nonstudent tickets were 5y. Translate. One equation comes from the fact that 750 tickets were sold. A second equation comes from the fact that the total receipts were $3066. 3x + 5y = 3066

2x + 6 = 8

Carry out. We solve the system of equations. x + y = 750, (1)

2x = 2 x=1 The solution is (1, −2). Since the system of equations has exactly one solution, it is consistent and independent. 4x + 2y + z = 4,

(1)

3x − y + 5z = 4,

(2)

z=4

20x + 12y − 12z = −8

(2)

Multiply equation (1) by −3 and add. 3x + 5y = 3066 2y = 816

(3)

12x − 4y + 20z = 16

3x + 5y = 3066

−3x − 3y = −2250

Multiply equations (2) and (3) by 4. 4x + 2y +

Back-substitute 3 for y and 2 for z in equation (1) and solve for x. 4x + 2 · 3 + 2 = 4

x + y = 750

y = −2 Back-substitute to find x. 2x − 3(−2) = 8

5x + 3y − 3z = −2

y=3

The solution is (−1, 3, 2).

Multiply equation (1) by 5 and equation (2) by −2 and add. 10x − 15y = 40

5.

(7)

2y − 17 · 2 = −28

x=y+4

5x − 2y = 9

(6)

Back-substitute 2 for z in equation (7) and solve for y.

(2)

Solve equation (1) for x.

4.

(1)

2y − 17z = −28

4x +

0=0 The equation 0 = 0 is true for all values of x and y. Thus the system of equations has infinitely many solutions. Solving either equation for y, we get y = 2x − 3, so the solutions are ordered pairs of the form (x, 2x − 3). Equivay+3 , lently, if we solve either equation for x, we get x = " 2 ! y+3 , y . Since so the solutions can also be expressed as 2 there are infinitely many solutions, the system of equations is consistent and dependent. x − y = 4,

2y +

Interchange equations (6) and (7).

4x − 6 = 4x − 6

3.

z=4

−10y + 17z = 4

(1)

y = 408 Substitute 408 for y in equation (1) and solve for x.

(4)

x + 408 = 750

(5)

x = 342

566

Chapter 8: Systems of Equations and Matrices Check. The number of tickets sold was 342 + 408, or 750. The total receipts were $3·342+$5·408 = $1026+$2040 = $3066. The solution checks. State. 342 student tickets and 408 nonstudent tickets were sold.

7. Familiarize. Let x, y, and z represent the number of orders that can be processed per day by Tricia, Maria, and Antonio, respectively.

11. The product AB is not defined because the number of columns of A, 3, is not equal to the number of rows of B, 2. 1 2 1 2 1 −1 3 2 −2 6 12. 2A = 2 = −2 5 2 −4 10 4 1 2 3 −4 13. C = −1 0 Write the augmented matrix. 2 1 3 −4 1 0 −1 0 0 1

Translate. Tricia, Maria, and Antonio can process 352 orders per day. x + y + z = 352

Interchange rows. 2 1 −1 0 0 1 3 −4 1 0

Tricia and Maria together can process 224 orders per day. x + y = 224

Multiply row 1 by 3 and add it to row 2. 2 1 −1 0 0 1 0 −4 1 3

Tricia and Antonio together can process 248 orders per day. x + z = 248

1 Multiply row 1 by −1 and row 2 by − . 4   1 0 0 −1  3  1 − 0 1 − 4 4   0 −1   C−1 =  1 3  − − 4 4

We have a system of equations: x + y + z = 352, x+y

= 224, + z = 248.

x

Carry out. Solving the system of equations, we get (120, 104, 128). Check. Tricia, Maria, and Antonio can process 120+104+ 128, or 352, orders per day. Together, Tricia and Maria can process 120 + 104, or 224, orders per day. Together, Tricia and Antonio can process 120 + 128, or 248, orders per day. The solution checks. State. Tricia can process 120 orders per day, Maria can process 104 orders per day, and Antonio can process 128 orders per day. 1 2 1 2 −5 1 3 −4 8. B + C = + −2 4 −1 0 =

1

−5 + 3 −2 + (−1)

=

1

−2 −3

−3 4

1 + (−4) 4+0

2

2

=

3(−5) + (−4)(−2) 3(1) + (−4)(4) −1(−5) + 0(−2) −1(1) + 0(4)

=

1

−7 5

−13 −1

2

49 14. a) M =  43 51 / b) N = 26 18 / c) NM = 3221

 10 13 12 11  8 12 0 23 0 660 812

d) The entries of NM represent the total cost, in cents, for each type of menu item served on the given day.      3 −4 2 x −8 15.  2 3 1  y  =  7  1 −5 −3 z 3

16.

3x + 2y + 6z = 2, x + y + 2z = 1,

9. A and C do not have the same order, so it is not possible to find A − C. 1 21 2 3 −4 −5 1 10. CB = −1 0 −2 4 1



2

2x + 2y + 5z = 3 Write an  3 2  1 1 2 2 Then,

equivalent matrix equation, AX = B.     6 x 2 2  y  =  1  5 z 3 

1 X = A−1 B =  −1 0

    2 −2 2 −2 3 0  1  =  1  −2 1 3 1

The solution is (−2, 1, 1). 3 3 3 3 17. 3 3 −5 3 = 3 · 7 − 8(−5) = 21 + 40 = 61 38 7 3

Chapter 8 Test

567

18. We will expand across the first row. 3 3 3 2 −1 4 3 3 3 3 −3 1 −2 3 3 3 3 5 3 −1 3 3 3 3 3 3 3 1+1 3 1 −2 3 1+2 3 −3 = 2(−1) 3 3 −1 3 + (−1)(−1) 3 5 3 3 −3 4(−1)1+3 33 5

3 1 33 33

Graph the system of inequalities and find the vertices, (25, 15), (25, 75), and (85, 15). Evaluate P = 3x + 4y at each vertex.

= 2 · 1[1(−1) − 3(−2)] + (−1)(−1)[−3(−1) − 5(−2)]+ 4 · 1[−3(3) − 5(1)]

= 2(5) + 1(13) + 4(−14) 19.

= −33

5x + 2y = −1,

7x + 6y = 1 3 3 35 23 3 3 = 5(6) − 7(2) = 16 D=3 7 63 3 3 −1 Dx = 33 1 3 35 Dy = 33 7

3 2 33 = −1(6) − (1)(2) = −8 63

3 −1 33 = 5(1) − 7(−1) = 12 1 3

−8 1 Dx = =− D 16 2 12 3 Dy y= = = D 16! 4 " 1 3 The solution is − , . 2 4 x=

(25, 15) : P = 3(25) + 4(15) = 135

3 −2 33 + −1 3

20.

(25, 75) : P = 3(25) + 4(75) = 375 (85, 15) : P = 3(85) + 4(15) = 315 The maximum profit of $375 occurs when 25 pound cakes and 75 carrot cakes are prepared. 23.

3x − 11 3x − 11 = x2 + 2x − 3 (x − 1)(x + 3) A B = + x−1 x+3

A(x + 3) + B(x − 1) (x − 1)(x + 3) Equate the numerators. =

3x − 11 = A(x + 3) + B(x − 1)

Let x = −3 : 3(−3) − 11 = 0 + B(−3 − 1) −20 = −4B 5=B

Let x = 1 : 3(1) − 11 = A(1 + 3) + 0 −8 = 4A −2 = A The decomposition is −

5 2 + . x−1 x+3

24. Solve: A(2) − B(−2) = C(2) − 8 A(−3) − B(−1) = C(1) − 8

A(4) − B(2) = C(9) − 8 or 2A + 2B − 2C = −8 −3A + B − C = −8

4A − 2B − 9C = −8

The solution is (1, −3, 2), so A = 1, B = −3, and C = 2. 21. Graph !the system of inequalities and find the vertices, " 5 , (3, 3), and (6, 0). (1, 0), 1, 3 Evaluate Q = 2x + 3y at each vertex. (1, 0) : 2(1) + 3(0) = 2 Minimum " ! " ! 5 5 : 2(1) + 3 =7 1, 3 3 (3, 3) : 2(3) + 3(3) = 15 Maximum (6, 0) : 2(6) + 3(0) = 12 22. Let x = the number of pound cakes prepared and y = the number of carrot cakes. Find the maximum value of P = 3x + 4y subject to x + y ≤ 100, x ≥ 25, y ≥ 15

Chapter 9

Analytic Geometry Topics y

Exercise Set 9.1

4

1. Graph (f) is the graph of x2 = 8y.

y 2 " !6x

2

2. Graph (c) is the graph of y 2 = −10x.

2

!4 !2

3. Graph (b) is the graph of (y − 2)2 = −3(x + 4).

4

x

!2

4. Graph (e) is the graph of (x + 1)2 = 5(y − 2).

!4

5. Graph (d) is the graph of 13x − 8y − 9 = 0. 2

6. Graph (a) is the graph of 41x + 6y 2 = 12. 7.

10.

x2 = 20y x2 = 4 · 5 · y

Writing x2 = 4py

Vertex: (0, 0) Focus: (0, 5)

[(0, p)]

Directrix: y = −5

(y = −p)

y 2 = −2x " ! 1 y2 = 4 − x 2 " ! 1 1 V : (0, 0), F : − , 0 , D : x = 2 2 y 4

y

2

4

x2

" 20y

2

4

y 2 " !2x

x

!2

11.

!4

x2 − 4y = 0

x2 = 4 · 1 · y

x = 16y 2

Writing x2 = 4py

Vertex: (0, 0) Focus:

V : (0, 0), F : (0, 4), D : y = −4

(0, 1)

[(0, p)]

Directrix: y = −1 (y = −p)

y

y 4

4

2

2 2

!4 !2 !2

4

x

2

!4 !2

x 2 " 16y

!4

!2 !4

y 2 = −6x " ! 3 y2 = 4 − x 2

!4

x2 = 4y

x2 = 4 · 4 · y

9.

x

!2

!4 !2

8.

2

!4 !2

2

Writing y 2 = 4px

Vertex: (0, 0) " ! 3 Focus: − ,0 2 " ! 3 3 = Directrix: x = − − 2 2

12.

4

x 2 ! 4y " 0

y 2 + 4x = 0 y 2 = −4x

[(p, 0)] (x = −p)

x

y 2 = 4(−1)x V : (0, 0), F : (−1, 0), D : x = 1

570

Chapter 9: Analytic Geometry Topics y

16.

4 2

y 2 # 4x " 0 2

!4 !2

4

x

!2 !4

13.

x = 2y 2 1 y2 = x 2 1 Writing y 2 = 4px y2 = 4 · · x 8 Vertex: (0, 0) " ! 1 ,0 Focus: 8 1 Directrix: x = − 8

y x " 2y 2 1

!2 !1

14. y =

1 2 x 2

V : (0, 0), F :

!

0,

2

x

" 1 1 , D:y=− 2 2 y 4 2

x !2 !4

y"

qx 2

15. Since the directrix, x = −4, is a vertical line, the equation is of the form (y − k)2 = 4p(x − h). The focus, (4, 0), is on the x-axis so the line of symmetry is the x-axis and p = 4. The vertex, (h, k), is the point on the x-axis midway between the directrix and the focus. Thus, it is (0, 0). We have (y − k)2 = 4p(x − h) (y − 0)2 = 4 · 4(x − 0) y = 16x. 2

Substituting

(x − h)2 = 4p(y − k) 1 (x − 0)2 = 4 · (y − 0) 4 x2 = y

17. Since the directrix, y = π, is a horizontal line, the equation is of the form (x−h)2 = 4p(y−k). The focus, (0, −π), is on the y-axis so the line of symmetry is the y-axis and p = −π. The vertex (h, k) is the point on the y-axis midway between the directrix and the focus. Thus, it is (0, 0). We have (x − h)2 = 4p(y − k)

(x − 0)2 = 4(−π)(y − 0)

18.

Substituting

x2 = −4πy

(y − k)2 = 4p(x − h) √ (y − 0)2 = 4(− 2)(x − 0) √ y 2 = −4 2x

19. Since the directrix, x = −4, is a vertical line, the equation is of the form (y − k)2 = 4p(x − h). The focus, (3, 2), is on the horizontal line y = 2, so the line of symmetry is y = 2. The vertex is the point on the line y = 2 that is midway between the directrix and the focus. That is, it " from (−4, 2) to (3, 2): " of the ! segment ! is the midpoint 1 1 −4 + 3 2 + 2 , , or − , 2 . Then h = − and the 2 2 2 2 directrix is x = h − p, so we have x = h−p 1 −4 = − − p 2 7 − = −p 2 7 = p. 2 Now we find the equation of the parabola. (y − k)2 = 4p(x − h) ! "$ ! "# 7 1 2 (y − 2) = 4 x− − 2 2 " ! 1 (y − 2)2 = 14 x + 2 20. Since the directrix, y = −3, is a horizontal line, the equation is of the form (x − h)2 = 4p(y − k). The focus (−2, 3), is on the vertical line x = −2, so the line of symmetry is x = −2. The vertex is the point on the line x = −2 that is midway between the directrix and the focus. That is, it is ! the midpoint "of the segment from (−2, 3) to (−2, −3): −2 − 2 3 − 3 , , or (−2, 0). Then k = 0 and the directrix 3 2 is y = k − p, so we have y = k−p −3 = 0 − p

3=p Now we find the equation of the parabola. (x − h)2 = 4p(y − k) [x − (−2)]2 = 4 · 3(y − 0) (x + 2)2 = 12y

Exercise Set 9.1 21.

571

y

(x + 2)2 =−6(y − 1) " ! 3 (y−1) [x−(−2)]2 =4 − 2

[(x−h)2 = 4p(y−k)]

2

Vertex: (−2, 1) [(h, k)] ! ! "" ! " 1 3 , or − 2, − Focus: − 2, 1 + − 2 2 !

3 2

−

"

=

5 2

x

!4 !6

(y = k − p)

x 2 # 2x # 2y # 7 " 0

y 24.

4

y 2 + 6y − x + 16 = 0

y 2 + 6y + 9 = x − 16 + 9

2 2

!4

4

(y + 3)2 = x − 7 ! " 1 [y − (−3)]2 = 4 (x − 7) 4

x

!2 !4

V : (7, −3) " ! " ! 29 1 , −3 F : 7 + , −3 , or 4 4 27 1 D : x=7− = 4 4

(x # 2)2 " !6(y ! 1) 22.

4

!2

[(h, k + p)] Directrix: y = 1 −

2

!4 !2

(y − 3)2 = −20(x + 2)

(y − 3)2 = 4(−5)[x − (−2)]

y

V : (−2, 3)

y 2 # 6y ! x # 16 " 0

2

F : (−2 − 5, 3), or (−7, 3)

4

D : x = −2 − (−5) = 3

8

16

20 x

!2 !4

y 12 8 4

25.

4 x

!16 !12 !8 !4

x2 = y + 2 1 (x − 0)2 = 4 · · [y − (−2)] 4 [(x − h)2 = 4p(y − k)]

!4 !8

(y !

23.

3)2

" !20(x # 2)

x2 + 2x + 2y + 7 = 0 x + 2x = −2y − 7 2

(x + 2x + 1) = −2y − 7 + 1 = −2y − 6 2

x2 − y − 2 = 0

(x + 1)2 = −2(y + 3) " ! 1 [y − (−3)] [x − (−1)]2 = 4 − 2 [(x − h)2 = 4p(y − k)]

Vertex: (−1, −3) [(h, k)] "" ! " ! ! 1 7 Focus: − 1, −3+ − , or − 1, − 2 2

Vertex: (0, −2) [(h, k)] " ! " ! 7 1 , or 0, − [(h, k + p)] Focus: 0, −2 + 4 4 9 1 (y = k − p) Directrix: y = −2 − = − 4 4

y 4 2 2

!4 !2

4

x

[(h, k+p)] Directrix: y = −3 −

!

−

1 2

"

=−

5 2

(y = k − p)

!4

x2 ! y ! 2 " 0

572 26.

Chapter 9: Analytic Geometry Topics V : (−3, 1) ! " ! " 1 5 F : − 3, 1 + , or − 3, 4 4 3 1 D : y =1− = 4 4

x2 − 4x − 2y = 0

x − 4x + 4 = 2y + 4 2

(x − 2)2 = 2(y + 2) ! " 1 [y − (−2)] (x − 2)2 = 4 2

V : (2, −2) " ! " ! 1 3 F : 2, −2 + , or 2, − 2 2 5 1 D : y = −2 − = − 2 2

y 6 4 2

y " x 2 # 6x # 10

2 2

!4 !2

4

29.

x

!2

y2 − y − x + 6 = 0 y2 − y 1 y2 − y + 4 "2 ! 1 y− 2 "2 ! 1 y− 2

!4

x 2 ! 4x ! 2y " 0

= x−6

= x−6+

y − 3 = x2 + 4x

y − 3 + 4 = x2 + 4x + 4 y + 1 = (x + 2)2

1 · [y − (−1)] = [x − (−2)]2 4 [(x − h)2 = 4p(y − k)]

Vertex: (−2, −1) [(h, k)] ! " ! " 3 1 Focus: − 2, −1 + , or − 2, − [(h, k + p)] 4 4 5 1 (y = k−p) Directrix: y = −1− = − 4 4

= x−

y 4

y

2

4

2

2

4

8

6

x

!2 2

!4

1 4

23 4 ! " 1 23 = 4· x− 4 4 [(y − k)2 = 4p(x − h)] " ! 23 1 , [(h, k)] Vertex: 4 2 ! " ! " 23 1 1 1 Focus: + , , or 6, [(h + p, k)] 4 4 2 2 22 11 23 1 − = or (x = h − p) Directrix: x = 4 4 4 2

y = x2 + 4x + 3

4·

x

!2

4

27.

2

!4 !4 !2

y

4

!4

x

!2

y2 ! y ! x # 6 " 0

!4

y " x 2 # 4x # 3 28.

y = x2 + 6x + 10 y − 10 + 9 = x2 + 6x + 9

y − 1 = (x + 3)2 ! " 1 4 (y − 1) = [x − (−3)]2 4

30.

y2 + y − x − 4 = 0

y2 + y 1 y2 + y + 4 "2 ! 1 y+ 2 # ! "$2 1 y− − 2

= x+4 = x+4+

1 4

17 4 # ! "$ 1 17 = 4· x− − 4 4 = x+

Exercise Set 9.1 V :

!

−

573

17 1 ,− 4 2

9 Then the equation of the parabola is y 2 = 4 · x, or 4 y 2 = 9x. ! " 9 b) The focus is at (p, 0), or , 0 , so the bulb should 4 9 1 be placed in., or 2 in., from the vertex. 4 4

"

!

" ! " 17 1 1 1 F : − + , − , or − 4, − 4 4 2 2 17 1 9 D: x=− − =− 4 4 2

33. We position a coordinate system with the origin at the vertex and the x-axis on the parabola’s axis of symmetry.

y 4

y

2 2

!2

4

(1.5, 2)

x

!2 !4

4 ft

x

1.5 ft

y2 # y ! x ! 4 " 0

31. a) The vertex is (0, 0). The focus is (4, 0), so p = 4. The parabola has a horizontal axis of symmetry so the equation is of the form y 2 = 4px. We have

The parabola the form y 2 = 4px and a point on the ! is of " 4 parabola is 1.5, , or (1.5, 2). 2

y 2 = 4px y2 = 4 · 4 · x

y 2 = 4px

y 2 = 16x

22 = 4 · p · (1.5)

b) We make a drawing.

4 = 6p 4 = p, or 6 2 =p 3

y 15 ( x, — ) 2

x

15 ft

Substituting

x

Since the focus is at (p, 0), or 8 in., from the vertex.

15 ( x, – — ) 2

34.

Substituting

y

(4, 0)

16x 16x

" 2 2 , 0 , the focus is ft, or 3 3

(18, y)

The depth of"the satellite dish at the vertex is x ! 15 is a point on the parabola. where x, 2 y2 = ! "2 15 = 2 225 = 4 225 = 64 33 3 = 64

!

(18, 0) x

15 for y 2

16x y 2 = 4px

x, or

y 2 = 4 · 4 · 18 y 2 = 288 √ y = 12 2

x

The depth of the satellite dish at the vertex is 3

33 ft. 64

2 32. a) The parabola is!of the " form y = 4px. A point on 6 , or (1, 3). the parabola is 1, 2

y 2 = 4px 32 = 4 · p · 1 9 =p 4

√ √ The width at the opening is 2 · 12 2, or 24 2 in. ≈ 34 in. 35.

Vertex: (0.867, 0.348)

Focus: (0.867, −0.190) Directrix: y = 0.887

36.

Vertex: (7.126, 1.180) Focus: (7.045, 1.180) Directrix: x = 7.207

574

Chapter 9: Analytic Geometry Topics

37. No; parabolas with a horizontal axis of symmetry fail the vertical-line test. 38. See page 769 of the text. 39. When we let y = 0 and solve for x, the only equation for 2 which x = is (h), so only equation (h) has x-intercept 3 ! " 2 ,0 . 3 40. Equations (a) - (f) are in the form y = mx + b and b = 7 in equation (d). When we solve equations (g) and (h) 1 1 7 for y we get y = 2x − and y = − x + , respectively. 4 2 3 Neither has b = 7, so only equation (d) has y-intercept (0, 7). 7 41. Note that equation (g) is equivalent to y = 2x − and 4 1 1 equation (h) is equivalent to y = − x + . When we look 2 3 at the equations in the form y = mx+b, we see that m > 0 for (a), (b), (f), and (g) so these equations have positive slope, or slant up front left to right. 42. The equation for which |m| is smallest is (b), so it has the least steep slant. 43. When we look at the equations in the form y = mx+b (See 1 1 Exercise 41.), only (b) has m = so only (b) has slope . 3 3 44. When we substitute 3 for x in each equation, we see that y = 7 only in equation (f) so only (f) contains the point (3, 7). 45. Parallel lines have the same slope and different yintercepts. When we look at the equations in the form y = mx + b (See Exercise 41.), we see that (a) and (g) represent parallel lines.

The equation of the parabola is (y − 1)2 = 4(−4)[x − (−2)], or

(y − 1)2 = −16(x + 2).

49. Position a coordinate system as shown below with the yaxis on the parabola’s axis of symmetry. y (–100, 50) (0, 10) x 20 40 60 80 100

The equation of the parabola is of the form (x − h)2 = 4p(y − k). Substitute 100 for x, 50 for y, 0 for h, and 10 for k and solve for p. (x − h)2 = 4p(y − k)

(100 − 0)2 = 4p(50 − 10)

10, 000 = 160p 250 =p 4 Then the equation is ! " 250 2 (y − 10), or x =4 4 2 x = 250(y − 10).

To find the lengths of the vertical cables, find y when x = 0, 20, 40, 60, 80, and 100. When x = 0 : 02 = 250(y − 10) 0 = y − 10

10 = y

When x = 20 : 202 = 250(y − 10) 400 = 250(y − 10)

46. The pairs of equations for which the product of the slopes is −1 are (a) and (h), (g) and (h), and (b) and (c). 47. A parabola with a vertical axis of symmetry has an equation of the type (x − h)2 = 4p(y − k).

1.6 = y − 10

11.6 = y When x = 40 :

4 = −4p

6.4 = y − 10

16.4 = y When x = 60 :

−1 = p The equation of the parabola is [x − (−1)]2 = 4(−1)(y − 2), or (x + 1)2 = −4(y − 2).

48. A parabola with a horizontal axis of symmetry has an equation of the type (y − k)2 = 4p(x − h). Find p by substituting (−2, 1) for (h, k) and (−3, 5) for (x, y). (5 − 1)2 = 4p[−3 − (−2)] 16 = 4p(−1)

16 = −4p

−4 = p

402 = 250(y − 10)

1600 = 250(y − 10)

Solve for p substituting (−1, 2) for (h, k) and (−3, 1) for (x, y). [−3 − (−1)]2 = 4p(1 − 2)

(100, 50)

602 = 250(y − 10)

3600 = 250(y − 10) 14.4 = y − 10 24.4 = y When x = 80 :

802 = 250(y − 10)

6400 = 250(y − 10) 25.6 = y − 10 35.6 = y

When x = 100, we know from the given information that y = 50. The lengths of the vertical cables are 10 ft, 11.6 ft, 16.4 ft, 24.4 ft, 35.6 ft, and 50 ft.

Exercise Set 9.2

575 9. Complete the square twice.

Exercise Set 9.2

x2 + y 2 + 6x − 2y = 6

x2 + 6x + y 2 − 2y = 6

1. Graph (b) is the graph of x + y = 5. 2

2

x2 + 6x + 9 + y 2 − 2y + 1 = 6 + 9 + 1 (x + 3)2 + (y − 1)2 = 16

2. Graph (f) is the graph of y = 20 − x . 2

2

[x − (−3)]2 + (y − 1)2 = 42

3. Graph (d) is the graph of x2 + y 2 − 6x + 2y = 6.

Center: (−3, 1)

4. Graph (c) is the graph of x2 + y 2 + 10x − 12y = 3.

Radius: 4

5. Graph (a) is the graph of x2 + y 2 − 5x + 3y = 0.

y

6. Graph (e) is the graph of x2 + 4x − 2 = 6y − y 2 − 6.

4

7. Complete the square twice.

2

x2 + y 2 − 14x + 4y = 11

!6

x2 − 14x + y 2 + 4y = 11

!4

!2

x2 − 14x + 49 + y 2 + 4y + 4 = 11 + 49 + 4 2

x 2 # y 2 # 6x ! 2y " 6

(x − 7)2 + [y − (−2)]2 = 82

Center: (7, −2)

10.

Radius: 8

x2 + y 2 − 4x + 2y = 4

x − 4x + 4 + y 2 + 2y + 1 = 4 + 4 + 1 2

y

(x − 2)2 + (y + 1)2 = 9

(x − 2)2 + [y − (−1)]2 = 32

8

Center: (2, −1)

4

Radius: 3 8

4

!8 !4

12

x

y 4

!8

2 !4

x 2 # y 2 ! 14x # 4y " 11 8.

x

!4

(x − 7) + (y + 2) = 64 2

2 !2

x2 + y 2 + 2x − 6y = −6

x2 + 2x + 1 + y 2 − 6y + 9 = −6 + 1 + 9 (x + 1) + (y − 3) = 4 2

!2

2

4

x

!2 !4

x 2 # y 2 ! 4x # 2y " 4

2

[x − (−1)]2 + (y − 3)2 = 22

Center: (−1, 3)

11. Complete the square twice. x2 + y 2 + 4x − 6y − 12 = 0

x2 + 4x + y 2 − 6y = 12

Radius: 2

x2 + 4x + 4 + y 2 − 6y + 9 = 12 + 4 + 9 (x + 2)2 + (y − 3)2 = 25

y

[x − (−2)]2 + (y − 3)2 = 52

4

Center: (−2, 3)

2 2

!4 !2

4

!2 !4

x 2 # y 2 # 2x ! 6y " !6

x

Radius: 5

576

Chapter 9: Analytic Geometry Topics 14.

y

x2 + y 2 − 2x + 6y + 1 = 0

x2 − 2x + y 2 + 6y = −1

x − 2x + 1 + y 2 + 6y + 9 = −1 + 1 + 9 2

4

(x − 1)2 + (y + 3)2 = 9

4

(x − 1)2 + [y − (−3)]2 = 32

2

Center: (1, −3) 2

!4 !2

4

Radius: 3

x

y

!4

2 !4

!2

x 2 # y 2 # 4x ! 6y ! 12 " 0

2

4

x

!2 !4

12.

x2 + y 2 − 8x − 2y − 19 = 0

!6

x2 − 8x + y 2 − 2y = 19

x 2 # y 2 ! 2x # 6y # 1 " 0

x2 − 8x + 16 + y 2 − 2y + 1 = 19 + 16 + 1 (x − 4)2 + (y − 1)2 = 36

(x − 4)2 + (y − 1)2 = 62

15. Complete the square twice. x2 + y 2 + 6x − 10y = 0

Center: (4, 1)

x2 + 6x + y 2 − 10y = 0

Radius: 6

x2 + 6x + 9 + y 2 − 10y + 25 = 0 + 9 + 25 (x + 3)2 + (y − 5)2 = 34 √ [x − (−3)]2 + (y − 5)2 = ( 34)2

y 8

4

!8 !4

8

Center: (−3, 5) √ Radius: 34 x

y

!8

8 4

x 2 # y 2 ! 8x ! 2y ! 19 " 0

2

x2 − 6x + y 2 − 8y = −16

x

!8

x − 6x + 9 + y − 8y + 16 = −16 + 9 + 16 2

8

!4

x + y − 6x − 8y + 16 = 0 2

4

!8

13. Complete the square twice.

2

(x − 3)2 + (y − 4)2 = 9

(x − 3) + (y − 4) = 3 2

2

16.

Center: (3, 4) Radius: 3

y 6 4 2 !2

2

4

6

x 2 # y 2 # 6x ! 10y " 0

2

x

!2

x 2 # y 2 ! 6x ! 8y # 16 " 0

x2 + y 2 − 7x − 2y 49 + y 2 − 2y + 1 x2 − 7x + 4 ! "2 7 x− + (y − 1)2 2 "2 ! 7 + (y − 1)2 x− 2 " ! 7 Center: ,1 2 √ 53 Radius: 2

=0 49 +1 = 4 = =

53 4 !√

53 2

"2

Exercise Set 9.2

577

y

19. Graph (c) is the graph of 16x2 + 4y 2 = 64.

4

20. Graph (a) is the graph of 4x2 + 5y 2 = 20.

2 2

!2

4

6

21. Graph (d) is the graph of x2 + 9y 2 − 6x + 90y = −225.

8 x

22. Graph (b) is the graph of 9x2 + 4y 2 + 18x − 16y = 11.

!2 !4

23.

x 2 # y 2 ! 7x ! 2y " 0

y2 x2 + 2 =1 2 2 1 a = 2, b = 1

17. Complete the square twice. x2 + y 2 − 9x = 7 − 4y

x2 − 9x + y 2 + 4y 81 + y 2 + 4y + 4 x2 − 9x + 4 "2 ! 9 + (y + 2)2 x− 2 ! "2 9 + [y − (−2)]2 x− 2 ! " 9 Center: , −2 2 √ 5 5 Radius: 2

Standard form

The major axis is horizontal, so the vertices are (−2, 0) 2 2 2 , we have and (2, 0). Since we know √ that c = a − b √ 2 c√= 4 − 1 = 3, so c = 3 and the foci are (− 3, 0) and ( 3, 0).

=7

= 7+

x2 y2 + =1 4 1

81 +4 4

125 4 ! √ "2 5 5 = 2

To graph the ellipse, plot the vertices. Note also that since b = 1, the y-intercepts are (0, −1) and (0, 1). Plot these points as well and connect the four plotted points with a smooth curve.

=

y 4

y

2

8

4

!4

4 4

!8 !4

!4

8

x x2 y2 # "1 4 1

!8

x 2 # y 2 ! 9x " 7 ! 4y 18.

x

!2

y 2 − 6y − 1 = 8x − x2 + 3

x2 − 8x + y 2 − 6y = 4

x2 − 8x + 16 + y 2 − 6y + 9 = 4 + 16 + 9 (x − 4)2 + (y − 3)2 = 29 √ (x − 4)2 + (y − 3)2 = ( 29)2

24.

x2 y2 y2 x2 + = 1, or 2 + 2 = 1 25 36 5 6 a = 6, b = 5 The major axis is vertical, so the vertices are (0, −6) and 2 2 2 have c2 = 36 √ − 25 = 11, so (0, 6). √ Since c = a − b , we √ c = 11 and the foci are (0, − 11) and (0, 11). y

Center: (4, 3) √ Radius: 29

4 2

!2

8

!4

4

!8 !4

2

!4 !2

y

x

!4 !8

y 2 ! 6y ! 1 " 8x ! x 2 # 3

x2 y2 # "1 25 36

4

x

578 25.

Chapter 9: Analytic Geometry Topics To graph that since √ the ellipse, plot the vertices. √ Note also√ b = 2, the y-intercepts are (0, − 2) and (0, 2). Plot these points as well and connect the four plotted points with a smooth curve.

16x2 + 9y 2 = 144 2

2

y x + =1 9 16

Dividing by 144

y2 x2 + =1 32 42 a = 4, b = 3

Standard form

y 4

The major axis is vertical, so the vertices are (0, −4) and 2 2 2 2 (0, 4). √ Since c = a − b , we √ have c =√16 − 9 = 7, so c = 7 and the foci are (0, − 7) and (0, 7).

2

!4

28. 2

4

5x2 + 7y 2 = 35 y2 x2 + =1 7 5

x

x2 y2 √ + √ =1 ( 7)2 ( 5)2 √ √ a = 7, b = 5

!2

√ The major axis is horizontal, so the vertices are (− 7, 0) √ 7, 0). Since c2 = a2 − √ b2 , we have √ c2 = 7 − 5 = 2, and ( √ so c = 2 and the foci are (− 2, 0) and ( 2, 0).

16x 2 # 9y 2 " 144 9x2 + 4y 2 = 36

y

y2 x2 + =1 4 9

4

x2 y2 + 2 =1 2 2 3 a = 3, b = 2

y

4

!4

x

!4

9x 2 # 4y 2 " 36

2x2 + 3y 2 = 6 2

x y + =1 3 2 y2 x2 √ + √ =1 2 ( 3) ( 2)2 √ √ a = 3, b = 2

√ The major axis is horizontal, so the vertices are (− 3, 0) √ and ( 3, 0). Since c2 = a2 − b2 , we have c2 = 3 − 2 = 1, so c = 1 and the foci are (−1, 0) and (1, 0).

4

x

!4

5x 2 # 7y 2 " 35

29.

4

2

2

!4 !2

The major axis is vertical, so the vertices are (0, −3) and 2 2 2 2 (0, 3). √ have c = √ 9 − 4 = 5, so √ Since c = a − b , we c = 5 and the foci are (0, − 5) and (0, 5).

27.

x

2x 2 # 3y 2 " 6

2

26.

4

!2

y

!4 !2

2

!4 !2

To graph the ellipse, plot the vertices. Note also that since b = 3, the x-intercepts are (−3, 0) and (3, 0). Plot these points as well and connect the four plotted points with a smooth curve.

4x2 + 9y 2 = 1 x2 y2 + =1 1 1 4 9 y2 x2 ! " 2 + ! "2 = 1 1 1 2 3 1 1 a= ,b= 2 3

" ! 1 The major axis is horizontal, so the vertices are − , 0 2 " ! 1 1 5 1 , 0 . Since c2 = a2 −b2 , we have c2 = − = , and 2√ 4 9 36 ! √ " !√ " 5 5 5 so c = and the foci are − , 0 and ,0 . 6 6 6 To graph the ellipse, plot the ! vertices. also that " Note ! " since 1 1 1 and 0, . Plot b = , the y-intercepts are 0, − 3 3 3

Exercise Set 9.2

579

these points as well and connect the four plotted points with a smooth curve.

y

The equation is

33. The vertices, (0, −8) and (0, 8), are on the y-axis, so the major axis is vertical and a = 8. Since the vertices are equidistant from the origin, the center of the ellipse is at the origin. The length of the minor axis is 10, so b = 10/2, or 5.

1

1

!1

Write the equation:

x

y2 x2 + =1 b2 a2

!1

y2 x2 + 2 =1 2 5 8

4x 2 # 9y 2 " 1 30.

x2 y2 + = 1. 20 36

y2 x2 + =1 25 64

25x2 + 16y 2 = 1 x2 y2 + =1 1 1 25 16 y2 x2 ! "2 + ! "2 = 1 1 1 5 4 1 1 a= ,b= 4 5 The major so the vertices are " axis !is vertical, " ! 1 1 and 0, . Since c2 = a2 − b2 , we have 0, − 4 4 ! " 1 9 3 3 1 2 − = , so c = and the foci are 0, − c = 400 20 20 !16 25 " 3 and 0, . 20

34. The vertices, (−5, 0) and (5, 0) are on the x-axis, so the major axis is horizontal and a = 5. Since the vertices are equidistant from the origin, the center of the ellipse is at the origin. The length of the minor axis is 6, so b = 6/2, y2 x2 + = 1. or 3. The equation is 25 9 35. The foci, (−2, 0) and (2, 0) are on the x-axis, so the major axis is horizontal and c = 2. Since the foci are equidistant from the origin, the center of the ellipse is at the origin. The length of the major axis is 6, so a = 6/2, or 3. Now we find b2 : c2 = a2 − b2 22 = 32 − b2 4 = 9 − b2

b2 = 5

Write the equation:

y

x2 y2 + 2 =1 2 a b !0.25

0.25

x2 y2 + =1 9 5

x

25x 2 # 16y 2 " 1

36. The foci, (0, −3) and (0, 3), are on the y-axis, so the major axis is vertical. The foci are equidistant from the origin, so the center of the ellipse is at the origin. The length of the major axis is 10, so a = 10/2, or 5. Find b2 : c2 = a2 − b2

31. The vertices are on the x-axis, so the major axis is horizontal. We have a = 7 and c = 3, so we can find b2 : c2 = a2 − b2

9 = 25 − b2

b2 = 16

32 = 72 − b2

b = 49 − 9 = 40 Write the equation: y2 x2 + 2 =1 a2 b

The equation is

2

2

2

x y + =1 49 40 32. The major axis is vertical; a = 6 and c = 4. c2 = a2 − b2 16 = 36 − b2 b2 = 20

37.

x2 y2 + = 1. 16 25

(y − 2)2 (x − 1)2 + =1 9 4 (y − 2)2 (x − 1)2 + =1 2 3 22

Standard form

The center is (1, 2). Note that a = 3 and b = 2. The major axis is horizontal so the vertices are 3 units left and right of the center: (1 − 3, 2) and (1 + 3, 2), or (−2, 2) and (4, 2).

580

Chapter 9: Analytic Geometry Topics √ We know that c2 =√a2 − b2 , so c2 = 9 − 4 = 5 and c = 5. Then the foci are 5 units left and right of the center: √ √ (1 − 5, 2) and (1 + 5, 2).

We know that c2 = a2 − b2 , √ so c2 = 36 − 25 = 11 and √ c = 11. Then the foci are 11 units below and above the vertex: √ √ (−3, 5 − 11) and (−3, 5 + 11).

To graph the ellipse, plot the vertices. Since b = 2, two other points on the graph are 2 units below and above the center:

To graph the ellipse, plot the vertices. Since b = 5, two other points on the graph are 5 units left and right of the center:

(1, 2 − 2) and (1, 2 + 2) or (1, 0) and (1, 4)

(−3 − 5, 5) and (−3 + 5, 5), or (−8, 5) and (2, 5)

Plot these points also and connect the four plotted points with a smooth curve.

Plot these points also and connect the four plotted points with a smooth curve.

y

y 8

2

4 2

!4 !2

4

x

4

!8

!2

8

x

!4

!4

!8

(x ! 1)2 (y ! 2)2 # "1 9 4 38.

(x − 1)2 (y − 2)2 + =1 1 4

(x # 3)2 (y ! 5)2 # "1 25 36 40.

(x − 1)2 (y − 2)2 + =1 2 1 22

Standard form

(y + 3)2 (x − 2)2 + =1 16 25 [y − (−3)]2 (x − 2)2 + =1 2 4 52

The center is (1, 2). Note that a = 2 and b = 1. The major axis is vertical so the vertices are 2 units below and above the center:

The center is (2, −3). Note that a = 5 and b = 4. The major axis is vertical so the vertices are 5 units below and above the center:

(1, 2 − 2) and (1, 2 + 2), or (1, 0) and (1, 4).

√ We know that c2 =√a2 − b2 , so c2 = 4 − 1 = 3 and c = 3. Then the foci are 3 units below and above the center: √ √ (1, 2 − 3) and (1, 2 + 3).

(2, −3 − 5) and (2, −3 + 5), or (2, −8) and (2, 2).

We know that c2 = a2 − b2 , so c2 = 25 − 16 = 9 and c = 3. Then the foci are 3 units below and above the center: (2, −3 − 3) and (2, −3 + 3), or (2, −6) and (2, 0).

Since b = 1, two points on the graph other than the vertices are (1 − 1, 2) and (1 + 1, 2) or (0, 2) and (2, 2).

Since b = 4, two points on the graph other than the vertices are (2 − 4, −3) and (2 + 4, −3), or (−2, −3) and (6, −3).

y

y

4

2

3

(x ! 1)2 (y ! 2)2 # "1 1 4

2

39.

2

!2

4

6

x

!2

1 !1

Standard form

!4

1

2

3

x

(y − 5)2 (x + 3)2 + =1 25 36 (y − 5)2 [x − (−3)]2 + =1 2 5 62

(x ! 2)2 (y # 3)2 # "1 16 25

Standard form

The center is (−3, 5). Note that a = 6 and b = 5. The major axis is vertical so the vertices are 6 units below and above the center: (−3, 5 − 6) and (−3, 5 + 6), or (−3, −1) and (−3, 11).

41.

3(x + 2)2 + 4(y − 1)2 = 192 (x + 2)2 (y − 1)2 + =1 64 48

(y − 1)2 [x − (−2)]2 + √ =1 2 8 ( 48)2

Dividing by 192 Standard form

Exercise Set 9.2

581

√ The √ center is (−2, 1). Note that a = 8 and b = 48, or 4 3. The major axis is horizontal so the vertices are 8 units left and right of the center:

43. Begin by completing the square twice. 4x2 + 9y 2 − 16x + 18y − 11 = 0

4x2 − 16x + 9y 2 + 18y = 11

(−2 − 8, 1) and (−2 + 8, 1), or (−10, 1) and (6, 1).

We know that c2 = a2 −b2 , so c2 = 64−48 = 16 and c = 4. Then the foci are 4 units left and right of the center: (−2 − 4, 1) and (−2 + 4, 1) or (−6, 1) and (2, 1).

√ To graph the ellipse, plot the vertices. Since b = 4 3 ≈ 6.928, two other points on the graph are about 6.928 units below and above the center: (−2, 1 − 6.928) and (−2, 1 + 6.928), or (−2, −5.928) and (−2, 7.928).

Plot these points also and connect the four plotted points with a smooth curve.

4(x2 − 4x) + 9(y 2 + 2y) = 11

4(x − 4x + 4) + 9(y 2 + 2y + 1) = 11 + 4 · 4 + 9 · 1 2

4(x − 2)2 + 9(y + 1)2 = 36 (y + 1)2 (x − 2)2 + =1 9 4

[y − (−1)]2 (x − 2)2 + =1 2 3 22 The center is (2, −1). Note that a = 3 and b = 2. The major axis is horizontal so the vertices are 3 units left and right of the center: (2 − 3, −1) and (2 + 3, −1), or (−1, −1) and (5, −1).

√ We know that c2 =√a2 − b2 , so c2 = 9 − 4 = 5 and c = 5. Then the foci are 5 units left and right of the center: √ √ (2 − 5, −1) and (2 + 5, −1).

y

To graph the ellipse, plot the vertices. Since b = 2, two other points on the graph are 2 units below and above the center:

4 4

!8 !4

8

x

(2, −1 − 2) and (2, −1 + 2), or (2, −3) and (2, 1).

!4

Plot these points also and connect the four plotted points with a smooth curve.

!8

y

3(x # 2)2 # 4(y ! 1)2 " 192 42.

4

4(x − 5)2 + 3(y − 4)2 = 48

2

(y − 4)2 (x − 5)2 + =1 12 16

2

!4 !2

(x − 5)2 (y − 4)2 √ + =1 2 42 ( 12)

4

x

Standard form √

The √ center is (5, 4). Note that a = 16 and b = 12, or 2 3. The major axis is vertical so the vertices are 4 units below and above the center: (5, 4 − 4) and (5, 4 + 4), or (5, 0) and (5, 8).

We know that c2 = a2 − b2 , so c2 = 16 − 12 = 4 and c = 2. Then the foci are 2 units below and above the center: (5, 4 − 2) and (5, 4 + 2), or (5, 2) and (5, 6). √ Since b = 2 3 ≈ 3.464, two points on the graph other than the vertices are about (5 − 3.464, 4) and (5 + 3.464, 4), or (1.536, 4) and (8.464, 4). y

!4

4x 2 # 9y 2 ! 16x # 18y ! 11 " 0 44. Begin by completing the square twice. x2 + 2y 2 − 10x + 8y + 29 = 0

x2 − 10x + 2(y 2 + 4y) = −29

x2 − 10x + 25 + 2(y 2 + 4y + 4) = −29 + 25 + 2 · 4 (x − 5)2 + 2(y + 2)2 = 4

(y + 2)2 (x − 5)2 + =1 4 2

(x − 5)2 [y − (−2)]2 √ + =1 2 2 ( 2)2

8

√ The center is (5, −2). Note that a = 2 and b = 2. The major axis is horizontal so the vertices are 2 units left and right of the center:

6 4 2 2

4

6

8

x

4(x ! 5)2 # 3(y ! 4)2 " 48

(5 − 2, −2) and (5 + 2, −2), or (3, −2) and (7, −2).

√ We know that c2 =√a2 − b2 , so c2 = 4 − 2 = 2 and c = 2. Then the foci are 2 units left and right of the center:

582

Chapter 9: Analytic Geometry Topics √ 2, −2) and (5 + 2, −2). √ Since b = 2 ≈ 1.414, two points on the graph other than the vertices are about (5 −

√

(5, −2 − 1.414) and (5, −2 + 1.414), or (5, −3.414) and (5, −0.586). 4

9(x2 + 6x) + 4(y 2 − 2y) = −49

9(x2 + 6x + 9) + 4(y 2 − 2y + 1) = −49+9·9+4·1 9(x + 3)2 + 4(y − 1)2 = 36

(y − 1)2 [x − (−3)]2 + =1 2 2 32 The center is (−3, 1). Note that a = 3 and b = 2. The major axis is vertical so the vertices are 3 units below and above the center:

2 2

x

!2 !4

(−3, 1 − 3) and (−3, 1 + 3), or (−3, −2) and (−3, 4).

x 2 # 2y 2 ! 10x # 8y # 29 " 0

45. Begin by completing the square twice. 4x2 + y 2 − 8x − 2y + 1 = 0

√ We know that c2 =√a2 − b2 , so c2 = 9 − 4 = 5 and c = 5. Then the foci are 5 units below and above the center: √ √ (−3, 1 − 5) and (−3, 1 + 5).

Since b = 2, two points on the graph other than the vertices are

4x2 − 8x + y 2 − 2y = −1

4(x2 − 2x) + y 2 − 2y = −1

(−3 − 2, 1) and (−3 + 2, 1), or (−5, 1) and (−1, 1).

4(x − 1)2 + (y − 1)2 = 4

y

4(x − 2x + 1) + y 2 − 2y + 1 = −1 + 4 · 1 + 1 2

9x2 + 4y 2 + 54x − 8y + 49 = 0

(y − 1)2 (x − 3)2 + =1 4 9

y

!2

46. Begin by completing the square twice.

(y − 1)2 (x − 1)2 + =1 1 4

4 2

(y − 1) (x − 1) + =1 12 22 The center is (1, 1). Note that a = 2 and b = 1. The major axis is vertical so the vertices are 2 units below and above the center: 2

2

(1, 1 − 2) and (1, 1 + 2), or (1, −1) and (1, 3).

√ We know that c2 =√a2 − b2 , so c2 = 4 − 1 = 3 and c = 3. Then the foci are 3 units below and above the center: √ √ (1, 1 − 3) and (1, 1 + 3).

To graph the ellipse, plot the vertices. Since b = 1, two other points on the graph are 1 unit left and right of the center: (1 − 1, 1) and (1 + 1, 1) or (0, 1) and (2, 1).

Plot these points also and connect the four plotted points with a smooth curve.

y 4 2 2

!4 !2

4

x

!2 !4

2

!6 !4 !2

x

!2 !4

9x 2 # 4y 2 # 54x ! 8y # 49 " 0

47. The ellipse in Example 4 is flatter than the one in Example 2, so the ellipse in Example 2 has the smaller eccentricity. We compute the eccentricities: In Example 2, c = 3 and a√= 5, so e = c/a = 3/5 = 0.6.√ In Example 4, c = 2 3 and a = 4, so e = c/a = 2 3/4 ≈ 0.866. These computations confirm that the ellipse in Example 2 has the smaller eccentricity. 48. Ellipse (b) is flatter than ellipse (a), so ellipse (a) has the smaller eccentricity. 49. Since the vertices, (0, −4) and (0, 4) are on the y-axis and are equidistant from the origin, we know that the major axis of the ellipse is vertical, its center is at the origin, and a = 4. Use the information that e = 1/4 to find c: c e= a 1 c = Substituting 4 4 c=1 Now c2 = a2 − b2 , so we can find b2 :

4x 2 # y 2 ! 8x ! 2y # 1 " 0

12 = 4 2 − b 2

1 = 16 − b2

b2 = 15

Write the equation of the ellipse:

Exercise Set 9.2

583

y2 x2 + 2 =1 2 b a

or 0.1. Then the distance from the sun to the other focus is twice this distance: 2(0.1 × 107 mi) = 0.2 × 107 mi

y2 x2 + =1 15 16 50. Since the vertices, (−3, 0) and (3, 0), are on the x-axis and are equidistant from the origin, we know that the major axis of the ellipse is horizontal, its center is at the origin, and a = 3. Find c: 7 c = a 10 7 c = 3 10 21 c= 10

= 2 × 106 mi

54. Position a coordinate system as shown. y (0, 1.5)

x (–2, 0)

F2 (2, 0)

F1

(0, –1.5)

a) We have an ellipse with a = 2 and b = 1.5. The foci 2 2 2 2 are at (±c, 0) where √ c = a − b , so c = 4 − 2.25 = 1.75 and c =√ 1.75. Then the string should be attached 2 − 1.75 ft, or about 0.7 ft from the ends of the board.

Now find b2 : c2 = a2 − b2 ! "2 21 = 32 − b 2 10 459 b2 = 100

b) The string should be the length of the major axis, 4 ft.

The equation of the ellipse is

y2 x2 + = 1. 9 459/100

51. From the figure in the text we see that the center of the ellipse is (0, 0), the major axis is horizontal, the vertices are (−50, 0) and (50, 0), and one y-intercept is (0, 12). Then a = 50 and b = 12. The equation is y2 x2 + 2 =1 2 a b y x2 + 2 =1 502 12

Vertices: (−1.017, −1.005), (5.023, −1.005)

56. Center: (−3.004, 1.002) Vertices: (−3.004, −1.970), (−3.004, 3.974) 57. Circles and ellipses are not functions. 58. No, the center of an ellipse is not part of the graph of the ellipse. Its coordinates do not satisfy the equation of the ellipse. 59. midpoint

y2 x2 + = 1. 2500 144

60. zero

52. Find the equation of the ellipse with center (0, 0), a = 1048/2 = 524, b = 898/2 = 449, and a horizontal major axis: y2 x2 + =1 2 524 4492 y2 x2 + =1 274, 576 201, 601

y

9.3

x

9.1 Sun

61. y-intercept 62. two different real-number solutions 63. remainder 64. ellipse 65. parabola

53. Position a coordinate system as shown below where 1 unit = 107 mi.

Focus

55. Center: (2.003, −1.005)

V

The length of the major axis is 9.3 + 9.1, or 18.4. Then the distance from the center of the ellipse (the origin) to V is 18.4/2, or 9.2. Since the distance from the sun to V is 9.1, the distance from the sun to the center is 9.2 − 9.1,

66. circle 67. The center of the ellipse is the midpoint of the segment connecting the vertices: " ! 3 + 3 −4 + 6 , , or (3, 1). 2 2 Now a is the distance from the origin to a vertex. We use the vertex (3, 6). % a = (3 − 3)2 + (6 − 1)2 = 5 Also b is one-half the length of the minor axis. % (5 − 1)2 + (1 − 1)2 4 = =2 b= 2 2

584

Chapter 9: Analytic Geometry Topics The vertices lie on the vertical line x = 3, so the major axis is vertical. We write the equation of the ellipse. (x − h)2 (y − k)2 + =1 b2 a2

(y − 1) (x − 3) + =1 4 25 " ! −1 − 1 −1 + 5 , , or (−1, 2) 68. Center: 2 2 % a = [1 − (−1)]2 + (−1 − 2)2 = 3 % (−3 − 1)2 + (2 − 2)2 4 = =2 b= 2 2 The vertices are on the line x = −1, so the major axis is vertical. The equation is 2

2

(y − 2)2 (x + 1)2 + = 1. 4 9 69. The center is the midpoint of the segment connecting the vertices: " ! −3 + 3 0 + 0 , , or (0, 0). 2 2 Then a = 3 and since the vertices are on the x-axis, the major axis is horizontal. The equation is of the form x2 y2 + 2 = 1. 2 a b 22 for y and solve for b2 . Substitute 3 for a, 2 for x, and 3 484 4 + 92 = 1 9 b 4 484 + 2 =1 9 9b 4b2 + 484 = 9b2 484 = 5b2 484 = b2 5 Then the equation is

x2 y2 + = 1. 9 484/5

70. a = 4/2 = 2; b = 1/2 The equation is

(y − 3)2 (x + 2)2 + = 1. 1/4 4

71. Position a coordinate system as shown. y (0, 14) (–25, 0)

(25, 0)

x

The equation of the ellipse is x2 y2 + 2 =1 2 25 14 y2 x2 + = 1. 625 196 A point 6 ft from the riverbank corresponds to (25 − 6, 0), or (19, 0) or to (−25 + 6, 0), or (−19, 0). Substitute either 19 or −19 for x and solve for y, the clearance. y2 192 + =1 625 196

361 y2 = 1− 196 625 ! " 361 y 2 = 196 1 − 625 y ≈ 9.1

The clearance 6 ft from the riverbank is about 9.1 ft.

Exercise Set 9.3 1. Graph (b) is the graph of

x2 y2 − = 1. 25 9

2. Graph (e) is the graph of

x2 y2 − = 1. 4 36

3. Graph (c) is the graph of

(y − 1)2 (x + 3)2 − = 1. 16 1

4. Graph (f) is the graph of

(y − 2)2 (x + 4)2 − = 1. 100 81

5. Graph (a) is the graph of 25x2 − 16y 2 = 400. 6. Graph (d) is the graph of y 2 − x2 = 9. 7. The vertices are equidistant from the origin and are on the y-axis, so the center is at the origin and the transverse axis is vertical. Since c2 = a2 + b2 , we have 52 = 32 + b2 so b2 = 16. x2 y2 The equation is of the form 2 − 2 = 1, so we have a b y2 x2 − = 1. 9 16 8. The vertices are equidistant from the origin and are on the x-axis, so the center is at the origin and the transverse axis is horizontal. Since c2 = a2 + b2 , we have 22 = 12 + b2 so x2 y2 − = 1. b2 = 3. The equation is 1 3 9. The asymptotes pass through the origin, so the center is the origin. The given vertex is on the x-axis, so the trans3 b verse axis is horizontal. Since x = x and a = 2, we a 2 x2 y2 have b = 3. The equation is of the form 2 − 2 = 1, so a b y2 x2 x2 y2 − = 1. we have 2 − 2 = 1, or 2 3 4 9

Exercise Set 9.3

585

10. The asymptotes pass through the origin, so the center is the origin. The given vertex is on the y-axis, so the transverse axis is vertical. We use the equation of an asymptote to find b. 5 a x= x b 4 5 3 x = x Substituting 3 for a b 4 12 =b 5 x2 y2 − = 1. The equation is 9 144/25 11.

y 4

4

x

!4

x2 y2 ! "1 1 9

13.

x2 y2 − =1 4 4 2

2

!4 !2

(x − 2)2 (y + 5)2 − =1 9 1 [y − (−5)]2 (x − 2)2 − =1 2 3 12

2

y x − 2 = 1 Standard form 22 2 The center is (0, 0); a = 2 and b = 2. The transverse axis is horizontal so the vertices are (−2, 0) and (2, √ 0). Since √ c2 = 4 + 4 = √ 8 and c = 8, or 2 2. c2 = a2 + b2 , we have √ Then the foci are (−2 2, 0) and (2 2, 0).

The center is (2, −5); a = 3 and b = 1. The transverse axis is horizontal, so the vertices are 3 units left and right of the center: (2 − 3, −5) and (2 + 3, −5), or (−1, −5) and (5, −5).

√ , we have c2 = 9 + 1 = 10 and c = 10. Since c2 = a2 + b2 √ Then the foci are 10 units left and right of the center: √ √ (2 − 10, −5) and (2 + 10, −5).

Find the asymptotes: b b y = x and y = − x a a 2 2 y = x and y = − x 2 2

Find the asymptotes: b b y − k=− (x − h) y − k= (x − h) and a a 1 1 y − (−5)= (x − 2) and y − (−5)=− (x − 2) 3 3 1 1 y + 5=− (x − 2), or y + 5= (x − 2) and 3 3 17 1 13 1 and y=− x − y= x − 3 3 3 3 Sketch the asymptotes, plot the vertices, and draw the graph.

y = x and y = −x To draw the graph sketch the asymptotes, plot the vertices, and draw the branches of the hyperbola outward from the vertices toward the asymptotes.

y 4 2

y

4

!4

Standard form

2

x

!2 2

!4 !2

!4

4

6

x

!2

x2 y2 ! "1 4 4 12.

x2 y2 − =1 1 9 y2 x2 − 2 = 1 Standard form 2 1 3 The center is (0, 0); a = 1 and b = 3. The transverse axis is horizontal so the vertices are (−1, 0) and (1, √0). Since 2 c2 = a2 + b2 , we have c = 1 + 9 = 10 and c = 10. Then √ √ the foci are (− 10, 0) and ( 10, 0). Find the asymptotes: b b y = x and y = − x a a 3 3 y = x and y = − x 1 1 y = 3x and y = −3x

(x ! 2)2 (y # 5)2 ! "1 9 1 14.

(x − 5)2 (y + 2)2 − =1 16 9 (x − 5)2 [y − (−2)]2 − =1 2 4 32

Standard form

The center is (5, −2); a = 4 and b = 3. The transverse axis is horizontal, so the vertices are 4 units left and right of the center: (5 − 4, −2) and (5 + 4, −2), or (1, −2) and (9, −2).

586

Chapter 9: Analytic Geometry Topics Since c2 = a2 + b2 , we have c2 = 16 + 9 = 25 and c = 5. Then the foci are 5 units left and right of the center:

y

(5 − 5, −2) and (5 + 5, −2), or (0, −2) and (10, −2).

2

Find the asymptotes: b b y − k = (x − h) and y − k = − (x − h) a a 3 3 y + 2 = (x − 5) and y + 2 = − (x − 5), or 4 4 23 3 7 3 and y = − x+ y = x− 4 4 4 4

6 !4 !6

(y # 3)2 (x # 1)2 ! "1 4 16

y 8

16.

4 8

!4

12

!8

Standard form

The center is (−2, −4); a = 5 and b = 4. The transverse axis is vertical, so the vertices are 5 units below and above the center:

(x ! 5)2 (y # 2)2 ! "1 16 9

(−2, −4 − 5) and (−2, −4 + 5), or (−2, −9) and (−2, 1). √ we have c2 = 25+16 = 41 and c = 41. Since c2 = a2 +b2 , √ Then the foci are 41 units below and above the center: √ √ (−2, −4 − 41) and (−2, −4 + 41).

(x + 1)2 (y + 3)2 − =1 4 16 [x − (−1)]2 [y − (−3)]2 − =1 2 2 42

(x + 2)2 (y + 4)2 − =1 25 16 [x − (−2)]2 [y − (−4)]2 − =1 52 42

x

!4

15.

x

!2

Standard form

Find the asymptotes: a a y − k = (x − h) and y − k = − (x − h) b b 5 5 y + 4 = (x + 2) and y + 4 = − (x + 2), or 4 4 5 3 5 13 y = x− and y = − x− 4 2 4 2

The center is (−1, −3); a = 2 and b = 4. The transverse axis is vertical, so the vertices are 2 units below and above the center: (−1, −3 − 2) and (1, −3 + 2), or (−1, −5) and (−1, −1). √ 2 2 2 2 Since √c = a + b , we have c√= 4 + 16 = 20 and c = 20, or 2 5. Then the foci are 2 5 units below and above of the center: √ √ (−1, −3 − 2 5) and (−1, −3 + 2 5).

Find the asymptotes: a a y − k=− (x − h) y − k= (x − h) and b b 2 2 y−(−3)= (x−(−1)) and y−(−3)=− (x−(−1)) 4 4 1 1 y+3= (x+1) and y+3=− (x+1), or 2 2 5 1 7 1 and y=− x − y= x − 2 2 2 2 Sketch the asymptotes, plot the vertices, and draw the graph.

y 4 4

!8

8

x

!8 !12

(y # 4)2 (x # 2)2 ! "1 25 16

17.

x2 − 4y 2 = 4 x2 y2 − =1 4 1

y2 x2 − 2 = 1 Standard form 2 2 1 The center is (0, 0); a = 2 and b = 1. The transverse axis is horizontal, so the vertices are (−2, 0) and (2,√0). Since 2 c2 = a2 + b2 , we √ have c =√4 + 1 = 5 and c = 5. Then the foci are (− 5, 0) and ( 5, 0). Find the asymptotes:

Exercise Set 9.3

587

b b x and y = − x a a 1 1 y = x and y = − x 2 2 Sketch the asymptotes, plot the vertices, and draw the graph.

Find the asymptotes: a a y = x and y = − x b b 3 3 y = x and y = − x 9 9 1 1 y = x and y = − x 3 3 Sketch the asymptotes, plot the vertices, and draw the graph.

y=

y

4 2

y

4

!4

16

x

12 8

!2 !4

4 !16

x 2 ! 4y 2 " 4

8 12 16 x

!8 !4 !8

18.

4x − y = 16 2

!12

2

!16

y2 x2 − =1 4 16

x2 y2 − 2 =1 Standard form 2 2 4 The center is (0, 0); a = 2 and b = 4. The transverse axis is horizontal, so the vertices are (−2, 0) and (2, 0). √ Since 4 + 16 = 20 √ and c = 20, or c2√= a2 + b2 , we have c2 = √ 2 5. Then the foci are (−2 5, 0) and (2 5, 0). Find the asymptotes: b b y = x and y = − x a a 4 4 y = x and y = − x 2 2 y = 2x and y = −2x y 4

4x 2 ! y 2 " 16

2 4

!4

x

9y 2 ! x2 " 81 20.

y 2 − 4x2 = 4 y2 x2 − =1 4 1

x2 y2 − 2 = 1 Standard form 2 2 1 The center is (0, 0); a = 2 and b = 1. The transverse axis is vertical, so the vertices are (0, −2) and (0,√2). Since have c2 = 4√+ 1 = 5 and c = 5. Then c2 = a2 + b2 , we √ the foci are (0, − 5) and (0, 5). Find the asymptotes: a a y = x and y = − x b b 2 2 y = x and y = − x 1 1 y = 2x and y = −2x y

!2 !4

4

y 2 ! 4x 2 " 4

19.

9y − x = 81 2

2

y2 x2 − =1 9 81

x2 y2 − = 1 Standard form 32 92 The center is (0, 0); a = 3 and b = 9. The transverse axis is vertical, so the vertices are (0, −3) and (0, 3). √ Since c = 90, or c2√= a2 + b2 , we have c2 = 9 +√81 = 90 and √ 3 10. Then the foci are (0, −3 10) and (0, 3 10).

2

!4 !2 !4

4

x

588 21.

Chapter 9: Analytic Geometry Topics x2 − y 2 = 2

23.

y2 x2 − =1 2 2 y2 x2 √ − √ =1 ( 2)2 ( 2)2

1 4

y 2 − x2 =

x2 y2 − =1 1/4 1/4 Standard form

x2 y2 − =1 2 (1/2) (1/2)2

√

√ The center is (0, 0); a = 2 and b = √2. The transverse √ axis is horizontal, so the vertices are (− 2, 0) and ( 2, 0). 2 2 2 2 Since c = a + b , we have c = 2 + 2 = 4 and c = 2. Then the foci are (−2, 0) and (2, 0).

Standard form

1 1 and b = . The transverse axis 2 2" ! " ! 1 1 and 0, . Since is vertical, so the vertices are 0, − 2 2 & 1 1 1 1 c2 = a2 + b2 , we have c2 = + = and c = , or 4 4 2 2 √ √ " ! ! √ " 2 2 2 . Then the foci are 0, − and 0, . 2 2 2 Find the asymptotes: a a and y = − x y= x b b The center is (0, 0); a =

Find the asymptotes: b b y = x and y = − x a a √ √ 2 2 y = √ x and y = − √ x 2 2 y=x and y = −x Sketch the asymptotes, plot the vertices, and draw the graph.

1/2 1/2 x and y = − x 1/2 1/2 y=x and y = −x y=

y

4

Sketch the asymptotes, plot the vertices, and draw the graph.

2 2

!4 !2

4

y

x

!2

1

!4

22.

x

!1

x2 − y 2 = 3

x2 y2 − =1 3 3 y2 x2 √ − √ =1 2 ( 3) ( 3)2

1

!1

x2 ! y2 " 2

y2 ! x2 " ~ Standard form

√ √ The center is (0, 0); a = 3 and b = √3. The transverse √ axis is horizontal, so the vertices are (− 3, 0) and ( 3,√0). Since c2 = a2 + b2 , √ we have c2 = √ 3 + 3 = 6 so c = 6. Then the foci are (− 6, 0) and ( 6, 0). Find the asymptotes: b b y = x and y = − x a a √ √ 3 3 y = √ x and y = − √ x 3 3 y=x and y = −x y 4 2 4

!4 !2 !4

x2 ! y2 " 3

x

24.

y 2 − x2 =

1 9

x2 y2 − =1 1/9 1/9 y2 x2 − =1 2 (1/3) (1/3)2

Standard form

1 1 and b = . The transverse axis 3 3" ! " ! 1 1 and 0, . Since is vertical, so the vertices are 0, − 3 3 √ 1 1 2 2 . Then c2 = a2 + b2 , we have c2 = + = and c = 9! 9√ "9 3 √ " ! 2 2 the foci are 0, − and 0, . 3 3 Find the asymptotes: a a and y = − x y= x b b The center is (0, 0); a =

1/3 1/3 x and y = − x 1/3 1/3 y=x and y = −x y=

Exercise Set 9.3

589 y 1

y2 ! x2 " › 1

!1

(−1 − 1, −2) and (−1 + 1, −2) or (−2, −2) and (0, −2) √ Since c2 = a2 + b2√ , we have c2 = 1 + 4 = 5 and c = 5. Then the foci are 5 units left and right of the center: √ √ (−1 − 5, −2) and (1 + 5, −2).

x

!1

25. Begin by completing the square twice. x2 − y 2 − 2x − 4y − 4 = 0 (x2 − 2x) − (y 2 + 4y) = 4

(x2 − 2x + 1) − (y 2 + 4y + 4) = 4 + 1 − 1 · 4 (x − 1)2 − (y + 2)2 = 1

[y − (−2)]2 (x − 1)2 − =1 12 12

Standard form

The center is (1, −2); a = 1 and b = 1. The transverse axis is horizontal, so the vertices are 1 unit left and right of the center: (1 − 1, −2) and (1 + 1, −2) or (0, −2) and (2, −2)

√ , we have c2 = 1 + 1 = 2 and c = 2. Since c2 = a2 + b2√ Then the foci are 2 units left and right of the center: √ √ (1 − 2, −2) and (1 + 2, −2). Find the asymptotes: b b y − k = − (x − h) y − k = (x − h) and a a 1 1 y − (−2) = (x − 1) and y − (−2) = − (x − 1) 1 1 y+2 = x−1 and y + 2 = −(x − 1), or

y = x−3 and y = −x − 1 Sketch the asymptotes, plot the vertices, and draw the graph.

y 2 4

!4 !2

6 x

!6

x 2 ! y 2 ! 2x ! 4y ! 4 " 0 26. Begin by completing the square twice. 4x2 − y 2 + 8x − 4y − 4 = 0 4(x2 + 2x) − (y 2 + 4y) = 4

4(x + 2x + 1) − (y 2 + 4y + 4) = 4 + 4 · 1 − 1 · 4 2

The center is (−1, −2); a = 1 and b = 2. The transverse axis is horizontal, so the vertices are 1 unit left and right of the center:

4(x + 1)2 − (y + 2)2 = 4

(y + 2)2 (x + 1)2 − =1 1 4

[y − (−2)]2 [x − (−1)]2 − =1 12 22

Standard form

Find the asymptotes: b b y − k = (x − h) and y − k = − (x − h) a a 2 2 y + 2 = (x + 1) and y + 2 = − (x + 1) 1 1 y + 2 = 2(x + 1) and y + 2 = −2(x + 1), or y = 2x

and

y = −2x − 4

y 2 2

!4 !4 !8

4

x

4x 2 ! y 2 # 8x ! 4y ! 4 " 0

27. Begin by completing the square twice. 36x2 − y 2 − 24x + 6y − 41 = 0

(36x2 − 24x) − (y 2 − 6y) = 41 " ! 2 2 36 x − x − (y 2 − 6y) = 41 3 " ! 2 1 1 −(y 2 −6y+9) = 41+36· −1 · 9 36 x2 − x+ 3 9 9 "2 ! 1 − (y − 3)2 = 36 36 x − 3 "2 ! 1 x− (y − 3)2 3 − =1 1 36 "2 ! 1 x− (y − 3)2 3 − = 1 Standard 2 1 62 form " ! 1 , 3 ; a = 1 and b = 6. The transverse The center is 3 axis is horizontal, so the vertices are 1 unit left and right of the center: " ! " ! " ! 1 2 1 − 1, 3 and + 1, 3 or − , 3 and 3 3 !3 " 4 ,3 . 3 √ Since c2 = a2 + b2 ,√we have c2 = 1 + 36 = 37 and c = 37. Then the foci are 37 units left and right of the center: ! " ! " 1 √ 1 √ − 37, 3 and + 37, 3 . 3 3 Find the asymptotes:

590

Chapter 9: Analytic Geometry Topics b (x − h) a ! " 1 6 x− y−3 = 1 3 ! " 1 y−3 = 6 x− 3 y = 6x + 1

y−k =

b and y − k = − (x − h) a ! " 6 1 and y − 3 = − x − 1 3 ! " 1 and y − 3 = −6 x − , or 3 and y = −6x + 5

y 6

2 2 x

!8 !6 !4

Sketch the asymptotes, plot the vertices, and draw the graph.

y

9x 2 ! 4y 2 # 54x # 8y # 41 " 0

29. Begin by completing the square twice. 9y 2 − 4x2 − 18y + 24x − 63 = 0

9(y 2 − 2y) − 4(x2 − 6x) = 63

9(y − 2y + 1) − 4(x2 − 6x + 9) = 63+9·1−4·9 2

!8 !4

4

8

9(y − 1)2 − 4(x − 3)2 = 36

x

(y − 1)2 (x − 3)2 − =1 4 9 (y − 1)2 (x − 3)2 − =1 2 2 32

36x 2 ! y 2 ! 24x # 6y ! 41 " 0 28. Begin by completing the square twice. 9x2 − 4y 2 + 54x + 8y + 41 = 0

9(x2 + 6x) − 4(y 2 − 2y) = −41

9(x2 + 6x + 9) − 4(y 2 − 2y + 1) = −41+9·9−4·1 9(x + 3)2 − 4(y − 1)2 = 36 (x + 3)2 (y − 1)2 − =1 4 9

(y − 1)2 [x − (−3)]2 − =1 2 2 32

Standard form

The center is (−3, 1); a = 2 and b = 3. The transverse axis is horizontal, so the vertices are 2 units left and right of the center: (−3 − 2, 1) and (−3 + 2, 1), or (−5, 1) and (−1, 1).

The center is (3, 1); a = 2 and b = 3. The transverse axis is vertical, so the vertices are 2 units below and above the center: (3, 1 − 2) and (3, 1 + 2), or (3, −1) and (3, 3).

√ , we have c2 = 4 + 9 = 13 and c = 13. Since c2 = a2 + b2 √ Then the foci are 13 units below and above the center: √ √ (3, 1 − 13) and (3, 1 + 13). Find the asymptotes: a y − k = (x − h) and y − k b 2 y − 1 = (x − 3) and y − 1 3 2 y = x − 1 and y 3 Sketch the asymptotes, plot the graph.

√ , we have c2 = 4 + 9 = 13 and c = 13. Since c2 = a2 + b2 √ Then the foci are 13 units left and right of the center: √ √ (−3 − 13, 1) and (−3 + 13, 1). Find the asymptotes: b b y − k = (x − h) and y − k = − (x − h) a a 3 3 y − 1 = (x + 3) and y − 1 = − (x + 3), or 2 2 11 3 7 3 and y = − x− y = x+ 2 2 2 2

Standard form

a = − (x − h) b 2 = − (x − 3), or 3 2 = − x+3 3 vertices, and draw the

y 6

2 2

!2

4

6

8 x

!4

9y 2 ! 4x 2 ! 18y # 24x ! 63 " 0

Exercise Set 9.3

591 Sketch the asymptotes, plot the vertices, and draw the graph.

30. Begin by completing the square twice. x2 − 25y 2 + 6x − 50y = 41

y

x2 + 6x + 9 − 25(y 2 + 2y + 1) = 41 + 9 − 25 · 1 (x + 3)2 − 25(y + 1)2 = 25

4

(y + 1)2 (x + 3)2 − =1 25 1

2

[x − (−3)]2 [y − (−1)]2 − =1 52 12 The center is (−3, −1); a = 5 and b = 1. The transverse axis is horizontal, so the vertices are 5 units left and right of the center:

!4 !2

(−3 − 5, −1) and (−3 + 5, −1), or (−8, −1) and (2, −1). √ Since c2 = a2 + b2 ,√we have c2 = 25 + 1 = 26 and c = 26. Then the foci are 26 units left and right of the center: √ √ (−3 − 26, −1) and (−3 + 26, −1).

x 2 ! y 2 ! 2x ! 4y " 4

Find the asymptotes: b b y − k = (x − h) and y − k = − (x − h) a a 1 1 y + 1 = (x + 3) and y + 1 = − (x + 3), or 5 5 2 1 8 1 and y = − x− y = x− 5 5 5 5

4 2 4

x

!4

x 2 ! 25y 2 # 6x ! 50y " 41

31. Begin by completing the square twice. x2 − y 2 − 2x − 4y = 4

(x2 − 2x + 1) − (y 2 + 4y + 4) = 4 + 1 − 4 (x − 1)2 − (y + 2)2 = 1

(x − 1)2 [y − (−2)]2 − =1 12 12

32. Begin by completing the square twice. 9y 2 − 4x2 − 54y − 8x + 41 = 0

9(y − 6y + 9)−4(x2 + 2x + 1) = −41+9 · 9−4·1 2

9(y − 3)2 − 4(x + 1)2 = 36 (x + 1)2 (y − 3)2 − =1 4 9

[x − (−1)]2 (y − 3)2 − =1 2 2 32

Standard form

Find the asymptotes: a a y − k = (x − h) and y − k = − (x − h) b b 2 2 y − 3 = (x + 1) and y − 3 = − (x + 1), or 3 3 11 2 7 2 and y = − x+ y = x+ 3 3 3 3

Standard form

y 6

The center is (1, −2); a = 1 and b = 1. The transverse axis is horizontal, so the vertices are 1 unit left and right of the center: (1 − 1, −2) and (1 + 1, −2), or (0, −2) and (2, −2). √ , we have c2 = 1 + 1 = 2 and c = 2. Since c2 = a2 + b2√ Then the foci are 2 units left and right of the center: √ √ (1 − 2, −2) and (1 + 2, −2). Find the asymptotes: b b y − k = − (x − h) y − k = (x − h) and a a 1 1 y − (−2) = (x − 1) and y − (−2) = − (x − 1) 1 1 y+2 = x−1 and y + 2 = −(x − 1), or and

x

(−1, 3 − 2) and (−1, 3 + 2), or (−1, 1) and (−1, 5). √ 2 + b2 , we have c2 = 4 + 9 and c = 13. Then Since c2 = a√ the foci are 13 units below and above the center: √ √ (−1, 3 − 13) and (−1, 3 + 13).

!2

y = x−3

4

The center is (−1, 3); a = 2 and b = 3. The transverse axis is vertical, so the vertices are 2 units below and above the center:

y

!12

2

y = −x − 1

4 2 2

!4 !2

x

!2

9y 2 ! 4x 2 ! 54y ! 8x # 41 " 0

592

Chapter 9: Analytic Geometry Topics Then the foci are 2 units left and right of the center:

33. Begin by completing the square twice. y − x − 6x − 8y − 29 = 0 2

2

(y 2 − 8y + 16) − (x2 + 6x + 9) = 29 + 16 − 9 (y − 4) − (x + 3) = 36 2

2

(y − 4)2 (x + 3)2 − =1 36 36 [x − (−3)]2 (y − 4)2 − =1 2 6 62

Standard form

(4 − 2, 1) and (4 + 2, 1), or (2, 1) and (6, 1).

Find the asymptotes: b b y − k = (x − h) and y − k = − (x − h) a a √ √ 2 2 y − 1 = √ (x − 4) and y − 1 = − √ (x − 4) 2 2 y−1 = x−4 and y − 1 = −(x − 4), or y = x−3

The center is (−3, 4); a = 6 and b = 6. The transverse axis is vertical, so the vertices are 6 units below and above the center:

y = x+7

and

y = −x + 1

Sketch the asymptotes, plot the vertices, and draw the graph. y

4

8 4 4

!4

2 4

8

x

!2 !4

x 2 ! y 2 " 8x ! 2y ! 13

35. The hyperbola in Example 3 is wider than the one in Example 2, so the hyperbola in Example 3 has the larger eccentricity. Compute the eccentricities: In Example 2,√c = 5 and a = 4, so e √ = 5/4, or 1.25. In Example 3, c = 5 and a = 1, so e = 5/1 ≈ 2.24. These computations confirm that the hyperbola in Example 3 has the larger eccentricity. 36. Hyperbola (b) is wider so it has the larger eccentricity.

12

!12

y = −x + 5

y

(−3, 4 − 6) and (−3, 4 + 6), or (−3, −2) and (−3, 10). √ 2 2 2 2 Since √c = a +b , we have c√= 36+36 = 72 and c = 72, or 6 2. Then the foci are 6 2 units below and above the center: √ √ (−3, 4 − 6 2) and (−3, 4 + 6 2). Find the asymptotes: a a y − k = (x − h) and y − k = − (x − h) b b 6 6 y − 4 = (x − (−3)) and y − 4 = − (x − (−3)) 6 6 y−4 = x+3 and y − 4 = −(x + 3), or

and

8

x

!4 !8

37. The center is the midpoint of the segment connecting the vertices: " ! 3−3 7+7 , , or (0, 7). 2 2 The vertices are on the horizontal line y = 7, so the transverse axis is horizontal. Since the vertices are 3 units left and right of the center, a = 3. Find c:

y 2 ! x 2 ! 6x ! 8y ! 29 " 0 34. Begin by completing the square twice. x2 − y 2 = 8x − 2y − 13

x2 − 8x − y 2 + 2y = −13

x2 − 8x + 16 − (y 2 − 2y + 1) = −13 + 16 − 1 (x − 4)2 − (y − 1)2 = 2

(y − 1)2 (x − 4)2 √ − √ = 1 Standard form 2 ( 2) ( 2)2 √ √ The center is (4, 1); a = 2 and b = √ 2. The transverse axis is horizontal, so the vertices are 2 units left and right of the center: √ √ (4 − 2, 1) and (4 + 2, 1). Since c2 = a2 + b2 , we have c2 = 2 + 2 = 4 and c = 2.

e=

5 c = a 3 c 5 = 3 3 c=5

Substituting 3 for a

Now find b2 : c2 = a2 + b2 52 = 32 + b2 16 = b2 Write the equation: (x − h)2 (y − k)2 − =1 2 a b2 (y − 7)2 x2 − =1 9 16

Exercise Set 9.3

593

38. The center is the midpoint of the segment connecting the vertices: " ! −1 − 1 3 + 7 , , or (−1, 5). 2 2 The vertices are on the vertical line x = −1, so the transverse axis is vertical. Since the vertices are 2 units below and above the center, a = 2. Find c: e=

c =4 a c =4 2 c=8

Then the diameter of the bottom of the tower is 2x2 ≈ 2(226.4) ≈ 453 ft.

64 = 4 + b2 60 = b2

39.

(y − 5)2 (x + 1)2 − = 1. 4 60 y

(0, 6)

Then the diameter of the top of the tower is 2x1 ≈ 2(137.4) ≈ 275 ft. Find x2 : 3002 x22 − =1 902 1302 ! " 3002 2 2 x2 = 90 1 + 1302 x2 ≈ 226.4

Now find b2 : c2 = a2 + b2

The equation is

Find x1 : 1502 x21 − =1 902 1302 ! " 1502 2 2 x1 = 90 1 + 1302 x1 ≈ 137.4

F1

Hyperbola

(0, 5)

41. Center: (−1.460, −0.957)

Vertices: (−2.360, −0.957), (−0.560, −0.957)

Asymptotes: y = −1.429x − 3.043, y = 1.429x + 1.129

42. Center: (1.023, −2.044)

Vertices: (2.07, −2.044), (−0.024, −2.044)

Asymptotes: y = x − 3.067, y = −x − 1.021

x

43. See the figure on page 768 of the text. (0, –6)

F2

Parabola

(0, –8)

One focus is 6 units above the center of the hyperbola, so c = 6. One vertex is 5 units above the center, so a = 5. Find b2 : c2 = a2 + b2 62 = 5 2 + b 2

45. a) The graph of f (x) = 2x − 3 is shown below. y 5 4 3 2 1

11 = b2 Write the equation: y2 x2 − 2 =1 2 a b

-5 -4 -3 -2 -1 0 -1

f(x) = 2x – 3 x 1 2 3 4 5

-2 -3 -4

y2 x2 − =1 25 11 40.

44. No, the asymptotes of a hyperbola are not part of the graph of the hyperbola. The coordinates of points on the asymptotes do not satisfy the equation of the hyperbola.

-5

y (x1, y1) x 450 ft

b) Replace f (x) with y: y = 2x − 3 Interchange x and y: x = 2y − 3

(x2, y2)

1 1 y2 , so y2 + y2 = 450. Then y2 = 300 and 2 2 1 y1 = · 300 = 150. 2 y1 =

Since there is no horizontal line that crosses the graph more than once, the function is one-to-one.

Solve for y: x + 3 = 2y x+3 =y 2

Replace y with f −1 (x): f −1 (x) =

x+3 2

594

Chapter 9: Analytic Geometry Topics

46. a) The graph of f (x) = x3 +2 is shown below. It passes the horizontal line test, so it is one-to-one.

√ 48. a) The graph of f (x) = x + 4 is shown below. It passes the horizontal line test, so it is one-to-one.

y

y

10 8

5 4

6 4 2

3 2 1

f(x) = x3 + 2 x

-10 -8 -6 -4 -2 0 -2

2 4 6 8 10

-4 -6 -8

-2 -3 -4

-10

-5

Replace y with f

−1

(x): f

47. a) The graph of f (x) =

−1

Solve for y: (x) =

√ 3

x−2

5 is shown below. x−1

3 2 1 -5 -4 -3 -2 -1 0 -1

x2 = y + 4 x2 − 4 = y

Replace y with f −1 (x): f −1 (x) = x2 − 4, x ≥ 0 49.

y 5 4

√

x+4 √ Interchange x and y: x = y + 4

x − 2 = y3 x−2 = y

1 2 3 4 5

b) Replace f (x) with y: y =

Interchange x and y: x = y 3 + 2 √ 3

x+4

x

-5 -4 -3 -2 -1 0 -1

b) Replace f (x) with y: y = x3 + 2 Solve for y:

f(x) =

x+y= x−y= 2x = x=

5, 7 12 6

(1) (2) Adding

Back-substitute in either equation (1) or (2) and solve for y. We use equation (1).

5 f(x) = x – 1

6+y = 5 y = −1

x 1 2 3 4 5

-2 -3 -4

The solution is (6, −1). 50.

-5

Since there is no horizontal line that crosses the graph more than once, the function is one-to-one. 5 b) Replace f (x) with y: y = x−1 5 Interchange x and y: x = y−1 Solve for y: x(y − 1) = 5 5 y−1 = x 5 y = +1 x 5 5+x Replace y with f −1 (x): f −1 = + 1, or x x

3x − 2y = 5x + 2y = 8x = x=

5, 3 8 1

(1) (2) Adding

Back-substitute and solve for y. 5 · 1 + 2y = 3

Using equation (2)

2y = −2 y = −1

The solution is (1, −1). 51.

2x − 3y = 7,

(1)

3x + 5y = 1

(2)

Multiply equation (1) by 5 and equation (2) by 3 and add to eliminate y. 10x − 15y = 9x + 15y = 19x = x=

35 3 38 2

Back-substitute and solve for y. 3 · 2 + 5y = 1

Using equation (2)

5y = −5 y = −1

The solution is (2, −1).

Exercise Set 9.4 52.

595

3x + 2y = −1 2x + 3y = 6

Write the equation of the hyperbola:

(1)

(x + 7)2 (y − 4)2 − =1 4 36

(2)

Multiply equation (1) by 3 and equation (2) by −2 and add. 9x + 6y = −3 −4x − 6y = −12 5x = −15 x = −3

55. S and T are the foci of the hyperbola, so c = 300/2 = 150. 0.186 mi 200 microseconds · = 37.2 mi, the 1 microsecond difference of the ships’ distances from the foci. That is, 2a = 37.2, so a = 18.6. Find b2 :

Back-substitute and solve for y. 3(−3) + 2y = −1

c2 = a2 + b2

Using equation (1)

1502 = 18.62 + b2

2y = 8

22, 154.04 = b2

y=4

Then the equation of the hyperbola is

The solution is (−3, 4). 53. The center is the midpoint of the segment connecting (3, −8) and (3, −2): " ! 3 + 3 −8 − 2 , , or (3, −5). 2 2 The vertices are on the vertical line x = 3 and are 3 units above and below the center so the transverse axis is vertical and a = 3. Use the equation of an asymptote to find b: a y − k = (x − h) b 3 y + 5 = (x − 3) b 9 3 y = x− −5 b b This equation corresponds to the asymptote y = 3x − 14, 3 so = 3 and b = 1. b Write the equation of the hyperbola: (x − h) (y − k) − =1 a2 b2 2

2

(y + 5)2 (x − 3)2 − =1 9 1 54. The center is the midpoint of the segment connecting the vertices: " ! −9 − 5 4 + 4 , , or (−7, 4). 2 2 The vertices are on the horizontal line y = 4 and are 2 units left and right of the center, so the transverse axis is horizontal and a = 2. Use the equation of an asymptote to find b: b y − k = (x − h) a b y − 4 = (x + 7) 2 7 b y = x+ b+4 2 2 This equation corresponds to the asymptote y = 3x + 25, b so = 3 and b = 6. 2

x2 y2 x2 y2 − = 1, or − = 1. 18.62 22, 154.04 345.96 22, 154.04

Exercise Set 9.4 1. The correct graph is (e). 2. The correct graph is (a). 3. The correct graph is (c). 4. The correct graph is (f). 5. The correct graph is (b). 6. The correct graph is (d). 7.

x2 + y 2 = 25,

(1)

y−x=1

(2)

y =x+1

(3)

First solve equation (2) for y. Then substitute x + 1 for y in equation (1) and solve for x. x2 + y 2 = 25 x2 + (x + 1)2 = 25 x + x2 + 2x + 1 = 25 2

2x2 + 2x − 24 = 0 x2 + x − 12 = 0

(x + 4)(x − 3) = 0

x+4 = 0

Multiplying by Factoring

or x − 3 = 0

x = −4 or

1 2

x=3

Principle of zero products

Now substitute these numbers into equation (3) and solve for y. y = −4 + 1 = −3 y = 3+1=4

The pairs (−4, −3) and (3, 4) check, so they are the solutions.

596 8.

Chapter 9: Analytic Geometry Topics y = 0 or y = 3 2 x = 2 − (0) = 2 3 2 x = 2 − (3) = 0 3 The pairs (2, 0) and (0, 3) check.

x2 + y 2 = 100, y−x=2 y =x+2 x2 + (x + 2)2 = 100 x + x2 + 4x + 4 = 100 2

2x2 + 4x − 96 = 0 x + 2x − 48 = 0 2

(x + 8)(x − 6) = 0

x = −8 or x = 6 y =6+2=8

The pairs (−8, −6) and (6, 8) check. 4x2 + 9y 2 = 36, (1) 3y + 2x = 6

First solve equation (2) for y. (3)

2 Then substitute − x + 2 for y in equation (1) and solve 3 for x. 4x2 + 9y 2 = 36 (2 ' 2 4x2 + 9 − x + 2 = 36 3 '4 ( 8 2 2 4x + 9 x − x + 4 = 36 9 3 4x2 + 4x2 − 24x + 36 = 36 8x − 24x = 0 2

x2 − 3x = 0

x = 0 or x = 3

x(x − 3) = 0

Now substitute these numbers in equation (3) and solve for y. 2 y = − ·0+2=2 3 2 y = − ·3+2=0 3 The pairs (0, 2) and (3, 0) check, so they are the solutions. 10.

(1)

y =x+5

(2)

2

We substitute x + 5 for y 2 in equation (1) and solve for x. x2 + y 2 = 25 x2 + x − 20 = 0

(x + 5)(x − 4) = 0 x+5 = 0

or

x = −5

(2)

3y = −2x + 6 2 y = − x+2 3

x2 + y 2 = 25,

x2 + (x + 5) = 25

y = −8 + 2 = −6

9.

11.

9x2 + 4y 2 = 36, 3x + 2y = 6 2 x = 2− y 3 ' 2 (2 9 2 − y + 4y 2 = 36 3 ' 4 ( 8 9 4 − y + y 2 + 4y 2 = 36 3 9 36 − 24y + 4y 2 + 4y 2 = 36 8y 2 − 24y = 0 y 2 − 3y = 0

y(y − 3) = 0

or

x−4 = 0

x=4

We substitute these numbers for x in either equation (1) or equation (2) and solve for y. Here we use equation (2). y 2 = −5 + 5 = 0 and y = 0. y 2 = 4 + 5 = 9 and y = ±3.

The pairs (−5, 0), (4, 3) and (4, −3) check. They are the solutions. 12. y = x2 , x = y2 x = (x2 )2 x = x4 0 = x4 − x

0 = x(x3 − 1)

0 = x(x − 1)(x2 + x + 1) x = 0 or

x = 1 or

x = 0 or

x = 1 or

√

12 − 4 · 1 · 1 2 √ 1 3 x=− ± i 2 2 x=

−1 ±

y = 02 = 0 y = 12 = 1 ' 1 y= − + 2 ' 1 y= − − 2

√ ( √ 3 2 1 3 i =− − i 2 2 2 √ ( √ 3 2 1 3 i =− + i 2 2 2 √ ( ' 1 √3 1 3 i, − − i , The pairs (0, 0), (1, 1), − + 2 2 2 2 √ √ ' 1 3 1 3 ( i, − + i check. and − − 2 2 2 2

Exercise Set 9.4 13.

597

x2 + y 2 = 9,

(1)

x −y =9

(2)

2

2

16.

xy = −7

y = −x − 6

Here we use the elimination method. (1) x2 + y 2 = 9 x −y =

9

x =

9

2

2

2x2

x(−x − 6) = −7

(2)

= 18

2

−x2 − 6x = −7

Adding

0 = x2 + 6x − 7

0 = (x + 7)(x − 1)

x = ±3

x = −7 or x = 1

y = −(−7) − 6 = 1

If x = 3, x2 = 9, and if x = −3, x2 = 9, so substituting 3 or −3 in equation (1) gives us

y = −1 − 6 = −7

x2 + y 2 = 9 9 + y2 = 9

y = 0. The pairs (3, 0) and (−3, 0) check. They are the solutions. y 2 − 4x2 = 4 4x2 + y 2 = 4

(1)

(2) Adding

y2 = 4 y = ±2

Substitute for y in equation (2). 4x2 + 4 = 4

2y = x + 4

(2)

First solve equation (2) for x. 2y − 4 = x

(3)

y2 = x + 3 y 2 = (2y − 4) + 3 y 2 = 2y − 1

y 2 − 2y + 1 = 0

(y − 1)(y − 1) = 0 y − 1 = 0 or y = 1 or

y−1 = 0 y=1

Now substitute 1 for y in equation (3) and solve for x.

4x2 = 0

2·1−4 = x

x=0 The pairs (0, 2) and (0, −2) check. 2x − 3 = y

(1)

(1)

4x2 + y 2 = 4 2y 2 = 8

y 2 − x2 = 9

y 2 = x + 3,

Then substitute 2y − 4 for x in equation (1) and solve for y.

(2)

−4x2 + y 2 = 4

15.

The pairs (−7, 1) and (1, −7) check. 17.

y2 = 0

14.

x + y = −6,

(1) (2)

Substitute 2x − 3 for y in equation (1) and solve for x. y 2 − x2 = 9

(2x − 3)2 − x2 = 9

4x2 − 12x + 9 − x2 = 9

3x2 − 12x = 0 x2 − 4x = 0

x(x − 4) = 0

x = 0 or x = 4

Now substitute these numbers into equation (2) and solve for y. If x = 0, y = 2 · 0 − 3 = −3. If x = 4, y = 2 · 4 − 3 = 5.

The pairs (0, −3) and (4, 5) check. They are the solutions.

−2 = x

The pair (−2, 1) checks. It is the solution. 18. y = x2 , 3x = y + 2 y = 3x − 2 3x − 2 = x2

0 = x2 − 3x + 2

0 = (x − 2)(x − 1)

x = 2 or x = 1

y =3·2−2=4 y =3·1−2=1

The pairs (2, 4) and (1, 1) check.

598 19.

Chapter 9: Analytic Geometry Topics √ √ 6 21 6 21 Substituting or − for x in equation (1) gives 7 7 us 36 · 21 + y2 = 4 49 108 y2 = 4 − 7 80 y2 = − 7 & & 80 5 = ±4i y=± − 7 7 √ 4i 35 y=± . Rationalizing the 7 denominator ' 6√21 4i√35 ( , , The pairs 7 7 ' 6√21 4i√35 ( ' 6√21 4i√35 ( ,− , − , , and 7 7 7 7 √ √ ' 6 21 4i 35 ( − ,− check. They are the solutions. 7 7

x2 + y 2 = 25, (1) xy = 12

(2)

First we solve equation (2) for y. xy = 12 12 y= x

12 Then we substitute for y in equation (1) and solve for x x. x2 + y 2 = 25 ' 12 (2 = 25 x2 + x 144 x2 + 2 = 25 x x4 + 144 = 25x2 Multiplying by x2 x4 − 25x2 + 144 = 0

u2 − 25u + 144 = 0

Letting u = x2

(u − 9)(u − 16) = 0

u = 9 or u = 16

We now substitute x2 for u and solve for x. x2 = 9 or x2 = 16 x = ±3 or

22.

20.

x = ±4

x − y = 16, (1) x + y 2 = 4 (2) x2 + x = 20 Adding

(1)

25x + 16y = 400

(2)

2

−16x − 16y = −400 Multiplying (1) by −16 25x2 + 16y 2 = 400 9x2 = 0 Adding x= 0

x = −5 or x = 4 y 2 = 4 − x Solving equation (2) for y 2 y 2 = 4 − (−5) = 9 and y = ±3

y = 4 − 4 = 0 and y = 0 2

The pairs (0, 5) and (0, −5) check. 23.

x2 + 4y 2 = 25, (1)

21.

x + y = 4,

(1)

16x + 9y = 144

(2)

2

2

−9x2 − 9y 2 = −36 Multiplying (1) by −9 16x2 + 9y 2 = 144 7x2 = 108 Adding x2 =

108 7 &

& 108 3 = ±6 x=± 7 7 √ 6 21 Rationalizing the dex=± nominator 7

(2)

First solve equation (2) for x. x = −2y + 7

(3)

Then substitute −2y + 7 for x in equation (1) and solve for y. x2 + 4y 2 = 25

The pairs (−5, 3), (−5, −3), and (4, 0) check. 2

Substituting in (1)

y = ±5

x + 2y = 7

(x + 5)(x − 4) = 0

2

2

02 + y 2 = 25

2

x2 + x − 20 = 0

2

2

Since y = 12/x, if x = 3, y = 4; if x = −3, y = −4; if x = 4, y = 3; and if x = −4, y = −3. The pairs (3, 4), (−3, −4), (4, 3), and (−4, −3) check. They are the solutions. 2

x2 + y 2 = 25,

(−2y + 7)2 + 4y 2 = 25 4y 2 − 28y + 49 + 4y 2 = 25 8y 2 − 28y + 24 = 0

2y 2 − 7y + 6 = 0

(2y − 3)(y − 2) = 0

3 or y = 2 2 Now substitute these numbers in equation (3) and solve for x. 3 x = −2 · + 7 = 4 2 x = −2 · 2 + 7 = 3 ' 3( The pairs 4, and (3, 2) check, so they are the solutions. 2 y=

Exercise Set 9.4 24.

599 ' 9( 11 x=2 − +5= 8 4 x = 2(−2) + 5 = 1 ' 11 9 ( ,− and (1, −2) check. The pairs 4 8

y 2 − x2 = 16,

2x − y = 1 y = 2x − 1

(2x − 1)2 − x2 = 16

4x2 − 4x + 1 − x2 = 16

27.

3x − 4x − 15 = 0

x2 + y 2 = 16,

2

y 2 − 2x2 = 10

(3x + 5)(x − 3) = 0 5 x = − or x = 3 3 ' 5( 13 −1=− y=2 − 3 3 y = 2(3) − 1 = 5 ' 5 13 ( and (3, 5) check. The pairs − , − 3 3 25.

x2 − xy + 3y 2 = 27,

= 32 Multiplying (1) by 2 = 10 = 42 Adding = 14√ = ± 14 √ √ Substituting 14 or − 14 for y in equation (1) gives us x2 + 14 = 16 x2 = 2

√ x=± 2 √ √ √ √ √ √ The pairs (− 2, − 14), (− 2, 14), ( 2, − 14), and √ √ ( 2, 14) check. They are the solutions.

First solve equation (2) for y. (3)

Then substitute x − 2 for y in equation (1) and solve for x. x2 − xy + 3y 2 = 27

28.

x2 − x(x − 2) + 3(x − 2)2 = 27

x − x2 + 2x + 3x2 − 12x + 12 = 27 2

3x2 − 10x − 15 = 0 % −(−10) ± (−10)2 − 4(3)(−15) x= 2·3 √ √ 10 ± 280 10 ± 100 + 180 = x= 6 6 √ √ 10 ± 2 70 5 ± 70 x= = 6 3 Now substitute these numbers in equation (3) and solve for y. √ √ −1 + 70 5 + 70 −2= y= 3 3 √ √ 5 − 70 −1 − 70 −2= y= 3 3 ' 5 + √70 −1 + √70 ( The pairs , and 3 3 √ √ ' 5 − 70 −1 − 70 ( , check, so they are the solutions. 3 3 26. 2y 2 + xy + x2 = 7, x − 2y = 5 x = 2y + 5 2y 2 + (2y + 5)y + (2y + 5)2 = 7 2y 2 + 2y 2 + 5y + 4y 2 + 20y + 25 = 7 8y + 25y + 18 = 0 2

(8y + 9)(y + 2) = 0 y=−

9 or y = −2 8

(2)

2x2 + 2y 2 −2x2 + y 2 3y 2 y2 y

(2)

x−2=y

−2x2 + y 2 = 10

Here we use the elimination method.

(1)

x−y =2

or

x2 + y 2 = 16, (1)

x2 + y 2 = 14, (1) x2 − y 2 = 4 (2) 2x2 = 18 Adding x2 = 9 x = ±3

9 + y 2 = 14

Substituting in equation (1)

y2 = 5

√ y=± 5

√ √ √ The√pairs (−3, − 5), (−3, 5), (3, − 5), and (3, 5) check. 29.

x2 + y 2 = 5, (1) xy = 2 (2) First we solve equation (2) for y. xy = 2 2 y= x

2 Then we substitute for y in equation (1) and solve for x x. x2 + y 2 = 5 ' 2 (2 x2 + =5 x 4 x2 + 2 = 5 x x4 + 4 = 5x2 Multiplying by x2 x4 − 5x2 + 4 = 0

u2 − 5u + 4 = 0

(u − 4)(u − 1) = 0

u = 4 or u = 1

Letting u = x2

600

Chapter 9: Analytic Geometry Topics !

We now substitute x2 for u and solve for x. x2 = 4 or x2 = 1 x = ±2

x = ±1

30.

x = −4y + 7

x + y = 20,

2y 2 + (−4y + 7)y = 5 2y 2 − 4y 2 + 7y = 5

0 = 2y 2 − 7y + 5

x2 +

' 8 (2 x

x2 +

0 = (2y − 5)(y − 1)

= 20

5 y = or y = 1 2 '5( + 7 = −3 x = −4 2 x = −4(1) + 7 = 3 ' 5( and (3, 1) check. The pairs − 3, 2

64 = 20 x2

x4 + 64 = 20x2 x4 − 20x2 + 64 = 0

u2 − 20u + 64 = 0

Letting u = x2

(u − 16)(u − 4) = 0 u = 16

or

u=4

x2 = 16

or

x2 = 4

x = ±4

or

33.

3x + y = 7 4x2 + 5y = 56

(2)

First solve equation (1) for a.

x = ±2

a = −b + 7

(3)

Then substitute −b + 7 for a in equation (2) and solve for b. (−b + 7)b = 4

(1) (2)

−b2 + 7b = 4

0 = b2 − 7b + 4 % −(−7) ± (−7)2 − 4 · 1 · 4 b= 2·1 √ 7 ± 33 b= 2 Now substitute these numbers in equation (3) and solve for a. √ ' 7 + √33 ( 7 − 33 +7= a=− 2 2 √ ' 7 − √33 ( 7 + 33 +7= a=− 2 2 ' 7 − √33 7 + √33 ( The pairs , and 2 2 ' 7 + √33 7 − √33 ( , check, so they are the 2 2 solutions.

First solve equation (1) for y. 3x + y = 7 y = 7 − 3x (3)

Next substitute 7 − 3x for y in equation (2) and solve for x. 4x2 + 5y = 56 4x2 + 5(7 − 3x) = 56

4x2 + 35 − 15x = 56 4x2 − 15x − 21 = 0

Using the quadratic formula, we find that √ √ 15 + 561 15 − 561 or x = . x= 8 8 Now substitute these numbers into equation (3) and solve for y. √ √ ! " 15 − 561 15 − 561 , y =7−3 , or If x = 8 8 √ 11 + 3 561 . 8 √ √ ! " 15 + 561 15 + 561 , y =7−3 , or If x = 8 8 √ 11 − 3 561 . 8 √ √ " ! 15 − 561 11 + 3 561 , and The pairs 8 8

a + b = 7, (1) ab = 4

y = 8/x, so if x = 4, y = 2; if x = −4, y = −2; if x = 2, y = 4; if x = −2, y = −4. The pairs (4, 2), (−4, −2), (2, 4), and (−2, −4) check. 31.

√ " 561 11 − 3 561 , check and are the solutions. 8 8

4y + x = 7

2

xy = 8 8 y= x

√

32. 2y 2 + xy = 5,

Since y = 2/x, if x = 2, y = 1; if x = −2, y = −1; if x = 1, y = 2; and if x = −1, y = −2. The pairs (2, 1), (−2, −1), (1, 2), and (−1, −2) check. They are the solutions. 2

15 +

34.

p + q = −4,

pq = −5

p = −q − 4

(−q − 4)q = −5

−q 2 − 4q = −5

0 = q 2 + 4q − 5

0 = (q + 5)(q − 1)

Exercise Set 9.4

601

q = −5 or q = 1

37.

p = −(−5) − 4 = 1

x2 + y 2 = 13,

(1)

xy = 6

(2)

2x + 6y + 13 = 0 2x = −6y − 13 −6y − 13 x= 2 −6y − 13 for x in equation (1) and solve for y. Substitute 2 x2 + y 2 + 6y + 5 = 0 "2 ! −6y − 13 + y 2 + 6y + 5 = 0 2

6 for y in equation (1) and solve for Then we substitute x x. x2 + y 2 = 13 ' 6 (2 x2 + = 13 x 36 x2 + 2 = 13 x x + 36 = 13x 4

36y 2 + 156y + 169 + y 2 + 6y + 5 = 0 4 36y 2 + 156y + 169 + 4y 2 + 24y + 20 = 0

Multiplying by x

2

2

x4 − 13x2 + 36 = 0

u2 − 13u + 36 = 0

40y 2 + 180y + 189 = 0

Letting u = x2

Using the quadratic formula, we find that √ √ −45 ± 3 15 −45 ± 3 15 . Substitute for y in y= 20 20 −6y − 13 and solve for x. x= 2 √ −45 + 3 15 If y = , then 20 √ " ! −45 + 3 15 √ −6 − 13 5 − 9 15 20 = . x= 2 20 √ −45 − 3 15 If y = , then 20 √ " ! −45 − 3 15 √ −6 − 13 5 + 9 15 20 = . x= 2 20 √ √ " ! 5 + 9 15 −45 − 3 15 The pairs , and 20 20 √ √ " ! 5 − 9 15 −45 + 3 15 , check and are the solutions. 20 20

(u − 9)(u − 4) = 0 u=4

We now substitute x2 for u and solve for x. or x2 = 4 x2 = 9 or

x = ±2

Since y = 6/x, if x = 3, y = 2; if x = −3, y = −2; if x = 2, y = 3; and if x = −2, y = −3. The pairs (3, 2), (−3, −2), (2, 3), and (−2, −3) check. They are the solutions. 36.

x2 + 4y 2 = 20, xy = 4 y=

4 x

' 4 (2

x2 + 4

x

= 20

64 = 20 x2 x4 + 64 = 20x2

x2 +

x4 − 20x2 + 64 = 0

u2 − 20u + 64 = 0

38. Letting u = x2

(u − 16)(u − 4) = 0 u = 16 or

u=4

x2 = 16 or x2 = 4 x = ±4 or

(1) (2) (3)

Solve equation (3) for x.

xy = 6 6 y= x

x = ±3

(2)

x2 + y 2 + 6y + 5 = 0 −x2 − y 2 + 2x + 8 = 0 2x + 6y + 13 = 0

First we solve Equation (2) for y.

or

x + y − 2x − 8 = 0 2

Using the elimination method, multiply equation (2) by −1 and add the result to equation (1).

The pairs (1, −5) and (−5, 1) check.

u=9

(1)

2

p = −1 − 4 = −5

35.

x2 + y 2 + 6y + 5 = 0

x = ±2

y = 4/x, so if x = 4, y = 1; if x = −4, y = −1; if x = 2, y = 2; and if x = −2, y = −2. The pairs (4, 1), (−4, −1), (2, 2), and (−2, −2) check.

2xy + 3y 2 = 7,

(1)

3xy − 2y = 4

(2)

2

6xy + 9y −6xy + 4y 2 13y 2 y2 y 2

= = = = =

21 −8 13 1 ±1

Multiplying (1) by 3 Multiplying (2) by − 2

Substitute for y in equation (1) and solve for x. When y = 1 : 2 · x · 1 + 3 · 12 = 7 2x = 4 x =2

602

Chapter 9: Analytic Geometry Topics When y = −1 : 2 · x · (−1) + 3(−1)2 = 7 −2x = 4

The pairs (2, 1) and (−2, −1) check. 39.

2a + b = 1,

(1)

b = 4 − a2

(2)

x = −2

The pairs (−5, −8) and (8, 5) check. They are the solutions. 42.

x+y =5 x = −y + 5

(−y + 5)y = 4

Equation (2) is already solved for b. Substitute 4 − a2 for b in equation (1) and solve for a.

−y 2 + 5y = 4

0 = y 2 − 5y + 4

2a + 4 − a2 = 1

0 = (y − 4)(y − 1)

0 = a2 − 2a − 3

a=3

or

y = 4 or y = 1

0 = (a − 3)(a + 1)

x = −4 + 5 = 1

a = −1

x = −1 + 5 = 4

Substitute these numbers in equation (2) and solve for b. b = 4 − 32 = −5

The pairs (1, 4) and (4, 1) check. 43.

b = 4 − (−1)2 = 3

xy = 4,

The pairs (3, −5) and (−1, 3) check. They are the solutions.

xy − y 2 = 2, 2xy − 3y 2 = 0 −2xy + 2y 2 2xy − 3y 2 −y 2 y2 y

40. 4x2 + 9y 2 = 36, x + 3y = 3 x = −3y + 3

(1) (2)

= −4 Multiplying (1) by −2 = 0 = −4 Adding = 4 = ±2

We substitute for y in equation (1) and solve for x.

4(−3y + 3)2 + 9y 2 = 36

When y = 2 : x · 2 − 22 = 2

4(9y 2 − 18y + 9) + 9y 2 = 36

2x − 4 = 2

2x = 6

36y 2 − 72y + 36 + 9y 2 = 36

x=3

45y − 72y = 0 2

When y = −2 : x(−2) − (−2)2 = 2

5y − 8y = 0 2

y(5y − 8) = 0 8 y = 0 or y = 5 x = −3 · 0 + 3 = 3 '8( 9 +3=− x = −3 5 5 ' 9 8( check. The pairs (3, 0) and − , 5 5 41.

a2 + b2 = 89,

(1)

a−b=3

(2)

a=b+3

(3)

First solve equation (2) for a. Then substitute b + 3 for a in equation (1) and solve for b. (b + 3)2 + b2 = 89 b + 6b + 9 + b2 = 89 2

2b2 + 6b − 80 = 0 b2 + 3b − 40 = 0

(b + 8)(b − 5) = 0

b = −8 or b = 5

Substitute these numbers in equation (3) and solve for a. a = −8 + 3 = −5 a = 5+3=8

−2x − 4 = 2

−2x = 6

x = −3

The pairs (3, 2) and (−3, −2) check. They are the solutions. 44.

4a2 − 25b2 = 0,

2a2 − 10b2 = 3b + 4

(1) (2)

4a2 − 25b2 = 0

−4a2 + 20b2 = −6b − 8 Multiplying (2) by −2 −5b2 = −6b − 8

0 = 5b2 − 6b − 8

0 = (5b + 4)(b − 2)

4 or b = 2 b=− 5 Substitute for b in equation (1) and solve for a. ' 4 (2 4 4a2 − 25 − =0 When b = − : 5 5 4a2 = 16 a2 = 4 a = ±2

Exercise Set 9.4 When b = 2:

603 4a2 − 25(2)2 = 0

46.

4a2 = 100

ab − b2 −ab + 2b2 b2 b

a2 = 25 a = ±5 ' 4( 4( ' The pairs 2, − , − 2, − , (5, 2) and (−5, 2) check. 5 5 45.

m2 − 3mn + n2 + 1 = 0,

3m2 − mn + 3n2

= 13

m2 − 3mn + n2 = −1

3m − mn + 3n = 13 2

2

(3)

(1) (2) Rewriting (1)

(2)

−3m2 +9mn− 3n2 = 3 Multiplying (3) by −3

3m2 − mn+ 3n2 = 13 8mn = 16 mn = 2 2 (4) n= m 2 Substitute for n in equation (1) and solve for m. m ! " ! "2 2 2 m2 − 3m + +1 = 0 m m 4 m2 − 6 + 2 + 1 = 0 m 4 m2 − 5 + 2 = 0 m m4 − 5m2 + 4 = 0 Multiplying by m2

47.

m =4

or

x2 + y 2 = 5,

(1)

x−y =8

(2) (3)

y 2 + 16y + 64 + y 2 = 5 2y 2 + 16y + 59 = 0 % −16 ± (16)2 − 4(2)(59) y= 2·2 √ −16 ± −216 y= 4 √ −16 ± 6i 6 y= 4 3 √ y = −4 ± i 6 2 Now substitute these numbers in equation (3) and solve for x. 3 √ 3 √ x = −4 + i 6 + 8 = 4 + i 6 2 2 3 √ 3 √ x = −4 − i 6 + 8 = 4 − i 6 2 2 " ! 3 √ 3 √ The pairs 4 + i 6, −4 + i 6 and 2 2 ! " 3 √ 3 √ 4 − i 6, −4 − i 6 check. They are the solutions. 2 2

u=1

m = ±1

Substitute for m in equation (4) and solve for n. 2 When m = 2, n = = 1. 2 2 = −1. When m = −2, n = −2 2 When m = 1, n = = 2. 1 2 = −2. When m = −1, n = −1 The pairs (2, 1), (−2, −1), (1, 2), and (−1, −2) check. They are the solutions.

Multiplying (2) by −1

Then substitute y + 8 for x in equation (1) and solve for y. (y + 8)2 + y 2 = 5

or m2 = 1

m = ±2 or

= −4 = 6 = √2 =± 2

Substitute for b in equation (1) and solve for a. √ √ √ When b = 2 : a( 2) − ( 2)2 = −4 √ a 2 = −2 √ 2 a = −√ = − 2 2 √ √ √ When b = − 2 : a(− 2)−(− 2)2 = −4 √ −a 2 = −2 √ −2 a= √ = 2 − 2 √ √ √ √ The pairs (− 2, 2) and ( 2, − 2) check.

x=y+8

(u − 4)(u − 1) = 0 2

(1) (2)

First solve equation (2) for x.

Substitute u for m2 . u2 − 5u + 4 = 0 u=4

ab − b2 = −4, ab − 2b2 = −6

48.

4x2 + 9y 2 = 36, y−x=8 y =x+8 4x2 + 9(x + 8)2 = 36 4x2 + 9(x2 + 16x + 64) = 36 4x2 + 9x2 + 144x + 576 = 36 13x2 + 144x + 540 = 0

604

Chapter 9: Analytic Geometry Topics % (144)2 − 4(13)(540) 2 · 13 √ −72 ± 6i 51 72 6 √ x= = − ± i 51 13 13 13 6 √ 32 6 √ 72 + i 51 y = − + i 51 + 8 = 13 13 13 13 6 √ 32 6 √ 72 − i 51 y = − − i 51 + 8 = 13 13 13 13 ! " 72 6 √ 32 6 √ The pairs − + i 51, + i 51 and 13 13 13 13 ! " 6 √ 32 6 √ 72 − i 51, − i 51 check. − 13 13 13 13 x=

49.

a + b = 14, √ ab = 3 5 2

√ √ √ When x = −i 3 : (−i 3)2 + (−i 3)y = 5 √ −i 3y = 8

−144 ±

2

8 y=− √ i 3 √ 8i 3 y=− 3 √ " √ " ! ! √ √ 8i 3 8i 3 and − i 3, . The pairs i 3, − 3 3 51.

Solve equation (2) for b. √ 3 5 b= a √ 3 5 for b in equation (1) and solve for a. Substitute a √ "2 ! 3 5 a2 + = 14 a 45 a2 + 2 = 14 a a4 + 45 = 14a2 Letting u = a2

(u − 9)(u − 5) = 0 u=9

a2 = 9

or

u=5

or a2 = 5

√ a = ±3 or a = ± 5 √ √ √ Since b √ = 3 5/a, if a = 3, b = √5; if a = −3, b = − 5; if a√= 5, b =√3; and √ if a = − √ 5, b = −3. The pairs (3, 5), (−3, − 5), ( 5, 3), (− 5, −3) check. They are the solutions. 50.

x2 + xy = 5,

(1)

2x2 + xy = 2

(2)

−x − xy = −5 Multiplying (1) by −1 2 2x2 + xy = x2 = √ −3 x = ±i 3

(2) Multiplying (1) by −4

9x2 + 4y 2 = 36 5x2 = −64 64 2 x =− 5 & −64 8i = ±√ x=± 5 5 √ 8i 5 x=± Rationalizing the 5 denominator √ √ 8i 5 8i 5 or − for x in equation (1) and Substituting 5 5 solving for y gives us 64 − + y 2 = 25 5 189 y2 = 5 & & 189 21 y=± = ±3 5 5 √ 3 105 y=± . Rationalizing the 5 denominator √ √ √ ! √ " ! " 8i 5 3 105 8i 5 3 105 , , − , , The pairs 5 5 5 5 √ √ √ √ " ! " ! 8i 5 3 105 8i 5 3 105 ,− , and − ,− 5 5 5 5 check.

(2)

u2 − 14u + 45 = 0

(1)

9x2 + 4y 2 = 36

−4x2 − 4y 2 = −100

(1)

a4 − 14a2 + 45 = 0

x2 + y 2 = 25,

They are the solutions.

2

Substitute for x in equation (1) and solve for y. √ √ √ When x = i 3 : (i 3)2 + (i 3)y = 5 √ i 3y = 8 8 y= √ i 3√ 8i 3 y=− 3

52.

x2 + y 2 = 1,

(1)

9x2 − 16y 2 = 144

(2)

16x2 + 16y 2 = 16 Multiplying (1) by 16 9x2 − 16y 2 = 144 25x2 = 160 160 x2 = 25 √ 4 10 x=± 5 Substituting for x in equation (1) and solving for y gives us

Exercise Set 9.4

605

160 + y2 = 1 25

54. 135 25 &

xy = 1

y2 = −

y=± −

y=

135 25

√ 3i 15 y=± . 5 √ " ! √ √ " ! √ 4 10 3i 15 4 10 3i 15 , , The pairs , ,− 5 5 5 5 √ √ " √ √ " ! ! 4 10 3i 15 4 10 3i 15 − , , and − ,− 5 5 5 5 check. 53.

5y 2 − x2 = 1, xy = 2

(1)

5y 4 − y 2 − 4 = 0

5u − u − 4 = 0 2

4 5 4 2 y =− 5 2i y = ±√ 5 √ 2i 5 y=± 5

Letting u = y

or

u=1

or

y =1

u2 − 6u − 7 = 0

2

y = ±1

Letting u = x2

(u − 7)(u + 1) = 0 or

u = −1

or x2 = −1 √ x = ± 7 or x = ±i √ √ √ √ Since y =√1/x, if x√= 7, y = 1/ 7 = 7/7; if x = − 7, y = 1/(− 7) = − 7/7; if x = i, y = 1/i = −i, if x = −i, y = 1/(−i) = i. √ " ! √ " ! √ √ 7 7 7, , − 7, − , (i, −i), and (−i, i) The pairs 7 7 check. 55. Familiarize. We first make a drawing. We let l and w represent the length and width, respectively. !

! 10!! ! !

2

y = ±1 √ 2 5 2i 5 ,x= √ = √ = Since x = 2/y, if y = 5 2i 5 i 5 √ √5 √ −i 5 2i 5 5 √ · √ = −i 5; if y = − , 5 i 5 −i 5 √ 2 √ = i 5; x= 2i 5 − 5 if y = 1, x = 2/1 = 2; if y = −1, x = 2/ − 1 = −2. √ " ! √ " ! √ √ 2i 5 2i 5 , i 5, − , (2, 1) and The pairs − i 5, 5 5 (−2, −1) check. They are the solutions. or

x4 − 6x2 − 7 = 0

x =7

or u − 1 = 0

or

! "2 1 =6 x 7 x2 − 2 = 6 x x4 − 7 = 6x2

x2 − 7

2

(5u + 4)(u − 1) = 0 u=−

1 x

u=7

(2)

Solve equation (2) for x. 2 x= y 2 Substitute for x in equation (1) and solve for y. y ! "2 2 =1 5y 2 − y 4 5y 2 − 2 = 1 y 5y 4 − 4 = y 2

5u + 4 = 0

x2 − 7y 2 = 6,

w

!

!

l Translate. The perimeter is 28 cm. 2l + 2w = 28, or l + w = 14 Using the Pythagorean theorem we have another equation. l2 + w2 = 102 , or l2 + w2 = 100 Carry out. We solve the system: l + w = 14, (1) l2 + w2 = 100

(2)

First solve equation (1) for w. w = 14 − l

(3)

Then substitute 14 − l for w in equation (2) and solve for l. l2 + w2 = 100 l2 + (14 − l)2 = 100

l2 + 196 − 28l + l2 = 100 2l2 − 28l + 96 = 0 l2 − 14l + 48 = 0

(l − 8)(l − 6) = 0

l = 8 or l = 6

If l = 8, then w = 14 − 8, or 6. If l = 6, then w = 14 − 6, or 8. Since the length is usually considered to be longer

606

Chapter 9: Analytic Geometry Topics than the width, we have the solution l = 8 and w = 6, or (8, 6).

59. Familiarize. We first make a drawing. Let l = the length and w = the width.

Check. If l = 8 and w = 6, then the√perimeter is 2·8+2·6, √ or 28. The length of a diagonal is 82 + 62 , or 100, or 10. The numbers check.

!

! 2 !! ! !

State. The length is 8 cm, and the width is 6 cm. 56. Let l and w represent the length and width, respectively. Solve the system:

!

!

l Translate.

2l + 2w = 6, √ l2 + w2 = ( 5)2 , or

w

Area: lw =

√

3

(1)

From the Pythagorean theorem: l2 + w2 = 22

l+w = 3

(2)

Carry out. We solve the system of equations.

l2 + w 2 = 5 The solutions are (1, 2) and (2, 1). Choosing the larger number as the length, we have the solution. The length is 2 m, and the width is 1 m. 57. Familiarize. We first make a drawing. Let l = the length and w = the width of the brochure.

w

l Translate. Area: lw = 20 Perimeter: 2l + 2w = 18, or l + w = 9

We first solve equation (1) for w. √ lw = 3 √ 3 w= l √ 3 for w in equation 2 and solve for Then we substitute l l. ' √ 3 (2 2 =4 l + l 3 l2 + 2 = 4 l l4 + 3 = 4l2 l4 − 4l2 + 3 = 0

u2 − 4u + 3 = 0

Letting u = l2

(u − 3)(u − 1) = 0

Carry out. We solve the system:

u = 3 or u = 1

Solve the second equation for l: l = 9 − w

We now substitute l2 for u and solve for l. or l2 = 1 l2 = 3 √ l = ± 3 or l = ±1

Substitute 9 − w for l in the first equation and solve for w. (9 − w)w = 20

9w − w2 = 20

0 = w − 9w + 20 2

0 = (w − 5)(w − 4)

w = 5 or w = 4

If w = 5, then l = 9 − w, or 4. If w = 4, then l = 9 − 4, or 5. Since length is usually considered to be longer than width, we have the solution l = 5 and w = 4, or (5, 4). Check. If l = 5 and w = 4, the area is 5 · 4, or 20. The perimeter is 2 · 5 + 2 · 4, or 18. The numbers check.

State. The length of the brochure is 5 in. and the width is 4 in.

58. Let l and w represent the length and width, respectively. We solve the system: lw = 2, 2l + 2w = 6 The solutions are (1, 2) and (2, 1). We choose the larger number to be the length, so the length is 2 yd and the width is 1 yd.

Measurements cannot be negative, so √ we only need √ to con√ = 1. Since w = 3/l, if l = 3, w = 1 sider l = 3 and l√ and if l = 1, w = 3. Length is usually considered √ to be longer than√width, so we have the solution l = 3 and w = 1, or ( 3, 1). √ √ √ Check. √ If2 l =2 3 and w = 1, 2the area is 3 · 1 = 3. Also ( 3) + 1 = 3 + 1 = 4 = 2 . The numbers check. √ State. The length is 3 m, and the width is 1 m. 60. Let l = the length and w = the width. Solve the system √ lw = 2, √ l2 + w2 = ( 3)2 . √ √ √ The solutions are ( 2, 1), (− √ 2, −1), (1, √2), and √ (−1, − 2). Only the pairs ( 2, 1) and (1, 2) have meaning in this problem. Since length is √ usually considered to be longer than width, the length is 2 m, and the width is 1 m.

Exercise Set 9.4

607

61. Familiarize. We make a drawing of the dog run. Let l = the length and w = the width.

Translate. The sum of the areas is 832 ft2 . ) *+ , )*+, ) *+ , ↓ ↓ ↓ = 832 x2 + y 2

w

The difference of the areas is 320 ft2 . *+ , )*+, ) *+ , ) ↓ ↓ ↓ = 320 x2 − y 2

l

Carry out. We solve the system of equations.

Since it takes 210 yd of fencing to enclose the run, we know that the perimeter is 210 yd.

x2 + y 2 = 832 x2 − y 2 = 320

Translate. Perimeter: 2l + 2w = 210, or l + w = 105

2x2

Area: lw = 2250

= 1152

Adding

x = 576 2

Carry out. We solve the system:

x = ±24

Solve the first equation for l: l = 105 − w

Since measurements cannot be negative, we consider only x = 24. Substitute 24 for x in the first equation and solve for y.

Substitute 105 − w for l in the second equation and solve for w. (105 − w)w = 2250

242 + y 2 = 832

105w − w2 = 2250

576 + y 2 = 832

0 = w − 105w + 2250 2

y 2 = 256

0 = (w − 30)(w − 75)

y = ±16

w = 30 or w = 75

Again, we consider only the positive value, 16. The possible solution is (24, 16).

If w = 30, then l = 105 − 30, or 75. If w = 75, then l = 105 − 75, or 30. Since length is usually considered to be longer than width, we have the solution l = 75 and w = 30, or (75, 30).

Check. The areas of the test plots are 242 , or 576, and 162 , or 256. The sum of the areas is 576 + 256, or 832. The difference of the areas is 576 − 256, or 320. The values check.

Check. If l = 75 and w = 30, the perimeter is 2·75+2·30, or 210. The area is 75(30), or 2250. The numbers check. State. The length is 75 yd and the width is 30 yd. 62. Let l and w represent the length and width, respectively. Solve the system: √ l2 + w2 = l + 1, √ l2 + w2 = 2w + 3 The solutions are (12, 5) and (0, −1). Only (12, 5) has meaning in this problem. It checks. The length is 12 ft and the width is 5 ft. 63. Familiarize. We let x = the length of a side of one test plot and y = the length of a side of the other plot. Make a drawing.

State. The lengths of the test plots are 24 ft and 16 ft. 64. Let p = the principal and r = the interest rate. Solve the system: pr = 7.5, (p + 25)(r − 0.01) = 7.5

The solutions are (125, 0.06) and (−1.50, −0.05). Only (125, 0.06) has meaning in this problem. The principal was $125 and the interest rate was 0.06, or 6%. 65. The correct graph is (b). 66. The correct graph is (e). 67. The correct graph is (d). 68. The correct graph is (f).

x

x Area: x2

y y Area: y 2

69. The correct graph is (a). 70. The correct graph is (c). 71.

Graph: x2 + y 2 ≤ 16, y<x

The solution set of x2 + y 2 ≤ 16 is the circle x2 + y 2 = 16 and the region inside it. The solution set of y < x is the half-plane below the line y = x. We shade the region common to the two solution sets.

608

Chapter 9: Analytic Geometry Topics 74.

y 5 3 2 1 !5

(!2!2,

!3 !2 !1 !1 !2 !3 !2!2)

Graph: x ≥ y 2 ,

x−y ≤2

(2!2, 2!2)

y 1 2 3

x

5

5 4 3 2 1

!5

!5 !4 !3 !2 !1 !1 !2

To find the points of intersection of the graphs we solve the system of equations

!3 !4

y = x.

√ The √ points √ of intersection are (−2 2, −2 2) and (2 2, 2 2). 72.

√

75.

Graph: x2 + y 2 ≤ 25, The solution set of x2 + y 2 ≤ 25 is the circle x2 + y 2 = 25 and the region inside it. The solution set of x − y > 5 is the half-plane below the line x − y = 5. We shade the region common to the two solution sets.

Graph: x + y ≤ 10, y>x

y

y

5 4

(!!5,

4 3 2 1

(!5, !5)

2 1 !2 !1 !1 !2 !!5)

x

x−y >5

2

!5 !4

1 2 3 4 5

(1, !1)

!5

x2 + y 2 = 16,

2

(4, 2)

1 2

4 5

x

(5, 0) 1 2 3 4

!4 !3 !2 !1 !1 !2 !3 !4

!4 !5

x

(0, !5)

73.

Graph: x ≤ y, 2

To find the points of intersection of the graphs we solve the system of equations

x+y ≥2

x2 + y 2 = 25,

The solution set of x2 ≤ y is the parabola x2 = y and the region inside it. The solution set of x + y ≥ 2 is the line x + y = 2 and the half-plane above the line. We shade the region common to the two solution sets. 76.

y (!2, 4)

5 4 3 2 1

!5 !4 !3 !2 !1 !1 !2 !3 !4 !5

x − y = 5.

The points of intersection are (0, −5) and (5, 0). Graph: x2 + y 2 ≥ 9, x−y >3 y

(1, 1) 1 2 3 4 5

5 4

x

!5 !4

To find the points of intersection of the graphs we solve the system of equations x2 = y, x + y = 2. The points of intersection are (−2, 4) and (1, 1).

(3, 0)

2 1 !2 !1 !1 !2 !4 !5

77.

1 2

4 5

x

(0, !3)

Graph: y ≥ x2 − 3, y ≤ 2x

The solution set of y ≥ x2 − 3 is the parabola y = x2 − 3 and the region inside it. The solution set of y ≤ 2x is the line y = 2x and the half-plane below it. We shade the region common to the two solution sets.

Exercise Set 9.4

609 80.

y 7 6 5 4 3 2 1

Graph: y ≤ 1 − x2 , y >x−1

(3, 6)

y

!5 !4 !3 !2 !1

1 2 3 4 5

5 4 3 2

x

(!1, !2) !2

(1, 0) !5 !4 !3 !2

To find the points of intersection of the graphs we solve the system of equations y = x2 − 3,

y = 2x.

The points of intersection are (−1, −2) and (3, 6). 78.

Graph: y ≤ 3 − x2 , y ≥x+1 5 4 3 2 1 !5 ! 4 !3

79.

1

x

3 4 5

Graph: y ≥ x2 ,

y <x+2

The solution set of y ≥ x2 is the parabola y = x2 and the region inside it. The solution set of y < x + 2 is the half-plane below the line y = x + 2. We shade the region common to the two solution sets. y

!5 !4 !3 !2 !1 !1 !2 !3 !4 !5

x

!3 !4 !5

81. Find the points of intersection of y1 = ln x+2 and y2 = x2 . They are (1.564, 2.448) and (0.138, 0.019). √ √ 82. Graph y1 = ln(x + 4), y2 = 6 − x2 , and y3 = − 6 − x2 and find the points of intersection. They are (1.720, 1.744) and (−2.405, 0.467).

84. Find the point of intersection of y1 = e−x + 1 and y2 = 2x + 5. It is (−0.841, 3.318).

(1, 2)

!1 !1 !2 !3 !4 !5

5 4 3 2 (!1, 1) 1

1 2 3 4 5

83. Find the point of intersection of y1 = ex − 1 and y2 = −3x + 4. It is (0.871, 1.388).

y

(!2, !1)

(!2, !3)

!1 !2

(2, 4)

1 2 3 4 5

x

To find the points of intersection of the graphs we solve the system of equations y = x2 , y = x + 2. The points of intersection are (−1, 1) and (2, 4).

85. Find the points of intersection of y1 = ex and y2 = x + 2. They are (1.146, 3.146) and (−1.841, 0.159). 86. Find the points of intersection of y1 = e−x and y2 = 3 − x. They are (2.948, 0.052) and (−1.505, 4.505). % 87. Graph y1 = 19, 380, 510.36 − x2 , % y2 = − 19, 380, 510.36 − x2 , and

y3 = 27, 941.25x/6.125 and find the points of intersection. They are (0.965, 4402.33) and (−0.965, −4402.33).

88. Find the points of intersection of y = (1660 − 2x)/2 (or y = 830 − x) and y = 35, 325/x. They are (785, 45) and (45, 785). & 14.5x2 − 64.5 , 89. Graph y1 = 13.5 & 14.5x2 − 64.5 y2 = − , and y3 = (5.5x − 12.3)/6.3 and 13.5 find the points of intersection. They are (2.112, −0.109) and (−13.041, −13.337).

90. Find the points of intersection of y = −15.6/(13.5x) and y2 = (5.6x − 42.3)/6.7. They are (7.366, −0.157) and (0.188, −6.157). & 56, 548 − 0.319x2 , 91. Graph y1 = 2688.7 & 56, 548 − 0.319x2 , y2 = − 2688.7 & 0.306x2 − 43, 452 y3 = , 2688.7 & 0.306x2 − 43, 452 and y4 = − and find the points of in2688.7 tersection. They are (400, 1.431), (−400, 1.431), (400, −1.431), and (−400, −1.431).

610

Chapter 9: Analytic Geometry Topics &

6408 − 18.465x2 , 92. Graph y1 = 788.723 & & 6408 − 18.465x2 106.535x2 − 2692 y2 = , y3 = , 788.723 788.723 & 106.535x2 − 2692 and find the points of interand y4 = 788.723 section. They are (8.532, 2.534), (8.532, −2.534), (−8.532, 2.534), and (−8.532, −2.534). 93. Equations of parabolas that open up or down can be expressed using function notation. The graphs of the other conic sections studied in the chapter fail the vertical-line test, so they are not functions. 94. Answers will vary. 95.

3x = 6 x=2 The solution is 2.

3h − k = 3

we find that (h, k) = (2, 3). Find r2 , substituting (2, 3) for (h, k) and (2, 4) for (x, y). We could also use (3, 3) for (x, y). (x − h)2 + (y − k)2 = r2 (2 − 2)2 + (4 − 3)2 = r2 0 + 1 = r2 1 = r2 The equation of the circle is (x − 2)2 + (y − 3)2 = 1.

h2 +4h+4+k 2 −6k+9 = h2 +8h+16+k 2 −2k+1 4h − 6k + 13 = 8h − 2k + 17

x ln 5 = ln 27 ln 27 x= ln 5 x ≈ 2.048

−4h − 4k = 4

h + k = −1

We get a second equation by substituting (h, k) in 5x + 8y = −2.

log3 x = 4

5h + 8k = −2

x = 34

We now solve the following system:

x = 81

h + k = −1,

The solution is 81. 98.

h − k = −1,

(h + 2)2 + (k − 3)2 = (h + 4)2 + (k − 1)2

5x = 27 ln 5x = ln 27

97.

Solving the system

100. Let (h, k) represent the point on the line 5x + 8y = −2 which is the center of a circle that passes through the points (−2, 3) and (−4, 1). The distance between (h, k) and (−2, 3) is the same as the distance between (h, k) and (−4, 1). This gives us one equation: % % [h−(−2)]2 +(k−3)2 = [h−(−4)]2 +(k−1)2

23x = 64 23x = 26

96.

If the center (h, k) is on the line 3x−y = 3, then 3h−k = 3.

5h + 8k = −2

log(x − 3) + log x = 1

The solution, which is the center of the circle, is (−2, 1).

log(x − 3)(x) = 1

Next we find the length of the radius. We can find the distance between either (−2, 3) or (−4, 1) and the center (−2, 1). We use (−2, 3). % r = [−2 − (−2)]2 + (1 − 3)2 % r = 02 + (−2)2 √ r= 4=2

x2 − 3x = 10

x2 − 3x − 10 = 0

(x − 5)(x + 2) = 0 x = 5 or x = −2 Only 5 checks.

99. (x − h)2 + (y − k)2 = r2

If (2, 4) is a point on the circle, then (2 − h)2 + (4 − k)2 = r2 .

If (3, 3) is a point on the circle, then (3 − h)2 + (3 − k)2 = r2 .

We can write the equation of the circle with center (−2, 1) and radius 2. (x − h)2 + (y − k)2 = r2 [x − (−2)]2 + (y − 1)2 = 22 (x + 2)2 + (y − 1)2 = 4

Thus

(2 − h)2 + (4 − k)2 = (3 − h)2 + (3 − k)2

4 − 4h + h2 + 16 − 8k + k 2 =

9 − 6h + h2 + 9 − 6k + k 2

−4h − 8k + 20 = −6h − 6k + 18 2h − 2k = −2 h − k = −1

101. The equation of the ellipse of the " form ! " ! is √ √ 1 y2 3 x2 and for (x, y) + = 1. Substitute 1, 3, a2 b2 2 2 to get two equations.

Exercise Set 9.4 ! √ "2 3 12 2 + = 1, or a2 b2 ! "2 1 √ ( 3)2 2 + = 1, or a2 b2 1 Substitute u for 2 and v a 3 u + v = 1, 4 or 1 3u + v = 1 4

611

1 3 + 2 =1 a2 4b 3 1 + 2 =1 a2 4b 1 for 2 . b 4u + 3v = 4, 12u + v = 4

1 Solving for u and v, we get u = , v = 1. Then 4 1 1 1 u = 2 = , so a2 = 4; v = 2 = 1, so b2 = 1. a 4 b Then the equation of the ellipse is y2 x2 x2 + = 1, or + y 2 = 1. 4 1 4 102.

2

2

y x − 2 =1 a2 b Substitute each ordered pair for (x, y). ! √ "2 3 5 − (−3)2 2 − = 1, a2 b2 ! √ "2 3 5 (−3)2 b2 − = 1, 2 a b2 "2 ! 3 − 02 2 − 2 =1 2 a b 45 9 − 2 = 1, (1) a2 4b 9 45 − 2 = 1, (2) a2 4b 9 =1 (3) 4a2 Note that equation (1) and equation (2) are identical. Multiply both sides of equation (3) by 4: 9 =4 a2 9 Substitute 4 for 2 in equation (1) and solve for b2 . a 45 4− 2 = 1 4b 16b2 − 45 = 4b2

12b2 = 45 45 15 , or b2 = 12 4 Solve equation (3) for a2 . 9 =1 4a2 9 = a2 4

The equation of the hyperbola is 103. (x − h)2 + (y − k)2 = r2

x2 y2 − = 1. 9/4 15/4

Standard form

Substitute (4, 6), (−6, 2), and (1, −3) for (x, y). (4 − h)2 + (6 − k)2 = r2

(1)

2

(2)

(1 − h)2 + (−3 − k)2 = r2

(3)

(−6 − h) + (2 − k) = r 2

2

Thus

(4 − h)2 + (6 − k)2 = (−6 − h)2 + (2 − k)2 , or

5h + 2k = 3 and

(4 − h)2 + (6 − k)2 = (1 − h)2 + (−3 − k)2 , or h + 3k = 7.

We solve the system 5h + 2k = 3, h + 3k = 7. 32 5 and k = . Substituting these 13 13 5365 . values in equation (1), (2), or (3), we find that r2 = 169 The equation of the circle is ! "2 ! "2 32 5365 5 x+ + y− = . 13 13 169 Solving we get h = −

104. Using (x − h)2 + (y − k)2 = r2 and the given points, we have (2 − h)2 + (3 − k)2 = r2 (1) (4 − h)2 + (5 − k)2 = r2

(0 − h)2 + (−3 − k)2 = r2

(2) (3)

Then equation (1) − equation (2) gives h + k = 7 and equation (2) − equation (3) gives h + 2k = 4. We solve this system: h + k = 7, h + 2k = 4. Then h = 10, k = −3, r = 10 and the equation of the circle is (x−10)2 +[y−(−3)]2 = 102 , or (x−10)2 +(y+3)2 = 100. 105. See the answer section in the text. 106. Let x and y represent the numbers. Solve: xy = 2, 1 33 1 + = . x y 8 The solutions are numbers are

!

" ! " 1 1 , 8 and 8, . In either case the 4 4

1 and 8. 4

107. Familiarize. Let x and y represent the numbers. Translate. The square of a certain number exceeds twice the square 1 of another number by . 8 1 2 2 x = 2y + 8

612

Chapter 9: Analytic Geometry Topics

The sum of the squares is

5 . 16

We let x and y represent the length and width of the base of the box, respectively. Then the dimensions of the metal sheet are x + 10 and y + 10.

5 16 Carry out. We solve the system. 1 (1) x2 − 2y 2 = , 8 5 (2) x2 + y 2 = 16 x2 + y 2 =

x2 − 2y 2 = 2x + 2y = 2

3x2

2

= x2 = x=

Solve the system (x + 10)(y + 10) = 340, x · y · 5 = 350.

The solutions are (10, 7) and (7, 10). The dimensions of the box are 10 in. by 7 in. by 5 in. 109. See the answer section in the text.

1 , 8 5 8 6 8 1 4 1 ± 2

110. Multiplying (2) by 2

x−y =a−b

(3)

Substitute for x in equation (1) and solve for y. (y + a − b)2 − y 2 = a2 − b2

y 2 + 2ay − 2by + a2 − 2ab + b2 − y 2 = a2 − b2

2ay − 2by = 2ab − 2b2 2y(a − b) = 2b(a − b) y=b

Substitute for y in equation (3) and solve for x. x=b+a−b=a

The pair (a, b) checks. 111.

"

5 5

x y

y + 10

y 5

x 5

5 x + 10

$

!

x3 + y 3 = 72,

(1)

x+y =6

(2)

Solve equation (2) for y: y = 6 − x

Substitute for y in equation (1) and solve for x. x3 + (6 − x)3 = 72

x + 216 − 108x + 18x2 − x3 = 72 3

18x2 − 108x + 144 = 0 x2 − 6x + 8 = 0

Multiplying by

1 18

(x − 4)(x − 2) = 0

x = 4 or x = 2

If x = 4, then y = 6 − 4 = 2. If x = 2, then y = 6 − 2 = 4.

The pairs (4, 2) and (2, 4) check.

112.

5

#

(2)

x=y+a−b

108. Make a drawing.

5

(1)

Solve equation (2) for x.

1 Substitute ± for x in (2) and solve for y. 2 ! "2 5 1 + y2 = ± 2 16 5 1 + y2 = 4 16 1 y2 = 16 1 y=± 4 ! " ! " ! " 1 1 1 1 1 1 We get , , − , , ,− and 2" 4 2 4 2 4 ! 1 1 − ,− . 2 4 "2 "2 ! ! 1 1 exceeds twice ± Check. It is true that ± 2 4 ! " 1 1 1 1 by : = 2 + 8 4 16 8 ! "2 ! "2 1 1 5 . The pairs check. Also ± + ± = 2 4 16 1 1 1 1 1 State. The numbers are and or − and or and 2 4 2 4 2 1 1 1 − or − and − . 4 2 4

5

x2 − y 2 = a2 − b2 ,

5 , 6 a b 13 + = b a 6 a+ b=

b=

5 − 6a 5 −a= 6 6

(1) (2) Solving equation (1) for b

Exercise Set 9.4

613

5 − 6a a 13 6 + = Substituting for b 5 − 6a a 6 in equation (2) 6 5 − 6a 13 6a + = 5 − 6a 6a 6

(x − 1)2 + (4 − x2 ) = 4

x − 2x + 1 + 4 − x2 = 4 2

−2x = −1 1 x= 2

1 for x in equation (3) and solve for y. 2 ! "2 1 y2 = 4 − 2 1 y2 = 4 − 4 15 y2 = 4 √ 15 y=± 2 √ " ! √ " ! 15 15 1 1 The pairs , and ,− check. They are 2 2 2 2 the solutions.

36a2 + 25 − 60a + 36a2 = 65a − 78a2

Substitute

150a2 − 125a + 25 = 0 6a2 − 5a + 1 = 0

(3a − 1)(2a − 1) = 0

1 1 or a = 3 2 Substitute for a and solve for b. #1$ 5−6 1 3 = 1. When a = , b = 3 6 2 #1$ 5−6 1 2 = 1. When a = , b = 2 6 3 #1 1$ #1 1$ The pairs , and , check. They are the solu3 2 2 3 tions. a=

113.

115.

32x−y = 1000 (x + y) log 5 = 2, Taking logarithms and

p2 + q 2 = 13, (1) 1 1 =− (2) pq 6 Solve equation (2) for p. p 1 =− q 6 6 − =p q Substitute −6/q for p in equation (1) and solve for q. # 6 $2 + q 2 = 13 − q 36 + q 2 = 13 q2 36 + q 4 = 13q 2 q 4 − 13q 2 + 36 = 0

u2 − 13u + 36 = 0

(2x − y) log 3 = 3

x log 5 + y log 5 = 2, (1)

or

u=4

x =9

or

x2 = 4

x = ±3 or

2x log 3 · log 5 − y log 3 · log 5 = 3 log 5 3x log 3 · log 5 = 2 log 3 + 3 log 5

2 log 3 + 3 log 5 3 log 3 · log 5 Substitute in (1) to find y. 2 log 3 + 3 log 5 · log 5 + y log 5 = 2 3 log 3 · log 5 2 log 3 + 3 log 5 y log 5 = 2 − 3 log 3 6 log 3 − 2 log 3 − 3 log 5 y log 5 = 3 log 3 4 log 3 − 3 log 5 y log 5 = 3 log 3 4 log 3 − 3 log 5 y= 3 log 3 · log 5 " ! 2 log 3 + 3 log 5 4 log 3 − 3 log 5 , checks. It is The pair 3 log 3 · log 5 3 log 3 · log 5 the solution. x=

x = ±2

Since p = −6/q, if q = 3, p = −2; if q = −3, p = 2; if q = 2, p = −3; and if q = −2, p = 3. The pairs (−2, 3), (2, −3), (−3, 2), and (3, −2) check. They are the solutions. 114.

x2 + y 2 = 4,

(1)

(x − 1) + y = 4

(2)

2

2

Solve equation (1) for y 2 . y =4−x 2

2

(3)

Substitute for 4 − x2 for y 2 in equation (2) and solve for x.

(2)

Multiply equation (1) by log 3 and equation (2) by log 5 and add. x log 3 · log 5 + y log 3 · log 5 = 2 log 3

(u − 9)(u − 4) = 0 2

simplifying

2x log 3 − y log 3 = 3

Letting u = q 2

u=9

5x+y = 100,

116.

ex − ex+y = 0, y

x−y

e −e

(1)

=0

(2)

Factor (1): e (1 − ey ) = 0 ex = 0

No solution

x

or

1 − ey = 0 y= 0

614

Chapter 9: Analytic Geometry Topics y " = −x sin θ + y cos θ √ = −0 · sin 60◦ + 3 cos 60◦ √ √ 1 3 = 0+ 3· = 2 2 ! √ " 3 3 , . The coordinates are 2 2

Substitute in (2). e0 − ex−0 = 0 1 − ex = 0 x=0

The solution is (0, 0).

5. We use the rotation of axes formulas to find x and y.

Exercise Set 9.5

x = x" cos θ − y " sin θ

= 1 · cos 45◦ − (−1) sin 45◦ √ √ 2 2 = + 2 2 √ 2 2 √ = = 2 2

1. We use the rotation of axes formulas to find x" and y " . x" = x cos θ + y sin θ √ √ = 2 cos 45◦ − 2 sin 45◦ √ √ √ 2 √ 2 = 2· − 2· 2 2 = 1−1=0 y " = −x sin θ + y cos θ √ √ = − 2 sin 45◦ − 2 cos 45◦ √ √ √ 2 √ 2 = − 2· − 2· 2 2 = −1 − 1 = −2

y = x" sin θ + y " cos θ = 1 · sin 45◦ − 1 · cos 45◦ √ √ 2 2 = − =0 2 2 √ The coordinates are ( 2, 0). 6.

The coordinates are (0, −2). 2.

x" = x cos θ + y sin θ = −1 · cos 45◦ + 3 sin 45◦ √ √ 2 2 =− +3· 2 2 √ 2 2 √ = = 2 2

y = x" sin θ + y " cos θ √ √ = −3 2 sin 45◦ + 2 cos 45◦ √ √ √ 2 √ 2 = −3 2 · + 2· 2 2 = −3 + 1 = −2

y " = −x sin θ + y cos θ

= −(−1) sin 45◦ + 3 cos 45◦ √ √ 2 2 = +3· 2 2 √ √ 4 2 = =2 2 2 √ √ The coordinates are ( 2, 2 2).

The coordinates are (−4, −2).

7. We use the rotation of axes formulas to find x and y. x = x" cos θ − y " sin θ

= 2 cos 30◦ − 0 · sin 30◦ √ √ 3 = 2· −0= 3 2

3. We use the rotation of axes formulas to find x" and y " . x" = x cos θ + y sin θ

y = x" sin θ + y " cos θ

= 0 · cos 30 + 2 sin 30 1 = 0+2· =1 2 ◦

◦

y " = −x sin θ + y cos θ

= −0 · sin 30◦ + 2 cos 30◦ √ 3 √ = 0+2· = 3 2 √ The coordinates are (1, 3). 4.

x" = x cos θ + y sin θ √ = 0 · cos 60◦ + 3 sin 60◦ √ √ 3 3 = = 0+ 3· 2 2

x = x" cos θ − y " sin θ √ √ = −3 2 cos 45◦ − 2 sin 45◦ √ √ √ 2 √ 2 = −3 2 · − 2· 2 2 = −3 − 1 = −4

= 2 sin 30◦ + 0 · cos 30◦ 1 = 2· +0=1 2 √ The coordinates are ( 3, 1). 8.

x = x" cos θ − y " sin θ

√ = −1 · cos 60◦ − (− 3) sin 60◦ √ 1 √ 3 = − + 3· 2 2 1 3 =− + =1 2 2

Exercise Set 9.5

615

y = x" sin θ + y " cos θ √ = −1 · sin 60◦ − 3 cos 60◦ √ 3 √ 1 =− − 3· 2 2 √ √ 2 3 =− 3 =− 2 √ The coordinates are (1, − 3). 9. 3x2 − 5xy + 3y 2 − 2x + 7y = 0 A = 3, B = −5, C = 3

B 2 − 4AC = (−5)2 − 4 · 3 · 3 = 25 − 36 = −11

Since the discriminant is negative, the graph is an ellipse (or circle). 10. 5x2 + 6xy − 4y 2 + x − 3y + 4 = 0

B 2 − 4AC = 62 − 4 · 5 · (−4) = 36 + 80 = 116

Since the discriminant is positive, the graph is a hyperbola.

11. x2 − 3xy − 2y 2 + 12 = 0

A = 1, B = −3, C = −2

B − 4AC = (−3) − 4 · 1 · (−2) = 9 + 8 = 17 2

2

Since the discriminant is positive, the graph is a hyperbola. 12. 4x2 + 7xy + 2y 2 − 3x + y = 0

B 2 − 4AC = 72 − 4 · 4 · 2 = 49 − 32 = 17

Since the discriminant is positive, the graph is a hyperbola.

13. 4x2 − 12xy + 9y 2 − 3x + y = 0 A = 4, B = −12, C = 9

B 2 − 4AC = (−12)2 − 4 · 4 · 9 = 144 − 144 = 0

Since the discriminant is zero, the graph is a parabola. 14. 6x2 + 5xy + 6y 2 + 15 = 0

B 2 − 4AC = 52 − 4 · 6 · 6 = 25 − 144 = −119

18. x2 + xy − y 2 − 4x + 3y − 2 = 0

B 2 − 4AC = 12 − 4 · 1 · (−1) = 1 + 4 = 5

Since the discriminant is positive, the graph is a hyperbola. 19. 3x2 + 2xy + 3y 2 = 16 A = 3, B = 2, C = 3 B 2 − 4AC = 22 − 4 · 3 · 3 = 4 − 36 = −32

Since the discriminant is negative, the graph is an ellipse (or circle). To rotate the axes we first determine θ. 3−3 A−C = =0 cot 2θ = B 2 ◦ ◦ Then 2θ = 90 and θ = 45 , so √ √ 2 2 and cos θ = . sin θ = 2 2 Now substitute in the rotation of axes formulas. x = x" cos θ − y " sin θ !√ " !√ " √ 2 2 2 " " " −y = (x − y " ) =x 2 2 2 y = x" sin θ + y " cos θ !√ " !√ " √ 2 2 2 " = x" + y" = (x + y " ) 2 2 2 Substitute for x and y in the given equation. &2 %√ &% √ & %√ 2 " 2 " 2 " (x − y " ) + 2 (x − y " ) (x + y " ) + 3 2 2 2 %√ &2 2 " 3 (x + y " ) = 16 2 After simplifying we have (y " )2 (x" )2 + = 1. 4 8 √ This √is the equation √ of an ellipse√with vertices" (0, − 8) and (0, 8), or (0, −2 2) and (0, 2 2) on the y -axis. The x" -intercepts are (−2, 0) and (2,0). We sketch the graph.

Since the discriminant is negative, the graph is an ellipse (or circle).

15. 2x2 − 8xy + 7y 2 + x − 2y + 1 = 0 A = 2, B = −8, C = 7

B 2 − 4AC = (−8)2 − 4 · 2 · 7 = 64 − 56 = 8

Since the discriminant is positive, the graph is a hyperbola.

y y#

(x#)2 (y#)2 $ %1 4 8

4

x#

2

45" 2

!4 !2

4

x

!2 !4

16. x2 + 6xy + 9y 2 − 3x + 4y = 0

B 2 − 4AC = 62 − 4 · 1 · 9 = 36 − 36 = 0

Since the discriminant is zero, the graph is a parabola.

17. 8x2 − 7xy + 5y 2 − 17 = 0 A = 8, B = −7, C = 5

B 2 − 4AC = (−7)2 − 4 · 8 · 5 = 49 − 160 = −111

Since the discriminant is negative, the graph is an ellipse (or circle).

20. 3x2 + 10xy + 3y 2 + 8 = 0 B 2 − 4AC = 102 − 4 · 3 · 3 = 100 − 36 = 64

Since the discriminant is positive, the graph is a hyperbola. To rotate the axes we first determine θ. 3−3 A−C cot 2θ = = =0 B 10 ◦ ◦ Then 2θ = 90 and θ = 45 , so √ √ 2 2 and cos θ = . sin θ = 2 2 Now substitute in the rotation of axes formulas.

616

Chapter 9: Analytic Geometry Topics x = x" cos θ − y " sin θ !√ " !√ " √ 2 2 2 " − y" = (x − y " ) = x" 2 2 2 y = x" sin θ + y " cos θ !√ " !√ " √ 2 2 2 " " " =x +y = (x + y " ) 2 2 2 After substituting for x and y in the given equation and simplifying, we have (x" )2 (y " )2 − = 1. 4 1 y y#

!

4

1

%1

45" 2

!4 !2

(x#)2

x#

2

4

y#

x# 45"

2 2

!4 !2

4

x

!2 !4

(x#)2 (y#)2 ! %1 9 6

√ √ 22. x2 + 2xy + y 2 + 4 2x − 4 2y = 0

B 2 − 4AC = 22 − 4 · 1 · 1 = 4 − 4 = 0

(y#)2 4

y

4

x

!2 !4

21. x2 − 10xy + y 2 + 36 = 0

A = 1, B = −10, C = 1

B 2 − 4AC = (−10)2 − 4 · 1 · 1 = 100 − 4 = 96

Since the discriminant is positive, the graph is a hyperbola. To rotate the axes we first determine θ. 1−1 A−C = =0 cot 2θ = B −10 ◦ ◦ Then 2θ = 90 and θ = 45 , so √ √ 2 2 and cos θ = . sin θ = 2 2 Now substitute in the rotation of axes formulas. x = x" cos θ − y " sin θ !√ " !√ " √ 2 2 2 " " " (x − y " ) −y = =x 2 2 2 y = x" sin θ + y " cos θ !√ " !√ " √ 2 2 2 " + y" = (x + y " ) = x" 2 2 2 Substitute for x and y in the given equation. &2 %√ &% √ & %√ 2 " 2 " 2 " " " " (x − y ) − 10 (x − y ) (x + y ) + 2 2 2 %√ &2 2 " (x + y " ) + 36 = 0 2 After simplifying we have (y " )2 (x" )2 − = 1. 9 6 This is the equation of a hyperbola with vertices (−3, √ 0) 6 " " " and (3, 0) on the x -axis. The asymptotes are y = − x 3 √ 6 " x . We sketch the graph. and y " = 3

Since the discriminant is zero, the graph is a parabola. To rotate the axes we first determine θ. 1−1 A−C = =0 cot 2θ = B 2 Then 2θ = 90◦ and θ = 45◦ , so √ √ 2 2 and cos θ = . sin θ = 2 2 Now substitute in the rotation of axes formulas. x = x" cos θ − y " sin θ !√ " !√ " √ 2 2 2 " " " =x −y = (x − y " ) 2 2 2 y = x" sin θ + y " cos θ !√ " !√ " √ 2 2 2 " (x + y " ) = x" + y" = 2 2 2 After substituting for x and y in the given equation and simplifying, we have (x" )2 = 4y " . y y#

(x#)2 % 4y#

4

x#

2

45" 2

!4 !2

4

x

!2 !4

√ √ 23. x2 − 2 3xy + 3y 2 − 12 3x − 12y = 0 √ A = 1, B = −2 3, C = 3 √ B 2 − 4AC = (−2 3)2 − 4 · 1 · 3 = 12 − 12 = 0

Since the discriminant is zero, the graph is a parabola. To rotate the axes we first determine θ. 1−3 −2 1 A−C √ = √ =√ = cot 2θ = B −2 3 −2 3 3 Then 2θ = 60◦ and θ = 30◦ , so √ 3 1 . sin θ = and cos θ = 2 2 Now substitute in the rotation of axes formulas. x = x" cos θ − y " sin θ √ √ x" 3 y " 3 1 − y" · = − = x" · 2 2 2 2

Exercise Set 9.5

617

y = x" sin θ + y " cos θ √ √ x" y" 3 1 3 = + = x" · + y " · 2 2 2 2 Substitute for x and y in the given equation. √ " ! "√ "2 ! √ "! " √ x" 3 y " x 3 y" x y" 3 − −2 3 − + + 2 2 2 2 2 2 √ "2 ! " ! √ " √ x" 3 y " x y" 3 − 12 3 3 + − − 2 2 2 2 √ " ! " y" 3 x + =0 12 2 2 After simplifying we have (y " )2 = 6x" . This is the equation of a parabola with vertex at (0, 0) of the x" y " -coordinate system and axis of symmetry y " = 0. We sketch the graph. y#

y (y#)2 % 6x#

4 2

30" 2

!4 !2

4

x# x

!2 !4

√ 24. 13x2 + 6 3xy + 7y 2 − 16 = 0 √ B 2 − 4AC = (6 3)2 − 4 · 13 · 7 = 108 − 364 = −256

Since the discriminant is negative, the graph is an ellipse or a circle. To rotate the axes we first determine θ. 6 1 13 − 7 A−C = √ = √ =√ cot 2θ = B 6 3 6 3 3 Then 2θ = 60◦ and θ = 30◦ , so √ 1 3 . sin θ = and cos θ = 2 2 Now substitute in the rotation of axes formulas. x = x" cos θ − y " sin θ √ √ 3 x" 3 y " " 1 " −y · = − =x · 2 2 2 2

y = x sin θ + y cos θ √ √ 3 x" y" 3 " " 1 = x · +y · = + 2 2 2 2 After substituting for x and y in the given equation and simplifying, we have "

"

(y " )2 (x" )2 + = 1. 1 4 y y#

4

(x#)2 (y#)2 $ %1 1 4 x# 30"

2 2

!4 !2 !2 !4

4

x

√ 25. 7x2 + 6 3xy + 13y 2 − 32 = 0 √ A = 7, B = 6 3, C = 13 √ B 2 − 4AC = (6 3)2 − 4 · 7 · 13 = 108 − 364 = −256

Since the discriminant is negative, the graph is an ellipse or a circle. To rotate the axes we first determine θ. 7 − 13 −6 1 A−C = √ = √ = −√ cot 2θ = B 6 3 6 3 3 Then 2θ = 120◦ and θ = 60◦ , so √ 1 3 and cos θ = . sin θ = 2 2 Now substitute in the rotation of axes formulas. x = x" cos θ − y " sin θ √ √ x" y" 3 1 3 = − = x" · − y " · 2 2 2 2 y = x" sin θ + y " cos θ √ √ 1 3 x" 3 y " = x" · + y" · = + 2 2 2 2 Substitute for x and y in the given equation. √ "2 √ "! " √ ! " ! " √ x" y" 3 y" 3 x 3 y" x +6 3 − − + + 7 2 2 2 2 2 2 "2 ! "√ x 3 y" + − 32 = 0 13 2 2 After simplifying we have

(y " )2 (x" )2 + = 1. 2 8 √ This √is the equation √ of an ellipse√with vertices" (0, − 8) and (0, 8), or (0, −2 2)√and (0, 2 2)√on the y -axis. The x" -intercepts are (− 2, 0) and ( 2, 0). We sketch the graph. y x#

4

y#

60"

2 2

!4 !2

4

x

!2 !4

(x#)2 2

$

(y#)2 %1 8

26. x2 + 4xy + y 2 − 9 = 0

B 2 − 4AC = 42 − 4 · 1 · 1 = 16 − 4 = 12

Since the discriminant is positive, the graph is a hyperbola. To rotate the axes we first determine θ. 1−1 A−C cot 2θ = = =0 B 4 Then 2θ = 90◦ and θ = 45◦ , so √ √ 2 2 and cos θ = . sin θ = 2 2 Now substitute in the rotation of axes formulas. x = x" cos θ − y " sin θ !√ " !√ " √ 2 2 2 " − y" = (x − y " ) = x" 2 2 2

618

Chapter 9: Analytic Geometry Topics y = x" sin θ + y " cos θ !√ " !√ " √ 2 2 2 " + y" = (x + y " ) = x" 2 2 2 After substituting for x and y in the given equation and simplifying, we have

y y#

2

y#

4

x

28. 5x2 − 8xy + 5y 2 = 8

B 2 − 4AC = (−8)2 − 4 · 5 · 5 = 64 − 100 = −36

x#

Since the discriminant is negative, the graph is an ellipse or a circle. To rotate the axes we first determine θ. A−C 5−5 cot 2θ = = =0 B −8 ◦ ◦ Then 2θ = 90 and θ = 45 , so √ √ 2 2 sin θ = and cos θ = . 2 2 Now substitute in the rotation of axes formulas. x = x" cos θ − y " sin θ !√ " !√ " √ 2 2 2 " " " −y = (x − y " ) =x 2 2 2

45" 2

!4 !2

4

!4

(x#)2 (y#)2 ! %1 3 9

2

x#

!2

" 2

y

30" 2

!4 !2

(x ) (y ) − = 1. 3 9 " 2

4

(x#)2 (y#)2 ! %1 2 8

4

x

!2 !4

√ 27. 11x2 + 10 3xy + y 2 = 32 √ A = 11, B = 10 3, C = 1 √ B 2 − 4AC = (10 3)2 − 4 · 11 · 1 = 300 − 44 = 256

Since the discriminant is positive, the graph is a hyperbola. To rotate the axes we first determine θ. 11 − 1 10 1 A−C √ = √ =√ = cot 2θ = B 10 3 10 3 3 Then 2θ = 60◦ and θ = 30◦ , so √ 3 1 . sin θ = and cos θ = 2 2 Now substitute in the rotation of axes formulas. x = x" cos θ − y " sin θ √ √ 3 x" 3 y " " 1 " =x · −y · = − 2 2 2 2 y = x" sin θ + y " cos θ √ √ x" y" 3 1 3 = + = x" · + y " · 2 2 2 2 Substitute for x and y in the given equation. √ " "2 "! " ! "√ ! √ √ x" 3 y " x y" 3 x 3 y" − − + + + 10 3 11 2 2 2 2 2 2 √ "2 ! " x y" 3 + = 32 2 2 After simplifying we have

(y " )2 (x" )2 − = 1. 2 8 √ This is√the equation of a hyperbola with vertices (− 2, 0) " " and √ x -axis. The asymptotes are y = √ ( 2, 0) on the 8 " 8 − √ x and y " = √ x" , or y " = −2x" and y " = 2x" . We 2 2 sketch the graph.

y = x" sin θ + y " cos θ !√ " !√ " √ 2 2 2 " (x + y " ) + y" = = x" 2 2 2 After substituting for x and y in the given equation and simplifying, we have (y " )2 (x" )2 + = 1. 81 9 y y#

8

x# 45"

4 4

!8 !4

8

x

!4 !8

29.

(x#)2 (y#)2 $ %1 81 9

√ √ 2x2 + 2 2xy + 2y 2 − 8x + 8y = 0 √ √ √ A = 2, B = 2 2, C = 2 √ √ √ B 2 − 4AC = (2 2)2 − 4 · 2 · 2 = 8 − 8 = 0 √

Since the discriminant is zero, the graph is a parabola. To rotate the axes we first determine θ. √ √ 2− 2 A−C √ = cot 2θ = =0 B 2 2 Then 2θ = 90◦ and θ = 45◦ , so √ √ 2 2 and cos θ = . sin θ = 2 2 Now substitute in the rotation of axes formulas. x = x" cos θ − y " sin θ !√ " !√ " √ 2 2 2 " " " −y = (x − y " ) =x 2 2 2

Exercise Set 9.5

619

y = x" sin θ + y " cos θ !√ " !√ " √ 2 2 2 " + y" = (x + y " ) = x" 2 2 2 Substitute for x and y in the given equation. %√ &2 %√ &% √ & √ √ 2 " 2 " 2 " 2 (x − y " ) + 2 2 (x − y " ) (x + y " ) + 2 2 2 √ √ %√ &2 √ 2 " " 2 " " 2 " " 2 (x +y ) −8· (x −y )+8· (x +y ) = 0 2 2 2 After simplifying we have 1 y " = − (x" )2 . 4 This is the equation of a parabola with vertex at (0, 0) of the x" y " -coordinate system and axis of symmetry x" = 0. We sketch the graph. y# % !~(x#)2

y 4

y#

x#

2

45" 2

!4 !2

4

x

!2 !4

√ √ 30. x2 + 2 3xy + 3y 2 − 8x + 8 3y = 0 √ B 2 − 4AC = (2 3)2 − 4 · 1 · 3 = 12 − 12 = 0

Since the discriminant is zero, the graph is a parabola. To rotate the axes we first determine θ. 1−3 −2 1 A−C cot 2θ = = √ = √ = −√ B 2 3 2 3 3 Then 2θ = 120◦ and θ = 60◦ , so √ 3 1 and cos θ = . sin θ = 2 2 Now substitute in the rotation of axes formulas. x = x" cos θ − y " sin θ √ √ 1 3 x" y" 3 = − = x" · − y " · 2 2 2 2 y = x" sin θ + y " cos θ √ √ 3 1 x" 3 y " = x" · + y" · = + 2 2 2 2 After substituting for x and y in the given equation and simplifying, we have √ (x" + 1)2 = −2 3y " + 1. y

60"

2 2

!4 !2

4

!2 !4

(x# $ 1)2 % !2!3 y# $ 1

Since the discriminant is positive, the graph is a hyperbola. To rotate the axes we first determine θ. 6 1 1 − (−5) A−C √ = √ =√ = cot 2θ = B 6 3 6 3 3 Then 2θ = 60◦ and θ = 30◦ , so √ 1 3 sin θ = and cos θ = . 2 2 Now substitute in the rotation of axes formulas. x = x" cos θ − y " sin θ √ √ 3 1 x" 3 y " = x" · − y" · = − 2 2 2 2 y = x" sin θ + y " cos θ √ √ 1 3 x" y" 3 = x" · + y " · = + 2 2 2 2 Substitute for x and y in the given equation. √ " "2 ! √ "! " ! "√ √ x" 3 y " x y" 3 x 3 y" − +6 3 − + − 2 2 2 2 2 2 √ √ " " ! " ! 2 x y" 3 x" 3 y " 5 + +8 − − 2 2 2 2 √ ! " " √ x y" 3 8 3 + − 48 = 0 2 2 After simplifying we have

(y " + 1)2 (x" )2 − = 1. 10 5 √ This is the equation of a hyperbola with vertices (− 10, 0) √ √ 5 and ( 10, 0) and asymptotes y " +1 = − √ x" and y " +1 = 10 √ 5 1 1 √ x" , or y " + 1 = − √ x" and y " + 1 = √ x" We sketch 10 2 2 the graph. y y#

4 2

30" 2

!4 !2 !2 !4

(x#)2 10

4

!

x# x

(y# $ 1)2 %1 5

√ √ 32. 3x2 − 2xy + 3y 2 − 6 2x + 2 2y − 26 = 0

B 2 − 4AC = (−2)2 − 4 · 3 · 3 = 4 − 36 = −32

x#

4

y#

√ √ 31. x2 + 6 3xy − 5y 2 + 8x − 8 3y − 48 = 0 √ A = 1, B = 6 3, C = −5 √ B 2 − 4AC = (6 3)2 − 4 · 1 · (−5) = 108 + 20 = 128

x

Since the discriminant is negative, the graph is an ellipse or a circle. To rotate the axes we first determine θ. 3−3 A−C = =0 cot 2θ = B −2 Then 2θ = 90◦ and θ = 45◦ , so √ √ 2 2 sin θ = and cos θ = . 2 2 Now substitute in the rotation of axes formulas.

620

Chapter 9: Analytic Geometry Topics x = x! cos θ − y ! sin θ !√ " !√ " √ 2 2 2 ! − y! = (x − y ! ) = x! 2 2 2 y = x! sin θ + y ! cos θ !√ " !√ " √ 2 2 2 ! ! ! =x +y = (x + y ! ) 2 2 2 After substituting for x and y in the given equation and simplifying, we have (y ! + 1)2 (x! − 1)2 + = 1. 16 8 y y"

(x" ! 1)2 (y"# 1)2 # $1 16 8

4

x"

2

45% 2

!4 !2

4

x

!2 !4

33. x2 + xy + y 2 = 24 A = 1, B = 1, C = 1 B 2 − 4AC = 12 − 4 · 1 · 1 = 1 − 4 = −3

Since the discriminant is negative, the graph is an ellipse or a circle. To rotate the axes we first determine θ. 1−1 A−C = =0 cot 2θ = B 1 ◦ ◦ Then 2θ = 90 and θ = 45 , so √ √ 2 2 and cos θ = . sin θ = 2 2 Now substitute in the rotation of axes formulas. x = x! cos θ − y ! sin θ !√ " !√ " √ 2 2 2 ! (x − y ! ) − y! = = x! 2 2 2 y = x! sin θ + y ! cos θ !√ " !√ " √ 2 2 2 ! = x! + y! = (x + y ! ) 2 2 2 Substitute for x and y in the given equation. $2 # √ $# √ $ #√ 2 ! 2 ! 2 ! (x − y ! ) + (x − y ! ) (x + y ! ) + 2 2 2 #√ $2 2 ! (x + y ! ) = 24 2 After simplifying we have (y ! )2 (x! )2 + = 1. 16 48 √ This is √ the equation of √ an ellipse with √ vertices! (0, − 48) and (0, 48, or (0, −4 3) and (0, 4 3) on the y -axis. The x! -intercepts are (−4, 0) and (4,0). We sketch the graph.

y 8

y"

x"

4

45% 4

!8 !4

8

x

!4 !8

(x")2 (y")2 # $1 16 48

√ 34. 4x2 + 3 3xy + y 2 = 55 √ B 2 − 4AC = (3 3)2 − 4 · 4 · 1 = 27 − 16 = 11

Since the discriminant is positive, the graph is a hyperbola. To rotate the axes we first determine θ. 4−1 3 1 A−C = √ = √ =√ cot 2θ = B 3 3 3 3 3 Then 2θ = 60◦ and θ = 30◦ , so √ 1 3 sin θ = and cos θ = . 2 2 Now substitute in the rotation of axes formulas. x = x! cos θ − y ! sin θ √ √ 3 1 x! 3 y ! = x! · − y! · = − 2 2 2 2 y = x! sin θ + y ! cos θ √ √ 3 x! y! 3 ! ! 1 = + = x · +y · 2 2 2 2 After substituting for x and y in the given equation and simplifying, we have (y ! )2 (x! )2 − = 1. 10 110 y y"

4

x" 30%

2 2

!4 !2

4

x

!2

(x")2 (y")2 ! $1 10 110

!4

√ √ 35. 4x2 − 4xy + y 2 − 8 5x − 16 5y = 0 A = 4, B = −4, C = 1

B 2 − 4AC = (−4)2 − 4 · 4 · 1 = 16 − 16 = 0

Since the discriminant is zero, the graph is a parabola. To rotate the axes we first determine θ. A−C 4−1 3 = =− cot 2θ = B −4 4 Since cot 2θ < 0, we have 90◦ < 2θ < 180◦ . We make a sketch.

y

5

4

2 -3

x

Exercise Set 9.5

621

3 From the sketch we see that cos 2θ = − . Using half-angle 5 formulas, we have & " ! ' '1 − − 3 % 1 − cos 2θ ( 2 5 = =√ sin θ = 2 2 5 and & ! " ' '1 + − 3 % ( 1 + cos 2θ 1 5 = =√ . cos θ = 2 2 5 Now substitute in the rotation of axes formulas. x = x! cos θ − y ! sin θ 1 2 x! 2y ! = x! · √ − y ! · √ = √ − √ 5 5 5 5 y = x! sin θ + y ! cos θ 2 1 2x! y! = x! · √ + y ! · √ = √ + √ 5 5 5 5 Substitute for x and y in the given equation. "2 ! ! "! ! " ! ! 2y ! x 2y ! 2x y! x √ +√ + −4 √ − √ 4 √ −√ 5 5 5 5 5 5 " " ! ! ! 2 ! ! ! √ y 2y 2x x √ +√ −8 5 √ − √ − 5 5 5 5 ! " ! √ 2x! y =0 16 5 √ + √ 5 5 After simplifying we have (y ! )2 = 8x! . This is the equation of a parabola with vertex (0, 0) of the x! y ! -coordinate system and axis of symmetry y ! = 0. 2 Since we know that sin θ = √ and 0◦ < θ < 90◦ , we can 5 use a calculator to find that θ ≈ 63.4◦ . Thus, the xy-axes are rotated through an angle of about 63.4◦ to obtain the x! y ! -axes. We sketch the graph.

y

25 24 2 x

7

7 From the sketch we see that cos 2θ = . Then 25 & ' 7 % ' 1 − cos 2θ ( 1 − 25 3 = = sin θ = 2 2 5 and & ' 7 % ' 1 + cos 2θ ( 1 + 25 4 = = . cos θ = 2 2 5 Now substitute in the rotation of axes formulas. 4 3 x = x! cos θ − y ! sin θ = x! − y ! 5 5 3 ! 4 ! ! ! y = x sin θ + y cos θ = x + y 5 5 After substituting for x and y in the given equation and simplifying, we have (y ! )2 = 20x! . 3 Since we know that sin θ = and 0◦ < θ < 90◦ , we can 5 use a calculator to find that θ ≈ 36.9◦ . y y"

(y")2 $ 20x" x"

4 2

36.9% 2

!4 !2

4

x

!2 !4

y x"

37. 11x2 + 7xy − 13y 2 = 621

4

2

!4 !2

A = 11, B = 7, C = −13

63.4%

2

y"

4

x

!2 !4

(y")2 $ 8x"

36. 9x2 − 24xy + 16y 2 − 400x − 300y = 0

B 2 − 4AC = (−24)2 − 4 · 9 · 16 = 576 − 576 = 0

Since the discriminant is zero, the graph is a parabola. To rotate the axes we first determine θ. 9 − 16 7 A−C = = cot 2θ = B −24 24 Since cot 2θ > 0, we have 0◦ < 2θ < 90◦ . We make a sketch.

B 2 − 4AC = 72 − 4 · 11 · (−13) = 49 + 572 = 621

Since the discriminant is positive, the graph is a hyperbola. To rotate the axes we first determine θ. 11 − (−13) 24 A−C = = cot 2θ = B 7 7 Since cot 2θ > 0, we have 0◦ < 2θ < 90◦ . We make a sketch.

y 25 2 24

7 x

24 . Using half-angle From the sketch we see that cos 2θ = 25 formulas, we have

622

Chapter 9: Analytic Geometry Topics

sin θ =

%

and

& ' 24 ' 1 − cos 2θ ( 1 − 25 1 = =√ 2 2 50

& ' 24 ' 1 + cos 2θ ( 1 + 25 7 = =√ . cos θ = 2 2 50 Now substitute in the rotation of axes formulas. x = x! cos θ − y ! sin θ 7x! y! 7 1 = x! · √ − y ! · √ = √ − √ 50 50 50 50

y

y"

(y ! )2 (x! )2 − = 1. 54 46 √ This is√the equation of√a hyperbola with √ vertices (− !54, 0) and ( 54, 0), or (−3 6, 0)%and (3 6, 0) on % the x -axis. 23 ! 23 ! ! ! The asymptotes are y = − x and y = x . Since 27 27 1 and 0◦ < θ < 90◦ , we can we know that sin θ = √ 50 use a calculator to find that θ ≈ 8.1◦ . Thus, the xy-axes are rotated through an angle of about 8.1◦ to obtain the x! y ! -axes. We sketch the graph. y" y 8 4

8.1% 4

!8 !4

x" 8

x

!4 !8

(x")2 (y")2 ! $1 54 46

38. 3x2 + 4xy + 6y 2 = 28 B 2 − 4AC = 42 − 4 · 3 · 6 = 16 − 72 = −56

Since the discriminant is negative, the graph is an ellipse or a circle. To rotate the axes we first determine θ. 3−6 3 A−C = =− cot 2θ = B 4 4 Proceeding as we did in Exercise 35, we find that 2y ! 2x! y! x! x = √ − √ , y ! = √ + √ , and θ ≈ 63.4◦ . 5 5 5 5 After substituting for x and y in the given equation and simplifying, we have (y ! )2 (x! )2 + = 1. 4 14

63.4%

2 2

!4 !2

4

x

!2

%

y = x! sin θ + y ! cos θ 1 7 x! 7y ! = x! · √ + y ! · √ = √ + √ 50 50 50 50 Substitute for x and y in the given equation. "2 ! ! "! ! " ! ! y! y! x 7y ! 7x 7x √ +√ +7 √ − √ − 11 √ − √ 50 50 50 50 50 50 "2 ! ! 7y ! x = 621 13 √ + √ 50 50 After simplifying we have

x"

4

(x")2 (y")2 # $1 4 14

!4

39. The procedure for rotation of axes would be done first when B $= 0. Then you would proceed as when B = 0. 40. Answers will vary. π radians 41. 120◦ = 120◦ · 180◦ ◦ 120 = π radians 180◦ 2π radians = 3 π radians 42. −315◦ = −315◦ · 180◦ −315◦ = π radians 180◦ 7π radians =− 4 π 180◦ π 43. radians = radians · 3 3 π radians π = · 180◦ 3π 1 = · 180◦ = 60◦ 3 3π 180◦ 3π 44. radians = radians · 4 4 π radians 3π ◦ = · 180 4π 3 = · 180◦ = 135◦ 4 45.

x! = x cos θ + y sin θ, y ! = y cos θ − x sin θ

First rewrite the system. x cos θ + y sin θ = x! , −x sin θ + y cos θ = y

!

(1) (2)

Multiply Equation (1) by sin θ and Equation (2) by cos θ and add to eliminate x. x sin θ cos θ + y sin2 θ = x! sin θ −x sin θ cos θ + y cos2 θ = y ! cos θ

y(sin2 θ + cos2 θ) = x! sin θ + y ! cos θ y = x! sin θ + y ! cos θ

Substitute x sin θ + y cos θ for y in Equation (1) and solve for x. !

!

Exercise Set 9.6

623

x cos θ + (x! sin θ + y ! cos θ) sin θ = x!

48.

x2 + y 2 = r2

x cos θ + x sin θ + y sin θ cos θ = x !

2

!

!

(x! cos θ − y ! sin θ)2 + (x! sin θ + y ! cos θ)2 = r2

x cos θ = x − x sin θ−

(x! )2 cos2 θ − 2x! y ! cos θ sin θ + (y ! )2 sin2 θ+

x cos θ = x (1 − sin θ)−

(x! )2 (cos2 θ + sin2 θ)+

!

2

!

y ! sin θ cos θ

(x! )2 sin2 θ + 2x! y ! sin θ cos θ + (y ! )2 cos2 θ = r2

2

!

y ! sin θ cos θ

x! y ! (−2 sin θ cos θ + 2 sin θ cos θ)+

x cos θ = x cos θ− !

2

(y ! )2 (sin2 θ + cos2 θ) = r2

y ! sin θ cos θ

(x ) · 1 + x! y ! · 0 + (y ! )2 · 1 = r2 ! 2

x = x cos θ−y sin θ !

!

Thus, we have x = x cos θ − y sin θ and y = x sin θ + y ! cos θ. !

!

46. A(x! cos θ−y ! sin θ)2+B(x! cos θ−y ! sin θ)(x! sin θ+y ! cos θ)+ C(x! sin θ+y ! cos θ)2 +D(x! cos θ−y ! sin θ)+ E(x! sin θ+y ! cos θ)+F = 0 A(x! )2 cos2 θ − 2Ax! y ! cos θ sin θ + A(y ! )2 sin2 θ + B(x! )2 cos θ sin θ + Bx! y ! (cos2 θ − sin2 θ) − B(y ! )2 sin θ cos θ + C(x! )2 sin2 θ + 2Cx! y ! sin θ cos θ + C(y ! )2 cos2 θ + Dx! cos θ − Dy ! sin θ + Ex! sin θ + Ey ! cos θ + F = 0 (x! )2 (A cos2 θ + B cos θ sin θ + C sin2 θ) + x! y ! (−2A cos θ sin θ + B(cos2 θ − sin2 θ) + 2C sin θ cos θ) + (y ! )2 (A sin2 θ − B sin θ cos θ + C cos2 θ) + x! (D cos θ + E sin θ) + y ! (−D sin θ + E cos θ) + F = 0 Thus we have an equation of the form A! (x! )2 + B ! x! y ! + C ! (y ! )2 + D! x! + E ! y ! + F ! = 0, where

1. Graph (b) is the graph of r =

3 . 1 + cos θ

2. Graph (e) is the graph of r =

4 . 1 + 2 sin θ

3. Graph (a) is the graph of r =

8 . 4 − 2 cos θ

4. Graph (f) is the graph of r =

12 . 4 + 6 sin θ

5. Graph (d) is the graph of r =

5 . 3 − 3 sin θ

6. Graph (c) is the graph of r = 7. r =

2(C − A) sin θ cos θ + B(cos θ − sin θ), 2

C ! = A sin2 θ − B sin θ cos θ + C cos2 θ, D! = D cos θ + E sin θ,

E ! = −D sin θ + E cos θ, and

F! = F.

A! + C ! = (A cos2 θ + B sin θ cos θ + C sin2 θ)+ (A sin2 θ − B sin θ cos θ + C cos2 θ)

1 1 + cos θ

b) Since e = 1 and ep = 1 · p = 1, we have p = 1. Thus the parabola has a vertical directrix 1 unit to the right of the pole. c) We find the vertex by letting θ = 0. When θ = 0, 1 1 1 = = . r= 1 + cos 0 1+1 2 " ! 1 ,0 . Thus, the vertex is 2 d) r=

1 1 + cos q

2

C(sin2 θ + cos2 θ)

= A+C

y 4

= A(cos2 θ + sin2 θ) + B(sin θ cos θ − sin θ cos θ)+ = A·1+B·0+C ·1

6 . 3 + 2 cos θ

ep with 1 + e cos θ e = 1, so the graph is a parabola.

B ! = −2A sin θ cos θ + B(cos2 θ − sin2 θ)+ 2C sin θ cos θ, or 2

Exercise Set 9.6

a) The equation is in the form r =

A! = A cos2 θ + B sin θ cos θ + C sin2 θ,

47.

(x! )2 + (y ! )2 = r2

!

2 x

!8 !6 !4 !2 !2 !4

624

Chapter 9: Analytic Geometry Topics 4 2 + cos θ a) We first divide numerator and denominator by 2: 2 r= 1 1 + cos θ 2 ep with The equation is in the form r = 1 + e cos θ 1 e= . 2 Since 0 < e < 1, the graph is an ellipse.

When θ =

8. r =

r=

3π 5 − 10 sin 2

=

Thus, the vertices are d)

y

r=

2 2

!4 !2

b) Since e = 1 and ep = 1 · p = 2, we have p = 4. 2 2 Thus the ellipse has a vertical directrix 4 units to the right of the pole. c) We find the vertices by letting θ = 0 and θ = π. When θ = 0, 4 4 4 = = . r= 2 + cos 0 2+1 3 When θ = π, 4 4 = = 4. r= 2 + cos π 2−1 " ! 4 Thus, the vertices are , 0 and (4, π). 3

3π , 2 15

!

15 15 = = 1. 5 − 10(−1) 15 − 3,

π 2

"

and

!

1,

" 3π . 2

15 5 − 10 sin q 4

x

!2 !4 !6

12 4 + 8 sin θ a) We first divide numerator and denominator by 4: 3 r= 1 + 2 sin θ ep with The equation is in the form r = 1 + e sin θ e = 2.

10. r =

Since e > 1, the graph is a hyperbola.

d)

y 5 4 3 2 1 !5 !4 !3 !2 !1 !1 !2 !3 !4 !5

r=

4 2 + cos q

1 2 3 4 5

3 . 2 3 Thus the hyperbola has a horizontal directrix 2 units above the pole. π and θ = c) We find the vertices by letting θ = 2 π 3π . When θ = , 2 2 12 12 12 r= π = 4 + 8 · 1 = 12 = 1. 4 + 8 sin 2 3π , When θ = 2 12 12 12 = = = −3. r= 3π 4 + 8(−1) −4 4 + 8 sin 2 " ! " ! π 3π and − 3, . Thus, the vertices are 1, 2 2 b) Since e = 2 and ep = 2 · p = 3, we have p =

x

15 5 − 10 sin θ a) We first divide numerator and denominator by 5: 3 r= 1 − 2 sin θ ep The equation is in the form r = with 1 − e sin θ e = 2.

9. r =

Since e > 1, the graph is a hyperbola. 3 . 2 3 Thus the hyperbola has a horizontal directrix 2 units below the pole. π and θ = c) We find the vertices by letting θ = 2 π 3π . When θ = , 2 2 15 15 15 r= π = 5 − 10 · 1 = −5 = −3. 5 − 10 sin 2 b) Since e = 2 and ep = 2 · p = 3, we have p =

d)

y 7 6 5 4 3 2 1 !5 !4 !3 !2 !1 !1 !2 !3

r=

12 4 + 8 sin q

1 2 3 4 5

x

Exercise Set 9.6

625

8 6 − 3 cos θ a) We first divide numerator and denominator by 6: 4/3 r= 1 1 − cos θ 2 ep with The equation is in the form r = 1 − e cos θ 1 e= . 2 Since 0 < e < 1, the graph is an ellipse.

11. r =

1 4 8 1 and ep = · p = , we have p = . 2 2 3 3 8 Thus the ellipse has a vertical directrix units 3 to the left of the pole.

b) Since e =

c) We find the vertices by letting θ = 0 and θ = π. When θ = 0, 8 8 8 = = . r= 6 − 3 cos 0 6−3·1 3 When θ = π, 8 8 8 = = . r= 6 − 3 cos π 6 − 3(−1) 9 " ! " ! 8 8 , 0 and ,π . Thus, the vertices are 3 9 d)

y 4

r=

2 2

!4 !2

8 6 − 3 cos q 4

x

!2 !4

6 2 + 2 sin θ a) We first divide numerator and denominator by 2: 3 r= 1 + sin θ ep The equation is in the form r = with 1 + e sin θ e = 1, so the graph is a parabola.

12. r =

b) Since e = 1 and ep = 1 · p = 3, we have p = 3. Thus the parabola has a horizontal directrix 3 units above the pole. π c) We find the vertex by letting θ = . When 2 π θ= , 2 6 3 6 6 r= π = 2 + 2 · 1 = 4 = 2. 2 + 2 sin 2 ! " 3 π Thus, the vertex is , . 2 2

d)

y 5 4 3 2 1 !5 !4 !3 !2 !1 !1 !2 !3 !4 !5

r=

6 2 + 2 sin q

1 2 3 4 5

x

20 10 + 15 sin θ a) We first divide numerator and denominator by 10: 2 r= 3 1 + sin θ 2 ep The equation is in the form r = with 1 + e sin θ 3 e= . 2 Since e > 1, the graph is a hyperbola.

13. r =

3 4 3 and ep = · p = 2, we have p = . 2 2 3 4 Thus the hyperbola has a horizontal directrix 3 units above the pole. π and θ = c) We find the vertices by letting θ = 2 3π π . When θ = , 2 2 20 20 4 20 r= π = 10 + 15 · 1 = 25 = 5 . 10 + 15 sin 2 3π , When θ = 2 20 20 20 r = π = 10 + 15(−1) = −5 = 10 + 15 sin 2 −4. " ! " ! 3π 4 π , and − 4, . Thus, the vertices are 5 2 2 b) Since e =

d)

y 6 4 2

r= 4

!8 !4 !2

20 10 + 15 sin q 8

x

626

Chapter 9: Analytic Geometry Topics 10 8 − 2 cos θ a) We first divide numerator and denominator by 8: 5/4 r= 1 1 − cos θ 4 ep with The equation is in the form r = 1 − e cos θ 1 e= . 4 Since 0 < e < 1, the graph is an ellipse.

14. r =

c) We find the vertices by letting θ = 0 and θ = π. When θ = 0, 9 9 9 r= = = = 1. 6 + 3 cos 0 6+3·1 9 When θ = π, 9 9 9 r= = = = 3. 6 + 3 cos π 6 + 3(−1) 3 Thus, the vertices are (1, 0) and (3, π). d)

y 4

1 5 1 b) Since e = and ep = · p = , we have p = 5. 4 4 4 Thus the ellipse has a vertical directrix 5 units to the left of the pole. c) We find the vertices by letting θ = 0 and θ = π. When θ = 0, 10 10 5 10 = = = . r= 8 − 2 cos 0 8−2·1 6 3 When θ = π, 10 10 10 = = = 1. r= 8 − 2 cos π 8 − 2(−1) 10 " ! 5 Thus, the vertices are , 0 and (1, π). 3 d)

y 2

r=

10 8 − 2 cos q

1 !2

1

!1

2

x

!1 !2

9 6 + 3 cos θ a) We first divide numerator and denominator by 6: 3/2 r= 1 1 + cos θ 2 ep The equation is in the form r = with 1 + e cos θ 1 e= . 2 Since 0 < e < 1, the graph is an ellipse.

15. r =

1 3 1 b) Since e = and ep = · p = , we have p = 3. 2 2 2 Thus the ellipse has a vertical directrix 3 units to the right of the pole.

r=

2

2

!4 !2

9 6 + 3 cos q 4

x

!2 !4

4 3 − 9 sin θ a) We first divide numerator and denominator by 3: 4/3 r= 1 − 3 sin θ ep with The equation is in the form r = 1 − e sin θ e = 3.

16. r =

Since e > 1, the graph is a hyperbola. 4 4 , we have p = . 3 9 4 Thus the hyperbola has a horizontal directrix 9 units below the pole. π and θ = c) We find the vertices by letting θ = 2 3π π . When θ = , 2 2 4 4 2 4 r= π = 3 − 9 · 1 = −6 = − 3 . 3 − 9 sin 2 3π When θ = , 2 4 1 4 4 = = . r= = 3π 3 − 9(−1) 12 3 3 − 9 sin 2 " ! " ! 2 π 1 3π Thus, the vertices are − , and , . 3 2 3 2 b) Since e = 3 and ep = 3 · p =

d)

y 5 4 3 2 1 !5 !4 !3 !2 !1 !1 !2 !3 !4 !5

r=

4 3 − 9 sin q

1 2 3 4 5

x

Exercise Set 9.6

627

3 2 − 2 sin θ a) We first divide numerator and denominator by 2: 3/2 r= 1 − sin θ ep The equation is in the form r = with 1 − e sin θ e = 1, so the graph is a parabola.

17. r =

3 3 , we have p = . 2 2 3 Thus the parabola has a horizontal directrix 2 units below the pole. c) We find the vertex by letting θ = 3π . When 2 3π θ= , 2 3 3 3 = r= = . 3π 2 − 2(−1) 4 2 − 2 sin 2 " ! 3 3π , . Thus, the vertex is 4 2

d)

y 2 1 3

!1

6 4 2 2 !2

r=

12 3 + 9 cos q

4 2 − cos θ a) We first divide numerator and denominator by 2: 2 r= 1 1 − cos θ 2 ep The equation is in the form r = with 1 − e cos θ 1 e= . 2 Since 0 < e < 1, the graph is an ellipse.

19. r =

1 1 b) Since e = and ep = · p = 2, we have p = 4. 2 2 Thus the ellipse has a vertical directrix 4 units to the left of the pole.

y

!4 !2

x

!2

b) Since e = 1 and ep = 1 · p =

d)

4

!1

r=

4

x 3 2 − 2 sin q

12 3 + 9 cos θ a) We first divide numerator and denominator by 3: 4 r= 1 + 3 cos θ ep The equation is in the form r = with 1 + e cos θ e = 3. Since e > 1, the graph is a hyperbola.

18. r =

c) We find the vertices by letting θ = 0 and θ = π. When θ = 0, 4 4 = = 4. r= 2 − cos 0 2−1 When θ = π, 4 4 4 = = . r= 2 − cos π 2 − (−1) 3 ! " 4 ,π . Thus, the vertices are (4, 0) and 3 d)

y 4

4 . 3 4 Thus the hyperbola has a vertical directrix 3 units to the right of the pole.

2

b) Since e = 3 and ep = 3 · p = 4, we have p =

c) We find the vertices by letting θ = 0 and θ = π. When θ = 0, 12 12 12 = = = 1. r= 3 + 9 cos 0 3+9·1 12 When θ = π, 12 12 12 r= = = = −2. 3 + 9 cos π 3 + 9(−1) −6 Thus, the vertices are (1, 0) and (−2, π).

!2

2

4

6

r=

4 2 − cos q

x

!2 !4

5 1 − sin θ a) The equation is in the form r =

20. r =

e = 1,

ep with 1 − e sin θ

so the graph is a parabola. b) Since e = 1 and ep = 1 · p = 5, we have p = 5. Thus the parabola has a horizontal directrix 5 units below the pole.

628

Chapter 9: Analytic Geometry Topics c) We find the vertex by letting θ = 3π . When 2 3π , θ= 2 5 5 5 r= = . = 3π 1 − (−1) 2 1 − sin 2 ! " 5 3π , . Thus, the vertex is 2 2 d)

3 8 − 4 cos θ a) We first divide numerator and denominator by 8: 3/8 r= 1 1 − cos θ 2 ep with The equation is in the form r = 1 − e cos θ 1 e= . 2 Since 0 < e < 1, the graph is an ellipse.

22. r =

y 4 3 2 1 !5 !4 !3 !2 !1 !1 !2 !3 !4 !5 !6

r=

5 1 − sin q

1 2 3 4 5

1 3 3 1 and ep = · p = , we have p = . 2 2 8 4 3 unit Thus the ellipse has a vertical directrix 4 to the left of the pole.

b) Since e = x

c) We find the vertices by letting θ = 0 and θ = π. When θ = 0, 3 3 3 = = . r= 8 − 4 cos 0 8−4·1 4 When θ = π, 3 3 1 3 = = = . r= 8 − 4 cos π 8 − 4(−1) 12 4 " ! " ! 3 1 , 0 and ,π . Thus, the vertices are 4 4

7 2 + 10 sin θ a) We first divide numerator and denominator by 2: 7/2 r= 1 + 5 sin θ ep with The equation is in the form r = 1 + e sin θ e = 5.

21. r =

d)

y

Since e > 1, the graph is a hyperbola. 7 7 , we have p = . 2 10 7 Thus the hyperbola has a horizontal directrix 10 unit above the pole. π and θ = c) We find the vertices by letting θ = 2 3π π . When θ = , 2 2 7 7 7 r= π = 2 + 10 · 1 = 12 . 2 + 10 sin 2 3π When θ = , 2 7 7 7 r= = =− . 3π 2 + 10(−1) 8 2 + 10 sin 2 " ! " ! 7 π 7 3π Thus, the vertices are , and − , . 12 2 8 2

1

r=

b) Since e = 5 and ep = 5 · p =

d)

r=

23.

1 1 + cos θ r + r cos θ = 1 r=

#

!4

x2

r = 1 − r cos θ

+ y2 = 1 − x

x2 + y 2 = 1 − 2x + x2

y 2 = −2x + 1, or

24.

y 2 + 2x − 1 = 0

4 2 + cos θ 2r + r cos θ = 4 r=

7 2 + 10 sin q

2 2

!2

4

x

!1

#

2 !4 !2

1

!1

y 4

3 8 − 4 cos q

x

x2

2r = 4 − r cos θ

+ y2 = 4 − x

4x2 + 4y 2 = 16 − 8x + x2

3x + 4y + 8x − 16 = 0 2

2

Exercise Set 9.6

629

15 5 − 10 sin θ 5r − 10r sin θ = 15

25.

9 6 + 3 cos θ 6r + 3r cos θ = 9

31.

r=

r=

5r = 10r sin θ + 15

6r = 9 − 3r cos θ

r = 2r sin θ + 3 '

+

x2

y2

= 2y + 3

2

'

x + y = 4y + 12y + 9

26.

2

2

12 4 + 8 sin θ 4r + 8r sin θ = 12

32.

r=

4 3 − 9 sin θ 3r − 9r sin θ = 4

3r = 9r sin θ + 4

3

'

2

33.

2

2r = 2r sin θ + 3

6r = 3r cos θ + 8

'

2

x2 + y 2 = 3x + 8

'

x2 + y 2 = 2y + 3

4x2 + 4y 2 = 4y 2 + 12y + 9

36x2 + 36y 2 = 9x2 + 48x + 64 27x + 36y − 48x − 64 = 0 6 r= 2 + 2 sin θ 2r + 2r sin θ = 6 2

28.

+ y 2 = 9y + 4

9x − 72y − 72y − 16 = 0 3 r= 2 − 2 sin θ 2r − 2r sin θ = 3 2

8 r= 6 − 3 cos θ 6r − 3r cos θ = 8 6

x2

9x2 + 9y 2 = 81y 2 + 72y + 16

2

x − 3y + 12y − 9 = 0 2

27.

x + y = 9 − 12y + 4y 2

+ y2 = 3 − x

r=

r = 3 − 2r sin θ

2

2

3x2 + 4y 2 + 6x − 9 = 0

4r = 12 − 8r sin θ x2 + y 2 = 3 − 2y

2r = 3 − r cos θ

4x + 4y 2 = 9 − 6x + x2

2

x2 − 3y 2 − 12y − 9 = 0

'

x2

4x2 = 12y + 9, or

2

4x − 12y − 9 = 0 2

34.

12 3 + 9 cos θ 3r + 9r cos θ = 12 r=

2r = 6 − 2r sin θ '

+

r = 3 − r sin θ

y2

= 3−y

x + y = 9 − 6y + y 2

2

'

2

x2 = −6y + 9, or

x + 6y − 9 = 0 2

29.

x2

3r = 12 − 9r cos θ

35.

20 r= 10 + 15 sin θ 10r + 15r sin θ = 20

2

x2

+

= 4 − 3y

4x + 4y = 16 − 24y + 9y

30.

2

2

4 2 − cos θ 2r − r cos θ = 4

2r = r cos θ + 4

2

2

10 8 − 2 cos θ 8r − 2r cos θ = 10 r=

36.

4

+

= x+5

16x2 + 16y 2 = x2 + 10x + 25 15x2 + 16y 2 − 10x − 25 = 0

2

r = r sin θ + 5

'

4r = r cos θ + 5 y2

+ y2 = x + 4

3x + 4y − 8x − 16 = 0 5 r= 1 − sin θ r − r sin θ = 5

8r = 2r cos θ + 10

x2

x2

4x2 + 4y 2 = x2 + 8x + 16

2

4x2 − 5y 2 + 24y − 16 = 0

'

+ y 2 = 4 − 3x

x + y 2 = 16 − 24x + 9x2 2

r=

'

2r = 4 − 3r sin θ y2

r = 4 − 3r cos θ

−8x2 + y 2 + 24x − 16 = 0

10r = 20 − 15r sin θ '

x2

x2 + y 2 = y + 5 x2 + y 2 = y 2 + 10y + 25 x2 = 10y + 25, or

x − 10y − 25 = 0 2

630

Chapter 9: Analytic Geometry Topics 7 2 + 10 sin θ 2r + 10r sin θ = 7

37.

r=

2

'

x2 + y 2 = 7 − 10y

4x2 + 4y 2 = 49 − 140y + 100y 2

4x − 96y + 140y − 49 = 0 2

38.

2r = 7 − 10r sin θ

2

3 8 − 4 cos θ 8r − 4r cos θ = 3 r=

8r = 4r cos θ + 3

8

'

x2 + y 2 = 4x + 3

64x2 + 64y 2 = 16x2 + 24x + 9 48x2 + 64y 2 − 24x − 9 = 0

39. e = 2, r = 3 csc θ

The equation of the directrix can be written 3 , or r sin θ = 3. r= sin θ This corresponds to the equation y = 3 in rectangular coordinates, so the directrix is a horizontal line 3 units above the polar axis. Using the table on page 824 of the text, we see that the equation is of the form ep . r= 1 + e sin θ Substituting 2 for e and 3 for p, we have 6 2·3 = . r= 1 + 2 sin θ 1 + 2 sin θ 2 , r = − sec θ 3 The equation of the directrix can be written −1 r= , or r cos θ = −1. cos θ This corresponds to the equation x = −1 in rectangular coordinates, so the directrix is a vertical line 1 unit to the left of the pole. Using the table on page 824 of the text, we see that the equation is of the form ep . r= 1 − e cos θ 2 Substituting for e and 1 for p, we have 3 2 2 ·1 2 3 3 = , or r= . 2 2 3 − 2 cos θ 1 − cos θ 1 − cos θ 3 3

40. e =

41. e = 1, r = 4 sec θ The equation of the directrix can be written 4 , or r cos θ = 4. r= cos θ This corresponds to the equation x = 4 in rectangular coordinates, so the directrix is a vertical line 4 units to the right of the pole. Using the table on page 824 of the text, we see that the equation is of the form ep . r= 1 + e cos θ Substituting 1 for e and 4 for p, we have

r=

1·4 4 = . 1 + 1 · cos θ 1 + cos θ

42. e = 3, r = 2 csc θ

The equation of the directrix can be written 2 , or r sin θ = 2. r= sin θ This corresponds to the equation y = 2 in rectangular coordinates, so the directrix is a horizontal line 2 units above the polar axis. Using the table on page 824 of the text, we see that the equation is of the form ep . r= 1 + e sin θ Substituting 3 for e and 2 for p, we have 6 3·2 = . r= 1 + 3 sin θ 1 + 3 sin θ 1 , r = −2 sec θ 2 The equation of the directrix can be written −2 r= , or r cos θ = −2. cos θ This corresponds to the equation x = −2 in rectangular coordinates, so the directrix is a vertical line 2 units to the left of the pole. Using the table on page 824 of the text, we see that the equation is of the form ep . r= 1 − e cos θ 1 Substituting for e and 2 for p, we have 2 1 ·2 1 2 2 = , or r= . 1 1 2 − cos θ 1 − cos θ 1 − cos θ 2 2

43. e =

44. e = 1, r = 4 csc θ The equation of the directrix can be written 4 , or r sin θ = 4. r= sin θ This corresponds to the equation y = 4 in rectangular coordinates, so the directrix is a horizontal line 4 units above the polar axis. Using the table on page 824 of the text, we see that the equation is of the form ep r= . 1 + e sin θ Substituting 1 for e and 4 for p, we have 4 1·4 = . r= 1 + 1 · sin θ 1 + sin θ 3 , r = 5 csc θ 4 The equation of the directrix can be written 5 r= , or r sin θ = 5. sin θ This corresponds to the equation y = 5 in rectangular coordinates, so the directrix is a horizontal line 5 units above the polar axis. Using the table on page 824 of the text, we see that the equation is of the form ep . r= 1 + e sin θ

45. e =

Exercise Set 9.6 Substituting

3 for e and 5 for p, we have 4

3 ·5 15/4 15 4 = or . r= 3 3 4 + 3 sin θ 1 + sin θ 1 + sin θ 4 4 4 , r = 2 sec θ 5 The equation of the directrix can be written 2 r= , or r cos θ = 2. cos θ This corresponds to the equation x = 2 in rectangular coordinates, so the directrix is a vertical line 2 units to the right of the pole. Using the table on page 824 of the text, we see that the equation is of the form ep . r= 1 + e cos θ 4 Substituting for e and 2 for p, we have 5 4 ·2 8/5 8 5 = , or r= . 4 4 5 + 4 cos θ 1 + cos θ 1 + cos θ 5 5

46. e =

631 51. f (x) = (x − 3)2 + 4

f (t) = (t − 3)2 + 4 = t2 − 6t + 9 + 4 = t2 − 6t + 13

Thus, f (t) = (t − 3)2 + 4, or t2 − 6t + 13.

52. f (2t) = (2t − 3)2 + 4 = 4t2 − 12t + 9 + 4 = 4t2 − 12t + 13 Thus, f (2t) = (2t − 3)2 + 4, or 4t2 − 12t + 13.

53. f (x) = (x − 3)2 + 4

f (t − 1) = (t − 1 − 3)2 + 4 = (t − 4)2 + 4 = t2 − 8t + 16 + 4 = t2 − 8t + 20 Thus, f (t − 1) = (t − 4)2 + 4, or t2 − 8t + 20.

54. f (t + 2) = (t + 2 − 3)2 + 4 = (t − 1)2 + 4 = t2 − 2t + 1 + 4 = t2 − 2t + 5 Thus, f (t + 2) = (t − 1)2 + 4, or t2 − 2t + 5.

55.

y

100 million

P

6

47. e = 4, r = −2 csc θ

x

The equation of the directrix can be written −2 , or r sin θ = −2. r= sin θ This corresponds to the equation y = −2 in rectangular coordinates, so the directrix is a horizontal line 2 units below the polar axis. Using the table on page 824 of the text, we see that the equation is of the form ep . r= 1 − e sin θ Substituting 4 for e and 2 for p, we have 8 4·2 = . r= 1 − 4 sin θ 1 − 4 sin θ

Since the directrix lies above the pole, the equation is of ep . The point P on the parabola the form r = 1 + e sin θ 8 has coordinates (1 × 10 , π/6). (Note that 100 million = 1×108 .) Since the conic is a parabola, we know that e = 1. We substitute 1 × 108 for r, 1 for e, and π/6 for θ and then find p. 1 × 108 =

p 1 + 0.5 p 1 × 108 = 1.5 1.5 × 108 = p

49. Each graph is an ellipse. The value of e determines the location of the center and the lengths of the major and minor axes. The larger the value of e, the farther the center is from the pole and the longer the axes. 50. Answers will vary.

π 6

1 × 108 =

48. e = 3, r = 3 csc θ

The equation of the directrix can be written 3 , or r sin θ = 3. r= sin θ This corresponds to the equation y = 3 in rectangular coordinates, so the directrix is a horizontal line 3 units above the polar axis. Using the table on page 824 of the text, we see that the equation is of the form ep . r= 1 + e sin θ Substituting 3 for e and 3 for p, we have 9 3·3 = . r= 1 + 3 sin θ 1 + 3 sin θ

1·p

1 + 1 · sin

The equation of the orbit is r = 56.

1.5 × 108 . 1 + sin θ

y

P 120 million 4 x

Since the directrix lies above the pole, the equation is of ep . The point P on the parabola the form r = 1 + e sin θ 8 has coordinates (1.2 × 10 , π/4). (Note that 120 million = 1.2 × 108 .) Since the conic is a parabola, we know that

632

Chapter 9: Analytic Geometry Topics e = 1. We substitute 1.2 × 108 for r, 1 for e, and π/4 for θ and then find p. 1.2 × 108 =

4

1·p

π 1 + 1 · sin 4 p 8 √ 1.2 × 10 = 2 1+ 2 p √ 1.2 × 108 = 2+ 2 2 2.0 × 108 ≈ p

The equation of the orbit is r =

2

x"

2

!4 !2

4

x

!2 !4

x = 4t 2, y = 2t, −1 $ t $ 1

To find an equivalent rectangular equation, we first solve y = 2t for t. y = 2t y =t 2

2.0 × 108 . 1 + sin θ

y Then we substitue for t in x = 4t2 . 2 ! "2 y x=4 2

Exercise Set 9.7 1.

y

3.

1 t, y " 6t ! 7; !1 # t # 6 2

x = 4·

40

y2 4

x = y2 Given that −1 ≤ t ≤ 1, we find the corresponding restrictions on y.

5

!5 !20

To find an equivalent rectangular equation, we first solve 1 x = t for t. 2 1 x= t 2 2x = t Then we substitute 2x for t in y = 6t − 7.

For t = −1: y = 2t = 2(−1) = −2. For t = 1: y = 2t = 2 · 1 = 2.

Then we have x = y 2 , −2 ≤ y ≤ 2.

Rather than express restrictions on y, we could express equivalent restrictions on x, 0 ≤ x ≤ 4. 4.

x " !"t , y " 2t % 3; 0 # t # 8 25

y = 6(2x) − 7

y = 12x − 7

Given that −1 ≤ t ≤ 6, we find the corresponding restrictions on x: 1 1 1 For t = −1 : x = t = (−1) = − . 2 2 2 1 1 For t = 6 : x = t = · 6 = 3. 2 2 1 Then we have y = 12x − 7, − ≤ x ≤ 3. 2 2.

x " t, y " 5 ! t ; !2 # t # 3 10

5

!5 !5

To find an equivalent rectangular equation, we substitute x for t in y = 5 − t: y = 5 − x

Given that −2 ≤ t ≤ 3 and x = t, the corresponding restrictions on x are −2 ≤ x ≤ 3. Then we have y = 5 − x, −2 ≤ x ≤ 3.

8

!2 !2

To find √ an equivalent rectangular equation, we first solve x = t for t: √ x= t x2 = t Then we substitute x2 for t in y = 2t + 3: y = 2x2 + 3 Given that 0 ≤ t ≤ 8, we find the corresponding restrictions on x: √ √ For t = 0 : x = t = 0 = 0. √ √ √ For t = 8 : x = t = 8, or 2 2. √ Then we have y = 2x2 + 3, 0 ≤ x ≤ 2 2. 5. To find an equivalent rectangular equation, we first solve x = t2 for t. x = t2 √ x=t (We choose the nonnegative square root because 0 ≤ t ≤ 4.) √ √ Then we substitute x for t in y = t:

Exercise Set 9.7

633

'√

&2 1 y= (x + 1) 2 1 y = (x + 1)2 4 Given that −3 ≤ t ≤ 3, we find the corresponding restrictions on x:

y=

x = (x1/2 )1/2 √ y = x1/4 , or 4 x

%

Given that 0 ≤ t ≤ 4, we find the corresponding restrictions on x: For t = 0 : x = t2 = (0)2 = 0. For t = 4 : x = t2 = 42 = 16. √ Then we have y = 4 x, 0 ≤ x ≤ 16. (This result could also be expressed as x = y 4 , 0 ≤ y ≤ 2.) 6. To find an equivalent rectangular equation, we first solve x = t3 + 1 for t: x = t3 + 1 √ 3

x − 1 = t3 x−1 = t

Then we substitute

√ 3

x − 1 for t in y = t: y =

√ 3

x−1

Given that −3 ≤ t ≤ 3, we find the corresponding restrictions on x: For t = −3 : x = t3 + 1 = (−3)3 + 1 = −26. For t = 3 : x = t3 + 1 = 33 + 1 = 28. √ Then we have y = 3 x − 1, −26 ≤ x ≤ 28.

7. To find an equivalent rectangular equation, we can substi1 1 : y= . tute x for t + 3 in y = t+3 x Given that −2 ≤ t ≤ 2, we find the corresponding restrictions on x: For t = −2 : x = t + 3 = −2 + 3 = 1. For t = 2 : x = t + 3 = 2 + 3 = 5. 1 Then we have y = , 1 ≤ x ≤ 5. x

8. To find an equivalent rectangular equation, we first solve x = 2t3 + 1 for 2t3 : x = 2t3 + 1 x − 1 = 2t3

Then we substitute x − 1 for 2t3 in y = 2t3 − 1: y = (x − 1) − 1

y = x−2

Given that −4 ≤ t ≤ 4, we find the corresponding restrictions on x: For t = −4 : x = 2t + 1 = 2(−4) + 1 = −127. 3

3

For t = 4 : x = 2t3 + 1 = 2 · 43 + 1 = 129.

Then we have y = x − 2, −127 ≤ x ≤ 129. 9. To find an equivalent rectangular equation, we first solve x = 2t − 1 for t: x = 2t − 1 x + 1 = 2t 1 (x + 1) = t 2

Then we substitute

For t = −3 : x = 2t − 1 = 2(−3) − 1 = −7. For t = 3 : x = 2t − 1 = 2 · 3 − 1 = 5. 1 Then we have y = (x + 1)2 , −7 ≤ x ≤ 5. 4

10. To find an equivalent rectangular equation, we first solve 1 x = t for t: 3 1 x= t 3 3x = t Then we substitute 3x for t in y = t: y = 3x. Given that −5 ≤ t ≤ 5, we find the corresponding restrictions on x: 1 5 1 For t = −5 : x = t = (−5) = − . 3 3 3 1 5 1 For t = 5 : x = t = · 5 = . 3 3 3 5 5 Then we have y = 3x, − ≤ x ≤ . 3 3 11. To find an equivalent rectangular equation, we first solve x = e−t for et : x = e−t 1 x= t e 1 et = x 1 1 Then we substitute for et in y = et : y = . x x Given that −∞ ≤ t ≤ ∞, we find the corresponding restrictions on x: As t approaches −∞, e−t approaches ∞. As t approaches ∞, e−t approaches 0. Thus, we see that x > 0. 1 Then we have y = , x > 0. x 12. To find an equivalent rectangular equation, we first solve x = 2 ln t for t2 : x = 2 ln t x = ln t2 ex = t2 Then we substitute ex for t2 in y = t2 : y = ex . Given that 0 < t < ∞, we find the corresponding restrictions on x. As t approaches 0, 2 ln t approaches −∞. As t approaches ∞, x = 2 ln t also approaches ∞. Then we have y = ex , −∞ < x < ∞.

1 (x + 1) for t in y = t2 : 2

634

Chapter 9: Analytic Geometry Topics

13. To find an equivalent rectangular equation, we first solve for cos t and sin t in the parametric equations: y x = cos t, = sin t 3 3 Using the identity sin2 θ + cos2 θ = 1, we can substitute to eliminate the parameter: sin2 t + cos2 t ! "2 ! "2 x y + 3 3 2 x y2 + 9 9 x2 + y 2

=1 =1 =1 =9

For 0 ≤ t ≤ 2π, −3 ≤ 3 cos t ≤ 3.

Then we have x2 + y 2 = 9, −3 ≤ x ≤ 3. 14. To find an equivalent rectangular equation, we first solve for cos t and sin t in the parametric equations: y x = cos t, = sin t 2 4 Using the identity sin2 θ + cos2 θ = 1, we can substitute to eliminate the parameter: sin2 t + cos2 t = 1 ! "2 ! "2 x y + =1 4 2 x2 y2 + =1 4 16 For 0 ≤ t ≤ 2π, −2 ≤ 2 cos t ≤ 2. Then we have

y2 x2 + = 1, −2 ≤ x ≤ 2. 4 16

15. To find an equivalent rectangular equation, we first solve y y = 2 sin t for sin t: = sin t. 2 Using the identity sin2 θ + cos2 θ = 1, we can substitute to eliminate the parameter: sin2 t + cos2 t = 1 ! "2 y + x2 = 1 2 y2 =1 x2 + 4 For 0 ≤ t ≤ 2π, −1 ≤ cos t ≤ 1. Then we have x2 +

2

y = 1, −1 ≤ x ≤ 1. 4

16. To find an equivalent rectangular equation, we first solve for cos t and sin t in the parametric equations: y x = cos t, = sin t 2 2 Using the identity sin2 θ + cos2 θ = 1, we can substitute to eliminate the parameter: sin2 t + cos2 t ! "2 ! "2 x y + 2 2 y2 x2 + 4 4 2 x + y2

=1 =1 =1 =4

For 0 ≤ t ≤ 2π, −2 ≤ 2 cos t ≤ 2.

Then we have x2 + y 2 = 4, −2 ≤ x ≤ 2.

17. To find an equivalent rectangular equation, we first solve x = sec t for cos t: x = sec t 1 x= cos t 1 cos t = x 1 1 Then we substitute for cos t in y = cos t: y = . x x π π For − < t < , 1 ≤ x < ∞. 2 2 1 Then we have y = , x ≥ 1. x 18. To find an equivalent rectangular equation, we first solve x = sin t for csc t: x = sin t 1 1 = x sin t 1 = csc t x 1 1 Then we substitute for csc t in y = csc t: y = . x x For 0 < t < π, 0 < x ≤ 1. 1 Then we have y = , 0 < x ≤ 1. x 19. To find an equivalent rectangular equation, we first solve for cos t and sin t in the parametric equations: x = 1 + 2 cos t

y = 2 + 2 sin t

x − 1 = 2 cos t y − 2 = 2 sin t y−2 x−1 = cos t = sin t 2 2 Using the identity sin2 θ + cos2 θ = 1, we can substitute to eliminate the parameter: !

sin2 t + cos2 t "2 ! "2 y−2 x−1 + 2 2 (x − 1)2 (y − 2)2 + 4 4 (x − 1)2 + (y − 2)2

=1 =1 =1 =4

For 0 ≤ t ≤ 2π, −1 ≤ 1 + 2 cos t ≤ 3.

Then we have (x − 1)2 + (y − 2)2 = 4, −1 ≤ x ≤ 3. 20. To find an equivalent rectangular equation, we first solve for sec t and tan t in the parametric equations: x = 2 + sec t x − 2 = sec t

y = 1 + 3 tan t

y − 1 = 3 tan t y−1 = tan t 3 Using the identity 1 + tan2 θ = sec2 θ, we can substitute to eliminate the parameter:

Exercise Set 9.7

635

1 + tan2 t = sec2 t ! "2 y−1 = (x − 2)2 1+ 3 1 = (x − 2)2 −

(y − 1)2 9

π , 3 < 2 + sec t < ∞. 2 (y − 1)2 = 1, x > 3. Then we have (x − 2)2 − 9 21. y = 4x − 3 For 0 < t <

Answers may vary.

If x = t, then y = 4t − 3. t t If x = , then y = 4 · − 3 = t − 3. 4 4 22. y = x2 − 1

Answers may vary. If x = t, then y = t2 − 1.

If x = t−2, then y = (t−2)2 −1 = t2 −4t+4−1 = t2 −4t+3.

23. y = (x − 2)2 − 6x

Answers may vary. If x = t, then y = (t − 2)2 − 6t.

If x = t + 2, then y = (t + 2 − 2)2 − 6(t + 2) = t2 − 6t − 12.

24. y = x3 + 3

Answers may vary. If x = t, then y = t3 + 3. √ √ If x = 3 t, then y = ( 3 t)3 + 3 = t + 3. 25. a) We substitute 7 for h, 80 for v0 , and 30◦ for θ in the parametric equations for projective motion. x = (v0 cos θ)t = (80 cos 30◦ )t √ " ! √ 3 = 80 · t = 40 3t 2 y = h + (v0 sin θ)t − 16t2

= 7 + (80 sin 30◦ )t − 16t2 ! " 1 = 7 + 80 · t − 16t2 2 = 7 + 40t − 16t2 b) The height of the ball at time t is given by y. When t = 1, y = 7 + 40 · 1 − 16 · 12 = 31 ft.

When t = 2, y = 7 + 40 · 2 − 16 · 22 = 23 ft. c) The ball hits the ground when y = 0, so we solve the equation y = 0 using the quadratic formula. 7 + 40t − 16t = 0

d) Since the ball is in the air for about 2.7 sec, the horizontal distance it travels is given by √ x = 40 3(2.7) ≈ 187.1 ft. e) To find the maximum height of the ball, we find the maximum value of y. At the vertex of the quadratic function represented by y we have 40 b =− = 1.25. t=− 2a 2(−16) When t = 1.25, y = 7 + 40(1.25) − 16(1.25)2 = 32 ft. ! " 1 t = 100t 26. a) x = (200 cos 60◦ )t = 200 · 2 √ " ! 3 ◦ 2 y = 0+(200 sin 60 )t−16t = 200 · t−16t2 = 2 √ 100 3t − 16t2 √ b) When t = 4, y = 100 3(4) − 16 · 42 ≈ 436.8 ft. √ When t = 8, y = 100 3(8) − 16 · 82 ≈ 361.6 ft. c) Solve y = 0. √ 100 3t − 16t2 = 0 √ t(100 3 − 16t) = 0 t = 0 or t ≈ 10.8

The projectile is in the air for about 10.8 sec. d) When t ≈ 10.8, x ≈ 100(10.8) ≈ 1080 ft. e) At the vertex of the quadratic function represented by y we have √ 100 3 b =− ≈ 5.4 t=− 2a 2(−16) When t ≈ 5.4, √ y ≈ 100 3(5.4) − 16(5.4)2 ≈ 468.7 ft.

27. Find the equivalent rectangular equation. For x = a cos t+ x−h and sin t = h and y = b sin t + k, we have cos t = a y−k . Then, substitute in the identity sin2 θ + cos2 θ = 1. b "2 ! "2 ! x−h y−k + =1 b a (y − k)2 (x − h)2 + =1 2 a b2 Thus, the given equations are parametric equations of an ellipse with center (h, k). 28. The graph of the equation is a circle centered at the origin with radius a. 29. Graph y = x3 .

2

−16t + 40t + 7 = 0 2

' 402 − 4(−16)(7) t= 2(−16) t ≈ −0.2 or t ≈ 2.7 −40 ±

The negative value for t has no meaning in this application. Thus the ball is in the air for about 2.7 sec.

x y −2 −8 −1 −1 0 0 1 1 2 8

y 4 2

y % x3 2

!4 !2 !2 !4

4

x

636

Chapter 9: Analytic Geometry Topics

30.

2.

y 4

x " y3

2 2

!4 !2

4

x

!2

31. Graph f (x) =

3. The statement is true. See page 788 in the text.

√

4. The statement is false. See Example 2 on page 800 in the text. x − 2.

x f (x) 2 0 3 1 6 2 11 3

5. 2x2 + xy + 2y 2 = 10 A = 2, B = 1, C = 2

y

B 2 − 4AC = 12 − 4 · 2 · 2 = −15 < 0, so the graph is an ellipse (or circle). The given statement is false.

f(x) " !x ! 2

4 2

6. Graph (d) is the graph of y 2 = 5x. 2

!2

4

6

x

!2

7. Graph (a) is the graph of y 2 = 9 − x2 .

!4

8. Graph (e) is the graph of 3x2 + 4y 2 = 12. 9. Graph (g) is the graph of 9y 2 − 4x2 = 36.

32.

10. Graph (b) is the graph of x2 + y 2 + 2x − 3y = 8.

y 4

11. Graph (f) is the graph of 4x2 + y 2 − 16x − 6y = 15.

2

12. Graph (h) is the graph of x2 − 8x + 6y = 0. 2

!4 !2

4

x

13. Graph (c) is the graph of

!2

f (x) "

3 x2 ! 1

14.

33. The curve is generated clockwise when the equations x = 3 cos t, y = −3 sin t are used. Alternatively, the equations x = 3 sin t, y = 3 cos t can be used.

15.

34. To find the equivalent rectangular equation we first solve the given equations for cos2 t and sin2 t, respectively. x = cos3 t

y = sin3 t

x2/3 = cos2 t

y 2/3 = sin2 t

Now substitute in the identity sin2 θ + cos2 θ = 1. y 2/3 + x2/3 = 1, or x2/3 + y 2/3 = 1

Chapter 9 Review Exercises 1.

(x − 2)2 (y + 3)2 + =1 4 9 2 2 [y − (−3)] (x − 2) + = 1 Standard form 22 32 The graph is an ellipse with center (2, −3), so the statement is false.

x + y2 = 1 y 2 = −x + 1

y 2 = −1 · (x − 1) " ! 1 (x − 1) (y − 0)2 = 4 − 4 This parabola has a horizontal axis of symmetry and p = 1 − < 0. Thus it opens to the left and the statement is 4 true.

(y − 1)2 (x + 3)2 − = 1. 16 25

(x − h)2 = 4p(y − k) ! " 3 (x − 0)2 = 4 − (y − 0) 2 x2 = −6y y 2 = −12x

y 2 = 4(−3)x

F : (−3, 0), V : (0, 0), D: x = 3 16.

x2 + 10x + 2y + 9 = 0 x2 + 10x = −2y − 9

x2 + 10x + 25 = −2y − 9 + 25

(x + 5)2 = −2y + 16 " ! 1 (y − 8) [x − (−5)]2 = 4 − 2

V : (−5, 8) ! ! "" ! " 1 15 F : − 5, 8 + − or − 5, 2 2 17 1 D : y =8+ = 2 2 17. Begin by completing the square twice. 16x2 + 25y 2 − 64x + 50y − 311 = 0

16(x2 − 4x) + 25(y 2 + 2y) = 311

16(x2 − 4x + 4) + 25(y 2 + 2y + 1) = 311 + 16 · 4 + 25 · 1 16(x − 2)2 + 25(y + 1)2 = 400 [y − (−1)]2 (x − 2)2 + =1 25 16

Chapter 9 Review Exercises The center is (2, −1). Note that a = 5 and b = 4. The major axis is horizontal so the vertices are 5 units left and right of the center: (2 − 5, −1) and (2 + 5, −1), or (−3, −1) 2 2 2 and (7, −1). √ We know that c = a − b = 25 − 16 = 9 and c = 9 = 3. Then the foci are 3 units left and right of the center: (2 − 3, −1) and (2 + 3, −1), or (−1, −1) and (5, −1).

637 2 20. The parabola ! is of"the form y = 4px. A point on the 2 , or (1.5, 1). parabola is 1.5, 2

y 2 = 4px 12 = 4 · p · 1.5 1 = 6p 1 =p 6

Since the focus is at (p, 0) = 0.167 ft, from the vertex. 21.

" 1 1 , 0 , the focus is ft, or 6 6

x2 − 16y = 0, (1) x2 − y 2 = 64

(2)

From equation (1) we have x2 = 16y. Substitute in equation (2).

18. The vertices (0, −4) and (0, 4) are on the y-axis, so the major axis is vertical and a = 4. Since the vertices are equidistant from the origin, the center of the ellipse is at the origin. The length of the minor axis is 6, so b = 6/2, or 3. The equation is

16y − y 2 = 64

0 = y 2 − 16y + 64

0 = (y − 8)2 0 = y−8

y2 x2 + = 1. 9 16 19. Begin by completing the square twice. 1 x2 − 2y 2 + 4x + y − = 0 8 ! " 1 1 2 2 (x + 4x) − 2 y − y = 2 8 " ! 1 1 1 1 = +4−2· (x2 + 4x + 4) − 2 y 2 − y + 2 16 8 16 "2 ! 1 (x + 2)2 − 2 y − =4 4 "2 ! 1 y − [x − (−2)]2 4 − =1 4 2 " ! 1 . The transverse axis is horizonThe center is − 2, 4 tal, so ! the vertices " are 2!units left and " right ! of the " cen1 1 1 − 2 − 2, ter: and − 2 + 2, , or − 4, and 4 4 4 ! " 1 . Since c2 = a2 + b2 , we have c2 = 4 + 2 = 6 and 0, 4√ √ c = 6.!Then the foci " are !6 units left and " right of the √ 1 √ 1 and − 2 + 6, . center: − 2 − 6, 4 4 Find the asymptotes: b b and y − k = − (x − h) y − k = (x − h) a a √ √ 1 1 2 2 (x + 2) and y − = − (x + 2) y− = 4 2 4 2

!

8=y

x2 − (8)2 = 64

Substituting in equation (2)

x = 128 √ √ x = ± 128 = ±8 2 √ √ The pairs (−8 2, 8) and (8 2, 8) check. 2

22.

4x2 + 4y 2 = 65, (1) 6x2 − 4y 2 = 25 (2) 10x2 = 90 Adding x2 = 9 √ x = ± 9 = ±3

4(±3)2 + 4y 2 = 65

Substituting in (1)

36 + 4y 2 = 65 4y 2 = 29 (

√ 29 ± 29 y=± = 4 2 √ " ! √ " ! √ " ! 29 29 29 The pairs 3, , − 3, , 3, − , and 2 2 2 √ " ! 29 check. − 3, − 2 23.

x2 − y 2 = 33,

(1)

x + y = 11

(2)

y = −x + 11

x2 − (−x + 11)2 = 33

x2 − (x2 − 22x + 121) = 33 x2 − x2 + 22x − 121 = 33

22x = 154 x=7

y = −7 + 11 = 4

The pair (7, 4) checks.

Substituting in (1)

638 24.

Chapter 9: Analytic Geometry Topics x2 = 0 + 3

x2 − 2x + 2y 2 = 8, (1)

2x + y = 6

x2 = 3

(2)

y = −2x + 6

x2 − 2x + 2(−2x + 6)2 = 8

√ x=± 3

Substituting in (1)

x2 = 1 + 3

x2 − 2x + 2(4x2 − 24x + 36) = 8

x2 = 4

x2 − 2x + 8x2 − 48x + 72 = 8

x = ±2

9x2 − 50x + 64 = 0

(x − 2)(9x − 32) = 0

x − 2 = 0 or 9x − 32 = 0 x = 2 or

√ √ The pairs ( 3, 0), (− 3, 0), (2, 1), and (−2, 1) check. 28.

2x + y = 3

9x = 32 32 x= 9

x = 2 or

(5x + 3)(x − 3) = 0

5x + 3 = 0

2x − x2 + 3 = 3

29.

The pairs (0, −3) and (2, 1) check. x + y = 25, (1) 2

x2 − y 2 = 7 2x2 = 32

(2) Adding

(2)

3x + 3y = 300

Multiplying (1) by 3

2x2 − 3y 2 = 5x2 = x2 = x=

Adding

2

−120 180 36 ±6

x2 = 16

y 2 = 64

x = ±4

y = ±8

The pairs (6, 8), (−6, 8), (6, −8), and (−6, −8) check. 30.

y2 = 9 y = ±3

The pairs (4, 3), (4, −3), (−4, 3) and (−4, −3) check. 27.

2x − 3y = −120

(±6)2 + y 2 = 100

(±4)2 + y 2 = 25

x − y = 3, 2

(1)

2

2

y = 22 − 3 = 1

26.

x2 + y 2 = 100, 2

y = 02 − 3 = −3

2

or x − 3 = 0

5x = −3 or x=3 3 x = − or x=3 5 " ! 21 3 +3= y = −2 − 5 5 y = −2 · 3 + 3 = −3 ! " 3 21 and (3, −3) check. The pairs − , 5 5

2x − (x2 − 3) = 3

x = 0 or x = 2

2

5x2 − 12x − 9 = 0

From equation (1) we have y = x2 − 3. Substitute in equation (2).

0 = x(x − 2)

2

y = x2 − 3

(1) (2)

From equation (2) we have x2 = y + 3. Substitute in equation (1). y + 3 − y2 = 3

0 = y2 − y

0 = y(y − 1)

y = 0 or y = 1

Substituting in (1)

x + 4x − 12x + 9 = 18 2

(2)

0 = x2 − 2x

Solving (2) for y

x2 + (−2x + 3)2 = 18

x2 − y = 3, (1)

2x − y = 3

(2)

y = −2x + 3

y = −2(2) + 6 = 2 ! " 10 32 +6=− y = −2 9 9 ! " 32 10 The pairs (2, 2) and ,− check. 9 9 25.

x2 + y 2 = 18, (1)

x2 + 2y 2 = 12, (1) xy = 4 4 y= x

(2)

Solving (2) for y. ! "2 4 = 12 Substituting in (1) x2 + 2 x 32 x2 + 2 = 12 x x4 + 32 = 12x2

x4 − 12x2 + 32 = 0

(x2 − 4)(x2 − 8) = 0

Chapter 9 Review Exercises x2 − 4 = 0

x =4 2

639

or x2 − 8 = 0

x2 = 8

or

√ x = ±2 2

x = ±2 or

4 =2 2 4 y= = −2 −2 √ 4 y= √ = 2 2 2 √ 4 √ =− 2 y= −2 2

y=

√ √ √ √ The pairs (2, 2), (−2, −2), (2 2, 2) and (−2 2, − 2) check. 31. Familiarize. Let x and y represent the numbers. Translate. The sum of the numbers is 11. x + y = 11 The sum of the squares of the numbers is 65. x2 + y 2 = 65 Carry out. We solve the system of equations. x + y = 11,

(1)

2

First we solve equation (1) for y.

2(l − 7)(l − 12) = 0

l − 7 = 0 or l − 12 = 0 l = 7 or

l = 12

84 84 = 12. If l = 12, then w = = 7. If l = 7, then w = 7 12 Since length is usually considered to be longer than width, we have l = 12 and w = 7, or (12, 7). Check. If l = 12 and w = 7, then the perimeter is 2 · 12 + 2 · 7, or 38, and the area is 12 · 7, or 84. The solution checks.

33. Familiarize. Let x and y represent the positive integers.

y = 11 − x

Translate. The sum of the numbers is 12.

Then substitute 11 − x for y in equation (2) and solve for x. x2 + (11 − x)2 = 65 x2 + 121 − 22x + x2 = 65 2x2 − 22x + 121 = 65 2x2 − 22x + 56 = 0 x2 − 11x + 28 = 0

Dividing by 2

(x − 4)(x − 7) = 0

x − 4 = 0 or x − 7 = 0 x = 4 or

x=7

If x = 4, then y = 11 − 4 = 7. If x = 7, then y = 11 − 7 = 4.

In either case, the possible numbers are 4 and 7. Check. 4 + 7 = 11 and 42 + 72 = 16 + 49 = 65. The answer checks. State. The numbers are 4 and 7. 32. Familiarize. Let l and w represent the length and width of the rectangle, in meters, respectively. Translate. The perimeter is 38 m. 2l + 2w = 38

x + y = 12 The sum of the reciprocals is

3 . 8

1 3 1 + = x y 8 Carry out. We solve the system of equations. x + y = 12, (1) 1 3 1 + = (2) x y 8 First solve equation (1) for y. y = 12 − x

Then substitute 12 − x for y in equation (2) and solve for x. 1 3 1 + = , LCD is 8x(12 − x) x 12 − x 8 " ! 1 1 3 + = 8x(12 − x) · 8x(12 − x) x 12 − x 8 8(12 − x) + 8x = x(12 − x) · 3 96 − 8x + 8x = 36x − 3x2 96 = 36x − 3x2

3x2 − 36x + 96 = 0 x2 − 12x + 32 = 0

The area is 84 m2 .

(x − 4)(x − 8) = 0

lw = 84 Carry out. We solve the system of equations: 2l + 2w = 38, (1) lw = 84

2l2 − 38l + 168 = 0

State. The length of the rectangle is 12 m, and the width is 7 m.

x + y = 65 (2) 2

First solve equation (2) for w. 84 w= l 84 Then substitute for w in equation (1) and solve for l. l 84 = 38 2l + 2 · l 168 = 38 2l + l 2 2l + 168 = 38l Multiplying by l

(2)

x − 4 = 0 or x − 8 = 0 x = 4 or

x=8

If x = 4, y = 12 − 4 = 8.

Dividing by 3

640

Chapter 9: Analytic Geometry Topics Since the length of a radius cannot be negative, we consider only x = 11. Substitute 11 for x in equation (1) and solve for y.

If x = 8, y = 12 − 8 = 4.

In either case, the possible numbers are 4 and 8. 2 1 3 1 1 Check. 4 + 8 = 12; + = + = . The answer 4 8 8 8 8 checks.

π · 112 + πy 2 = 130π 121π + πy 2 = 130π

State. The numbers are 4 and 8.

πy 2 = 9π

34. Familiarize. Let x and y represent the length of a side of the larger and smaller squares, in centimeters, respectively.

y2 = 9 y = ±3

Translate. The perimeter of the larger square is 12 cm more than the perimeter of the smaller square.

Again, we consider only the positive solution. Check. If the radii are 11 ft and 3 ft, the sum of the areas is π · 112 + π · 32 = 121π + 9π = 130π ft2 . The difference of the areas is 121π − 9π = 112π ft2 . The answer checks.

4x = 4y + 12 The area of the larger square exceeds the area of the other square by 39 cm2 .

State. The radius of the larger circle is 11 ft, and the radius of the smaller circle is 3 ft.

x2 = y 2 + 39 We have a system of equations. 4x = 4y + 12, (1) x2 = y 2 + 39

36.

Graph: y ≤ 4 − x2 , x−y ≤2

(2)

The solution set of y ≤ 4 − x2 is the parabola y = 4 − x2 and the region inside it. The solution set of x − y ≤ 2 is the line x − y = 2 and the half-plane above it. We shade the region common to the two solution sets.

Carry out. We solve the system of equations. First solve equation (1) for x. x=y+3 Then substitute y + 3 for x in equation (2) and solve for y. (y + 3)2 = y 2 + 39

y 5 3 2 1

y 2 + 6y + 9 = y 2 + 39 6y + 9 = 39

!5 !4 !3

6y = 30 y=5

(!3, !5)

If y = 5, then x = 5 + 3 = 8.

35. Familiarize. Let x = the radius of the larger circle and let y = the radius of the smaller circle. We will use the formula for the area of a circle, A = πr2 . Translate. The sum of the areas is 130π ft2 . πx2 + πy 2 = 130π

x − y = 2.

The points of intersection are (−3, −5) and (2, 0). 37.

Graph: x2 + y 2 ≤ 16, x+y <4

The solution set of x2 + y 2 ≤ 16 is the circle x2 + y 2 = 16 and the region inside it. The solution set of x + y < 4 is the half-plane below the line x + y = 4. We shade the region common to the two solution sets. y

πx2 − πy 2 = 112π

5

We have a system of equations. πx2 + πy 2 = 130π, (1) (2)

Carry out. We add. πx2 + πy 2 = 130π πx − πy = 2πx2 = x2 = x= 2

2

112π 242π 121 ±11

(0, 4)

3 2 1 !5

!3 !2 !1 !1 !2 !3 !5

Dividing by 2π

x

3 4 5

y = 4 − x2 ,

The difference of the areas is 112π ft2 .

πx2 − πy 2 = 112π

(2, 0) 1

To find the points of intersection of the graphs we solve the system of equations

Check. The perimeters of the square are 4 · 8, or 32 cm, and 4 · 5, or 20 cm. The perimeter of the larger square is 12 cm more than the perimeter of the smaller square. The areas are 82 , or 64 cm2 , and 52 , or 25 cm2 . The area of the larger square exceeds the area of the smaller square by 39 cm2 . The solution checks. State. The perimeters of the square are 32 cm and 20 cm.

!1 !1 !2 !3 !4 !5

(4, 0) 1 2 3

5

x

Chapter 9 Review Exercises 38.

Graph: y ≥ x2 − 1, y<1

The solution set of y ≥ x2 −1 is the parabola y = x2 −1 and the region inside it. The solution set of y < 1 is the halfplane below the line y = 1. We shade the region common to the two solution sets. y 5 4 3 2

(!!2, 1)

Substitute for x and y in the given equation. %√ &2 %√ &% √ & 2 " 2 " 2 " 5 (x − y " ) − 2 (x − y " ) (x + y " ) + 2 2 2 &2 %√ 2 " (x + y " ) − 24 = 0 5 2 After simplifying we have 4(x" )2 + 6(y " )2 − 24 = 0, or

(!2, 1) 1 2 3 4 5

!5 !4 !3 !2 !1

641

x

!2 !3 !4 !5

(x" )2 (y " )2 + = 1. 6 4 √ This is √ the equation "of an ellipse "with vertices (− 6, 0) and ( 6, 0) on the x -axis. The y -intercepts are (0, −2) and (0,2). y

To find the points of intersection of the graphs we solve the system of equations y = x2 − 1,

y#

x ≤ −1

The solution set of x2 + y 2 ≤ 9 is the circle x2 + y 2 = 9 and the region inside it. The solution set of x ≤ −1 is the line x = −1 and the half-plane to the left of it. We shade the region common to the two solution sets. y

(!1, 2!2)

5 4 2 1

!5 !4

!2 !1 !1

1 2

4 5

x

!2

(!1, !2!2)

!4 !5

40. 5x2 − 2xy + 5y 2 − 24 = 0 A = 5, B = −2, C = 5

B 2 − 4AC = (−2)2 − 4 · 5 · 5 = 4 − 100 = −96

Since the discriminant is negative, the graph is an ellipse (or circle). To rotate the axes we first determine θ. 5−5 A−C = =0 cot 2θ = B −2 Then 2θ = 90◦ and θ = 45◦ , so √ √ 2 2 sin θ = and cos θ = . 2 2 Now substitute in the rotation of axes formulas. x = x" cos θ − y " sin θ !√ " !√ " √ 2 2 2 " − y" = (x − y " ) = x" 2 2 2 y = x" sin θ + y " cos θ !√ " !√ " √ 2 2 2 " = x" + y" = (x + y " ) 2 2 2

45" 2

4

x

!2

√ √ The points of intersection are (− 2, 1) and ( 2, 1). Graph: x2 + y 2 ≤ 9,

x#

2 !4 !2

y = 1.

39.

4

(x#)2 (y#)2 $ %1 6 4

!4

41. x2 − 10xy + y 2 + 12 = 0

B 2 − 4AC = (−10)2 − 4 · 1 · 1 = 100 − 4 = 96

Since the discriminant is positive, the graph is a hyperbola. To rotate the axes we first determine θ. 1−1 A−C = =0 cot 2θ = B −10 ◦ ◦ Then 2θ = 90 and θ = 45 , so √ √ 2 2 sin θ = and cos θ = . 2 2 Now substitute in the rotation of axes formulas. x = x" cos θ − y " sin θ !√ " !√ " √ 2 2 2 " " " −y = (x − y " ) =x 2 2 2 y = x" sin θ + y " cos θ !√ " !√ " √ 2 2 2 " (x + y " ) + y" = = x" 2 2 2 After substituting for x and y in the given equation and simplifying, we have (y " )2 (x" )2 − = 1. 3 2 y y#

4

x#

2

45" 2

!4 !2 !2 !4

(x#)2 3

4

!

x (y#)2 %1 2

642

Chapter 9: Analytic Geometry Topics

√ 42. 5x2 + 6 3xy − y 2 = 16 √ A = 5, B = 6 3, C = −1 √ B 2 − 4AC = (6 3)2 − 4 · 5 · (−1) = 108 + 20 = 128

Since the discriminant is positive, the graph is a hyperbola. To rotate the axes we first determine θ. 6 1 5 − (−1) A−C √ = √ =√ = cot 2θ = B 6 3 6 3 3 Then 2θ = 60◦ and θ = 30◦ , so √ 1 3 sin θ = and cos θ = . 2 2 Now substitute in the rotation of axes formulas. x = x" cos θ − y " sin θ √ √ 3 1 x" 3 y " = x" · − y" · = − 2 2 2 2 y = x" sin θ + y " cos θ √ √ 1 3 x" y" 3 = x" · + y " · = + 2 2 2 2 Substitute for x and y in the given equation. √ " "2 ! √ "! " ! "√ √ x" 3 y " x y" 3 x 3 y" − +6 3 − + − 5 2 2 2 2 2 2 √ " ! " 2 x y" 3 + = 16 2 2 After simplifying we have

(y " )2 (x" )2 − = 1. 2 4 √ This is the equation of a hyperbola with vertices (− 2, 0) √ 2 2 and ( 2, 0) and asymptotes y " = − √ x" and y " = √ x" , 2 2 √ √ or y " = − 2x" and y " = 2x" . y y#

4

30" 2

4

x

!4

√

2x +

√

2y = 0

B − 4AC = 2 − 4 · 1 · 1 = 4 − 4 = 0 2

(x" )2 = −y " . y y#

2

Since the discriminant is zero, the graph is a parabola. To rotate the axes we first determine θ. 1−1 A−C = =0 cot 2θ = B 2 ◦ ◦ Then 2θ = 90 and θ = 45 , so √ √ 2 2 and cos θ = . sin θ = 2 2 Now substitute in the rotation of axes formulas.

(x#)2 $ !y# x#

4 2

45" 2

!4 !2

4

x

!2 !4

6 3 − 3 sin θ We first divide the numerator and denominator by 3: 2 r= 1 − sin θ ep This equation is in the form r = with e = 1. 1 − e sin θ Since e = 1, the graph is a parabola.

44. r =

y 6 4 2 2

!4 !2 !2

x#

!2

43. x2 + 2xy + y 2 −

y = x" sin θ + y " cos θ !√ " !√ " √ 2 2 2 " " " =x +y = (x + y " ) 2 2 2 After substituting for x and y in the given equation and simplifying, we have

(x#)2 (y#)2 ! $1 2 4

2 !4 !2

x = x" cos θ − y " sin θ !√ " !√ " √ 2 2 2 " − y" = (x − y " ) = x" 2 2 2

r=

4

x 6 3 − 3 sin q

Since e = 1 and ep = 1 · p = 2, we have p = 2.

Thus the parabola has a horizontal directrix 2 units below the pole. 3π 3π . When θ = , We find the vertex by letting θ = 2 2 6 6 = r= = 1. 3π 3 − 3(−1) 3 − 3 sin 2 " ! 3π . Thus the vertex is 1, 2 8 2 + 4 cos θ We first divide the numerator and denominator by 2: 4 r= 1 + 2 cos θ ep Thus the equation is in the form r = with 1 + e cos θ e = 2. Since e > 1, the graph is a hyperbola.

45. r =

Chapter 9 Review Exercises y

18 9 + 6 sin θ We first divide the numerator and denominator by 9: 2 r= 2 1 + sin θ 3 ep with Thus the equation is in the form r = 1 + e sin θ 2 e= . 3 Since 0 < e < 1, the graph is an ellipse.

47. r =

4

r=

2 4

!4

643

8

8 2 + 4 cos q 12

x

!2 !4

Since e = 2 and ep = 2 · p = 4, we have p = 2. Thus the hyperbola has a vertical directrix 2 units to the right of the pole.

y

We find the vertices by letting θ = 0 and θ = π. When θ = 0, 8 4 8 = = . r= 2 + 4 cos 0 2+4 3 When θ = π, 8 8 = = −4. r= 2 + 4 cos π 2−4 ! " 4 , 0 and (−4, π). The vertices are 3

2

!6

4 2 6

x

48.

!2 !4

r=

4 2 − cos q

6 3 − 3 sin θ 3r − 3r sin θ = 6 r=

r = r sin θ + 2 #

We find the vertices by letting θ = 0 and θ = π.

x2

+ y2 = y + 2

x + y 2 = y 2 + 4y + 4 2

x2 − 4y − 4 = 0

When θ = 0, 4 4 = = 4. 2 − cos 0 2−1 When θ = π, 4 4 4 r= = = . 2 − cos π 2 − (−1) 3 " ! 4 ,π . The vertices are (4, 0) and 3

18 9 + 6 sin q

3r = 3r sin θ + 6

1 1 Since e = and ep = · p = 2, we have p = 4. Thus the 2 2 ellipse has a vertical directrix 4 units to the left of the pole.

r=

r=

2 2 Since e = and ep = · p = 2, we have p = 3. Thus the 3 3 ellipse has a horizontal directrix 3 units above the pole. 3π π . We find the vertices by letting θ = and θ = 2 2 π When θ = , 2 18 6 18 r= π = 9 + 6 = 5. 9 + 6 sin 2 3π When θ = , 2 18 18 = = 6. r= 3π 9−6 9 + 6 sin 2 " ! " ! 6 π 3π The vertices are , and 6, . 5 2 2

y

4

x

!4

4 2 − cos θ We first divide the numerator and denominator by 2: 2 r= 1 1 − cos θ 2 ep Then the equation is in the form r = with 1 − e cos θ 1 e= . 2 Since 0 < e < 1, the graph is an ellipse.

2

4

!2

46. r =

!2

2

!4 !2

49.

8 2 + 4 cos θ 2r + 4r cos θ = 8 r=

2r = −4r cos θ + 8 #

x2

r = −2r cos θ + 4

+ y 2 = −2x + 4

x2 + y 2 = 4x2 − 16x + 16

0 = 3x2 − y 2 − 16x + 16

644

Chapter 9: Analytic Geometry Topics 4 2 − cos θ 2r − r cos θ = 4

50.

r=

2r = r cos θ + 4

2

'

x2 + y 2 = x + 4

4x2 + 4y 2 = x2 + 8x + 16

3x2 + 4y 2 − 8x − 16 = 0

18 9 + 6 sin θ 9r + 6r sin θ = 18

51.

r=

9

'

3

'

x2

9r = −6r sin θ + 18

+ y 2 = −6y + 18

x2 + y 2 = −2y + 6

9x2 + 9y 2 = 4y 2 − 24y + 36

9x2 + 5y 2 + 24y − 36 = 0

1 , r = 2 sec θ 2 The equation of the directrix can be written 2 , or r cos θ = 2. r= cos θ This corresponds to the equation x = 2 in rectangular coordinates, so the directrix is a vertical line 2 units to the right of the pole. Using the table on page 824 in the text, we see that the equation is of the form ep . r= 1 + e cos θ 1 Substituting for e and 2 for p, we have 2 1 ·2 1 2 2 = , or . r= 1 1 2 + cos θ 1 + cos θ 1 + cos θ 2 2

52. e =

53. e = 3, r = −6 csc θ

The equation of the directrix can be written 6 , or r sin θ = −6. r=− sin θ This corresponds to the equation y = −6 in rectangular coordinates, so the directrix is a horizontal line 6 units below the pole. Using the table on page 824 in the text, we see that the equation is of the form ep . r= 1 − e sin θ Substituting 3 for e and 6 for p, we have 18 3·6 = . r= 1 − 3 sin θ 1 − 3 sin θ

54. e = 1, r = −4 sec θ

The equation of the directrix can be written 4 , or r cos θ = −4. r=− cos θ This corresponds to the equation x = −4 in rectangular coordinates, so the directrix is a vertical line 4 units to the left of the pole. Using the table on page 824 in the text, we see that the equation is of the form

ep . 1 − e cos θ Substituting 1 for e and 4 for p, we have 4 1·4 = . r= 1 − 1 · cos θ 1 − cos θ r=

55. e = 2, r = 3 csc θ

The equation of the directrix can be written 3 , or r sin θ = 3. r= sin θ This corresponds to the equation y = 3 in rectangular coordinates, so the directrix is a horizontal line 3 units above the pole. Using the table on page 824 in the text, we see that the equation is of the form ep . r= 1 + e sin θ Substituting 2 for e and 3 for p, we have 2·3 6 r= = . 1 + 2 sin θ 1 + 2 sin θ 56. x = t, y = 2 + t; −3 ≤ t ≤ 3 y 6 4 2 !4

2

!2

x

4

!2

x = t, y = 2 + t; −3 & t & 3

To find an equivalent rectangular equation, we substitute x for t in y = 2 + t: y = 2 + x. Given that −3 ≤ t ≤ 3, we find the corresponding restrictions on y: For t = −3: x = t = −3. For t = 3: x = t = 3.

Then we have y = 2 + x, −3 ≤ x ≤ 3. √ 57. x = t, y = t − 1; 0 ≤ t ≤ 9 y 6 4 2 1

!1

2

3

x

!2

x = √t, y = t − 1; 0 ' t ' 9

To find √ an equivalent rectangular equation, we first solve x = t for t. √ x= t x2 = t Then we substitute x2 for t in y = t − 1: y = x2 − 1. Given that 0 ≤ t ≤ 9, we find the corresponding restrictions on x:

Chapter 9 Review Exercises For t = 0: x = For t = 9: x =

√

√

t= t=

√ √

645 61. y = x2 + 4

0 = 0. 9 = 3.

Answers may vary.

Then we have y = x2 − 1, 0 ≤ x ≤ 3. 58. x = 2 cos t, y = 2 sin t; 0 ≤ t ≤ 2π y

If x = t, y = t2 + 4. If x = t − 2, y = (t − 2)2 + 4 = t2 − 4t + 8. 62. a) We substitute 0 for h, 150 for v0 , and 45◦ for θ in the parametric equations for projectile motion. x = (v0 cos θ)t

4

= (150 cos 45◦ )t √ " ! √ 2 t = 75 2t = 150 · 2

2 2

!4 !2

4

x

!2

y = h + (v0 sin θ)t − 16t2

!4

= 0 + (150 sin 45◦ )t − 16t2 √ " ! √ 2 t − 16t2 = 75 2t − 16t2 = 150 · 2

x = 2 cos t, y = 2 sin t; 0 ' t ' 2p

To find an equivalent rectangular equation, we first solve for cos t and sin t in the parametric equations: y x = cos t, = sin t 2 2 Using the identity sin2 θ + cos2 θ = 1, we can eliminate the parameter:

b) The height of the projectile at time t is given by y. √ When t = 3, y = 75 2(3) − 16 · 32 ≈ 174.2 ft. √ When t = 6, y = 75 2(6) − 16 · 62 ≈ 60.4 ft.

c) The ball hits the ground when y = 0, so we solve the equation y = 0. √ 75 2t − 16t2 = 0 √ t(75 2 − 16t) = 0

sin2 t + cos2 t = 1 ! "2 ! "2 x y + =1 2 2

t = 0 or t ≈ 6.6 Time t = 0 corresponds to the time before the projectile is launched. The projectile hits the ground when t ≈ 6.6, so it is in the air for about 6.6 sec.

x2 y2 + =1 4 4 x2 + y 2 = 4 59. x = 3 sin t, y = cos t; 0 ≤ t ≤ 2π

d) Since the projectile is in the air for about 6.6 sec, the horizontal distance it travels is given by √ x = 75 2(6.6) ≈ 700 ft.

y 4

e) The maximum height of the projectile is the maximum value of the quadratic function represented by y. At the vertex of that function we have √ 75 2 b =− ≈ 3.3 t=− 2a 2(−16) When t ≈ 3.3, √ y ≈ 75 2(3.3) − 16(3.3)2 ≈ 175.8 ft.

2 2

!4 !2

4

x

!2 !4

x = 3 sin t, y = 2 cos t; 0 ' t ' 2p

To find an equivalent rectangular equation, we first solve x = sin t. x = 3 sin t for sin t: 3 Using the identity sin2 θ + cos2 θ = 1, we can eliminate the parameter: sin t + cos t = 1 ! "2 x + y2 = 1 3 2

2

x2 + y2 = 1 9 60. y = 2x − 3

Answers may vary. If x = t, then y = 2t − 3.

If x = t + 1, then y = 2(t + 1) − 3 = 2t − 1.

63.

y 2 − 4y − 12x − 8 = 0

y 2 − 4y = 12x + 8

y − 4y + 4 = 12x + 8 + 4 2

(y − 2)2 = 12x + 12

(y − 2)2 = 12(x + 1)

(y − 2)2 = 12[x − (−1)]

The vertex of this parabola is (−1, 2), so answer B is correct. 64. A straight line can intersect an ellipse at 0 points, 1 point, or 2 points but not at 4 points, so answer D is correct. 65. Although we can always visualize the real-number solutions, we cannot visualize the imaginary-number solutions.

646

Chapter 9: Analytic Geometry Topics

66. The equation of a circle can be written as (y − k)2 (x − h)2 + =1 2 a b2 where a = b = r, the radius of the circle. In an ellipse, a > b, so a circle is not a special type of ellipse. 67. The vertices are (0, −3) and (0, 3), so the center is (0, 0), and the major axis is vertical. y2 x2 + 2 =1 a2 3 √ " ! 1 3 3 and solve for a. Substitute − , 2 2 "2 ! √ "2 ! 1 3 3 − 2 2 + =1 a2 32 3 1 + =1 4a2 4 1 + 3a2 = 4a2 Multiplying by 4a2 1 = a2

If x =

8 4 7 7 ,y= =4· = . 7 8/7 8 2

8 7 and . 2 7 7 8 1 1 2 7 16 49 65 Check. · = 4; + = + = + = . 2 7 7/2 8/7 7 8 56 56 56 The answer checks. 8 7 State. The numbers are and . 2 7 In either case the possible numbers are

69. Using (x − h)2 + (y − k)2 = r2 and the given points, we have (10 − h)2 + (7 − k)2 = r2 (1) (−6 − h)2 + (7 − k)2 = r2 (−8 − h) + (1 − k) = r 2

2

(2) (3)

Subtract equation (2) from equation (1). (10 − h)2 − (−6 − h)2 = 0 64 − 32h = 0 h=2

Subtract equation (3) from equation (2).

1=a y2 = 1. The equation of the ellipse is x + 9 2

(−6 − h)2 − (−8 − h)2 + (7 − k)2 − (1 − k)2 = 0 20 − 4h − 12k = 0

20 − 4(2) − 12k = 0

68. Familiarize. Let x and y represent the numbers. Translate. The product of the numbers is 4. xy = 4 The sum of the reciprocals is

2

65 . 56

1 65 1 + = x y 56 Carry out. We solve the system of equations. xy = 4, (1) 1 1 65 + = (2) x y 56 First solve equation (1) for y. 4 y= x 4 Then substitute for y in equation (2) and solve for x. x 1 1 65 + = x 4/x 56 65 1 x + = x 4 56 " ! 1 x 65 + = 56x · 56x x 4 56 56 + 14x2 = 65x 14x − 65x + 56 = 0

k=1

Substitute h = 2, k = 1 into equation (1). (10 − 2)2 + (7 − 1)2 = r2 100 = r2 10 = r The equation is (x − 2)2 + (y − 1)2 = 100. 70. A and B are the foci of the hyperbola, so c =

400 = 200. 2

0.186 mi = 55.8 mi, the differ300 microseconds · 1 microsecond ence of the ship’s distances from the foci. That is, 2a = 55.8, so a = 27.9. Find b2 . c2 = a2 + b2 2002 = 27.92 + b2 39, 221.59 = b2 Then the equation of the hyperbola is y2 x2 − = 1. 778.41 39, 221.59

Chapter 9 Test

2

(2x − 7)(7x − 8) = 0

2x − 7 = 0 or 7x − 8 = 0

2x = 7 or 7x = 8 8 7 or x= x= 2 7 4 2 8 7 =4· = . If x = , y = 2 7/2 7 7

1. Graph (c) is the graph of 4x2 − y 2 = 4. 2. Graph (b) is the graph of x2 − 2x − 3y = 5. 3. Graph (a) is the graph of x2 + 4x + y 2 − 2y − 4 = 0. 4. Graph (d) is the graph of 9x2 + 4y 2 = 36.

Chapter 9 Test 5.

647 9.

x2 = 12y x2 = 4 · 3y

9x2 + 16y 2 = 144 y2 x2 + =1 16 9 x2 y2 + 2 =1 2 4 3 a = 4, b = 3

V : (0, 0), F : (0, 3), D : y = −3

The center is (0, 0). The major axis is horizontal, so the 2 2 2 vertices are (−4, 0) and (4, √ 0). Since c = a − b√, we have 2 c√ = 16 − 9 = 7, so c = 7 and the foci are (− 7, 0) and ( 7, 0). 6.

y 2 + 2y − 8x − 7 = 0

y 2 + 2y = 8x + 7

y + 2y + 1 = 8x + 7 + 1 2

(y + 1)2 = 8x + 8 [y − (−1)]2 = 4(2)[x − (−1)]

V : (−1, −1)

F : (−1 + 2, −1) or (1, −1) D : x = −1 − 2 = −3

10.

(y − 2)2 (x + 1)2 + =1 4 9 (y − 2)2 [x − (−1)]2 + =1 22 32 The center is (−1, 2). Note that a = 3 and b = 2. The major axis is vertical, so the vertices are 3 units below and above the center: (−1, 2 − 3) and (−1, 2 + 3) or (−1, −1) and (−1, 5).

7.

√ We know that c2 =√a2 − b2 , so c2 = 9 − 4 = 5 and c = 5. Then the foci are 5 units below and above the center: √ √ (−1, 2 − 5) and (−1, 2 + 5).

(x − h)2 = 4p(y − k)

(x − 0)2 = 4 · 2(y − 0) x2 = 8y

8. Begin by completing the square twice. x2 + y 2 + 2x − 6y − 15 = 0

x2 + 2x + y 2 − 6y = 15

(x2 + 2x + 1) + (y 2 − 6y + 9) = 15 + 1 + 9 (x + 1)2 + (y − 3)2 = 25

[x − (−1)]2 + (y − 3)2 = 52

Center: (−1, 3), radius: 5

11. The vertices (0, −5) and (0, 5) are on the y-axis, so the major axis is vertical and a = 5. Since the vertices are equidistant from the center, the center of the ellipse is at 4 the origin. The length of the minor axis is 4, so b = = 2. 2 x2 y2 The equation is + = 1. 4 25 12.

4x2 − y 2 = 4

y2 x2 − =1 1 4 y2 x2 − 2 =1 2 1 2 The center is (0, 0); a = 1 and b = 2.

648

Chapter 9: Analytic Geometry Topics The transverse axis is horizontal, so the vertices are (−1, 0) and √ (1, 0). Since c2 = a2 + b2 ,√we have c2 √ = 1 + 4 = 5 and c = 5. Then the foci are (− 5, 0) and ( 5, 0).

14.

2y 2 − x2 = 18

x2 y2 − =1 9 18 y2 x2 − √ =1 2 3 (3 2)2

Find the asymptotes: b b y = x and y = − x a a 2 2 y = x and y = − x 1 1 y = 2x and y = −2x

√ h = 0, k = 0, a = 3, b = 3 2 a a and y − k = − (x − h) y − k = (x − h) b b 3 3 y − 0 = √ (x − 0) and y − 0 = − √ (x − 0) 3 2 3 2 √ √ 2 2 y= x and y=− x 2 2 2 15. The parabola ! is of"the form y = 4px. A point on the 18 parabola is 6, , or (6, 9). 2

y 2 = 4px 92 = 4 · p · 6 13.

81 = 24p 27 =p 8

(x + 1) (y − 2) − =1 4 9 2

2

[x − (−1)]2 (y − 2)2 − =1 22 32 The center is (−1, 2); a = 2 and b = 3.

Since the focus is at (p, 0) = from the vertex.

The transverse axis is vertical, so the vertices are 2 units below and above the center:

16.

Find the asymptotes: a and y − k = y − k = (x − h) b 2 y − 2 = (x − (−1)) and y − 2 = 3 2 y − 2 = (x + 1) and y − 2 = 3 8 2 and y= y = x+ 3 3

a − (x − h) b 2 − (x − (−1)) 3 2 − (x + 1) 3 2 4 − x+ 3 3

(2)

2x2 − 3y 2 = −10

√ , we have c2 = 4 + 9 = 13 and c = 13. Since c2 = a2 + b2 √ Then the foci are 13 units below and above the center: √ √ (−1, 2 − 13) and (−1, 2 + 13).

" 27 27 , 0 , the focus is in. 8 8

2x2 − 3y 2 = −10, (1) x2 + 2y 2 = 9

(−1, 2 − 2) and (−1, 2 + 2) or (−1, 0) and (−1, 4).

!

−2x2 − 4y 2 = −18 − 7y 2 = −28

Multiplying (2) by −2 Adding

y2 = 4

y = ±2

x + 2(±2)2 = 9 2

Substituting into (2)

x +8 = 9 2

x2 = 1 x = ±1

The pairs (1, 2), (1, −2), (−1, 2) and (−1, −2) check. 17.

x2 + y 2 = 13, (1) x+y =1

(2)

First solve equation (2) for y. y =1−x

Then substitute 1 − x for y in equation (1) and solve for x. x2 + (1 − x)2 = 13 x2 + 1 − 2x + x2 = 13

2x2 − 2x − 12 = 0

2(x2 − x − 6) = 0

2(x − 3)(x + 2) = 0

x = 3 or x = −2

If x = 3, y = 1 − 3 = −2. If x = −2, y = 1 − (−2) = 3. The pairs (3, −2) and (−2, 3) check.

Chapter 9 Test 18.

649 2700 = 210 l 5400 = 210 2l + l 2 2l + 5400 = 210l

x + y = 5, (1) xy = 6

2l + 2 ·

(2)

First solve equation (1) for y. y = −x + 5

Multiplying by l

2l2 − 210l + 5400 = 0

Then substitute −x + 5 for y in equation (2) and solve for x.

2(l2 − 105l + 2700) = 0

x(−x + 5) = 6

2(l − 45)(l − 60) = 0

−x2 + 5x − 6 = 0

l = 45 or l = 60

x = 2 or x = 3

2700 2700 = 60. If l = 60, then w = = If l = 45, then w = 45 60 45. Since length is usually considered to be longer than width, we have l = 60 and w = 45.

If x = 2, y = −2 + 5 = 3. If x = 3, y = −3 + 5 = 2.

Check. Perimeter: 2 · 60 + 2 · 45 = 210 ft

−1(x2 − 5x + 6) = 0

−1(x − 2)(x − 3) = 0

The pairs (2, 3) and (3, 2) check.

Area: 60 · 45 = 2700 ft2 The solution checks.

19. Familiarize. Let l and w represent the length and width of the rectangle, in feet, respectively.

State. The dimensions of the playground are 60 ft by 45 ft.

Translate. The perimeter is 18 ft. 2l + 2w = 18

(1)

21.

From the Pythagorean theorem, we have √ (2) l2 + w2 = ( 41)2 Carry out. We solve the system of equations. We first solve equation (1) for w. 2l + 2w = 18

Graph: y ≥ x2 − 4, y < 2x − 1

The solution set of y ≥ x2 − 4 is the parabola y = x2 − 4 and the region inside it. The solution set of y < 2x − 1 is the half-plane below the line y = 2x − 1. We shade the region common to the two solution sets.

2w = 18 − 2l

y

w = 9−l

Then substitute 9 − l for w in equation (2) and solve for l. √ l2 + (9 − l)2 = ( 41)2 l2 + 81 − 18l + l2 = 41

!1 !1

1

3 4 5

x

!5

2(l2 − 9l + 20) = 0 2(l − 4)(l − 5) = 0

To find the points of intersection of the graphs we solve the system of equations

l = 4 or l = 5

If l = 4, then w = 9 − 4 = 5. If l = 5, then w = 9 − 5 = 4. Since length is usually considered to be longer than width, we have l = 5 and w = 4. Check. The perimeter is 2 · 5√+ 2 · 4, or 18 ft. The length √ of a diagonal is 52 + 42 , or 41 ft. The solution checks. State. The dimensions of the garden are 5 ft by 4 ft. 20. Familiarize. Let l and w represent the length and width of the playground, in feet, respectively. Translate. Area: lw = 2700

!5 !4 !3

(!1, !3) !3

2l2 − 18l + 40 = 0

Perimeter: 2l + 2w = 210

(3, 5)

5 4 3 2 1

(1)

(2)

Carry out. We solve the system of equations. First solve equation (2) for w. 2700 w= l 2700 for w in equation (1) and solve for Then substitute l l.

y = x2 − 4,

y = 2x − 1.

The points of intersection are (−1, −3) and (3, 5). 22. 5x2 − 8xy + 5y 2 = 9

A = 5, B = −8, C = 5

B 2 − 4AC = (−8)2 − 4 · 5 · 5 = 64 − 100 = −36

Since the discriminant is negative, the graph is an ellipse (or a circle). To rotate the axes we first determine θ. 5−5 A−C = =0 cot 2θ = B −8 Then 2θ = 90◦ and θ = 45◦ , so √ √ 2 2 sin θ = and cos θ = . 2 2 Now substitute in the rotation of axes formulas. x = x" cos θ − y " sin θ !√ " !√ " √ 2 2 2 " − y" = (x − y " ) = x" 2 2 2

650

Chapter 9: Analytic Geometry Topics y = x! sin θ + y ! cos θ !√ " !√ " √ 2 2 2 ! + y! = (x + y ! ) = x! 2 2 2 Substitute for x and y in the given equation. #√ $2 #√ $# √ $ 2 ! 2 ! 2 ! 5 (x − y ! ) − 8 (x − y ! ) (x + y ! ) + 2 2 2 $2 #√ 2 ! (x + y ! ) = 9 5 2 After simplifying we have

right of the pole. Using the table on page 824 in the text, we see that the equation is of the form ep r= . 1 + e cos θ Substituting 2 for e and 3 for p, we have 2·3 6 r= = . 1 + 2 cos θ 1 + 2 cos θ 25.

y 18

(x! )2 + (y ! )2 = 1. 9 This is the equation of an ellipse with vertices (−3, 0) and (3, 0) on the x! -axis. The y ! -intercepts are (0, −1) and (0,1). We sketch the graph. !2

y y'

2 !4

2

4

x

!2 !4

23. r =

2 1 − sin θ

ep with e = 1. 1 − e sin θ Since e = 1, the graph is a parabola. The equation is of the form r =

y 2

r=

2 1 − sin q

1 1

!2 !1

2

3

4

x

26. x = 3 cos θ, y = 3 sin θ; 0 ≤ θ ≤ 2π

45"

!2

2

x = √t, y = t + 2; 0 # t # 16

x'

4

1

x

!1 !2

To find an equivalent rectangular equation, we first solve for cos θ and sin θ in the parametric equations: y x = cos θ, = sin θ 3 3 Using the identity sin2 θ + cos2 θ = 1, we can substitute to eliminate the parameter: sin2 θ + cos2 θ ! " 2 ! "2 y x + 3 3 y2 x2 + 9 9 x2 + y 2

=1 =1 =1 =9

For 0 ≤ θ ≤ 2π, −3 ≤ 3 cos θ ≤ 3.

Then we have x2 + y 2 = 9, −3 ≤ x ≤ 3. 27. y = x − 5

Answers may vary.

Since e = 1 and ep = 1 · p = 2, we have p = 2. Thus the parabola has a horizontal directrix 2 units below the pole. 3π . We find the vertex by letting θ = 2 3π , When θ = 2 2 2 r= = = 1. 3π 1 − (−1) 1 − sin 2 " ! 3π . The vertex is 1, 2 24. e = 2, r = 3 sec θ The equation of the directrix can be written 3 , or r cos θ = 3. r= cos θ This corresponds to the equation x = 3 in rectangular coordinates, so the directrix is a vertical line 3 units to the

If x = t, y = t − 5.

If x = t + 5, y = t + 5 − 5 = t.

28. a) We substitute 10 for h, 250 for v0 , and 30◦ for θ in the parametric equations for projectile motion. x = (v0 cos θ)t = (250 cos 30◦ )t √ " ! √ 3 = 250 · t = 125 3t 2 y = h + (v0 sin θ)t − 16t2

= 10 + (250 sin 30◦ )t − 16t2 ! " 1 = 10 + 250 · t − 16t2 2 = 10 + 125t − 16t2

b) The height of the projectile at time t is given by y. When t = 1, y = 10 + 125 · 1 − 16 · 12 = 119 ft.

When t = 3, y = 10 + 125 · 3 − 16 · 32 = 241 ft.

Chapter 9 Test

651

c) The ball hits the ground when y = 0, so we solve the equation y = 0. 10 + 125t − 16t2 = 0

−16t2 + 125t + 10 = 0

' (125)2 − 4(−16)(10) 2(−16) t ≈ −0.1 or t ≈ 7.9 t=

−125 ±

The negative value for t has no meaning in this application. Thus we see that the projectile is in the air for about 7.9 sec. d) Since the projectile is in the air for about 7.9 sec, the horizontal distance it travels is given by √ x = 125 3(7.9) ≈ 1710.4 ft.

e) The maximum height of the projectile is the maximum value of the quadratic function represented by y. At the vertex of that function we have 125 b =− = 3.90625. t=− 2a 2(−16) When t = 3.90625, y = 10 + 125(3.90625) − 16(3.90625)2 ≈ 254.1 ft.

29. Use the midpoint formula to find the center. " ! 1 + 5 1 + (−3) , = (3, −1) (h, k) = 2 2 Use the distance formula to find the radius. √ 1' 1' r= (1 − 5)2 + (1−(−3))2 = (−4)2 + (4)2 = 2 2 2 2 Write the equation of the circle. (x − h)2 + (y − k)2 = r2

√ (x − 3)2 + [y − (−1)]2 = (2 2)2 (x − 3)2 + (y + 1)2 = 8

Chapter 10

Sequences, Series, and Combinatorics an =

n2 − 1 , n2 + 1

a1 =

12 − 1 = 0, 12 + 1

a1 = 4 · 1 − 1 = 3,

a2 =

22 − 1 3 = , 22 + 1 5

a3 = 4 · 3 − 1 = 11,

a3 =

32 − 1 8 4 = = , 32 + 1 10 5

a10 = 4 · 10 − 1 = 39;

a4 =

42 − 1 15 = ; 42 + 1 17

5.

Exercise Set 10.1 1.

an = 4n − 1

a2 = 4 · 2 − 1 = 7,

a4 = 4 · 4 − 1 = 15;

2.

a15 = 4 · 15 − 1 = 59

a1 = (1 − 1)(1 − 2)(1 − 3) = 0,

a10 =

a3 = (3 − 1)(3 − 2)(3 − 3) = 0,

a15 =

a2 = (2 − 1)(2 − 2)(2 − 3) = 0, a4 = (4 − 1)(4 − 2)(4 − 3) = 3 · 2 · 1 = 6;

a10 = (10 − 1)(10 − 2)(10 − 3) = 9 · 8 · 7 = 504;

6.

a15 = (15−1)(15−2)(15−3) = 14 · 13 · 12 = 2184

n ,n≥2 n−1 The first 4 terms are a2 , a3 , a4 , and a5 : 2 = 2, a2 = 2−1 3 3 = , a3 = 3−1 2 4 4 = , a4 = 4−1 3 5 5 a5 = = ; 5−1 4 10 10 = ; a10 = 10 − 1 9 15 15 = a15 = 15 − 1 14

a2 =

3. an =

4.

a3 = a4 = a10 = a15 = 7.

a3 = (−1)3 32 = −9, a4 = (−1)4 42 = 16;

a4 = 4 − 1 = 15,

a10 = (−1)10 102 = 100;

a5 = 52 − 1 = 24, 2

a15 = 152 − 1 = 224

an = (−1)n n2 a2 = (−1)2 22 = 4,

2

a6 = 6 − 1 = 35;

152 − 1 224 112 = = 152 + 1 226 113 ! "1−1 1 − = 1, 2 "2−1 ! 1 1 − =− , 2 2 ! "3−1 1 1 − = , 2 4 ! "4−1 1 1 − =− , 2 8 ! "10−1 1 1 − =− ; 2 512 "15−1 ! 1 1 − = 2 16, 384

a1 = (−1)1 12 = −1,

a3 = 32 − 1 = 8,

a10 = 102 − 1 = 99;

a1 =

102 − 1 99 = ; 102 + 1 101

8.

a15 = (−1)15 152 = −225 a1 = (−1)1−1 (3 · 1 − 5) = −2, a2 = (−1)2−1 (3 · 2 − 5) = −1, a3 = (−1)3−1 (3 · 3 − 5) = 4,

a4 = (−1)4−1 (3 · 4 − 5) = −7;

a10 = (−1)10−1 (3 · 10 − 5) = −25, a15 = (−1)15−1 (3 · 15 − 5) = 40

654 9.

Chapter 10: Sequences, Series, and Combinatorics an = 5 +

(−2)n+1 2n

a1 = 5 +

(−2)1+1 4 = 5 + = 7, 21 2

a2 = 5 +

(−2)2+1 −8 =5+ = 3, 22 4

a3 = 5 +

(−2)3+1 16 =5+ = 7, 23 8

a4 = 5 +

(−2)4+1 −32 =5+ = 3; 24 16

a10

(−2)10+1 −1 · 211 = 5+ =5+ = 3; 10 2 210

(−2)15+1 216 = 5 + 15 = 7 15 2 2 2·1−1 1 a1 = 2 = , 1 +2·1 3 2·2−1 3 = , a2 = 2 2 +2·2 8 2·3−1 5 1 a3 = 2 = = , 3 +2·3 15 3 2·4−1 7 = ; a4 = 2 4 +2·4 24 19 2 · 10 − 1 = ; a10 = 2 10 + 2 · 10 120 2 · 15 − 1 29 = a15 = 2 15 + 2 · 15 255 an = 5n − 6

a15 = 5 + 10.

11.

a8 = 5 · 8 − 6 = 40 − 6 = 34

22. 5n − 7

2 3 4 5 6 , , , , ,. . . 3 4 5 6 7 These are fractions in which the denominator is 1 greater than the numerator. Also, each numerator is 1 greater than the preceding numerator. The general term might be n+1 . n+2 √ 24. 2n 23.

25. 1 · 2, 2 · 3, 3 · 4, 4 · 5, . . .

These are the products of pairs of consecutive natural numbers. The general term might be n(n + 1).

26. −1 − 3(n − 1), or −3n + 2, or −(3n − 2) 27. 0, log 10, log 100, log 1000, . . . We can see a pattern if we write the sequence as log 1, log 10, log 100, log 1000, . . . The general term might be log 10n−1 . This is equivalent to n − 1. 28. ln en+1 , or n + 1 29. 1, 2, 3, 4, 5, 6, 7, . . . S3 = 1 + 2 + 3 = 6 S7 = 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 30. S2 = 1 − 3 = −2

S8 = 1 − 3 + 5 − 7 + 9 = 5

31. 2, 4, 6, 8, . . .

12. a7 = (3 · 7 − 4)(2 · 7 + 5) = 17 · 19 = 323 13.

a6 = (2 · 6 + 3) = 225 14. a12 = (−1)

12−1

[4.6(12) − 18.3] = −36.9

an = 5n2 (4n − 100)

a11 = 5(11)2 (4 · 11 − 100) = 5(121)(−56) = 16. a80 17.

S5 = 2 + 4 + 6 + 8 + 10 = 30

an = (2n + 3)

2

15.

S4 = 2 + 4 + 6 + 8 = 20

2

−33, 880 "2 ! "2 ! 1 81 6561 = 1+ = = 80 80 6400

an = ln en a67 = ln e67 = 67

1000 = 2 − 10 = −8 100 19. 2, 4, 6, 8, 10,. . .

18. a100 = 2 −

These are the even integers, so the general term might be 2n.

32. S1 = 1

1 1 1 1 5269 + + + = 4 9 16 25 3600 5 # 1 1 1 1 1 1 = + + + + 33. 2k 2·1 2·2 2·3 2·4 2·5 k=1 1 1 1 1 1 = + + + + 2 4 6 8 10 30 20 15 12 60 = + + + + 120 120 120 120 120 137 = 120 1 1 1 1 1 1 43, 024 34. + + + + + = 3 5 7 9 11 13 45, 045 6 # 2i = 20 + 21 + 22 + 23 + 24 + 25 + 26 35. S5 = 1 +

i=0

20. 3n 21. −2, 6, −18, 54, . . .

We can see a pattern if we write the sequence as −1 · 2 · 1, 1 · 2 · 3, −1 · 2 · 9, 1 · 2 · 27, . . .

The general term might be (−1)n 2(3)n−1 .

36. 37.

√

= 1 + 2 + 4 + 8 + 16 + 32 + 64

7+

10 #

√

= 127 √ √ 9 + 11 + 13 ≈ 12.5679

ln k = ln 7 + ln 8 + ln 9 + ln 10 =

k=7

ln(7 · 8 · 9 · 10) = ln 5040 ≈ 8.5252

Exercise Set 10.1

655

38. π + 2π + 3π + 4π = 10π ≈ 31.4159 39.

8 #

k=1

40.

41.

48.

k 1 2 3 4 = + + + + k+1 1+1 2+1 3+1 4+1 6 7 8 5 + + + 5+1 6+1 7+1 8+1 1 2 3 4 5 6 7 8 = + + + + + + + 2 3 4 5 6 7 8 9 15, 551 = 2520

1−1 2−1 3−1 4−1 5−1 + + + + 1+3 2+3 3+3 4+3 5+3 1 2 3 4 = 0+ + + + 5 6 7 8 1 1 3 1 = 0+ + + + 5 3 7 2 307 = 210 5 # (−1)i

=

(−1)k+1 3k

52.

k=0

2

2 2 2 2 + + + + + 1 12 + 1 22 + 1 32 + 1 2 2 2 + + 42 + 1 52 + 1 62 + 1 2 2 2 2 2 = 2+1+ + + + + 5 10 17 26 37 2 1 2 2 1 + + = 2+1+ + + 5 5 17 13 37 157, 351 = 40, 885 =

02

46. 1·2+2·3+3·4+4·5+5·6+6·7+7·8+8·9+9·10+10·11 = 440 47.

5 #

k=0

(k 2 − 2k + 3)

= (02 − 2 · 0 + 3) + (12 − 2 · 1 + 3)+

∞ #

= 43

23 23 +1

+

24 24 +1

+

7k

k=1

53. 2 − 4 + 8 − 16 + 32 − 64

This is a sum of powers of 2 with alternating signs. Sigma notation is 6 6 # # (−1)k+1 2k, or (−1)k−1 2k k=1

54.

5 #

k=1

3k

k=1

1 2 3 4 5 6 55. − + − + − + 2 3 4 5 6 7 This is a sum of fractions in which the denominator is one greater than the numerator. Also, each numerator is 1 greater than the preceding numerator and the signs alternate. Sigma notation is 6 #

(22 − 2 · 2 + 3) + (32 − 2 · 3 + 3)+

(42 − 2 · 4 + 3) + (52 − 2 · 5 + 3) = 3 + 2 + 3 + 6 + 11 + 18

+

k=1

= 3 − 6 + 9 − 12 + 15 − 18 + 21 − 24

k 2 +1

22 22 +1

This is a sum of multiples of 5, and it is an infinite series. Sigma notation is ∞ # 5k.

(−1)8 3 · 7 + (−1)9 3 · 8

6 #

+

51. 5 + 10 + 15 + 20 + 25+ . . .

(−1)5 3 · 4 + (−1)6 3 · 5 + (−1)7 3 · 6+

45.

21 21 +1

50. (−2)0 + (−2)2 + (−2)4 + (−2)6 = 1 + 4 + 16 + 64 = 85

= (−1)2 3 · 1 + (−1)3 3 · 2 + (−1)4 3 · 3+

44. 4 − 42 + 43 − 44 + 45 − 46 + 47 − 48 = −52, 428

+

210 +1 1 2 4 8 16 32 64 128 = + + + + + + + + 2 3 5 9 17 33 65 129 256 512 1024 + + 257 513 1025 ≈ 9.736

k=1

= −12

2

20 +1

210

= −1

43.

i=0 0

2i 2i + 1

26 27 28 29 25 + + + + + 25 +1 26 + 1 27 + 1 28 + 1 29 + 1

i=1

8 #

10 #

49.

= (−1)1 + (−1)2 + (−1)3 + (−1)4 + (−1)5 = −1 + 1 − 1 + 1 − 1 42. −1 + 1 − 1 + 1 − 1 + 1 = 0

1 1 1 1 1 + + + + + 1·2 2·3 3·4 4·5 5·6 1 1 1 1 1 + + + + 6 · 7 7 · 8 8 · 9 9 · 10 10 · 11 1 1 1 1 1 1 1 = + + + + + + + 2 6 12 20 30 42 56 1 1 1 + + 72 90 110 10 = 11

k=1

56.

5 #

k=1

1 k2

(−1)k

k . k+1

656

Chapter 10: Sequences, Series, and Combinatorics a8 = $1000(1.062)8 ≈ $1618.07

57. 4 − 9 + 16 − 25 + . . . + (−1)n n2

This is a sum of terms of the form (−1)k k 2 , beginning with k = 2 and continuing through k = n. Sigma notation is n # (−1)k k 2 . k=2

58.

n #

(−1)k+1 k 2 , or

k=3

59.

n #

a9 = $1000(1.062)9 ≈ $1718.39

a10 = $1000(1.062)10 ≈ $1824.93

b) a20 = $1000(1.062)20 ≈ $3330.35

68. Find each term by multiplying the preceding term by 0.75: $5200, $3900, $2925, $2193.75, $1645.31,

(−1)k−1 k 2

$1233.98, $925.49, $694.12, $520.59, $390.44

k=3

1 1 1 1 + + + +. . . 1·2 2·3 3·4 4·5 This is a sum of fractions in which the numerator is 1 and the denominator is a product of two consecutive integers. The larger integer in each product is the smaller integer in the succeeding product. It is an infinite series. Sigma notation is ∞ # 1 . k(k + 1)

69. Find each term by multiplying the preceding term by 2. Find 17 terms, beginning with a1 = 1, since there are 16 fifteen minute periods in 4 hr. 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16,384, 32,768, 65,536 70. Find each term by adding $0.30 to the preceding term: $8.30, $8.60, $8.90, $9.20, $9.50, $9.80,

k=1

60.

∞ #

k=1

61.

1 k(k + 1)2

a1 = 4,

$10.10, $10.40, $10.70, $11.00 71.

ak+1 = 1 +

1 ak

1 1 5 = 1 , or 4 4 4 1 4 4 9 1 + = 1 + = 1 , or 5 5 5 5 4 5 5 14 1 1 + = 1 + = 1 , or 9 9 9 9 5 √ √ 256, a2 = 256 = 16, a3 = 16 = 4, √ 4=2 √ 6561, ak+1 = (−1)k ak √ (−1)1 6561 = −81 √ (−1)2 −81 = 9i √ √ (−1)3 9i = −3 i

a4 = 62.

a1 = a4 =

63.

a1 = a2 = a3 = a4 =

65.

a1 = 2,

(Given)

a5 = a4 + a3 = 3 + 2 = 5 a6 = a5 + a4 = 5 + 3 = 8 a7 = a6 + a5 = 8 + 5 = 13 72.

Q

n

Un

1

.41421

2

.31784

3

.26795

4

.23607

5

.21342

6

.19626

7

.18268

8

.17157

9

.16228

10 .15435

ak+1 = ak + ak−1

a2 = 3

73. See the answer section in the text.

a3 = 3 + 2 = 5

74.

a4 = 5 + 3 = 8 66.

a2 = 1

a4 = a3 + a2 = 2 + 1 = 3

64. a1 = e , a2 = ln e = Q, a3 = ln Q, a4 = ln(ln Q) Q

(Given)

a3 = a2 + a1 = 1 + 1 = 2

a2 = 1 + a3 =

a1 = 1

a1 = −10, a2 = 8, a3 = 8 − (−10) = 18,

a4 = 18 − 8 = 10

67. a) a1 = $1000(1.062)1 = $1062

n

Un

1

2

2

1.5

3

1.4167

4

1.4142

a2 = $1000(1.062)2 ≈ $1127.84

5

1.4142

a3 = $1000(1.062)3 ≈ $1197.77

6

1.4142

a4 = $1000(1.062)4 ≈ $1272.03

7

1.4142

a5 = $1000(1.062)5 ≈ $1350.90

8

1.4142

a6 = $1000(1.062)6 ≈ $1434.65

9

1.4142

a7 = $1000(1.062)7 ≈ $1523.60

10 1.4142

Exercise Set 10.1

657 We have a system of equations.

75. See the answer section in the text.

y + g = 228.1, (1)

76. a) Using the linear regression feature on a graphing calculator, we get an = 18.9n + 145.6, where an is in billions of dollars and n is the number of years after 2000.

y = g + 17.7

Carry out. We use the substitution method. g + 17.7 + g = 228.1

b) a5 ≈ $240 billion

2g + 17.7 = 228.1

a8 ≈ $297 billion

2g = 210.4

a10 ≈ $335 billion

g = 105.2 Back-substitute to find y.

77. a) Using a quadratic regression feature on a graphing calculator, we have an = −13.21231156n2 + 52, 871.41834n−52, 893, 240.07, where an is in thousands.

y = 105.2 + 17.7 = 122.9 Check. 105.2 + 122.9 = 228.1 million, and 122.9 million is 17.7 million more than 105.2 million. The answer checks.

b) a1998 ≈ 253 thousand

State. Yahoo was visited 122.9 million times, and Google was visited 105.2 million times.

a2000 ≈ 350 thousand a2002 ≈ 342 thousand a2005 ≈ 131 thousand

78. a) a1 = 41, a2 = 43, a3 = 47, a4 = 53; the terms fit the pattern 41, 41 + 2, 41 + 6, 41 + 12, · · · , or 41 + 1 · 0, 41 + 2 · 1, 41 + 3 · 2, 41 + 4 · 3, · · · , or 41 + n(n − 1). b) Yes; n2 − n + 41 = n(n − 1) + 41 = 41 + n(n − 1).

79. For a Fibonacci sequence, as n → ∞, an+1 /an → τ ≈ 1.618033989 where τ is a ratio called the golden section. It is the ratio that results when a line is divided so that the ratio of the larger part to the smaller part is equal to the ratio of the whole to the larger part. 80.

3x − 2y = 3,

2x + 3y = −11

(1) (2)

Multiply equation (1) by 3 and equation (2) by 2 and add. 9x − 6y = 4x + 6y = 13x = x=

9 −22 −13 −1

Back-substitute to find y. 2(−1) + 3y = −11

Using equation (2)

−2 + 3y = −11 3y = −9 y = −3

The solution is (−1, −3). 81. Familiarize. Let y = the number of times Yahoo was visited in April 2006, and let g = the number of times Google was visited, in millions. Translate. The total number of visits was 228.1 million. y + g = 228.1 Yahoo was visited 17.7 million more times than Google. y = g + 17.7

(2)

82.

x2 + y 2 − 6x + 4y = 3

x − 6x + 9 + y 2 + 4y + 4 = 3 + 9 + 4 2

(x − 3)2 + (y + 2)2 = 16

Center: (3, −2); radius: 4

83. We complete the square twice. x2 + y 2 + 5x − 8y = 2

x2 + 5x + y 2 − 8y = 2 25 25 + y 2 − 8y + 16 = 2 + + 16 x2 + 5x + 4 4 ! "2 97 5 + (y − 4)2 = x+ 2 4 "%2 ! √ "2 $ ! 97 5 + (y − 4)2 = x− − 2 2 √ ! " 5 97 The center is − , 4 and the radius is . 2 2 1 84. an = n log 1000n 2 1 1 1 3 a1 = 1 log 10001 = log 103 = · 3 = 2 2 2 2 1 1 1 a2 = 2 log 10002 = log (103 )2 = log 106 = 2 4 4 3 1 ·6= 4 2 1 1 1 a3 = 3 log 10003 = log (103 )3 = log 109 = 2 8 8 9 1 ·9= 8 8 1 1 a4 = 4 log 10004 = log (103 )4 = 2 16 1 3 1 log 1012 = · 12 = 16 16 4 1 1 a5 = 5 log 10005 = log (103 )5 = 2 32 1 1 15 log 1015 = · 15 = 32 32 32 171 3 3 9 3 15 = S5 = + + + + 2 2 8 4 32 32

658 85.

Chapter 10: Sequences, Series, and Combinatorics 9. 2, 6, 10, . . .

an = in a1 = i

a1 = 2, d = 4, and n = 12

a2 = i2 = −1

an = a1 + (n − 1)d

a3 = i3 = −i a4 = i4 = 1

10.

a5 = i5 = i4 · i = i;

86.

a1 = ln 1 = 0

11. 3,

a3 = ln(1 · 2 · 3) = ln 6

a4 = ln(1 · 2 · 3 · 4) = ln 24

87.

88.

= ln(2 · 6 · 24 · 120) = ln 34, 560 ≈ 10.450

12.

a1 = $1200, d = $964.32 − $1200 = −$235.68, and n = 13

Sn = ln 1 + ln 2 + ln 3 + · · · + ln n

= ln(1 · 2 · 3 · · · n) " ! " ! " ! 1 1 1 1 1 Sn = 1 − + − + − + ···+ 2 2 3 3 4 " ! " ! 1 1 1 1 − + − n−1 n n n+1 n 1 = 1− = n+1 n+1

7 5 , ,. . . 3 3

2 a1 = 3, d = − , and n = 14 3 an = a1 + (n − 1)d & 2' 26 17 =3− a14 = 3 + (14 − 1) − =− 3 3 3

a2 = ln(1 · 2) = ln 2

a5 = ln(1 · 2 · 3 · 4 · 5) = ln 120 S5 = 0 + ln 2 + ln 6 + ln 24 + ln 120;

a1 = 7, d = −3, and n = 17

a17 = 7 + (17 − 1)(−3) = 7 + 16(−3) = 7 − 48 = −41

S5 = i − 1 − i + 1 + i = i

an = ln(1 · 2 · 3 · · · n)

a12 = 2 + (12 − 1)4 = 2 + 11 · 4 = 2 + 44 = 46

a13 = $1200 + (13 − 1)(−$235.68) =

$1200 + 12(−$235.68) = $1200 − $2828.16 = −$1628.16

13. $2345.78, $2967.54, $3589.30, . . . a1 = $2345.78, d = $621.76, and n = 10 an = a1 + (n − 1)d

14.

a10 = $2345.78 + (10 − 1)($621.76) = $7941.62 106 = 2 + (n − 1)(4) 106 = 2 + 4n − 4

Exercise Set 10.2

108 = 4n 1. 3, 8, 13, 18, . . .

27 = n

a1 = 3 d=5

(8 − 3 = 5, 13 − 8 = 5, 18 − 13 = 5)

2. a1 = $1.08, d = $0.08, ($1.16 − $1.08 = $0.08)

The 27th term is 106. 15.

Let an = 1.67, and solve for n.

3. 9, 5, 1, −3, . . .

1.67 = 0.07 + (n − 1)(0.05)

a1 = 9

d = −4

1.67 = 0.07 + 0.05n − 0.05

(5−9 = −4, 1−5 = −4, −3−1 = −4)

1.65 = 0.05n

4. a1 = −8, d = 3 (−5 − (−8) = 3)

3 9 15 , , 3, ,. . . 2 4 4 3 a1 = 2 ! " 3 9 3 3 9 3 − = ,3 − = d= 4 4 2 4 4 4 ! " 3 1 1 3 1 − =− 6. a1 = , d = − 5 2 10 5 2

33 = n

5.

7.

8.

a1 = $316 d = −$3 ($313 − $316 = −$3, $310 − $313 = −$3, $307 − $310 = −$3)

a1 = 0.07, d = 0.05, and n = 11

a11 = 0.07 + (11 − 1)(0.05) = 0.07 + 0.5 = 0.57

a1 = 0.07, d = 0.05 an = a1 + (n − 1)d

The 33rd term is 1.67. 16.

−296 = 7 + (n − 1)(−3) −296 = 7 − 3n + 3 −306 = −3n 102 = n

The 102nd term is −296. 17.

2 3 an = a1 + (n − 1)d a1 = 3, d = −

Let an = −27, and solve for n.

Exercise Set 10.2

659

& 2' −27 = 3 + (n − 1) − 3 −81 = 9 + (n − 1)(−2)

25. 5 + 8 + 11 + 14 + . . . Note that a1 = 5, d = 3, and n = 20. First we find a20 : an = a1 + (n − 1)d

−81 = 9 − 2n + 2

a20 = 5 + (20 − 1)3

−92 = −2n 46 = n

Then

The 46th term is −27. an = a1 + (n − 1)d 33 = a1 + (8 − 1)4

33 = a1 + 28

Substituting 33 for a8 , 8 for n, and 4 for d

5 = a1 (Note that this procedure is equivalent to subtracting d from a8 seven times to get a1 : 33 − 7(4) = 33 − 28 = 5) 20.

26 = 8 + (11 − 1)d 26 = 8 + 10d 18 = 10d 1.8 = d

21.

S20

26. 1 + 2 + 3 + . . . + 299 + 300. 300 (1 + 300) = 150(301) = 45, 150 S300 = 2 27. The sum is 2 + 4 + 6+ . . . +798 + 800. This is the sum of the arithmetic sequence for which a1 = 2, an = 800, and n = 400. n Sn = (a1 + an ) 2 400 (2 + 800) = 200(802) = 160, 400 S400 = 2 28. 1 + 3 + 5 + . . . + 197 + 199. 100 S100 = (1 + 199) = 50(200) = 10, 000 2

an = a1 + (n − 1)d

−507 = 25 + (n − 1)(−14)

−507 = 25 − 14n + 14 −546 = −14n 39 = n

22. We know that a17 = −40 and a28 = −73. We would have to add d eleven times to get from a17 to a28 . That is, −40 + 11d = −73 11d = −33 d = −3.

Since a17 = −40, we subtract d sixteen times to get to a1 . a1 = −40 − 16(−3) = −40 + 48 = 8

We write the first five terms of the sequence:

29. The sum is 7 + 14 + 21 + . . . + 91 + 98. This is the sum of the arithmetic sequence for which a1 = 7, an = 98, and n = 14. n Sn = (a1 + an ) 2 14 (7 + 98) = 7(105) = 735 S14 = 2 30. 16 + 20 + 24 + . . . + 516 + 520 127 S127 = (16 + 520) = 34, 036 2 31. First we find a20 : an = a1 + (n − 1)d a20 = 2 + (20 − 1)5

8, 5, 2, −1, −4 95 25 23. + 15d = 3 6 45 15d = 6 1 d= 2 & 25 1 ' 25 1 a1 = − 16 = −8= 3 2 3 3 The first five terms of the sequence are 24.

n (a1 + an ) 2 20 = (5 + 62) 2 = 10(67) = 670.

Sn =

18. a20 = 14 + (20 − 1)(−3) = 14 + 19(−3) = −43 19.

= 5 + 19 · 3 = 62

= 2 + 19 · 5 = 97

Then

n (a1 + an ) 2 20 (2 + 97) = 2 = 10(99) = 990.

Sn = S20

32. 1 5 4 11 7 , , , , . 3 6 3 6 3

a14 = 11 + (14 − 1)(−4) = 11 + 13(−4) = −41 14 [11 + (−41)] = 7(−30) = −210 S14 = 2

33.

a32 = 7 + (32 − 1)(−3) = 7 + (31)(−3) = −86 32 [7 + (−86)] = 16(−79) = −1264 S32 = 2 40 #

(2k + 3)

k=1

Write a few terms of the sum: 5 + 7 + 9+ . . . +83

660

Chapter 10: Sequences, Series, and Combinatorics This is a series coming from an arithmetic sequence with a1 = 5, n = 40, and a40 = 83. Then n Sn = (a1 + an ) 2 40 S40 = (5 + 83) 2

38. First find

k=101

20 #

This is equivalent to a series coming from an arithmetic sequence with a1 = 112.34, n = 100, and a100 = 225.2. 100 (112.34 + 225.2) = 16, 877 S100 = 2 " 5 ! # k+4 Next find . 10 k=1 ! " 5 1 9 S5 = + = 3.5 2 2 10 Then 16, 877 − 3.5 = 16, 873.5.

8k

k=5

40 + 48 + 56 + 64+ . . . +160 This is equivalent to a series coming from an arithmetic sequence with a1 = 40, n = 16, and n16 = 160. 16 S16 = (40 + 160) = 1600 2 35.

19 # k−3 4

k=0

Write a few terms of the sum: 1 3 1 1 − − − + 0 + + . . . +4 4 2 4 4 Since k goes from 0 through 19, there are 20 terms. Thus, this is equivalent to a series coming from an arithmetic 3 sequence with a1 = − , n = 20, and a20 = 4. Then 4 n Sn = (a1 + an ) 2 ! " 20 3 − +4 S20 = 2 4 65 13 = . = 10 · 4 2 36.

50 #

k=2

(2000 − 3k) 1994 + 1991 + 1988+ . . . +1850

This is equivalent to a series coming from an arithmetic sequence with a1 = 1994, n = 49, and n49 = 1850. 49 S49 = (1994 + 1850) = 94, 178 2 57 # 7 − 4k 37. 13 k=12

Write a few terms of the sum: 41 45 49 221 − −. . . − − − 13 13 13 13 Since k goes from 12 through 57, there are 46 terms. Thus, this is equivalent to a series coming from an arithmetic 41 221 . Then sequence with a1 = − , n = 46, and a46 = − 13 13 n Sn = (a1 + an ) 2 ! " 46 41 221 − − S46 = 2 13 13 " ! 6026 262 =− . = 23 − 13 13

(1.14k − 2.8).

112.34 + 113.48+ . . . +225.2

= 20(88) = 1760 34.

200 #

39. Familiarize. We go from 50 poles in a row, down to six poles in the top row, so there must be 45 rows. We want the sum 50 + 49 + 48 + . . . + 6. Thus we want the sum of an arithmetic sequence. We will use the formula n Sn = (a1 + an ). 2 Translate. We want to find the sum of the first 45 terms of an arithmetic sequence with a1 = 50 and a45 = 6. Carry out. Substituting into the formula, we have 45 (50 + 6) S45 = 2 45 · 56 = 1260 = 2 Check. We can do the calculation again, or we can do the entire addition: 50 + 49 + 48 + . . . + 6. State. There will be 1260 poles in the pile. 40.

a25 = 5000 + (25 − 1)(1125) = 5000 + 24 · 1125 = 32, 000

S25 =

25 25 (5000 + 32, 000) = (37, 000) = 2 2 $462, 500

41. We first find how many plants will be in the last row. Familiarize. The sequence is 35, 31, 27, . . . . It is an arithmetic sequence with a1 = 35 and d = −4. Since each row must contain a positive number of plants, we must determine how many times we can add −4 to 35 and still have a positive result. Translate. We find the largest integer x for which 35 + x(−4) > 0. Then we evaluate the expression 35 − 4x for that value of x. Carry out. We solve the inequality. 35 − 4x > 0

35 > 4x 35 >x 4 3 8 >x 4 The integer we are looking for is 8. Thus 35 − 4x = 35 − 4(8) = 3.

Exercise Set 10.2 Check. If we add −4 to 35 eight times we get 3, a positive number, but if we add −4 to 35 more than eight times we get a negative number.

661 46. Yes; d = 0.6080−0.5908 = 0.6252−0.6080 = . . . = 0.7112 − 0.6940 = 0.0172

State. There will be 3 plants in the last row.

47. Yes; d = 6 − 3 = 9 − 6 = 3n − 3(n − 1) = 3.

Next we find how many plants there are altogether.

48. The first formula can be derived from the second by substituting a1 + (n − 1)d for an . When the first and last terms of the sum are known, the second formula is the better one to use. If the last term is not known, the first formula allows us to compute the sum in one step without first finding an .

Familiarize. We want to find the sum 35+31+27+. . .+3. We know a1 = 35 an = 3, and, since we add −4 to 35 eight times, n = 9. (There are 8 terms after a1 , for a total of 9 n terms.) We will use the formula Sn = (a1 + an ). 2 Translate. We want to find the sum of the first 9 terms of an arithmetic sequence in which a1 = 35 and a9 = 3.

49.

Carry out. Substituting into the formula, we have 9 S9 = (35 + 3) 2 9 = · 38 = 171 2 Check. We can check the calculations by doing them again. We could also do the entire addition:

(50 + 51) = 101 + 101 + 101 + . . . + 101 )* + ( 50 addends of 101

= 50 · 101 = 5050

35 + 31 + 27 + . . . + 3.

A formula for the first n natural numbers is n (1 + n). 2

State. There are 171 plants altogether. 42.

a8 = 10 + (8 − 1)(2) = 10 + 7 · 2 = 24 marchers 8 S8 = (10 + 24) = 4 · 34 = 136 marchers 2

50.

21x − 6y = 2x + 6y = 23x = x=

Carry out. First we find a31 .

44. Yes; d = 48 − 16 = 80 − 48 = 112 − 80 = 144 − 112 = 32. a10 = 16 + (10 − 1)32 = 304 10 (16 + 304) = 1600 ft S10 = 2

45. Familiarize. We have arithmetic sequence with a1 = 28, d = 4, and n = 20. n Translate. We want to find Sn = (a1 + an ) where an = 2 a1 + (n − 1)d, a1 = 28, d = 4, and n = 20. Carry out. First we find a20 .

a20 = 28 + (20 − 1)4 = 104 20 Then S20 = (28 + 104) = 10 · 132 = 1320. 2 Check. We can do the calculations again, or we can do the entire addition: 28 + 32 + 36+ . . . 104. State. There are 1320 seats in the first balcony.

(1) (2)

Multiply equation (1) by 3 and equation (2) by 2 and add. 12 34 46 2

Back-substitute to find y.

a31 = 10 + (31 − 1)10 = 10 + 30 · 10 = 310 31 31 Then S31 = (10 + 310) = · 320 = 4960. 2 2 Check. We can do the calculation again, or we can do the entire addition: State. A total of 4960/c, or $49.60 is saved.

7x − 2y = 4,

x + 3y = 17

43. Familiarize. We have a sequence 10, 20, 30, . . . It is an arithmetic sequence with a1 = 10, d = 10, and n = 31. n Translate. We want to find Sn = (a1 + an ) where an = 2 a1 + (n − 1)d, a1 = 10, d = 10, and n = 31.

10 + 20 + 30+ . . . +310.

1 + 2 + 3 + . . . + 100 = (1 + 100) + (2 + 99) + (3 + 98) + . . .+

2 + 3y = 17

Using equation (2)

3y = 15 y=5 The solution is (2, 5). 51.

2x + y + 3z = 12 x − 3y − 2z = −1

5x + 2y − 4z = −4

We will use Gauss-Jordan elimination with matrices. First we write the augmented matrix.   2 1 3 12    1 −3 −2 −1  5 2 −4 −4 Next we interchange the first two rows.   1 −3 −2 −1   1 3 12   2 5

2 −4

−4

Now multiply the first row by −2 and add it to the second row. Also multiply the first row by −5 and add it to the third row.

662

Chapter 10: Sequences, Series, and Combinatorics 

1

−3

  0 0

−2 −1

7

56.

 14  1

7

17



6

Multiply the second row by

a2 = $8760 + (−$798.23) = $7961.77 a3 = $8760 + 2(−$798.23) = $7163.54 a4 = $8760 + 3(−$798.23) = $6365.31

1 . 7

a5 = $8760 + 4(−$798.23) = $5567.08 a6 = $8760 + 5(−$798.23) = $4768.85





1 −3 −2 −1   1 1 2   0 1 0 17 6

a7 = $8760 + 6(−$798.23) = $3970.62 a8 = $8760 + 7(−$798.23) = $3172.39

Multiply the second row by 3 and add it to the first row. Also multiply the second row by −17 and add it to the third row.   5 1 0 1   1 2   0 1 0 0 −11 −33 Multiply the third row by − 

1 . 11

0

1

5

  0 0

1

1

0

1

 2  3

52.

10p + q = 40 − 3q + 2d

10p + 4q − 40 = 2d 5p + 2q − 20 = d.

= 60 − 5p − 5q.

It is

y2 x2 + =1 16 9 Vertices: (−4, 0), (4, 0) c2 = a2 − b2 = 16 − 9 = 7 √ c= 7 √ √ Foci: (− 7, 0), ( 7, 0) 53. The vertices are on the y-axis, so the transverse axis is vertical and a = 5. The length of the minor axis is 4, so b = 4/2 = 2. The equation is y2 x2 + = 1. 4 25 54. Let x = the first number in the sequence and let d = the common difference. Then the three numbers in the sequence are x, x + d, and x + 2d. Solve: x + x + 2d = 10, x(x + d) = 15. We get x = 3 and d = 2 so the numbers are 3, 3 + 2, and 3 + 2 · 2, or 3, 5, and 7. n (1 + 2n − 1) = n2 2

57. Let d = the common difference. Then a4 = a2 + 2d, or

= 40 − 3q − 5p − 2q + 20

9x2 + 16y 2 = 144

55. Sn =

a10 = $8760 + 9(−$798.23) = $1575.93 10 S10 = ($8760 + $1575.93) = $51, 679.65 2

a1 = 40 − 3q − (5p + 2q − 20)

Multiply the third row by −1 and add it to the first row and also to the second row.   2 1 0 0    0 1 0 −1  0 0 1 3 Now we can read the solution from the matrix. (2, −1, 3).

a9 = $8760 + 8(−$798.23) = $2374.16

Also, a1 = a2 − d, so we have



1

a1 = $8760

58. P (x) = x4 + 4x3 − 84x2 − 176x + 640 has at most 4 zeros because P (x) is of degree 4. By the rational roots theorem, the possible zeros are ±1, ±2, ±4, ±5, ±8, ±10, ±16, ±20, ±32, ±40, ±64, ±80, ±128, ±160, ±320, ±640

Using the graph of the function and synthetic division, we find that two zeros are −4 and 2. Also by synthetic division we determine that ±1 and −2 are not zeros. Therefore we determine that d = 6 in the arithmetic sequence. Possible arithmetic sequences: −4, 2, 8, 14

a)

−4, 2, 8

b)

c) −16, −10, −4, 2

The solution cannot be (a) because 14 is not a possible zero. Checking −16 by synthetic division we find that −16 is not a zero. Thus (b) is the only possible arithmetic sequence which contains all four zeros. Synthetic division confirms that −10 and 8 are also zeros. The zeros are −10, −4, 2, and 8. 59. 4, m1 , m2 , m3 , 12 We look for m1 , m2 , and m3 such that 4, m1 , m2 , m3 , 12 is an arithmetic sequence. In this case, a1 = 4, n = 5, and a5 = 12. First we find d: an = a1 + (n − 1)d 12 = 4 + (5 − 1)d 12 = 4 + 4d 8 = 4d 2=d Then we have

Exercise Set 10.3

663 10 11 8 m8 = 200 11 6 m9 = 225 11 4 m10 = 250 11

m1 = a1 + d = 4 + 2 = 6

m7 = 175

m2 = m1 + d = 6 + 2 = 8 m3 = m2 + d = 8 + 2 = 10 60. −3, m1 , m2 , m3 , 5

We look for m1 , m2 , and m3 such that −3, m1 , m2 , m3 , 5 is an arithmetic sequence. In this case, a1 = −3, n = 5, and a5 = 5. First we find d: an = a1 + (n − 1)d 5 = −3 + (5 − 1)d

Find n:

n (1 + 50) 2 18 = n

459 =

2=d

m2 = m1 + d = −1 + 2 = 1, m3 = m2 + d = 1 + 2 = 3.

61. 4, m1 , m2 , m3 , m4 , 13 We look for m1 , m2 , m3 , and m4 such that 4, m1 , m2 , m3 , m4 , 13 is an arithmetic sequence. In this case a1 = 4, n = 6, and a6 = 13. First we find d. an = a1 + (n − 1)d 13 = 4 + (6 − 1)d

9 = 5d 4 1 =d 5 Then we have

4 4 =5 , 5 5 4 4 8 3 m2 = m1 + d = 5 + 1 = 6 = 7 , 5 5 5 5 3 4 7 2 m3 = m2 + d = 7 + 1 = 8 = 9 , 5 5 5 5 2 4 6 1 m4 = m3 + d = 9 + 1 = 10 = 11 . 5 5 5 5 62. 27, m1 , m2 , . . . ,m9 , m10 , 300

Find d: 50 = 1 + (18 − 1)d 49 =d 17 The sequence has a total of 18 terms, so we insert 16 arith49 . metic means between 1 and 50 with d = 17 & $5200 − $1100 ' 64. a) at = $5200 − t 8 at = $5200 − $512.50t b) a0 = $5200 − $512.50(0) = $5200

a1 = $5200 − $512.50(1) = $4687.50 a2 = $5200 − $512.50(2) = $4175

a3 = $5200 − $512.50(3) = $3662.50

m1 = a1 + d = 4 + 1

300 = 27 + (12 − 1)d

273 = 11d 273 = d, or 11 9 =d 24 11

a4 = $5200 − $512.50(4) = $3150

a7 = $5200 − $512.50(7) = $1612.50 65.

a8 = $5200 − $512.50(8) = $1100 m=p+d m=q−d 2m = p + q Adding p+q m= 2

Exercise Set 10.3

m1 = 27 + 24

1. 2, 4, 8, 16, . . . 4 8 16 = 2, = 2, =2 2 4 8 r=2

m2 =

2. r = −

m3 = m4 = m5 = m6 =

9 9 = 51 11 11 9 9 7 51 + 24 = 76 11 11 11 7 9 5 76 + 24 = 101 , 11 11 11 5 9 3 101 + 24 = 126 , 11 11 11 3 9 1 126 + 24 = 151 , 11 11 11 9 10 1 = 175 , 151 + 24 11 11 11

8 , 11 6 = 225 , 11 4 = 250 , 11 2 = 275 11 = 200

63. 1, 1 + d, 1 + 2d, . . . , 50 has n terms and Sn = 459.

8 = 4d

Then we have m1 = a1 + d = −3 + 2 = −1,

9 11 9 + 24 11 9 + 24 11 9 + 24 11 + 24

1 6 =− 18 3

3. 1, −1, 1, −1, . . . −1 1 −1 = −1, = −1, = −1 1 −1 1 r = −1 4. r =

−0.8 = 0.1 8

664

5.

Chapter 10: Sequences, Series, and Combinatorics 2 4 8 16 ,− , ,− ,. . . 3 3 3 3 8 16 4 − − 3 = −2, 3 = −2, 3 = −2 2 4 8 − 3 3 3 r = −2

1 15 = 75 5 0.6275 0.06275 7. = 0.1, = 0.1 6.275 0.6275

6. r =

r = 0.1 1 2 1 x = 8. r = 1 x x 5a2 5a3 5a a a a 9. 2 = , 4 = , 8 = 5a 5a 5 2 2 2 2 4 a r= 2 10. r =

$858 = 1.1 $780

11. 2, 4, 8, 16, . . . 4 , or 2. 2 We use the formula an = a1 rn−1 . a1 = 2, n = 7, and r = a7 = 2(2)

7−1

= 2 · 2 = 2 · 64 = 128 6

12. a9 = 2(−5) = 781, 250 √ 13. 2, 2 3, 6, . . . 9−1

a1 = 2, n = 9, and r =

√ √ 2 3 , or 3 2

an = a1 rn−1 √ √ a9 = 2( 3)9−1 = 2( 3)8 = 2 · 81 = 162 14. a57 = 1(−1)57−1 = 1 15.

7 7 ,− ,. . . 625 25

7 − 7 25 a1 = , n = 23, and r = = −25. 7 625 625 an = a1 rn−1 7 7 a23 = (−25)23−1 = (−25)22 625 625 7 = 2 · 252 · 2520 = 7(25)20 , or 7(5)40 25 16. a5 = $1000(1.06)5−1 ≈ $1262.48

17. 1, 3, 9, . . .

3 a1 = 1 and r = , or 3 1 an = a1 rn−1 an = 1(3)n−1 = 3n−1

& 1 'n−1 52 18. an = 25 = n−1 = 53−n 5 5 19. 1, −1, 1, −1, . . . −1 a1 = 1 and r = = −1 1 an = a1 rn−1 an = 1(−1)n−1 = (−1)n−1 20. an = (−2)n 21.

1 1 1 , , ,. . . x x2 x3

1 2 1 1 x a1 = and r = = 1 x x x an = a1 rn−1 1 & 1 'n−1 1 1 1 1 an = = · n−1 = 1+n−1 = n x x x x x x & a 'n−1 22. an = 5 2 23. 6 + 12 + 24 + . . . 12 a1 = 6, n = 7, and r = , or 2 6 n a1 (1 − r ) Sn = 1−r 6(1 − 27 ) 6(1 − 128) 6(−127) = = = 762 S7 = 1−2 −1 −1 2 & 1 '10 3 & 1 ' 16 1 − − 16 1 − 1024 = = 24. S10 = & 12 ' 3 1− − 2 2 & 1023 ' 16 1024 = 341 , or 10 21 3 32 32 2 1 1 1 25. − + −. . . 18 6 2 1 − 1 18 1 6 =− · , n = 9, and r = = −3 a1 = 1 18 6 1 18 a1 (1 − rn ) Sn = 1−r 3 1 12 1−(−3)9 (1 + 19, 683) = 18 S9 = 18 1 − (−3) 4 1 & 1 ' 4921 (19, 684) 1 18 = (19, 684) = 4 18 4 18 n−1 an = a1 r 26. ! "n−1 1 1 − = (−8) · − 32 2 n=9 $! "9 % 1 −8 − −1 171 2 =− S9 = 1 32 − −1 2

Exercise Set 10.3

665

27.

4+2+1+. . . 424 414 1 4 4 4 4 |r| = 4 4 = 4 4 = , and since |r| < 1, the series 4 2 2 does have a sum. a1 4 2 4 = =4· =8 = S∞ = 1 1 1−r 1 1− 2 2 434 3 4 4 28. |r| = 4 4 = < 1, so the series has a sum. 7 7 7 7 49 S∞ = = = 3 4 4 1− 7 7 29.

25 + 20 + 16 + . . . 4 20 4 4 4 4 4 4 4 4 4 |r| = 4 4 = 4 4 = , and since |r| < 1, the 25 5 5 series does have a sum. a1 25 25 5 = = = 25 · = 125 S∞ = 4 1 1−r 1 1− 5 5 4 −10 4 1 4 4 < 1, so the series has a sum. 30. |r| = 4 4= 100 10 100 100 1000 S∞ = = = & 11 1' 11 1− − 10 10 31. 8 + 40 + 200 + . . . 4 40 4 4 4 |r| = 4 4 = |5| = 5, and since |r| > 1 the series does not 8 have a sum. 4 3 4 4 14 1 4 4 4 4 32. |r| = 4 4 = 4 − 4 = < 1, so the series has a sum. −6 2 2 −6 −6 2 S∞ = & 1 ' = 3 = −6 · = −4 3 1− − 2 2 33. 0.6 + 0.06 + 0.006+ . . . 4 4 4 0.06 4 4 = |0.1| = 0.1, and since |r| < 1, the series does |r| = 44 0.6 4 have a sum. a1 0.6 0.6 6 2 = = = = S∞ = 1−r 1 − 0.1 0.9 9 3 34.

10 #

S11

36.

200(1.08)k

k=0

a1 = 200, |r| = 1.08, n = 51

200[1 − (1.08)51 ] ≈ 124, 134.354 1 − 1.08 ∞ ! "k−1 # 1 37. 2 k=1 4 4 414 1 a1 = 1, |r| = 44 44 = 2 2 a1 1 1 = =2 S∞ = = 1 1 1−r 1− 2 2 S51 =

38. Since |r| = |2| > 1, the sum does not exist. 39.

∞ #

12.5k

k=1

Since |r| = 12.5 > 1, the sum does not exist. 40. Since |r| = 1.0625 > 1, the sum does not exist. 41.

∞ #

$500(1.11)−k

k=1

$500 1 ; |r| = |1.11−1 | = 1.11 1.11 $500 $500 a1 1.11 = 1.11 ≈ $4545.45 = = 1 0.11 1−r 1− 1.11 1.11

a1 = $500(1.11)−1 , or

S∞

42.

3

k

∞ #

$1000(1.06)−k

k=1

k=0

a1 =

a1 = 1, |r| = 3, n = 11

1(1 − 311 ) = 88, 573 1−3 ! "k 11 # 2 15 35. 3 k=1 4 424 2 2 4 a1 = 15 · or 10; |r| = 44 = , n = 11 3 3 3

50 #

$ ! "11 % $ % 2048 2 10 1 − 10 1 − 3 177, 147 = = 2 1 1− 3 3 175, 099 ·3 = 10 · 177, 147 1, 750, 990 38, 569 = , or 29 59, 049 59, 049

S11 =

S∞

43.

∞ #

$1000 1 , |r| = 1.06 1.06

$1000 $1000 1.06 = = 1.06 ≈ $16, 666.66 0.06 1 1− 1.06 1.06 16(0.1)k−1

k=1

a1 = 16, |r| = |0.1| = 0.1 a1 16 16 160 S∞ = = = = 1−r 1 − 0.1 0.9 9

666

Chapter 10: Sequences, Series, and Combinatorics

! "k−1 ∞ # 8 1 44. 3 2 k=1

8 1 , |r| = 3 2 8 8 16 3 3 = = = 1 1 3 1− 2 2

a1 =

S∞

45. 0.131313 . . . = 0.13 + 0.0013 + 0.000013+ . . . This is an infinite geometric series with a1 = 0.13. 4 4 4 0.0013 4 4 = |0.01| = 0.01 < 1, so the series has a limit. |r| = 44 0.13 4 a1 0.13 0.13 13 S∞ = = = = 1−r 1 − 0.01 0.99 99

46. 0.2222 = 0.2 + 0.02 + 0.002 + 0.0002+ . . . 4 4 4 0.02 4 4 = |0.1| = 0.1 |r| = 44 0.2 4 0.2 0.2 2 = = S∞ = 1 − 0.1 0.9 9

47. We will find fractional notation for 0.9999 and then add 8. 0.9999 = 0.9 + 0.09 + 0.009 + 0.0009+. . . This is an infinite geometric series with a1 = 0.9. 4 4 4 0.09 4 4 = |0.1| = 0.1 < 1, so the series has a limit. |r| = 44 0.9 4 a1 0.9 0.9 = = =1 S∞ = 1−r 1 − 0.1 0.9 Then 8.9999 = 8 + 1 = 9. 48. 0.16 = 0.16 + 0.0016 + 0.000016+ . . . 4 4 4 0.0016 4 4 = |0.01| = 0.01 |r| = 44 0.16 4 0.16 0.16 16 S∞ = = = 1 − 0.01 0.99 99 16 610 Then 6.16 = 6 + = . 99 99 49. 3.4125125 = 3.4 + 0.0125125 We will find fractional notation for 0.0125125 and then add 17 34 , or . 3.4, or 10 5 0.0125125 = 0.0125 + 0.0000125+ . . . This is an infinite geometric series with a1 = 0.0125. 4 4 4 0.0000125 4 4 = |0.001| = 0.001 < 1, so the series has |r| = 44 0.0125 4 a limit. a1 0.0125 0.0125 125 = = = S∞ = 1−r 1 − 0.001 0.999 9990 17 125 33, 966 125 34, 091 Then + = + = 5 9990 9990 9990 9990

50.

12.7809809 = 12.7 + 0.0809809 0.0809809 = 0.0809 + 0.0000809 + . . . 4 4 4 0.0000809 4 4 = |0.001| = 0.001 |r| = 44 0.0809 4

0.0809 0.0809 809 = = 1 − 0.001 0.999 9990 127 809 127, 682 809 = + = = Then 12.7 + 9990 10 9990 9990 63, 841 4995 S∞ =

51. The amount earned is $0.01 + $0.01(2) + $0.01(22 ) + $0.01(23 ) + . . . + $0.01(227 ). This is a geometric series with a1 = $0.01, r = 2, and n = 28. a1 (1 − rn ) Sn = 1−r S28 =

$0.01(1 − 228 ) = $2, 684, 354.55 1−2

52. a) The rebound distances form a geometric sequence: & 1 '2 & 1 '3 1 × 16, × 16, × 16, . . . , 4 4 4 & 1 '2 1 × 4, × 4, . . . or 4, 4 4 The height of the 6th rebound is the 6th term of the sequence. We will use the formula an = a1 rn−1 , with a1 = 4, 1 r = , and n = 6: 4 & 1 '6−1 1 = ft a6 = 4 4 256 a 4 16 4 1 b) S∞ = = = = ft, or 5 ft 1 3 1−r 3 3 1− 4 4 53. a) The rebound distances form a geometric sequence: 0.6 × 200, (0.6)2 × 200, (0.6)3 × 200, . . . ,

or 120, 0.6 × 120, (0.6)2 × 120, . . .

The total rebound distance after 9 rebounds is the sum of the first 9 terms of this sequence. a1 (1 − rn ) with We will use the formula Sn = 1−r a1 = 120, r = 0.6, and n = 9. 120[1 − (0.6)9 ] ≈ 297 1 − 0.6 The bungee jumper has traveled about 297 ft upward after 9 rebounds. S9 =

b) S∞ =

a1 120 = = 300 ft 1−r 1 − 0.6

54. a) a1 = 100, 000, r = 1.03. The population in 15 years will be the 16th term of the sequence 100, 000, (1.03)100, 000, (1.03)2 100, 000, . . . a16 = 100, 000(1.03)16−1 ≈ 155, 797

b) Solve: 200, 000 = 100, 000(1.03)n−1 n ≈ 24 yr

55. The amount of the annuity is the geometric series $1000 + $1000(1.032) + $1000(1.032)2 + . . . + $1000(1.032)17 , where a1 = $1000, r = 1.032, and n = 18. Using the formula

Exercise Set 10.3 Sn = we have

667

a1 (1 − rn ) 1−r

Using the formula a1 S∞ = 1−r we have $13, 000, 000, 000 S∞ = 1 − 0.85 S∞ ≈ $86, 666, 666, 667.

$1000[1 − (1.032)18 ] 1 − 1.032 ≈ $23, 841.50.

S18 = S18

The amount of the annuity is $23,841.50. 56. a) We have a geometric sequence with a1 = P and r = 1 + i. Then P [1 − (1 + i)N ] SN = V = 1 − (1 + i) = =

The total effect on the economy is $86,666,666,667. 5, 000, 000(0.3) ≈ 2, 142, 857 1 − 0.7 2, 142, 857 ≈ 0.429, so this is about 42.9% of the popula5, 000, 000 tion.

60. S∞ =

P [1 − (1 + i)N ] −i P [(1 + i)N − 1] i

b) We have a geometric sequence with a1 = P and i r =1+ . n The number of terms is nN . Then $ ! "nN % i P 1− 1+ n ! " SnN = V = i 1− 1+ n "nN % $ ! i P 1− 1+ n = i − n "nN $! % i P 1+ −1 n = i n

61. Left to the student 62. Left to the student 63. Answers may vary. One possibility is given. Casey invests $900 at 8% interest, compounded annually. How much will be in the account at the end of 40 years? 64. S1 = 2, S2 = 2.5, S3 = 2.6, S4 = 2.7083, S5 = 2.716, S6 = 2.71805 2.718 < S∞ < 2.719 since the terms from a7 on will cause changes in Sn in the fourth decimal place and beyond. 65. f (x) = x2 , g(x) = 4x + 5 (f ◦ g)(x) = f (g(x)) = f (4x + 5) = (4x + 5)2 =

16x2 + 40x + 25

(g ◦ f )(x) = g(f (x)) = g(x2 ) = 4x2 + 5 66. f (x) = x − 1, g(x) = x2 + x + 3

(f ◦ g)(x) = f (g(x)) = f (x2 + x + 3) = x2 + x + 3 − 1 =

57. The amounts owed at the beginning of successive years form a geometric sequence:

x2 + x + 2

(g ◦ f )(x) = g(f (x)) = g(x − 1) =

$120, 000, (1.12)$120, 000, (1.12)2 $120, 000, . . . The amount to be repaid at the end of 13 years is the amount owed at the beginning of the 14th year.

(x − 1)2 + (x − 1) + 3 = x2 − 2x + 1 + x − 1 + 3 = x2 − x + 3 67.

with a1 = 120, 000, r = 1.12, Use the formula an = a1 r and n = 14: a14 = 120, 000(1.12)14−1 n−1

ln 5x = ln 35

a14 ≈ $523, 619.17.

At the end of 13 years, $523,619.17 will be repaid. 58. We have a sequence 0.01, 2(0.01), 22 (0.01), 23 (0.01), . . . . The thickness after 20 folds is given by the 21st term of the sequence. a21 = 0.01(2)21−1 = 10, 485.76 in. 59. The total effect on the economy is the sum of an infinite geometric series $13, 000, 000, 000 + $13, 000, 000, 000(0.85)+ $13, 000, 000, 000(0.85)2 + . . . with a1 = $13, 000, 000, 000 and r = 0.85.

5x = 35

68.

x ln 5 = ln 35 ln 35 x= ln 5 x ≈ 2.209 log2 x = −4 x = 2−4 = x=

1 16

1 24

69. See the answer section in the text. 70. The sequence is not geometric; a4 /a3 )= a3 /a2 .

668

Chapter 10: Sequences, Series, and Combinatorics 1 . 2 We find the sum of the infinite geometric series 256+128+ 64 + . . .. a1 S∞ = 1−r 256 S∞ = 1 1− 2 S∞ = 512

71. a) If the sequence is arithmetic, then a2 − a1 = a3 − a2 .

x + 7 − (x + 3) = 4x − 2 − (x + 7) 13 x= 3 13 22 The three given terms are +3= , 3 3 34 13 46 13 +7= , and 4 · −2= . 3 3 3 3 12 , or 4, so the fourth term is Then d = 3 46 12 58 + = . 3 3 3 b) If the sequence is geometric, then a2 /a1 = a3 /a2 . 4x − 2 x+7 = x+3 x+7 11 x=− or x = 5 3 11 For x = − : The three given terms are 3 2 11 10 11 − +3=− , − +7= , and 3 3 3 3 " ! 11 50 4 − −2=− . 3 3 Then r = −5, so the fourth term is 250 50 . − (−5) = 3 3 For x = 5: The three given terms are 5 + 3 = 8, 3 5 + 7 = 12, and 4 · 5 − 2 = 18. Then r = , so the 2 3 fourth term is 18 · = 27. 2

72. Sn =

This is a geometric sequence with a1 = 256 and r =

The sum of the areas is 512 cm2 .

Exercise Set 10.4 1. n2 < n3 12 < 13 , 22 < 23 , 32 < 33 , 42 < 43 , 52 < 53 The first statement is false, and the others are true. 2. 12 − 1 + 41 is prime, 22 − 2 + 41 is prime, 33 − 3 + 41 is prime, 42 − 4 + 41 is prime, 52 − 5 + 41 is prime. Each of these statements is true. The statement is false for n = 41; 412 − 41 + 41 is not prime. n(n − 3) diagonals. 2 3(3 − 3) A polygon of 3 sides has diagonals. 2 4(4 − 3) diagonals. A polygon of 4 sides has 2 5(5 − 3) diagonals. A polygon of 5 sides has 2 6(6 − 3) A polygon of 6 sides has diagonals. 2 7(7 − 3) diagonals. A polygon of 7 sides has 2 Each of these statements is true.

3. A polygon of n sides has

1(1 − xn ) 1 − xn = 1−x 1−x

73. x2 − x3 + x4 − x5 + . . .

This is a geometric series with a1 = x2 and r = −x.

Sn = 74.

x2 (1 − (−x)n ) x2 (1 − (−x)n ) a1 (1 − rn ) = = 1−r 1 − (−x) 1+x

an + 1 (an + 1) = r, so = r2 ; Thus a21 , a22 , a23 , . . . is a an (an )2 geometric sequence with the common ratio r2 . 2

4. The sum of the angles of a polygon of 3 sides is (3 − 2) · 180◦ .

The sum of the angles of a polygon of 4 sides is (4 − 2) · 180◦ . The sum of the angles of a polygon of 5 sides is (5 − 2) · 180◦ .

75. See the answer section in the text. 76. Let the arithmetic sequence have the common difference d = an+1 −an . Then for the sequence 5a1 , 5a2 , 5a3 , . . . , we 5an+1 have an = 5an+1 −an = 5d . Thus, we have a geometric 5 sequence with the common ratio 5d . 77. The length of a side of the first square is 16 cm. The length of a side of the next square is the length of the hypotenuse √ of a right triangle with legs 8 cm and 8 cm, or 8 2 cm. The length of a side of the next square is the √ length of 2 cm and the hypotenuse of a right triangle with legs 4 √ 4 2 cm, or 8 cm. The areas of the squares form a sequence: √ (16)2 , (8 2)2 , (8)2 , . . . , or 256,

128,

64, . . ..

The sum of the angles of a polygon of 6 sides is (6 − 2) · 180◦ . The sum of the angles of a polygon of 7 sides is (7 − 2) · 180◦ . Each of these statements is true.

5. See the answer section in the text. 6.

Sn : 4 + 8 + 12 + ... + 4n = 2n(n + 1) S1 : 4 = 2 · 1 · (1 + 1)

Sk : 4 + 8 + 12 + ... + 4k = 2k(k + 1) Sk+1 : 4 + 8 + 12 + ... + 4k + 4(k + 1) = 2(k + 1)(k + 2)

Exercise Set 10.4

669

1) Basis step: Since 2 · 1 · (1 + 1) = 2 · 2 = 4, S1 is true.

1) Basis step: Since 31 < 31+1 , or 3 < 9, S1 is true.

2) Induction step: Let k be any natural number. Assume Sk . Deduce Sk+1 . Starting with the left side of Sk+1 , we have 4 + 8 + 12 + ... + 4k +4(k + 1) ( )* + = 2k(k + 1) +4(k + 1)

3k < 3k+1 3 ·3 < 3 k

By Sk

= (k + 1)(2k + 4)

3k+1 < 3k+2

14.

7. See the answer section in the text. 3n(n + 1) 8. Sn : 3 + 6 + 9 + ... + 3n = 2 3 · 1(1 + 1) 2

3k(k + 1) 2 Sk+1 : 3 + 6 + 9 + ... + 3k + 3(k + 1) = 3(k + 1)(k + 2) 2 1) Basis step: Since 3 · 1(1 + 1) = 3 · 2 = 3, S1 is true. 2 2 2) Induction step: Let k be any natural number. Sk : 3 + 6 + 9 + ... + 3k =

·3

Multiplying by 3

1 1 1 n + + ... + = 1·2 2·3 n(n + 1) n+1 1 1 = S1 : 1·2 1+1 1 1 1 k Sk : + + ... + = 1·2 2·3 k(k + 1) k+1 1 1 1 1 + +...+ + Sk+1 : 1·2 2·3 k(k + 1) (k + 1)(k + 2) k+1 = k+2 Sn :

1 1 1 = = , S1 is true. 1+1 2 1·2 2) Induction step: Let k be any natural number. Assume Sk . Deduce Sk+1 . 1 1 1 k + + ... + = (Sk ) 1·2 2·3 k(k + 1) k+1 1 1 1 1 + +...+ + = 1·2 2·3 k(k + 1) (k + 1)(k + 2) k 1 + k+1 (k+1)(k+2) " ! 1 on both sides Adding (k + 1)(k + 2)

+3(k + 1) +3(k + 1)

Sk

1) Basis step: Since

Assume Sk . Deduce Sk+1 . Starting with the left side of Sk+1 , we have 3 + 6 + 9 + ... + 3k ( )* + 3k(k + 1) = 2

k+1

13. See the answer section in the text.

= 2(k + 1)(k + 2)

S1 : 3 =

2) Induction step: Let k be any natural number. Assume Sk . Deduce Sk+1 .

By Sk

=

3k(k + 1) + 6(k + 1) 2

=

(k + 1)(3k + 6) 2

=

3(k + 1)(k + 2) 2

=

k(k + 2) + 1 (k + 1)(k + 2)

9. See the answer section in the text.

=

k 2 + 2k + 1 (k + 1)(k + 2)

10.

Sn : 2 ≤ 2n

(k + 1))(k + 1) (k + 1)(k + 2) k+1 = k+2 =

S1 : 2 ≤ 21

Sk : 2 ≤ 2k

Sk+1 : 2 ≤ 2k+1

1) Basis step: Since 2 ≤ 2, S1 is true.

2) Induction step: Let k be any natural number. Assume Sk . Deduce Sk+1 . 2 ≤ 2k

2 · 2 ≤ 2k · 2

2 < 2 · 2 ≤ 2k+1 2 ≤ 2k+1

Sk Multiplying by 2 (2 < 2 · 2)

11. See the answer section in the text. 12.

Sn : 3n < 3n+1 S1 : 31 < 31+1 Sk : 3 < 3 k

k+1

Sk+1 : 3k+1 < 3k+2

15. See the answer section in the text. 16.

Sn : x ≤ xn S1 : x ≤ x

Sk : x ≤ xk

Sk+1 : x ≤ xk+1

1) Basis step: Since x = x, S1 is true. 2) Induction step: Let k be any natural number. Assume Sk . Deduce Sk+1 . x ≤ xk

Sk

x · x ≤ x · x Multiplying by x, x > 1 k

x ≤ x · x ≤ xk · x x ≤ xk+1

670

Chapter 10: Sequences, Series, and Combinatorics 2) Induction step:

17. See the answer section in the text. 18.

Sn : 12 + 22 + 32 + ... + n2 = S1 : 1 2 =

14 + 24 + ... + k 4 =

n(n + 1)(2n + 1) 6

k(k + 1)(2k + 1)(3k 2 + 3k − 1) 30 14 + 24 + ... + k 4 + (k + 1)4

1(1 + 1)(2 · 1 + 1) 6

Sk : 12 + 22 + 32 + ... + k 2 =

k(k+1)(2k+1) 6

Sk+1 : 12+22+ ... +k 2+(k+1)2 = (k+1)(k+1+1)(2(k+1)+1) 6 1(1 + 1)(2 · 1 + 1) 1) Basis step: 12 = is true. 6 2) Induction step: Let k be any natural number. Assume Sk . Deduce Sk+1 . k(k + 1)(2k + 1) 12 + 22 + ... + k 2 = 6 12 +22 + ... +k 2 +(k+1)2

k(k + 1)(2k + 1)(3k 2 + 3k − 1) + (k + 1)4 30

=

k(k + 1)(2k + 1)(3k 2 + 3k − 1) + 30(k + 1)4 30

=

(k + 1)(6k 4 + 39k 3 + 91k 2 + 89k + 30) 30

=

(k + 1)(k + 2)(2k + 3)(3k 2 + 9k + 5) 30

= [(k+1)(k+1+1)(2(k+1)+1)(3(k+1)2 + 3(k+1)−1)]/30 21. See the answer section in the text. n(3n + 1) 2

=

k(k+1)(2k+1) + (k+1)2 6

=

k(k+1)(2k+1) + 6(k+1)2 6

S1 : 2 =

=

(k+1)(2k 2 +7k+ 6) 6

Sk : 2 + 5 + 8 + ... + 3k − 1 =

=

(k+1)(k+2)(2k+3) 6

Sk+1 : 2+5+ ... +(3k−1)+(3(k+1)−1) =

(k+1)(k+1+1)(2(k+1)+1) = 6 19. See the answer section in the text. 20.

=

Sn : 14 + 24 + 34 + ... + n4 = n(n + 1)(2n + 1)(3n2 + 3n − 1) 30 1(1 + 1)(2 · 1 + 1)(3 · 12 + 3 · 1 − 1) 30 Sk : 14 + 24 + 34 + ... + k 4 = S1 : 1 4 =

k(k + 1)(2k + 1)(3k 2 + 3k − 1) 30 Sk+1 : 14 +24 + ... +k 4 +(k+1)4 = (k+1)(k+1+1)(2(k+1)+1)(3(k+1)2 −3(k+1)−1) 30 1) Basis step: 1(1 + 1)(2 · 1 + 1)(3 · 12 + 3 · 1 − 1) 14 = 30 is true.

22.

Sn : 2 + 5 + 8 + ... + 3n − 1 = 1(3 · 1 + 1) 2

k(3k + 1) 2

(k+1)(3(k+1)+1) 2 1(3 · 1 + 1) 1) Basis step: 2 = is true. 2 2) Induction step: Let k be any natural number. Assume Sk . Deduce Sk+1 . k(3k + 1) 2 + 5 + ... + 3k − 1 = 2 2+5+ ... +(3k−1)+(3(k+1)−1) = k(3k+1) +(3(k+1)−1) 2 =

3k 2 +k+6k+6−2 2

=

3k 2 + 7k + 4 2

=

(k + 1)(3k + 4) 2

=

(k+1)(3(k+1)+1) 2

23. See the answer section in the text.

Exercise Set 10.4 24.

Sn

!

1 1+ 1

671

"!

1 1+ 2

"!

1 1+ 3

"

!

1 ··· 1 + n

"

=

x+

n+1 1 S1 : 1 + = 1 + 1 1 " ! " ! 1 1 ··· 1 + =k+1 Sk 1 + 1 k " ! "! " ! 1 1 1 · · · 1+ 1+ = (k+1)+1 Sk+1 1+ 1 k k+1 " ! 1) Basis step: 1 + 1 = 1 + 1 is true. 1 2) Induction step: Let k be any natural number. Assume Sk . Deduce Sk+1 . " ! " ! 1 1 · · · 1+ = k+1 1+ 1 k ! " ! "! " ! " 1 1 1 1 1+ · · · 1+ 1+ = (k+1) 1+ 1 k k+1 k+1 " ! 1 Multiplying by 1 + k+1 ! " k+1+1 = (k+1) k+1 = (k + 1) + 1 25. See the answer section in the text. 26. We can prove an infinite sequence of statements Sn by showing that a basis statement S1 is true and then that for all natural numbers k, if Sk is true, then Sk+1 is true. 27. Two possibilities are n < n2 and n2 ≤ 2n . The basis step is false for the first and the induction step fails for the second. 28.

2x − 3y = 1, (1)

3x − 4y = 3

(2)

Multiply equation (1) by 4 and multiply equation (2) by −3 and add. 8x − 12y = 4 −9x + 12y = −9 −x = −5 x=5 Back-substitute to find y. We use equation (1). 2 · 5 − 3y = 1 10 − 3y = 1

−3y = −9 y=3

The solution is (5, 3). 29.

x + y + z = 3, (1) 2x − 3y − 2z = 5, (2)

3x + 2y + 2z = 8 (3) We will use Gaussian elimination. First multiply equation (1) by −2 and add it to equation (2). Also multiply equation (1) by −3 and add it to equation (3).

y+ z=

3

−5y − 4z = −1 −y − z = −1

Now multiply the last equation above by 5 to make the y-coefficient a multiple of the y-coefficient in the equation above it. x + y + z = 3 (1) − 5y − 4z = −1 (4) − 5y − 5z = −5 (5)

Multiply equation (4) by −1 and add it to equation (3). x+ y+ z=

3 (1)

− 5y − 4z = −1 (4) − z = −4 (6)

Now solve equation (6) for z. −z = −4 z=4

Back-substitute 4 for z in equation (4) and solve for y. −5y − 4 · 4 = −1 −5y − 16 = −1 −5y = 15

y = −3

Finally, back-substitute −3 for y and 4 for z in equation (1) and solve for x. x−3+4 = 3 x+1 = 3

x=2 The solution is (2, −3, 4). 30. Let h = the number of hardback books sold and p = the number of paperback books sold. Solve:

h + p = 80, 24.95h + 9.95p = 1546

h = 50, p = 30 31. Familiarize. Let x, y, and z represent the amounts invested at 1.5%, 2%, and 3%, respectively. Translate. We know that simple interest for one year was $104. This gives us one equation: 0.015x + 0.02y + 0.03z = 104 The amount invested at 2% is twice the amount invested at 1.5%: y = 2x, or −2x + y = 0

There is $400 more invested at 3% than at 2%: z = y + 400, or −y + z = 400

We have a system of equations:

0.015x + 0.02y + 0.03z = 104, −2x + −

y y+

=

0

z = 400

Carry out. Solving the system of equations, we get (800, 1600, 2000).

672

Chapter 10: Sequences, Series, and Combinatorics Check. Simple interest for one year would be 0.015($800)+0.02($1600)+0.03($2000), or $12+$32+$60, or $104. The amount invested at 2%, $1600, is twice $800, the amount invested at 1.5%. The amount invested at 3%, $2000, is $400 more than $1600, the amount invested at 2%. The answer checks.

37. See the answer section in the text. 38.

Starting with the left side of Sk+1 , we have z1 z2 · · · zk zk+1 = z1 z2 · · · zk · zk+1

State. Martin invested $800 at 1.5%, $1600 at 2%, and $2000 at 3%. 32. Sn : a1 + a1 r + a1 r2 + ... + a1 rn−1 = S1 : a1 =

a1 − a1 r 1−r

a1 − a1 r 1−r

n

= z 1 z 2 · · · z k · z k+1

40.

a1 (1 − r) a1 − a1 r = is true. 1−r 1−r 2) Induction step: Let n be any natural number. Assume Sk . Deduce Sk+1 . 1) Basis step: a1 =

a1 +a1 r+...+a1 r

k−1

= k 2 + k + 2(k + 1) By Sk , 2 is a factor of k 2 + k; hence 2 is a factor of the right-hand side, so 2 is a factor of (k + 1)2 + (k + 1). 41. See the answer section in the text. 42. a) The least number of moves for 1 disk(s) is 1 = 21 − 1, 2 disk(s) is 3 = 22 − 1,

a1 − a1 rk = 1−r

a1 + a+1 r+ ... +a1 rk−1 +a1 rk =

= =

3 disk(s) is 7 = 23 − 1,

4 disk(s) is 15 = 24 − 1; etc.

a1 −a1 rk + a1 rk 1−r Adding a1 rk

b) Let Pn be the least number of moves for n disks. We conjecture and must show: Sn : Pn = 2n − 1.

a1−a1 rk+a1 rk−a1 rk+1 1−r a1 − a1 r 1−r

1) Basis step: S1 is true by substitution.

k+1

2) Induction step: Assume Sk for k disks: Pk = 2k − 1. Show: Pk+1 = 2k+1 − 1. Now suppose there are k + 1 disks on one peg. Move k of them to another peg in 2k − 1 moves (by Sk ) and move the remaining disk to the free peg (1 move). Then move the k disks onto it in (another) 2k −1 moves. Thus the total moves Pk+1 is 2(2k − 1) + 1 = 2k+1 − 1: Pk+1 = 2k+1 − 1.

33. See the answer section in the text. Sn : 2n + 1 < 3n S2 : 2 · 2 + 1 < 32 Sk : 2k + 1 < 3k

Sk+1 : 2(k + 1) + 1 = 3k+1 1) Basis step: 2 · 2 + 1 < 32 is true. 2) Induction step: Let k be any natural number greater than or equal to 2. Assume Sk . Deduce Sk+1 . 2k + 1 < 3k 3(2k + 1) < 3 · 3k

Multiplying by 3

6k + 3 < 3k+1 2k + 3 < 6k + 3 < 3k+1

(2k < 6k)

k+1

Exercise Set 10.5 1. 6 P6 = 6! = 6 · 5 · 4 · 3 · 2 · 1 = 720 2. 4 P3 = 4 · 3 · 2 = 24, or 4! 4! 4·3·2·1 = = = 24 4 P3 = (4 − 3)! 1! 1

3. Using formula (1), we have 10 P7

35. See the answer section in the text. 36.

S1 : z 1 = z 1 Sk : z k = z k zk · z = zk · z z k · z = z k+1 z k+1 = z k+1

S1 : 2 is a factor of 12 + 1. (k + 1)2 + (k + 1) = k 2 + 2k + 1 + k + 1

a1 − a1 rk 1−r : a1 + a1 r + ... + a1 rk−1 + a1 r(k+1)−1 =

2(k + 1) + 1 < 3

By Sk

Sk : 2 is a factor of k 2 + k.

a1 − a1 rk+1 1−r

34.

By S2

39. See the answer section in the text.

Sk : a1 + a1 r + a2 r2 + ... + a1 rk−1 = Sk+1

S 2 : z 1 z2 = z1 · z 2

S k : z 1 z2 · · · z k = z 1 z 2 · · · z k

Using formula (2), we have 10! 10! 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3! = = = 10 P7 = (10 − 7)! 3! 3! 604, 800.

If z = a + bi, z = a − bi. Multiplying both sides of Sk by z

= 10 · 9 · 8 · 7 · 6 · 5 · 4 = 604, 800.

4.

= 10 · 9 · 8 = 720, or 10! 10! 10 · 9 · 8 · 7! = = = 720 10 P3 = (10 − 3)! 7! 7! 10 P3

Exercise Set 10.5 5. 5! = 5 · 4 · 3 · 2 · 1 = 120 6. 7! = 7 · 6 · 5 · 4 · 3 · 2 · 1 = 5040 7. 0! is defined to be 1. 8. 1! = 1 9. 10.

9 · 8 · 7 · 6 · 5! 9! = = 9 · 8 · 7 · 6 = 3024 5! 5! 9! 9 · 8 · 7 · 6 · 5 · 4! = = 9 · 8 · 7 · 6 · 5 = 15, 120 4! 4!

11. (8 − 3)! = 5! = 5 · 4 · 3 · 2 · 1 = 120 12. (8 − 5)! = 3! = 3 · 2 · 1 = 6 13.

14.

10 · 9 · 8 · 7! 10 · 3 · 3 · 4 · 2 10! = = = 7!3! 7!3 · 2 · 1 3·2·1 10 · 3 · 4 = 120 7! 7 · 6 · 5! 7! = = = 42 (7 − 2)! 5! 5!

15. Using formula (2), we have 8! 8! = = 1. 8 P0 = (8 − 0)! 8! 16.

13 P1

= 13

52 P5

31. There are 5 P5 choices for the order of the rock numbers and 4 P4 choices for the order of the speeches, so we have 5 P5 ·4 P4 = 5!4! = 2880. 32. BUSINESS: 1 B, 1 U, 3 S’s, 1 I, 1 N, 1 E, a total of 8. 8! = 1! · 1! · 3! · 1! · 1! · 1! 8! 8 · 7 · 6 · 5 · 4 · 3! = = = 8 · 7 · 6 · 5 · 4 = 6720 3! 3! BIOLOGY: 1 B, 1 I, 2 0’s, 1 L, 1 G, 1 Y, a total of 7. 7! = 1! · 1! · 2! · 1! · 1! · 1! 7! 7 · 6 · 5 · 4 · 3 · 2! = = = 7 · 6 · 5 · 4 · 3 = 2520 2! 2! MATHEMATICS: 2 M’s, 2 A’s, 2 T’s, 1 H, 1 E, 1 I, 1 C, 1 S, a total of 11. 11! = 2! · 2! · 2! · 1! · 1! · 1! · 1! · 1! 11 · 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2! 11! = = 2! · 2! · 2! 2! · 2! · 2! 11 · 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 = 2·1·2·1 = 4, 989, 600

(Using formula (1))

17. Using a calculator, we find 52 P4 = 6, 497, 400 18.

673

33. The first number can be any of the eight digits other than 0 and 1. The remaining 6 numbers can each be any of the ten digits 0 through 9. We have

= 311, 875, 200

8 · 106 = 8, 000, 000

Accordingly, there can be 8,000,000 telephone numbers within a given area code before the area needs to be split with a new area code.

19. Using formula (1), we have n P3 = n(n − 1)(n − 2). Using formula (2), we have n! n(n − 1)(n − 2)(n − 3)! = = n P3 = (n − 3)! (n − 3)! n(n − 1)(n − 2).

20.

n P2

= n(n − 1)

21. Using formula (1), we have n P1 = n. Using formula (2), we have n(n − 1)! n! = = n. n P1 = (n − 1)! (n − 1)! 22.

n P0

=

n! n! = =1 (n − 0)! n!

23. 6 P6 = 6! = 720 24. 4 P4 = 4! = 24

25. 9 P9 = 9! = 362, 880 26. 8 P8 = 8! = 40, 320 27. 9 P4 = 9 · 8 · 7 · 6 = 3024 28. 8 P5 = 8 · 7 · 6 · 5 · 4 = 6720 29. Without repetition: 5 P5 = 5! = 120 With repetition: 55 = 3125 30. 7 P7 = 7! = 5040

34.

24! = 16, 491, 024, 950, 400 3!5!9!4!3!

35. a2 b3 c4 = a · a · b · b · b · c · c · c · c

There are 2 a’s, 3 b’s, and 4 c’s, for a total of 9. We have 9! 2! · 3! · 4! 9 · 8 · 7 · 6 · 5 · 4! 9·8·7·6·5 = = 1260. = 2 · 1 · 3 · 2 · 1 · 4! 2·3·2

36. a) 4 P4 = 4! = 24

b) There are 4 choices for the first coin and 2 possibilities (head or tail) for each choice. This results in a total of 8 choices for the first selection. Likewise there are 6 choices for the second selection, 4 for the third, and 2 for the fourth. Then the number of ways in which the coins can be lined up is 8 · 6 · 4 · 2, or 384. 37. a) 6 P5 = 6 · 5 · 4 · 3 · 2 = 720 b) 65 = 7776

c) The first letter can only be D. The other four letters are chosen from A, B, C, E, F without repetition. We have 1 ·5 P4 = 1 · 5 · 4 · 3 · 2 = 120.

674

Chapter 10: Sequences, Series, and Combinatorics d) The first letter can only be D. The second letter can only be E. The other three letters are chosen from A, B, C, F without repetition. We have

45.

(x + 3)(x − 2) = 0 x+3 = 0

1 · 1 ·4 P3 = 1 · 1 · 4 · 3 · 2 = 24. 38. There are 80 choices for the number of the county, 26 choices for the letter of the alphabet, and 9999 choices for the number that follows the letter. By the fundamental counting principle we know there are 80 · 26 · 9999, or 20,797,920 possible license plates. 39. a) Since repetition is allowed, each of the 5 digits can be chosen in 10 ways. The number of zip-codes possible is 10 · 10 · 10 · 10 · 10, or 100,000. b) Since there are 100,000 possible zip-codes, there could be 100,000 post offices.

40. 109 = 1, 000, 000, 000

x = −3 or

47. f (x) = x3 − 4x2 − 7x + 10

We use synthetic division to find one factor of the polynomial. We try x − 1. 4 1 4 1 −4 −7 10 1 −3 −10 1 −3 −10 0 x3 − 4x2 − 7x + 10 = 0

(x − 1)(x2 − 3x − 10) = 0 (x − 1)(x − 5)(x + 2) = 0

x − 1 = 0 or x − 5 = 0 or x + 2 = 0

42. Put the following in the form of a paragraph.

x = 1 or

48.

x = −2

= 7 ·n P4 n! n! = 7· (n − 5)! (n − 4)! n! n! = 7(n − 5)! (n − 4)! 7(n − 5)! = (n − 4)! n P5

The denominators must be the same. 7(n − 5)! = (n − 4)(n − 5)! 7 = n−4

11 = n 49.

4x − 9 = 0

4x = 9 9 x = , or 2.25 4 9 The solution is , or 2.25. 4

x = 5 or

The solutions are −2, 1, and 5.

The number of arrangements possible is 15!. 15! ≈ 41, 466 yr. The time is 31, 536, 000

44.

x=2

46. 2x2 − 3x − 1 = 0 5 −(−3) ± (−3)2 − 4 · 2 · (−1) x= 2·2 √ 3 ± 17 = 4 √ √ √ 3 + 17 3 − 17 3 ± 17 The solutions are and , or . 4 4 4

b) Since more than 285 million social security numbers are possible, each person can have a social security number.

43. For each circular arrangement of the numbers on a clock face there are 12 distinguishable ordered arrangements on a line. The number of arrangements of 12 objects on a line is 12 P12 , or 12!. Thus, the number of circular permutations 12! 12 P12 = = 11! = 39, 916, 800. is 12 12 In general, for each circular arrangement of n objects, there are n distinguishable ordered arrangements on a line. The total number of arrangements of n objects on a line is n Pn , or n!. Thus, the number of circular permutations is n! n(n − 1)! = = (n − 1)!. n n

or x − 2 = 0

The solutions are −3 and 2.

41. a) Since repetition is allowed, each digit can be chosen in 10 ways. There can be 10 · 10 · 10 · 10 · 10 · 10 · 10 · 10 · 10, or 1,000,000,000 social security numbers.

First find the number of seconds in a year (365 days): ! · 60 sec = " hr 60 min ! · 124day 365 days !· 1" hr 1 min ! 31,536,000 sec.

x2 + x − 6 = 0

n P4 n! (n − 4)! n! (n − 4)! n!

= 8 ·n−1 P3 (n − 1)! = 8· (n − 1 − 3)! (n − 1)! = 8· (n − 4)! = 8 · (n − 1)! Multiplying by (n−4)!

n(n − 1)! = 8 · (n − 1)! n=8

50.

= 9 ·n−1 P4 (n − 1)! n! = 9· (n − 5)! (n − 1 − 4)! n! (n − 1)! = 9· (n − 5)! (n − 5)! n! = 9(n − 1)! n P5

n(n − 1)! = 9(n − 1)! n=9

Dividing by (n − 1)!

Exercise Set 10.6 51.

= 8 ·n P3 n! n! = 8· (n − 4)! (n − 3)!

675 6.

n P4

(n − 3)! = 8(n − 4)!

(n − 3)(n − 4)! = 8(n − 4)! n−3 = 8

n = 11

Multiplying by (n − 4)!(n − 3)! n!

7.

Dividing by (n − 4)!

52. n! = n(n − 1)(n − 2)(n − 3)(n − 4) · · · 1 = n(n − 1)(n − 2)[(n − 3)(n − 4) · · · 1] =

n(n − 1)(n − 2)(n − 3)!

53. There is one losing team per game. In order to leave one tournament winner there must be n − 1 losers produced in n − 1 games. 54. 2 losses for each of (n−1) losing teams means 2n−2 losses. The tournament winner will have lost at most 1 game; thus at most there are (2n − 2) + 1 or (2n − 1) losses requiring 2n − 1 games.

Exercise Set 10.6 1.

13 C2

13! 2!(13 − 2)! 13 · 12 · 11! 13! = = 2!11! 2 · 1 · 11! 13 · 12 13 · 6 · 2 = = 2·1 2·1 =

= 78 2.

3.

4.

5.

9 C6

8. 9.

10.

8 8

5 P3

3!

'

8! 8!(8 − 8)! 8! 8! = = 8!0! 8! · 1 =1 =

5·4·3 3! 5·4·3 5·2·2·3 = = 3·2·1 3·2·1 = 5 · 2 = 10 =

10 · 9 · 8 · 7 · 6 = = 252 5! 5·4·3·2·1 & ' 6! 6 = 0 0!(6 − 0)! 6! 6! = = 0!6! 6! · 1 =1

10 P5

& 6 ' 6 = =6 1 1

& 6 ' 6·5 = = 15 2 2·1 &6' 6! 12. = 3 3!(6 − 3)! 6 · 5 · 4 · 3! 6! = = 3!3! 3! · 3 · 2 · 1 = 20 11.

13.

& n ' & n ' = , so r n−r & ' & ' & ' & ' & ' 7 7 7 7 7 + + + + + 0 1 2 3 4 & 7 ' & 7 ' & 7 ' + + 5 6 7 2& 7 ' & 7 ' & 7 ' & 7 '3 + + + =2 0 1 2 3 % $ 7! 7! 7! 7! + + + =2 7!0! 6!1! 5!2! 4!3! = 2(1 + 7 + 21 + 35) = 2 · 64 = 128

14.

& 6 ' & 6 ' & 6 ' & 6 ' & 6 ' + + + + + 0 1 2 3 4 & ' & ' 6 6 + 5 6 2& ' & ' & '3 & ' 6 6 6 6 =2 + + + 0 1 2 3 = 2(1 + 6 + 15) + 20 = 64

9! 6!(9 − 6)! 9 · 8 · 7 · 6! 9! = = 6!3! 6! · 3 · 2 · 1 =

= 84 & 13 ' 13! = 11 11!(13 − 11)! 13! = 11!2! = 78 (See Exercise 1.) & ' 9! 9 = 3 3!(9 − 3)! 9! = 3!6! = 84 (See Exercise 2.) &7' 7! = 1 1!(7 − 1)! 7! 7 · 6! = = 1!6! 1 · 6! =7

&

676

Chapter 10: Sequences, Series, and Combinatorics

15. We will use form (1). 52! 52 C4 = 4!(52 − 4)! 52 · 51 · 50 · 49 · 48! = 4 · 3 · 2 · 1 · 48! 52 · 51 · 50 · 49 = 4·3·2·1 = 270, 725 16. We will use form (1). 52! 52 C5 = 5!(52 − 5)! 52 · 51 · 50 · 49 · 48 · 47! = 5 · 4 · 3 · 2 · 1 · 47! = 2, 598, 960

24. Playing all other teams once: 9 C2 = 36 Playing all other teams twice: 2 ·9 C2 = 72 25.

13 C10

13! 10!(13 − 10)! 13 · 12 · 11 · 10! 13! = = 10!3! 10! · 3 · 2 · 1 13 · 3 · 2 · 2 · 11 13 · 12 · 11 = = 3·2·1 3·2·1 = 286 =

26. Using the fundamental counting principle, we have 58 C6 ·42 C4 . ! " ! " 10 5 · Using the fundamen27. 10 C7 ·5 C3 = 7 3 tal counting principle 5! 10! · 7!(10 − 7)! 3!(5 − 3)! 10 · 9 · 8 · 7! 5 · 4 · 3! · = 7! · 3! 3! · 2! 10 · 9 · 8 5 · 4 = · = 120 · 10 = 1200 3·2·1 2·1 =

17. We will use form (2). & 27 '

11 27 · 26 · 25 · 24 · 23 · 22 · 21 · 20 · 19 · 18 · 17 = 11 · 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 13, 037, 895

18. We will use form (2). & ' 37 8 37 · 36 · 35 · 34 · 33 · 32 · 31 · 30 = 8·7·6·5·4·3·2·1 = 38, 608, 020 n(n − 1)! n! = =n 1!(n − 1)! 1!(n − 1)!

19.

&

n 1

'

=

20.

&

n 3

'

n! = = 3!(n − 3)!

28. Since two points determine a line and no three of these 8 points are colinear, we need to find the number of combinations of 8 points taken 2 at a time, 8 C2 . &8' 8! = 8 C2 = 2 2!(8 − 2)! 8 · 7 · 6! 4·2·7 = = 2 · 1 · 6! 2·1 = 28 Thus 28 lines are determined. Since three noncolinear points determine a triangle, we need to find the number of combinations of 8 points taken 3 at a time, 8 C3 . &8' 8! = 8 C3 = 3 3!(8 − 3)! 8 · 7 · 6 · 5! 8·7·3·2 = = 3 · 2 · 1 · 5! 3·2·1 = 56

n(n − 1)(n − 2)(n − 3)! n(n − 1)(n − 2) = 3 · 2 · 1 · (n − 3)! 6 & m ' m! m! = =1 21. = m m!(m − m)! m!0! 22.

& t ' t! t(t − 1)(t − 2)(t − 3)(t − 4)! = = = 4 4!(t−4)! 4 · 3 · 2 · 1 · (t − 4)!

t(t − 1)(t − 2)(t − 3) 12 23! 23. 23 C4 = 4!(23 − 4)! 23! 23 · 22 · 21 · 20 · 19! = = 4!19! 4 · 3 · 2 · 1 · 19! 23 · 22 · 21 · 20 23 · 2 · 11 · 3 · 7 · 4 · 5 = = 4·3·2·1 4·3·2·1 = 8855

Thus 56 triangles are determined. = 2, 598, 960

29.

52 C5

30.

52 C13

31. a)

= 635, 013, 559, 600

31 P2

= 930

b) 31 · 31 = 961 c)

31 C2

= 465

32. Order is considered in a combination lock. 33. Choosing k objects from a set of n objects is equivalent to not choosing the other n − k objects.

Exercise Set 10.6 34.

677

−17 = 2x 17 − =x 2

2x − 3 = 0 or x + 1 = 0 2x = 3 or 3 or x= 2

The solutions are

n3 − n = n2 − n 6 n3 − n = 6n2 − 6n

x = −1

x = −1

n3 − 6n2 + 5n = 0

3 and −1. 2

n(n2 − 6n + 5) = 0

36. x2 + 5x + 1 = 0 √ −5 ± 52 − 4 · 1 · 1 x= 2·1 √ −5 ± 21 = 2 √ √ −5 − 21 −5 + 21 and , or The solutions are 2 2 √ −5 ± 21 . 2 x3 + 3x2 − 10x = 24

x + 3x2 − 10x − 24 = 0

We use synthetic division to find one factor of the polynomial on the left side of the equation. We try x − 3. 4 3 4 1 3 −10 −24 3 18 24 1 6 8 0 Now we have: (x − 3)(x2 + 6x + 8) = 0 x − 3 = 0 or x + 2 = 0 x = 3 or

The solutions are −4, −2, and 3.

x = −4

38. 4 C3 ·4 C2 = 24 39. There are 13 diamonds, and we choose 5. We have 13 C5 = 1287. n C4 n C2

Playing twice: 2 ·n C2

n = 0 or n = 5 or n = 1

Only 5 checks. The solution is 5. ! " n =6 43. n−2 n! =6 (n − (n − 2))!(n − 2)! n! =6 2!(n − 2)! n(n − 1)(n − 2)! =6 2 · 1 · (n − 2)!

n(n − 1) =6 2 n(n − 1) = 12

n2 − n = 12

n2 − n − 12 = 0

(n − 4)(n + 3) = 0

or x + 4 = 0

x = −2 or

n(n − 5)(n − 1) = 0

n = 4 or n = −3

(x − 3)(x + 2)(x + 4) = 0

41. Playing once:

! " n 2

(n + 1)(n)(n − 1) = n(n − 1) 6

(2x − 3)(x + 1) = 0

40.

= 2·

(n + 1)(n)(n − 1)(n − 2)! n(n − 1)(n − 2)! = 2· (n − 2)!3 · 2 · 1 (n − 2)! · 2 · 1

2x2 − x − 3 = 0

3

"

(n + 1)! n! = 2· (n − 2)!3! (n − 2)!2!

17 . 2

2x2 − x = 3

37.

n+1 3

n! (n + 1)! = 2· (n + 1 − 3)!3! (n − 2)!2!

The solution is − 35.

!

42.

3x − 7 = 5x + 10

Only 4 checks. The solution is 4. ! " ! " n n−1 = 2· 44. 3 2 (n − 1)! n! = 2· (n − 3)!3! (n − 1 − 2)!2! n! (n − 1)! = 2· (n − 3)!3! (n − 3)!2! n! 3! n! n(n − 1)! (n − 1)! n

(n − 1)! 2! = 3!(n − 1)! = 2· =6 =6

This number checks. The solution is 6.

678 45.

Chapter 10: Sequences, Series, and Combinatorics !

n+2 4

"

= 6·

! " n 2

Exercise Set 10.7

n! (n + 2)! = 6· (n + 2 − 4)!4! (n − 2)!2!

1. Expand: (x + 5)4 .

(n + 2)! n! = 6· (n − 2)!4! (n − 2)!2! (n + 2)! n! = 6· 4! 2!

4! ·

We have a = x, b = 5, and n = 4.

Multiplying by (n−2)!

1

n! (n + 2)! = 4! · 6 · 4! 2! (n + 2)! = 72 · n!

(n + 2)(n + 1)n! = 72 · n! (n + 2)(n + 1) = 72

Dividing by n!

n = −10 or n = 7

Only 7 checks. The solution is 7. n! = 2!(n − 2)! n(n − 1)(n − 2)! n(n − 1) = 2 · 1 · (n − 2)! 2 Diagonals: The n line segments that form the sides of the n-gon are not diagonals. Thus, the number of diagonals is n(n − 1) −n= n C2 − n = 2 2 2 n − 3n n(n − 3) n − n − 2n = = , n ≥ 4. 2 2 2 Let Dn be the number of diagonals on an n-gon. Prove the result above for diagonals using mathematical induction. n(n − 3) Sn : Dn = , for n = 4, 5, 6, ... 2 4·1 S4 : D4 = 2 =

k(k − 3) 2

(k + 1)(k − 2) 2 1) Basis step: S4 is true (a quadrilateral has 2 diagonals). Sk+1 : Dk+1 =

2) Induction step: Assume Sk . Observe that when an additional vertex Vk+1 is added to the k-gon, we gain k segments, 2 of which are sides of the (k + 1)-gon, and a former side V1 Vk becomes a diagonal. Thus the additional number of diagonals is k − 2 + 1, or k − 1. Then the new total of diagonals is Dk + (k − 1), or Dk+1 = Dk + (k − 1) k(k − 3) = + (k − 1) 2 (k + 1)(k − 2) = 2 47. See the answer section in the text.

4

1

(x + 5)

(n + 10)(n − 7) = 0

Sk : Dk =

6

= 1 · x4 + 4 · x3 · 5 + 6 · x2 · 52 +

n2 + 3n − 70 = 0

n C2

4

4

n2 + 3n + 2 = 72

46. Line segments:

Pascal’s triangle method: Use the fifth row of Pascal’s triangle.

(by Sk )

4 · x · 53 + 1 · 54

= x + 20x3 + 150x2 + 500x + 625 4

Factorial notation method: (x + 5)4 & ' & ' & ' 4 4 4 = x4 + x3 · 5 + x2 · 52 + 0 1 2 & 4 ' & 4 ' 54 x · 53 + 4 3 4! 4 4! 3 4! 2 2 x + x ·5+ x ·5 + 0!4! 1!3! 2!2! 4! 4 4! x · 53 + 5 3!1! 4!0! = x4 + 20x3 + 150x2 + 500x + 625 =

2. Expand: (x − 1)4 .

Pascal’s triangle method: Use the 5th row of Pascal’s triangle. 1

4

6

4

1

(x − 1)

4

= 1 · x4 + 4 · x3 (−1) + 6x2 (−1)2 + 4x(−1)3 + 1 · (−1)4

= x4 − 4x3 + 6x2 − 4x + 1

Factorial notation method: (x − 1)4 & 4 ' & 4 ' & 4 ' = x4 + x3 (−1) + x2 (−1)2 + 0 1 2 & 4 ' & 4 ' (−1)4 x(−1)3 + 4 3 = x4 − 4x3 + 6x2 − 4x + 1 3. Expand: (x − 3)5 .

We have a = x, b = −3, and n = 5.

Pascal’s triangle method: Use the sixth row of Pascal’s triangle. 1

5

10

10

5

1

(x − 3)5

= 1 · x5 + 5x4 (−3) + 10x3 (−3)2 + 10x2 (−3)3 + 5x(−3)4 + 1 · (−3)5

= x − 15x4 + 90x3 − 270x2 + 405x − 243 5

Exercise Set 10.7

679

Factorial notation method: (x − 3)5 & & & 5 ' 5 5 ' 4 5 ' 3 = x + x (−3) + x (−3)2 + 0 1 2 & & & 5 ' 2 5 ' 5 ' x (−3)3 + x(−3)4 + (−3)5 3 4 5 5! 5 5! 4 5! 3 = x + x (−3) + x (9)+ 0!5! 1!4! 2!3! 5! 5! 2 5! x (−27) + x(81) + (−243) 3!2! 4!1! 5!0! = x5 − 15x4 + 90x3 − 270x2 + 405x − 243

5! 5 5! 4 5! 3 2 x + x (−y) + x (y )+ 0!5! 1!4! 2!3! 5! 5! 5! 2 x (−y 3 ) + x(y 4 ) + (−y 5 ) 3!2! 4!1! 5!0! = x5 − 5x4 y + 10x3 y 2 − 10x2 y 3 + 5xy 4 − y 5 =

6. Expand: (x + y)8 . Pascal’s triangle method: Use the ninth row of Pascal’s triangle. 1

36

84

126

126

84

36

9

1

(x + 2)

9

= 1 · x9 + 9x8 · 2 + 36x7 · 22 + 84x6 · 23 +

126x5 · 24 + 126x4 · 25 + 84x3 · 26 + 36x2 · 27 +

9x · 28 + 1 · 29

= x9 + 18x8 + 144x7 + 672x6 + 2016x5 + 4032x4 + 5376x3 + 4608x2 + 2304x + 512 Factorial notation method: (x + 2)9 & 9 ' & 9 ' & 9 ' = x9 + x8 · 2+ x7 · 22+ 0 1 2 & ' & ' & 9 ' 9 9 x5 · 24 + x4 · 25 + x6 · 23 + 4 5 3 & 9 ' & 9 ' & 9 ' x3 · 26 + x2 · 27 + x · 28 + 6 7 8 & 9 ' 29 9 = x9 + 18x8 + 144x7 + 672x6 + 2016x5 + 4032x4 + 5376x3 + 4608x2 + 2304x + 512 5. Expand: (x − y)5 .

We have a = x, b = −y, and n = 5.

Pascal’s triangle method: We use the sixth row of Pascal’s triangle. 1 5 (x − y)5

10

56

70

56

28

8

1

(x + y)

= x8 + 8x7 y + 28x6 y 2 + 56x5 y 3 + 70x4 y 4 + 56x3 y 5 + 28x2 y 6 + 8xy 7 + y 8

Pascal’s triangle method: Use the 10th row of Pascal’s triangle. 9

28 8

4. Expand: (x + 2)9 .

1

8

10

5

1

= 1 · x5 + 5x4 (−y) + 10x3 (−y)2 + 10x2 (−y)3 + 5x(−y)4 + 1 · (−y)5

= x5 − 5x4 y + 10x3 y 2 − 10x2 y 3 + 5xy 4 − y 5

Factorial notation method: (x − y)5 ! " ! " ! " 5 5 5 = x5 + x4 (−y) + x3 (−y)2 + 0 1 2 ! " ! " ! " 5 5 5 x2 (−y)3 + x(−y)4 + (−y)5 3 4 5

Factorial notation method: (x + y)8 & 8 ' & 8 ' & 8 ' = x8 + x7 y + x6 y 2 + 0 1 2 & 8 ' & 8 ' & 8 ' x5 y 3 + x4 y 4 + x3 y 5 + 3 4 5 & 8 ' & 8 ' & 8 ' xy 7 + y8 x2 y 6 + 7 8 6 = x8 + 8x7 y + 28x6 y 2 + 56x5 y 3 + 70x4 y 4 + 56x3 y 5 + 28x2 y 6 + 8xy 7 + y 8 7. Expand: (5x + 4y)6 . We have a = 5x, b = 4y, and n = 6. Pascal’s triangle method: Use the seventh row of Pascal’s triangle. 1

6

15

20

15

6

1

(5x + 4y)

6

= 1 · (5x)6 + 6 · (5x)5 (4y) + 15(5x)4 (4y)2 +

20(5x)3 (4y)3 + 15(5x)2 (4y)4 + 6(5x)(4y)5 +

1 · (4y)6

= 15, 625x6 + 75, 000x5 y + 150, 000x4 y 2 + 160, 000x3 y 3 +96, 000x2 y 4 +30, 720xy 5 +4096y 6 Factorial notation method: (5x + 4y)6 & 6 ' & 6 ' = (5x)6 + (5x)5 (4y)+ 0 1 & 6 ' & 6 ' (5x)3 (4y)3 + (5x)4 (4y)2 + 2 3 & 6 ' & 6 ' & ' 6 (5x)(4y)5 + (4y)6 (5x)2 (4y)4 + 5 6 4 =

6! 6! (15, 625x6 ) + (3125x5 )(4y)+ 0!6! 1!5! 6! 6! (625x4 )(16y 2 ) + (125x3 )(64y 3 )+ 2!4! 3!3! 6! 6! (25x2 )(256y 4 ) + (5x)(1024y 5 )+ 4!2! 5!1! 6! (4096y 6 ) 6!0!

680

Chapter 10: Sequences, Series, and Combinatorics = 15, 625x6 + 75, 000x5 y + 150, 000x4 y 2 + 160, 000x3 y 3 + 96, 000x2 y 4 + 30, 720xy 5 + 4096y 6

8. Expand: (2x − 3y)5 .

Pascal’s triangle method: Use the sixth row of Pascal’s triangle. 1

5

10

10

5

1

(2x − 3y)

5

= 1 · (2x)5 + 5(2x)4 (−3y) + 10(2x)3 (−3y)2 + 10(2x)2 (−3y)3 + 5(2x)(−3y)4 + 1 · (−3y)5

= 32x5 − 240x4 y + 720x3 y 2 − 1080x2 y 3 + 810xy 4 − 243y 5

Factorial notation method: (2x − 3y)5 & ' & ' 5 5 = (2x)5 + (2x)4 (−3y)+ 0 1 & 5 ' & 5 ' (2x)2 (−3y)3+ (2x)3 (−3y)2 + 2 3 & 5 ' & 5 ' (−3y)5 (2x)(−3y)4 + 5 4 = 32x − 240x y + 720x y − 1080x y + 5

4

3 2

2 3

810xy 4 − 243y 5

9. Expand:

!

2t +

1 t

"7

.

1 , and n = 7. t Pascal’s triangle method: Use the eighth row of Pascal’s triangle. We have a = 2t, b =

1

7 21 35 "7 ! 1 2t + t

35

21

7

1

! " ! "2 1 1 + 21(2t)5 + t t ! "3 ! "4 ! "5 1 1 1 + 35(2t)3 + 21(2t)2 + 35(2t)4 t t t ! "7 ! "6 1 1 +1· 7(2t) t t 1 1 = 128t7 + 7 · 64t6 · + 21 · 32t5 · 2 + t t 1 1 1 35 · 16t4 · 3 + 35 · 8t3 · 4 + 21 · 4t2 · 5 + t t t 1 1 7 · 2t · 6 + 7 t t = 128t7 + 448t5 + 672t3 + 560t + 280t−1 + = 1 · (2t)7 + 7(2t)6

84t−3 + 14t−5 + t−7

Factorial notation method: ! "7 1 2t + t ! " & 7 ' & 7 ' 1 = + (2t)7 + (2t)6 0 1 t ! "2 & ! "3 & 7 ' 1 1 7 ' + (2t)4 + (2t)5 3 2 t t ! "4 & ! "5 & 7 ' 1 1 7 ' (2t)3 + + (2t)2 4 5 t t ! "6 & & ' ' ! 1 "7 1 7 7 (2t) + 6 7 t t ! " ! " 7! 7! 1 1 7! = (128t7 )+ (64t6 ) + (32t5 ) 2 + 0!7! 1!6! t 2!5! t ! " ! " 1 1 7! 7! (16t4 ) 3 + (8t3 ) 4 + 3!4! t 4!3! t ! " ! " ! " 1 1 1 7! 7! 7! (4t2 ) 5 + (2t) 6 + 5!2! t 6!1! t 7!0! t7 = 128t7 + 448t5 + 672t3 + 560t + 280t−1 + 84t−3 + 14t−5 + t−7 "4 1 . y Pascal’s triangle method: Use the fifth row of Pascal’s triangle.

10. Expand:

!

3y −

1 4 6 4 "4 ! 1 3y − y

1

= 1 · (3y)4 + 4(3y)3

!

−

1 y

"

+ 6(3y)2

!

−

1 y

"2

+

! "3 ! "4 1 1 4(3y) − +1· − y y = 81y 4 − 108y 2 + 54 − 12y −2 + y −4

Factorial notation method: "4 ! 1 3y − y " ! & 4 ' & 4 ' 1 = (3y)4 + (3y)3 − + 0 1 y "2 & ! "3 ! & 4 ' 1 1 4 ' + (3y) − + (3y)2 − 2 3 y y & ' ! 1 "4 4 − 4 y = 81y 4 − 108y 2 + 54 − 12y −2 + y −4 11. Expand: (x2 − 1)5 .

We have a = x2 , b = −1, and n = 5.

Pascal’s triangle method: Use the sixth row of Pascal’s triangle.

Exercise Set 10.7 1

5

10

681 10

5

1

(x2 − 1)5

= 1 · (x2 )5 + 5(x2 )4 (−1) + 10(x2 )3 (−1)2 + 10(x2 )2 (−1)3 + 5(x2 )(−1)4 + 1 · (−1)5

= x10 − 5x8 + 10x6 − 10x4 + 5x2 − 1

Factorial notation method: (x2 − 1)5 & 5 ' & 5 ' = (x2 )5 + (x2 )4 (−1)+ 0 1 & & 5 ' 5 ' 2 2 (x ) (−1)3 + (x2 )3 (−1)2 + 2 3 & ' & 5 5 ' (x2 )(−1)4 + (−1)5 4 5

5! 10 5! 8 5! 6 (x ) + (x )(−1) + (x )(1)+ 0!5! 1!4! 2!3! 5! 2 5! 5! 4 (x )(−1) + (x )(1) + (−1) 3!2! 4!1! 5!0! 10 8 6 4 2 = x − 5x + 10x − 10x + 5x − 1 =

12. Expand: (1 + 2q 3 )8 . Pascal’s triangle method: Use the ninth row of Pascal’s triangle. 1

8

28

56

70

56

28

8

1

(1 + 2q 3 )8 = 1 · 18 +8 · 17 (2q 3 )+28 · 16 (2q 3 )2 +56 · 15 (2q 3 )3+ 70 · 14 (2q 3 )4 + 56 · 13 (2q 3 )5 + 28 · 12 (2q 3 )6 + 8 · 1(2q 3 )7 + 1 · (2q 3 )8

= 1 + 16q 3 + 112q 6 + 448q 9 + 1120q 12 + 1792q 15 + 1792q 18 + 1024q 21 + 256q 24 Factorial notation method: (1 + 2q 3 )8 & 8 ' & 8 ' & 8 ' = (1)8 + (1)7 (2q 3 )+ (1)6 (2q 3 )2+ 0 1 2 & 8 ' & 8 ' (1)4 (2q 3 )4+ (1)5 (2q 3 )3 + 4 3 & ' & ' 8 8 (1)2 (2q 3 )6 + (1)3 (2q 3 )5 + 5 6 & 8 ' & 8 ' (2q 3 )8 (1)(2q 3 )7 + 7 8 = 1 + 16q 3 + 112q 6 + 448q 9 + 1120q 12 + 1792q 15 + 1792q 18 + 1024q 21 + 256q 24 √ 13. Expand: ( 5 + t)6 . √ We have a = 5, b = t, and n = 6. Pascal’s triangle method: We use the seventh row of Pascal’s triangle: 1

6

15

20

15

6

1

√ √ √ ( 5 + t)6 = 1 · ( 5)6 + 6( 5)5 (t)+ √ √ 15( 5)4 (t2 ) + 20( 5)3 (t3 )+ √ 5 √ 2 4 15( 5) (t ) + 6 5t + 1 · t6 √ √ = 125 + 150 5 t + 375t2 + 100 5 t3 + √ 75t4 + 6 5 t5 + t6 Factorial notation method: ! " ! " √ 6 √ 6 6 √ 5 ( 5 + t)6 = ( 5) + ( 5) (t)+ 0 1 ! " ! " 6 √ 3 3 6 √ 4 2 ( 5) (t )+ ( 5) (t ) + 3 2 ! " ! " 6 √ 2 4 6 √ ( 5) (t ) + ( 5)(t5 )+ 4 5 ! " 6 (t6 ) 6 √ 6! 6! 6! (125) + (25 5)t + (25)(t2 )+ = 0!6! 1!5! 2!4! 6! 6! √ (5 5)(t3 ) + (5)(t4 )+ 3!3! 4!2! 6! √ 6! 6 ( 5)(t5 ) + (t ) 5!1! 6!0! √ √ = 125 + 150 5 t + 375t2 + 100 5 t3 + √ 75t4 + 6 5 t5 + t6 √ 14. Expand: (x − 2)6 .

Pascal’s triangle method: Use the seventh row of Pascal’s triangle. 1

6

15 20 15 6 1 √ 6 2) √ √ √ 6 = 1 · x +6x5 (− 2)+15x4 (− 2)2 +20x3 (− 2)3+ √ √ √ 15x2 (− 2)4 + 6x(− 2)5 + 1 · (− 2)6 √ √ √ = x6 −6 2x5 +30x4 −40 2x3 +60x2 −24 2x+8 (x −

Factorial notation method: √ (x − 2)6 & 6 ' & 6 ' √ = x6 + (x5 )(− 2)+ 0 1 & 6 ' & 6 ' √ √ (x3 )(− 2)3 + (x4 )(− 2)2 + 3 2 & 6 ' & 6 ' √ √ (x2 )(− 2)4 + (x)(− 2)5 + 4 5 & 6 ' √ (− 2)6 6 √ √ √ = x6 −6 2x5 +30x4 −40 2x3 +60x2 −24 2x+8 15. Expand:

&

a−

2 '9 . a

9

36

2 We have a = a, b = − , and n = 9. a Pascal’s triangle method: Use the tenth row of Pascal’s triangle. 1

84

126

126

84

36

9

1

682

Chapter 10: Sequences, Series, and Combinatorics &

a−

& 2' & 2 '2 2 '9 = 1 · a9 + 9a8 − + 36a7 − + a a a & 2 '3 & 2 '4 84a6 − + 126a5 − + a a & 2 '5 & 2 '6 + 84a3 − + 126a4 − a a & 2 '7 & 2 '8 & 2 '9 +9a − +1 · − 36a2 − a a a = a9 − 18a7 + 144a5 − 672a3 + 2016a− 4032a−1 + 5376a−3 − 4608a−5 + 2304a−7 − 512a−9

Factorial notation method: & 2 '9 a− a ! " ! " & ! " & 2' 2 '2 9 9 9 = a9 + a8 − + a7 − + 0 1 2 a a ! " & ! " & 2 '3 2 '4 9 9 + a5 − + a6 − 4 3 a a ! " ! " & & 2 '6 2 '5 9 9 + a3 − + a4 − 6 5 a a ! " & ! " & 2 '7 2 '8 9 9 a2 − + a − + 7 8 a a ! "& 2 '9 9 − 9 a 9! 9 9! 8 & 2 ' 9! 7 & 4 ' = + a + a − + a 9!0! 8!1! a 7!2! a2 9! 6 & 8' 9! 5 & 16 ' a − 3 + a + 6!3! a 5!4! a4 9! 3 & 64 ' 9! 4 & 32 ' + a − 5 + a 4!5! a 3!6! a6 9! 2 & 128 ' 9! & 256 ' a − 7 + a + 2!7! a 1!8! a8 & ' 512 9! − 9 0!9! a 9 = a − 9(2a7 ) + 36(4a5 ) − 84(8a3 ) + 126(16a)− 126(32a−1 ) + 84(64a−3 ) − 36(128a−5 )+

9(256a−7 ) − 512a−9 = a9 − 18a7 + 144a5 − 672a3 + 2016a − 4032a−1 + 5376a−3 − 4608a−5 + 2304a−7 − 512a−9

16. Expand: (1 + 3)n Use the factorial notation method. (1 + 3)n & n ' & n ' & = (1)n + (1)n−1 3 + 0 1 & n ' & n (1)n−3 33 + · · · + 3 n−2 & n ' & n ' 3n (1)3n−1 + n n−1

n ' n−2 2 (1) 3 + 2 ' (1)2 3n−2 +

= 1 + 3n + &

n n−2

&

'

& n ' 2 n ' 3 3 + 3 + ···+ 2 3

3n−2 + 3n−1 n + 3n

√ √ 17. ( 2 + 1)6 − ( 2 − 1)6 √ First, expand ( 2 + 1)6 . & & √ 6 '√ 6 6 '√ 5 ( 2+1)6 = ( 2) + ( 2) (1)+ 0 1 & & ' ' 6 √ 4 2 6 √ 3 3 ( 2) (1) + ( 2) (1) + 2 3 & 6 '√ & 6 '√ ( 2)2 (1)4 + ( 2)(1)5 + 5 4 & ' 6 (1)6 6 √ 6! 6! 6! = ·8+ ·4 2+ · 4+ 6!0! 5!1! 4!2! √ 6! 6! 6! √ 6! ·2 2+ ·2+ · 2+ 3!3! 2!4! 1!5! 0!6! √ √ √ = 8 + 24 2 + 60 + 40 2 + 30 + 6 2 + 1 √ = 99 + 70 2 √ Next, expand ( 2 − 1)6 . √ ( 2 − 1)6 & ' & ' 6 √ 6 6 √ 5 = ( 2) + ( 2) (−1)+ 0 1 & 6 '√ & 6 '√ ( 2)3 (−1)3 + ( 2)4 (−1)2 + 3 2 & 6 '√ & 6 '√ ( 2)2 (−1)4 + ( 2)(−1)5 + 4 5 & 6 ' (−1)6 6 √ √ 6! 6! 6! 6! = ·8− ·4 2+ ·4− · 2 2+ 6!0! 5!1! 4!2! 3!3! 6! 6! 6! √ ·2− · 2+ 2!4! 1!5! 0!6! √ √ √ = 8 − 24 2 + 60 − 40 2 + 30 − 6 2 + 1 √ = 99 − 70 2 √ √ ( 2 + 1)6 − ( 2 − 1)6 √ √ = (99 + 70 2) − (99 − 70 2) √ √ = 99 + 70 2 − 99 + 70 2 √ = 140 2 √ √ √ 18. (1 − 2)4 = 1 · 14 + 4 · 13 (− 2) + 6 · 12 (− 2)2 + √ √ 4 · 1 · (− 2)3 + 1 · (− 2)4 √ √ = 1 − 4 2 + 12 − 8 2 + 4 √ = 17 − 12 2 √ √ √ (1 + 2)4 = 1 + 4 2 + 12 + 8 2 + 4 Using the result above √ = 17 + 12 2 √ √ √ √ (1 − 2)4 + (1 + 2)4 = 17 − 12 2 + 17 + 12 2 = 34

Exercise Set 10.7

683

19. Expand: (x−2 + x2 )4 .

21. Find the 3rd term of (a + b)7 . First, we note that 3 = 2 + 1, a = a, b = b, and n = 7. Then the 3rd term of the expansion of (a + b)7 is ! " 7! 5 2 7 a7−2 b2 , or a b , or 21a5 b2 . 2 2!5!

We have a = x−2 , b = x2 , and n = 4. Pascal’s triangle method: Use the fifth row of Pascal’s triangle. 1 4 6 (x−2 + x2 )4 = 1 · (x

) + 4(x

−2 4

4(x

−2

4

1.

) (x ) + 6(x

−2 3

2

) (x ) +

−2 2

2 2

)(x ) + 1 · (x ) 2 3

2 4

= x−8 + 4x−4 + 6 + 4x4 + x8 Factorial notation method: (x−2 + x2 )4 ! " ! " 4 4 = (x−2 )4 + (x−2 )3 (x2 )+ 0 1 ! " ! " 4 4 (x−2 )(x2 )3 + (x−2 )2 (x2 )2 + 3 2 ! " 4 (x2 )4 4

22.

23. Find the 6th term of (x − y)10 .

First, we note that 6 = 5 + 1, a = x, b = −y, and n = 10. Then the 6th term of the expansion of (x − y)10 is & 10 ' x5 (−y)5 , or −252x5 y 5 . 5

24.

First, we note that 12 = 11+1, a = a, b = −2, and n = 14. Then the 12th term of the expansion of (a − 2)14 is ! " 14! 3 14 a (−2048) a14−11 · (−2)11 = 11 3!11! = 364a3 (−2048)

4! −8 4! −6 2 4! −4 4 (x ) + (x )(x ) + (x )(x )+ 4!0! 3!1! 2!2! 4! 8 4! −2 6 (x )(x ) + (x ) 1!3! 0!4! −8 −4 4 = x + 4x + 6 + 4x + x8 & 1 √ '6 √ − x . x Pascal’s triangle method: We use the seventh row of Pascal’s triangle: 1 6 15 20 15 6 1 & 1 √ '6 √ − x x & 1 '6 & 1 '5 √ = 1· √ +6 √ (− x)+ x x & 1 '3 √ & 1 '4 √ (− x)2 + 20 √ (− x)3 + 15 √ x x & 1 ' √ & 1 '2 √ √ (− x)4 + 6 √ (− x)5 +1 · ( x)6 15 √ x x = x−3 − 6x−2 + 15x−1 − 20 + 15x − 6x2 + x3

Factorial notation method: & 1 √ '6 √ − x x ! "& ! "& 1 '6 1 '5 √ 6 6 √ √ = + (− x)+ 0 1 x x ! "& ! " & 1 '3 √ 1 '4 √ 2 6 6 √ √ (− x) + (− x)3 + 3 2 x x ! "& ! "& 1 ' √ 1 '2 √ 4 6 6 √ √ (− x)5 + (− x) + 5 4 x x ! " √ 6 (− x)6 6 = x−3 − 6x−2 + 15x−1 − 20 + 15x − 6x2 + x3

& 9 ' p5 (−2q)4 = 2016p5 q 4 4

25. Find the 12th term of (a − 2)14 .

=

20. Expand:

& 8 ' x3 y 5 = 56x3 y 5 5

26.

!

12 10

"

= −745, 472a3 x12−10 (−3)10 = 3, 897, 234x2

27. Find the 5th term of (2x3 −

√

y)8 .

√ First, we note that 5 = 4 + 1, a = 2x3 , b = − y, and √ 3 n = 8. Then the 5th term of the expansion of (2x − y)8 is ! " √ 8 (2x3 )8−4 (− y)4 4 8! √ (2x3 )4 (− y)4 4!4! = 70(16x12 )(y 2 ) =

= 1120x12 y 2 ! "& ' & ' 1 7−3 b 3 35 −5 7 28. = b 3 b2 3 27 29. The expansion of (2u − 3v 2 )10 has 11 terms so the 6th term is the middle term. Note that 6 = 5 + 1, a = 2u, b = −3v 2 , and n = 10. Then the 6th term of the expansion of (2u − 3v 2 )10 is ! " 10 (2u)10−5 (−3v 2 )5 5 10! (2u)5 (−3v 2 )5 5!5! = 252(32u5 )(−243v 10 ) =

= −1, 959, 552u5 v 10 ! " √ 5 √ 5−2 √ 2 30. 3rd term: ( x) ( 3) = 30x x 2 ! " √ 5 √ 5−3 √ 3 4th term: ( x) ( 3) = 30x 3 3

684

Chapter 10: Sequences, Series, and Combinatorics

31. The number of subsets is 27 , or 128 32. 26 , or 64 33. The number of subsets is 224 , or 16,777,216. 34. 226 , or 67,108,864 35. The term of highest degree of (x5 + 3)4 is the first term, or & 4! 20 4 ' 5 4−0 0 x = x20 . (x ) 3 = 0 4!0! Therefore, the degree of (x5 + 3)4 is 20. 36. The term of highest degree of (2 − 5x3 )7 is the last term, or & 7 ' (−5x3 )7 = −78, 125x21 . 7 Therefore, the degree of (2 − 5x3 )7 is 21. 37. We use factorial notation. Note that a = 3, b = i, and n = 5. (3 + i)5 & 5 ' & 5 ' & 5 ' = (35 ) + (34 )(i) + (33 )(i2 )+ 0 1 2 & 5 ' & 5 ' & 5 ' (32 )(i3 ) + (3)(i4 ) + (i5 ) 3 4 5

5! 5! 5! (243) + (81)(i) + (27)(−1)+ 0!5! 1!4! 2!3! 5! 5! 5! (9)(−i) + (3)(1) + (i) 3!2! 4!1! 5!0! = 243 + 405i − 270 − 90i + 15 + i

=

38.

= −12 + 316i

(1 + i)6 & ' & ' & ' 6 6 6 = 16 + 15 · i + 14 · i2 + 0 1 2 & 6 ' & 6 ' & 6 ' 13 · i3 + 12 · i4 + 1 · i5 + 3 4 5 & ' 6 6 i 6 = 1 + 6i − 15 − 20i + 15 + 6i − 1

= −8i

39. We use factorial notation. Note that √ a = 2, b = −i, and n = 4. & 4 '√ & 4 '√ √ ( 2−i)4 = ( 2)4 + ( 2)3 (−i)+ 0 1 & 4 '√ & 4 '√ ( 2)(−i)3 + ( 2)2 (−i)2 + 3 2 & 4 ' (−i)4 4

4! 4! √ (4) + (2 2)(−i)+ 0!4! 1!3! 4! √ 4! (2)(−1)+ ( 2)(i)+ 2!2! 3!1! 4! (1) 4!0! √ √ = 4 − 8 2i − 12 + 4 2i + 1 √ = −7 − 4 2i !√ "11 3 1 − i 2 2 ! √ "11 & ! √ "10 ! " & ' 3 3 1 11 11 ' = + − i + 0 1 2 2 2 & 11 '! √3 "9 ! 1 "2 − i + 2 2 2 ' ! √ 3 "8 ! 1 "3 & 11 − i + 3 2 2 & 11 '! √3 "7 ! 1 "4 − i + 4 2 2 & 11 '! √3 "6 ! 1 "5 − i + 5 2 2 & 11 '! √3 "5 ! 1 "6 − i + 6 2 2 & 11 '! √3 "4 ! 1 "7 − i + 7 2 2 & ' ! √ 3 "3 ! 1 "8 11 − i + 8 2 2 & 11 '! √3 "2 ! 1 "9 − i + 9 2 2 & 11 '! √3 "! 1 "10 & 11 '! 1 "11 − i + − i 11 10 2 2 2 √ √ 243 3 2673 4455 3 13, 365 = − i− + i+ 2048 2048 2048 2048 √ √ 8910 3 12, 474 4158 3 2970 − i− + i+ 2048 2048 2048 2048 √ √ 495 3 165 11 3 1 − i− + i 2048 2048 2048 2048 √ 1024 3 1024 + i = 2048 2048 √ 3 1 = + i 2 2 =

40.

Exercise Set 10.7 41.

685

& n ' & n ' an (−b)0 + an−1 (−b)1+ 0 1 & ' n an−2 (−b)2 +· · ·+ 2 & & n ' 1 n ' 0 a (−b)n−1 + a (−b)n n−1 n & n ' & n ' = (−1)0 an b0 + (−1)1 an−1 b1+ 0 1 & n ' (−1)2 an−2 b2 +· · ·+ 2 & n ' (−1)n−1 a1 bn−1 + n−1 & n ' (−1)n a0 bn n n & # n ' = (−1)k an−k bk k

(a − b)n =

k=0

42.

(x + h) − x h 13

13

715x9 h4 + 1287x8 h5 + 1716x7 h6 + 1716x6 h7 + 1287x5 h8 + 715x4 h9 +286x3 h10 +78x2 h11+ 13xh12 +h13 − x13 )/h

= 13x12 + 78x11 h + 286x10 h2 + 715x9 h3 + 1287x8 h4 + 1716x7 h5 + 1716x6 h6 + 1287x5 h7 + 715x4 h8 + 286x3 h9 + 78x2 h10 + 13xh11 + h12 (x + h)n − xn h & n ' & n ' & n ' xn + xn−1 h+· · ·+ hn −xn 0 1 n = h & n ' & n ' & n ' xn−2 h + · · · + hn−1 = xn−1 + 2 n 1 n & # n ' n−k k−1 = h x k k=1

44. In expanding (a + b)n , it would probably be better to use Pascal’s triangle when n is relatively small. When n is large, and many rows of Pascal’s triangle must be computed to get to the (n + 1)st row, it would probably be better to use factorial notation. In addition, factorial notation allows us to write a particular term of the expansion more efficiently than Pascal’s triangle. 45. The array of numbers that is known as Pascal’s triangle appeared as early as 1303 in a work of the Chinese algebraist Chu Sh¨ı-ki´e. Because Pascal developed many of the triangle’s properties and then found many applications of these properties, this array became known as Pascal’s triangle. 46.

(f + g)(x) = f (x) + g(x) = (x + 1) + (2x − 3) = 2

x2 + 2x − 2

(f g)(x) = f (x)g(x) = (x2 + 1)(2x − 3) =

2x3 − 3x2 + 2x − 3

48. (f ◦ g)(x) = f (g(x)) = f (2x − 3) = (2x − 3)2 + 1 = 4x2 − 12x + 9 + 1 = 4x2 − 12x + 10 49.

(g ◦ f )(x) = g(f (x)) = g(x2 + 1) = 2(x2 + 1) − 3 =

2x2 + 2 − 3 = 2x2 − 1 50.

8 & # 8 ' 8−k k x 3 =0 k

k=0

The left side of the equation is sigma notation for (x + 3)8 , so we have: (x + 3)8 = 0 x + 3 = 0 Taking the 8th root on both sides x = −3 51.

4 & 4 & # # 4 ' 4 ' 4−k (−1)k x4−k 6k = x (−6)k , so k k

k=0

= (x13 + 13x12 h + 78x11 h2 + 286x10 h3 +

43.

47.

k=0

the left side of the equation is sigma notation for (x − 6)4 . We have: (x − 6)4 = 81 x − 6 = ±3

x−6 = 3

x=9

or or

Taking the 4th root on both sides x − 6 = −3 x=3

The solutions are 9 and 3. If we also observe that (3i)4 = 81, we also find the imaginary solutions 6 ± 3i. & 3x2 1 '12 is − 52. The (k + 1)st term of 2 3x ! "& 2' 3x 12−k & 1 'k 12 . In the term which does not − k 2 3x contain x, the exponent of x in the numerator is equal to the exponent of x in the denominator. 2(12 − k) = k 24 − 2k = k

24 = 3k 8=k

Find the (8 + 1)st, or 9th term: ! "& 2' & 1 '8 12! & 34 x8 '& 1 ' 55 3x 4 12 = − = 8 2 3x 4!8! 24 38 x8 144 53. The expansion of (x2 − 6y 3/2 )6 has 7 terms, so the 4th term is the middle term. ! " 6! 6 6 (x )(−216y 9/2 ) = (x2 )3 (−6y 3/2 )3 = 3 3!3! −4320x6 y 9/2 ! " & 5 (p2 )2 − 3 54. ! " & 5 (p2 )3 − 2

'3 1 √ 1 √ p3q − p7 q 3 q 2 8 = = − ' 5 2 1 1 √ 2p p8 3 q 2 p3q 4 2

686

Chapter 10: Sequences, Series, and Combinatorics

&√ 1 '7 55. The (k + 1)st term of 3 x − √ is x ! " & √ 1 1 'k 7 . The term containing 1/6 is the ( 3 x)7−k − √ k x x term in which the sum of the exponents is −1/6. That is, & 1' &1' 1 (7 − k) + − (k) = − 3 2 6 1 7 k k − − =− 3 3 2 6 5k 15 − =− 6 6 k=3 Find the (3 + 1)st, or 4th term. ! " √ & 7! 4/3 1 '3 7 = (x )(−x−3/2 ) = ( 3 x)4 − √ 3 4!3! x 35 −35x−1/6 , or − 1/6 . x 56. The total number of subsets of a set of 7 bills is 27 . This includes the empty set. Thus, 27 −1, or 127, different sums of money can be formed. 57.

100 C0 +100 C1 + · · · +100 C100

is the total number of subsets of a set with 100 members, or 2100 .

+n C1 + ... +n Cn is the total number of subsets of a set with n members, or 2n . " 23 ! # 23 59. (loga x)23−k (loga t)k = k

58.

n C0

k=0

60.

(loga x + loga t)23 = [loga (xt)]23 " 15 ! # 15 30−2k i k k=0

= i30 + 15i28 + 105i26 + 455i24 + 1365i22 + 3003i20 + 5005i18 + 6435i16 + 6435i14 + 5005i12 + 3003i10 + 1365i8 + 455i6 + 105i4 + 15i2 + 1 = −1 + 15 − 105 + 455 − 1365 + 3003 − 5005+ 6435 − 6435 + 5005 − 3003 + 1365 − 455+ 105 − 15 + 1

=0

61. See the answer section in the text.

2. The total number of gumdrops is 7 + 8 + 9 + 4 + 5 + 6, or 39. 8 Lemon: 39 5 Lime: 39 3 9 = Orange: 39 13 6 2 Grape: = 39 13 7 Strawberry: 39 0 =0 Licorice: 39 3. The company can expect 78% of the 15,000 pieces of advertising to be opened and read. We have: 78%(15, 000) = 0.78(15, 000) = 11, 700. 4. a) B: 136/9136 ≈ 1.5%

C: 273/9136 ≈ 3.0%

D: 286/9136 ≈ 3.1%

E: 1229/9136 ≈ 13.5% F: 173/9136 ≈ 1.9%

G: 190/9136 ≈ 2.1% H: 399/9136 ≈ 4.4% I: 539/9136 ≈ 5.9% J: 21/9136 ≈ 0.2%

K: 57/9136 ≈ 0.6% L: 417/9136 ≈ 4.6%

M: 231/9136 ≈ 2.5% N: 597/9136 ≈ 6.5%

O: 705/9136 ≈ 7.7% P: 238/9136 ≈ 2.6% Q: 4/9136 ≈ 0.04%

R: 609/9136 ≈ 6.7% S: 745/9136 ≈ 8.2% T: 789/9136 ≈ 8.6%

U: 240/9136 ≈ 2.6% V: 113/9136 ≈ 1.2%

Exercise Set 10.8 1. a) We use Principle P . 18 , or For 1: P = 100 24 For 2: P = , or 100 23 , or For 3: P = 100 23 For 4: P = , or 100 12 , or For 5: P = 100

b) Opinions may vary, but it seems that people tend not to select the first or last numbers.

W: 127/9136 ≈ 1.4% 0.18 0.24 0.23 0.23 0.12

X: 20/9136 ≈ 0.2%

Y: 124/9136 ≈ 1.4% 3566 b) 853 + 1229 + 539 + 705 + 240 = ≈ 39% 9136 9136 c) In part (b) we found that there are 3566 vowels. Thus, there are 9136 − 3566, or 5570 consonants. 5570 ≈ 61% P = 9136 (We could also find this by subtracting the probability of a vowel occurring from 100%: 100% − 39% = 61%)

Exercise Set 10.8 5. a) The consonants with the 5 greatest numbers of occurrences are the 5 consonants with the greatest probability of occurring. They are T, S, R, N, and L. b) E is the vowel with the greatest number of occurrences, so E is the vowel with the greatest probability of occurring.

11. The number of ways of selecting 5 cards from a deck of 52 cards is 52 C5 . Three sevens can be selected in 4 C3 ways and 2 kings in 4 C2 ways. 4 C3 ·4 C2 , or P (drawing 3 sevens and 2 kings) = 52 C5 1 . 108, 290

c) Yes

12. Since there are only 4 aces in the deck, P (5 aces) = 0.

6. a) 52 1 4 , or b) 52 13 13 1 c) , or 52 4 4 1 d) , or 52 13 1 26 , or e) 52 2 2 4+4 f) , or 52 13 2 1 g) , or 52 26 7. a) Since there are 14 equally likely ways of selecting a marble from a bag containing 4 red marbles and 10 green marbles, we have, by Principle P , 2 4 = . P (selecting a red marble) = 14 7 b) Since there are 14 equally likely ways of selecting a marble from a bag containing 4 red marbles and 10 green marbles, we have, by Principle P , 5 10 = . P (selecting a green marble) = 14 7 c) Since there are 14 equally likely ways of selecting a marble from a bag containing 4 red marbles and 10 green marbles, we have, by Principle P , 0 = 0. P (selecting a purple marble) = 14 d) Since there are 14 equally likely ways of selecting a marble from a bag containing 4 red marbles and 10 green marbles, we have, by Principle P , P (selecting a red or a green marble) = 4 + 10 = 1. 14 8.

687

10 C2 ·10

20 C4

C2

=

45 · 45 135 = 4845 323

9. The total number of coins is 7 + 5 + 10, or 22 and the total number of coins to be drawn is 4 + 3 + 1, or 8. The number of ways of selecting 8 coins from a group of 22 is 22 C8 . Four dimes can be selected in 7 C4 ways, 3 nickels in 5 C3 ways, and 1 quarter in 10 C1 ways. P (selecting 4 dimes, 3 nickels, and 1 quarter) = 350 7 C4 ·5 C3 ·10 C1 , or 31, 977 22 C8 1 1 ≈ 0.000009% = 10. 10, 737, 573 47 C6

13. The number of ways of selecting 5 cards from a deck of 52 cards is 52 C5 . Since 13 of the cards are spades, then 5 spades can be drawn in 13 C5 ways 1287 13 C5 = = P (drawing 5 spades) = 2, 598, 960 52 C5 33 66, 640 14.

4 C4 ·4

C1

52 C5

=

1 649, 740

15. a) HHH, HHT, HTH, HTT, THH, THT, TTH, TTT b) Three of the 8 outcomes have exactly one head. 3 Thus, P (exactly one head) = . 8 c) Seven of the 8 outcomes have exactly 0, 1, or 2 7 heads. Thus, P (at most two heads) = . 8 d) Seven of the 8 outcomes have 1, 2, or 3 heads. Thus, 7 P (at least one head) = . 8 e) Three of the 8 outcomes have exactly two tails. 3 Thus, P (exactly two tails) = . 8 16.

18 9 , or 38 19

17. The roulette wheel contains 38 equally likely slots. Eighteen of the 38 slots are colored black. Thus, by Principle P, 9 18 = . P (the ball falls in a black slot) = 38 19 18.

1 38

19. The roulette wheel contains 38 equally likely slots. Only 1 slot is numbered 0. Then, by Principle P , 1 P (the ball falls in the 0 slot) = . 38 20.

1 2 = 38 19

21. The roulette wheel contains 38 equally likely slots. Thirtysix of the slots are colored red or black. Then, by Principle P, 18 36 = . P (the ball falls in a red or a black slot) = 38 19 22.

1 38

688

Chapter 10: Sequences, Series, and Combinatorics

23. The roulette wheel contains 38 equally likely slots. Eighteen of the slots are odd-numbered. Then, by Principle P, P (the ball falls in a an odd-numbered slot) = 9 18 = . 38 19 24. The dartboard can be thought of as having 18 areas of equal size. Of these, 6 are red, 4 are green, 3 are blue, and 5 are yellow. 1 6 = P (red) = 18 3 2 4 = P (green) = 18 9 3 1 P (blue) = = 18 6 5 P (yellow) = 18 25. a), b) Answers may vary 26. From the choices of the 26 letters of the alphabet or a space, the probability of selecting each of the 50 let1 . Then the probability that the ters or spaces is 27 given passage could have been written by a monkey is ! "50 1 ≈ 2.7 × 10−72 . Thus, it is extremely unlikely that 27 the given passage, much less an entire novel, could be written by a monkey. 27. Answers may vary. 28. zero

30. function; domain; range; domain; range 31. zero 32. combination 33. inverse variation 34. factor 35. geometric sequence 36. a) 4 4 52 C5

=

A K Q J 10 9 8 7 6 5 4 3 2 A straight flush can be any of the following combinations in the same suit. K Q J 10 9 Q

J

4 ≈ 1.54 × 10−6 2, 598, 960

10

9

8

J

10

9

8

7

10

9

8

7

6

9

8

7

6

5

8

7

6

5

4

7

6

5

4

3

6

5

4

3

2

5

4

3

2

A

Remember a straight flush does not include A K Q J 10 which is a royal flush. a) Since there are 9 straight flushes per suit, there are 36 straight flushes in all 4 suits. b) Since 2,598,960, or 52 C5 , poker hands can be dealt from a standard 52-card deck and 36 of those hands are straight flushes, the probability of getting a 36 straight flush is , or about 1.39 × 10−5 . 2, 598, 960 38. a) There are 13 ways to choose a denomination. Then there are 48 ways to choose one of the 48 cards remaining after 4 cards of the same denomination are chosen. Thus there are 13 · 48, or 624, four of a kind hands. b)

29. one-to-one

b)

37. Consider a suit

624 52 C5

=

624 ≈ 2.4 × 10−4 2, 598, 960

39. a) There are 13 ways to select a denomination. Then from that denomination there are 4 C3 ways to pick 3 of the 4 cards in that denomination. Now there are 12 ways to select any one of the remaining 12 denominations and 4 C2 ways to pick 2 cards from the 4 cards in that denomination. Thus the number of full houses is (13 ·4 C3 ) · (12 ·4 C2 ) or 3744. b)

3744

52 C5

=

3744 ≈ 0.00144 2, 598, 960

40. a) There ! " are 13 ways to select a denomination and then 4 ways to choose 3 of the 4 cards in that de3 ! " 48 nomination. Now there are ways to choose 2 2 cards from the 12 remaining denominations (4 · 12, or 48 cards). But these combinations include the 3744 hands in a full house like Q-Q-Q-4-4 (Exercise 31), so these must be subtracted. Thus ! " ! the " number 4 48 of three of a kind hands is 13 · · − 3744, 3 2 or 54,912. b)

54, 912 54, 912 ≈ 0.0211 = 2, 598, 960 52 C5

Chapter 10 Review Exercises

689

" 13 5 ways to choose 5 cards from that suit. But these combinations include hands with all the cards in sequence, so we subtract the 4 royal flushes (Exercise 28) and the 36 flushes (Exercise 29). Thus ! straight " 13 there are 4 · − 4 − 36, or 5108 flushes. 5

41. a) There are 4 ways to select a suit and then

!

5108 5108 ≈ 0.00197 = C 2, 598, 960 52 5 ! " 42. a) There are 13 ways to select 2 denominations 2 from the 13 denominations. Then in each denomina! " 4 tion there are ways to choose 2 of the 4 cards. 2! " 44 Finally there are ways to choose the fifth card 1 from the 11 remaining denominations (4 · 11, or 44 cards). Thus the number of two pairs hands is ! " ! " ! " ! " 13 4 4 44 · · · , or 123,552. 2 2 2 1

5.

a1 = a2 = a3 =

b)

b)

a4 = a11 = a23 =

k=1

(−1)1+1 31 (−1)2+1 32 (−1)3+1 33 (−1)4+1 34 + + + 31 − 1 32 − 1 33 − 1 34 − 1 3 9 27 81 − = − + 2 8 26 80 417 = 1040 =

43. a) There are 10 sets of 5 consecutive cards: A K Q J 10 Q

J

10

9 8.

.

9.

. 5

7 #

k=1

. 4

3

2

A

In each of these 10 sets there are 4 ways to choose (from 4 suits) each of the 5 cards. These combinations include the 4 royal flushes and the 36 straight flushes, both of which consist of 5 cards of the same suit in sequence. Thus there are 10 · 4 · 4 · 4 · 4 · 4 − 4 − 36, or 10,200 straights. 10, 200 10, 200 b) = ≈ 0.00392 2, 598, 960 52 C5

!

6. (−1)n+1 (n2 + 1) 4 # (−1)k+1 3k 7. 3k − 1

123, 552 123, 552 = ≈ 0.0475 2, 598, 960 52 C5

K

an =

" n2 (−1) n4 + 1 ! 2 " 1 1 (−1)1 4 =− 1 +1 2 ! 2 " 4 2 = (−1)2 4 2 +1 17 ! 2 " 9 3 =− (−1)3 4 3 +1 82 ! 2 " 4 16 (−1)4 4 = 4 +1 257 " ! 112 121 (−1)11 =− 114 + 1 14, 642 ! " 529 232 (−1)23 =− 4 23 + 1 279, 842 n

10.

(k 2 − 1)

an = a1 + (n − 1)d 13 3 1 3 − = , and n = 10 a1 = , d = 4 12 4 3 3 1 3 1 3 15 a10 = + (10 − 1) = + 9 · = + 3 = 4 3 4 3 4 4 an = a1 + (n − 1)d a1 = a − b, d = a − (a − b) = b

11.

a6 = a − b + (6 − 1)b = a − b + 5b = a + 4b an = a1 + (n − 1)d

a18 = 4 + (18 − 1)3 = 4 + 51 = 55 n (a1 + an ) 2 18 (4 + 55) = 531 S18 = 2 12. 1 + 2 + 3 + . . . +199 + 200 n Sn = (a1 + an ) 2 200 (1 + 200) = 20, 100 S200 = 2 13. a1 = 5, a17 = 53 Sn =

Chapter 10 Review Exercises 1. The statement is true. See page 844 in the text. 2. An infinite geometric series has a sum if |r| < 1, so the statement is false. 3. The statement is true. See page 881 in the text. 4. The total number of subsets of a set with n elements is 2n , so the statement is false.

53 = 5 + (17 − 1)d 53 = 5 + 16d 48 = 16d 3=d a3 = 5 + (3 − 1)3 = 11

690

Chapter 10: Sequences, Series, and Combinatorics

14. d = 3, a10 = 23 23 = a1 + (10 − 1)3 23 = a1 + 27

15.

−4 = a1

a1 = −2, r = 2, an = −64

an = a1 rn−1

−64 = −2 · 2n−1 −64 = −2n −26 = −2n 6=n

Sn =

a1 (1 − rn ) 1−r

−2(1 − 26 ) = 2(1 − 64) = −126 1−2 1 31 r = , n = 5, Sn = 2 2

Sn = 16.

a1 (1 − rn ) 1−r ! ! "5 " 1 a1 1 − 2 = 1 1− 2" ! 1 a1 1 − 32 = 1 2 31 a1 = 32 1 2 31 2 = a1 · 32 1 31 = a1 16 = a1

Sn = 31 2 31 2 31 2 31 2 31 2 8

an = a1 rn−1 ! "5−1 ! "4 1 1 1 1 a5 = 8 =8 =8· = 2 2 16 2 4 4 4 27.5 4 4 4 4 = 41.14 = 1.1 > 1, the sum does not ex17. Since |r| = 44 2.5 4 ist. 4 4 4 4 0.0027 4 4 4 = 40.014 = 0.01 < 1, the series has a 18. Since |r| = 44 0.27 4 sum. 0.27 0.27 27 3 = = = S∞ = 1 − 0.01 0.99 99 11 4 1 4 4 4− 4 4 4 4 14 1 4 19. Since |r| = 44 16 44 = 44 − 44 = < 1, the series has a sum. 3 3 4 4 2 S∞ =

1−

1 1 3 3 "= 2 = · = 4 1 2 4 8 − 3 3

1 2 !

20. 2.43 = 2 + 0.43 + 0.0043 + 0.000043+ . . . . We will notation for 0.43 and then add 2. 4 4 find fraction 4 0.0043 4 4 4 = |0.01| = 0.01 < 1, so the series has a limit. |r| = 4 0.43 4

0.43 0.43 43 = = 1 − 0.01 0.99 99 198 43 241 43 = + = Then 2 + 99 99 99 99 21. 5, m1 , m2 , m3 , m4 , 9 S∞ =

We look for m1 , m2 , m3 , and m4 , such that 5, m1 , m2 , m3 , m4 , 9 is an arithmetic sequence. In this case, a1 = 5, n = 6, and a6 = 9. First we find d: an = a1 + (n − 1)d 9 = 5 + (6 − 1)d

4 = 5d 4 =d 5 Then we have:

m1 = a1 + d = 5 + 4 5 3 m 3 = m2 + d = 6 5 2 m 4 = m3 + d = 7 5 m 2 = m1 + d = 5

4 4 =5 5 5 4 3 + =6 5 5 4 2 + =7 5 5 4 1 + =8 5 5

22. The distances the ball drops are given by a geometric sequence: ! "2 ! "3 ! "4 ! "5 3 3 3 3 3 30, · 30, · 30, · 30, · 30, · 30. 4 4 4 4 4 Then the total distance the ball drops is the sum of these 6 terms. The total rebound distance is this sum less 30 ft, the distance the ball drops initially. To find the total distance the ball drops we will use the formula a1 (1 − rn ) 3 with a1 = 30, r = , and n = 6. Sn = 1−r 4 ! ! "6 " 3 30 1 − 50, 505 4 = S6 = 3 512 1− 4 50, 505 − 30, or Then the total rebound distance is 12 35, 145 , and the total distance the ball has traveled is 512 50, 505 34, 145 42, 825 + , or , or about 167.3 ft. 512 512 256 23. The amount of the annuity is a geometric series $2000 + $2000(1.028) + $2000(1.028)2 + . . . $2000(1.028)17 , where a1 = $2000, n = 18, and r = 1.028. a1 (1 − rn ) , we have Using the formula Sn = 6 7 1−r 18 2000 1 − (1.028) ≈ $45, 993.04. S18 = 1 − 1.028

Chapter 10 Review Exercises

691 31 − 1 2 = = 1 is true. 2 2 2. Induction step: Assume Sk . Add 3k to both sides.

24. a), b) We have an arithmetic sequence with a1 = 10, d = 2, and n = 365. n We want to find an = a1 + (n − 1)d and Sn = (a1 + an ) 2 where a1 = 10, d = 2, and n = 365. First we find a365 . c, or $7.38 a365 = 10 + (365 − 1)(2) = 738/ 365 Then S365 = (10 + 738) = 136, 510/ c or $1365.10. 2 You will receive $7.38 on the 365th day. The sum of all the gifts is $1365.10. 25. The total economic effect is given by the infinite geometric series $24, 000, 000, 000 + $24, 000, 000, 000(0.73)+ $24, 000, 000, 000(0.73)2 + . . .. Since |r| = |0.73| = 0.73 < 1, the series has a sum. a1 Using the formula S∞ = with 1−r a1 = $24, 000, 000, 000 and r = 0.73, we have 24, 000, 000, 000 ≈ $88, 888, 888, 889. S∞ = 1 − 0.73 26. Sn : 1 + 4 + 7 + . . . + (3n − 2) = S1 : 1 =

1. Basis step:

1 + 3 + . . . + 3k−1 + 3k 3k − 1 3k − 1 2 + 3k = + 3k · 2 2 2 3k+1 − 1 3 · 3k − 1 = = 2 2 ! "! " ! " 1 1 1 28. Sn : 1 − 1 1− ... 1 − = 2 3 n n " ! 1 1 S2 : 1 − = 2 2 ! "! " ! " 1 1 1 1 Sk : 1 − 1− ... 1 − = 2 3 k k "! " ! "! " ! 1 1 1 1 1 1− . . . 1− 1− = Sk+1 : 1− 2 3 k k+1 k+1 1. Basis step: S2 is true by substitution. =

2. Induction step: Assume Sk . Deduce Sk+1 . Starting with the left side of Sk+1 , we have "! " ! "! " ! 1 1 1 1 1− 1− 1− ... 1 − 2 3 k k+1 ( )* +! " 1 1 · 1− By Sk = k k+1 ! " k+1−1 1 = · k k+1 1 k = · k k+1 1 = . Simplifying k+1

n(3n − 1) 2

1(3 − 1) 2

k(3k − 1) 2 : 1 + 4 + 7 + . . . + (3k − 2) + [3(k + 1) − 2]

Sk : 1 + 4 + 7 + . . . + (3k − 2) = Sk+1

= 1 + 4 + 7 + . . . + (3k − 2) + (3k + 1) (k + 1)(3k + 2) = 2 2 1(3 − 1) = = 1 is true. 1. Basis step: 2 2 2. Induction step: Assume Sk . Add 3k + 1 to both sides. = = = = = 27.

1 + 4 + 7 + . . . + (3k − 2) + (3k + 1) k(3k − 1) + (3k + 1) 2 k(3k − 1) 2(3k + 1) + 2 2 3k 2 − k + 6k + 2 2 3k 2 + 5k + 2 2 (k + 1)(3k + 2) 2

Sn : 1 + 3 + 32 + . . . + 3n−1 = S1 : 1 =

31 − 1 2

3k − 1 2 3k+1 − 1 2 (k+1)−1 : 1 + 3 + 3 + ... + 3 = 2

Sk : 1 + 3 + 32 + . . . + 3k−1 = Sk+1

3n − 1 2

29. 6! = 720 30. 9 · 8 · 7 · 6 = 3024 ! " 15! 15 = 6435 31. = 8 8!(15 − 8)!

32. 24 · 23 · 22 = 12, 144 33.

9! = 3780 1!4!2!2!

34. 3 · 4 · 3 = 36 35. a) 6 P5 =

6! = 720 (6 − 5)!

b) 65 = 7776 5! = 120 c) 5 P4 = (5 − 4)! 3! d) 3 P2 = =6 (3 − 2)! 36. 28 = 256 37. (m + n)7

Pascal’s triangle method: Use the 8th row of Pascal’s triangle. 1 7 21 35 35 21 7 1

692

Chapter 10: Sequences, Series, and Combinatorics (m + n)7 = m7 + 7m6 n + 21m5 n2 + 35m4 n3 +

Factorial notation method: ! "8 ! " ! " ! " ! " ! "2 1 1 1 8 8 8 + a6 a+ = a8 + a7 + 0 1 2 a a a ! " ! "3 ! " ! "4 1 1 8 8 + a4 + a5 4 3 a a ! " ! "5 ! " ! "6 1 1 8 8 + a2 + a3 5 6 a a ! " ! "7 ! "! "8 1 1 8 8 + a 8 7 a a = a8 + 8a6 + 28a4 + 56a2 + 70+

35m3 n4 + 21m2 n5 + 7mn6 + n7 Factorial notation method: ! " ! " ! " ! " 7 7 7 7 (m+n)7 = m7 + m6 n+ m5 n2 + m4 n3+ 0 1 2 3 ! " ! " ! " ! " 7 7 7 7 3 4 2 5 6 m n + mn + n7 m n + 5 6 7 4 = m7 + 7m6 n + 21m5 n2 + 35m4 n3 + 35m3 n4 + 21m2 n5 + 7mn6 + n7 38. Expand: (x −

√

2)5

Pascal’s triangle method: Use the 6th row. 1 5 10 10 5 1 √ √ √ (x − 2)5 = x5 + 5x4 (− 2) + 10x3 (− 2)2 + √ √ √ 10x2 (− 2)3 + 5x(− 2)4 + (− 2)5 √ √ √ = x5 − 5 2x4 + 20x3 − 20 2x2 + 20x − 4 2

56a−2 + 28a−4 + 8a−6 + a−8 41. Expand: (1 + 5i)6 Pascal’s triangle method: Use the 7th row. 1 6 15 20 15 6 1 (1 + 5i)6 = 16 + 6(1)5 (5i) + 15(1)4 (5i)2 + 20(1)3 (5i)3 +

Factorial notation method: ! " ! " ! " √ 5 5 5 4 √ 5 3 √ 2 x + x (− 2)+ x (− 2) + (x− 2)5 = 0 1 2 ! " ! " ! " √ √ 5 5 5 2 √ 3 x(− 2)4 + (− 2)5 x (− 2) + 4 5 3 √ √ √ = x5 − 5 2x4 + 20x3 − 20 2x2 + 20x − 4 2

15(1)2 (5i)4 + 6(1)(5i)5 + (5i)6 = 1 + 30i − 375 − 2500i + 9375+ 18, 750i − 15, 625

= −6624 + 16, 280i

Factorial notation method: ! " ! " ! " 6 6 6 6 6 5 1 + (1) (5i) + (1)4 (5i)2 + (1 + 5i) = 0 1 2 ! " ! " 6 6 (1)2 (5i)4 + (1)3 (5i)3 + 4 3 ! " ! " 6 6 5 (5i)6 (1)(5i) + 6 5 = 1 + 30i − 375 − 2500i + 9375+

39. Expand: (x2 − 3y)4

Pascal’s triangle method: Use the 5th row. 1 4 6 4 1 (x2 − 3y)4 = (x2 )4 + 4(x2 )3 (−3y) + 6(x2 )2 (−3y)2 + 4(x2 )(−3y)3 + (−3y)4

= x8 − 12x6 y + 54x4 y 2 − 108x2 y 3 + 81y 4

Factorial notation method: ! " ! " 4 4 (x2 )4 + (x2 )3 (−3y)+ (x2 − 3y)4 = 0 1 ! " ! " 4 4 (x2 )(−3y)3 + (x2 )2 (−3y)2 + 3 2 ! " 4 (−3y)4 4 = x8 − 12x6 y + 54x4 y 2 − 108x2 y 3 + 81y 4 "8 1 a Pascal’s triangle method: Use the 9th row.

40. Expand:

!

a+

1 8 28 56 70 56 28 8 1 "8 ! " ! "2 ! "3 ! 1 1 1 1 + 28a6 = a8 + 8a7 + 56a5 + a+ a a a a ! "4 ! "5 ! "6 1 1 1 + 56a3 + 28a2 + 70a4 a a a ! "7 ! "8 1 1 + 8a a a = a8 + 8a6 + 28a4 + 56a2 + 70+ 56a−2 + 28a−4 + 8a−6 + a−8

18, 750i − 15, 625

= −6624 + 16, 280i 42. Find 4th term of (a + x)12 . ! " 12 a9 x3 = 220a9 x3 3

43. Find 12th term of (2a − b)18 . ! " ! " 18 18 7 11 128a7 b11 (2a) (−b) = − 11 11 44. Of 36 possible combinations, we can get a 10 with (4, 6), (6, 4) or (5, 5). 1 3 = Probability = 36 12 Since we cannot get a 10 on one die, the probability of getting a 10 on one die is 0. 45. Of 52 cards, 13 are clubs. 1 13 = . Probability = 52 4 46.

4 C2 ·4

C1

52 C3

=

6·4 6 = 22, 100 5525

Chapter 10 Review Exercises 47.

693

86 86 = ≈ 0.42 86 + 97 + 23 206 97 97 = ≈ 0.47 B: 86 + 97 + 23 206 23 23 = ≈ 0.11 C: 86 + 97 + 23 206

c) Yes; if d is the common difference of a1 , a2 , . . . , an , then each term of b1 , b2 , . . . , bn is obtained by adding 7d to the previous term and b1 , b2 , . . . , bn is an arithmetic sequence.

A:

d) No (unless an is constant) e) No (unless an is constant) f) No (unless an is constant)

48. 12, 10, 8, 6, . . . d = 10 − 12 = −2

S25 = 12 + (25 − 1)(−2) = 12 + 24(−2) = 12 − 48 = −36

57. f (x) = x4 − 4x3 − 4x2 + 16x = x(x3 − 4x2 − 4x + 16) We know that 0 is a zero.

Consider x3 − 4x2 − 4x + 16.

Answer B is correct.

Possibilities for p/q: ±1, ±2, ±4, ±8, ±16 4 −2 4 1 −4 −4 16 −2 12 −16 1 −6 8 0

49. There are 3 pairs that total 4: 1 and 3, 2 and 2, 3 and 1. There are 6 · 6, or 36, possible outcomes. Thus, we have 1 3 , or . Answer A is correct. 36 12 50.

n

Un

1

0.3

2

2.5

3

13.5

4

68.5

5

343.5

6

1718.5

7

8593.5

8

42968.5

9

214843.5

10 1074218.5 51. a) an = 0.9694927954n + 46.17184726, where an is in millions and n is the number of years after 1980 b) In 2010, n = 2010 − 1980 = 30. a30 ≈ 75.257 million

52. A list of 9 candidates for an office is to be narrowed down to 4 candidates. In how many ways can this be done? 53. Someone who has managed several sequences of hiring has a considerable income from the sales of the people in the lower levels. However, with a finite population, it will not be long before the salespersons in the lowest level have no one to hire and no one to sell to.

f (x) = x(x + 2)(x2 − 6x + 8)

= x(x + 2)(x − 2)(x − 4) The zeros are −2, 0, 2, and 4.

1 3 58. r = − , S∞ = 3 8 a1 S∞ = 1−r 3 a !1 " = 1 8 1− − 3 1 a1 = 2 ! " 1 1 1 a2 = − =− 2 3 6 ! " 1 1 1 a3 = − − = 6 3 18 ! " 10 # 10 59. (−1)k (log x)10−k (log y)k k k=0

60.

n! (n − 1)! = 3· 6!(n − 6)! 5!(n − 1 − 5)!

54. S1 fails for both (a) and (b).

b) Yes; if d is the common difference of a1 , a2 , . . . , an , then it is also the common difference of b1 , b2 , . . . , bn and consequently b1 , b2 , . . . , bn is an arithmetic sequence.

n! 3(n − 1)! = 6!(n − 6)! · 6!(n − 6)! 5!(n − 6)! n! = 18(n − 1)! 18(n − 1)! n! = (n − 1)! (n − 1)! n = 18 ! " n = 36 n−1 n! = 36 (n − 1)![n − (n − 1)]!

6!(n − 6)! ·

ak+1 bk+1 ak+1 bk+1 55. = r1 , = r2 , so = r1 r2 , a constant. ak bk ak bk 56. a) If all of the terms of a1 , a2 , . . . , an are positive or if they are all negative, then b1 , b2 , . . . , bn is an arithmetic sequence whose common difference is |d|, where d is the common difference of a1 , a2 , . . . , an .

= (log x − log y)10 "10 ! x = log y ! " ! " n n−1 = 3· 6 5

61.

n(n − 1)! = 36 (n − 1)!1! n = 36

694 62.

Chapter 10: Sequences, Series, and Combinatorics " 5 ! # 5 95−k ak = 0 k

k=0

8. a1 = 8, a21 = 108, n = 21 an = a1 + (n − 1)d

(9 + a)5 = 0

108 = 8 + (21 − 1)d

9+a = 0

100 = 20d

a = −9

5=d Use an = a1 + (n − 1)d again to find a7 .

Chapter 10 Test 1.

an = (−1)n (2n + 1)

a7 = 8 + (7 − 1)(5) = 8 + 30 = 38

9. a1 = 17, d = 13 − 17 = −4, n = 20 First find a20 : an = a1 + (n − 1)d

a21 = (−1)21 [2(21) + 1] = −43 2.

an = a1 = a2 = a3 = a4 = a5 =

3.

4 #

n+1 n+2 1+1 = 1+2 2+1 = 2+2 3+1 = 3+2 4+1 = 4+2 5+1 = 5+2

a20 = 17 + (20 − 1)(−4) = 17 − 76 = −59

2 3 3 4 4 5 5 6 6 7

Now find S20 : n Sn = (a1 + an ) 2 20 (17 − 59) = 10(−42) = −420 S20 = 2 25 # (2k + 1) 10. k=1

a1 = 2 · 1 + 1 = 3

a25 = 2 · 25 + 1 = 51 n Sn = (a1 + an ) 2 25 25 S25 = (3 + 51) = · 54 = 675 2 2

(k 2 + 1) = (12 + 1) + (22 + 1) + (32 + 1) + (42 + 1)

k=1

= 2 + 5 + 10 + 7

11.

6 #

4k

∞ #

2k

k=1

5.

k=1

6.

7.

1 an+1 = 2 + an 1 a1 = 2 + = 2 + 1 = 3 1 1 1 a2 = 2 + = 2 3 3 1 3 3 a3 = 2 + = 2 + = 2 7 7 7 3 1 7 7 =2+ a4 = 2 + =2 17 17 17 7 d=5−2=3

an = a1 + (n − 1)d

a15 = 2 + (15 − 1)3 = 44

−5 1 =− 10 2

an = a1 rn−1 "11−1 ! 1 5 a11 = 10 − = 2 512

= 34 4.

a1 = 10, r =

12. r = 0.2, S4 = 1248 a1 (1 − rn ) Sn = 1−r

a1 (1 − 0.24 ) 1 − 0.2 0.9984a1 1248 = 0.8 a1 = 1000 1248 =

13.

8 #

2k

k=1

a1 = 21 = 2, r = 2, n = 8 S8 =

2(1 − 28 ) = 510 1−2

14. a1 = 18, r =

1 < 1, the series has a sum. 3 18 18 3 = = 18 · = 27 = 2 1 2 1− 3 3

Since |r| = S∞

1 6 = 18 3

Chapter 10 Test

695

15. 0.56 = 0.56 + 0.0056 + 0.000056 + . . . 4 4 4 0.0056 4 4 4 = |0.01| = 0.01 < 1, so the series has a sum. |r| = 4 0.56 4 0.56 0.56 56 S∞ = = = 1 − 0.01 0.99 99

16.

a1 = $10, 000

=

20.

15 P6

21.

21 C10

22.

!

n 4

15! = 3, 603, 600 (15 − 6)!

= "

21! = 352, 716 10!(21 − 10)!

a2 = $10, 000 · 0.80 = $8000

=

a4 = $6400 · 0.80 = $5120

=

a3 = $8000 · 0.80 = $6400 a5 = $5120 · 0.80 = $4096

a6 = $4096 · 0.80 = $3276.80

23. 6 P4 =

17. We have an arithmetic sequence $8.50, $8.75, $9.00, $9.25, and so on with d = $0.25. Each year there are 12/3, or 4 raises, so after 4 years the sequence will have the original hourly wage plus the 4 · 4, or 16, raises for a total of 17 terms. We use the formula an = a1 + (n − 1)d with a1 = $8.50, d = $0.25, and n = 17. a17 = $8.50+(17−1)($0.25) = $8.50+16($0.25) = $8.50+ $4.00 = $12.50 At the end of 4 years Tamika’s hourly wage will be $12.50. 18. We use the formula Sn = r = 1.056, and n = 18. S18 = 19.

=

a1 (1 − rn ) with a1 = $2500, 1−r

28 C4

26.

12 C8 ·8

=

2 + 5 + 8 + . . . + (3k − 1) +[3(k + 1) − 1] )* + ( k(3k + 1) 2

=

3k 2 k + + 3k + 2 2 2

=

7k 3k 2 + +2 2 2

3k 2 + 7k + 4 = 2 (k + 1)(3k + 4) = 2 =

(k + 1)[3(k + 1) + 1] 2

+ [3k + 3 − 1] By Sk

5! = 60 (5 − 3)!

28! = 20, 475 4!(28 − 4)! C4 =

12! 8! · = 34, 650 8!(12 − 8)! 4!(8 − 4)!

27. Expand: (x + 1)5 . Pascal’s triangle method: Use the 6th row. 1 5 10 10 5 1 (x+1)5 = x5 +5x4 ·1+10x3 ·12 +10x2 ·13 +5x·14 +15 = x5 + 5x4 + 10x3 + 10x2 + 5x + 1

Factorial notation method: ! " ! " ! " 5 5 5 x5 + x4 · 1 + x3 · 12 + (x + 1)5 = 0 1 2 ! " ! " ! " 5 5 5 x · 14 + 15 x2 · 13 + 4 5 3

1(3 · 1 + 1) 2

(k + 1)[3(k + 1) + 1] 2 1·4 1(3 · 1 + 1) 1) Basis step: = = 2, so S1 is true. 2 2 2) Induction step:

=

25.

Sk : 2 + 5 + 8 + . . . + (3k − 1) = Sk+1

n(n − 1)(n − 2)(n − 3) 24

6! = 360 (6 − 4)!

b) 5 P3 =

n(3n + 1) Sn : 2 + 5 + 8 + . . . + (3n − 1) = 2

k(3k + 1) 2 : 2 + 5 + 8 + . . . + (3k − 1) + [3(k + 1) − 1] =

n(n − 1)(n − 2)(n − 3)(n − 4)! 4!(n − 4)!

24. a) 64 = 1296

2500[1 − (1.056)18 ] = $74, 399.77 1 − 1.056

S1 : 2 =

n! 4!(n − 4)!

= x5 + 5x4 + 10x3 + 10x2 + 5x + 1 28. Find 5th term of (x − y)7 . ! " 7 x3 (−y)4 = 35x3 y 4 4 29. 29 = 512 30.

8 4 8 = = 6+8 14 7

31.

6 C1 ·5

32.

6 · 10 · 4 C2 ·4 C5 48 = = C 5005 1001 15 6 n P7 = 9 ·n P6 n! n! = 9· (n − 7)! (n − 6)!

n! (n − 6)! n! (n − 6)! · = 9· · (n − 7)! n! (n − 6)! n! (n − 6)(n − 7)! =9 (n − 7)! n−6 = 9

n = 15