Instructor’s Solutions Manual of College Algebra And Trigonometry

INSTRUCTOR’S SOLUTIONS MANUAL Richard N. Aufmann Palomar College Vernon C. Barker Palomar College Richard D. Nation Palo...

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INSTRUCTOR’S SOLUTIONS MANUAL Richard N. Aufmann Palomar College Vernon C. Barker Palomar College Richard D. Nation Palomar College Christine S. Verity

COLLEGE ALGEBRA AND TRIGONOMETRY SIXTH EDITION

Aufmann/Barker/Nation

HOUGHTON MIFFLIN COMPANY BOSTON NEW YORK

Editor-in-Chief/Publisher: Richard Stratton Sponsoring Editor: Molly Taylor Marketing Manager: Jennifer Jones Editorial Associate: Andrew Lipsett Marketing Associate: Mary Legere Editorial Assistant: Anthony D’Aries

Copyright © 2008 by Houghton Mifflin Company. All rights reserved. Houghton Mifflin Company hereby grants you permission to reproduce the Houghton Mifflin material contained in this work in classroom quantities, solely for use with the accompanying Houghton Mifflin textbook. All reproductions must include the Houghton Mifflin copyright notice, and no fee may be collected except to cover the cost of duplication. If you wish to make any other use of this material, including reproducing or transmitting the material or portions thereof in any form or by any electronic or mechanical means including any information storage or retrieval system, you must obtain prior written permission from Houghton Mifflin Company, unless such use is expressly permitted by federal copyright law. If you wish to reproduce material acknowledging a rights holder other than Houghton Mifflin Company, you must obtain permission from the rights holder. Address inquiries to College Permissions, Houghton Mifflin Company, 222 Berkeley Street, Boston, MA 02116-3764. Printed in the U.S.A. ISBN 13: 978-0-618-82508-0 ISBN 10: 0-618-82520-7 1 2 3 4 5 6 7 8 9-XX-11 10 09 08 07

Contents Chapter P

Preliminary Concepts

Chapter 1

Equations and Inequalities

36

Chapter 2

Functions and Graphs

98

Chapter 3

Polynomial and Rational Functions

177

Chapter 4

Exponential and Logarithmic Functions

238

Chapter 5

Trigonometric Functions

303

Chapter 6

Trigonometric Identities and Equations

361

Chapter 7

Applications of Trigonometry

436

Chapter 8

Topics in Analytic Geometry

496

Chapter 9

Systems of Equations and Inequalities

598

Chapter 10

Matrices

691

Chapter 11

Sequences, Series, and Probability

762

1

Projects

818

Additional College Trigonometry Solutions

871

Correlation Chart

Correlation Chart College Trigonometry, 6e Solutions/Projects Chapter 1 Section 1.1 Section 1.2* Section 1.3* Section 1.4* Section 1.5* Section 1.6* Section 1.7 Exploring Concepts with Technology Chapter 1 Assessing Concepts Chapter 1 Review Exercises Chapter 1 Quantitative Reasoning Chapter 1 Test Chapter 2 Chapter 2 up through Chapter 2 Test Cumulative Review Exercises Chapter 3 Chapter 3 up through Chapter 3 Test Cumulative Review Exercises Chapter 4 Section 4.1 Section 4.2 Section 4.3 Exploring Concepts with Technology Chapter 4 Review Exercises Chapter 4 Assessing Concepts Chapter 4 Test Cumulative Review Exercises Chapter 5 Section 5.1 Exercises 1-62 Section 5.1* Exercises 63-72 Section 5.2 Section 5.3 Exploring Concepts with Technology Chapter 5 Assessing Concepts Chapter 5 Review Exercises Chapter 5 Quantitative Reasoning Chapter 5 Test Cumulative Review Exercises Chapter 6 Sections 6.1-6.6 Section 6.7, Exercises 1-18 Section 6.7, Exercises 19-40 Chapter 6 Connecting Concepts Chapter 6 Review Exercises Chapter 6 Test Cumulative Review Exercises Chapter 7 Chapter 7 up through Chapter 7 Test Cumulative Review Exercises

College Algebra and Trigonometry, 6e Solutions/Projects See Additional College Trigonometry Solutions Section 2.1 Section 2.2 Section 2.5 Section 2.6 Section 4.1 Section 2.7 Chapter 2 Exploring Concepts with Technology See Additional College Trigonometry Solutions

Chapter 5 up through Chapter 5 Test See Additional College Trigonometry Solutions Chapter 6 up through Chapter 6 Test See Additional College Trigonometry Solutions Section 7.1 Section 7.2 Section 7.3 Chapter 7 Exploring Concepts with Technology Exercises 1-45 Chapter 7 Review Exercises See Additional College Trigonometry Solutions

Section P.6 Exercises 1-62 See Additional College Trigonometry Solutions Section 7.4 Section 7.5

See Additional College Trigonometry Solutions

Chapters 8.1-8.6 Section 8.7, Exercises 1-18

See Additional College Trigonometry Solutions

Section 4.2-4.7 up through Chapter 4 Test See Additional College Trigonometry Solutions

* See the Additional College Trigonometry Solutions for solutions to the Prepare for Section exercises.

Solutions

Chapter P

Preliminary Concepts Section P.1 1.

− 15 : rational, real; 0: integer, rational, real; –44: integer, rational, real; π : irrational, real; 3.14: rational, real; 5.05005000500005…: irrational, real;

2.

81 = 9 : integer, rational, prime, real; 53: integer, rational, prime, real

5 : irrational, real; 5 : rational, real; 31: integer, rational, prime, real; −2 1 : rational, real; 4.235653907493: rational, real; 7 7 2

51: integer, rational, real; 0.888… = 0.8 = 8 : rational, real 9 3.

Let x = 1, 2, 3, 4. Then {2x | x is a positive integer} = {2, 4, 6, 8}

4.

Let x = 0, 1, 2, 3. (We could have used x = –3, –2, –1, 0.) Then {|x| | x is an integer} = {0, 1, 2, 3}

5.

Let x = 1, 2, 3, 4. (Recall 0 is not a natural number.) Then {y | y = 2x + 1, x is a natural number} = {3, 5, 7, 9}

6.

Let x = 0, 1, 2, 3. (We could have used x = −3, −2, −1, 0.) Then {y | y = x2 − 1} = {−1, 0, 3, 8}

7.

Let x = 0, 1, 2, 3. (We could have used x = −3, −2, −1, 0.) Then {z | z = |x|, x is an integer}= {0, 1, 2, 3}

8.

Let x = −1, −2, −3, −4. Then {z | z = |x| − x, x is a negative integer}= {2, 4, 6, 8}

9.

A ∪ B = {−3, −2, −1, 0, 1, 2, 3, 4, 6}

11.

A ∩ C = {0, 1, 2, 3}

14.

(A ∩ C) = {0, 1, 2, 3} B ∪ (A ∩ C) = {−2, 0, 1, 2, 3, 4, 6}

16.

(A ∩ B) ∪ (A ∩ C) = {−2, 0, 2} ∪ {0, 1, 2, 3} = {−2, 0, 1, 2, 3}

17.

(B ∪ C) ∩ (B ∪ D) = {−2, 0, 1, 2, 3, 4, 5, 6} ∩ {−3, −2, −1, 0, 1, 2, 3, 4, 6} = {−2, 0, 1, 2, 3, 4, 6}

18.

(A ∩ C) ∪ (B ∩ D) = {0, 1, 2, 3} ∪ ∅ = {0, 1, 2, 3}

19.

10. 12.

C ∩ D = {1, 3}

21.

23.

{x | −3 < x < 3}

24.

26.

28.

{x | −5 ≤ x ≤ −1}

{x | x ≥ 2}

{x | x < 4} 27.

( −∞, − 1)

(3, 5) 29.

[−2, ∞) 30.

[0, 1]

[−1, 5)

B∩D=∅

(B ∪ C) = {−2, 0, 1, 2, 3, 4, 5, 6} D ∩ (B ∪ C) = {1, 3}

{x |1 ≤ x ≤ 5}

{x | −2 < x < 3}

25.

13. 15.

20.

22.

C ∪ D = {−3, −1, 0, 1, 2, 3, 4, 5, 6}

(−4, 5]

31.

–5

32.

–(4)2 = –16

33.

3(4) = 12

34.

|−3| − |−7| = 3 − 7 = −4

35.

π2 + 10

36.

10 – π2

Copyright © Houghton Mifflin Company. All rights reserved.

2

Chapter P: Preliminary Concepts

37.

x−4 + x+5 = 4− x+ x+5=9

38.

39.

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

40.

|x + 1| + |x – 3| = (x + 1) + (x – 3) = 2x – 2

41.

|x–3|

45.

m−n

46.

p −8

47.

a−4 7

49.

x+2 >4

50.

y+3 >6

42.

|a––2|=|a+2|

x + 6 + x − 2 = x + 6 + x − 2 = 2x + 4

43.

44.

x − −2 = 4 x+2 =4

51.

52.

53.

54.

55.

56.

57.

58.

59.

60.

61.

62.

63.

64.

65.

66.

z −5 =1

67.

−( −2)3 = −( −8) = 8

68.

−( −2)2 = −(4) = −4

69.

2(3)( −2)( −1) = 12

70.

−3(3)( −1) = 9

71.

−2(3)2 ( −2)2 = −2(9)(4) = −72

72.

2( −2)3 ( −1)2 = 2( −8)(1) = −16

73.

3( −2) − ( −1)[3 − ( −2)]2 = 3( −2) − ( −1)[3 + 2]2 = (3)( −2) − ( −1)[5]2 = (3)( −2) − ( −1)(25) = −6 + 25 = 19

74.

[ −1 − 2( −2)]2 − 3( −1)3 = ( −1 + 4)2 − 3( −1)3 = (3)2 − 3( −1) = 9 + 3 = 12

75.

32 + ( −2)2 9 + 4 13 = = = 13 3 + ( −2) 1 1

76.

2(3)( −2)2 ( −1)4

77.

3( −2) 2( −1) −6 −2 − = − = − 2 − 1 = −3 3 3 −2 −2

78.

(3 + 1)2 (3 − 1)2 = 42 ⋅ 22 = 16 ⋅ 4 = 64

79.

( ab2 )c = a (b2 c ) Associative property of multiplication

80.

2 x − 3 y = −3 y + 2 x Commutative property of addition

81.

4(2a − b) = 8a − 4b Distributive property

82.

6 + (7 + a) = 6 + (a + 7) Commutative property of addition

83.

(3x ) y = y (3x ) Commutative property of multiplication

84.

4ab + 0 = 4ab Identity property of addition

[ −2 − ( −1)]4

=

2(3)(4)(1) ( −2 + 1) 4


= 24 4 = 24 ( −1)

Section P.1

3

85.

1 ⋅ (4 x ) = 4 x Identity property of multiplication

86.

7(a + b) = 7(b + a) Commutative property of addition

87.

x2 + 1 = x2 + 1 Reflexive property of equality

88.

If a + b = 2, then 2 = a + b Symmetric property of equality

89.

If 2x + 1 = y and 3x – 2 = y, then 2x+ 1 = 3x – 2 Transitive property of equality

90.

4x + 2y = 7 and x= 3, then 4(3) = 2y = 7 Substitution property of equality

91.

4⋅ 1 =1 4 Inverse property of multiplication

92.

ab + ( −ab) = 0 Inverse property of addition

93.

3(2 x )

−2(4 y ) −8 y

95.

3(2 + x )

3x−1x 4 2 3x−2x 4 4 1x 4

99.

94.

6x

97.

2a + 5a 3 6 4a + 5a 6 6 9a 6 3a 2

101.

5 − 3(4 x − 2 y )

98.

5(4 r − 7t ) − 2(10r + 3t ) 20r − 35t − 20r − 6t 20r − 20r − 35t − 6t −41t

107.

Area = 1 bh = 1 (3 in)(4 in) = 6 in 2 2 2

≈ 66 Heart rate is about 66 beats per minute.

4 + 2(2a − 3) 4 + 4a − 6 4a − 2

3(2a − 4b) − 4( a − 3b) 6a − 12b − 4a + 12b 6a − 4a − 12b + 12b 2a

105.

5a − 2[3 − 2(4a + 3)] 5a − 2(3 − 8a − 6) 5a − 2( −8a − 3) 5a + 16a + 6 21a + 6

106.

6 + 3[2 x − 4(3x − 2)] 6 + 3(2 x − 12 x + 8) 6 + 3( −10 x + 8) 6 − 30 x + 24 30 x + 30

= −0.5(110) + 120(110) − 2000 = −0.5(12100) + 120(110) − 2000 = −6050 + 13200 − 2000 = 5150 The profit for selling 110 bicycles is $5150.

= 65 + 53 41

100.

103.

= −0.5(110)2 + 120(110) − 2000

111. Heart rate = 65 + 53 4t + 1 53 = 65 + 4(10) + 1

2 + 3(2 x − 5) 2 + 6 x − 15 6 x − 13

7 − 2(5n − 8m) 7 − 10n + 16m 16m − 10n + 7

109. Profit = = −0.5 x 2 + 120 x − 2000 2

−2(4 + y ) −2 y − 8

3x + 6

102.

5 − 12 x + 6 y −12 x + 6 y + 5 104.

96.

108. V = lwh = (40 ft)(30 ft)(12 ft) = 14,400 ft 3 110. Circulation = n 2 − n + 1 = 122 − 12 + 1 144 − 12 + 1 = 133 ≈ 11.5 The circulation of the magazine after 12 months is approximately 11.5 thousand or 11,500 subscriptions.

112. BMI = 7052w h 705(160) 112800 = = ≈ 23 4900 (70)2 The body mass index (BMI) of a person who weighs 160 pounds and is 5 feet 10 inches (70 inches) tall is about 23.


4


113. Height = −16t 2 + 80t + 4 = −16(2)2 + 80(2) + 4 = −16(4) + 80(2) + 4 = −64 + 160 + 4 = 100 After 2 seconds, the ball will have a height of 100 feet.

114. Concentration = 50t t +1 50(24) = 24 + 1 1200 = = 48 25 After 24 minutes, the concentration will be 48 grams per liter.

.......................................................

Connecting Concepts

115. For any set A, A ∪ A = A . 116. For any set A, A∩ A = A .

117. For any set A, A ∩ ∅ = ∅.

119. If A and B are two sets and a A ∪ B = A, then all elements of B are contained in A. So B is a subset of A.

120. If A and B are two sets and a A ∩ B = B, then all elements of B are contained in A. So B is a subset of A.

121. No. (8 ÷ 4) ÷ 2 = 2 ÷ 2 = 1 8 ÷ (4 ÷ 2) = 8 ÷ 2 = 4

122. No. 5–3=2 3 – 5 = –2

123. All but the multiplicative inverse property

124. All

125.

x+7 x+7 = x+7 = x+7 = x+7 = x + x −1 x − ( x − 1) 1 x + x −1

126.

x+3 x+3 x+3 x+3 = = = = x+3 1 1 1 1 1 1 1⎞ ⎛ 1⎞ ⎛ + x− + x+ −⎜ x − ⎟ + ⎜ x + ⎟ x− + x+ 2 2 2 2 2 2 2⎠ ⎝ 2⎠ ⎝

127. |x – 2| < |x – 6|

128. |x – a| < |x – b|

131. 2 < |x – 4| < 7

132. b < |x – a| < c

129. |x – 3| > |x + 7|

....................................................... PS1. 22 ⋅ 23 = 4 ⋅ 8 = 32 3

Alternate method: 2 ⋅ 2 = 2

2 +3

5

= 2 = 32

130. |x| > |x – 5|

Prepare for Section P.2 PS2.

2

118. For any set A, A ∪ ∅ = A.

43 = 43−5 = 4 −2 = 1 = 1 45 42 16 3 Alternate method: 45 = 51−3 = 12 = 1 16 4 4 4

PS3. (23 ) 2 = 82 = 64

PS4. 3.14(105 ) = 3.14(100,000) = 314,000

Alternate method: (23 )2 = 23(2) = 26 = 64

Alternate method: To multiply by 105 , move the decimal point 5 places to the right. 000. = 314,000 Thus, 3.14(105 ) = 314 → → → → →

PS5. False

34 ⋅ 32 = 36 , not 96 .

PS6. False

(3 + 4)2 = 72 = 49 but 32 + 42 = 9 + 16 = 25.


Section P.2

5

Section P.2 0

1.

−53 = −(53 ) = −125

2.

( −5)3 = −125

3.

⎛ 2⎞ =1 ⎜ ⎟ ⎝ 3⎠

4.

−60 = −(60 ) = −1

5.

4 −2 = 12 = 1 16 4

6.

3−4 = 14 = 1 81 3

7.

1 = 25 = 32 2 −5

8.

1 = 33 = 27 3−3

9.

2 −3 = ⎛ 2 ⎞ ⎜ ⎟ 6 −3 ⎝ 6 ⎠

10.

4 −2 = 23 = 8 = 1 2 −3 42 16 2

11.

−2 x 0 = −2

12.

x0 = 1 4 4

−3

= ⎛⎜ 1 ⎞⎟ ⎝ 3⎠

−3

3

= ⎛⎜ 3 ⎞⎟ = 33 = 27 ⎝1⎠

13.

2 x −4 = 2 ( x −4 ) = 24 x

14.

3 y −2 = 3 ( y −2 ) = 32 y

15.

5 = 5z 6 z −6

16.

8 = 8z5 x −5

17.

( x 3 y 2 )( xy 5 ) = x 3+1 y 2+5 = x 4 y 7

18.

(uv 6 )(u 2 v ) = u1+ 2 v 6+1 = u 3v 7

19.

( −2ab4 )( −3a 2 b5 ) = ( −2)( −3)a1+ 2 b4+5 = 6a 3b9

20.

(9 xy 2 )( −2 x 2 y 5 ) = (9)( −2) x1+ 2 y 2+5 = −18 x 3 y 7

21.

16a 7 = 16 a 7 −1 = 8a 6 2a 2

22.

24 z 8 = 24 z 8−3 = −8 z 5 −3z 3 −3

23.

6a 4 = 6 a 4 −8 = 3 a −4 = 3 4 8a 8 8 4a 4

24.

12 x 3 = 12 x 3− 4 = 3 x −1 = 3 4 4x 16 x 4 16

25.

12 x 3 y 4

2 y2 = 12 x 3−5 y 4 −2 = 2 x −2 y 2 = 2 18 3 18 x y 3x

26.

5v 4 w −3 = 5 v 4 −8 w −3 = 1 v −4 w −3 = 1 10 2 10v8 2 v 4 w3

27.

36a −2 b3 = 36 a −2 −1b3− 4 = 12a −3b −1 = 12 3 a 3b 3ab4

28.

−48ab10 = −48 a1− 4 b10−3 = 3 a −3b7 = 3b7 2 2a 3 −32a 4 b3 −32

29.

( −2m 3n 2 )( −3mn 2 )2 = ( −2m3n 2 )(9m 2 n 4 )

30.

(2a 3b2 )3 ( −4a 4 b2 ) = (8a 9 b6 )( −4a 4 b2 )

5 2

31.

= ( −2)(9)m 3+ 2 n 2 + 4

= (8)( −4)a 9 + 4 b6+ 2

= −18m5n 6

= −32a13b8

( x −2 y )2 ( xy ) −2 = ( x −4 y 2 )( x −2 y −2 )

32.

( x −1 y 2 ) −3 ( x 2 y −4 ) −3 = ( x 3 y −6 )( x −6 y12 )

= x −4 − 2 y 2 − 2

= x 3−6 y −6+12

= x −6 y 0 = 16 x

= x −3 y 6 =

y6 x3


6

33.

35.

Chapter P: Preliminary Concepts 2

⎛ 3a 2 b3 ⎞ 9a 4 b6 ⎜ 4 4⎟ = 36a 8b8 ⎝ 6a b ⎠ = 9 a 4 −8 b 6 −8 36 = 1 a −4 b −2 4 = 14 2 4a b

34.

( −4 x 2 y 3 ) 2

36.

2 3

(2 xy )

=

9

= 8c 125

16 x 4 y 6 3 6

( −3a 2 b3 )2

4 6 = 9a 3b 12 ( −2ab ) −8a b = − 9 a 4 −3b6−12 8 = − 9 ab −6 8 = − 9a6 8b 4 3

8x y

= 2 x 4 −3 y 6−6 = 2x

37.

3

3 6 9 ⎛ 2ab2 c3 ⎞ = 8a b 3c 6 ⎜ 2 ⎟ ⎝ 5ab ⎠ 125a b = 8 a 3−3b6−6 c9 125 = 8 a 0 b0 c 9 125

2

⎛ a −2 b ⎞ a −4 b2 ⎜ 3 −4 ⎟ = 6 −8 a b ⎝a b ⎠

38.

= a −4 −6 b2 −( −8)

⎛ x −3 y −4 ⎞ ⎜⎜ x −2 y ⎟⎟ ⎝ ⎠

−2

=

x 6 y8 x 4 y −2

= x 6− 4 y 8−( −2)

= a −4 −6 b2 +8

= x 6 − 4 y 8+ 2

= a −10 b10

= x 2 y10

10 = b10 a

39.

2,011,000,000,000 = 2.011 × 1012

40.

49,100,000,000 = 4.91 × 1010

41.

0.000000000562 = 5.62 × 10−10

42.

0.000000402 = 4.02 × 10−7

43.

3.14 × 107 = 31,400,000

44.

4.03 × 109 = 4,030,000,000

45.

−2.3 × 10−6 = −0.0000023

46.

6.14 × 10−8 = 0.0000000614

47.

(3 × 1012 )(9 × 10−5 ) = (3)(9) × 1012 −5

48.

(8.9 × 10−5 )(3.4 × 10−6 ) = (8.9)(3.4) × 10−5−6

49.

= 27 × 107

= 30.26 × 10−11

= 2.7 × 108

= 3.026 × 10−10

9 × 10−3 = 9 × 10−3−8 6 × 108 6

50.

2.5 × 108 = 2.5 × 108−10 5 5 × 1010 = 0.5 × 10−2

= 1.5 × 10−11

= 5 × 10−3

51.

(3.2 × 10−11 )(2.7 × 1018 ) 1.2 × 10

−5

=

(3.2)(2.7) × 10−11+18−( −5) 1.2

52.

(6.9 × 1027 )(8.2 × 10−13 ) 4.1 × 1015

=

(6.9)(8.2) × 1027 −13−15 4.1

= 7.2 × 10−11+18+5

= 13.8 × 10−1

= 7.2 × 1012

= 1.38 × 100


Section P.2

53.

7

(4.0 × 10−9 )(8.4 × 105 ) −6

18

(3.0 × 10 )(1.4 × 10 )

=

(4.0)(8.4) × 10−9 + 5+ 6−18 (3.0)(1.4)

= 8 × 10

54.

(7.2 × 108 )(3.9 × 10−7 ) (2.6 × 10

−10

−8

)(1.8 × 10 )

−16

=

(7.2)(3.9) × 108−7 −( −10)−( −8) (2.6)(1.8)

= 6 × 108−7 +10+8 = 6 × 1019

3

56.

−163/ 2 = − 16 = −43 = −64

3

57.

−642 / 3 = − 3 64 = −42 = −16

59.

1 = 1 = 1 = 1 9 −3/ 2 = 3/ 9 2 ( 9 )3 33 27

60.

32 −3/ 5 =

55.

43/ 2 = 4 = 23 = 8

58.

1254 / 3 = 3 125 = 54 = 625

61.

⎛4⎞ ⎜ ⎟ ⎝9⎠

63.

⎛1⎞ ⎜ ⎟ ⎝8⎠

65.

(4a 2 / 3b1/ 2 )(2a1/ 3b3 / 2 ) = (4)(2)a 2 / 3+1/ 3b1/ 2 + 3/ 2 = 8a 3 / 3b4 / 2 = 8ab2

66.

(6a 3/ 5b1/ 4 )( −3a1/ 5b3/ 4 ) = (6)( −3)a 3 / 5+1/ 5b1/ 4 +3/ 4 = −18a 4 / 5b

67.

( −3x 2 / 3 )(4 x1/ 4 ) = ( −3)(4) x 2 / 3+1/ 4 = −12 x8 /12 + 3/12 = −12 x11/12

68.

( −5 x1/ 3 )( −4 x1/ 2 ) = ( −5)( −4) x1/ 3+1/ 2 = 20 x 2 / 6+ 3/ 6 = 20 x 5 / 6

69.

(81x8 y12 )1/ 4 = 811/ 4 x8 / 4 y12 / 4 = 4 81x 2 y 3 = 3x 2 y 3

70.

(27 x 3 y 6 )2 / 3 = 272 / 3 x 6 / 3 y12 / 3 = 3 27 x 2 y 4 = 32 x 2 y 4 = 9 x 2 y 4

71.

16 z 3/ 5 = 16 z 3/ 5−1/ 5 = 4 z 2 / 5 12 3 12 z1/ 5

73.

(2 x 2 / 3 y1/ 2 )(3x1/ 6 y1/ 3 ) = (2)(3) x 2 / 3+1/ 6 y1/ 2+1/ 3 = 6 x 5 / 6 y 5 / 6

74.

x1/ 3 y 5 / 6

4

1/ 2

= 4= 4=2 9 9 3

−4 / 3

4

= 84 / 3 = 3 8 = 24 = 16

⎛ 16 ⎞ ⎜ ⎟ ⎝ 25 ⎠

64.

⎛ 8 ⎞ ⎜ ⎟ ⎝ 27 ⎠

3

1 = 1 = 1 =1 323/ 5 ( 5 32 )3 23 8

3

3 ⎛ ⎞ = ⎜ 16 ⎟ = ⎛⎜ 4 ⎞⎟ = 43 = 64 125 5 ⎝ 25 ⎠ ⎝ 5 ⎠

−2 / 3

= ⎜⎛ 27 ⎟⎞ ⎝ 8 ⎠

2/3

2

2

2 ⎛3 ⎞ ⎛ ⎞ = ⎜ 3 27 ⎟ = ⎜ 327 ⎟ = ⎜⎛ 3 ⎟⎞ = 9 4 ⎝2⎠ ⎝ 8 ⎠ ⎝ 8 ⎠

2

x

2 / 3 1/ 6

y

72.

= x1/ 3− 2 / 3 y 5 / 6−1/ 6 = x −1/ 3 y 4 / 6 =

x1/ 3

9a 3/ 4 b = 9a 3 / 4 − 2 / 3b1− 2 = 3a 9 /12 −8 /12 b −1 = 3a1/12 3 b 3a 2 / 3b2

76.

12 x1/ 6 y1/ 4 16 x

3 / 4 1/ 2

y

=

12 x1/ 6−3/ 4 y1/ 4 −1/ 2 3x 2 /12 −9 /12 y1/ 4 − 2 / 4 3x −7 /12 y −1/ 4 3 = = = 7 /12 16 4 4 4x y1/ 4

45 = 32 ⋅ 5 = 3 5

77. 3

3

6a 2 / 3 = 6a 2 / 3−1/ 3 = 2a1/ 3 = 2a1/ 3 9 3 3 9a1/ 3

y2 / 3

75.

80.

3/ 2

62.

2

135 = 33 ⋅ 5 = 33 5

78. 81.

3

3

75 = 52 ⋅ 3 = 5 3

79.

3

24 = 23 ⋅ 3 = 2 3 3

−135 = 3 ( −3)3 ⋅ 5

82.

3

−250 = 3 ( −5)3 ⋅ 2

= −3 3 5


= −5 3 2

8


83.

24 x 2 y 3 = 22 x 2 y 2 ⋅ 6 y = 2 xy 6 y

84.

16a 3 y 7 = 3 23 a 3 y 6 ⋅ 3 2 y = 2ay 2 3 2 y

86.

18 x 2 y 5 = 32 x 2 y 4 ⋅ 2 y = 3 x y 2 2 y 3

3

3

3

54m 2 n 7 = 33 n 6 ⋅ 2m 2 n = 3n 2 2m 2 n

85.

3

87.

2 32 − 3 98 = 2 16 ⋅ 2 − 3 49 ⋅ 2 = 2(4) 2 − 3(7) 2 = 8 2 − 21 2 = −13 2

88.

5 3 32 + 2 3 108 = 5 3 8 ⋅ 4 + 2 3 27 ⋅ 4 = 5 3 8 ⋅ 3 4 + 2 3 27 ⋅ 3 4 = 5 23 ⋅ 3 4 + 2 33 ⋅ 3 4 = 5(2) 3 4 + 2(3) 3 4

3

3

= 10 3 4 + 6 3 4 = 16 3 4 89.

4

4

−8 4 48 + 2 4 243 = −8 4 16 ⋅ 3 + 2 4 81 ⋅ 3 = −8 4 16 ⋅ 4 3 + 2 4 81 ⋅ 4 3 = −8 24 ⋅ 4 3 + 2 34 ⋅ 4 3 = −8(2) 4 3 + 2(3) 4 3 = −16 4 3 + 6 4 3 = −10 4 3

90.

3

3

3

3

2 3 40 − 33 135 = 2 3 8 ⋅ 5 − 33 27 ⋅ 5 = 2 3 8 ⋅ 3 5 − 33 27 ⋅ 3 5 = 2 23 ⋅ 3 5 − 3 33 ⋅ 3 5 = 2 23 ⋅ 3 5 − 3 33 ⋅ 3 5 = 2(2) 3 5 − 3(3) 3 5 = 4 3 5 − 9 3 5 = −5 3 5

91.

4 3 32 y 4 + 3 y 3 108 y = 4 3 8 y 3 ⋅ 4 y + 3 y 3 27 ⋅ 4 y = 4 3 8 y 3 ⋅ 3 4 y + 3 y 3 27 ⋅ 3 4 y 3

= 4 3 23 y 3 ⋅ 3 4 y + 3 y 33 ⋅ 3 4 y = 4(2 y ) 3 4 y + 3 y (3) 3 4 y = 8 y 3 4 y + 9 y 3 4 y = 17 y 3 4 y 92.

3

3

3

3

3

3

−3x 54 x 4 + 2 16 x 7 = −3x 27 x 3 ⋅ 2 x + 2 8 x 6 ⋅ 2 x = −3x 27 x 3 ⋅ 3 2 x + 2 8 x 6 ⋅ 3 2 x 3

3

= −3x 33 x 3 ⋅ 3 2 x + 2 23 x 6 ⋅ 3 2 x = −3x (3x ) 3 2 x + 2(2 x 2 ) 3 2 x = −9 x 2 3 2 x + 4 x 2 3 2 x = −5 x 2 3 2 x 93.

3

x 3 8 x 3 y 4 − 4 y 3 64 x 6 y = x 3 8 x 3 y 3 ⋅ y − 4 y 3 64 x 6 ⋅ y = x 3 8 x 3 y 3 ⋅ 3 y − 4 y 64 x 6 ⋅ 3 y 3

= x 3 23 x 3 y 3 ⋅ 3 y − 4 y 43 x 6 ⋅ 3 y = x (2 xy ) 3 y − 4 y (4 x 2 ) 3 y = 2 x 2 y 3 y − 16 x 2 y 3 y = −14 x 2 y 3 y 94.

4 a 5b − a 2 ab = 4 a 4 ⋅ ab − a 2 ab = 4 a 4 ⋅ ab − a 2 ab = 4a 2 ab − a 2 ab = 3a 2 ab

95.

( 5 + 3)( 5 + 4) = 5 + 4 5 + 3 5 + (3)(4) = 5 + 7 5 + 12 = 17 + 7 5

96.

( 7 + 2)( 7 − 5) = 7 − 5 7 + 2 7 + (2)( −5) = 7 − 3 7 − 10 = −3 − 3 7

97.

( 2 − 3)( 2 + 3) = 2 + 3 2 − 3 2 + ( −3)(3) = 2 − 9 = −7

98.

(2 7 + 3)(2 7 − 3) = (2 7)2 − 3(2 7) + 3(2 7) + (3)( −3) = 4(7) − 6 7 + 6 7 − 9 = 28 − 9 = 19

99.

(3 z − 2)(4 z + 3) = (3)(4) z + 3(3 z ) − 2(4 z ) + ( −2)(3) = 12 z + 9 z − 8 z − 6 = 12 z + z − 6

100.

(4 a − b )(3 a + 2 b ) = (4 a )(3 a ) + (4 a ) ⋅ (2 b ) − b (3 a ) − b (2 b )

2

2

2

2

2

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

2

= 12a + 8 ab − 3 ab − 2b = 12a + 5 ab − 2b 101.

2

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


Section P.2

9 2

102.

(3 5 y − 4)2 = 32 5 y + 2(3 5 y )( −4) + ( −4)2 = 9(5 y ) − 24 5 y + 16 = 45 y − 24 5 y + 16

103.

( x − 3 + 2)2 = x − 3 + 2( x − 3)(2) + 22 = x − 3 + 4 x − 3 + 4 = x + 4 x − 3 + 1

104.

( 2 x + 1 − 3)2 = 2 x + 1 + 2( 2 x + 1)( −3) + ( −3)2 = 2 x + 1 − 6 2 x + 1 + 9 = 2 x − 6 2 x + 1 + 10

2

2

3x = 3 x ⋅ 3 = 3 x 3 = 3 x 3 = 3 x 3 = x 3 2 3 3 3 3 3 3

105.

2 = 2 ⋅ 2=2 2=2 2= 2 2= 2 2 2 2 2 2 2 2

107.

5 = 5 = 5 ⋅ 2 = 5 ⋅ 2 = 5 ⋅ 2 = 10 = 10 18 2 6 2 ⋅ 32 2 ⋅ 32 22 ⋅ 32 22 ⋅ 32 2 ⋅ 3

108.

7 = 7 = 7 ⋅ 2 ⋅ 5 = 7 ⋅ 2 ⋅ 5 = 7 ⋅ 2 ⋅ 5 = 7 ⋅ 2 ⋅ 5 = 70 40 20 23 ⋅ 5 23 ⋅ 5 2 ⋅ 5 24 ⋅ 52 22 ⋅ 5 24 ⋅ 52 3

3

109.

3 = 3 ⋅ 22 = 3 2 2 = 33 4 3 3 3 2 2 3 2 3 22 2

110.

2 = 2 = 2 ⋅ 3 2 = 23 2 = 23 2 = 2 3 2 = 3 2 3 2 2 4 3 22 3 2 2 3 2 3 23 2

111.

4 3

112.

106.

8x

2

=

4 3

3 2

2 x

=

4 3

2 x

2

=

3 3 3 = 2 = 2 ⋅3x =2 x =2 x 3 2 3 2 3 3 x x x x x 2 x

4

3

2

3 3 3 4 2 3 4 2 3 4 4 4 2 = 2 = 2 ⋅ 2 y = 2 2 y = 2 4y = 2 4y = 4y 4 4y 4 22 y 4 22 y 4 22 y 3 4 24 y 4 2y 2y y

113.

3( 3 − 4) 3( 3 − 4) 3( 3 − 4) 3 3 − 12 3 = 3 ⋅ 3−4 = = = = = − 3 3 − 12 2 2 3 16 13 13 − − 3+4 3 + 4 3 − 4 ( 3 + 4)( 3 − 4) 3 −4

114.

2( 5 + 2) 2( 5 + 2) 2( 5 + 2) 2 5 + 4 2 = 2 ⋅ 5+2 = = = = =2 5+4 2 5−4 1 5−2 5 − 2 5 + 2 ( 5 − 2)( 5 + 2) 5 − 22 3

115.

3( 5 − 1) 3( 5 − 1) 6 6 6 = = = 3 = 3 ⋅ 5 −1 = = 2 2 5 + 2 2( 5 + 1) 2 ( 5 + 1) 5 +1 5 + 1 5 − 1 ( 5 + 1)( 5 − 1) 5 −1 =

116.

−7(3 2 + 5) −7(3 2 + 5) −7(3 2 + 5) −7(3 2 + 5) −7 = −7 ⋅ 3 2 + 5 = = = = (9)(2) − 25 18 − 25 3 2 − 5 3 2 − 5 3 2 + 5 (3 2 − 5)(3 2 + 5) (3 2)2 − 52 =

117.

3( 5 − 1) 3 5 − 3 = 5 −1 4

−7(3 2 + 5) −7 (3 2 + 5) = =3 2 +5 −7 −7

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


10

118.


5 = y− 3

5 ⋅ y− 3

y+ 3

=

y+ 3

5( y + 3) ( y − 3)( y + 3)

=

5 y +5 3 2

( y ) − ( 3)

119.

$8.1 × 1012 = $8.1 × 1012 −8 per person ≈ $2.72 × 104 per person 2.98 × 108 people 2.98

120.

4.7 × 1021 bacteria ⋅

670 femtograms 1 × 10−15 grams ⋅ 1 bacteria 1 femtogram

121.

−15 4.7 × 1021 ⋅ 670 ⋅ 1 × 10 = (4.7)(6.70)(1) × 1021+ 2 −15 1 1 = 3.149 × 109

2

=

5 y +5 3 y−3

1 seed ⋅ 1 ounce 3.2 × 10−8 ounce package 1 1 ⋅1 = × 108 −8 3.2 3.2 × 10

= 0.3125 × 108

= 3.13 × 107 seeds per package

They weigh 3.149 × 109 grams. 122.

800 nm ⋅ 1 × 10−9 m = (8 × 102 )(1 × 10−9 ) = 8 × 10−7 m 1 1 nm 1 frequency = wavelength 1 = = 1.25 × 108 cycles per second 8 × 10−7

123.

Red shift =

λr − λs λs −7

× 10 = 5.13 × 10 − 5.06 5.06 × 10−7 =

−7

(5.13 − 5.06) × 10−7 5.06 × 10−7 −7

= 0.07 ⋅ 10−7 5.06 10 = 1.38 × 10−2 124.

5.2 AU = 5.2(9.3 × 107 miles)

125.

= (5.2)(9.3) × 107 miles = 48.36 × 107 miles = 4.84 × 108 miles

1 sec ⋅ 1.5 × 1011 m ⋅ 1 min 60 sec 3 × 108 m 11

1 ⋅ 1.5 × 1011 ⋅ 1 = (1)(1.5)(1) × 10 60 3 × 108 3(60) × 108 = 1.5 × 1011−8 180 ≈ 0.008 × 103 ≈ 8 minutes

126.

1 gram ⋅ 1 atom 6.023 × 1023 atoms 1 ⋅ 1 = 1 × 10−23 6.023 6.023 × 1023 ≈ 0.1660302175 × 10−23

≈ 1.66 × 10−24

One hydrogen atom weighs approximately 1.66 × 10−24 gram. 127. Evaluate R when x = 20.

R = 1250 x (2 −0.007 x ) = 1250(20)(2 −0.007(20) ) = 25,000(2 −0.14 ) ≈ 25,000(0.907519) ≈ 22,688 When the company sells 20 thousand phones, the revenue is $22,688.


Section P.2

128. a. b.

11

Evaluate A when t = 4. A = 2(10−0.0078t ) = 2(10−0.0078(4) ) = 2(10−0.0312 ) ≈ 2(.930679) ≈ 1.86 Four hours after taking 2 milligrams of digoxin, about 1.86 milligrams remain in the patient’s blood. Determine the sum of the amounts of medication remaining at 6:00 PM from each of the two doses. For the 1:00 PM dose, use t = 5; for the 5:00 PM dose, use t = 1. A = 2(10−0.0078(5) ) + 2(10−0.0078(1) ) = 2(10−0.039 ) + 2(10−0.0078 ) ≈ 2(.9141) + 2(0.9822) ≈ 1.828 + 1.964 ≈ 3.79 At 6:00 PM, the amount of digoxin remaining in the patient’s blood is 3.79 milligrams.

129. Evaluate P when n = 50. P = 6.5(20.016354 n ) = 6.5(20.016354(45) ) = 6.5(20.73593 ) ≈ 6.5(1.665) ≈ 10.8

In 2050, the world’s population will be approximately 10.8 billion. 130. One hour is 60 minutes. Evaluate P when t = 60. P = 90 − 3t 2 / 3 = 90 − 3(60)2 / 3 ≈ 90 − 3(15.3261) ≈ 44.02

Thus, the percent of the students who remembered the number after 1 hour was 44.02%. 131. a.

Evaluate P when d = 10. P = 102 − d / 40 = 102 −10 / 40 = 102 −0.25 = 101.75 ≈ 56 The amount of light that will pass to a depth of 10 feet below the ocean’s surface is about 56%.

b.

Evaluate P when d = 25. P = 102 − d / 40 = 102 − 25 / 40 = 102−0.625 = 101.375 ≈ 24 The amount of light that will pass to a depth of 25 feet below the ocean’s surface is about 24%.

.......................................................

Connecting Concepts

x x y 132. If 2 x = y , then 2 x − 4 = 2 x ⋅ 2 −4 = 2 4 = 2 4 = 4 2 2 2

133. No, if a and b are nonzero numbers and a < b, then the statement a −1 < b −1 is not a true statement. Let a = 2 and b = 3. Then a < b, but a −1 = 2 −1 = 1 and b −1 = 3−1 = 1 . 1 > 1 so a −1 > b −1. 3 2 3 2 134.

450 ⋅ 5100 = 450 ⋅ 550 ⋅ 550 = (4 ⋅ 5 ⋅ 5)50 = 10050 = (102 )50 = 10100 10100 is 1 followed by 100 zeros, thus has 101 digits.

135.

a 2 / 5a p = a 2 2+ p=2 5 p =2− 2 5 8 p= 5

137.

x −3 / 4 = x 4 x3 p − 3 − 3p = 4 4 −3 p = 19 4 p = − 19 12

136.

138.

b −3/ 4 b2 p = b3 − 3 + 2p = 3 4 2p = 3+ 3 4 15 ⎛ p = ⎜ ⎞⎟ 1 ⎝ 4 ⎠2 p = 15 8

( x 4 x 2 p )1/ 2 = x (4 + 2 p) 1 = 1

2 2 + p =1 p = −1


12

Chapter P: Preliminary Concepts 2 2 4 + h − 2 = 4 + h − 2 ⋅ 4 + h + 2 = ( 4 + h − 2)( 4 + h + 2) = ( 4 + h ) − 2 = 4 + h − 4 h h 4+h +2 h ( 4 + h + 2) h( 4 + h + 2) h ( 4 + h + 2)

139.

=

h 1 h = = 4+h +2 h( 4 + h + 2) h ( 4 + h + 2)

2 2 9 + h − 3 = 9 + h − 3 ⋅ 9 + h + 3 = ( 9 + h − 3)( 9 + h + 3) = ( 9 + h ) − 3 = 9 + h − 9 h h h( 9 + h + 3) h( 9 + h + 3) h( 9 + h + 3) 9+h +3

140.

=

h h 1 = = h( 9 + h + 3) h ( 9 + h + 3) 9+h +3

141.

2 2 2 2 2 n 2 + 1 − n = n 2 + 1 − n ⋅ n 2 + 1 + n = ( n + 1 − n )( n + 1 + n ) = ( n + 1) − n = n 2 + 1 − n 2 = 1 1 1 n2 + 1 + n n2 + 1 + n n2 + 1 + n n2 + 1 + n n2 + 1 + n

142.

2 2 2 2 2 n 2 + n − n = n 2 + n − n ⋅ n 2 + n + n = ( n + n − n )( n + n + n ) = ( n + n ) − n = n 2 + n − n 2 = n 2 2 2 2 2 1 1 n +n +n n +n +n n +n +n n +n +n n +n +n

143.

(

2

2

)

2

= 2

( 2 )( 2 )

= 2

( 2 )2

2

= 2 =2

.......................................................

Prepare for Section P.3

PS1. −3(2a − 4b) −6a + 12b

PS2. 5 − 2(2 x − 7) 5 − 4 x + 14 −4 x + 19

PS3. 2 x 2 + 3x − 5 + x 2 − 6 x − 1

PS4. 4 x 2 − 6 x − 1 − 5 x 2 + x

2 x 2 + x 2 + 3x − 6 x − 5 − 1

4 x 2 − 5x 2 − 6 x + x − 1

(2 + 1) x 2 + (3 − 6) x − (5 + 1)

(4 − 5) x 2 + ( −6 + 1) x − 1

3x 2 − 3x − 6

− x 2 − 5x − 1

PS6. 12 + 15 = 27 4 4 = 6 3 ≠ 18 4 False.

?

PS5.

4 − 3x − 2 x 2 =− 2 x 2 − 4 x + 4 ?

−2 x 2 − 3x + 4 =− 2 x 2 − 4 x + 4 False.

Section P.3 1.

D

2.

E

3.

H

4.

F

5.

6.

I

7.

B

8.

C

9.

J

10. A

11.

a. b. c. d. e.

x2 + 2x − 7 2 1, 2, −7 1 x2, 2x , − 7

12.

a. b. c. d. e.

−12x4 − 3x2 − 11 4 −12, − 3, − 11 −12 −12x4, − 3x2, − 11

13.

a. b. c. d. e.


x3 − 1 3 1, −1 1 x3, − 1

G

Section P.3

14.

a. b. c. d. e.

17.

13

4x2 − 2x + 7 2 4, −2, 7 4 4x2, −2x, 7

2x4 + 3x3 + 4x2 + 5 4 2, 3, 4, 5 2 2x4, 3x3, 4x2 , 5

15.

a. b. c. d. e.

3

18.

20.

6

21.

23.

(3x2 + 4x + 5) + (2x2 + 7x – 2) = 5x2 + 11x + 3

25.

(4w3 – 2w + 7) + (5w3 + 8w2 – 1) = 9w3 + 8w2 – 2w + 6

26.

(5x4 – 3x2 + 9) + (3x3 – 2x2 – 7x + 3) = 5x4 + 3x3 – 5x2 – 7x + 12

27.

(r2 – 2r – 5) – (3r2 – 5r + 7) = r2 – 2r – 5 – 3r2 + 5r – 7 = –2r2 + 3r – 12

28.

(7s2 – 4s + 11) – (–2s2 + 11s – 9) = 7s2 – 4s + 11 + 2s2 – 11s + 9 = 9s2 – 15s + 20

29.

(u3 – 3u2 – 4u + 8) – (u3 – 2u + 4) = u3 – 3u2 – 4u + 8 – u3 + 2u – 4 = –3u2 – 2u + 4

30.

(5v4 – 3v2 + 9) – (6v4 + 11v2 – 10) = 5v4 – 3v2 + 9 – 6v4 – 11v2 + 10 = –v4 – 14v2 + 19

31.

(2x2 + 7x – 8) (4x – 5) = 8x3 – 10x2 + 28x2 – 35x – 32x + 40 = 8x3 + 18x2 – 67x + 40

32.

(3x2 – 8x – 5) (5x – 7) = 15x3 – 21x2 – 40x2 + 56x – 25x + 35 = 15x3 – 61x2 + 31x + 35

33.

16.

a. b. c. d. e.

3

19.

5

2

22.

4

3x 2 − 2 x + 5

24.

−5x3 + 3x2 + 7x − 1 3 −5, 3, 7, −1 −5 −5x3, 3x2 , 7x, − 1

(5y2 – 7y + 3) + (2y2 + 8y + 1) = 7y2 + y + 4

34.

2y3 − 3 y + 4

2 x2 − 5x + 2

2 y2 − 5 y + 7

+ 6 x 2 − 4 x + 10

+ 14y3

− 21 y + 28

3

2

4

3

2

4 y5

4

3

2

4 y 5 − 10 y 4 + 8 y 3 + 23 y 2 − 41 y + 28

− 15 x + 10 x − 25 x + 6 x − 4 x + 10 x

− 10 y

6 x − 19 x + 26 x − 29 x + 10

4

2

+ 15 y − 20 y − 6 y3 + 8 y 2

35.

(2x + 4) (5x + 1) = 10x2 + 2x + 20x + 4 = 10x2 + 22x + 4

36.

(5x − 3) (2x + 7) = 10x2 + 35x − 6x − 21 = 10x2 + 29x − 21

37.

(y + 2) (y + 1) = y2 + y + 2y + 2 = y2 + 3y + 2

38.

(y + 5) (y + 3) = y2 + 3y + 5y + 15 = y2 + 8y + 15

39.

(4z − 3) (z − 4) = 4z2 − 16z − 3z + 12 = 4z2 – 19z + 12

40.

(5z − 6) (z − 1) = 5z2 − 5z − 6z + 6 = 5z2 – 11z + 6

41.

(a + 6) (a − 3) = a2 − 3a + 6a – 18 = a2 + 3a − 18

42.

(a − 10) (a + 4) = a2 + 4a − 10a − 40 = a2 – 6a − 40

43.

(5x − 11y) (2x − 7y) = 10x2 − 35xy − 22xy + 77y2 = 10x2 – 57xy + 77y2

44.

(3a − 5b) (4a − 7b) = 12a2 − 21ab − 20ab + 35b2 = 12a2 – 41ab + 35b2

45.

(9x + 5y) (2x + 5y) = 18x2 + 45xy + 10xy + 25y2 = 18x2 + 55xy + 25y2

46.

(3x − 7z) (5x − 7z) = 15x2 − 21xz − 35xz + 49z2 = 15x2 – 56xz + 49z2

47.

(3p + 5q) (2p − 7q) = 6p2 − 21pq + 10pq – 35q2 = 6p2 – 11pq − 35q2

48.

(2r − 11s) (5r + 8s) = 10r2 + 16rs – 55rs – 88s2 = 10r2 – 39rs − 88s2

49.

(4d − 1)2 − (2d − 3)2 = (16d 2 − 8d + 1) – (4d 2 – 12d + 9) = 16d 2 − 8d + 1 – 4d 2 + 12d − 9 = 12d 2 + 4d − 8

50.

(5c − 8)2 − (2c − 5)2 = (25c2 − 80c + 64) – (4c2 – 20c + 25) = 25c2 − 80c + 64 – 4c2 + 20c − 25 = 21c2 − 60c + 39 Copyright © Houghton Mifflin Company. All rights reserved.

14


51.

52.

r 2 − rs + s 2 r+s

r 2 + rs + s 2 r−s

+ r 2 s − rs 2 + s3

− r 2 s − rs 2 − s3

r 3 − r 2 s + rs 2 r3

53.

r 3 + r 2 s + rs 2 + s3

r3

(3c – 2)(4c + 1)(5c – 2) = (12c2 – 5c – 2)(5c – 2)

54.

− s3

(4d – 5)(2d – 1)(3d – 4) = (8d2 – 14d + 5)(3d – 4)

12c 2 − 5c − 2

8d 2 − 14d + 5

5c − 2

3d − 4

2

2

− 24c + 10c + 4 3

− 32d + 56d − 20

2

3

24d − 42d 2 + 15d

60c − 25c − 10c 60c3 − 49c 2

24d 3 − 74d 2 + 71d − 20

+4

55.

(3x + 5)(3x – 5) = 9x2 – 25

56.

(4x2 – 3y)(4x2 + 3y) = 16x4 – 9y2

57.

(3x2 – y) 2 = 9x4 – 6x2y + y2

58.

(6x + 7y)2 = 36x2 + 84xy + 49y2

59.

(4w + z) 2 = 16w2 + 8wz + z2

60.

(3 x − 5 y 2 )2 = 9 x 2 − 30 xy 2 + 25 y 4

61.

[(x + 5) + y][(x + 5) – y] = (x + 5)2 – y2 = x2 + 10x + 25 – y2

62.

[(x – 2y) + 7][(x – 2y) – 7] = (x – 2y)2 – 49 = x2 – 4xy + 4y2 – 49

63.

x2 + 7x – 1 = 32 + 7(3) – 1 = 9 + 21 – 1 = 29

64.

x2 – 8x + 2 = 42 – 8(4) + 2 = 16 − 32 + 2 = −14

65.

−x2 + 5x – 3 = −(−2)2 + 5(−2) – 3 = −4 − 10 – 3 = −17

66.

−x2 – 5x + 4 = −(−5)2 – 5(−5) + 4 = −25 + 25 + 4 = 4

67.

3x3 − 2x2 – x + 3 = 3(−1) 3 − 2(−1)2 – (−1) + 3 = 3(−1) – 2(1) + 1 + 3 = −3 – 2 +1 + 3 = −1

68.

5x3 − x2 + 5x − 3 = 5(−1)3 − (−1)2 + 5 (−1) − 3 = 5(−1) – (1) – 5 − 3 = −5 – 1 – 5 – 3 = −14

69.

1 – x5 = 1 – (−2)5 = 1 – (−32) = 1 + 32 = 33

71. a. b.

73. a.

b.

70.

Substitute the given value of v into 0.016v2. Then simplify 0.016 v2 0.016(10)2 = 1.6 The air resistance is 1.6 pounds. 0.016 v2 0.016(15)2 = 3.6 The air resistance is 3.6 pounds.

72.

Substitute the given value of h and r into π r2h. Then simplify π r2h π (3)2 (8) = 72π The volume is 72π in3 . π r2h π (5)2 (12) = 300π The volume 300π cm3.

74.

1 – x3 – x5 = 1 – 23 − 25 = 1 – 8 – 32 = −39

a. b.

a.

b.

Substitute the given value of v into 0.015v2 + v + 10. Then simplify. 0.015v2 + v + 10 0.015(30)2 + 30 + 10 = 53.5 The safe distance is 53.5 feet. 0.015v2 + v + 10 0.015(55)2 + 55 + 10 = 110.375 The safe distance is 110.375 feet. Substitute the given value of v into −0.02v2 + 1.5v + 2. Then simplify. 0.02v2 + 1.5v + 2 −0.02 (45)2 + 1.5(45) + 2 = 29 The fuel efficiency is 29 miles per gallon. 0.02v2 + 1.5v + 2 −0.02 (60)2 + 1.5(60) + 2 = 20 The fuel efficiency is 20 miles per gallon.


Section P.3

15

75.

Substitute the given value of v into 0.005x2 – 0.32x + 12. a. 0.005x2 – 0.32x + 12 0.005(20)2 – 0.32(20) + 12 = 7.6 The reaction time is 7.6 hundredths of a second or 0.076 seconds. b. 0.005x2 – 0.32x + 12 0.005(50)2 – 0.32(50) + 12 = 8.5 The reaction time is 8.5 hundredths of a second or 0.085 seconds.

76.

1 3 1 2 1 1 1 1 n − n + n = (21)3− (21)2 + (21) = 1330 committees 3 6 3 6 2 2

77.

1 2 1 1 1 n − n= (150)2 − (150) = 11,175 chess matches 2 2 2 2

78.

a. b. c.

4.3 × 10−6 (1000)2 − 2.1 × 10−4 (1000) = 4.09 sec 4.3 × 10−6 (5000)2 − 2.1 × 10−4 (5000) = 106.45 sec 4.3 × 10−6 (10,000)2 − 2.1 × 10−4 (10,000) = 427.9 sec

79.

a. b.

1.9 × 10−6 (4000)2 − 3.9 × 10−3 (4000) = 14.8 sec 1.9 × 10−6 (8000)2 − 3.9 × 10−3 (8000) = 90.4 sec

80.

a.

velocity = 6r 2 − 10r 3

81.

Evaluate −16t 2 + 4.7881t + 6 when t = 0.5 height = −16t 2 + 4.7881t + 6

= −10r 3 + 6r 2 m/s

= −16(0.5)2 + 4.7881(0.5)t + 6 = 4.39 Yes. The ball is approximately 4.4 feet high when it crosses home plate.

b.

Evaluate −10r 3 + 6r 2 when r = 0.35. velocity = −10(0.35)3 + 6(0.35)2 = −10(0.042875) + 6(0.1225) = −0.42875 + 0.735 = 0.30625 The velocity of the air in a cough when the radius of the trachea is 0.35 centimeters is 0.31 m/s.

82.

a.

Evaluate 0.0002t 3 − 0.0114t 2 + 0.0158t + 104 when t = 0

b.

Evaluate 0.0002t 3 − 0.0114t 2 + 0.0158t + 104 when t = 25

Temp = 0.0002t 3 − 0.0114t 2 + 0.0158t + 104

Temp = 0.0002t 3 − 0.0114t 2 + 0.0158t + 104

= 0.0002(0)3 − 0.0114(0)2 + 0.0158(0) + 104

= 0.0002(25)3 − 0.0114(25)2 + 0.0158(25) + 104

= 104 The patient’s temperature was 104°F before taking the medication.

83.

= 100.395 The patient’s temperature was 100.4°F 25 minutes after taking the medication.

211 ⋅ 36 ⋅ 53 ⋅ 72 ⋅ 11 ⋅ 13 = 2 ⋅ 3 ⋅ (22 ) ⋅ 5 ⋅ (2 ⋅ 3) ⋅ 7 ⋅ (23 ) ⋅ (32 ) ⋅ (2 ⋅ 5) ⋅ 11 ⋅ (22 ⋅ 3) ⋅ 13 ⋅ (2 ⋅ 7) ⋅ (3 ⋅ 5) = 2 ⋅ 3 ⋅ 4 ⋅ 5 ⋅ 6 ⋅ 7 ⋅ 8 ⋅ 9 ⋅ 10 ⋅ 11 ⋅ 12 ⋅ 13 ⋅ 14 ⋅ 15 = 15! n = 15

....................................................... 3

3

2

3

3

2

2

3

3

2

3

(a − b) = a − 3a b + 3ab − b

3

2

(a + b) = a + 3a b + 3ab + b

87.

(y + 2) = y + 3y (2) + 3y(2) + 2 = y + 6y + 12y + 8

3

2

85.

84.

2

3

Connecting Concepts 86.

3

3

2

(x – 1) = x – 3x + 3x − 1


16


3

2

3

3

2

2

3

3

2

2

88.

(2x – 3y) = (2x) + 3(2x) (−3y) + 3(2x)(−3y) + (−3y) = 8x – 36x y + 54xy – 27y

89.

(3x + 5y) = (3x) + 3(3x) 5y + 3(3x)(5y) + (5y) = 27x + 135x y + 225xy + 125y

2

3

3

2

2

3

3

.......................................................

PS2. ( −12 x 4 )3x 2 = ( −12)(3) x 4 + 2 = −36 x 6

PS1. 6 x 3 = 3x 3−1 = 3x 2 2x

x 6 = ( x 2 )3

PS3. a.

Prepare for Section P.4

b.

x 6 = ( x 3 )2

PS5. −3(5a − ?) = −15a + 21 = −3(5a − 7) Thus, ? = 7

PS4. 6a 3b4 ⋅ ? = 18a 3b7 = 6a 3b4 (3b3 )

Thus, ? = 3b3 PS6. 2 x (3x − ?) = 6 x 2 − 2 x = 2 x (3x − 1) Thus, ? = 1

Section P.4 2

2

1.

5x + 20 = 5(x + 4)

3.

−15x − 12x = −3x (5x + 4)

5.

l0x y + 6xy − 14xy = 2xy(5x + 3 − 7y)

7.

(x − 3)(a + b) + (x − 3)(a + 2b) = (x − 3)(a + b + a + 2b) = (x − 3)(2a + 3b)

8.

(x − 4)(2a − b) + (x + 4)(2a − b) = (2a − b)(x − 4 + x + 4) = (2a − b)(2x)

9.

x + 7x + 12 = (x + 3)(x + 4)

11.

a – 10a – 24 = (a – 12)(a + 2)

13.

6x + 25x + 4 = (6x + 1)(x + 4)

15.

51x – 5x − 4 = (17x + 4)(3x – 1)

17.

6x + xy – 40y = (3x + 8y)(2x – 5y)

19.

x + 6x + 5 = (x + 5)(x + 1)

21.

6x + 23x + 15 = (6x + 5)(x + 3)

23.

b – 4ac = 26 − 4(8)(15) = 196 = 14 The trinomial is factorable over the integers

25.

b – 4ac = (−5) – 4(4)(6) = −71 The trinomial is not factorable over the integers.

27.

b – 4ac = (−14) – 4(6)(5) = 76 The trinomial is not factorable over the integers.

2

2

2

2

2

2

2

2

4

2

4

2

2

2

2

2

2

2

2

2

2

2

2

2

2.

8x + 12x − 40 = 4(2x + 3x − 10)

4.

−6y – 54y = −6y (y + 9)

6.

6a b – 12a b + 72ab = 6ab(a b − 2a + 12b )

2

3 2

2

3

2

2

2

10.

x + 9x + 20 = (x + 4)(x + 5)

12.

b + 12b – 28 = (b + 14)(b – 2)

14.

8a – 26a + 15 = (4a – 3)(2a – 5)

16.

57y + y − 6 = (19y – 6)(3y + 1)

18.

8x + 10xy – 25y = (4x – 5y)(2x + 5y)

20.

x + 11x + 18 = (x + 9)(x + 2)

22.

9x + 10x + 1 = (9x + 1)(x + 1)

24.

b – 4ac = 8 – 4(16)(−35) = 2304 = 48 The trinomial is factorable over the integers.

26.

b – 4ac = 8 – 4(6)(− 3) = 136 The trinomial is not factorable over the integers.

28.

b – 4ac = (–4) – 4(10)(–5) = 216 The trinomial is not factorable over the integers.

2

2

2

2

4

2

2

4

2

2

2

2

2

2

2

2

2

2

2

2


Section P.4

17

29.

x4 – x2 – 6 = (x2 – 3)(x2 + 2)

30.

x4 + 3x2 + 2 = (x2 + 1)(x2 + 2)

31.

x2y2 – 2xy – 8 = (xy – 4)(xy + 2)

32.

2x2y2 + xy – 1 = (2xy – 1)(xy + 1)

33.

3x4 + 11x2 – 4 = (3x2 – 1)(x2 + 4)

34.

2x4 + 3x2 – 9 = (2x2 – 3)(x2 + 3)

35.

3x6 + 2x3 – 8 = (3x3 – 4)(x3 + 2)

36.

8x6 – 10x3 – 3 = (4x3 + 1)(2x3 – 3)

37.

x – 9 = (x –3)(x + 3)

38.

x – 64 = (x – 8)(x + 8)

39.

4a – 49 = (2a – 7)(2a + 7)

40.

81b2 – 16c2 = (9b – 4c)(9b + 4c)

41.

1 – 100x2 = (1 − 10x)(1 + 10x)

42.

1 − 121y2 = (1 − 11y)(1 + 11y)

43.

x − 9 = (x2 − 3)(x2 + 3)

44.

y − 196 = (y2 – 14)(y2 + 14)

45.

(x + 5) − 4 = (x + 5 − 2)(x + 5 + 2) = (x + 3)(x + 7)

46.

(x − 3) − 16 = (x − 3 − 4)(x − 3 + 4) = (x − 7)(x + 1)

47.

x + 10x + 25 = (x + 5)

48.

y + 6y + 9 = (y + 3)

49.

a − 14a + 49-= (a − 7)

50.

b − 24b + 144 = (b − 12)

51.

4x + 12x + 9 = (2x + 3)

52.

25y + 40y + 16 = (5y + 4)

53.

z + 4 z w + 4w = (z + 2w )

54.

9x − 30x y + 25y

55.

x − 8 = (x – 2 )(x + 2x + 4)

56.

b + 64 = (b + 4)(b − 4b + 16)

57.

8x − 27y = (2x – 3y)(4x + 6xy + 9y )

58.

64u −27v = (4u – 3v)(16u + 12uv + 9v )

59.

8 − x = (2 − x )(4 + 2x + x )

60.

1+y

61.

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

62.

(y + 3) + 8 = ((y + 3) + 2)((y + 3) − 2(y + 3) + 4) = (y + 5)(y + 6y + 9 – 2y − 6 + 4) = (y + 5)(y + 4y + 7)

63.

3x + x + 6x + 2 = x (3x + 1) + 2(3x + 1) = (3x + 1)(x + 2)

64.

18w + 15w + 12w + 10 = 3w (6w + 5) + 2(6w + 5) = (6w + 5)(3w + 2)

65.

ax – ax + bx – b = ax(x – 1) + b(x – 1) = (x – 1)(ax + b)

66.

a y – ay + ac – cy = ay (a – y) + c(a – y) = (a – y)(ay + c)

67.

6w + 4w – 15w – 10 = 2w (3w + 2) – 5(3w + 2) = (3w + 2)(2w – 5)

68.

10z − l5z – 4z + 6 = 5z (2z – 3) – 2(2z – 3) = (2z – 3)(5z – 2)

69.

18x – 2 = 2(9x – 1) = 2(3x – 1)(3x + 1)

70.

4bx + 32b = 4b(x + 8) = 4b(x + 2)(x – 2x + 4)

71.

16x – 1 = (4x – 1)(4x + 1) = (2x – 1)(2x + 1)(4x + 1)

2

2

4

2

2

2

2

2

2

2

4

2 2

4

3

2

22

2

3

3

2

6

3

2

2

2

4

3

2

2

3

2

2

2

2

3

2

4

2

2

2

2

2

2

2

4

2 2

3

4

2

2

3

3

12

2

4

4

2

2

2

2

2

2 2

3

3

2

3

2

2

2

2

2

2

2

3 4

2

2

3

2

2

2

8

= (1 + y )(1 – y + y )

2

2

22

= (3x − 5y )

2


2

18


2

2

2

72.

81y – 16 = (9y – 4)(9y + 4) = (3y – 2)(3y + 2)(9y + 4)

73.

12ax – 23axy + 10ay = a(12x – 23xy + 10y ) = a(3x – 2y)(4x – 5y)

74.

6ax – 19axy – 20ay

75.

3bx + 4bx – 36x – 4b = bx (3x + 4) – b(3x + 4) = (3x + 4)(bx – b) = b(3x + 4)(x – 1) = b(3x + 4)(x – 1)(x + 1)

76.

2 x6 − 2 = 2( x6 − 1) = 2(x – 1)(x + 1) = 2(x – 1)(x + x + 1)(x + 1)(x – x + 1)

77.

72bx + 24bxy + 2by = 2b(36x + 12xy + y ) = 2b(6x + y)

78.

64y – 16y z + yz = y(64y – 16yz + z ) = y(8y – z)

79.

(w – 5) + 8 = [(w – 5) + 2][(w – 5)2 – 2(w – 5) + 4] = (w – 3)(w – 10w + 25 – 2w + 10 + 4) = (w−3)(w – 12w + 39)

80.

5xy + 20y – 15x – 60 = 5(xy + 4y – 3x – 12) = 5[y(x + 4) – 3(x + 4)] = 5(x + 4)(y − 3)

81.

x + 6xy + 9y – 1 = (x + 3y) – 1 = (x + 3y – 1)(x + 3y + 1)

82.

4y – 4yz + z – 9 = (2y – z) – 9 = (2y – z – 3)(2y – z + 3)

83.

8x + 3x – 4 is not factorable over the integers.

85.

5x(2x – 5) − (2x – 5) = (2x – 5) [5x − (2x – 5)] = (2x − 5) (5x − 2x + 5) = (2x − 5) (3x + 5)

86.

6x(3x + 1) − (3x + l) = (3x + 1) [6x − (3x + 1)] = (3x + 1) (6x – 3x – 1) = (3x + 1) (3x – 1)

87.

4x2 + 2x – y – y = 4x – y + 2x – y = (2x – y)(2x + y) + (2x – y) = (2x – y)(2x + y + l)

88.

a +a+b−b

2

2

2 3

2

2

2

2

= a(6x – 19xy – 20y ) = a(6x + 5y)(x – 4y)

2

2

2

3

2

3

2

3

2

2

2

2

2

2

2

2

2

2

2

2

3

2

2

2

2

2

2

2

2

2

84.

2

16x + 81 is not factorable over the integers.

2

3

2

2

2

3

4

3

3

3

2

2

2

2

2

2

2

= a − b + a + b = (a − b)(a + b) + (a + b) = (a + b)(a – b + 1)

....................................................... 2

2

2

89.

x + kx + 16 = (x + 4) = x + 8x + 16, thus k = 8

90.

36x + kx + 100y2 = (6x + 10y)

91.

x + 16x + k = (x +

92.

x – 14xy + ky = (x −

93.

x

94.

x

2

2 2

2

− 1 = (x

4n

− 2x

2n

2n

– 1)(x

+ 1= (x

2

2

= 36x + 120x + 100y2, thus k = 120

2

k ) = x + 2x k + k ⇒ 16x = 2x k ⇒ 8 = k ⇒ k = 64

2

4n

Connecting Concepts

2n

2n

k y)

2

2

= x – 2xy k + ky

2

⇒ −14xy = −2xy k ⇒ 7 =

k ⇒

n n 2n + 1) = (x − 1)(x + 1)(x + 1)

– 1)(x

2n

n

n

n

n

n

2 n

2

– 1) = (x – 1)(x + 1)(x – 1)(x + 1) = (x – 1) (x + 1)


k = 49

Section P.5

19 2

2

2

2

95.

A = π R − π r = π (R − r ) = π (R – r)(R + r)

97.

A = (2r) − π r = r (4 − π )

2

2

2

2

2

2

2

96.

A = π r + (2r) = π r + 4r = r2 ( π + 4)

98.

A = x − y = (x – y)(x + y)

2

2

....................................................... PS1. 1 +

Prepare for Section P.5 −1 PS2. ⎛ w ⎞ ⎛ y ⎞ ⎜ ⎟ ⎜ ⎟ ⎝x⎠ ⎝z⎠

1 = 1+ 1 ⋅⎛ 3⎞ ⎜ ⎟ 2−1 2 − 1 ⎝ 3⎠ 3 3 1⋅ 3 =1+ 2⋅3− 1 ⋅3 3 =1+ 3 = 1+ 3 6 −1 5 = 1 3 or 8 5 5

−1

⎛ = ⎛⎜ x ⎞⎟ ⎜ ⎝ w ⎠⎝

z⎞ ⎟ y⎠

= xz wy

PS3. x2 + 2x – 3 = (x + 3)(x – 1) x2 + 7x + 12 = (x + 4)(x + 3) The common factor is x + 3.

PS4. (2 x − 3)(3x + 2) − (2 x − 3)( x + 2) = (2 x − 3)[(3x + 2) − ( x + 2)] = (2 x − 3)(2 x ) = 2 x (2 x − 3)

PS5. x2 – 5x – 6 = (x – 6)(x + 1)

PS6. x3 – 64 = (x – 4)(x2 + 4x + 16)

Section P.5 1.

x 2 − x − 20 ( x + 4)( x − 5) x + 4 = = 3 x − 15 3( x − 5) 3 x3 − 9 x

3. 3

2

x + x − 6x

4.

x3 + 125

2 x3 − 50 x 5.

a3 + 8 a2 − 4

6.

7.

=

=

=

x ( x 2 + x − 6)

2 x( x 2 − 25)

2 x 2 − 5 x − 12 2

2x + 5x + 3

=

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

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

=

( x + 5)( x 2 − 5 x + 25) x 2 − 5 x + 25 = 2 x( x − 5)( x + 5) 2 x( x − 5)

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

y 3 − 27 y2 + 3y + 9 ( y − 3)( y 2 + 3 y + 9) ( y − 3)( y 2 + 3 y + 9) = = =− − ( y − 8)( y − 3) y −8 − y 2 + 11 y − 24 − ( y 2 − 11 y + 24) x 2 + 3x − 40

=

( x − 5)( x + 8) x+8 =− − ( x − 5)( x + 2) x+2

2 x3 − 6 x 2 + 5 x − 15 9 − x2

9.

=

( x + 5)( x 2 − 5 x + 25)

− ( x 2 − 3 x − 10) 8.

x ( x 2 − 9)

2.

4 y 3 − 8 y 2 + 7 y − 14 − y 2 − 5 y + 14

=

=

2 x 2 ( x − 3) + 5( x − 3) − ( x 2 − 9)

=

4 y 2 ( y − 2) + 7( y − 2) − ( y 2 + 5 y − 14)

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

=

( y − 2)(4 y 2 + 7) 4 y2 + 7 =− − ( y + 7)( y − 2) y+7


20

10.


x3 − x 2 + x 3

x +1

x( x 2 − x + 1)

=

2

( x + 1)( x − x + 1)

=

x x +1

12.

4 3 4 ⎛ 12 x 2 y ⎞⎛ 25 x 2 z 3 ⎞ ⎟ = − 12 ⋅ 25 x yz = − 4 x ⎟⎜ − ⎜ ⎜ 5 z 4 ⎟⎜ 15 y 2 ⎟ yz 5 ⋅ 15 y 2 z 4 ⎠ ⎠⎝ ⎝

14.

⎛ 4r 2 s ⎞ ⎜ ⎟ ⎜ 3t 3 ⎟ ⎝ ⎠

15.

x 2 + x 3 x 2 + 19 x + 28 x( x + 1) (3x + 7)( x + 4) x(3x + 7) = ⋅ ⋅ = 2x + 3 2 x + 3 ( x + 4)( x + 1) 2x + 3 x2 + 5x + 4

−1

2

x + 7 x + 12 17.

18.

x 2 − 4 x − 21

⋅

2

x − 4x

y3 − 8

⋅

y2 + 3y 3

2

12 y 2 + 28 y + 15 2

6 y + 35 y + 25 z 2 − 81 z 2 − 16 21.

a2 + 9 2

a − 64 22.

⎛ 6 p2 ⎞ ⎜ ⎟ ⎜ 5q 2 ⎟ ⎝ ⎠

−1

2

2 2 ⎛ 2p ⎞ ⎜ ⎟ = 5q ⋅ 4 p = 10 ⎜ 3q 2 ⎟ 6 p 2 9q 4 27 q 2 ⎝ ⎠

=

( x − 4)( x + 4) ( x + 3)( x − 7) x − 7 ⋅ = x( x − 4) x ( x + 3)( x + 4)

⋅

y + y − 6 y + 2y + 4y

20.

13.

2 x 2 + 16 x + 30 3( x − 15) 2( x 2 + 8 x + 15) 3( x − 15) 2( x + 3)( x + 5) x+3 ⋅ = = ⋅ = 2 2 + + − + + 6 x 9 3 ( 2 x 3 ) 2 ( x 5 )( x 5 ) 3 ( 2 x 3 ) 2 x+3 2 x − 50 2( x − 25) 3x − 15

2

19.

⎛ 4a ⎞⎛ 6b ⎞ 24ab 8 ⎟⎜ ⎟ = − ⎜⎜ − =− 2 ⎟⎜ 4 ⎟ 4 2 3a b a3b ⎝ 3b ⎠⎝ a ⎠

3 3 2 ⎛ 6rs3 ⎞ ⎜ ⎟ = 3t ⋅ 6rs = 9s t ⎜ 5t 2 ⎟ 4r 2 s 5t 2 10r ⎝ ⎠

x 2 − 16

16.

11.

÷

÷

÷

2

3 y + 11 y − 20

z 2 + 5 z − 36

=

2

a + 5a − 24

4x − 9 y

÷

=

=

a2 + 9 ( a − 3)( a + 8) a2 + 9 (a − 3)(a + 8) 1 ⋅ = ⋅ = 2 (a − 8)(a + 8) a (a − 3) + 9(a − 3) (a − 8)(a + 8) (a − 3)(a 2 + 9) a − 8

3x 2 − xy − 2 y 2 2

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

( z − 9)( z + 9) ( z + 9)( z − 4) ( z − 9)( z + 9)( z + 9) ( z − 9)( z + 9)2 ⋅ = = ( z − 4)( z + 4) ( z − 5)( z + 4) ( z + 4)( z − 5)( z + 4) ( z + 4) 2 ( z − 5)

a3 − 3a 2 + 9a − 27

2

y ( y + 3) ( y − 2)( y 2 + 2 y + 4) =1 ⋅ ( y − 2)( y + 3) y ( y 2 + 2 y + 4)

2 y2 − y − 3

z 2 − z − 20

6 x 2 + 13xy + 6 y 2 2

=

2 x + xy − 3 y

2

=

(3 x + 2 y )(2 x + 3 y ) (2 x + 3 y )( x − y ) 2 x + 3 y ⋅ = (2 x − 3 y )(2 x + 3 y ) (3 x + 2 y )( x − y ) 2 x − 3 y 2 s + 5t −2s + 3t 2 s + 5t − 2 s + 3t 8t + = = =2 4t 4t 4t 4t

23.

p + 5 2 p − 7 p + 5 + 2p − 7 3p − 2 + = = r r r r

25.

7x 8 x 2 − 32 x 8 x( x − 4) x x( x + 3) + 7 x( x − 5) x 2 + 3x + 7 x 2 − 35 x + = = = = x−5 x+3 ( x − 5)( x + 3) ( x − 5)( x + 3) ( x − 5)( x + 3) ( x − 5)( x + 3)

26.

2x 5x 2 x( x − 7) + 5 x(3 x + 1) 2 x 2 − 14 x + 15 x 2 + 5 x 17 x 2 − 9 x x(17 x − 9) + = = = = 3x + 1 x − 7 (3 x + 1)( x − 7) (3x + 1)( x − 7) (3 x + 1)( x − 7) (3x + 1)( x − 7)

27.

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

24.


Section P.5

21

28.

6 x − 5 3 x − 8 (6 x − 5) − (3 x − 8) 6 x − 5 − 3x + 8 3x + 3 3( x + 1) = = = = − x −3 x −3 x−3 x−3 x−3 x−3

29.

4z 5z 4 z ( z − 5) + 5 z (2 z − 3) 4 z 2 − 20 z + 10 z 2 − 15 z 14 z 2 − 35 z 7 z (2 z − 5) + = = = = 2z − 3 z − 5 (2 z − 3)( z − 5) (2 z − 3)( z − 5) (2 z − 3)( z − 5) ( 2 z − 3)( z − 5)

30.

3 y − 1 2 y − 5 (3 y − 1)( y − 3) − (2 y − 5)(3 y + 1) (3 y 2 − 10 y + 3) − (6 y 2 − 13 y − 5) − = = 3y + 1 y − 3 (3 y + 1)( y − 3) (3 y + 1)( y − 3) =

31.

3 y 2 − 10 y + 3 − 6 y 2 + 13 y + 5 −3 y 2 + 3 y + 8 = (3 y + 1)( y − 3) (3 y + 1)( y − 3)

x ( x + 4) − (3x − 1)( x − 3) x − x 3x − 1 3x − 1 = − = ( x − 3)( x + 3)( x + 4) x 2 − 9 x 2 + 7 x + 12 ( x − 3)( x + 3) ( x + 3)( x + 4)

=

32.

2 ( x 2 + 4 x ) − (3x 2 − 10 x + 3) x 2 + 4 x − 3x 2 + 10 x − 3 = = −2 x + 14 x − 3 ( x − 3)( x + 3)( x + 4) ( x − 3)( x + 3)( x + 4) ( x − 3)( x + 3)( x + 4)

( m − n )( m − n ) + (3m − 5n )( m − 3n ) 3m − 5n m−n m−n + 2 3m − 5n 2 = + = 2 m n m n m n m n ( 2 )( 3 ) ( )( 2 ) ( m + 2n )( m − 3n )( m − n ) + − − + m − mn − 6n m + mn − 2n 2

2 2 2 2 2 2 = m − 2mn + n + 3m − 14mn + 15n = 4m − 16mn + 16n ( m + 2n )( m − 3n )( m − n ) ( m + 2n )( m − 3n )( m − n )

=

33.

4( m 2 − 4mn + 4n 2 ) 4( m − 2n ) 2 = ( m + 2n )( m − 3n )( m − n ) ( m + 2n )( m − 3n )( m − n )

1 + 2 ⋅ 3x 2 + 11x − 4 = 1 + 2 ⋅ (3x − 1)( x + 4) = 1 + 2( x + 4) = 1( x − 5) + x[2( x + 4)] x 3x − 1 x −5 x 3x − 1 x x−5 x ( x − 5) ( x − 5) 2 2 (2 x − 1)( x + 5) = x − 5 + 2 x + 8x = 2 x + 9 x − 5 = x ( x − 5) x ( x − 5) x ( x − 5)

34.

2 2 − 3 ⋅ y − 1 = 2 − 3 ⋅ ( y − 1)( y + 1) = 2 − 3( y − 1) = 2( y + 4) − 3( y − 1) y = 2( y + 4) − y[3( y − 1)] y y +1 y + 4 y y +1 y+4 y y+4 y ( y + 4) y ( y + 4)

=

35.

2 y + 8 − 3 y 2 + 3 y −3 y 2 + 5 y + 8 3y2 − 5 y − 8 (3 y − 8)( y + 1) = =− =− y ( y + 4) y ( y + 4) y ( y + 4) y ( y + 4)

q +1 2q q + 5 q +1 2q q − 3 q + 1 2q ( q + 1)( q + 5) − 2q (q − 3) ÷ − = − ⋅ = − = (q − 3)(q + 5) q −3 q −3 q −3 q −3 q −3 q+5 q−3 q+5 =

36.

q 2 +6q + 5 − 2q 2 + 6q − q 2 +12q + 5 = (q − 3)(q + 5) (q − 3)(q + 5)

p p p+2 p p ( p + 3)( p − 4) p p( p + 3) p ( p + 2) + p ( p + 3)( p + 5) + = + ⋅ = + = ÷ p + 5 p − 4 p 2 − p − 12 p + 5 p − 4 p+2 p+5 p+2 ( p + 5)( p + 2)

=

p 2 + 2 p + p( p 2 + 8 p + 15) p 2 + 2 p + p3 + 8 p 2 + 15 p = ( p + 5)( p + 2) ( p + 5)( p + 2)

=

p3 + 9 p 2 + 17 p p ( p 2 + 9 p + 17) = ( p + 5)( p + 2) ( p + 5)( p + 2)


22

37.


1 1 1 1 + 1 + 1 = + + x 2 + 7 x + 12 x 2 − 9 x 2 − 16 ( x + 3)( x + 4) ( x − 3)( x + 3) ( x − 4)( x + 4) 1( x − 3)( x − 4) + 1( x − 4)( x + 4) + 1( x − 3)( x + 3) = ( x + 3)( x + 4)( x − 3)( x − 4) 2 2 2 3x 2 − 7 x − 13 = x − 7 x + 12 + x − 16 + x − 9 = ( x + 3)( x + 4)( x − 3)( x − 4) ( x + 3)( x + 4)( x − 3)( x − 4)

38.

2 5 2 3 5 + 3 − = + − a 2 − 3a + 2 a 2 − 1 a 2 + 3a − 10 ( a − 1)( a − 2) ( a − 1)( a + 1) ( a − 2)( a + 5) 2( a + 1)( a + 5) + 3( a − 2)( a + 5) − 5( a − 1)( a + 1) = ( a − 1)( a − 2)( a + 1)( a + 5) =

2( a 2 + 6a + 5) + 3( a 2 + 3a − 10) − 5( a 2 − 1) ( a − 1)( a − 2)( a + 1)( a + 5)

2 2 2 = 2a + 12a + 10 + 3a + 9a − 30 − 5a + 5 ( a − 1)( a − 2)( a + 1)( a + 5)

=

3(7a − 5) 21a − 15 = ( a − 1)( a − 2)( a + 1)( a + 5) ( a − 1)( a − 2)( a + 1)( a + 5)

39.

1 ⎞ ⎛ x 2 ⎞⎛ 3 x 1 ⎞ ⎛ x + 2 ⎞⎛ 3x − 1 ⎞ ( x + 2)(3 x − 1) ⎛ 2 ⎞⎛ ⎟= ⎟⎜ ⎜1 + ⎟⎜ 3 − ⎟ = ⎜ + ⎟⎜ − ⎟ = ⎜ x ⎠⎝ x ⎠ ⎝ x x ⎠⎝ x x ⎠ ⎝ x ⎠⎝ x ⎠ ⎝ x2

40.

1 ⎞⎛ 2 ⎞ ⎛ 4 z 1 ⎞⎛ 4 z 2 ⎞ ⎛ 4 z − 1 ⎞⎛ 4 z + 2 ⎞ (4 z − 1)(4 z + 2) (4 z − 1)2(2 z + 1) 2(4 z − 1)(2 z + 1) ⎛ = = + ⎟=⎜ − ⎟⎜ ⎟= ⎟⎜ ⎜ 4 − ⎟⎜ 4 + ⎟ = ⎜ z z⎠ ⎝ z z ⎠⎝ z z ⎠ ⎝ z ⎠⎝ z ⎠ ⎠⎝ ⎝ z2 z2 z2

41.

44.

45.

1 ⎛⎜ 4 + 1 ⎞⎟ x 4x + 1 x⎠ x =⎝ = 1 1⎞ x −1 ⎛ 1− ⎜1 − ⎟ x x x ⎠ ⎝ 4+

42.

2 ⎛⎜ 3 − a =⎝ 3 ⎛ 5+ ⎜5 + a ⎝ 3−

2⎞ ⎟a 3a − 2 a⎠ = 3⎞ 5a + 3 ⎟a a⎠

43.

x −2 y = y−x

2 2 2 2 2 2 3( x − 3) 2 3+ 3+ 3+ 3+ 3+ + x−3 = x−3 = x−3 = x−3 = x−3 = x−3 = x−3 x−3 1 1 1 2x + 1 4(2 x + 1) x x x 4+ 4+ 4+ 4 +1÷ 4 + 1⋅ 4+ + 1 2x 1 2x + 1 + + + +1 2 1 2 1 2 1 2 x x x x x + 2+ x x x x 3( x − 3) + 2 3x − 9 + 2 3x − 7 3x − 7 9 x + 4 3x − 7 2 x + 1 (3x − 7)(2 x + 1) − 3 x = = x−3 = x−3 = ÷ = ⋅ = 4(2 x + 1) + x 8 x + 4 + x 9 x + 4 x − 3 2 x + 1 x − 3 9 x + 4 ( x − 3)(9 x + 4) 2x + 1 2x + 1 2x + 1 3+

1 1 1 1 1 1 5( x + 2) 1 5− 5− 5− 5− 5− − x+2 = x+2 = x+2 = x+2 = x+2 = x+2 = x+2 x+2 3 3 3 3x 1( x + 3) 3x x+3 x 1+ 1+ 1+ 1+ 3÷ 1 + 3⋅ 1+ + 3 1( x ) 3 x+3 3 3 3 3 x x x x + + + +3 1+ + x x x x 5( x + 2) − 1 5 x + 10 − 1 5 x + 9 x+2 x + 2 = x + 2 = 5 x + 9 ÷ 4 x + 3 = 5 x + 9 ⋅ x + 3 = (5 x + 9)( x + 3) = = 1( x + 3) + 3x x + 3 + 3x 4 x + 3 x + 2 x+3 x + 2 4 x + 3 ( x + 2)(4 x + 3) x+3 x+3 x+3 5−


⎛x ⎞ ⎜⎜ − 2 ⎟⎟ y y ⎝ ⎠ = x − 2y ( y − x )y y ( y − x)

Section P.5

46.

23

1( x + h )2 1 − 1( x + h )2 1 − 1( x 2 + 2 xh + h 2 ) 1 − x 2 − 2 xh − h 2 1 1 1 − − ( x + h )2 ( x + h )2 ( x + h )2 ( x + h )2 ( x + h)2 ( x + h )2 = = = = h h h h h 2 2 = 1 − x − 2 xh2 − h ( x + h)

1+

48.

r−

r 1 r+ 3

=r−

=

1−

51.

2−

54.

55.

56.

( x + h)

h

h( x + h)

r r 3r + 1 ⎞ 3 ⎞ 3r ⎛ ⎛ =r− = r − ⎜r ÷ ⎟ = r −⎜r ⋅ ⎟=r− 3r 1 3r + 1 r r 3 3 1 3 + +1 ⎝ ⎠ ⎝ ⎠ + 3 3 3

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

⎛ 1 ⎞ ⎜⎜1 − ⎟⎟ x2 ⎠ ⋅ x =⎝ 1 ⎛ 1⎞ 1+ ⎜1 + ⎟ x x⎠ ⎝

49.

53.

1

⎛ 1 ⎞ 1 ⎜1 + ⎟ b − 2 = ⎝ b − 2 ⎠ ⋅ (b − 2)(b + 3) = 1(b − 2)(b + 3) + 1(b + 3) = b2 + b − 6 + b + 3 = b2 + 2b − 3 = (b + 3)(b − 1) (b − 2)(b + 2) ⎛ 1 − 1 ⎞ (b − 2)(b + 3) 1(b − 2)(b + 3) − 1( b − 2) b2 + b − 6 − b + 2 b2 − 4 1− 1 ⎜ ⎟ b + 3 ⎝ b + 3⎠

47.

52.

2 2 2 2 ÷ h = 1 − x − 2 xh2 − h ⋅ 1 = 1 − x − 2 xh2− h

1

2

x2

x2 − 1

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

50.

1 1 1 b+a ab ab = = = 1÷ = 1⋅ = 1 1 1b 1a b+a ab b+a b+a + + a b ab ab ab

(m) m m m m m =2− =2− =2− =2− m =2− ⋅ = 2 − m2 1 1 1( ) 1 1 m m m m m − − + − 1 m − ⎛ 1⎞ m 1− 1+ + ⎜ ⎟ m m m −m m m ⎝m⎠

⎛ x + h +1 − x ⎞ ⎜ ⎟ x + 1 ⎠ ⋅ ( x + h )( x + 1 = ( x + h + 1)( x + 1) − x ( x + h ) = x 2 + x + xh + h + x + 1 − x 2 − xh = 2 x + h + 1 ⎝ x+h (h) ( x + h )( x + 1) h ( x + h )( x + 1) h( x + h )( x + 1) h( x + h )( x + 1) ⎛1 x−4⎞ ⎜ − ⎟ 2 2 ⎝ x x + 1 ⎠ ⋅ x( x + 1) = x + 1 − x( x − 4) = x + 1 − x + 4 x = − x + 5 x + 1 x x( x + 1) x( x) x2 x2 x +1

⎛ 2 3y − 2 ⎞ ⎜⎜ − ⎟ y − 1 ⎟⎠ y ( y − 1) 2( y − 1) − y (3 y − 2) 2 y − 2 − 3 y 2 + 2 y − 3 y 2 + 4 y − 2 ⎝y ⋅ = = = y ( y − 1) y( y) ⎛ y ⎞ y2 y2 ⎜⎜ ⎟⎟ ⎝ y −1 ⎠ 2 ⎞ ⎛ 1 − ⎟ ⎜ x − 1 − 2x − 6 − x−7 + − 1 ⎠ ( x + 3)( x − 1) 1( x − 1) − 2( x + 3) 3 x x ⎝ ⋅ = = = 3 ⎞ ( x + 3)( x − 1) x( x + 3) + 3( x − 1) x 2 + 3 x + 3 x − 3 x 2 + 6 x − 3 ⎛ x + ⎟ ⎜ ⎝ x −1 x + 3 ⎠

x+2 x+2 + 1 + 1 2 − + x x x + 1 ( x − 1)( x + 1)(2 x + 1) ( x + 2)(2 x + 1) + 1( x − 1)(2 x + 1) ( 1)( 1) + x 1 x −1 = ⋅ = 1 1 ( x − 1)( x + 1)(2 x + 1) x x x ( x + 1) + 1( x + 1)(2 x + 1) + + 2 x 2 − x − 1 x − 1 (2 x + 1)( x − 1) x − 1 2

2 2 2 (2 x + 1) = 2 x 2+ 5 x + 2 +22 x − x − 1 = 4 x2 + 4 x + 1 = x + x + 2 x + 3x + 1 3x + 4 x + 1 (3x + 1)( x + 1)


24


x 2 + 3 x − 10

57.

( x − 2)( x + 5) x+5 x+5 x+5 x + 5 2x − 3 2x − 3 ( x − 2)( x + 3) ÷ = = x+3 = = ⋅ = ( x + 5)( x − 6) x+5 x + 3 2x − 3 x + 3 x + 5 x+3 (2 x − 3)( x − 6) 2 x − 3

x2 + x − 6 x 2 − x − 30 2 x 2 − 15 x + 18

58.

59.

2 y 2 + 11 y + 15

(2 y + 5)( y + 3) 2y + 5 2y + 5 2y + 5 2y + 5 y − 7 y−7 ( y + 3)( y − 7) ÷ = = = = ⋅ =1 2 y y y (3 − 2)(2 + 5) 2 + 5 y−7 y−7 y − 7 2y + 5 6 y + 11 y − 10 (3 y − 2)( y − 7) y−7 3 y 2 − 23 y + 14 y 2 − 4 y − 21

1 1 1b 1a b+a + + a −1 + b −1 a b ab ab b+a = = = ab = a−b a−b a−b a−b ab

60.

e

61.

−2

−f ef

−1

1 2 = e

a −1b − ab −1 a 2 + b2

63.

(a + b )

− 2 −1

a.

ef

1 f

1f =

2

e f

− ef

1e2 2

e f

b+a 1 a+b ⋅ = ab a − b ab(a − b)

f − e2 =

e2 f ef

=

f − e2 e2 f

÷ a 2 + b2

b2 − a 2 b2 − a 2 1 (b − a)(b + a ) ⋅ = = 2 2 ab a +b ab(a 2 + b 2 ) ab(a 2 + b 2 )

⎛ 1 ⎞ ⎟⎟ = ⎜⎜ a + b2 ⎠ ⎝

−1

⎛ ab 2 1 ⎞⎟ =⎜ + ⎜ b2 b 2 ⎟⎠ ⎝

−1

⎛ ab 2 + 1 ⎞ ⎟ =⎜ ⎜ b2 ⎟ ⎝ ⎠

−1

=

2 2 290 = = 2÷ 1 1 110 + 180 180(110) + 180 110 180(110) = 2⋅

f − e2 1 f − e2 ⋅ = e2 f ef e3 f 2

÷ ef =

(b)b a (a ) b 2 a 2 b 2 − a 2 b a − − − b2 − a 2 b)a b( a) ab ab ( = a b = = = ab = ab a 2 + b2 a2 + b2 a 2 + b2 a 2 + b2 =

62.

−

÷ ( a − b) =

b2 ab 2 + 1

b.

2v v 2v v 2 2 = = 1 2 = 1 2 1 + 1 v2 + v1 v2 + v1 v1 + v2 v1 v2 v1v2

b.

(v + v ) 2 c2 (v + v ) v1 + v2 = 1 2 ⋅ c2 = 2 1 2 vv vv c + v1v2 1 + 1 22 ⎜⎛ 1 + 1 2 ⎟⎞ c c c2 ⎠ ⎝

66.

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

(180)(110) ≈ 136.55 mph (to the nearest hundredth) 290

8 8 v1 + v2 = 1.2 × 10 8+ 2.4 × 10 8 ≈ 3.4 × 108 v1v2 (1.2 × 10 )(2.4 × 10 ) 1+ 2 1+ c (6.7 × 108 )2

64.

a.

65.

1 + 1 = x + 1 + x = 2x + 1 x x + 1 x ( x + 1) x ( x + 1)

67.

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


Section P.5

68.

25

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

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

4 3 2 4 2 4 3 2 4 16 = x + 4 x + 4 x +2 x − 8 x2 + 16 +2 x − 4 x + 4 x = 2 3x + x ( x − 2) ( x + 2) x ( x − 2)2 ( x + 2)2

.......................................................

Connecting Concepts

69.

( x + 5) − x( x + 5) −1 x + 5 ( x + 5) 2 − x x 2 + 10 x + 25 − x x 2 + 9 x + 25 = = = ⋅ x+5 x+5 ( x + 5) 2 ( x + 5) 2 ( x + 5) 2

70.

( y + 2) + y 2 ( y + 2)−1 y + 2 ( y + 2) 2 + y 2 y 2 + 4 y + 4 + y 2 2 y 2 + 4 y + 4 2( y 2 + 2 y + 2) = = = = ⋅ y+2 y+2 ( y + 2)2 ( y + 2) 2 ( y + 2)2 ( y + 2) 2

71.

72.

73.

74.

1 − 4 xy 1 − 4y 2 x (1 − 4 xy ) x x = = ⋅ x2 = −1 −1 (1 − 2 xy )(1 + 2 xy ) ( x − 2 y )( x + 2 y ) ⎜⎛ 1 − 2 y ⎟⎞ ⎜⎛ 1 + 2 y ⎟⎞ ⎛ 1 − 2 xy ⎞ ⎛ 1 + 2 xy ⎞ x ⎝x ⎠⎝ x ⎠ ⎜⎝ x ⎟⎠ ⎜⎝ x ⎟⎠ x −1 − 4 y

x+ y ⋅ x− y

1 − x+ y x = ⋅ x −1 + y −1 x − y 1 + x x −1 − y −1

1 ⎡ ⎢1 − (1 + i ) n R⎢ ⎢ i ⎢ ⎣⎢

1 y xy x + y y − x ⋅ = ⋅ = −1 1 xy x − y y + x y

⎡ (1 + i ) n − 1 ⎤ ⎤ ⎢ ⎥ ⎥ n n ⎡ ⎤ ⎥ = R ⎢ (1 + i ) ⎥ = R ⎢ (1 + i ) − 1 ⎥ ⎢ ⎥ ⎥ i ⎢⎣ i (1 + i ) n ⎥⎦ ⎢ ⎥ ⎥ ⎢ ⎥ ⎦⎥ ⎣ ⎦

1 1 RR R R1R2 R3 ⋅ 1 2 3 = = 1 1 1 ⎛ ⎞ R R R R R R1R3 + R1R2 + 1 1 1 1 2 3 2 3 + + ⎜ ⎟ + + ⎟ R1 R2 R3 ⎜ R R R 2 3⎠ ⎝ 1

....................................................... PS1. (2 − 3x )(4 − 5 x ) = 8 − 10 x − 12 x + 15 x 2

Prepare for Section P.6 PS2. (2 − 5 x )2 = 22 + 2(2)( −5 x ) + ( −5 x ) 2

= 15 x 2 − 22 x + 8

= 4 − 20 x + 25 x 2 = 25 x 2 − 20 x + 4

PS3. PS4.

96 = 16 ⋅ 6 = 4 6

( 2 + 3 5 )( 3 − 4 5 ) = 6 − 8

5 + 9 5 − 12 ( 5 ) = 6 + 5 − 60 = −54 + 5 2

PS5. 5 + 2 = 5 + 2 ⋅ 3 + 2 = 15 + 8 2 + 2 = 17 + 8 2 9−2 7 3− 2 3− 2 3+ 2


26


PS6. a.

81 − x 2 is a difference of perfect squares with integer coefficients, which does factor over the integers.

b.

81 − x 2 = (9 − x )(9 + x )

9 + z 2 is a sum of perfect squares with integer coefficients. If there is no common factor, sums of perfect squares with integer coefficients are not factorable over the integers.

Section P.6 1.

−81 = i 81 = 9i

2.

−64 = i 64 = 8i

3.

−98 = i 98 = 7i 2

4.

−27 = i 27 = 3i 3

5.

16 + −81 = 4 + i 81 = 4 + 9i

6.

25 + −9 = 5 + i 9 = 5 + 3i

7.

5 + −49 = 5 + i 49 = 5 + 7i

8.

6 − −1 = 6 − i 1 = 6 − i

9.

8 − −18 = 8 − i 18 = 8 − 3i 2

10.

11 + −48 = 11 + i 48 = 11 + 4i 3

11.

(5 + 2i ) + (6 − 7i ) = 5 + 2i + 6 − 7i = (5 + 6) + (2i − 7i ) = 11 − 5i

12.

(4 − 8i ) + (5 + 3i ) = 4 − 8i + 5 + 3i = (4 + 5) + ( −8i + 3i ) = 9 − 5i

13.

( −2 − 4i ) − (5 − 8i ) = −2 − 4i − 5 + 8i = ( −2 − 5) + ( −4i + 8i ) = −7 + 4i

14.

(3 − 5i ) − (8 − 2i ) = 3 − 5i − 8 + 2i = (3 − 8) + ( −5i + 2i ) = −5 − 3i

15.

(1 − 3i ) + (7 − 2i ) = 1 − 3i + 7 − 2i = (1 + 7) + ( −3i − 2i ) = 8 − 5i

16.

(2 − 6i ) + (4 − 7i ) = 2 − 6i + 4 − 7i = (2 + 4) + ( −6i − 7i ) = 6 − 13i

17.

( −3 − 5i ) − (7 − 5i ) = −3 − 5i − 7 + 5i = ( −3 − 7) + ( −5i + 5i ) = −10

18.

(5 − 3i ) − (2 + 9i ) = 5 − 3i − 2 − 9i = (5 − 2) + ( −3i − 9i ) = 3 − 12i

19.

8i − (2 − 8i ) = 8i − 2 + 8i = −2 + (8i + 8i ) = −2 + 16i

22.

( −3i )(2i ) = −6i 2 = −6( −1) =6

24.

20.

23.

5i ⋅ 8i = 40i 2 = 40( −1) = −40

−50 ⋅ −2 = i 50 ⋅ i 2 = 5i 2 ⋅ i 2

25.

3(2 + 5i ) − 2(3 − 2i ) = 6 + 15i − 6 + 4i = (6 − 6) + (15i + 4i ) = 19i

27.

(4 + 2i )(3 − 4i ) = 4(3 − 4i ) + (2i )(3 − 4i )

2

= 6i ( 3) = 6( −1)(3) = −18

26.

21.

= 5i 2 ( 2)2 = 5( −1)(2) = −10

−12 ⋅ −27 = i 12 ⋅ i 27 = 2i 3 ⋅ 3i 3 2

3 − (4 − 5i ) = 3 − 4 + 5i = (3 − 4) + 5i = −1 + 5i

3i (2 + 5i ) + 2i (3 − 4i ) = 6i + 15i 2 + 6i − 8i 2 = 6i + 15( −1) + 6i − 8( −1) = 6i − 15 + 6i + 8 = ( −15 + 8) + (6i + 6i ) = −7 + 12i

= 12 − 16i + 6i − 8i 2 = 12 − 16i + 6i − 8( −1) = 12 − 16i + 6i + 8 = (12 + 8) + ( −16i + 6i ) = 20 − 10i


Section P.6

28.

27

(6 + 5i )(2 − 5i ) = 6(2 − 5i ) + 5i (2 − 5i )

29.

( −3 − 4i )(2 + 7i ) = −3(2 + 7i ) − 4i (2 + 7i )

2

= −6 − 21i − 8i − 28i 2 = −6 − 21i − 8i − 28( −1) = −6 − 21i − 8i + 28 = ( −6 + 28) + ( −21i − 8i ) = 22 − 29i

= 12 − 30i + 10i − 25i = 12 − 30i + 10i − 25( −1) = 12 − 30i + 10i + 25 = (12 + 25) + ( −30i + 10i ) = 37 − 20i 30.

( −5 − i )(2 + 3i ) = −5(2 + 3i ) − i (2 + 3i )

31.

(4 − 5i )(4 + 5i ) = 4(4 + 5i ) − 5i (4 + 5i )

2

= 16 + 20i − 20i − 25i 2 = 16 + 20i − 20i − 25( −1) = 16 + 20i − 20i + 25 = (16 + 25) + (20i − 20i ) = 41

= −10 − 15i − 2i − 3i = −10 − 15i − 2i − 3( −1) = −10 − 15i − 2i + 3 = ( −10 + 3) + ( −15i − 2i ) = −7 − 17i 32.

(3 + 7i )(3 − 7i ) = 3(3 − 7i ) + 7i (3 − 7i )

33.

= 9 − 21i + 21i − 49i 2 = 9 − 21i + 21i − 49( −1) = 9 − 21i + 21i + 49 = (9 + 49) + ( −21i + 21i ) = 58

34.

(3 + −4 )(2 − −9 ) = (3 + i 4 )(2 − i 9) = (3 + 2i )(2 − 3i ) = 3(2 − 3i ) + 2i (2 − 3i ) = 6 − 9i + 4i − 6i 2 = 6 − 9i + 4i − 6( −1) = 6 − 9i + 4i + 6 = (6 + 6) + ( −9i + 4i ) = 12 − 5i

(5 + 2 −16)(1 − −25) = (5 + 2i 16)(1 − i 25) = [5 + 2i (4)][1 − i (5)] = (5 + 8i )(1 − 5i ) = 5(1 − 5i ) + 8i (1 − 5i ) = 5 − 25i + 8i − 40i 2 = 5 − 25i + 8i − 40( −1) = 5 − 25i + 8i + 40 = (5 + 40) + ( −25i + 8i ) = 45 − 17i

35.

(3 + 2 −18)(2 + 2 −50) = (3 + 2i 18)(2 + 2i 50) = [3 + 2i (3 2 )][2 + 2i (5 2 )] = (3 + 6i 2 )(2 + 10i 2) = 3(2 + 10i 2 ) + 6i 2(2 + 10i 2 ) = 6 + 30i 2 + 12i 2 + 60i 2 ( 2)2 = 6 + 30i 2 + 12i 2 + 60( −1)(2) = 6 + 30i 2 + 12i 2 − 120 = (6 − 120) + (30i 2 + 12i 2 ) = −114 + 42i 2

36.

(5 − 3 −48)(2 − 4 −27 ) = (5 − 3i 48)(2 − 4i 27 ) = [5 − 3i (4 3)][2 − 4i (3 3)] = (5 − 12i 3)(2 − 12i 3) = 5(2 − 12i 3) − 12i 3(2 − 12i 3) = 10 − 60i 3 − 24i 3 + 144i 2 ( 3) 2 = 10 − 60i 3 − 24i 3 + 144( −1)(3) = 10 − 60i 3 − 24i 3 − 432 = (10 − 432) + ( −60i 3 − 24i 3) = −422 − 84i 3 4

−8 = − 8 = −4 = −4 ⋅ i = −4i = −4i = 4i −1 2i 2i i i i i2

37.

6 = 6 ⋅ i = 6i = 6i = −6i i i i i 2 −1

39.

6 + 3i = 6 + 3i ⋅ i = 6i + 3i 2 = 6i + 3( −1) = 6i − 3 = 3 − 6i i i i −1 −1 i2

40.

4 − 8i = 4(1 − 2i ) = 4 (1 − 2i ) = 1 − 2i = 1 − 2i ⋅ i = i − 2i 2 = i − 2( −1) = i + 2 = −2 − i 4i 4i 4i i i i −1 −1 i2

41.

1(7 − 2i ) 1 = 1 ⋅ 7 − 2i = = 7 − 2i = 7 − 2i = 7 − 2i = 7 − 2i = 7 − 2 i 7 + 2i 7 + 2i 7 − 2i (7 + 2i )(7 − 2i ) 49 − 4i 2 49 − 4( −1) 49 + 4 53 53 53

38.


28


42.

5(3 − 4i ) 5 = 5 ⋅ 3 − 4i = = 15 − 20i = 15 − 20i = 15 − 20i = 15 − 20i = 15 − 20 i = 3 − 4 i 3 + 4i 3 + 4i 3 − 4i (3 + 4i )(3 − 4i ) 9 − 16i 2 9 − 16( −1) 9 + 16 25 25 25 5 5

43.

2i = 2i ⋅ 1 − i = 2i (1 − i ) = 2i − 2i 2 = 2i − 2( −1) = 2i + 2 = 2 + 2i = 2 + 2 i = 1 + i 1 + i 1 + i 1 − i (1 + i )(1 − i ) 1 − ( −1) 1+1 2 2 2 1 − i2

44.

5i = 5i ⋅ 2 + 3i = 5i (2 + 3i ) = 10i + 15i 2 = 10i + 15( −1) = 10i − 15 = −15 + 10i = − 15 + 10 i 2 − 3i 2 − 3i 2 + 3i (2 − 3i )(2 + 3i ) 4 − 9( −1) 4+9 13 13 13 4 − 9i 2

45.

5 − i = 5 − i ⋅ 4 − 5i = (5 − i )(4 − 5i ) = 5(4 − 5i ) − i (4 − 5i ) = 20 − 25i − 4i + 5i 2 4 + 5i 4 + 5i 4 − 5i (4 + 5i )(4 − 5i ) 4(4 − 5i ) + 5i (4 − 5i ) 16 − 20i + 20i − 25i 2 20 − 25i − 4i + 5( −1) 20 − 25i − 4i − 5 (20 − 5) + ( −25i − 4i ) 15 − 29i 15 29 = = = = = − i 16 − 25( −1) 16 + 25 16 + 25 41 41 41

46.

4 + i = 4 + i ⋅ 3 − 5i = (4 + i )(3 − 5i ) = 4(3 − 5i ) + i (3 − 5i ) = 12 − 20i + 3i − 5i 2 = 12 − 20i + 3i − 5( −1) 3 + 5i 3 + 5i 3 − 5i (3 + 5i )(3 − 5i ) 9 − 25( −1) 9 − 25i 2 9 − 25i 2 12 − 20i + 3i − 5( −1) 12 − 20i + 3i + 5 (12 + 5) + ( −20i + 3i ) 17 − 17i 17 17 = = = = = − i = 1 − 1i 9 − 25( −1) 9 + 25 34 34 34 34 2 2

47.

2 32 + 2(3)(2i ) + (2i )2 9 + 12i + 4i 2 9 + 12i + 4( −1) 3 + 2i = 3 + 2i ⋅ 3 + 2i = (3 + 2i ) = = = 3 − 2i 3 − 2i 3 + 2i (3 − 2i )(3 + 2i ) 9 − 4( −1) 32 − (2i )2 9 − 4i 2 = 9 + 12i − 4 = 5 + 12i = 5 + 12 i 9+4 13 13 13

48.

8 − i = 8 − i ⋅ 2 − 3i = (8 − i )(2 − 3i ) = 8(2 − 3i ) − i (2 − 3i ) = 16 − 24i − 2i + 3i 2 = 16 − 24i − 2i + 3( −1) 2 + 3i 2 + 3i 2 − 3i (2 + 3i )(2 − 3i ) 4 − 9( −1) 22 − (3i )2 4 − 9i 2 = 16 − 24i − 2i − 3 = 13 − 26i = 13 − 26i = 1 − 2i 4+9 13 13 13

49.

−7 + 26i = −7 + 26i ⋅ 4 − 3i = ( −7 + 26i )(4 − 3i ) = −7(4 − 3i ) + 26i (4 − 3i ) = −28 + 21i + 104i − 78i 2 4 + 3i 4 + 3i 4 − 3i (4 + 3i )(4 − 3i ) 42 − (3i )2 16 − 9i 2 =

50.

−4 − 39i = −4 − 39i ⋅ 5 + 2i = ( −4 − 39i )(5 + 2i ) = −4(5 + 2i ) − 39i (5 + 2i ) = −20 − 8i − 195i − 78i 2 = 5 + 2i 5 − 2i 5 − 2i 5 + 2i (5 − 2i )(5 + 2i ) 5 + 2i 52 − (2i )2 25 − 4i 2 =

51.

−28 + 21i + 104i − 78( −1) −28 + 21i + 104i + 78 50 + 125i 50 125 = = = + i = 2 + 5i 16 − 9( −1) 16 + 9 25 25 25

−20 − 8i − 195i − 78( −1) −20 − 8i − 195i + 78 58 − 203i 58 203i = = = − = 2 − 7i 25 − 4( −1) 25 + 4 29 29 29

(3 − 5i )2 = 33 + 2(3)( −5i ) + ( −5i )2

52.

(2 + 4i )2 = 22 + 2(2)(4i ) + (4i )2

= 9 − 30i + 25i 2

= 4 + 16i + 16i 2

= 9 − 30i + 25( −1)

= 4 + 16i + 16( −1)

= 9 − 30i − 25

= 4 + 16i − 16

= −16 − 30i

= −12 + 16i


Section P.6

53.

55.

29

(1 + 2i )3 = (1 + 2i )(1 + 2i )2

54.

= (1 + 2i )[12 + 2(1)(2i ) + (2i ) 2 ]

= (2 − i )[22 + 2(2)( −i ) + ( −i )2 ]

= (1 + 2i )[1 + 4i + 4i 2 ]

= (2 − i )[4 − 4i + i 2 ]

= (1 + 2i )[1 + 4i + 4( −1)]

= (2 − i )[4 − 4i − 1]

= (1 + 2i )[1 + 4i − 4]

= (2 − i )(3 − 4i )

= (1 + 2i )( −3 + 4i )

= 2(3 − 4i ) − i (3 − 4i )

= 1( −3 + 4i ) + 2i ( −3 + 4i )

= 6 − 8i − 3i + 4i 2

= −3 + 4i − 6i + 8i 2

= 6 − 8i − 3i + 4( −1)

= −3 + 4i − 6i − 8

= 6 − 8i − 3i − 4

= −11 − 2i

= 2 − 11i

Use the Powers of i Theorem. The remainder of 15 ÷ 4 is 3.

56.

i15 = i 3 = −i 57.

(2 − i )3 = (2 − i )(2 − i )2

Use the Powers of i Theorem. The remainder of 66 ÷ 4 is 2. i 66 = i 2 = −1

Use the Powers of i Theorem. The remainder of 40 ÷4 is 0.

58.

−i 40 = −(i 0 ) = −1

Use the Powers of i Theorem. The remainder of 51 ÷4 is 3. −i 51 = −(i 3 ) = −( −i ) = i

59.

Use the Powers of i Theorem. The remainder of 25 ÷4 is 1. 1 = 1 = 1 ⋅ i = i = i = −i i 25 i i i i 2 −1

60.

Use the Powers of i Theorem. The remainder of 83 ÷4 is 3. 1 = 1 = 1 ⋅i = i = i =i i83 i 3 i 3 i i 4 1

61.

Use the Powers of i Theorem. The remainder of 34 ÷4 is 2. 1 = 1 = 1 = −1 i −34 = 34 i i 2 −1

62.

Use the Powers of i Theorem. The remainder of 52 ÷4 is 0. 1 = 1 = 1 =1 i −52 = 52 i i0 1

63.

Use a = 3, b = −3, c = 3.

64.

Use a = 2, b = 4, c = 4. 2 −b + b2 − 4ac = −(4) + (4) − 4(2)(4) 2a 2(2)

2

−b + b2 − 4ac = −( −3) + ( −3) − 4(3)(3) 2a 2(3)

= −4 + 16 − 32 = −4 + −16 4 4 = −4 + i 16 = −4 + 4i 4 4 4 4 i − = + = −1 + i 4 4

= 3 + 9 − 36 = 3 + −27 6 6 = 3 + i 27 = 3 + 3i 3 6 6 3 3 3 1 = + i = + 3i 6 6 2 2 65.

Use a = 2 b = 6, c = 6.

66. 2

−b + b2 − 4ac = −(6) + (6) − 4(2)(6) 2a 2(2) = −6 + 36 − 48 = −6 + −12 4 4 − + 6 2 3 i − + 6 12 i = = 4 4 = −6 + 2i 3 = − 3 + 3 i 4 4 2 2

Use a =2, b = 1, c = 3. 2 −b + b2 − 4ac = −(1) + (1) − 4(2)(3) 2a 2(2)

= −1 + 1 − 24 4 1 − + −23 = −1 + i 23 = 4 4 = − 1 + 23 i 4 4


30

67.


Use a = 4, b = −4, c = 2.

68. 2

Use a = 3, b = −2, c = 4. 2 −b + b2 − 4ac = −( −2) + ( −2) − 4(3)(4) 2a 2(3)

−b + b − 4ac = −( −4) + ( −4) − 4(4)(2) 2a 2(4) 2

= 2 + 4 − 48 = 2 + −44 6 6 + + i i 2 44 2 2 11 = = 6 6 = 2 + 2i 11 = 1 + 11 i 6 6 3 3

= 4 + 16 − 32 = 4 + −16 8 8 + 4 i 16 + 4 4 i = = 8 8 i 4 4 1 1 = + = + i 8 8 2 2

.......................................................

Connecting Concepts

69.

x 2 + 16 = x 2 + 42 = ( x + 4i )( x − 4i )

70.

x 2 + 9 = x 2 + 32 = ( x + 3i )( x − 3i )

71.

z 2 + 25 = z 2 + 52 = ( z + 5i )( z − 5i )

72.

z 2 + 64 = z 2 + 82 = ( z + 8i )( z − 8i )

73.

4 x 2 + 81 = (2 x )2 + 92 = (2 x + 9i )(2 x − 9i )

74.

9 x 2 + 1 = (3x )2 + 12 = (3x + i )(3x − i )

75.

If x = 1 + 2i, then x 2 − 2 x + 5 = (1 + 2i )2 − 2(1 + 2i ) + 5 = 1 + 4i + 4i 2 − 2 − 4i + 5 = 1 + 4i + 4(−1) − 2 − 4i + 5 = 1 + 4i − 4 − 2 − 4i + 5 = (1 − 4 − 2 + 5) + (4i − 4i ) = 0

76.

If x = 1 − 2i, then x 2 − 2 x + 5 = (1 − 2i )2 − 2(1 − 2i ) + 5 = 1 − 4i + 4i 2 − 2 + 4i + 5 = 1 − 4i + 4( −1) − 2 + 4i + 5 = 1 − 4i − 4 − 2 + 4i + 5 = (1 − 4 − 2 + 5) + ( −4i + 4i ) = 0

77.

Verify that ( −1 + i 3)3 = 8 . ( −1 + i 3)3 = ( −1 + i 3)( −1 + i 3)2 = ( −1 + i 3)[( −1)2 + 2( −1)(i 3) + (i 3)2 ] = ( −1 + i 3)[1 − 2i 3 + 3i 2 ] = ( −1 + i 3)[1 − 2i 3 + 3( −1)] = ( −1 + i 3)[1 − 2i 3 − 3] = ( −1 + i 3)( −2 − 2i 3) = −1( −2 − 2i 3) + i 3( −2 − 2i 3) = 2 + 2i 3 − 2i 3 − 2i 2 ( 3)2 = 2 + 2i 3 − 2i 3 − 2( −1)(3) = 2 + 2i 3 − 2i 3 + 6 = (2 + 6) + (2i 3 − 2i 3) =8

Verify that ( −1 − i 3)3 = 8 . ( −1 − i 3)3 = ( −1 − i 3)( −1 − i 3)2 = ( −1 − i 3)[( −1)2 + 2( −1)( −i 3) + ( −i 3)2 ] = ( −1 − i 3)[1 + 2i 3 + 3i 2 ] = ( −1 − i 3)[1 + 2i 3 + 3( −1)] = ( −1 − i 3)[1 + 2i 3 − 3] = ( −1 − i 3)( −2 + 2i 3) = −1( −2 + 2i 3) − i 3( −2 + 2i 3) = 2 − 2i 3 + 2i 3 − 2i 2 ( 3)2 = 2 − 2i 3 + 2i 3 − 2( −1)(3) = 2 − 2i 3 + 2i 3 + 6 = (2 + 6) + ( −2i 3 + 2i 3) =8 78.

⎡

⎤

2

Verify that ⎢ 2 (1 + i ) ⎥ = i . ⎣ 2 ⎦ 2

2

⎡ 2 ⎤ 2 2 2 2 1 1 1 ⎢ 2 (1 + i ) ⎥ = 2 (1 + i ) = 4 (1 + 2i + i ) = 2 [1 + 2i + ( −1)] = 2 (1 + 2i − 1) = 2 (2i ) = i 2 ⎣ ⎦ 79.

i + i 2 + i 3 + i 4 + ... + i 28 = 7(i + i 2 + i 3 + i 4 ) = 7(i + ( −1) + ( −i ) + 1) = 7(0) = 0

80.

i + i 2 + i 3 + i 4 + ... + i100 = 25(i + i 2 + i 3 + i 4 ) = 25(i + ( −1) + ( −i ) + 1) = 25(0) = 0


Chapter Review

31

.......................................................

Exploring Concepts with Technology

Can You Trust Your Calculator? 1.

Iteration 1 3 5 10 15 19 20

4p – 3p2 1.25 0.95703125 0.8198110957 0.3846309658 0.5610061236 1.218765181 0.4188950245

p + 3p(1 – p) 1.25 0.95703125 0.8198110957 0.3846309658 0.5610061236 1.218765181 0.4188950251

.......................................................

Assessing Concepts

1.

True

2.

False

3.

Any real number a between 0 and 1

4.

Any negative integer n

5.

−3x 2 + 6 x − 7

6.

b

7.

c

8.

a

9.

e

10.

c

11.

b

12.

e

.......................................................

Chapter Review

1.

Integer, rational number, real number, prime number [P.1]

2.

Irrational number, real number [P.1]

3.

Rational number, real number [P.1]

4.

Rational number, real number [P.1]

5.

A ∪ B = {1, 2, 3, 5, 7, 11} [P.1]

6.

A ∩ B = {5} [P.1]

7.

Distributive property [P.1]

8.

Commutative property of addition [P.1]

9.

Associative property of multiplication [P.1]

10.

Closure property of addition [P.1]

11.

Identity property of addition [P.1]

12.

Identity property of multiplication [P.1]

13.

Symmetric property of equality [P.1]

14.

Transitive property of equality [P.1]

15.

−4 < x < 2 [P.1]

17.

[−3, 2) [P.1]

16.

(−4, 2] 19.

| 7 | = 7 [P.1]

23.

| −3 – 14 | = 17 [P.1]

26.

( 22 ⋅ 3−2 )2 −1 3

3 2

20.

x < −1

or x > 3 [P.1]

18.

(−∞, −1] ∪ (3, ∞ )

−3 ≤ x < 2

|2 – π | = −(2 – π) = π – 2, 21. because π > 2 [P.1]

| 4 – π | = 4 − π, [P.1] because 4 > π

24.

5 − ( − 2 ) = 5 + 2 [P.1]

(−1, ∞) [P.1] x > −1

25.

22.

|−11| =11 [P.1]

−52 + (−11) = −25 − 11 = −36 [P.1]

4 −4 = 2−13 3 = 24 −33−4 −( −1) = 24 −33−4 +1 = 213−3 = 23 = 2 [P.1] 27 3 2 3


32


2

27.

(3 x 2 y )(2 x3 y ) 2 = 3 x 2 y ⋅ 4 x 6 y 2 = 12 x8 y 3 [P.2]

28.

2 8 4⎞ ⎛ ⎛ 2a 2b3c − 2 ⎞ ⎟ = ⎜ 2ab ⎟ = 4a b [P.2] ⎜ − 2 1 ⎜ 3c ⎟ ⎟ ⎜ 3ab 9c 4 ⎠ ⎝ ⎠ ⎝

29.

251 / 2 = 25 = 5 [P.2]

30.

− 27 2 / 3 = − 3 27

31.

x 2 / 3 ⋅ x3 / 4 = x 2 / 3 + 3 / 4 = x8 /12 + 9 /12 = x17 /12 [P.2]

32.

⎛ 8 x5 / 4 ⎞ ⎜ 1/ 2 ⎟ ⎝ x ⎠

33.

⎛ x2 y ⎞ ⎜⎜ x1/ 2 y −3 ⎟⎟ ⎝ ⎠

34.

( x1/ 2 − y1/ 2 )( x1/ 2 + y1/ 2 ) = x − y

2/3

= (8 x 5 / 4 − 1/ 2 )

1/ 2

2/3

= (8 x 5 / 4 −

= ( x 2 − 1/ 2 y1 − ( −3) )

1/ 2

)

2/4 2/3

= ( x4 / 2

12a3b = 4a 2 ⋅ 3ab = 2a 3ab [P.2]

38.

18 x3 y 5 = 9 x 2 y 4 ⋅ 2 xy = 3 xy 2 2 xy [P.2]

39.

54 xy 3 = 10 x

40.

−

24 xyz 3 15 z

5y

42.

3 9y

=

6

9 y2 ⋅ 3y

=−

5y

⋅

8 xy 5z 3

=−

3

3y2

3 y 3 3y2

3 2

5

=

2 2 xy z 5z

=

⋅

)

3y 3y 5

⋅

2/3

− 1/ 2 1 + 3 1/ 2

y

[P.2]

36.

27 y 3 = 5

= (8 x 3 / 4 )

5 5

=

( )2 = −(3)2 = −9 [P.2]

= 82 / 3 x (3/ 4)(2 / 3) = (23 )2 / 3 x (3/ 4)(2 / 3) = 22 x1 / 2 = 4 x1 / 2 [P.2]

= ( x3 / 2 y 4 )

1/ 2

= x (3/ 2)(1/ 2) y 4(1/ 2) = x 3/ 4 y 2 [P.2]

35.

48a 2b7 = 16a 2b6 ⋅ 3b = 4ab3 3b [P.2]

37.

72 x 2 y = 36 x 2 ⋅ 2 y = 6 x 2 y [P.2]

3 y 15 y [P.2] 5

2 10 xyz 5z [P.2] =− 5z 5z 2

5 y3 3 y 2 53 3 y 2 [P.2] = 3y 3

− 250 xy 6 = 3 − 125 y 6 ⋅ 2 x = −5 y 2 3 2 x [P.2]

7x

41.

=

7x 3

⋅

3 2

2 x

3 2

=

3

2 x2

43.

3

− 135 x 2 y 7 = 3 − 27 y 6 ⋅ 5 x 2 y = −3 y 2 3 5 x 2 y [P.2]

45.

620,000 = 6.2 × 10 [P.2]

2x2

2 x

5

44.

3

46.

0.0000017 = 1.7 × 10

49.

(2a + 3a – 7) + (−3a – 5a + 6) =[2a + (−3a2)] + [3a + (−5a)] + [(−7) + 6] = −a – 2a − 1 [P.3]

50.

(5b 2 − 11) − (3b 2 − 8b − 3) = 5b 2 − 11 − 3b 2 + 8b + 3 [P.3]

−6

2

2

47.

[P.2]

4

3.5 × 10 = 35,000 [P.2]

2

= 2b 2 + 8b − 8

7 x 3 4 x 73 4 x = [P.2] 2x 2

48.

4.31 × 10−7 = 0.000000431 [P.2]

2

51.

2 x 2 + 3x − 5 [P.3] 3x2 − 2 x + 4 + 8 x 2 + 12 x − 20 − 4 x3 − 6 x 2 + 10 x 6 x 4 + 9 x3 − 15 x 2 6 x 4 + 5 x3 − 13 x 2 + 22 x − 20

52.

(3 y − 5)3 = (3 y − 5)2 (3 y − 5) = (9 y 2 − 30 y + 25)(3 y − 5) = 27 y 3 − 45 y 2 − 90 y 2 +150 y + 75 y −125 = 27 y 3 −135 y 2 + 225 y −125 [P.3]


Chapter Review

33

53.

3x 2 + 30 x + 75 = 3( x 2 + 10 x + 25) = 3( x + 5) 2 [P.4]

54.

25 x 2 − 30 xy + 9 y 2 = (5 x − 3 y ) 2 [P.4]

55.

20a 2 − 4b 2 = 4(5a 2 − b 2 ) [P.4]

56.

16a3 + 250 = 2(8a3 + 125) = 2(2a + 5)( 4a 2 − 10a + 25) [P.4]

57.

6 x 2 − 19 x + 10

=

2

2 x + 3 x − 20 58.

4 x3 − 25 x

=

4

8 x + 125 x

(3 x − 2)(2 x − 5) 3x − 2 = [P.5] (2 x − 5)( x + 4) x+4

x(4 x 2 − 25) 3

x(8 x + 125)

=

x(2 x − 5)(2 x + 5) 2

x(2 x + 5)(4 x − 10 x + 25)

=

2x − 5 2

4 x − 10 x + 25

[P.5]

59.

10 x 2 + 13 x − 3 6 x 2 + 5 x + 1 (2 x + 3)(5 x − 1) (2 x + 1)(3 x + 1) 2 x + 3 = ⋅ = ⋅ [P.5] 6 x 2 − 13 x − 5 10 x 2 + 3 x − 1 (2 x − 5)(3 x + 1) (2 x + 1)(5 x − 1) 2 x − 5

60.

15 x 2 + 11x − 12 25 x 2 − 9

61.

÷

3x 2 + 13 x + 12 10 x 2 + 11x + 3

=

15 x 2 + 11x − 12 10 x 2 + 11x + 3 (5 x − 3)(3x + 4) (5 x + 3)(2 x + 1) 2 x + 1 ⋅ = ⋅ = [P.5] x+3 25 x 2 − 9 3 x 2 + 13 x + 12 (5 x − 3)(5 x + 3) (3 x + 4)( x + 3)

x ( x + 4) + 2 x ( x + 3) x 2 + 4 x + 2 x 2 + 6 x 2x 2x x + x = + = = 2 x − 9 x + x − 12 ( x − 3)( x + 3) ( x + 4)( x − 3) ( x − 3)( x + 3)( x + 4) ( x − 3)( x + 3)( x + 4) x (3x + 10) 3x 2 + 10 x = = ( x − 3)( x + 3)( x + 4) ( x − 3)( x + 3)( x + 4) 2

62.

63.

3x (2 x − 1) − x ( x + 4) x x 3x 3x − = − = [P.5] x 2 + 7 x + 12 2 x 2 + 5 x − 3 ( x + 3)( x + 4) (2 x − 1)( x + 3) ( x + 3)( x + 4)(2 x − 1) 2 2 x (5 x − 7) 5x 2 − 7 x = 6 x − 3x − x − 4 x = = ( x + 3)( x + 4)(2 x − 1) ( x + 3)( x + 4)(2 x − 1) ( x + 3)( x + 4)(2 x − 1) ⎛ 2+ 1 ⎜2 + x−5 = ⎝ ⎛3− 3− 2 x − 5 ⎜⎝

64.

1

2+

65.

[P.5]

3 1+ 4 x

=

1 ⎞ ⎟ x − 5 ⎠ ⋅ x − 5 = 2( x − 5) + 1 = 2 x − 10 + 1 = 2 x − 9 [P.5] 2 ⎞ x − 5 3( x − 5) − 2 3x − 15 − 2 3 x − 17 ⎟ x −5⎠

1

2+

3 x+4 x x

=

1 2+

3 x+4 x

=

x+4 x + 4 = x + 4 [P.5] 1 1 1 = = ⋅ x+4 = = 3x x + 4 2( x + 4) + 3x 2 x + 8 + 3x 5 x + 8 x x 4 ⎛ + ⎞ ⎛ ⎞ 2 + 2 + ⎜3÷ ⎟ 2 + ⎜ 3⋅ ⎟ x+4 x ⎠ ⎝ ⎝ x+4⎠

5 + −64 = 5 + 8i [P.6]

66.

2 − −18 = 2 − i 18

[P.6]

= 2 −i 9⋅2 = 2 − 3i 2 67.

(2 − 3i ) + (4 + 2i ) = 2 − 3i + 4 + 2i [P.6] = (2 + 4) + ( −3i + 2i ) =6−i

68.

(4 + 7i ) − (6 − 3i ) = 4 + 7i − 6 + 3i [P.6] = (4 − 6) + (7i + 3i ) = −2 + 10i

69.

2i (3 − 4i ) = 6i − 8i 2

70.

(4 − 3i )(2 + 7i ) = 4(2 + 7i ) − 3i (2 + 7i ) [P.6]

= 6i − 8( −1) = 6i + 8 = 8 + 6i

[P.6]

= 8 + 28i − 6i − 21i 2 = 8 + 22i − 21( −1) = 8 + 22i + 21 = 29 + 22i


34

71.


72.

(3 + i )2 = 32 + 2(3)(i ) + i 2 [P.6] = 9 + 6i + ( −1) = 8 + 6i

Use the Powers of i Theorem. [P.6] The remainder of 345 ÷ 4 is 1 i 345 = i1 = i

73.

4 − 6i = 2(2 − 3i ) = 2 (2 − 3i ) = 2 − 3i = 2 − 3i ⋅ i = 2i − 3i 2 = 2i − 3( −1) = 2i + 3 = −3 − 2i [P.6] −1 −1 i i i 2i 2i 2i i2

74.

2 − 5i = 2 − 5i ⋅ 3 − 4i = (2 − 5i )(3 − 4i ) = 2(3 − 4i ) − 5i (3 − 4i ) = 6 − 8i − 15i + 20i 2 = 6 − 8i − 15i + 20( −1) [P.6] 3 + 4i 3 + 4i 3 − 4i (3 + 4i )(3 − 4i ) 9 − 16( −1) (3)2 − (4i )2 9 − 16i 2 − − − − − i i i 6 8 15 20 14 23 14 23 = = =− − i 9 + 16 25 25 25

....................................................... QR1. Evaluate z =

λ0 − λs when λ0 = 390.5 × 10–9 λs

Quantitative Reasoning QR2. Evaluate z =

and λs = 375.4 × 10–9

λ0 − λs when λ0 = 412.3 × 10–9 λs

and λs = 401.5 × 10–9

−9

× 10 z = 390.5 × 10 − 375.4 375.4 × 10−9

−9

−9 = 15.1 × 10 −9 ≈ 0.040 375.4 × 10

−9 × 10−9 = 10.8 × 10−9 ≈ 0.027 z = 412.3 × 10 − 401.5 −9 401.5 × 10 401.5 × 10−9

2 2 ⎡ ⎤ ⎡ ⎤ QR3. Evaluate v = c ⎢ ( z + 1) − 1 ⎥ when c = 3 × 105 and z = 0.032. QR4. Evaluate v = c ⎢ ( z + 1) − 1 ⎥ when c = 3 × 105 and z = 0.041. 2 2 ⎣ ( z + 1) + 1 ⎦ ⎣ ( z + 1) + 1 ⎦

⎡ (0.032 + 1)2 − 1 ⎤ 5 ⎡ 0.065024 ⎤ v = 3 × 105 ⎢ ≈ 9446 ⎥ = 3 × 10 ⎢ 2 ⎣ 2.065024 ⎥⎦ (0.032 + 1) + 1 ⎣ ⎦ The relative speed is 9446 kilometers per second.

⎡ (0.041 + 1)2 − 1 ⎤ 5 ⎡ 0.083681 ⎤ v = 3 × 105 ⎢ ≈ 12,048 ⎥ = 3 × 10 ⎢ 2 ⎣ 2.083681 ⎥⎦ (0.041 + 1) + 1 ⎣ ⎦ The relative speed is 12,048 kilometers per second.

....................................................... 1. 3.

2. A ∪ B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} [P.1]

Distributive property [P.1] −12 − (−5) = −12 + 5 = −7 = 7 [P.1]

2

5.

Chapter Test

2 3 1 2 6 ( 2a −1bc −2 ) 22 a −2 b2 c −4 = = 2 ⋅ 2 3⋅ 32 ⋅ 4b c 3 ba a c ( 3−1 b )( 2−1 ac −2 ) ( 3−1 b )( 2−3 a 3c −6 )

4. a ( −2 x 0 y −2 )

2

( −3x 2 y −1 )

−2

= ( 4 y −4 )( 3−2 x −4 y 2 ) =

4 [P.2] 9 x4 y2

5 2 2 = 2 ⋅ 35⋅ bc = 96bc [P.2] 5 a a

6.

0.00137 = 1.37 × 10−3 [P.2]

7.

x1 / 3 y −3 / 4

x5 / 6 = x1 / 3 − ( −1 / 2) y −3 / 4 − 3 / 2 = x1 / 3 +1 / 2 y −3 / 4 − 3 / 2 = x 2 / 6 +3 / 6 y −3 / 4 −6 / 4 = x5 / 6 y −9 / 4 = [P.2] x −1 / 2 y3 / 2 y9 / 4

8.

3x3 81xy 4 − 2 y 3 3 x 4 y = 3 x3 27 y 3 ⋅ 3 xy − 2 y 3 x3 ⋅ 3 xy = 3 x ⋅ 3 y 3 3 xy − 2 y ⋅ x3 3 xy = 9 xy3 3 xy − 2 xy3 3 xy = 7 xy3 3 xy [P.2]

x

9. 4

2 x3

=

x 4

2 x3

⋅

4 3

2 x

4 3

2 x

4

=

x 23 x 4 4 4

2 x

=

x4 8x 4 8x = [P.2] 2x 2

10.

3 = x +2

3 x −2 3 x −6 [P.2] ⋅ = x−4 x +2 x −2


Chapter Test

35

11.

( x − 2 y )( x 2 − 2 x + y ) = x3 − 2 x 2 + xy − 2 x 2 y + 4 xy − 2 y 2 = x3 − 2 x 2 + 5 xy − 2 x 2 y − 2 y 2 [P.3]

12.

If y = −3, 3 y 3 − 2 y 2 − y + 2 = 3(−3)3 − 2(−3) 2 − (−3) + 2 = 3(−27) − 2(9) + 3 + 2 = −81 − 18 + 3 + 2 = −94 [P.4]

13.

7 x 2 + 34 x − 5 = (7 x − 1)( x + 5) [P.4]

14.

3ax − 12bx − 2a + 8b = (3ax − 12bx) − (2a − 8b) = 3x(a − 4b) − 2(a − 4b) = (a − 4b)(3x − 2) [P.4]

15.

16 x 4 − 2 xy 3 = 2 x (8 x 3 − y 3 ) = 2 x (2 x − y )(4 x 2 + 2 xy + y 2 ) [P.4]

16.

x 2 − 2 x − 15 = ( x − 5)( x + 3) = ⎛ x − 5 ⎞⎛ x + 3 ⎞ = −1 ⋅ ⎛ x + 3 ⎞ = − x + 3 [P.5] ⎜ ⎟⎜ ⎟ ⎜ ⎟ (5 − x )(5 + x ) ⎝ 5 − x ⎠⎝ x + 5 ⎠ x+5 ⎝ x + 5⎠ 25 − x 2

17.

x ( x − 3) − 2( x + 3) 2 x x [P.5] − 2 2 = − = x + x − 6 x − 5 x + 6 ( x − 2)( x + 3) ( x − 2)( x − 3) ( x − 2)( x + 3) ( x − 3) 2 ( x − 6)( x + 1) x 2 − 5x − 6 = x − 3x − 2 x − 6 = = ( x − 2)( x + 3) ( x − 3) ( x − 2)( x + 3) ( x − 3) ( x − 2)( x + 3) ( x − 3)

18.

2

2 x ( x + 2) 2 x 2 + 3x − 2 ÷ 2 x 2 − 7 x + 3 = 2 x 2 + 3x − 2 ⋅ x 3 − 3x 2 = (2 x − 1)( x + 2) ⋅ x ( x − 3) [P.5] = 2 3 2 2 2 (2 x − 1)( x − 3) x ( x − 3) x −3 2x − 7x + 3 x − 3x x − 3x x − 3x

19.

2 3 ⋅ a 2 − b2 − 5 = 3 ⋅ ( a − b)( a + b) − 5 = 3( a − b) − 5 = 3a ( a − b) − 5(2a − b) = 3a − 3ab − 10a + 5b [P.5] a + b 2a − b a a + b a a (2a − b) 2a − b 2a − b a a ( 2a − b)

20.

x−

21.

7 + −20 = 7 + 2i 5 [P.6]

23.

(2 + 5i )(1 − 4i ) = 2(1 − 4i ) + 5i (1 − 4i ) [P.6]

x = x − x = x − x = x − x ÷ 2x + 1 = x − x ⋅ 2 = x − 2x 2x + 1 2x + 1 2 2x + 1 2x + 1 x+1 2 2 2 2 [P.5] 2 2 2 x (2 x + 1) x (2 x − 1) 2 2 + 2 2 + − 2 2 − x x x x x x x x x = − = − = = = 2x + 1 2x + 1 2x + 1 2x + 1 2x + 1 2x + 1 2x + 1 22.

[P.6] (4 − 3i ) − (2 − 5i ) = 4 − 3i − 2 + 5i = (4 − 2) + ( −3i + 5i ) = 2 + 2i

= 2 − 8i + 5i − 20i 2 = 2 − 8i + 5i − 20( −1) = 2 − 8i + 5i + 20 = (2 + 20) + ( −8i + 5i ) = 22 − 3i 24.

3 + 4i = 3 + 4i ⋅ 5 + i = (3 + 4i )(5 + i ) = 3(5 + i ) + 4i (5 + i ) = 15 + 3i + 20i + 4i 2 = 15 + 3i + 20i + 4( −1) [P.6] 5−i 5 − i 5 + i (5 − i )(5 + i ) 25 − ( −1) 52 − i 2 25 − i 2 (15 − 4) + (3i + 20i ) 11 + 23i 11 23 = 15 + 3i + 20i − 4 = = = + i 25 + 1 26 26 26 26

25.

Use the Powers of i Theorem. [P.6] The remainder of 97 ÷ 4 is 1.

i 97 = i1 = i


Chapter 1

Equations and Inequalities Section 1.1 1.

5.

8.

2 x + 10 = 40 2 x = 40 − 10 2 x = 30 x = 15

−3 y + 20 = 2

2.

y=6

2( x − 3) − 5 = 4( x − 5) 2 x − 6 − 5 = 4 x − 20 2 x − 11 = 4 x − 20 2 x − 4 x = −20 + 11 −2 x = −9 x=9 2

6.

6(5s − 11) − 12(2 s + 5) = 0 30s − 66 − 24 s − 60 = 0 6s − 126 = 0

9.

6s = 126 s = 126 6 s = 21

11.

3.

−3 y = 2 − 20 −3 y = −18

5 x + 2 = 2 x − 10 5 x − 2 x = −10 − 2 3x = −12 x = −4

5( x − 4) − 7 = −2( x − 3) 5 x − 20 − 7 = −2 x + 6 5 x − 27 = −2 x + 6 5 x + 2 x = 6 + 27 7 x = 33 x = 33 7

7.

3x+1 = 2 4 2 3 12 ⋅ ⎛⎜ 3 x + 1 ⎞⎟ = 12 ⋅ ⎛⎜ 2 ⎞⎟ 2⎠ ⎝4 ⎝ 3⎠ 9x + 6 = 8 9x = 8 − 6 9x = 2 x=2 9

10.

1 x + 7 − 1 x = 19 2 4 2 4 ⋅ ⎛⎜ 1 x + 7 − 1 x ⎞⎟ = 4 ⋅ ⎛⎜ 19 ⎞⎟ 4 ⎠ ⎝2 ⎝ 2⎠ 2 x + 28 − x = 38 x + 28 = 38 x = 38 − 28 x = 10

13.

4.

4 x − 11 = 7 x + 20 4 x − 7 x = 20 + 11 −3x = 31 x = − 31 3

4(2 r − 17) + 5(3r − 8) = 0 8r − 68 + 15r − 40 = 0 23r − 108 = 0 23r = 108 r = 108 23

x −5= 1 4 2 4 ⋅ ⎛⎜ x − 5 ⎞⎟ = 4 ⎛⎜ 1 ⎞⎟ ⎝4 ⎠ ⎝2⎠ x − 20 = 2 x = 2 + 20 x = 22

0.2 x + 0.4 = 3.6 0.2 x = 3.6 − 0.4

2 x −5= 1 x −3 3 2 2 ⎛ ⎞ 6 ⋅ ⎜ x − 5 ⎟ = 6 ⋅ ⎛⎜ 1 x − 3 ⎞⎟ ⎝2 ⎠ ⎝3 ⎠ 4 x − 30 = 3x − 18 4 x − 3x = −18 + 30 x = 12

12.

14.

0.04 x − 0.2 = 0.07 0.04 x = 0.07 + 0.2 0.04 x = 0.27 x = 6.75

15.

x + 0.08(60) = 0.20(60 + x ) x + 4.8 = 12 + 0.20 x x − 0.20 x = 12 − 4.8 0.80 x = 7.2 x=9

16.

6(t + 1.5) = 12t 6t + 9 = 12t 6t − 12t = −9 −6t = −9 t=3 2

17.

3( x + 5)( x − 1) = (3x + 4)( x − 2)

18.

5( x + 4)( x − 4) = ( x − 3)(5 x + 4)

19.

5[ x − (4 x − 5)] = 3 − 2 x 5( x − 4 x + 5) = 3 − 2 x 5( −3x + 5) = 3 − 2 x −15 x + 25 = 3 − 2 x −15 x + 2 x = 3 − 25 −13x = −22 x = 22 13

3 ( x 2 + 4 x − 5) = 3 x 2 − 2 x − 8

5 ( x 2 − 16 ) = 5 x 2 − 11x − 12

3x 2 + 12 x − 15 = 3x 2 − 2 x − 8

5 x 2 − 80 = 5 x 2 − 11x − 12

12 x + 2 x = −8 + 15

11x = −12 + 80

14 x = 7

11x = 68

x=1 2

x = 68 11

0.2 x = 3.2 x = 16


Section 1.1

37

20.

6[3 y − 2( y − 1)] − 2 + 7 y = 0 6(3 y − 2 y + 2) − 2 + 7 y = 0 18 y − 12 y + 12 − 2 + 7 y = 0 13 y + 10 = 0 13 y = −10 y = − 10 13

21.

40 − 3x = 6 x + 7 5 8 ⎛ ⎞ ⎛ 6x + 7 ⎞ 40 3 x − 40 ⋅ ⎜ ⎟ = 40 ⋅ ⎜ ⎟ ⎝ 5 ⎠ ⎝ 8 ⎠ 8(40 − 3x ) = 5(6 x + 7) 320 − 24 x = 30 x + 35 −24 x − 30 x = 35 − 320 −54 x = −285 x = 95 18

22.

12 + x = 5 x − 7 + 2 3 −4 ⎛ 5x − 7 + 2 ⎞ 12 x + ⎛ ⎞ 12 ⋅ ⎜ ⎟ = 12 ⋅ ⎜ ⎟ ⎝ −4 ⎠ ⎝ 3 ⎠ −3(12 + x ) = 4(5 x − 7) + 24 −36 − 3x = 20 x − 28 + 24 −36 − 3x = 20 x − 4 −3x − 20 x = −4 + 36 −23x = 32 x = − 32 23

23.

−3( x − 5) = −3x + 15

24.

2 x + 1 = 6x + 1 3 3 1 ⎛ ⎞ ⎛ 3 ⋅ ⎜ 2 x + ⎟ = 3 ⋅ ⎜ 6 x + 1 ⎟⎞ 3⎠ ⎝ ⎝ 3 ⎠ 6x + 1 = 6x + 1 Identity

25.

2 x + 7 = 3( x − 1) 2 x + 7 = 3x − 3 2 x − 3x = −3 − 7 − x = −10 x = 10 Conditional equation

4[2 x − 5( x − 3)] = 6 4[2 x − 5 x + 15] = 6 4[ −3x + 15] = 6 −12 x + 60 = 6 −12 x = −54 x=9 2 Conditional equation

27.

4x + 8 = x + 8 4 4 x + 8 = 4( x + 8)

28.

3[ x − (4 x − 1)] = −3(2 x − 5)

3[ x − 2( x − 5)] − 1 = −3x + 29 3[ x − 2 x + 10] − 1 = −3x + 29 3[ − x + 10] − 1 = −3x + 29 −3x + 30 − 1 = −3x + 29 −3x + 29 = −3x + 29 Identity

30.

−3x + 15 = −3x + 15 Identity

26.

29.

32.

3[ x − 4 x + 1] = −6 x + 15 3[ −3x + 1] = −6 x + 15

4 x + 8 = 4 x + 32

−9 x + 3 = −6 x + 15 −3x = 12

8 = 32 Contradiction

33.

3( x − 4) + 7 = 3x − 5 3x − 12 + 7 = 3x − 5

x = −4 Conditional equation

4[3( x − 5) + 7] = 12 x − 32 4[3x − 15 + 7] = 12 x − 32 4[3x − 8] = 12 x − 32 12 x − 32 = 12 x − 32 Identity

34.

x =4 x=4

31.

2x − 8 = −x + 9 3x = 17 x = 17 3 Conditional equation

x =7 x=7

or x = –4

or

x = −7

3x − 5 = 3x − 5 Identity 35.

36.

x −5 = 2 x−5 = 2

x − 5 = −2

or

x=7

38.

x −8 = 3

x=3

39.

2 x = 24 x = 12

or

2 x − 3 = −21 2 x = −18 x = −9

x=5

40.

2 x + 6 = 10 2 x + 6 = 10 2x = 4 x=2

or

2 x − 5 = 11 2 x − 5 = 11 2 x = 16 x =8

x − 8 = −3

or

x = 11

2 x − 3 = 21 2 x − 3 = 21

37.

x −8 = 3

or

2x − 5 = −11 2x = −6 x = −3

2 x + 14 = 60

2x + 6 = −10

2 x + 14 = 60

2x = −16 x = −8

2 x = 46 x = 23


or

2 x + 14 = −60 2 x = −74 x = −37

38

41.

43.

Chapter 1: Equations and Inequalities

x−4 =8 2 x−4 =8 2 x − 4 = 8(2)

42. x − 4 = −8 2 x − 4 = −8(2)

x+3 =6 4 x+3=6 4 x + 3 = 6(4)

x − 4 = 16

x − 4 = −16

x + 3 = 24

x + 3 = −24

x = 20

x = −12

x = 21

x = −27

or

44.

2 x + 5 = −8 2x + 5 ≥ 0

−17 ≥ 0 Contradiction. There is no solution. 46.

2 x + 3 + 4 = 34 2 x + 3 = 30 x + 3 = 15 x + 3 = 15 x = 12

47.

or

x = a+b 2

or

3 x − 5 − 16 = 2 3 x − 5 = 18 x−5 =6 x −5=6 x = 11

x + 3 = −15 x = −18

48.

2 x − a = b, b > 0

2x − a = b 2x = a + b

4 x − 1 = −17 4 x −1 ≥ 0

−8 ≥ 0 Contradiction. There is no solution. 45.

x + 3 = −6 4 x + 3 = −6(4)

or

2x − a = − b 2x = a − b

x − 5 = −6 x = −1

3 x − d = c, c > 0 x−d = c 3 x−d = c 3

x = a−b 2

or

or

x=d+c 3

x−d=−c 3

x=d −c 3

49.

1.6 x + 1.87 = Revenue 1.6 x + 1.87 = 10 1.6 x = 8.13 x = 8.13 ≈ 5 1.6 2000 + 5 = 2005 The revenue first exceeded $10 billion in 2005.

50.

93.8 x + 542.8 = Number of megawatts (MW) 93.8 x + 542.8 = 1200 93.8 x = 657.2 x = 657.2 ≈ 7 93.8 2000 + 7 = 2007 The energy will exceeded 1200 MW in 2007.

51.

d = 210 − 50t

52.

m = − 1 s − 55 + 25 2 22 = − 1 s − 55 + 25 2 −3 = − 1 s − 55 2 6 = s − 55

60 = 210 − 50t or − 60 = 210 − 50t −150 = −50t − 270 = −50t t =3 t = 5.4 5.4 hours = 5 hours 24 minutes Ruben will be exactly 60 miles from Barstow after 3 hours and after 5 hours and 24 minutes.

53.

45x + 550 = Cost 45 x + 550 = 3800 45 x = 3250 x ≈ 72 Rounded to the nearest yard, 72 sq yards can be carpeted for $3800.

−6 = s − 55 6 = s − 55 or 49 = s 61 = s Kate can drive at 49 mph or 61 mph to obtain gas mileage of exactly 22 miles per gallon. 54.

1.75x + 8.00 = Retail price 1.75 x + 8.00 = 156.75 1.75 x = 148.75 x = 85 The wholesale price of the coat is $85.00.


Section 1.1

55.

39

100 −

42,000 t = Percent remaining 500,000

42,000 t 500,000 21 100 − t 250 21 − t 250 t

100 −

56.

2650 − 475t 2650 − 475t − 475t t 3.5 hours

58.

max = 0.85(220 − a ) 153 = 0.85(220 − a )

= 25 = 25

= miles remaining = 1000 = −1650 ≈ 3.5 to the nearest tenth

= −75

≈ 892.857142857 seconds ⎛ 1 min ⎞ 892.857142857 sec ⋅ ⎜ ⎟ ≈ 15 min ⎝ 60 sec ⎠ 57.

max = 0.85(220 − a ) min=0.65(220 − a ) = 0.65(220 − 25) = 0.85(220 − 25) = 0.65(195) = 0.85(195) = 126.75 = 165.75

153 = 187 − 0.85a − 34 = −0.85a 40 = a

The maximum exercise heart rate for a person who is 25 years of age is 166 beats per minute (to the nearest beat).

A person should have a maximum exercise heart rate of 153 beats per minute at age 40.

The minimum exercise heart rate for a person who is 25 years of age is 127 beats per minute (to the nearest beat). 59.

ax + b = c ax = c − b

60.

x = c−b, a ≠ 0 a

61.

if x + 4 ≥ 0 x ≥ −4 x+4= x+4 an identity

63.

62.

x+4 = x+4

ax + b = cx + d ax − cx = d − b x(a − c) = d − b x = d −b, a −c ≠ 0 a−c x −1 = x −1 if x − 1 ≥ 0

if x + 4 < 0 x < −4

x ≥1 x −1 = x −1

x + 4 = − ( x + 4)

x + 4 = −x − 4 2 x = −8 x = −4

an identity

if x − 1 < 0 x −2 4 x > −3 3 x>− 4

⎧ ⎨x x ≤ ⎩

and and and

4 x + 1 ≤ 17 4 x ≤ 16 x≤4

⎧ ⎧ ⎫ 3⎫ 3 ⎨ x x > − ⎬ ∩ { x x ≤ 4} = ⎨ x − < x ≤ 4 ⎬ 4⎭ 4 ⎩ ⎩ ⎭

x + 4 > 3x + 16

4.

−2 x > 12 x < −6

{ x x > 7}

{x x < 4}

5.

3.

3x > 21 x>7

{ x x < −6}

7.

⎧ 5⎫ ⎨x x < − ⎬ 3⎭ ⎩

−4(3x − 5) > 2( x − 4) −12 x + 20 > 2 x − 8 −14 x > −28

8.

x −16 2 x > −21 21 x>− 2

{

5x + 6 < 2 x + 1 3 x < −5 x 4 x < −1

or or or

4x + 1 < 5 4x < 4 x 4 2 x < −3 or 2x > 5 3 5 or x 2 2 3⎞ ⎛5 ⎞ ⎛ ⎜ − ∞, − ⎟ ∪ ⎜ , ∞ ⎟ 2⎠ ⎝2 ⎠ ⎝ 19.

4 − 5 x ≥ 24 4 − 5 x ≤ −24 or 4 − 5 x ≥ 24 − 5 x ≤ −28 − 5 x ≥ 20 28 x≥ x ≤ −4 5 28 ( −∞, − 4] ∪ ⎢⎡ , ∞ ⎞⎟ ⎣5 ⎠

x − 10 ≥ 2 or x − 10 ≥ 2

x≤8

x ≥ 12

(−∞, − 8] ∪ [12, ∞) 22.

2x − 5 ≥1 2x − 5 ≤ −1 or 2 x − 5 ≥ 1

−14 ≤ 3 x − 10 ≤ 14 − 4 ≤ 3 x ≤ 24 4 x ≤8 − ≤ 3 ⎡ 4 ⎤ ⎢− 3 , 8⎥ ⎦ ⎣ 23.

2 x < 16 x 3}

3x − 1 ≤ 5 3x ≤ 6 x≤2

(1, 8)

(−∞, − 8] ∪ [2, ∞ ) 21.

or or or

2x − 9 < 7 2< 1
4

x +1 > 4 x>3

{ x x ≤1}∪{ x x > 3} ={ x x ≤1

{x x < −1}∪ {x x < 1} = {x x < 1} 17.

0 ≤ 2 x + 6 ≤ 54 −6 ≤ 2 x ≤ 48 −3 ≤ x ≤ 24

2x ≤ 4 x≤2

2x ≥ 6 x≥3

(−∞, 2] ∪ [3, ∞)

24.

3− 2x ≤ 5 −5 ≤ 3 − 2 x ≤ 5 − 8 ≤ − 2x ≤ 2 4 ≥ x ≥ −1

[−1, 4]


Section 1.5

25.

73

26.

x −5 ≥ 0 (Note: The absolute value of any real number is greater than or equal to 0.) (−∞, ∞)

27.

(−∞, ∞)

28.

x−4 ≤0

x2 + 7 x > 0 x ( x + 7) > 0 The product x( x + 7) is positive. x = 0 is a critical value. x + 7 = 0 ⇒ x = −7 is a critical value.

2x + 7 ≤ 0 2x + 7 = 0 2 x = −7 7 x=− 2 ⎧ 7⎫ ⎨− ⎬ ⎩ 2⎭

(Note: No absolute value is less than 0.) x−4=0 x=4 {4}

29.

x −7 ≥ 0

30.

x ( x + 7)

x2 − 5x ≤ 0 x( x − 5) ≤ 0 The product x( x − 5) is negative or zero. x = 0 is a critical value. x − 5 = 0 ⇒ x = 5 is a critical value. x( x − 5) [0, 5]

(−∞, − 7) ∪ (0, ∞) 31.

33.

x 2 − 16 ≤ 0 ( x − 4)( x + 4) ≤ 0 The product ( x − 4)( x + 4) is negative or zero. x − 4 = 0 ⇒ x = 4 is a critical value. x + 4 = 0 ⇒ x = −4 is a critical value.

32.

x 2 − 49 > 0 ( x − 7)( x + 7) > 0 The product ( x − 7)( x + 7) is positive. x = 7 is a critical value. x + 7 = 0 ⇒ x = −7 is a critical value.

( x − 4)( x + 4)

( x − 7)( x + 7)

[−4, 4]

(−∞, − 7) ∪ (7, ∞)

x 2 + 7 x + 10 < 0 ( x + 5)( x + 2) < 0 The product ( x + 5)( x + 2) is negative. x + 5 = 0 ⇒ x = −5 is a critical value. x + 2 = 0 ⇒ x = −2 is a critical value.

34.

x2 + 5x + 6 < 0 ( x + 3)( x + 2) < 0 The product ( x + 3)( x + 2) is negative. x + 3 = 0 ⇒ x = −3 is a critical value. x + 2 = 0 ⇒ x = −2 is a critical value.

( x + 5)( x + 2)

( x + 3)( x + 2)

(−5, −2)

(−3, −2)


74

35.


x 2 − 3x ≥ 28

36.

x 2 + x − 30 < 0 ( x + 6)( x − 5) < 0 The product ( x + 6)( x − 5) is negative. x + 6 = 0 ⇒ x = −6 is a critical alue. x − 5 = 0 ⇒ x = 5 is a critical value.

x 2 − 3 x − 28 ≥ 0 ( x − 7)( x + 4) ≥ 0 The product ( x − 7)( x + 4) is positive or zero. x − 7 = 0 ⇒ x = 7 is a critical value. x + 4 = 0 ⇒ x = −4 is a critical value.

( x + 6)( x − 5)

( x − 7)( x + 4)

(−6, 5)

(−∞, − 4] ∪ [7, ∞) 37.

x+4 0 x+3 x−2 is positive. x+3 x − 2 = 0⇒ x = 2

The quotient

The quotient

x −1 = 0 ⇒ x = 1 The critical values are −4 and 1.

x + 3 = 0 ⇒ x = −3 The critical values are 2 and −3.

x+4 x −1

x−2 x+3

(−∞, − 3) ∪ (2, ∞)

(−4, 1) 39.

x 2 < − x + 30

x −5 ≥ 3 x +8 x −5 −3≥ 0 x +8 x − 5 − 3( x + 8) ≥0 x +8 x − 5 − 3x − 24 ≥ 0 x +8 −2 x − 29 ≥ 0 x +8 −2 x − 29 The quotient is positive or zero. x+8 29 − 2 x − 29 = 0 ⇒ x = − 8 x + 8 = 0 ⇒ x = −8 29 The critical values are − and − 8. 2

40.

x − 4 ≤1 x+6 x − 4 −1 ≤ 0 x+6 x − 4 − 1( x + 6) ≤0 x+6 x−4− x−6 ≤0 x+6 −10 ≤ 0 x+6 −10 is negative or zero. The quotient x+6 x + 6 = 0 ⇒ x = −6 The critical value is –6. −10 x+6

The denominator cannot equal zero ⇒ x ≠ −6.

−2 x − 29 x+8

(−6, ∞)

The denominator cannot equal zero ⇒ x ≠ −8. ⎞ ⎡ 29 ⎢− 2 , − 8 ⎟ ⎠ ⎣


Section 1.5

41.

75

x ≥4 2x + 7 x −4≥0 2x + 7 x − 4(2 x + 7) ≥0 2x + 7 x − 8 x − 28 ≥ 0 2x + 7 −7 x − 28 ≥ 0 2x + 7 −7 x − 28 The quotient is positive or zero. 2x + 7 −7 x − 28 = 0 ⇒ x = −4

42.

25 16 5 3x − 5 = 0 ⇒ x = 3 25 5 The critical values are and . 16 3 16 x − 25 = 0 ⇒ x =

7 2x + 7 = 0 ⇒ x = − 2 7 The critical values are − 4 and − . 2

−7 x − 28 2x + 7 The denominator cannot equal zero ⇒ x ≠ −

16 x − 25 3x − 5

7. 2

The denominator cannot equal zero ⇒ x ≠

7⎞ ⎡ ⎢− 4, − 2 ⎟ ⎣ ⎠ 43.

x ≤ −5 3x − 5 x +5≤0 3x − 5 x + 5(3x − 5) ≤0 3x − 5 x + 15 x − 25 ≤ 0 3x − 5 16 x − 25 ≤ 0 3x − 5 16 x − 25 The quotient is negative or zero. 3x − 5

⎡ 25 5 ⎞ ⎢ 16 , 3 ⎟ ⎠ ⎣

( x + 1)( x − 4) 0 x+5 The quotient

x ( x − 4) > 0 is positive. x+5

x=0 x−4 = 0⇒ x = 4 x + 5 = 0 ⇒ x = −5 The critical values are 0, 4, and –5.

( x + 1)( x − 4) x−2

x ( x − 4) x+5

(−∞, − 1) ∪ (2, 4)

(−5, 0) ∪ (4, ∞)


5 . 3

76

45.


x+2 ≤2 x−5 x+2 −2≤ 0 x−5 x + 2 − 2( x − 5) ≤0 x−5 x + 2 − 2 x + 10 ≤0 x −5 − x + 12 ≤0 x−5 − x + 12 The quotient is negative or zero. x −5 − x + 12 = 0 ⇒ x = 12

46.

x−5 = 0 ⇒ x = 5 The critical values are 12 and 5.

x−2 = 0⇒ x = 2 The critical values are 9 and 2.

− x + 12 x−5

47.

3x + 1 ≥4 x−2 3x + 1 −4≥ 0 x−2 3 x + 1 − 4( x − 2) ≥0 x−2 3x + 1 − 4 x + 8 ≥0 x−2 − x+9 ≥0 x−2 −x + 9 The quotient is positive or zero. x−2 −x + 9 = 0 ⇒ x = 9

−x + 9 x−2

The denominator cannot equal zero ⇒ x ≠ 5.


(−∞, 5) ∪ [12, ∞)

(2, 9]

6 x 2 − 11x − 10 >0 x (3 x + 2)(2 x − 5) >0 x (3x + 2)(2 x − 5) is positive. The quotient x 2 3x + 2 = 0 ⇒ x = − 3 5 2x − 5 = 0 ⇒ x = 2 x=0 2 5 The critical values are − , , and 0. 3 2 (3x + 2)(2 x − 5) x

⎛ 2 ⎞ ⎛5 ⎞ ⎜ − , 0⎟ ∪ ⎜ , ∞⎟ ⎝ 3 ⎠ ⎝2 ⎠

48.

3x2 − 2 x − 8 ≥0 x −1 (3 x + 4)( x − 2) ≥0 x −1 (3 x + 4)( x − 2) is positive or zero. The quotient x −1 4 3x + 4 = 0 ⇒ x = − 3 x−2 =0⇒ x = 2 x −1 ⇒ x = 1 4 The critical values are − , 2, and 1. 3 (3 x + 4)( x − 2) x −1 The denominator cannot equal zero ⇒ x ≠ 1. ⎡ 4 ⎞ ⎢− 3 , 1⎟ ∪ [2, ∞) ⎣ ⎠


Section 1.5

49.

77

x2 − 6 x + 9 ≤0 x−5 ( x − 3)( x − 3) ≤0 x−5 ( x − 3)( x − 3) The quotient is negative or zero. x −5 x−3 = 0 ⇒ x = 3 x−5 = 0 ⇒ x = 5 The critical values are 3 and 5.

50.

( x − 3)( x − 3) x−5

x 2 + 10 x + 25 ≥0 x +1 ( x + 5)( x + 5) ≥0 x +1 ( x + 5)( x + 5) The quotient is positive or zero. x +1 x + 5 = 0 ⇒ x = −5 x + 1 = 0 ⇒ x = −1 The critical values are −5 and −1. ( x + 5)( x + 5) x +1


The denominator cannot equal zero ⇒ x ≠ −1.

(−∞, 5)

{−5} ∪ (−1, ∞)

51.

Plan A: 5 + 0.01x Plan B: 1 + 0.08x 5 + 0.01x < 1 + 0.08x 4 < 0.07x 57.1 < x Plan A is less expensive if you use more than 57 checks.

52.

Company A: 19 + 0.12m Company B: 12 + 0.21m 19 + 0.12m < 12 + 0.21m 7 < 0.09m 77.7 < m Company A is less expensive if you drive at least 78 miles.

53.

Let h = the height of the package. length + girth ≤ 130 length + 2(width) + 2(height) ≤ 130 34 + 2(22) + 2h ≤ 130 34 + 44 + 2h ≤ 130 78 + 2h ≤ 130 2h ≤ 52 h ≤ 26 The height must be more than 0 but less than or equal to 26 inches.

54.

−0.05 x + 1.73 < 1.25 −0.05 x < −0.48 x > 9.6 9.6 years after 2000 is in the year 2009.

55.

17.1895 x + 95.2065 > 600 17.1895 x > 504.7935 x > 29.366 29 months after September 2004 is in January 2007.

56.

Plan A: 15 + 1.49x 57. Plan B: 1.99x 1.99x < 15 + 1.49x 0.50x < 15 x < 30 Plan B is less expensive if fewer than 30 videos are rented.

F ≤ 104 9 68 ≤ C + 32 ≤ 104 5 9 36 ≤ C ≤ 72 5 5 5 ⎛9 ⎞ 5 (36) ≤ ⎜ C ⎟ ≤ (72) 9 9 ⎝5 ⎠ 9 ≤ 40° 20° ≤ C

58.

1.63 − μ < 2.33 1.79 −4.1707 < 1.63 − μ < 4.1707 −167.2 < − μ < −158.8 μ 167.2 > > 158.8 μ 158.8 < < 167.2 lb

59.

190 − μ < 2.575 2.45 −6.30875 < 1.90 − μ < 6.30875 −196.30875 < − μ < −183.69125 196.30875 > μ > 183.69125 183.7 < μ < 196.3 lb

63 < x + ( x + 2) + ( x + 4) < 81 63 < 3x + 6 < 81 < 75 57 < 3x < 25 19 < x x must be odd, thus x = 21 or x = 23. Therefore, the numbers are {21, 23, 25} or {23, 25, 27}.

−2.33
0 The product is positive. 2x = 0 ⇒ x = 0 210 − x = 0 ⇒ x = 210 Critical values are 0 and 210.

312 x − 3 x 2 ≥ 5925 3(− x 2 + 104 x − 1975) ≥ 0 3(− x + 25)( x − 79) ≥ 0 Critical values are 25 and 79.

2 x(210 − x)

3(− x + 25)( x − 79)

($0, $210)

[$25, $79]

−3x 2 + 312 x − 5925 ≥ 0

14.25 x + 350,000 < 50 x 14.25 x + 350,000 < 50 x −35.75 x < −350,000 x > 9790.2 At least 9791 books must be published.

66.

2 C = 0.00014 x + 12 x + 400,000 < 30

x 0.00014 x 2 + 12 x + 400,000 < 30 x 0.00014 x 2 − 18 x + 400,000 < 0

Solve 0.00014 x 2 − 18 x + 400,000 = 0 to find the critical values. −( −18) ± ( −18)2 − 4(0.00014)(400,000) 18 ± 324 − 224 18 ± 100 18 ± 10 = = = 2(0.00014) 0.00028 0.00028 0.00028 x = 18 + 10 x = 18 − 10 or 0.00028 0.00028 28 8 = = 0.00028 0.00028 = 100,000 ≈ 28,571.4 The critical values are 100,000 and ≈ 28,571.4. Since x is a non-negative integer, the intervals are (0, 26,571.4) , (28,571.4, 100,000) , and (100,000, ∞ ) . x=

Test 1:

0.00014(1)2 + 12(1) + 400,000 < 30 ⇒ 400,012.00014 < 30 , which is false. 1

Test 50,000:

0.00014(50,000)2 + 12(50,000) + 400,000 < 30 ⇒ 27 < 30 , which is true. 50,000

0.00014(150000)2 + 12(150000) + 400,000 < 30 ⇒ 35.6 < 30 , which is false 150000 The company should manufacture from 28,572 to 99,999 pairs of running shoes.

Test 150,000:


Section 1.5

79

....................................................... 67.

69.

28 − 0.15 ≤ C ≤ 28 + 0.15 27.85 ≤ 2π r ≤ 28.15 27.85 28.15 ≤ r ≤ 2π 2π 4.432 ≤ r ≤ 4.480 The radius of the cylinder must be between 4.432 inches and 4.480 inches.

( x − 3) 2


735 ≤ πr h ≤ 765 735

πr

( x − 3) 2

( x − 4) 2 3

≤

h ≤

765

πr 2

( x − 1) 2

≥0 ( x − 4) 4 The quotient is positive or zero. x −1 = 0 ⇒ x = 1 x−4=0⇒ x = 4 Critical values are 1 and 4. ( x − 1) 2

2

( x − 4) 4

(−∞, 3) ∪ (3, 6) ∪ (6, ∞) 71.

2

735 765 ≤ h ≤ 4π 4π 58.5 ≤ h ≤ 60.9 The height of the beaker must be between 58.5 cm and 60.9 cm.

( x − 6)2 The quotient is positive. x−3 = 0 ⇒ x = 3 x−6 = 0⇒ x = 6 Critical values are 3 and 6.

( x − 6)

≤ 750 + 15

2

70.

>0

750 − 15 ≤ V

(−∞, 4) ∪ (4, ∞) 72.

≥0

( x + 3) The quotient is positive or zero. x−4 = 0⇒ x = 4 x + 3 = 0 ⇒ x = −3 Critical values are 4 and −3. ( x − 4) 2 ( x + 3)3

2x − 7 ≥0 ( x − 1) 2 ( x + 2)2 The quotient is positive or zero. 7 2x − 7 = 0 ⇒ x = 2 x −1 = 0 ⇒ x = 1 x + 2 = 0 ⇒ x = −2 7 Critical values are , 1, −2. 2 2x − 7

Denominator not 0 ⇒ x ≠ −3. (−3, ∞)

( x − 1)2 ( x + 2) 2 ⎡7 ⎞ ⎢ 2 , ∞⎟ ⎣ ⎠

73.

74.

1< x < 5 if x ≥ 0 if x < 0

1< x < 5

1 < −x < 5 − 1 > x > −5 (−5, − 1) ∪ (1, 5)

75.

2< x −3 (−3, − 2) ∪ (2, 3)

3≤ x < 7 if x ≥ 0 if x < 0

3≤ x −7 (−7, − 3] ∪ [3, 7)


80

76.


77.

0< x ≤3 if x ≥ 0

0< x≤3

0 < x −α < δ , if x − a ≥ 0

δ >0 0
x−a a> x

α
x ≥ −3 [−3, 0) ∪ (0, 3]

if x − a < 0

a −δ

( a − δ , a ) ∪ ( a, a + δ ) 78.

79.

0< x −5 < 2 if x − 5 ≥ 0 if x − 5 < 0

0< x−5 x 5< 0 < −( x − 5) 0> x−5 x 5>

48, v0 = 64, s0 = 0

2

−16t + 64t > 48 2

−16t + 64t − 48 > 0 −16(t 2 − 4t + 3) > 0 −16(t − 1)(t − 3) > 0 The product is positive. The critical values are 1 and 3. (t − 1)(t − 3)

1 second < t < 3 seconds The ball is higher than 48 ft between 1 and 3 seconds. 80.

s = −16t 2 + v0t + s0 ,

s > 96, t > 0, v0 = 80, s0 = 32

2

−16t + 80t + 32 > 96 −16t 2 + 80t − 64 > 0 −16(t 2 − 5t + 4) > 0 −16(t − 1)(t − 4) > 0 The product is positive. The critical values are 1 and 4. (t − 1)(t − 4)

1 second < t < 4 seconds The ball is higher than 96 ft between 1 and 4 seconds.

.......................................................

Prepare for Section 1.6 PS2. 20 = k 1.52 45 = k

PS1. 1820 = k (28) 65 = k

PS4. k 4.5 ⋅ 32 82 (12.5) 4.5 2⋅ 32 = 28.125 8

PS3. k 3 52

(225) 32 = 27 5 PS5. The area becomes 4 times as large.

PS6. No. The volume becomes 9 times as large.


Section 1.6

81

Section 1.6 1.

d = kt

2.

r = ks 2

3.

y=

k x

4.

p=

5.

m = knp

6.

t = krs3

7.

V = klwh

8.

u=

A = ks 2

10.

A = khr 2

11.

y = kx 64 = k ⋅ 48 64 = k 48 4 =k 3

14.

m = kn 92 = k ⋅ 23 92 = k 23 4=k

15.

9.

13.

17.

T = krs 2

18.

210 = k ⋅ 30 ⋅ 52 210 = k 30 ⋅ 52 7 =k 25 0.28 = k

21.

23.

u = kv w k 0.04 = ⋅ 8 0.04

s = k ⋅q 34 = k ⋅ 51 2 =k 3 p = 2 ⋅ 93 3 p = 62 semester hours

24.

p = kd 187.5 = k ⋅ 3 62.5 = k p = 62.5 ⋅ 7 p = 437.5 lb/ft 2

km1m2

r = kt 2

T = ktra 2

16.

C = kr 94.2 = k ⋅ 15 94.2 = k 15 6.28 = k

2

2 4 ⋅ 32 = k 2 4 ⋅ 36 1 =k 81

19.

V = klwh

3 t = kr s

20.

240 = k ⋅ 8 ⋅ 6 ⋅ 5

3 10 = k ⋅ 5 0.09 10 0.09 = k 53 2(0.3) =k 8 .06 = k 25 0.024 = k

240 = k 8⋅6⋅5 1= k

22.

d = k ⋅w 6 = k ⋅ 80 6 = k ⋅ 80 6 =k 80 3 =k 40 Therefore d =

25.

kv w2

12.

d2

144 = k ⋅ 108 144 = k 1082

0.04 0.04 = k 8 (0.04))0.2) =k 8 0.001 = k

V = kT 0.85 = k ⋅ 270 0.85 = k 270 0.17 = k 54 0.17 0.17 Thus V = T = ⋅ 324 = (0.17)6 = 1.02 liters 54 54

F=

k q

3 ⋅ 100 = 7.5 inches 40

j = k ⋅ d3 6 = k ⋅ (4)3 3 =k 32 3 p = ⋅ (5)3 32

p ≈ 11.7 fl oz


26.

r = kv 2 140 = k ⋅ 602 140 = k 602 7 =k 180 7 Thus r = ⋅ 652 180 r ≈ 164.3 ft.

82

27.


T =k l 1.8 = k 3 1.8 = k 3 1.03923 ≈ k a. T = 1.8 10 3

b.

= 1.8 30 3 = 0.6 30 ≈ 3.3 seconds

28.

A = kd 2

T =k l T = l k 2 = l 1.03923 4 = l 1.039232 3.7 ft ≈ l

29.

r=k t 30 = k 64 1920 = k r = 1920 48 r = 40 revolutions per minute

30.

f =k l 144 = k 20 2880 = k f = 2880 18 f = 160 vibrations per second

32.

l = k2 d 50 = k 2 10 5000 = k Thus I = 5000 d2 I = 5000 = 5000 225 152 I ≈ 22.2 footcandles

33.

a.

64 = k ⋅ 20 2 6 = k ⋅ 80 64 = k 400 4 =k 25 100 = 4 ⋅ d 2 25 625 = d 2 d = 25 ft

31.

l = k2 d 28 = k2 8 28 ⋅ 64 = k 1792 = k = 1792 l = 1792 16 42 l = 112 decibels

V = kr 2 h V1 = k (3r ) 2 h = 9( kr 2 h ) = 9V Thus the new volume is 9 times the original volume.

b.

V2 = kr 2 (3h ) = 3( kr 2 h ) = 3V Thus the new volume is 3 times the original volume.

c.

V3 = k (3r ) 2 (3h ) = k 9r 2 ⋅ 3 ⋅ h = 27( kr 2 h ) = 27V Thus the new volume is 27 times the original volume.


Section 1.6

34.

83

L = kwd 2

35.

200 = k ⋅ 2 ⋅ 62 200 = k 2 ⋅ 62 25 = k 9 Thus L = 25 ⋅ 4 ⋅ 4 2 9 1600 = 9 ≈ 178 lb 37.

V = knT P k (3n )T V1 = 1p 2

( )

6 ⋅ 102 = k 24 600 = 37.5 = k 16

⎛ ⎞ = 6 ⎜ knT ⎟ ⎝ p ⎠ = 6V Thus the new volume is 6 times larger than the original volume.

For Randy Johnson, ERA = kr i k (67) 2.32 = (260) 9.00 = k For Tom Glavine, 9(74) ERA = (224.2) = 2.97

38.

2 L = kbd l 2 800 = k ⋅ 4 ⋅ 8 12 800 ⋅ 12 = k 4 ⋅ 82 37.5 = k

37.5(3.5)(6) 2 16 ≈ 295 pounds

Thus L =

4 L = k ⋅ d2 h 4 6 = k ⋅ 22 10

36.

4

Thus L = 37.5 2⋅ 3 14 ≈ 15.5 tons 39.

2 F = kws r 2 k ⋅ 2800 = 1800 ⋅ 45 425 2800 ⋅ 425 = k 1800 ⋅ 452 14 ⋅ 425 = k 9 ⋅ 452 0.3264746 ≈ k

....................................................... 40.

S = kwd 3 when d = 10, w = 182 − 102 ≈ 15 ⇒ S ≈ k (15)(10)3 = 15,000k d = 12, w = 182 − 122 ≈ 13.4 ⇒ S ≈ k (13.4)(12)3 = 23,155k d = 14, w = 182 − 142 ≈ 11.3 ⇒ S ≈ k (11.3)(14)3 = 31,007k d = 16, w = 182 − 162 ≈ 8.2 ⇒ S ≈ k (8.2)(16)3 = 33,587k The strongest beam occurs when d = 16 inches.

41.

T = kd 3/ 2

365 = k ⋅ 933/ 2 365 = k 933 / 2 Thus

⋅ d 3/ 2 686 = 365 933/ 2

686 ⋅ 933 / 2 = d 3/ 2 365 ⎛ 686 ⋅ 933/ 2 ⎞ ⎜ ⎟ 365 ⎝ ⎠

(0.3264746) ⋅ 1800 ⋅ 552 450 ≈ 3950 pounds

Thus F =

2/3

=d 2/3

=d 93 ⎛⎜ 686 ⎞⎟ ⎝ 365 ⎠ 142 million miles ≈ d


Connecting Concepts

84


....................................................... 1.

No. The solution set of x = 3 is {3} but the solution

Assessing Concepts 2.

a 0 [1.5] ( x + 1)( x − 3) > 0 The product is positive. x + 1 = 0 ⇒ x = −1 x−3= 0⇒ x =3 Critical values are –1 and 3.

61 ≤ 9 C + 32 ≤ 95 [1.5] 5 29 ≤ 9 C ≤ 63 5 145 ≤ C ≤ 35 9 ⎡ 145 , 35⎤ ⎣⎢ 9 ⎦⎥

34.

30 < 5 ( F − 32) < 65 [1.5] 9 54 < F − 32 < 117 86 < F (86, 149)

x 3 − 7 x 2 + 12 x ≤ 0 [1.5]

36.

< 149

x 3 + 4 x 2 − 21x > 0 [1.5]

x ( x 2 − 7 x + 12) ≤ 0 x ( x − 3)( x − 4) ≤ 0.

x ( x 2 + 4 x − 21) > 0 x ( x + 7)( x − 3) > 0.

The product is negative or zero.

The product is positive.

x=0 x−3=0⇒ x =3

x=0 x + 7 = 0 ⇒ x = −7

x−4=0⇒ x =4 The critical values are 0, 3, and 4.

x−3=0⇒ x =3 The critical values are 0, − 7, and 3.

x( x − 3)( x − 4)

x( x + 7)( x − 3)

(−∞, 0] ∪ [3, 4]

(−7, 0) ∪ (3, ∞)

x+3 > 0 [1.5] x−4 The quotient is positive. x + 3 = 0 ⇒ x = −3

38.

x−4 =0⇒ x = 4 The critical values are − 3 and 4. x+3 x−4

(−∞, − 3) ∪ (4, ∞)

x( x − 5) ≤ 0 [1.5] x+7 The quotient is negative or zero. x=0 x−5 = 0 ⇒ x = 5 x + 7 = 0 ⇒ x = −7 The critical values are 0, 5 and − 7. x( x − 5) x+7

Denominator ≠ 0 ⇒ x ≠ −7. (−∞, − 7) ∪ [0, 5]


Chapter Review

39.

89

40.

2 x ≤ 10 [1.5] 3− x 2 x − 10 ≤ 0 3− x 2 x − 10(3 − x ) ≤0 3− x 2 x − 30 + 10 x ≤ 0 3− x 12 x − 30 ≤ 0 3− x The quotient is negative or zero. 12 x − 30 = 0 ⇒ x = 5 2 3− x = 0 ⇒ x = 3 The critical values are 5 and 3. 2 12 x − 30 3− x

x ≥ 1 [1.5] 5− x x −1 ≥ 0 5− x x − (5 − x ) ≥0 5− x x −5+ x ≥ 0 5− x 2x − 5 ≥ 0 5− x The quotient is positive or zero. 2x − 5 = 0 ⇒ x = 5 2 5− x = 0⇒ x = 5 The critical values are 5 and 5. 2 2x − 5 5− x Denominator ≠ 0 ⇒ x ≠ 5.

Denominator ≠ 0 ⇒ x ≠ 3.

( 41.

⎡5 , 5 ⎢⎣ 2

−∞, 5 ⎤⎥ ∪ (3, ∞) 2⎦

42.

3x − 4 < 2 [1.5] −2 < 3 x − 4 < 2 2 < 3x < 6 2< x 1.

If x − a < 0, then 0 < x − a < −b a > x > a − b.

(1, 2) ∪ (2, 3)

[ a − b, a ) ∪ ( a , a + b ]

V = π r 2 h [1.2]

46.

V =h π r2

A [1.2] 1 + rt P(1 + rt ) = A P + Prt = A Prt = A − P P=

2 A = hb1 + hb2 2 A − hb2 = hb1

t = A− P Pr 48.

P = 2(l + w) [1.2] P = 2l + 2 w P − 2l = 2 w P − 2l = w 2

49.

e = mc 2 [1.2] e =m c2

A = h (b1 + b2 ) [1.2] 2 2 A = h (b1 + b2 )

47.

2 A − hb2 = b1 h 50.

F =G

m1m2 s2

Fs 2 = Gm1m2 Fs 2 = m 1 Gm2


[1.2]

90

51.

54.


Let x = the number [1.2] 1 1 1 x− x = 4+ x 2 4 5 1 ⎞ 1 ⎞ ⎛ ⎛1 20⎜ x − x ⎟ = 20⎜ 4 + x ⎟ 5 ⎠ 2 4 ⎝ ⎝ ⎠ 10 x − 5 x = 80 + 4 x 5 x = 80 + 4 x x = 80

52.

d = rt d = 8t 6(7 − t ) = 8t

P = 54 54 = 2l + 2 w 54 = 2(2 w − 9) + 2 w 54 = 4 w − 18 + 2 w 72 = 6 w 12 = w 2 w − 9 = 2(12) − 9 = 24 − 9 = 15 width = 12 ft, length = 15 ft [1.2]

Let x = cost last year Cost = last year + raise Let x = the number. 21 = x + 0.05 x 21 = 1.05 x 21 =x 1.05 20 = x

55.

The cost last year was $20.00 . [1.2]

57.

53.

4%

x

6%

5500 − x

42 − 6t = 8t 42 = 14t 3=t

d = 8(3) = 24 nautical miles [1.2] 56.

0.04 x + 0.06(5500 − x) = 295 0.04 x + 330 − 0.06 x = 295 − 0.02 x = −35 x = 1750 5500 − 1750 = 3750 $1750 in the 4% account $3750 in the 6% account [1.2]

Let x = monthly maintenance cost per owner 18 x = 24( x − 12) 18 x = 24 x − 288 − 6 x = −288 x = 48 18 x = 864 The total monthly maintenance cost is $864. [1.2]

58.

d = 6( 7 − t )

Let x = price of battery x + 20 = price of calculator x + x + 20 = 21 2 x + 20 = 21 2x = 1 x = 0.50 x + 20 = 20.50 Price of calculator is $20.50. [1.2] Price of battery is $0.50.

P = 40 A = 96 40 = 2l + 2 w 20 = l + w l = 20 − w 96 = lw 96 = (20 − w) w 96 = 20w − w2 w2 − 20 w + 96 = 0 ( w − 12)( w − 8) = 0 w = 12 or w = 8 l = 20 − 12 or l = 20 − 8 l =8 l = 12

Length = 8 in. and width = 12 in., or length = 12 in. and width = 8 in. [1.2]


Chapter Review

91

59.

Part completed In 1 hour

Time Mason

x−9

Apprentic e

x

60.

1 x−9 1 x

1 ⎞ ⎛1 6⎜ + ⎟ =1 ⎝ x x−9⎠

Let x = number of adult tickets 4526 − x = number of student tickets 8 x + 2(4526 − x) = 33,196 8 x + 9052 − 2 x = 33,196 6 x = 24,144 x = 4024 4526 − x = 502 4024 adult tickets, 502 student tickets [1.2]

1 ⎞ ⎛1 6 x( x − 9)⎜ + ⎟ = 1x( x − 9) ⎝ x x−9⎠ 6( x − 9) + 6 x = x 2 − 9 x 6 x − 54 + 6 x = x 2 − 9 x 0 = x 2 − 21x + 54 0 = ( x − 18)( x − 3) x = 18 or x = 3 (Note : x = 3 ⇒ mason's time = − 6 hours. Thus x ≠ 3.) Apprentice takes 18 hours to build the wall. [1.4] 61.

62.

R = 72 x − 2 x 2 , R > 576 72 x − 2 x 2 > 576 0 > 2 x 2 − 72 x + 576 2 x 2 − 72 x + 576 < 0 x 2 − 36 x + 288 < 0 ( x − 24)( x − 12) < 0 The product is negative. x − 24 = 0 ⇒ x = 24 x − 12 = 0 ⇒ x = 12

1 d 4 V = 144 h=

Critical values are 24 and 12.

d = 2r ⇒ r =

(x − 24)(x − 12)

d 2

1 V = π r 2h 3

(12, 24)

2

1 ⎛d ⎞ ⎛1 ⎞ 144 = π ⎜ ⎟ ⎜ d ⎟ 3 ⎝2⎠ ⎝4 ⎠ 144 =

π d3 48

144(48) = d3 π d ≈ 13 ft 63.

The revenue is greater than $576 when the price is between $12 and $24. [1.5]

[1.2]

−1.96 < x − 50 < 1.96 5 −9.8 < 1.63 − x < 9.8 40.2 < − x < 59.8 x 41 < < 59, where x is an integer

64.

63.8 − μ < 1.645 0.45 −0.74025 < 63.8 − μ < 0.74025 −64.54025 < − μ < −63.05975 μ 64.5 > > 63.1 μ 63.1 < < 64.5 lb −1.645
38.0 40.0 > μ μ < 40.0 lb 38.0
12 x≤8 x < −4 {x | x ≤ 8} ∪ {x | x < −4} = {x | x ≤ 8} [1.5]

11.

a.

12.

x 2 + x − 12 ≥ 0 x +1 ( x + 4)( x − 3) ≥0 x +1 The quotient is positive or zero. x + 4 = 0 ⇒ x = −4 x−3=0⇒ x =3 x + 1 = 0 ⇒ x = −1 Critical values are –4, 3, and −1. ( x + 4)( x − 3) x +1

b.

2x − 1 < 9 and −3x + 1 ≤ 7 2 x < 10 −3x ≤ 6 x 18 + 0.10 x 0.08 x > 8 x > 100 If you drive more than 100 miles, then company A is less expensive. [1.5]

−4 x = −60 x = 15 The assistant takes 15 hours to cover the parking lot. [1.4]

18.

0.5 = −0.0002348 x 2 + 0.0375 x

19.

2

0.0002348 x − 0.0375 x + 0.5 = 0

200(2 x 2 + 25) = 4500 x

−( −0.0375) ± ( −0.0375)2 − 4(0.0002348)(0.5) x= 2(0.0002348)

400 x 2 − 4500 x + 5000 = 0 4 x 2 − 45 x + 50 = 0 ( x − 10)(4 x − 5) = 0

= 0.0375 ± 0.00094 0.00047 0.0375 ± 0.0306 = 0.0004696 x ≈ 145.0 or x ≈ 14.7 More than 14.7 ft but less than 145.0 ft from a side line. [1.5] 20.

x 200 = 4500 2 x 2 + 25

x − 10 = 0 4 x − 5 = 0 x = 10 x = 1.25 More than 1.25 mi but less than 10 mi from the city center. [1.5]

v= k d 4= k 3000 k = 4 3000 = 40 30 v = 40 30 = 40 30 50 2500 v = 4 30 ≈ 4.4 miles/second [1.6] 5

.......................................................

Cumulative Review

1.

4 + 3( −5) = 4 − 15 = −11 [P.1]

3.

(3x − 5) 2 − ( x + 4)( x − 4) = (9 x 2 − 30 x + 25) − ( x 2 − 16) [P.3] 4.

2.

0.00017 = 1.7 × 10−4 [P.2] 8 x 2 + 19 x − 15 = (8 x − 5)( x + 3) [P.4]

= 9 x 2 − 30 x + 25 − x 2 + 16 = 8 x 2 − 30 x + 41 5.

7 x − 3 − 5 = 7 x − 3 − 5 x + 20 = 2 x + 17 [P.5] x−4 x−4 x−4

6.

a 2 / 3 ⋅ a1/ 4 = a 2 / 3 + 1/ 4 = a11/12 [P.2]

7.

(2 + 5i )(2 − 5i ) = 4 − 25i 2 = 4 + 25 = 29 [P.6]

8.

2(3x − 4) + 5 = 17 [1.1] 2(3x − 4) = 12 6 x = 20 x = 10 3


96


10.

2 x 2 − 4 x = 3 [1.3]

9.

2x − 4x − 3 = 0 x= 11.

−( −4) ± ( −4) − 4(2)( −3) 4 ± 2 10 2 ± 10 = = 2(2) 4 2 x = 3+ 9 − x

[1.4]

( x − 3)2 = ( 9 − x )

2

x2 − 6x + 9 = 9 − x x 2 − 5 x = x ( x − 5) = 0 x = 0 or x = 5 x 3 − 36 x = x ( x 2 − 36) = x ( x + 6)( x − 6) = 0 The solutions are 0, –6, 6. [1.4]

2 x − 6 = −4

2 x = 10 or x=5

2

x −3= 9− x

12.

2 x − 6 = 4 [1.1] 2x − 6 = 4

2

2x = 2 x =1

Check 0: 0 = 3+ 9 − 0

Check 5: 5 = 3+ 9 −5

0 = 3+ 9 0 = 3+ 3 0=6 No

5 = 3+ 4 5 = 3+ 2 5=5 The solution is 5.

13.

2 x 4 − 11x 2 + 15 = 0 Let u = x 2 . 2u 2 − 11u + 15 = (2u − 5)(u − 3) = 0 u−3=0 or 2u − 5 = 0 u=3 u = x2 = 5 2 x2 = 3 10 5 =± x=± x=± 3 2 2 The solutions are − 10 , 2

14.

3x − 1 > 2 or −3x + 5 ≥ 8 3x > 3 −3x ≥ 3 x >1 x ≤ −1 The solution is {x | x ≤ −1 or x > 1} . [1.5]

16.

x − 2 ≥ 4 ⇒ x − 2 − 4 ≥ 0 ⇒ x − 2 − 4(2 x − 3) ≥ 0 ⇒ x − 2 − 8 x + 12 ≥ 0 ⇒ −7 x + 10 ≥ 0 2x − 3 2x − 3 2x − 3 2x − 3 2x − 3 2x − 3 Solve −7 x + 10 = 0 and 2 x − 3 − 0 to find the critical values. −7 x + 10 = 0 2x − 3 = 0

x = 10 7

15.

3 . [1.4]

x−6 ≥2 ⇒ x−6≥2 or x − 6 ≤ −2 x≥8 x≤4 The solution is ( −∞, 4] ∪ [8, ∞) . [1.5]

x=3 2

The critical values are 10 and 3 . The intervals are ⎛⎜ −∞, 10 ⎞⎟ , ⎛⎜ 10 , 3 ⎞⎟ and ⎛⎜ 3 , ∞ ⎞⎟ 7 2 7 ⎠ ⎝ 7 2⎠ ⎝2 ⎠ ⎝ 0 2 2 2 − − 10 : ≥4⇒ ≥ 4 ⇒ ≥ 4 , which is false. Test 0, in the interval −∞, 7 2(0) − 3 3 −3

(

10 , − 3, 2

)

Test 1.45, in the interval ⎛⎜ 10 , 3 ⎞⎟ : 1.45 − 2 ≥ 4 ⇒ −0.55 ≥ 4 ⇒ 5.5 ≥ 4 , which is true. −0.1 ⎝ 7 2 ⎠ 2(1.45) − 3 Test 2, in the interval ⎛⎜ 3 , ∞ ⎞⎟ : 2 − 2 ≥ 4 ⇒ 0 ≥ 4 ⇒ 0 ≥ 4 , which is false. 1 ⎝2 ⎠ 2(2) − 3 3 The denominator cannot equal zero ⇒ x ≠ . 2 ⎧ ⎫ The solution is ⎨ x 10 ≤ x < 3 ⎬ . [1.5] 2⎭ ⎩ 7


Cumulative Review

97

17.

18.

w

P = R−C = 200 x − 0.004 x 2 − (65 x + 320,000) = −0.004 x 2 + 135 x − 320,000 Profits must be greater than or equal to 600,000.

w +16 Perimeter = 2(Length) + 2(Width) 200 = 2( w + 16) + 2 w 200 = 2 w + 32 + 2 w 168 = 4 w 42 = w w = 42 w + 16 = 58 The width is 42 feet; the length is 58 feet. [1.2]

−0.004 x 2 + 135 x − 320,000 ≥ 600,000 −0.004 x 2 + 135 x − 920,000 ≥ 0 x=

−135 ± (135)2 − 4( −0.004)( −920,000) 2( −0.004)

= −135 ± 3505 −0.008 = 9475 or 24,275 9475 to 24,275 printers should be manufactured. [1.5]

19.

Let x = the score on the fourth test. 80 ≤ 86 + 72 + 94 + x < 90 and 0 ≤ x ≤ 100 4 252 x < 90 + 80 ≤ 4 320 ≤ 252 + x < 360 68 ≤ x < 108 [68, 108) ∩ [0, 100] = [68, 100] The fourth test score must be from 68 to 100. [1.5]

20.

600 p ≥ 100 100 − p 600 p ≥ 100(100 − p ) 600 p ≥ 10,000 − 100 p 700 p ≥ 10,000

p ≥ 14.3

and

600 p ≤ 180 100 − p 600 p ≤ 180(100 − p ) 600 p ≤ 18,000 − 180 p 780 p ≤ 18,000

p ≤ 23.1 They can expect to ticket from 14.3% to 23.1% of the speeders. [1.5]


Chapter 2

Functions and Graphs Section 2.1 1.

3.

2.

a.

$31,500

b.

Increase from 2004 to 2005 33.5 − 32.9 = 0.6 Increase from 2005 to 2006 33.50 + 0.6 = 34.1 The per capita income for 2006 would be $34,100. Percent increase from 2004 to 2005 33.5 − 32.9 = 1.824% 32.9 Percent increase from 2005 to 2006 33.50(1.01824) = 34.111 The per capita income for 2006 would be $34,111.

c.

4.

a. b.

c.

When the cost of a game is $22, 50 million games can be sold. The projected numbers of sales decreases as the price of this game increases.

p

R = p⋅N

8 8 ⋅ 80 = 640 15 15 ⋅ 70 = 1050 22 22 ⋅ 60 = 1320 27 27 ⋅ 50 = 1350 31 31 ⋅ 40 = 1240 34 34 ⋅ 30 = 1020 36 36 ⋅ 20 = 720 37

d.

5.

d = ( −8 − 6)2 + (11 − 4)2

6.

37 ⋅ 10 = 370

The revenue increases to a certain point and then decreases as the price of the game increases.

d = ( −10 − ( −5))2 + (14 − 8)2

7.

d = ( −10 − ( −4))2 + (15 − ( −20))2

= ( −14)2 + (7)2

= ( −5)2 + (6)2

= ( −6)2 + (35)2

= 196 + 49

= 25 + 36

= 36 + 1225

= 245

= 61

= 1261

=7 5


Section 2.1

8.

99

d = (36 − 40)2 + (20 − 32)2

9.

d = (0 − 5)2 + (0 − (−8))2

d = (5 − 0)2 + (13 − 0)2

10.

= (−4)2 + (−12)2

= (−5) 2 + (8)2

= 52 + 132

= 16 + 144

= 25 + 64

= 25 + 169

= 160

= 89

= 194

= 4 10 11.

d = ( 12 − 3)2 + ( 27 − 8)2

12.

d = (6 − 125)2 + (2 5 − 20)2

= (2 3 − 3)2 + (3 3 − 2 2) 2

= (6 − 5 5)2 + (2 5 − 2 5)2

= ( 3) 2 + (3 3 − 2 2)2

= (6 − 5 5)2 + 02

= 3 + (27 − 12 6 + 8)

= (6 − 5 5)2 = 6 − 5 5 = 5 5 − 6

= 3 + 27 − 12 6 + 8

Note: for another form of the solution,

= 38 − 12 6

d = (6 − 5 5)2 = 36 − 60 5 + 125 = 161 − 60 5

13.

d = ( − a − a ) 2 + ( −b − b ) 2

14.

d = (a − (a − b))2 + (a + b − b)2

= (−2a )2 + (−2b) 2

= ( a − a + b) 2 + ( a ) 2

= 4a 2 + 4b 2

= b2 + a 2

= 4(a 2 + b 2 )

= a 2 + b2

= 2 a 2 + b2

15.

d = (−2 x − x)2 + (3 x − 4 x)2 with x < 0

= (−3 x) 2 + ( − x)2

= 9 x2 + x2

= 9 x2 + x2

= 10 x 2

= 10 x 2 (Note: since x < 0, x 2 = − x)

(4 − x )2 + (6 − 0)2 = 10

(

d = (−2 x − x)2 + (3 x − 4 x)2 with x > 0

= (−3x)2 + ( − x)2

= − x 10

17.

16.

(4 − x )2 + (6 − 0)2

)

2

= 102

= x 10 (since x > 0, x 2 = x)

18.

(5 − 0)2 + ( y − ( −3)2 = 12

(

)

2

= 122

25 + y 2 + 6 y + 9 = 144

16 − 8 x + x 2 + 36 = 100

y 2 + 6 y − 110 = 0

x 2 − 8 x − 48 = 0 ( x − 12)( x + 4) = 0 x = 12 or x = −4 The points are (12, 0), ( − 4, 0).

(5) 2 + ( y + 3)2

−6 ± 62 − 4(1)( −110) 2(1) − ± + 440 6 36 y= 2 y = −6 ± 476 2 − ± 6 2 119 y= 2 y = −3 ± 119 y=

The points are (0, − 3 + 119), (0, − 3 − 119). Copyright © Houghton Mifflin Company. All rights reserved.

100

19.

22.

Chapter 2: Functions and Graphs

⎛ x + x2 y1 + y2 ⎞ M =⎜ 1 , 2 ⎟⎠ ⎝ 2

20.

⎛ x + x2 y1 + y2 ⎞ M =⎜ 1 , ⎟ 2 ⎠ ⎝ 2

= ⎛⎜ 1 + 5 , −1 + 5 ⎞⎟ 2 ⎠ ⎝ 2

= ⎛⎜ −5 + 6 , −2 + 10 ⎞⎟ 2 ⎠ ⎝ 2

= ⎛⎜ 6 , 4 ⎞⎟ ⎝2 2⎠ = (3, 2)

= ⎛⎜ 1 , 8 ⎞⎟ ⎝2 2⎠

21.

M = ⎛⎜ 6 + 6 , −3 + 11 ⎞⎟ 2 ⎠ ⎝ 2 = ⎛⎜ 12 , 8 ⎞⎟ ⎝ 2 2⎠ = (6, 4)

= ⎛⎜ 1 , 4 ⎞⎟ ⎝2 ⎠

⎛ 4 + ( −10) 7 + 7 ⎞ , M =⎜ ⎟ 2 2 ⎠ ⎝

23.

1.75 + ( −3.5) 2.25 + 5.57 ⎞ , M = ⎛⎜ ⎟ 2 2 ⎝ ⎠

24.

= ⎛⎜ − 1.75 , 7.82 ⎞⎟ 2 ⎠ ⎝ 2 = ( −0.875, 3.91)

= ⎛⎜ −6 , 14 ⎞⎟ ⎝ 2 2⎠ = ( −3, 7)

⎛ −8.2 + ( −2.4) , 10.1 + ( − 5.7) ⎞ ⎜ ⎟ 2 2 ⎝ ⎠ = ⎛⎜ − 10.6 , 4.4 ⎞⎟ 2 ⎠ ⎝ 2 = ( −5.3, 2.2)

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

37.

38.

39.

⎛ 12 ⎞ Intercepts: ⎜ 0, ⎟, (6, 0 ) ⎝ 5⎠


Section 2.1

40.

101

15 ⎞ ⎛ Intercepts: ⎜ 0,− ⎟, (5, 0) 4⎠ ⎝

41.

(0, 5 ), (0, − 5 ), (5, 0)

(0, 6 ), (0, − 6 ), (−6, 0)

42.

x = y2 − 6

x = − y2 + 5

43.

(0, 4), (0, −4), (−4, 0)

44.

(0, ± 2),( ± 2, 0)

45.

x2 + y 2 = 4

x = y3 − 2

x = | y | −4 46.

(0, 3 2 ), (−2, 0)

(0, 0)

47.

(0, ± 4), (±4, 0)

(0, ± 2), (±8, 0)

48.

| x − 4y | = 8 x2 = y 2

| x | + | y |= 4

49.

center (0, 0), radius 6

50.


51.


52.


53.

center (−2, −5), radius 5

54.

center (−3, −5), radius 11

55.


56.


57.

( x − 4)2 + ( y − 1) 2 = 22

60.

( x − 0 )2 + ⎛⎜ y − 2 ⎞⎟ = ( 11 ) 3⎠ ⎝

2

2

2

2

2

58.

( x − 5) 2 + ( y + 3) 2 = 42

59.

1⎞ ⎛ 1⎞ ⎛ ⎜x− ⎟ +⎜y − ⎟ = 2⎠ ⎝ 4⎠ ⎝

61.

( x − 0) 2 + ( y − 0) 2 = r 2

62.

( x − 0) 2 + ( y − 0) 2 = r 2

( 5 )2

(−3 − 0)2 + (4 − 0)2 = r 2

(5 − 0)2 + (12 − 0)2 = r 2

(−3)2 + 42 = r 2

52 + 122 = r 2

9 + 16 = r 2

25 + 144 = r 2 169 = 132 = r 2

25 = 52 = r 2 ( x − 0) 2 + ( y − 0)2 = 52

( x − 0)2 + ( y − 0)2 = 132


102

63.


( x + 2)2 + ( y − 5)2 = r 2

64.

(1 + 2)2 + (7 − 5)2 = r 2

( x − 1)2 + ( y − 3)2 = r 2

32 + 22 = r 2

(4 − 1) 2 + ( −1 − 3)2 = r 2

9 + 4 = r2

( 13 ) = r 2 2 ( x + 2)2 + ( y − 5)2 = ( 13 )

32 + (−4)2 = r 2

2

13 =

9 + 16 = r 2 25 = 52 = r 2 ( x − 1)2 + ( y − 3)2 = 52 65.

x2 − 6x

+ y 2 = −5

2

66.

2

2

x 2 − 14 x

( x − 3)2 + ( y − 2)2 = 12 center (3, 2), radius 1

+ y2 + 8 y

2

= −56

68.

2

2

4 x2 + 4 x

x2 + x +

70.

+ y 2 = 63 4 2 63 + y = +1 4 4

1 4

2

2

2

2

+ y2 + 3 y = 15 2 4 2 2 1 3 9 15 x −x+ +y + y + = +1+9 4 2 4 4 4 4 x2 − x

⎛x − 1⎞ + ⎛ y + 3⎞ = ⎛ 5⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 2⎠ ⎝ 2⎠ ⎝ ⎝2⎠ center ⎛⎜ 1 , − 3 ⎞⎟ , radius 5 2⎠ 2 ⎝2

( )

( x − 0)2 + ⎛⎜ y − 1 ⎞⎟ = 2 3⎠ ⎝ ⎛ 1⎞ center ⎜ 0, ⎟, radius 2 ⎝ 3⎠

⎛ 1 ⎞ center ⎜ − ,0 ⎟ , radius 4 ⎝ 2 ⎠

2

= 17

x 2 + ⎛⎜ y − 1 ⎞⎟ = 2 3⎠ ⎝

⎛ x + 1 ⎞ + ( y − 0) 2 = 42 ⎜ ⎟ 2⎠ ⎝

2

9 x2 + 9 y2 − 6 y

x2 + y2 − 2 y = 17 3 9 2 2 2 1 17 x +y − y + = +1 3 9 9 9

⎛ x + 1 ⎞ + y 2 = 16 ⎜ ⎟ 2⎠ ⎝

71.

= −25

2

( x − 5)2 + ( y + 1)2 = 12 center (5, −1), radius 1

+ 4 y 2 = 63

x2 + x

+ y2 + 2 y

x − 10 x + 25 + y + 2 y + 1 = −25 + 25 + 1

2

( x − 7) + ( y + 4) = 3 center (7, −4), radius 3 69.

x 2 − 10 x 2

x − 14 x + 49 + y + 8 y + 16 = −56 + 49 + 16 2

= −12

2

x − 6 x + 9 + y − 4 y + 4 = −12 + 9 + 4

2

( x − 3) + y = 2 center (3, 0), radius 2 67.

+ y2 − 4 y

2

x − 6 x + 9 + y = −5 + 9 2

x2 − 6x

2

72.

2

= − 25 4 2 2 9 25 25 x + 3x + + y − 5y + = − + 9 + 25 4 4 4 4 4 x 2 + 3x

+ y2 − 5 y

2

2

⎛ x + 3⎞ + ⎛ y − 5⎞ = ⎛ 3⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 2⎠ ⎝ 2⎠ ⎝ ⎝2⎠ center ⎛⎜ − 3 , 5 ⎞⎟ , radius 3 2 ⎝ 2 2⎠


2

Section 2.1

73.

103

d = ( −4 − 2)2 + (11 − 3)2

74.

= 36 + 64

d = (−3 − 7) 2 + (5 − (−2)) 2 = 100 + 49 = 149

Since the diameter is

= 100

149 , the radius is

149 . 2

⎛ 7 + ( − 3) (−2) + 5 ⎞ ⎛ 3 ⎞ Center is ⎜ , ⎟ = ⎜ 2, ⎟ 2 2 ⎝ ⎠ ⎝ 2⎠

= 10 Since the diameter is 10, the radius is 5. The center is the midpoint of the line segment from (2,3) to (-4,11). ⎛ 2 + (-4) 3 + 11 ⎞ , ⎟ = (−1,7) center ⎜ 2 ⎠ ⎝ 2

2 ⎛ 3⎞ 149 ⎞ ⎛ ( x − 2)2 + ⎜ y − ⎟ = ⎜⎜ ⎟⎟ 2⎠ ⎝ ⎝ 2 ⎠

2

( x + 1) 2 + ( y − 7) 2 = 5 2

75.

Since it is tangent to the x-axis, its radius is 11. 2

2

76.

2

Since it is tangent to the y-axis, its radius is 2. ( x + 2) 2 + ( y − 3) 2 = 2 2

( x − 7) + ( y − 11) = 11

.......................................................

Connecting Concepts

77.

78.

79.

80.

81.

82.

83.

84.

85.

87.

86.

⎛ x + 5 y +1⎞ , ⎜ ⎟ = (9, 3) 2 ⎠ ⎝ 2 therefore x + 5 = 9 2 x + 5 = 18 x = 13

88.

and

y +1 =3 2 y +1= 6 y=5

therefore x + 4 = −2 2 x + 4 = −4 x = −8

Thus (13, 5) is the other endpoint. 89.

⎛ x + ( −3) y + (−8) ⎞ , ⎜ ⎟ = (2, − 7) 2 2 ⎝ ⎠

⎛ x + 4 y + (−6) ⎞ , ⎜ ⎟ = (−2, 11) 2 ⎝ 2 ⎠ and

y + ( −6) = 11 2 y − 6 = 22 y = 28

Thus (−8, 28) is the other endpoint. 90.

⎛ x + 5 y + (−4) ⎞ , ⎜ ⎟ = (0, 0) 2 ⎝ 2 ⎠ y−4 =0 2 y−4=0 y=4

y −8 = −7 therefore x − 3 = 2 and 2 2 y − 8 = −14 x−3= 4 x=7 y = −6

therefore x + 5 = 0 2 x+5=0 x = −5

Thus (7, −6) is the other endpoint.

Thus (−5, 4) is the other endpoint.


and

104


(3 − x )2 + (4 − y )2 = 5

91.

( −5 − x )2 + (12 − y )2 = 13

92.

(3 − x )2 + (4 − y )2 = 52

( −5 − x )2 + (12 − y )2 = 132

9 − 6 x + x 2 + 16 − 18 y + y 2 = 25

25 + 10 x + x 2 + 144 − 24 y + y 2 = 169

x2 − 6x + y2 − 8 y = 0 93.

x 2 + 10 x + y 2 − 24 y = 0

(4 − x ) 2 + (0 − y )2 + ( −4 − x )2 + (0 − y ) 2 = 10 (4 − x )2 + (0 − y ) 2 = 100 − 20 ( −4 − x )2 + (0 − y )2 + ( −4 − x )2 + ( − y )2 16 − 8 x + x 2 + y 2 = 100 − 20 ( −4 − x )2 + ( − y )2 + 16 + 8 x + x 2 + y 2 −16 x − 100 = −20 ( −4 − x )2 + ( − y )2 4 x + 25 = 5 ( −4 − x )2 + ( − y )2 16 x 2 + 200 x + 625 = 25 ⎡( −4 − x )2 + ( − y )2 ⎤ ⎣ ⎦ 16 x 2 + 200 x + 625 = 25 ⎡16 + 8 x + x 2 + y 2 ⎤ ⎣ ⎦ 16 x 2 + 200 x + 625 = 400 + 200 x + 25 x 2 + 25 y 2

Simplifying yields 9 x 2 + 25 y 2 = 225 . (0 − x )2 + (4 − y )2 − (0 − x )2 + ( −4 − y )2 = 6

94.

(

x 2 + (4 − y )2 − x 2 + (4 + y )2

)

2

= 62

x 2 + (4 − y )2 − 2 x 2 + (4 − y )2 x 2 + (4 + y 2 + x 2 + (4 + y )2 = 36 x 2 + 16 − 8 y + y 2 − 2 x 2 + (4 − y )2 x 2 + (4 + y 2 + x 2 + 16 + 8 y + y )2 = 36 2 x 2 + 2 y 2 − 4 = 2 x 2 + (4 − y )2 x 2 + (4 + y )2 ( x 2 + y 2 − 2)2 =

(

x 2 + (4 − y )2 x 2 + (4 + y )2

)

2

x 4 + x 2 y 2 − 2 x 2 + x 2 y 2 + y 4 − 2 y 2 − 2 x 2 − 2 y 2 + 4 = ( x 2 + (4 − y )2 )( x 2 + (4 + y ) 2 ) x 4 + x 2 y 2 − 4 x 2 − 4 y 2 + y 4 + 4 = ( x 2 + 16 − 8 y + y 2 )( x 2 + 16 + 8 y + y 2 ) x 4 + 2 x 2 y 2 − 4 x 2 − 4 y 2 + y 4 + 4 = x 4 + 16 x + 8 x 2 y + x 2 y 2 + 16 x 2 + 256 + 128 y + 16 y 2 −8 x 2 y − 128 y − 64 y 2 − 8 y 3 + x 2 y 2 + 16 y 2 + 8 y 3 + y 4 −36 x 2 + 28 y 2 = 252

95.

The center is (-3,3). The radius is 3. ( x + 3) 2 + ( y − 3) 2 = 32

96.

or

− 9 x 2 + 7 y 2 = 63.

⎛ 5 ⎞⎟ 5 5 ,− The center is ⎜ − . The radius is . ⎟ ⎜ 2 2 2 ⎠ ⎝ 2

2

⎞ ⎛ ⎞ ⎞ ⎛ ⎛ ⎜x+ 5 ⎟ +⎜ y+ 5 ⎟ =⎜ 5 ⎟ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝

....................................................... PS1. x 2 + 3x − 4 2

( −3) + 3( −3) − 4 = 9 − 9 − 4 = −4

2

Prepare for Section 2.2 PS2. D ={−3, −2, −1, 0, 2} R ={1, 2, 4, 5}


Section 2.2

105

PS4. 2 x − 6 ≥ 0 2x ≥ 6 x≥3

PS3. d = (3 − ( −4))2 + ( −2 − 1)2 = 49 + 9 = 58

PS5.

PS6. a = 3x + 4, a = 6 x − 5

x2 − x − 6 = 0 ( x + 2)( x − 3) = 0 x+2=0 x = −2 –2, 3

3x + 4 = 6 x − 5 9 = 3x 3= x a = 3(3) + 4 = 13

x−3=0 x=3

Section 2.2 1.

Given f ( x ) = 3 x − 1, a. f (2) = 3(2) − 1 = 6 −1 =5 b. f ( −1) = 3( −1) − 1 = −3 − 1 = −4 c. f (0) = 3(0) − 1 = 0 −1 = −1 ⎛2⎞ ⎛2⎞ d. f ⎜ ⎟ = 3⎜ ⎟ − 1 ⎝3⎠ ⎝3⎠ = 2 −1 =1 e. f ( k ) = 3( k ) − 1 = 3k − 1 f. f ( k + 2) = 3( k + 2) −1 = 3k + 6 −1 = 3k + 5

2.

Given g ( x) = 2 x 2 + 3, a.

g (3) = 2(3)2 + 3 =18 + 3 = 21

3.

Given A( w) = w 2 + 5 , a.

= 5

b.

g (−1) = 2(−1)2 + 3 = 2+3 =5

b.

c.

g (0) = 2(0)2 + 3 = 0+3 =3

c.

2

d.

e.

⎛ ⎞ ⎛ ⎞ g ⎜ 1 ⎟ = 2⎜ 1 ⎟ +3 ⎝2⎠ ⎝2⎠ = 1 +3 2 =7 2

f.

A(2) = (2)2 + 5 = 9 =3 A(−2) = (−2)2 + 5 = 9 =3

d.

A(4) = 42 + 5 = 21

e.

A(r +1) = (r +1)2 + 5 = r 2 + 2r +1+ 5 = r 2 + 2r + 6

g (c) = 2(c)2 + 3 = 2c2 + 3

A(0) = (0)2 + 5

f.

g (c + 5) = 2(c + 5)2 + 3 2

= 2c + 20c + 50 + 3 = 2c 2 + 20c + 53


A(−c) = (−c)2 + 5 = c2 + 5

106

4.


Given J (t ) = 3t 2 − t , a.

b.

5.

J (−4) = 3(−4)2 − (−4) = 48 + 4 = 52

Given f ( x ) = a.

J (0) = 3(0)2 − (0) = 0−0 =0

c.

d.

1 ⎛1⎞ ⎛1⎞ J ⎜ ⎟ = 3⎜ ⎟ − 3 ⎝ 3⎠ ⎝3⎠ 1 1 = − 3 3 =0

J (−c) = 3(−c)2 − (−c)

d.

J ( x +1) = 3( x +1)2 − ( x +1)

e.

2

= 3x + 6 x + 3− x −1 = 3x 2 + 5 x + 2 f.

f.

J ( x + h) = 3( x + h)2 − ( x + h)

1 1 = −2 2

⎛ ⎞ f ⎜− 3 ⎟= 1 ⎝ 5⎠ −3 5 1 = 3

= 3c 2 + c e.

1 1 = 2 2

f ( −2) =

b.

2

c.

f ( 2) =

1 , x

5

=1÷ 3 =1⋅ 5 = 5 5 3 3 1 1 f ( 2) + f ( −2) = + = 1 2 2 1 1 2 f ( c + 4) = = 2 2 c +4 c +4 f (2 + h) =

1 2+h

= 3x 2 + 6 xh + 3h2 − x − h 6.

Given T ( x) = 5, a. T (−3) = 5 b. T ( 0) = 5 c. d. e. f.

⎛2⎞ T⎜ ⎟ = 5 ⎝7⎠ T (3) + T (1) = 5 + 5 = 10 T ( x + h) = 5 T (3k + 5) = 5

7.

x , x

Given s ( x) =

8.

a.

s ( 4) =

4 4 = =1 4 4

a.

b.

s (5) =

5 5 = =1 5 5

b.

c.

s ( −2 ) =

−2 −2 = = −1 2 −2

d.

s (−3) =

−3 −3 = = −1 3 −3

e.

Since t > 0, t = t. s (t ) =

f.

c.

d.

t t = =1 t t

Since t < 0, t = −t.

e.

t t = = −1 t −t

f.

s (t ) =

x , x+4 0 0 r ( 0) = = =0 0+4 4 −1 −1 1 r ( −1) = = =− −1+ 4 3 3 −3 −3 = = −3 r ( −3) = 1 −3+ 4 ⎛1⎞ 1 ⎜ ⎟ 2 ⎛1⎞ 2 =⎝ ⎠ r⎜ ⎟ = 1 9⎞ ⎛ ⎝2⎠ +4 ⎜ ⎟ 2 ⎝2⎠ 1 9 1 2 1 = ÷ = ⋅ = 2 2 2 9 9 0.1 0.1 1 r (0.1) = = = 0.1 + 4 4.1 41 10.000 r (10,000) = 10,000 + 4 10,000 2500 = = 10,004 2501

Given r ( x) =


Section 2.2

9.

107

a.

Since x = −4 < 2, use P ( x ) = 3 x + 1. P ( −4) = 3( −4) + 1 = −12 + 1 = −11

b.

Since x = 5 ≥ 2, use P ( x) = − x 2 + 11.

b.

P 5 = − 5 + 11 = −5 + 11 = 6 Since x = c < 2, use P ( x ) = 3 x + 1. P (c) = 3c + 1 Since k ≥ 1, then x = k + 1 ≥ 2,

c.

so use P ( x) = − x 2 + 11.

d.

c. d.

10.

Since t = −4 and 0 ≤ t ≤ 5, use Q (t ) = 4. Q ( 0) = 4 Since t = e and 6 < e < 7, then 5 < t ≤ 8, so use Q (t ) = −t + 9. Q (e ) = − e + 9 Since t = n and 1 < n < 2, then 0 ≤ t ≤ 5, so use Q(t ) = 4 Q ( 0) = 4

a.

( ) ( )2

Since t = m 2 + 7 and 1 < m ≤ 2,

P ( k + 1) = −( k + 1)2 + 11 = −( k 2 + 2k + 1) + 11

then 12 < m 2 ≤ 22

= − k 2 − 2k − 1 + 11

12 + 7 < m 2 + 7 ≤ 22 + 7

= − k 2 − 2k + 10

1 + 7 < m2 + 7 ≤ 4 + 7 8 < m 2 + 7 ≤ 11 thus 8 < t ≤ 11, so use Q(t ) = t − 7

Q( m 2 + 7) =

(m2 + 7)− 7

= m 2 = m = m since m > 0

11.

2x+3y = 7 3 y = −2 x + 7 y = − 2 x + 7 , y is a function of x. 3 3

12.

5x + y = 8 y = −5 x + 8, y is a function of x.

13.

− x + y2 = 2

14.

x2 − 2 y = 2

y2 = x + 2

−2 y = − x 2 + 2 y = 1 x 2 − 1, y is a function of x. 2

y = ± x + 2, y is a not function of x.

15.

y = 4 ± x , y is not a function of x since for each x > 0 there are two values of x.

16.

x2 + y2 = 9 y2 = 9 − x2 y = ± 9 − x 2 , y is a not function of x.

17.

y = 3 x , y is a function of x.

18.

y = x + 5, y is a function of x.

19.

y2 = x2

20.

y 3 = x3 3

y = x3 = x, y is a function of x.

y = ± x 2 , y is a not function of x.

21.

Function; each x is paired with exactly one y.

22.

Not a function; 5 is paired with 10 and 8.

23.


24.


25.


26.


27.

f ( x) = 3x − 4

Domain is the set of all real numbers.

28.

f ( x) = −2 x + 1


29.

f ( x) = x2 + 2


30.

f ( x ) = 3x 2 + 1



108


{

4 x+2

31.

f ( x) =

33.

f ( x) = 7 + x

35.

f ( x) = 4 − x 2

37.

f ( x) =

}

Domain is x x ≠ −2 .

1 x+4

{

}

Domain is x x ≥ −7 .

{

}

Domain is x − 2 ≤ x ≤ 2 .

{

}

Domain is x x > −4 .

39.

{

6 x−5

32.

f ( x) =

34.

f ( x) = 4 − x

36.

f ( x ) = 12 − x 2

38.

f ( x) =

}

Domain is x x ≠ 5 .

1 5− x

{

}

Domain is x x ≤ 4 .

{

}

Domain is x − 2 3 ≤ x ≤ 2 3 .

{

}

Domain is x x < 5 .

40.

Domain: the set of all real numbers Domain: the set of all real numbers 41.

42.

Domain: the set of all real numbers 43.

44.

Domain: { x

− 6 ≤ x ≤ 6}

Domain: { x 0 ≤ x ≤ 4}

45.

46.

Domain: { x

47.

Domain: the set of all real numbers

− 3 ≤ x ≤ 3}

int ⎣⎡102 (2.3458) + 0.5⎦⎤ 10

2

Domain:

[ ] = int 235.08 = 235 = 2.35 100 100

48.

{x

}

0≤ x≤4

int [10(34.567) + 0.5] int [346.17] 346 = = = 34.6 10 10 10


Section 2.2

109

49.

int ⎣⎡103 (34.05622) + 0.5⎦⎤

51.

int ⎡⎣104 (0.08951) + 0.5⎤⎦

53.

a.

103

104

b.

=

int [34,056.72] 34,056 = = 34.056 1000 1000

[ ] = int 895.6 = 895 = 0.0895 10,000 10,000

C (2.8) = 0.39 − 0.34int(1− 2.8) = 0.39 − 0.34int( −1.8) = 0.39 − 0.34( −2) = 0.39 + 0.68 = $1.07 c(w)

50.

int ⎡⎣100 (109.83) + 0.5⎤⎦

52.

int ⎡⎣103 (2.98245) + 0.5⎤⎦

54.

a.

Domain: [0, ∞)

b.

T (31,250) = 0.25(31,250 − 30,650) + 4220

100

[ ] = int 110.33 = 110 = 110 1 100

103

[ ] = int 2982.95 = 2982 = 2.982 1000 1000

= 0.25(600) + 4220 = 150 + 4220 = $4370

c.

T (78,900) = 0.28(78,900 − 74,200) + 15,107.50 = 0.28(4700) + 15,107.50 = 1316 + 15,107.50 = $16,423.50

55.

a. b. c. d.

Yes; every vertical line intersects the graph in one point. Yes; every vertical line intersects the graph in one point. No; some vertical lines intersect the graph at more than one point. Yes; every vertical line intersects the graph in at most one point.

56.

a. b. c. d.

Yes; every vertical line intersects the graph in at most one point. No; some vertical lines intersect the graph at more than one point. No; a vertical line intersects the graph at more than one point. Yes; every vertical line intersects the graph in one point.

57.

Decreasing on (−∞, 0] ; increasing on [0, ∞)

58.

Decreasing on (−∞, ∞)

59.

Increasing on (−∞, ∞)

60.

Increasing on (−∞, 2] ; decreasing on [2, ∞)

61.

Decreasing on (−∞, − 3] ; increasing on [ −3, 0] ; decreasing on [0, 3] ; increasing on [3, ∞)

62.

Increasing on (−∞, ∞)

64.

Constant on (−∞, ∞)

65.

Decreasing on (−∞, 0] ; constant on [0, 1]; increasing on [1, ∞)

66.

Constant on (−∞, 0] ; decreasing on [0, 3] ; constant on [3, ∞)

67.

g and F are one-to-one since every horizontal line intersects the graph at one point. f, V, and p are not one-to-one since some horizontal lines intersect the graph at more than one point.

68.

s is one-to-one since every horizontal line intersects the graph at one point. t, m, r and k are not one-to-one since some horizontal lines intersect the graph at more than one point.

63.

Constant on (−∞, 0] ; increasing on [0, ∞)


110

69.


a.

2l + 2w = 50 2w = 50 − 2l w = 25 − l

b.

A = lw A = l (25 − l )

70.

a.

4 = 12 l d +l 4( d + l ) = 12l 4d + 4l = 12l 4d = 8l 1d =l 2 l (d ) = 1 d 2

b.

Domain: [0, ∞)

c.

l (8) = 1 (8) = 4 ft 2

A = 25l − l 2

71.

v(t ) = 80,000 − 6500t ,

0 ≤ t ≤ 10

73.

a.

C ( x) = 5(400) + 22.80 x = 2000 + 22.80 x

b.

R ( x) = 37.00 x

c.

P ( x) = 37.00 x − C ( x) = 37.00 − [2000 + 22.80 x] = 37.00 x − 2000 − 22.80 x = 14.20 x − 2000

72.

v(t ) = 44,000 − 4200t ,

74.

a.

V = lwh V = (30 − 2 x)(30 − 2 x)( x) V = (900 −120 x + 4 x 2 )( x) V = 900 x −120 x 2 + 4 x3

V = lwh ⇒ the domain of V is dependent on the domains of l, w, and h. Length, width and height must be positive values ⇒ 30 − 2 x > 0 and x > 0. −2 x > −30 x < 15 Thus, the domain of V is {x | 0 < x < 15}.

b.

Note x is a natural number. 75.

0≤t ≤8

15 = 15 − h 3 r 15 5= −h r 5r =15 − h h =15 − 5r h(r ) =15 − 5r

76.

r =2 h 4 r=2h 4 1 r= h 2 1 V = π r 2h 3

a.

b.

2

⎛ ⎞ ⎛ ⎞ V = 1 π ⎜ 1 h ⎟ h = 1 π ⎜ 1 h2 ⎟ h 3 ⎝2 ⎠ 3 ⎝4 ⎠ V = 1 π h3 12 77.

d = (3t ) 2 + (50) 2

78.

d = 9t 2 + 2500 meters, 0 ≤ t ≤ 60

t= t=

d r 1 + x2 3 − x + hours 2 8


Section 2.2

79.

111

d = (45 − 8t )2 + (6t )2 miles where t is the number of hours after 12:00 noon

80.

A = xy

a.

A( x ) = x ⎛⎜ − 1 x + 4 ⎞⎟ ⎝ 2 ⎠ 2 1 A( x ) = − x + 4 x 2 b.

Circle C = 2π r x = 2π r r= x 2π

a.

Square C = 4s 20 − x = 4s s = 5− x 4

⎛ ⎞ Area = π r 2 = π ⎜ x ⎟ ⎝ 2π ⎠

2

82.

⎛ ⎞ Area = s 2 = ⎜ 5 − x ⎟ ⎝ 4⎠

2

= 25 − 5 x + x 2 16

2 2 Total Area = x + 25 − 5 x + x 4π 2 16 ⎛ 1 1⎞ 2 5 =⎜ + ⎟ x − x + 25 2 ⎝ 4π 16 ⎠

83.

x

0

4

8

12

16

20

Total Area

25

17.27

14.09

15.46

21.37

31.83

Domain: [0, 20].

a.

Left side triangle 2

2

c = 20 + (40 − x )

2

2

c = 30 + x

2

c = 900 + x 2

c = 400 + (40 − x )2

6

7

Area

3.5

6

8

6

3.5

Domain: (2, ∞)

b.

84.

Right side triangle 2

4

2 = x x−2

b.

c.

2

mPB = 0 − 2 = −2 x−2 x−2 0− y − y = m AB = x −0 x mPB = m AB −2 = − y x−2 x 2x = y x−2 Area = 1 bh = 1 xy 2 2 x 1 2 = x 2 x−2

a.

2

2

=x 4π

1

Domain: [0, 8].

c. 81.

x

Total length = 900 + x 2 + 400 + (40 − x )2

p

40

50

60

75

90

f(p)

4900

4300

3800

3200

2800

answers accurate to nearest 100 feet

b.

0

10

20

30

40

Total Length

74.72

67.68

64.34

64.79

70

Domain: [0, 40].

c. 85.

x

x

5

10

12.5

15

20

Y(x)

275

375

385

390

394

answers accurate to nearest apple

86.

x

100

200

500

750

1000

C(x)

57,121

59,927

65,692

69,348

72,507

answers accurate to nearest dollar


112

87.


f (c) = c2 − c − 5 = 1

88.

g ( c ) = −2c 2 + 4c − 1 = −4 = −2 c 2 + 4 c + 3 = 0

c2 − c − 6 = 0 ( c − 3)( c + 2) = 0 c−3=0 c=3

or

c=

c+2=0 c = −2

−4 ± 42 − 4( −2)(3) 2( −2)

c = −4 ± 16 + 24 −4 c = −4 ± 40 −4 − ± 4 2 10 c= −4 c = 2 ± 10 2

89.

1 is not in the range of f(x), since x −1 only if x + 1 = x − 1 or 1 = −1. 1= x +1

91.

Set the graphing utility to “dot” mode.

90.

0 is not in the range of g(x), since 1 only if ( x − 3)(0) = 1 or 0 = 1. 0= x−3

WINDOW FORMAT Xmin=-4.7 Xmax=4.7 Xscl=1 Ymin=-5 Ymax=2 Yscl=1

92.

Set the graphing utility to “dot” mode. WINDOW FORMAT Xmin=-5 Xmax=5 Xscl=1 Ymin=-5 Ymax=3 Yscl=1

93.

WINDOW FORMAT Xmin=-4.7 Xmax=4.7 Xscl=1 Ymin=-5 Ymax=1 Yscl=1

94.

WINDOW FORMAT Xmin=-5 Xmax=5 Xscl=1 Ymin=-5 Ymax=5 Yscl=1


Section 2.2

113

95.

96.

....................................................... 97.

f ( x) 32 = (9 − 3) − ( 4 − 2) = 6 − 2 = 4

Connecting Concepts f ( x) 74 = (−21 + 2) − (−12 + 2)

98.

= −19 − (−10) = −19 + 10 = −9

99.

f ( x) 02 = (16 − 12 − 2) − 0 = 2

101. a.

f ( x) 80 = 0 − 8 = − 8 = −2 2

100.

f (1, 7) = 3(1) + 5(7) − 2 = 3 + 35 − 2 = 36

b.

f (0, 3) = 3(0) + 5(3) − 2 = 13

c.

f ( −2, 4) = 3( −2) + 5(4) − 2 = 12

d.

f (4, 4) = 3(4) + 5(4) − 2 = 30

e.

f ( k , 2k ) = 3( k ) + 5(2k ) − 2 = 13k − 2

f.

f ( k + 2, k − 3) = 3( k + 2) + 5( k − 3) − 2 = 3k + 6 + 5k − 15 − 2 = 8k − 11

102. a.

g (3, − 4) = 2(3)2 − −4 + 3 = 18 − 4 + 3 = 17

b.

g ( −1, 2) = 2( −1)2 − 2 + 3 = 2 − 2 + 3 = 3

c.

g (0, − 5) = 2(0)2 − −5 + 3 = −2

d.

g ⎛⎜ 1 , − 1 ⎞⎟ = 2 1 2 ⎝2 4⎠

e.

g ( c, 3c ) = 2( c ) 2 − 3c + 3 = 2c 2 − 3c + 3

f.

()

2

− − 1 + 3 = 1 − 1 + 3 = 13 4 2 4 4 2

g ( c + 5, c − 2) = 2( c + 5) − c − 2 + 3

( 3c = 3c since c > 0) (Since c < 0, c − 2 < 0)

= 2c 2 + 20c + 50 − ( −( c − 2)) + 3 2

(Thus c − 2 = − c + 2)

2

= 2c + 20c + 50 + c − 2 + 3 = 2c + 21c + 51 103.

5 + 8 + 11 = 12 2 A(5, 8, 11) = 12(12 − 5)(12 − 8)(12 − 11)

s=

104.

C (18, 11) = 15(18) + 14(11) = 270 + 154 = $424

= 12(7)(4)(1) = 336 = 4 21


114

105.


a 2 + 3a − 3 = a

106.

2

a + 2a − 3 = 0 ( a − 1)( a + 3) = 0 a =1

or

a =a a+5 a = a ( a + 5)

107.

108.

a = a 2 + 5a

a = −3

0 = a 2 + 4a 0 = a ( a + 4) a = 0 or a = −4

....................................................... PS1. d = 5 − ( −2) = 7

PS3.

Prepare for Section 2.3 PS2. The product of any number and its negative reciprocal is –1. −7 ⋅ 1 = − 1 7

−4 − 4 = −8 2 − ( −3) 5

PS4.

y − 3 = −2( x − 3) y − 3 = −2 x + 6 y = −2 x + 9

PS5. 3x − 5 y = 15 −5 y = −3x + 15

PS6.

y = 3x−3 5

y = 3x − 2(5 − x ) 0 = 3x − 2(5 − x ) 0 = 3x − 10 + 2 x 10 = 5 x 2=x

Section 2.3 1.

y −y 7−4 3 3 m= 2 1 = = =− x2 − x1 1 − 3 − 2 2

4.

m=

4−4 0 = =0 2 − (−3) 5

6.

m=

0−0 0 = =0 3−0 3

9.

7 1 − m= 2 2 = 7 − (−4) 3

11.

m=

13.

m=

2.

m=

1− 4 −3 3 = =− 5 − ( −2) 7 7 5.

7.

6 2 = 3⋅ 3 = 9 19 19 19 3

m=

3.

m=

2−0 1 =− 0−4 2

m = 4 − 0 = 4 undefined 0−0 0

−6 −2 − 4 =6 = − 4 − (−3) − 1

8.

m=

4 − (−1) 5 = − 3 − (−5) 2

2 − 4 −2 8 = =− 7 1 5 5 − 4 2 4

10.

m=

f (3 + h) − f (3) f (3 + h) − f (3) = 3+ h −3 h

12.

m=

f ( − 2 + h ) − f ( −2 + h ) 0 = =0 − 2 + h − (−2) h

f ( h ) − f ( 0) f ( h ) − f ( 0) = h−0 h

14.

m=

f ( a + h ) − f ( a ) f ( a + h) − f ( a ) = a+h−a h


Section 2.3

15.

115

m=2 y-intercept (0, –4)

m = –1 y-intercept (0, 1)

16.

17.

m = −1

3

18.

m= 2 3

y-intercept (0, –2)

y-intercept (0, 4)

19.

m=0 y-intercept (0, 3)

20.


21.


22.


23.


24.

m=1 y-intercept (0, –4)

25.

m= −3

26.

m= −2

4

y-intercept (0, 4)

27.

Use y = mx + b with m = 1, b = 3. y = x+3

29.

Use y = mx + b with m = y=

31.

28.

3 1 ,b= . 4 2

30.

3 1 x + 4 2

Use y = mx + b with m = 0, b = 4. y=4

32.

Use y = mx + b with m = −2, b = 5. y = −2 x + 5 2 3 Use y = mx + b with m = − , b = . 3 4 2 3 y=− x + 3 4

Use y = mx + b with m = y=

33.

y − 2 = −4( x − (−3)) y − 2 = −4 x −12 y = −4 x −10

34.

y +1= −3( x + 5) y = −3x −15 −1 y = −3x −16

35.

3

y-intercept (0, –2)

1 , b = −1. 2

1 x −1 2

4 −1 −1 − 3 3 3 = =− 4 −4 3 y − 1 = − ( x − 3) 4 3 9 4 y = − x+ + 4 4 4 3 13 y = − x+ 4 4

m=


36.

−8 − (−6) 2−5 −2 2 = = −3 3 2 y − (−6) = ( x − 5) 3 2 10 y+6 = x− 3 3 2 10 y = x− −6 3 3 2 28 y = x− 3 3

m=

116

37.

41.

45.

47.


m = −1 − 11 2−7 = −12 = 12 −5 5 12 y − 11 = ( x − 7) 5 12 y − 11 = x − 84 5 5 12 84 y= x− + 55 5 5 5 12 29 = x− 5 5 f ( x) = 1 − 4 x = 3 −4 x = 2 x=−1 2

38.

42.

−4 − 6 −3 − (−5) −10 = = −5 2 y − 6 = −5( x + 5) y − 6 = −5 x − 25 y = −5 x − 25 + 6 y = −5 x − 19

m=

f ( x) = 2 x + 2 = 4 3 2x = 2 3 x = 2 ⎛⎜ 3 ⎞⎟ ⎝2⎠ x=3

f ( x ) = 3x − 12 3x − 12 = 0 3x = 12

39.

f ( x ) = 2 x + 3 = −1

40.

f ( x ) = 4 − 3x = 7

2 x = −4 x = −2

43.

f ( x) = 3 − x = 5 2 x − =2 2 x = 2( −2)

−3 x = 3 x = −1

44.

f ( x ) = 4 x − 3 = −2 4x = 1 x=1 4

x = −4 46.

f ( x ) = −2 x − 4 −2 x − 4 = 0 −2 x = 4

x=4 The x-intercept of the graph of f ( x ) is (4,0).

x = −2 The x-intercept of the graph of f ( x ) (x) is (−2,0).

Xmin = − 4, Xmax = 6, Xscl=2, Ymin = −12.2, Ymax = 2, Yscl = 2

Xmin = − 4, Xmax = 6, Xscl=2, Ymin = −12.2, Ymax = 2, Yscl = 2

f ( x) = 1 x + 5 4

48.

f ( x) = − 1 x + 2 3

1 x+5 = 0 4 1 x = −5 4 x = −20 The x-intercept of the graph of f ( x ) is (−20,0).

− 1 x+2=0 3 − 1 x = −2 3 x =6 The x-intercept of the graph of f ( x ) is (6,0).

Xmin = −30, Xmax = 30, Xscl = 10, Ymin = −10, Ymax = 10, Yscl = 1



Section 2.3

49.

117

Algebraic method: f1( x) = f 2 ( x) 4x+5 = x +6 3 x =1 x=1 3 Graphical method: Graph y = 4 x + 5 and y = x+6 They intersect at x =

50.

1 1 , y=6 . 3 3

Xmin = −6, Xmax = 6, Xscl = 2, Ymin = −7, Ymax =7.8, Yscl = 2

Xmin = −7.8, Xmax = 7.8, Xscl = 2, Ymin = −2 ,Ymax = 10, Yscl = 2 51.

Algebraic method:

f1 ( x ) = f 2 ( x )

f1( x) = f 2 ( x) −2 x −11= 3 x + 7 −5 x =18 x = − 18 5 Graphical method: Graph y = −2 x −11 and y = 3 x + 7. They intersect at x = −3.6, y = −3.8.

Algebraic method:

52.

2 x − 4 = − x + 12 3x = 16 x = 16 3 Graphical method: Graph y = 2 x − 4 and y = − x + 12

f1 ( x ) = f 2 ( x )

Algebraic method:

1 x+5= 2 x−7 2 3 1 ⎛ ⎞ 6 ⎜ x + 5 ⎟ = 6 ⎜⎛ 2 x − 7 ⎟⎞ ⎝2 ⎠ ⎝3 ⎠ 3x + 30 = 4 x − 42 72 = x Graphical method: Graph y = 1 x + 5 and 2 y = 2 x−7 3 They intersect at x = 72, y = 41.

1 2 They intersect at x = 5 , y = 6 . 3 3

Xmin = − 4, Xmax = 10, Xscl = 2, Ymin = −2, Ymax = 10, Yscl = 2 Xmin = − 20, Xmax = 120, Xscl = 20, Ymin = −20, Ymax = 100, Yscl = 20 53.

55.

m = 1505 −1482 = 2.875 28 − 20 The value of the slope indicates that the speed of sound in water increases 2.875 m for a one-degree increase in temperature.

a.

b.

m = 29 −13 ≈1.45 20 − 9 H (c) −13 =1.45(c − 9) H (c) =1.45c H (18) =1.45(18) ≈ 26 mpg

54.

56.

m = 4 −1 = 0.04 100 − 25 The value of the slope indicates that the file is being downloaded at 0.04 megabytes per second.

a.

b.

m = 799.1− 675.7 = 24.68 2005 − 2000 C (t ) − 675.7 = 24.68(t − 2000) C (t ) = 24.68t − 48,684.3 900 = 24.68t − 48,684.3 49584.3 = 24.68t 2009.1≈ t The debt will exceed $900 billion in 2009.


118

57.

59.


m=

b.

60,000 = 2500t − 4,962,000 5,022,000 = 2500t 2008.8 = t The number of jobs will exceed 60,000 in 2008.

a.

b.

c. 61.

63,000 − 38,000 = 2500 2010 − 2000 N (t ) − 63,000 = 2500(t − 2010) N (t ) = 2500t − 4,962,000

a.

m = 240 −180 = 30 18 −16 B (d ) −180 = 30(d −16) B (d ) = 30d − 300 The value of the slope means that a 1-inch increase in the diameter of a log 32 ft long results in an increase of 30 board-feet of lumber that can be obtained from the log. B (19) = 30(19) − 300 = 270 board feet

Line A represents Michelle Line B represents Amanda Line C represents the distance between Michelle and Amanda.

58.

a.

b. c.

60.

a.

b. c.

62.

a. b.

c. 63.

a.

b.

65.

Find the slope of the line. 180 − 110 70 m= = ≈ 1.842 108 − 70 38 Use the point-slope formula to find the equation. y − y1 = m( x − x1)

64.

a.

T (180) = −10(180) + 2350 = 550o F After 3 hours, the temperature will be 550°F. m = 1640 − 800 = 42 60 − 40 E (T ) − 800 = 42(T − 40) E (T ) = 42T − 880 The value of the slope means that an additional 42 acre-feet of water evaporate for a one degree increase in temperature. E (75) = 42(75) −880 = 2270 acre-feet

m AB = 1− 9 = −4o F 8− 6 m AB = 1− 9 = −4o F 8−6 mDE = −4 − 5 = 9o F 5−6 The temperature changed most rapidly between points D and E. The temperature remained constant (zero slope) between points C and D.

Find the slope of the line. m = 11.2 − 76.5 = −65.3 ≈ −0.87 75 − 0 75 Use the point-slope formula to find the equation. y − y1 = m( x − x1) y − 76.5 = −0.87( x − 0) y − 76.5 = −0.87 x y = −0.87 x + 76.5

y − 110 = 1.842( x − 70) y − 110 = 1.842 x − 128.94 y = 1.842 x − 18.94 y = 1.842(90) − 18.94 y = 165.78 − 18.94 y = 146.84 ≈ 147

b.

P( x ) = 92.50 x − (52 x + 1782) P( x ) = 92.50 x − 52 x − 1782 P( x ) = 40.50 x − 1782

m = 2200 − 2150 = −10 15 − 20 T (t ) − 2200 = −10(t −15) T (t ) = −10t + 2350 The value of the slope means that the temperature is decreasing at a rate of 10 degrees per minute.

66.

y = −0.87( 25) + 76.5 y = −21.75 + 76.5 y = 54.75 ≈ 55 years

P( x ) = 124 x − (78.5 x + 5005) P( x ) = 124 x − 78.5 x − 5005 P( x ) = 45.5 x − 5005 45.5 x − 5005 = 0

40.50 x − 1782 = 0

45.5 x = 5005

40.50 x = 1782 x = 1782 40.50 x = 44, the break-even point

x = 5005 45.5 x = 110, the break-even point


Section 2.3

67.

119

P( x ) = 259 x − (180 x + 10,270) P( x ) = 259 x − 180 x − 10, 270 P( x ) = 79 x − 10,270

68.

79 x − 10,270 = 0

6210 x − 1,602,180 = 0 6210 x = 1,602,180 1,602,180 x= 6210 x = 258, the break-even point

79 x = 10.270 x = 10.270 79 x = 130, the break-even point

69.

a. b. c. d.

C (0) = 8(0) + 275 = 0 + 275 = $275 C (1) = 8(1) + 275 = 8 + 275 = $283 C (10) = 8(10) + 275 = 80 + 275 = $355 The marginal cost is the slope of C ( x) = 8 x + 275, which is $8 per unit.

70.

71.

a. b. c.

C (t ) = 19,500.00 + 6.75t R (t ) = 55.00t P (t ) = R(t ) − C (t ) P (t ) = 55.00t − (19,500.00 + 6.75t ) P (t ) = 55.00t −19,500.00 − 6.75t P (t ) = 48.25t −19,500.00 48.25t =19,500.00 19,500.00 t= 48.25 t = 404.1451 days ≈ 405 days

72.

d.

P( x ) = 14,220 x − (8010 x + 1,602,180) P( x ) = 14,220 x − 8010 x − 1,602,180 P( x ) = 6210 x − 1,602,180

R (0) = 210(0) = $0 R (1) = 210(1) = $210 R (10) = 210(10) = $2100 The marginal revenue is the slope of R ( x) = 210 x, which is $210 per unit.

a. b. c. d.

m=

a.

b.

c.

117,500 − 98,000 19,500 = = 6.5 35,000 − 32,000 2000 P ( s ) − 98,000 = 6.5( s − 32,000) P ( s ) = 6.5s − 208,000 + 98,000 P ( s ) = 6.5s −110,000 P (50,000) = 6.5(50,000) −110,000 = 325,000 −110,000 = $215,000 Let 6.5s − 110,000 = 0. Then 6.5s = 110,000 s=

73.

75.

3 The graph of 3x + y = −24 has m = − . 4 3 y − 3 = − ( x − 1) 4 3 3 y = − x+ +3 4 4 3 15 y =− x+ 4 4

74.

The graph of y = − 1 x + 6 has m = − 1 . 2 2 1 y − 10 = − ( x − 4) 2 y = − 1 x + 2 + 10 2 y = − 1 x + 12 2

76.

110,000 ≈ 16,924 subscribers 6.5

The graph of x + y = 10 has m = −1. y + 1 = (−1)( x − 2) y = −x + 2 −1 y = −x +1

The graph of y = 5 x + 5 has m = 5 . 2 2 5 y + 2 = ( x − 10) 2 y = 5 x − 25 − 2 2 y = 5 x − 27 2


120


77.

The graph of y = 3 x + 6 has m = 3 . 2 2 Thus we use a slope of − 2 . 3 2 y − 7 = − ( x + 9) 3 2 y = − x−6+7 3 2 y = − x +1 3

78.

The graph of y = − 2 x + 3 has m = − 2 . 3 3 Thus we use a slope of 3 . 2 3 y + 6 = ( x − 4) 2 y = 3 x−6−6 2 y = 3 x − 12 2

79.

The graph of y = 4 x − 9 has m = 4 . 7 7 4 y − 3 = ( x − 4) 7 y = 4 x − 16 + 3 7 7 4 5 y= x+ 7 7

80.

The graph of y = 9 x + 2 has m = 9 . 7 7 7 Thus we use a slope of − . 9 y + 3 = − 7 ( x − 1) 9 y = −7 x+ 7 −3 9 9 7 y = − x − 20 9 9

81.

The graph of y = 1 x + 9 has m = 1 . 2 2 Thus we use a slope of –2. y + 1 = −2( x + 3) y = −2 x − 6 − 1 y = −2 x − 7

82.

The graph of y = − 5 x + 7 has m = − 5 . 4 4 5 y + 5 = − ( x + 2) 4 y = −5 x − 5 −5 4 2 5 15 y=− x− 4 2

83.

The graph of x + y = 4 has m = −1. Thus we use a slope of 1. y − 2 = 1( x − 1) y = x −1+ 2 y = x +1

84.

The graph of 2 x − y = 7 has m = 2. 1 Thus we use a slope of − . 2 1 y − 4 = − ( x + 3) 2 1 3 8 y =− x− + 2 2 2 1 5 y =− x+ 2 2


Section 2.3

85.

121

The equation of the line through (0,0) and P(3,4) has 4 slope . 3 The path of the rock is on the line through P(3,4) with 3 slope − , so y − 4 = − 3 ( x − 3). 4 4 y −4 = − 3 x+ 9 4 4 y = − 3 x+ 9 +4 4 4 y = − 3 x + 25 4 4 The point where the rock hits the wall at y = 10 is the point 3 25 of intersection of y = − x + and y = 10. 4 4

86.

3 25 − x+ = 10 4 4 −3 x + 25 = 40 −3 x = 15 x = −5 feet Therefore the rock hits the wall at (−5, 10). The x-coordinate is –5.

87.

a.

The path of the rock is on the line through P ( 15, 1) with slope − 15 so y −1= − 15( x − 15) y −1= − 15 x +15 y = − 15 x +15 +1 y = − 15 x +16 The point of impact with the wall at y = 14 is the point of

intersection of y = − 15 x + 16 and y = 14 intersect. − 15 x + 16 = 14 − 15 x = −2 2 x= ≈ 0.52 feet 15 ⎛ 2 ⎞ , 14 ⎟ . Therefore, the rock hits the wall at ⎜ 15 ⎝ ⎠ 2 The x-coordinate is or approximately 0.52. 15

h = 1 so Q (2 + h, [ 2 + h ] + 1) = Q (3, 32 + 1) = Q (3, 10) 2

m=

b.

The equation of the line through (0,0) and 1 . P ( 15 , 1) has slope 15

10 − 5 5 = =5 3− 2 1

h = 0.1 so Q (2 + h, [ 2 + h ] + 1) = Q (2.1, 2.12 + 1) = Q (2.1, 5.41) 2

m=

5.41 − 5 0.41 = = 4.1 2.1 − 2 0.1

c.

h = 0.01 so Q (2 + h, [ 2 + h ] + 1) = Q (2.01, 2.012 + 1) = Q (2.01, 5.0401)

d.

5.0401 − 5 0.0401 = = 4.01 2.01 − 2 0.01 As h approaches 0, the slope of PQ seems to be approaching 4.

e.

x1 = 2, y1 = 5, x2 = 2 + h, y2 = [2 + h]2 + 1

2

m=

2

[ 2 + h] + 1 − 5 = (4 + 4h + h2 ) + 1 − 5 = 4h + h2 = 4 + h y −y m= 2 1= x2 − x1 (2 + h) − 2 h h


122

88.


a.

h = 1 so Q ( −1 + h, 3[ −1 + h ] ) = Q (0, 0) 2

m=

b.

0−3 −3 = = −3 0 − ( −1) 1

h = 0.1 so Q ( −1 + h, 3[ −1 + h ] ) = Q ( −0.9, 3( −0.9) 2 ) = Q ( −0.9, 2.43) 2

m=

2.43 − 3 −0.57 = = −5.7 0 . 9 ( 1 ) 0. 1 − − −

c.

h = 0.01 so Q ( −1 + h, 3[ −1 + h ] ) = Q ( −0.99, 3( −0.99)2 ) = Q ( −0.99, 2.9403)

d.

2.9403 − 3 −0.0597 = = −5.97 0.01 − 0.99 − (−1) As h approaches 0, the slope of PQ seems to be approaching −6.

e.

x1 = −1, y1 = 3, x2 = −1 + h, y2 = 3[ −1 + h ]

2

m=

2

y2 − y1 3[ −1 + h ] − 3 3(1 − 2h + h 2 ) − 3 3 − 6h + 3h 2 − 3 −6h + 3h 2 = = = = = −6 + 3h x2 − x1 ( −1 + h ) − ( −1) h h h 2

m=

89.

m=

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

90.

m=

4( x + h )2 − 4 x 2 4( x 2 + 2 xh + h 2 ) − 4 x 2 4 x 2 + 8 xh + 4h 2 − 4 x 2 8 xh + 4h 2 h(8 x + 4h ) = = = = = 8 x + 4h x+h−x h h h h

.......................................................

Connecting Concepts

y − y1 y2 − y1 ( x − x1 ) , the two-point form. for m in the point-slope form y − y1 = m( x − x1 ) to yield y − y1 = 2 x2 − x1 x2 − x1

91.

Substitute

92.

y − 0 = b − 0 ( x − a) 0−a y = b ( x − a) −a y = − bx + b a bx + y = b Then divide by b to produce x + y = 1. a a b

93.

y − 1 = 3 − 1 ( x − 5) 4−5 y − 1 = 2 ( x − 5) −1 y = −2( x − 5) y − 1 = −2 x + 10 y = −2 x + 10 + 1 y = −2 x + 11

94.

y − 7 = 6 − 7 ( x − 2) −1 − 2 y − 7 = −1 ( x − 2) −3 1 y − 7 = ( x − 2) 3 1 y−7= x− 2 3 3 y = 1 x − 2 + 21 3 3 3 1 19 y= x+ 3 3

95.

x y + = 1 with a = 3 and b = 5. a b x y + =1 3 5 ⎛x y⎞ 15⎜ + ⎟ = 15(1) ⎝3 5⎠ 5 x + 3 y = 15 Use


Section 2.3

96.

98.

123

x y + = 1 with a = −2 and b = 7. a b x y + =1 −2 7 y⎞ ⎛ x 14⎜ + ⎟ = 14(1) ⎝−2 7⎠ − 7 x + 2 y = 14 Use

x y + = 1 Since (−3, 10) is on the line, − b 2b −3 + 10 = 1 − b 2b 2b ⎛⎜ 3 + 10 ⎞⎟ = 2b(1) ⎝ b 2b ⎠ 6 + 10 = 2b 16 = 2b 8=b x + y =1 −8 16 −2 x + y = 16

100. The slope of the radius from (0, 0) to (x, y) is 0.5, so

97.

99.

x y + = 1 with b = 3a. a b x + y =1 Since (5, 2) is on the line, a 3a 5 + 2 =1 a 3a ⎛ ⎞ 3a ⎜ 5 + 2 ⎟ = 3a (1) ⎝ a 3a ⎠ 15 + 2 = 3a 17 = 3a 17 = a 3 y x Thus + =1 ⎛ 17 ⎞ 3 ⎛ 17 ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ 3⎠ ⎝ 3⎠ 3x + y = 1 17 17 3x + y = 17 Use

3(1 + h )3 − 3 3(1 + 3h + 3h 2 + h 3 ) − 3 = h 1+ h −1 2 3 = 3 + 9h + 9h + 3h − 3 h 2 3 = 9h + 9h + 3h h

=

h (9 + 9h + 3h 2 ) h

= 9 + 9h + 3h 2

y−0 y = = 0.5 thus y = 0.5 x. x−0 x

Substitute y = 0.5x into x 2 + y 2 = 25. x 2 + (0.5 x ) 2 = 25 x 2 + 0.25 x 2 = 25 1.25 x 2 = 25 x 2 = 20 x = ± 20 = ±2 5 If x = 2 5, then y = 0.5(2 5) = 5. If x = −2 5, then y = 0.5(−2 5) = − 5. The points are (2 5,

5) and ( − 2 5, − 5).


124


101. The slope of the line through (3, 9) and (x, y) is

15 y − 9 15 , so = . 2 x−3 2

2( y − 9) = 15( x − 3)

Therefore

2 y − 18 = 15 x − 45 Substitute y = x 2 into this equation.

2 y − 15 x + 27 = 0 2x 2 − 15 x + 27 = 0 (2 x − 9)( x − 3) = 0

x = 9 or x = 3 2

If x =

2

81 9 ⎛ 9 81 ⎞ ⎛9⎞ , y = x2 = ⎜ ⎟ = ⇒ ⎜ , ⎟. 4 2 ⎝2 4 ⎠ ⎝2⎠

If x = 3, y = x 2 = (3)2 = 9 ⇒ (3, 9 ), but this is the point itself. ⎛ 9 81 ⎞ 15 ⎛ 9 81 ⎞ The point ⎜ , ⎟ is on the graph of y = x 2 , and the slope of the line containing (3, 9) and ⎜ , ⎟ is . ⎝2 4 ⎠ 2 ⎝2 4 ⎠ 3 y−2 3 102. The slope of the line through (3, 2) and (x, y) is , so = . 8 x−3 8 Therefore 8( y − 2) = 3( x − 3).

8 y − 16 = 3x − 9 8 y = 3x + 7

Substitute y = x + 1 into this equation.

8 x + 1 = 3x + 7

(8

x + 1 ) = ( 3x + 7 ) 2

2

64( x + 1) = 9 x 2 + 42 x + 49 64 x + 64 = 9 x 2 + 42 x + 49 0 = 9 x 2 − 22 x − 15 0 = (9 x + 5)( x − 3) x = − 5 or x = 3 9 5 If x = − , y = 9 If x = 3, y =

5 9 x +1 = − + = 9 9

4 2 ⎛ 5 2⎞ = ⇒ ⎜ − , ⎟. 9 3 ⎝ 9 3⎠

x + 1 = 3 + 1 = 4 = 2 ⇒ ( 3, 2 ) , but this is the point itself.

⎛ 5 2⎞ The point ⎜ − , ⎟ is on the graph of y = ⎝ 9 3⎠

⎛ 5 2⎞ 3 x + 1, and the slope of the line containing (3, 2) and ⎜ − , ⎟ is . ⎝ 9 3⎠ 8

....................................................... PS1. 3x 2 + 10 x − 8 = (3x − 2)( x + 4) PS3.

f ( −3) = 2( −3)2 − 5( −3) − 7 = 18 + 15 − 7 = 26

Prepare for Section 2.4 PS2. x 2 − 8 x = x 2 − 8 x + 16 = ( x − 4)2 PS4.

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

x −1= 0 x =1


Section 2.4

125

PS5. x 2 + 3x − 2 = 0 x=

53 = −16t 2 + 64t + 5

PS6.

16t 2 − 64t + 48 = 0

−3 ± (3)2 − 4(1)( −2) 2(1)

t 2 − 4t + 3 = 0

= −3 ± 17 2

(t − 1)(t − 3) = 0 t =1, 3

Section 2.4 1.

d

2.

f

3.

b

4.

h

5.

g

6.

e

7.

c

8.

a

9.

f ( x) = ( x 2 + 4 x) +1

10.

= ( x 2 + 4 x + 4) +1− 4

= ( x 2 + 6 x + 9) −1− 9

= ( x + 2)2 − 3 standard form, vertex (−2, −3), axis of symmetry x = −2

11.

f ( x) = ( x 2 + 6 x) −1

f ( x ) = ( x 2 −8 x) + 5

= ( x + 3)2 −10 standard form, vertex (−3, −10), axis of symmetry x = −3

12.

= ( x 2 −8 x +16) + 5 −16

f ( x) = ( x 2 −10 x) + 3 = ( x 2 −10 x + 25) + 3 − 25

= ( x − 4)2 −11 standard form, vertex (4, −11), axis of symmetry x = 4

= ( x − 5)2 − 22 standard form, vertex (5, −22), axis of symmetry x = 5


126

13.


f ( x) = ( x 2 + 3 x) +1

14.

⎛ ⎞ = ⎜ x 2 + 3 x + 9 ⎟ +1− 9 4⎠ 4 ⎝

⎛ ⎞ = ⎜ x 2 + 7 x + 49 ⎟ + 2 − 49 4 ⎠ 4 ⎝

2

2

⎛ ⎞ =⎜ x+ 3 ⎟ + 4 − 9 ⎝ 2⎠ 4 4 2

⎛ ⎞ =⎜ x+ 3 ⎟ ⎝ 2⎠ ⎛ 3 vertex ⎜ − , − ⎝ 2

15.

⎛ ⎞ = ⎜ x + 7 ⎟ + 8 − 49 ⎝ 2⎠ 4 4 2

⎛ ⎞ = ⎜ x + 7 ⎟ − 41 standard form, 4 ⎝ 2⎠ 7 ⎛ 7 41 ⎞ vertex ⎜ − , − ⎟ , axis of symmetry x = − 2 4 2 ⎠ ⎝

− 5 standard form, 4 5⎞ 3 ⎟ , axis of symmetry x = − 2 4⎠

f ( x) = − x 2 + 4 x + 2

16.

f ( x) = − x 2 − 2 x + 5

= −( x 2 − 4 x ) + 2

= −( x 2 + 2 x ) + 5

= −( x 2 − 4 x + 4) + 2 + 4

= −( x 2 + 2 x +1) + 5 +1

= −( x − 2)2 + 6 standard form, vertex (2, 6), axis of symmetry x = 2

17.

f ( x ) = ( x 2 + 7 x) + 2

f ( x) = −3 x 2 + 3 x + 7

= −( x +1)2 + 6 standard form, vertex (−1, 6), axis of symmetry x = −1

18.

2

f ( x) = −2 x 2 − 4 x + 5

= −3( x −1x) + 7

= −2( x 2 + 2 x) + 5

⎛ ⎞ = −3 ⎜ x 2 −1x + 1 ⎟ + 7 + 3 4⎠ 4 ⎝

= −2( x 2 + 2 x +1) + 5 + 2 = −2( x +1)2 + 7 standard form, vertex (−1, 7), axis of symmetry x = −1

2

⎛ ⎞ = −3 ⎜ x − 1 ⎟ + 28 + 3 4 4 ⎝ 2⎠ 2

⎛ ⎞ = −3 ⎜ x − 1 ⎟ + 31 4 ⎝ 2⎠

standard form,

1 ⎛ 1 31 ⎞ vertex ⎜ , ⎟ , axis of symmetry x = 2 ⎝2 4 ⎠


Section 2.4

19.

21.

23.

25.

x=

127

10 −b = =5 2a 2(1)

20.

x=

6 −b = =3 2a 2(1)

y = f (5) = (5) 2 − 10(5) = 25 − 50 = −25 vertex (5, − 25)

y = f (3) = (3)2 − 6(3) = 9 − 18 = −9 vertex (3, − 9)

f ( x) = ( x − 5)2 − 25

f ( x) = ( x − 3) 2 − 9

x=

0 −b = =0 2a 2(1)

22.

x=

0 −b = =0 2a 2(1)

y = f (0) = (0) 2 − 10 = −10 vertex (0, − 10)

y = f (0) = (0) 2 − 4 = −4 vertex (0, − 4)

f ( x) = x 2 − 10

f ( x) = x 2 − 4

x=

−b −6 −6 = = =3 2a 2(−1) − 2

24.

x=

−b −4 −4 = = =2 2a 2(−1) − 2

y = f (3) = −(3) 2 + 6(3) + 1 − 9 + 18 + 1 = 10 vertex (3, 10)

y = f (2) = −(2) 2 + 4(2) + 1 = −4 + 8 + 1 =5 vertex (2, 5)

f ( x) = −( x − 3)2 + 10

f ( x) = −( x − 2) 2 + 5

x=

3 3 −b = = 2a 2(2) 4 2

⎛3⎞ ⎛3⎞ ⎛3⎞ y = f ⎜ ⎟ = 2⎜ ⎟ − 3⎜ ⎟ + 7 ⎝4⎠ ⎝4⎠ ⎝4⎠ ⎛ 9⎞ 9 = 2⎜ ⎟ − + 7 ⎝ 16 ⎠ 4 9 9 = − +7 8 4 9 18 56 = − + 8 8 8 47 = 8 ⎛ 3 47 ⎞ vertex ⎜ , ⎟ ⎝4 8 ⎠ ⎛ f(x) = 2⎜ x − ⎝

2

3⎞ 47 ⎟ + 4⎠ 8

26.

x=

10 10 5 −b = = = 2a 2(3) 6 3 2

⎛5⎞ ⎛5⎞ ⎛5⎞ y = f ⎜ ⎟ = 3⎜ ⎟ − 10⎜ ⎟ + 2 ⎝3⎠ ⎝3⎠ ⎝3⎠ ⎛ 25 ⎞ 50 = 3⎜ ⎟ − +2 ⎝ 9 ⎠ 3 25 50 = − +2 3 3 25 50 6 = − + 3 3 3 19 =− 3 ⎛ 5 19 ⎞ vertex ⎜ , − ⎟ 3⎠ ⎝3 2

5⎞ 19 ⎛ f(x) = 3⎜ x − ⎟ − 3 3 ⎝ ⎠


128

27.


x=

−b −1 1 = = 2a 2(−4) 8

28.

2

6 6 3 −b = = =− 2a 2(−5) − 10 5 2

⎛1⎞ ⎛1⎞ ⎛1⎞ y = f ⎜ ⎟ = −4⎜ ⎟ + ⎜ ⎟ + 1 ⎝8⎠ ⎝8⎠ ⎝8⎠ 1 1 ⎛ ⎞ = −4⎜ ⎟ + + 1 ⎝ 64 ⎠ 8 1 1 = − + +1 16 8 1 2 16 =− + + 16 16 16 17 = 16 ⎛ 1 17 ⎞ vertex ⎜ , ⎟ ⎝ 8 16 ⎠

⎛ 3⎞ ⎛ 3⎞ ⎛ 3⎞ y = f ⎜ − ⎟ = −5⎜ − ⎟ − 6⎜ − ⎟ + 3 ⎝ 5⎠ ⎝ 5⎠ ⎝ 5⎠ 9 18 ⎛ ⎞ = −5⎜ ⎟ + +3 ⎝ 25 ⎠ 5 9 18 =− + +3 5 5 9 18 15 =− + + 5 5 5 24 = 5 ⎛ 3 24 ⎞ vertex ⎜ − , ⎟ ⎝ 5 5 ⎠

2

2

1⎞ 17 ⎛ f(x) = −4⎜ x − ⎟ + 8⎠ 16 ⎝ 29.

x=

3⎞ 24 ⎛ f(x) = −5⎜ x + ⎟ + 5⎠ 5 ⎝

f ( x) = x 2 − 2 x −1

30.

f ( x) = − x 2 − 6 x − 2

= ( x 2 − 2 x) −1

= −( x 2 + 6 x ) − 2

= ( x 2 − 2 x +1) −1−1

= −( x 2 + 6 x + 9) − 2 + 9

= ( x −1)2 − 2 vertex (1, −2) The y-value of the vertex is −2. The parabola opens up since a =1> 0. Thus the range is y y ≥ −2 .

{

f ( x) = 2 = x 2 − 2 x −1 0 = x2 − 2 x −3 0 = ( x − 3)( x +1) x − 3 = 0 or x + 1 = 0 x=3 x = −1

}

= −( x + 3)2 + 7 vertex (−3, 7) The y-value of the vertex is 7. The parabola opens down since a = −1 < 0. Thus the range is y y ≤ 7 .

{

}

f ( x) = 3 = − x 2 − 6 x − 2 x2 + 6 x + 5 = 0 ( x + 5)( x +1) = 0 x + 5 = 0 or x + 1 = 0 x = −5 x = −1


Section 2.4

31.

129

f ( x) = −2 x 2 + 5 x −1

32.

⎛ ⎞ = −2 ⎜ x 2 − 5 x ⎟ −1 2 ⎠ ⎝ ⎛ ⎞ ⎛ ⎞ = −2 ⎜ x 2 − 5 x + 25 ⎟ −1+ 2 ⎜ 25 ⎟ 2 16 ⎠ ⎝ ⎝ 16 ⎠

f ( x) = 2 x 2 + 6 x − 5 = 2( x 2 + 3 x) − 5 ⎛ ⎞ ⎛ ⎞ = 2 ⎜ x2 + 3x + 9 ⎟ − 5− 2 ⎜ 9 ⎟ 4⎠ ⎝ ⎝4⎠ 2

2

⎛ ⎞ = 2 ⎜ x + 3 ⎟ −5− 9 2 ⎝ 2⎠

2

⎛ ⎞ = 2 ⎜ x + 3 ⎟ − 10 − 9 2 2 ⎝ 2⎠

⎛ ⎞ = −2 ⎜ x − 5 ⎟ − 8 + 25 ⎝ 4⎠ 8 8

2

⎛ ⎞ = −2 ⎜ x − 5 ⎟ + 17 8 ⎝ 4⎠

2

⎛ ⎞ = 2 ⎜ x + 3 ⎟ − 19 2 ⎝ 2⎠ ⎛ 3 19 ⎞ vertex ⎜ − , − ⎟ 2⎠ ⎝ 2

⎛ 5 17 ⎞ vertex ⎜ , ⎟ ⎝4 8 ⎠ 17 . 8 The parabola opens down since a = −2 < 0. ⎧ 17 ⎫ Thus the range is ⎨ y y ≤ ⎬. 8⎭ ⎩

The y-value of the vertex is

19 . 2 The parabola opens up since a =2 > 0. ⎧ 19 ⎫ Thus the range is ⎨ y y ≥ − ⎬. 2⎭ ⎩

The y-value of the vertex is −

f ( x) = 2 = −2 x 2 + 5 x −1 2 x2 −5x + 3 = 0 (2 x − 3)( x −1) = 0 2 x − 3 = 0 or x − 1 = 0 3 x= x =1 2

f ( x) =15 = 2 x 2 + 6 x − 5 0 = 2 x 2 + 6 x − 20 0 = 2( x 2 + 3 x −10) 0 = 2( x − 2)( x + 5) x − 2 = 0 or x + 5 = 0

x=2 33.

f ( x) = x 2 + 3x + 6

34.

x = −5

f ( x) = −2 x 2 − x +1

2

⎛ ⎞ = −2 ⎜ x 2 + 1 x ⎟ +1 2 ⎠ ⎝ ⎛ ⎞ ⎛ ⎞ = −2 ⎜ x 2 + 1 x + 1 ⎟ +1+ 2 ⎜ 1 ⎟ 2 16 ⎠ ⎝ ⎝ 16 ⎠

2

⎛ ⎞ = −2 ⎜ x + 1 ⎟ + 8 + 1 ⎝ 4⎠ 8 8

2

⎛ ⎞ = −2 ⎜ x + 1 ⎟ + 9 ⎝ 4⎠ 8

= ( x 2 + 3 x) + 6 ⎛ ⎞ = ⎜ x2 + 3x + 9 ⎟ + 6 − 9 4⎠ 4 ⎝ ⎛ ⎞ = ⎜ x + 3 ⎟ + 6− 9 4 ⎝ 2⎠

2

⎛ ⎞ = ⎜ x + 3 ⎟ + 24 − 9 4 4 ⎝ 2⎠

2

⎛ ⎞ = ⎜ x + 3 ⎟ + 15 4 ⎝ 2⎠ ⎛ 3 15 ⎞ vertex ⎜ − , ⎟ ⎝ 2 4⎠

⎛ 1 9⎞ vertex ⎜ − , ⎟ ⎝ 4 8⎠

15 . 4 The parabola opens up since a =1 > 0. ⎧ 15 ⎫ Thus the range is ⎨ y y ≥ ⎬. 4⎭ ⎩


⎧ 15 ⎫ No, 3 ∉ ⎨ y y ≥ ⎬. 4⎭ ⎩

9 . 8 The parabola opens down since a = −2 < 0. ⎧ 9⎫ Thus the range is ⎨ y y ≤ ⎬. 8⎭ ⎩


⎧ Yes, −2 ∈ ⎨ y y ≤ ⎩

9⎫ ⎬. 8⎭


130

35.


f ( x) = x 2 + 8 x

36.

2

= −( x 2 + 6 x )

= ( x + 8 x +16) −16 2

= −( x 2 + 6 x + 9) + 9

= ( x + 4) −16 minimum value of –16 when x = −4

37.

f ( x) = − x 2 + 6 x + 2

= −( x + 3)2 + 9 maximum value of 9 when x = −3

38.

= −( x 2 −10 x) − 3

= −( x 2 − 6 x + 9) + 2 + 9

= −( x 2 −10 x + 25) − 3 + 25

f ( x) = 2 x 2 + 3 x +1

= −( x − 5)2 + 22 maximum value of 22 when x = 5

40.

2

2

⎛ ⎞ = 3 ⎜ x + 1 ⎟ − 12 − 1 ⎝ 6 ⎠ 12 12

2

⎛ ⎞ = 3 ⎜ x + 1 ⎟ − 13 6 ⎝ ⎠ 12

⎛ ⎞ = 2⎜ x+ 3 ⎟ + 8 − 9 ⎝ 4⎠ 8 8

2

⎛ ⎞ = 2⎜ x+ 3 ⎟ − 1 ⎝ 4⎠ 8 minimum value of −

1 3 when x = − 8 4

f ( x) = 5 x 2 −11

minimum value of −

42.

= 5( x 2 ) −11

13 1 when x = − 12 6

f ( x) = 3 x 2 − 41 = 3( x 2 ) − 41

= 5( x − 0)2 −11 minimum value of –11 when x = 0 43.

f ( x) = 3 x 2 + x −1 ⎛ ⎞ = 3 ⎜ x 2 + 1 x ⎟ −1 3 ⎠ ⎝ ⎛ 2 1 ⎞ ⎛ ⎞ = 3 ⎜ x + x + 1 ⎟ −1− 3 ⎜ 1 ⎟ 3 36 ⎠ ⎝ ⎝ 36 ⎠

⎛ ⎞ = 2 ⎜ x 2 + 3 x ⎟ +1 2 ⎠ ⎝ ⎛ 2 3 ⎞ ⎛ ⎞ = 2 ⎜ x + x + 9 ⎟ +1− 2 ⎜ 9 ⎟ 2 16 ⎠ ⎝ ⎝ 16 ⎠

41.

f ( x) = − x 2 +10 x − 3

= −( x 2 − 6 x ) + 2 = −( x − 3)2 +11 maximum value of 11 when x = 3

39.

f ( x) = − x 2 − 6 x

f ( x) = − 1 x 2 + 6 x +17 2 1 = − ( x 2 −12 x) +17 2 = − 1 ( x 2 −12 x + 36) +17 +18 2 1 = − ( x − 6)2 + 35 2 maximum value of 35 when x = 6

= 3( x − 0)2 − 41 minimum value of –41 when x = 0 44.

f ( x) = − 3 x 2 − 2 x + 7 4 5 ⎛ 2 8 ⎞ 3 = − ⎜ x + x ⎟+7 4⎝ 15 ⎠ ⎛ 2 8 ⎞ 3 = − ⎜ x + x + 16 ⎟ + 7 + 4 4⎝ 15 225 ⎠ 75 2

⎛ ⎞ = − 3 ⎜ x + 4 ⎟ + 529 4 ⎝ 15 ⎠ 75 529 4 4 maximum value of when x = − =7 75 75 15


Section 2.4

45.

131

3 2 3 x + 27 = − ( x − 0) 2 + 27 64 64 The maximum height of the arch is 27 feet. h(10) = − 3 (10)2 + 27 64 = − 3 (100) + 27 64 = − 75 + 27 16 = − 75 + 432 16 16 357 = = 22 5 feet 16 16 3 2 h( x) = 8 = − x + 27 64 8 − 27 = − 3 x 2 64 −19 = − 3 x 2 64

h( x) = −

a. b.

c.

46.

l + w = 240 a. w = 240 − l b. A = l (240 − l ) A = 240l − l 2 c.

A = −l 2 + 240l A = −(l 2 − 240l ) A = −(l 2 − 240l +1202 ) +1202

A = −(l −120) 2 +1202 Thus l = 120 and w = 120 produce the greatest area.

64( −19) = −3x 2 64( −19) = x2 −3 64( −19) =x −3 8 19 = x 3 8 19 3 = x 3 8 57 = x 3 20.1 ≈ x h( x) = 8 when x ≈ 20.1 feet

47.

a.

b.

c.

3w + 2l = 600 3w = 600 − 2l w = 600 − 2l 3 A = w⋅l

48.

⎛ ⎞ A = ⎜ 600 − 2l ⎟ l ⎝ 3 ⎠ = 200l − 2 l 2 3 2 2 A = − (l − 300l ) 3 A = − 2 (l 2 − 300l +1502 ) +15,000 3 In standard form, A = − 2 (l −150)2 +15,000 3

The maximum area of 15,000 ft 2 is produced when 600 − 2(150) l = 150 ft and the width w = = 100 ft . 3

4 w + 2l =1200 2l =1200 − 4 w l = 1200 − 4 w 2 l = 600 − 2w A = w(600 − 2w) A = 600w − 2 w2 A = −2w2 + 600w A = −2( w2 − 300w) A = −2( w2 − 300w +1502 ) + 2⋅1502 A = −2( w −150)2 + 45,000 1200 − 4(150) = 300. 2 Thus the dimensions that yield the greatest enclosed area are w = 150 ft and l = 300 ft.

Thus when w = 150, the length l =


132

49.


T (t ) = −0.7t 2 + 9.4t + 59.3

a.

50.

a.

9.4 ⎞ ⎛ t ⎟ + 59.3 = −0.7⎜ t 2 − 0.7 ⎠ ⎝ 94 ⎞ ⎛ = −0.7⎜ t 2 − t ⎟ + 59.3 7 ⎠ ⎝

⎛ ⎞ = −0.6 ⎜ t 2 − 32.1 t ⎟ − 350 0.6 ⎠ ⎝

2 2 ⎛ 94 ⎡ 47 ⎤ ⎡ 47 ⎤ ⎞ = −0.7⎜ t 2 − t + ⎢ ⎥ ⎟ + 59.3 + 0.7 ⎢ ⎥ ⎜ 7 ⎣7 ⎦ ⎣ 7 ⎦ ⎟⎠ ⎝

= −0.6(t − 26.75)2 + 79.3375

= −0.6(t 2 − 53.5t ) − 350 = −0.6 [t 2 − 53.5t + (26.75) 2 ] − 350 + 0.6(26.75) 2 2

≈ −0.6 ( t − 27 ) + 79 b.

The maximum number of larvae will survive at 27°C. The maximum number of larvae that will survive is 79.

c.

N(t) = 0 = −0.6t 2 + 32.1t − 350

2

⎛ 47 ⎞ ≈ −0.7⎜ t − ⎟ + 90.857 7 ⎠ ⎝ 2

5⎞ ⎛ ≈ −0.7⎜ t − 6 ⎟ + 91 7⎠ ⎝ The temperature is a maximum when 47 5 t= = 6 hours after 6:00 A.M. 7 7 5 Note (60 minutes) ≈ 43 minutes. 7 Thus the temperature is a maximum at 12:43 P.M.

t=

d.

51.

t = − b = − 82.86 = 0.14814 2a 2( −279.67)

52.

h( x) = −0.002 x 2 − 0.03 x + 8 h(39) = −0.002(39) − 0.03(39) + 8 = 3.788 > 3 Solve for x using quadratic formula. −0.002 x 2 − 0.03 x + 8 = 0

h(t ) = −9.8t 2 +100t h(t ) = −9.8(t − 5.1)2 + 254.9 The maximum height is 255 m.

54.

2

−32.1 ± 1030.41 − 840 −1.2 −32.1 ± 191.41 −32.1 ± 13.8 t= ≈ −1.2 −1.2 −32.1 + 13.8 −32.1 − 13.8 t= or t = −1.2 −1.2 = 15.25 ≈ 15 = 38.25 ≈ 38 Thus the x-intercepts to the nearest whole number for N(t) are (15, 0) and (38, 0). When the temperature is less than 15°C or greater than 38°C, none of the larvae survive.

h(t ) = −9.8(t 2 −10.2t )

E (0.14814) = −279.67(0.14814)2 + 82.86(0.14814) ≈ 6.1 The maximum energy is 6.1 joules.

53.

−32.1 ± (−32.1) 2 − 4(−0.6)( −350) 2(−0.6)

t=

The maximum temperature is approximately 91°F.

b.

N (t ) = −0.6t 2 + 32.1t − 350

h( x) = −0.0009 x 2 + 6 h(60.5) = −0.0009(60.5)2 + 6 ≈ 2.7 Since 2.7 is less than 5.4 and greater than 2.5, yes, the pitch is a strike.

x 2 +15 x − 4000 = 0 x= =

−15 ± (15)2 − 4(1)(−4000) 2(1) −15 ± 16, 225 2

, use positive value of x

x ≈ 56.2 Yes, the conditions are satisfied.


Section 2.4

55.

a.

133

E (v) = −0.018v 2 +1.476v + 3.4 ⎛ ⎞ = −0.018 ⎜ v 2 − 1.476 v ⎟ + 3.4 0.018 ⎠ ⎝ = −0.018 (v 2 − 82v) + 3.4

)

(

= −0.018 v 2 − 82v + 412 + 3.4 + 0.018(41) 2 2

b. 56.

= −0.018 (v − 41) + 33.658 The maximum fuel efficiency is obtained at a speed of 41 mph. The maximum fuel efficiency for this car, to the nearest mile per gallon, is 34 mpg.

h( x) = −0.0002348 x 2 + 0.0375 x ⎛ ⎞ = −0.0002348 ⎜ x 2 − 0.0375 x ⎟ 0.0002348 ⎠ ⎝ 2 2 ⎛ ⎡ ⎤ ⎞ ⎡ ⎤ = −0.0002348 ⎜ x 2 − 0.0375 x + ⎢ 1 ⋅ 0.0375 ⎥ ⎟ + 0.0002348 ⎢ 1 ⋅ 0.0375 ⎥ ⎜ 0.0002348 ⎣ 2 0.0002348 ⎦ ⎟⎠ ⎣ 2 0.0002348 ⎦ ⎝ 2 ⎛ ⎡ ⎤ ⎞ ≈ −0.0002348 ⎜ x 2 − 0.0375 x + ⎢ 1 ⋅ 0.0375 ⎥ ⎟ +1.5 ⎜ 0.0002348 ⎣ 2 0.0002348 ⎦ ⎟⎠ ⎝ The maximum height of the field, to the nearest tenth of a foot, is 1.5 feet.

57.

Let y = 0, then 0 = x 2 + 6 x 0 = x( x + 6) x = 0 or x + 6 = 0 x = −6 The x-intercepts are (0, 0) and (−6, 0).

58.

Let x = 0, then f ( x ) = 02 + 6(0) = 0 The y-intercept is (0, 0). 59.

Let y = 0, then 0 = −3 x 2 + 5 x − 6 x=

Let x = 0, then f ( x ) = −02 + 4(0) = 0 The y-intercept is (0, 0). 60.

63.

Let y = 0, then 0 = 2 x 2 + 3 x + 4

−5 ± 52 − 4(−3)(−6)

x=

2(−3) 2

61.

Let y = 0, then 0 = − x 2 + 4 x 0 = x ( − x + 4) x = 0 or −x + 4 = 0 x=4 The x-intercepts are (0, 0) and (4, 0).

−3 ± 32 − 4(2)(4) 2(2) 2

Since the discriminant 5 − 4( −3)(−6) = −47 is negative, there are no x-intercepts.

Since the discriminant 3 − 4(2)(4) = −23 is negative, there are no x-intercepts.

Let x = 0, then f ( x ) = −3(0)2 + 5(0) − 6 = −6 The y-intercept is (0, −6).

Let x = 0, then f ( x ) = 2(0) 2 + 3(0) + 4 = 4 The y-intercept is (0, 4).

−

b 296 =− = 740 2a 2(−0.2)

62.

−

b 810 =− = 675 2a 2(−0.6)

R (740) = 296(740) − 0.2(740) 2 = 109,520

R (675) = 810(675) − 0.6(675) 2 = 273,375

Thus, 740 units yield a maximum revenue of $109,520.

Thus, 675 units yield a maximum revenue of $273,375.

−

b 1.7 =− = 85 2a 2(−0.01)

64.

P (85) = −0.01(85) 2 + 1.7(85) − 48 = 24.25

Thus, 85 units yield a maximum profit of $24.25.

−

1.68 b = 11,760 =− 2a ⎛ 1 ⎞ ⎟⎟ 2⎜⎜ − ⎝ 14,000 ⎠

P (11,760) = −

(11,760) 2 + 1.68(11,760) − 4000 = 5878.40 14,000

Thus, 11,760 units yield a maximum profit of $5878.40. Copyright © Houghton Mifflin Company. All rights reserved.

134

65.


P ( x) = R ( x) − C ( x ) = x(102.50 − 0.1x ) − (52.50 x +1840)

P ( x) = R ( x) − C ( x) = x(210 − 0.25 x) − (78 x + 6399)

66.

= −0.1x 2 + 50 x −1840 The break-even points occur when R ( x) = C ( x) or P(x) = 0.

= −0.25 x 2 +132 x − 6399 − b = − 132 = 264 2a 2(−0.25)

Thus, 0 = −0.1x 2 + 50 x − 1840

P (264) = −0.25(264) 2 +132(264) − 6399 = $11,025, the maximum profit The break-even points occur when P(x) = 0.

−50 ± 502 − 4( −0.1)( −1840) x= 2( −0.1) −50 ± 1764 −0.2 −50 ± 42 = −0.2 x = 40 or x = 460

Thus, 0 = −0.25 x 2 + 132 x − 6399

=

−132 ± 1322 − 4(−0.25)(−6399) −132 ± 11025 = −0.5 2(−0.25) −132 ± 105 = ⇒ x = 54 or x = 474 −0.5

x=

The break-even points occur when x = 40 or x = 460.

The break-even points occur when x = 54 or x = 474. 67.

Let x = the number of people that take the tour. a. R ( x) = x(15.00 + 0.25(60 − x)) = x(15.00 +15 − 0.25 x) b.

= −0.25 x 2 + 30.00 x P ( x) = R ( x) − C ( x) = (−0.25 x 2 + 30.00 x ) − (180 + 2.50 x)

c.

= −0.25 x 2 + 27.50 x −180 b 27.50 − =− = 55 2a 2(−0.25)

P (55) = −0.25(55) 2 + 27.50(55) − 180

d. 68.

= $576.25 The maximum profit occurs when x = 55.

Let x = the number of parcels. a. b.

69.

R ( x) = xp = x(22 − 0.01x) = −0.01x 2 + 22 x P ( x) = R ( x) − C ( x) = (−0.01x 2 + 22 x ) − (2025 + 7 x)

c.

= −0.01x 2 +15 x − 2025 b 15 − =− = 750 2a 2(−0.01)

P (750) = −0.01(750) 2 + 15(750) − 2025

d. e.

= $3600 p(750) = 22 − 0.01(750) = $14.50 The break-even points occur when R ( x) = C ( x).

h(t ) = −16t 2 + 128t − b = − 128 = 4 seconds a. 2a 2( −16)

b.

h(4) = −16(4) 2 + 128(4) = 256 feet

c.

0 = −16t 2 +128t 0 = −16t (t − 8) −16t = 0 or t − 8 = 0 t =0 t =8 The projectile hits the ground at t = 8 seconds.

− 0.01x 2 + 22 x = 2025 + 7 x − 0.01x 2 + 15 x − 2025 = 0

− (15) ± 152 − 4(−0.01)(−2025) 2(−0.01) x = 150 or x = 1350 are the break-even points. Thus the minimum number of parcels the air freight company must ship to break even is 150. x=


Section 2.4

70.

135

h(t ) = −16t 2 + 64t + 80 b 64 a. − =− =2 2a 2(−16)

b.

c.

h(2) = −16(2)2 + 64(2) + 80 = 144 feet b − = − 64 2a 2( −16) = 2 seconds 0 = −16t 2 + 64t + 80 0 = −16(t 2 − 4t − 5) 0 = −16(t − 5)(t +1) t − 5 = 0 or t +1 = 0 t =5 t = −1 No The projectile has height 0 feet at t = 5 seconds.

71.

y ( x) = −0.014 x 2 + 1.19 x + 5 − b = − 1.19 2a 2( −0.014) = 42.5

y (42.5) = −0.014(42.5)2 + 1.19(42.5) + 5 = 30.2875 ≈ 30 feet

72.

h(t ) = −204.8t 2 + 256t − b = − 256 2a 2( −204.8) = 0.625 h(0.625) = −204.5(0.625) 2 + 256(0.625) = 80.1171875 ≈ 80 inches


136


73.

74.

y = a ( x − h) 2 + k y = a ( x − 0)2 + 6 y = ax 2 + 6 500 = a (2100)2 + 6

The perimeter is 48 = π r + h + 2r + h . Solve for h. 48 − π r − 2r = 2h 1 (48 − π r − 2r ) = h 2

494 = a (2100)2 494 = a 21002 0.000112018 ≈ a y = 0.000112018 x 2 + 6

Area = semicircle + rectangle 1 A = π r 2 + 2rh 2 1 2 ⎛1⎞ = π r + 2r ⎜ ⎟(48 − π r − 2r ) 2 ⎝2⎠ 1 2 = π r + r ( 48 − π r − 2r ) 2 1 2 = π r + 48r − π r 2 − 2r 2 2 ⎞ ⎛1 = ⎜ π − π − 2 ⎟ r 2 + 48r ⎝2 ⎠ ⎛ 1 ⎞ = ⎜ − π − 2 ⎟ r 2 + 48r ⎝ 2 ⎠

Graph the function A to find that its maximum occurs when r ≈ 6.72 feet.

Xmin = 0, Xmax = 14, Xscl = 1 Ymin = −50, Ymax = 200, Yscl = 50 1 (48 − π r − 2r ) 2 1 ≈ (48 − π(6.72) − 2(6.72)) 2 ≈ 6.72 feet

h=

Hence the optimal window has its semicircular radius equal to its height. Note: Using calculus it can be shown that the exact 48 value of r = h = . π+4


Section 2.4

137

....................................................... 75.

f ( x) = x 2 − ( a + b) x + ab a. x-intercepts occur when y = 0.

Connecting Concepts

0 = x 2 − (a + b) x + ab

f ( x) = ax 2 + bx + c a. a < 0, b and c any real numbers b. a > 0, b and c any real numbers

0 = ( x − a)( x − b)

c.

76.

b 2 − 4ac > 0

x − a = 0 or x − b = 0

b.

77.

x=a x=b Thus the x-intercepts are (a, 0) and (b, 0). b ( a + b) a + b − = = which is the x-coordinate of 2a 2(1) 2 the midpoint of the segment joining (a, 0) and (b, 0).

Let f ( x) = ax 2 + bx + c. We know f ( 2) = a ( 2) 2 + b ( 2) + c = 1

78.

f (−3) = a (−3) 2 + b(−3) + c = 2

(1)

This implies c = 4 and from Equation (1) we have 4a + 2b + 4 = 1 or 4a + 2b = −2 ( 2) The x-value of the vertex is 2, and by the vertex formula we −b 2a

, which implies b = −4a.

9 ⎛⎜ b ⎞⎟ − 3b − 5 = 2 ⎝6⎠ 3 b − 3b − 5 = 2 2 − 3b = 7 2 b = 7 ⋅ 2 = − 14 3 −3 b a= 6 − 14 a= 3 6 a = − 14 = − 7 18 9 7 14 Hence f ( x) = − x 2 − x − 5. 9 3

Substituting –4a for b in Equation (2) gives us 4a + 2(−4a ) = −3 4a − 8a = −3 − 4a = −3 3 a= 4 3 Substituting for a in Equation (2) gives us 4 ⎛3⎞ 4⎜ ⎟ + 2b = −3 ⎝4⎠ 3 + 2b = −3 2b = −6 b = −3 Thus the desired quadratic function is 3 f ( x ) = x 2 − 3 x + 4. 4 79.

P = 32 = 2 x + 2w 16 = x + w a. w = 16 − x b. Area A = xw A = x(16 − x) A =16 x − x 2

81.

(1)

f (0) = 0 + 0 + c = −5, which implies c = −5. −b Now the vertex is (−3, 2), so − 3 = or 6a = b or 2a b a = . Thus, substituting in Equation (1) gives us 6 9( a ) − 3b − 5 = 2

f( 0 ) = a (0) 2 + b(0) + c = 4

have 2 =

Let f ( x) = ax 2 + bx + c. We know a < 0.

80.

A = 16 x − x 2 attains its maximum when x = − b = − 16 = 8. 2a 2( −1) Now x = 8 implies w = 16 − x w = 16 − 8 w=8 Thus the rectangle with perimeter 32 inches that has the largest area is the square with each side of length 8 inches.

The discriminant is b 2 − 4(1)(−1) = b 2 + 4, which is always positive. Thus the equation has two real zeros for all values of b.


138


82.

The discriminant is b 2 − 4(−1)(1) = b 2 + 4, which is always positive. Thus the equation has two real zeros for all values of b.

83.

Increasing the constant c increases the height of each point on the graph by c units.

84.

Decreasing the coefficient a shrinks the graph of the parabola toward the x-axis.

85.

Let x = one number. Then 8 – x = the other number. P = x (8 − x ) = 8 x − x 2 , vertex at x = −b = −8 = 4. 2a −2 Thus, x = 4 and 8 − x = 4. The numbers are 4 and 4 .

86.

Let x = one number. Let x + 12 = the other number. P = x ( x + 12) = x 2 + 12x, vertex at x = −b = −12 = −6. 2a 2 Thus, x = −6 and x + 12 = −6 + 12 = 6. The numbers are –6 and 6.

87.

x1 = x, y1 = x3 , x2 = x + h, y2 = ( x + h)3 y −y ( x + h)3 − x3 x3 + 3hx 2 + 3h 2 x + h3 − x3 3hx 2 + 3h 2 x + h3 h(3 x 2 + 3hx + h 2 ) m= 2 1− = = = = 3x 2 + 3hx + h 2 x2 − x1 x+h−x h h h

88.

x1 = x, y1 = 4 x3 + x, x2 = x + h, y2 = 4( x + h)3 + ( x + h) y − y 4( x + h)3 + ( x + h) − (4 x3 + x) 4( x3 + 3hx 2 + 3h 2 x + h3 ) + x + h − 4 x3 − x = m= 2 1 = x2 − x1 x+h− x h 3 2 2 3 3 = 4 x +12hx +12h x + 4h + x + h − 4 x − x h 2 2 3 = 12hx +12h x + 4h + h h

=

h(12 x 2 +12hx + 4h 2 +1) h

=12 x 2 +12hx + 4h 2 +1

....................................................... PS1.

PS3.

Prepare for Section 2.5

f ( x) = x2 + 4 x − 6 − b = − 4 = −2 2a 2(1) x = −2

PS2.

f ( −2) = 2( −2)3 − 5( −2) = −16 + 10 = −6

PS4.

f (3) =

3(3)4 (3) 2 + 1

f ( −3) =

= 243 = 24.3 10

3( −3)4

( −3)2 + 1 f (3) = f ( −3)

− f (2) = −[2(2)3 − 5(2)] = −[16 − 10] = −6 f ( −2) = − f (2)

= 243 = 24.3 10

f ( −2) − g ( −2) = ( −2) 2 − [ −2 + 3] = 4 − 1 = 3 f ( −1) − g ( −1) = ( −1)2 − [ −1 + 3] = 1 − 2 = −1 f (0) − g (0) = (0) 2 − [0 + 3] = 0 − 3 = −3 f (1) − g (1) = (1)2 − [1 + 3] = 1 − 4 = −3 f (2) − g (2) = (2)2 − [2 + 3] = 4 − 5 = −1

PS5.

− a + a = 0, b + b = b 2 2 midpoint is (0, b)

PS6.

− a + a = 0, −b + b = 0 2 2 midpoint is (0, 0)


Section 2.5

139

Section 2.5 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

a. No

b. Yes

14.

a. Yes

b. No

15.

a. No

b. No

16.

a. No

b. No

17.

a. Yes

b. Yes

18.

a. Yes

b. Yes

19.

a. Yes

b. Yes

20.

a. No

b. No

21.

a. Yes

b. Yes

22.

Not symmetric with respect to the origin since (− y ) = (− x) + 1 does not simplify to the original equation y = x + 1.

23.

No, since (− y ) = 3(− x) − 2 simplifies to (−y) = −3x – 2, which is not equivalent to the original equation y = 3 x − 2.

24.

Yes, since (− y ) = (− x)3 − (− x) simplifies to − y = − x 3 + x, which is equivalent to the original equation y = x 3 − x.

25.

Yes, since (− y ) = −( − x)3 implies − y = x 3 or y = − x 3 , which is the original equation.

26.

Yes, since (− y ) =

27.

Yes, since (− x) 2 + (− y ) 2 = 10 simplifies to the original equation.

28.

Yes, since (− x) 2 − (− y ) 2 = 4 simplifies to the original equation.

29.

Yes, since − y =

30.

Yes, since − y = − x simplifies to the original equation.

9 9 is equivalent to the original equation y = . − x ( ) x

−x simplifies to the original equation. −x


140


31.

32.

symmetric with respect to the y-axis

33.

symmetric with respect to the x-axis

34.

symmetric with respect to the origin

35.

36.


symmetric with respect to the origin 37.


38.

39.

symmetric with respect to the line x = 2

symmetric with respect to the line x = 4

symmetric with respect to the line x = 2 40.

41.

42.

no symmetry

symmetric with respect to the x-axis, y-axis, and origin

symmetric with respect to the line x = 2 43.

Even since g (− x) = (− x) 2 − 7 = x 2 − 7 = g ( x).

44.

Even, since h(− x) = (− x) 2 + 1 = x 2 + 1 = h( x).

45.

Odd, since F (− x) = (− x)5 + (− x)3

46.

Neither, since G (− x) ≠ G ( x) and G (− x) ≠ −G ( x).

5

3

= −x − x = − F ( x). 47.

Even

48.

Even

49. Even

50.

Neither

51. Even

52.

Even

53.

Even

54. Neither

55.

Neither

56. Odd

57.

58.


Section 2.5

59.

61.

a.

f ( x+ 2)

b.

f ( x) + 2

a.

f ( x + 3) ( −2 − 3, 5) = ( −5, 5) (0 − 3, − 2) = ( −3, − 2) (1− 3, 0) = ( −2, 0) f ( x ) +1 ( −2, 5 +1) = ( −2, 6) (0, − 2 +1) = (0, −1) (1, 0 +1) = (1, 1)

62.

a.

f (− x)

64.

b.

− f ( x)

a.

f (− x) ( − −1, 3) = (1, 3) ( −2, − 4) − f ( x) ( −1, − 3) (2, − −4) = (2, 4)

b.

63.

65.

141

b.

60.

a.

g ( x−1)

b.

g ( x) −1

a.

g ( x − 2) ( −3 + 2, −1) = ( −1, −1) (1+ 2, − 3) = (3, − 3) (4 + 2, 2) = (6, 2) g ( x) − 2 ( −3, −1− 2) = ( −3, − 3) (1, − 3 − 2) = (1, − 5) (4, 2 − 2) = (4, 0)

b.

66.

a.

− g ( x)

b.

g (− x)

a.

− g ( x) (4, − −5) = (4, 5) ( −3, − 2) g (− x) ( −4, − 5) ( − − 3, 2) = (3, 2)

b.


142


67.

68.

69.

70.

1 y = − n( x ) + 1 2 1 y = − m( x) + 3 2 71.

a.

72.

b.

73.

b.

a.

74.

b.

75.

a.

a.

b.

76.

77.


Section 2.5

143

78.

79.

81.

80.

82.

83. a.

b.

c.

84. a.

b.

c.

....................................................... 85.

a. b.

f ( x) =

2 2

( x + 1) + 1

f ( x) = −

+1

2 ( x − 2) 2 + 1


a.

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

b.

f ( x ) = − ⎣⎡( x − 3) 2 + ( x − 3) ⎦⎤ − 2 = − ⎡⎣ ( x − 3) x − 1 ⎤⎦ − 2 f ( x ) = (3 − x ) x − 1 − 2


144


.......................................................

Prepare for Section 2.6 2 3 2 2 PS2. (3x − x + 2)(2 x − 3) = 6 x − 2 x + 4 x − 9 x + 3x − 6

PS1. (2 x 2 + 3x − 4) − ( x 2 + 3x − 5) = x 2 + 1

= 6 x 3 − 11x 2 + 7 x − 6 PS3.

f (3a ) = 2(3a )2 − 5(3a ) + 2

PS4.

f (2 + h ) = 2(2 + h )2 − 5(2 + h ) + 2 = 2h 2 + 8h + 8 − 5h − 10 + 2

= 18a 2 − 15a + 2

= 2h 2 + 3h PS6. 2 x − 8 = 0 x=4 Domain: x > 4 or [4, ∞)

PS5. Domain: all real numbers except x = 1

Section 2.6 1.

f ( x) + g ( x) = ( x 2 − 2 x −15) + ( x + 3)

2.

= x 2 − x −12 Domain all real numbers

= x 2 + x − 30 Domain all real numbers

f ( x) − g ( x) = ( x 2 − 2 x −15) − ( x + 3)

f ( x) − g ( x) = ( x 2 − 25) − ( x − 5)

= x 2 − 3 x −18 Domain all real numbers

= x 2 − x − 20 Domain all real numbers

f ( x) g ( x) = ( x 2 − 2 x −15)( x + 3)

f ( x) g ( x) = ( x 2 − 25)( x − 5)

= x3 + x 2 − 21x − 45 Domain all real numbers

= x3 − 5 x 2 − 25 x +125 Domain all real numbers

f ( x) / g ( x) = ( x 2 − 2 x −15) /( x + 3) = x − 5 Domain { x | x ≠ −3} 3.

f ( x ) + g ( x ) = (2 x + 8) + ( x + 4) = 3x +12 Domain all real numbers f ( x) − g ( x ) = (2 x + 8) − ( x + 4) = x + 4 Domain all real numbers f ( x) g ( x) = (2 x + 8)( x + 4)

f ( x) / g ( x) = ( x 2 − 25) /( x − 5) = x + 5 Domain { x | x ≠ 5} 4.

= 2 x 2 +16 x + 32 Domain all real numbers f ( x) / g ( x) = (2 x + 8) /( x + 4) = [ 2( x + 4)] /( x + 4)

f ( x ) + g ( x ) = ( x3 − 2 x 2 + 7 x ) + x = x3 − 2 x 2 + 8 x Domain all real numbers f ( x ) − g ( x ) = ( x3 − 2 x 2 + 7 x ) − x = x3 − 2 x 2 + 6 x Domain all real numbers

f ( x) + g ( x) = (5 x −15) + ( x − 3) = 6 x −18 Domain all real numbers f ( x) − g ( x) = (5 x −15) − ( x − 3) = 4 x −12 Domain all real numbers f ( x) g ( x) = (5 x −15)( x − 3) = 5 x 2 − 30 x + 45 Domain all real numbers f ( x) / g ( x) = (5 x −15) /( x − 3) = [5( x − 3)] /( x − 3)

= 2 Domain { x | x ≠ −4}

5.

f ( x) + g ( x) = ( x 2 − 25) + ( x − 5)

= 5 Domain { x | x ≠ 3}

6.

f ( x) + g ( x) = ( x 2 − 5 x − 8) + (− x) = x 2 − 6 x −8 Domain all real numbers f ( x) − g ( x) = ( x 2 − 5 x − 8) − (− x) = x 2 − 4 x −8 Domain all real numbers f ( x) g ( x) = ( x 2 − 5 x − 8)(− x)

f ( x ) g ( x ) = ( x3 − 2 x 2 + 7 x ) x = x 4 − 2 x3 + 7 x 2 Domain all real numbers f ( x ) / g ( x ) = ( x3 − 2 x 2 + 7 x ) / x = x 2 − 2 x + 7 Domain { x | x ≠ 0}

= − x3 + 5 x 2 + 8 x Domain all real numbers f ( x) / g ( x) = ( x 2 − 5 x −8 x) /(− x) = − x + 5 + 8 Domain { x | x ≠ 0} x


Section 2.6

7.

145

f ( x ) + g ( x ) = (4 x − 7) + (2 x 2 + 3x − 5) = 2 x 2 + 7 x −12 Domain all real numbers f ( x ) − g ( x ) = (4 x − 7) − (2 x 2 + 3x − 5) = −2 x 2 + x − 2 Domain all real numbers f ( x ) g ( x ) = (4 x − 7)(2 x 2 + 3x − 5) = 6 x 3 −10 x 2 +12 x 2 − 20 x − 21x + 35 = 6 x 3 + 2 x 2 − 41x + 35 Domain all real numbers f ( x ) / g ( x ) = (4 x − 7) /(2 x 2 + 3x − 5) =

8.

⎧ ⎫ 4x −7 Domain ⎨ x | x ≠1, x ≠ − 5 ⎬ ⎩ ⎭ 2 2 x + 3x − 5 2

f ( x ) + g ( x ) = (6 x +10) + (3x 2 + x −10) = 3x 2 + 7 x Domain all real numbers f ( x ) − g ( x ) = (6 x +10) − (3x 2 + x −10) = −3x 2 + 5 x + 20 Domain all real numbers f ( x ) g ( x ) = (6 x +10)(3x 2 + x −10) =18 x 3 + 6 x 2 − 60 x + 30 x 2 +10 x −100 =18 x 3 + 36 x 2 − 50 x −100 Domain all real numbers f ( x ) / g ( x ) = (6 x + 10) /(3x 2 + x − 10) =

9.

{

f ( x) + g ( x) =

x−3 + x

Domain { x | x ≥ 3}

f ( x) − g ( x) =

x−3 − x


f ( x) g ( x) = x x − 3

11.

}

6 x + 10 Domain x | x ≠ −2, x ≠ 5 3 3x 2 + x − 10


f ( x) / g ( x) =

x−3 +x x

f ( x) + g ( x) =

4 − x 2 + 2 + x Domain {x | −2 ≤ x ≤ 2}

f ( x) − g ( x) =

4 − x 2 − 2 − x Domain {x | −2 ≤ x ≤ 2}


10.

f ( x) + g ( x) =

x − 4 − x Domain {x | x ≥ 4}

f ( x) − g ( x) =

x − 4 + x Domain {x | x ≥ 4}

f ( x) g ( x) = − x x − 4


x−4 x


f ( x) / g ( x) = −

⎛ ⎞ f ( x) g ( x) = ⎜ 4 − x 2 ⎟(2 + x ) Domain {x | −2 ≤ x ≤ 2} ⎝ ⎠

12.

f ( x) / g ( x) =

4 − x2 2+ x

f ( x) + g ( x) =

x2 − 9 + x − 3

Domain {x | x ≤ −3 or x ≥ 3}

f ( x) − g ( x ) =

x2 − 9 − x + 3

Domain {x | x ≤ −3 or x ≥ 3}

Domain {x | −2 ≤ x ≤ 2}

⎛ ⎞ f ( x) g ( x) = ⎜ x 2 − 9 ⎟( x − 3) Domain {x | x ≤ −3 or x ≥ 3} ⎝ ⎠

f ( x) / g ( x) =

x2 − 9 x−3

Domain {x | x ≤ −3 or x ≥ 3}


146

13.


( f + g )( x) = x 2 − x − 2

14.

( f + g )(5) = (5) 2 − (5) − 2 = 25 − 5 − 2 = 18

16.

2

⎛1⎞ ⎛1⎞ ⎛1⎞ ( f + g)⎜ ⎟ = ⎜ ⎟ − ⎜ ⎟ − 2 ⎝2⎠ ⎝2⎠ ⎝ 2⎠ 1 1 = − −2 4 2 9 =− 4

= 49 + 7 − 2 = 54

( f + g )( x) = x 2 − x − 2

17.

2

( f − g )( x) = x 2 − 5 x + 6

( f − g )( x) = x 2 − 5 x + 6

18.

( f − g )( −3) = (−3) 2 − 5(−3) + 6

( f − g )( x) = x 2 − 5 x + 6

( f + g )( x) = x 2 − x − 2

15.

( f + g )(−7) = (−7)2 − (−7) − 2

⎛2⎞ ⎛2⎞ ⎛ 2⎞ ( f + g)⎜ ⎟ = ⎜ ⎟ − ⎜ ⎟ − 2 ⎝3⎠ ⎝3⎠ ⎝ 3⎠ 4 2 = − −2 9 3 20 =− 9 19.

( f + g )( x) = x 2 − x − 2

( f − g )(24) = (24)2 − 5(24) + 6

= 9 + 15 + 6

= 576 − 120 + 6

= 30

= 462

20.

( f − g )( x) = x 2 − 5 x + 6 ( f − g )(0) = (0)2 − 5(0) + 6 =6

( f − g )(−1) = (−1)2 − 5(−1) + 6 = 1+ 5 + 6 = 12

21.

)

(

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

22.

( fg )( x) = 2 x3 − 10 x 2 + 16 x − 8 ( fg )( −3) = 2(−3)3 − 10(−3)2 + 16(−3) − 8

= 2 x3 − 6 x 2 + 4 x − 4 x 2 + 12 x − 8

= −54 − 90 − 48 − 8

= 2 x3 − 10 x 2 + 16 x − 8

= −200

( fg )(7) = 2(7)3 − 10(7)2 + 16(7) − 8 = 686 − 490 + 112 − 8 = 300

23.

( fg )( x) = 2 x3 − 10 x 2 + 16 x − 8 3

24.

( fg )( −100) = 2(−100)3 − 10(−100)2 + 16( −100) − 8 = −2,000,000 − 100,000 − 1600 − 8

2

⎛2⎞ ⎛ 2⎞ ⎛2⎞ ⎛ 2⎞ ( fg ) ⎜ ⎟ = 2 ⎜ ⎟ − 10 ⎜ ⎟ + 16 ⎜ ⎟ − 8 ⎝5⎠ ⎝5⎠ ⎝5⎠ ⎝5⎠ 16 40 32 = − + −8 125 25 5 −384 = = −3.072 125 25.

⎛ ⎜ ⎝ ⎛ ⎜ ⎝

f ⎞ x2 − 3x + 2 ⎟ ( x) = 2x − 4 g⎠ 1 1 f ⎞ ⎟ ( x) = x − 2 2 g⎠

⎛f ⎞ 1 1 ⎜ ⎟ (−4) = ( −4 ) − 2 2 g ⎝ ⎠ 1 = −2 − 2 1 5 = −2 or − 2 2

( fg )( x) = 2 x3 − 10 x 2 + 16 x − 8

= −2,101,608

26.

⎛ f ⎞ 1 1 ⎜ ⎟ ( x) = x − 2 2 ⎝g⎠ ⎛f ⎞ 1 1 ⎜ ⎟ (11) = (11) − 2 2 ⎝g⎠ 11 1 = − 2 2 10 = =5 2


Section 2.6

27.

147

⎛f ⎞ 1 1 ⎜ ⎟ ( x) = x − 2 2 ⎝g⎠

28.

⎛ f ⎞⎛ 1 ⎞ 1 ⎛ 1 ⎞ 1 ⎜ ⎟⎜ ⎟ = ⎜ ⎟ − ⎝ g ⎠⎝ 2 ⎠ 2 ⎝ 2 ⎠ 2 1 1 = − 4 2 1 =− 4

29.

31.

⎛ f ⎞⎛ 1 ⎞ 1 ⎛ 1 ⎞ 1 ⎜ ⎟⎜ ⎟ = ⎜ ⎟ − ⎝ g ⎠⎝ 4 ⎠ 2 ⎝ 4 ⎠ 2 1 1 = − 8 2 3 =− 8

f ( x + h) − f ( x) [ 2( x + h) + 4] − (2 x + 4) = h h 2 x + 2(h) + 4 − 2 x − 4 = h 2 h = h =2 2 f ( x + h) − f ( x) [ ( x + h) − 6] − ( x − 6) = h h

=

x 2 + 2 x( h) + ( h) 2 − 6 − x 2 + 6 h

2 x ( h) + h 2 h = 2x + h

30.

32.

f ( x + h) − f ( x) [ 4( x + h) − 5] − (4 x − 5) = h h 4 x + 4(h) − 5 − 4 x + 5 = h 4(h) = h =4 2 2 ⎡ ⎤ f ( x + h) − f ( x) ⎢⎣( x + h) +11⎥⎦ − ( x +11) = h h

=

x 2 + 2 xh + (h)2 +11− x 2 −11 h

= 2 xh + h h = 2x + h

=

33.

⎛f ⎞ 1 1 ⎜ ⎟ ( x) = x − 2 2 ⎝g⎠

2

f ( x + h) − f ( x) 2( x + h)2 + 4( x + h) − 3− (2 x 2 + 4 x − 3) = h h 2 2 2 = 2 x + 4 xh + 2h + 4 x + 4h − 3 − 2 x − 4 x + 3 h 2 = 4 xh + 2h + 4h h = 4 x + 2h + 4

34.

f ( x + h) − f ( x) 2( x + h)2 − 5( x + h) − (2 x 2 − 5 x + 7) = h h 2 2 2 = 2 x + 4 xh + 2h − 5 x − 5h + 7 − 2 x + 5 x − 7 h 2

= 4 xh + 2h − 5h h = 4 x + 2h − 5

35.

f ( x + h) − f ( x) −4( x + h)2 + 6 − (−4 x 2 + 6) = h h 2 2 2 = −4 x − 8 xh − 4h + 6 + 4 x − 6 h

= −8 xh − 4h h = −8 x − 4h

2


148

36.


f ( x) = −5 x 2 − 4 x f ( x + h) − f ( x) −5( x + h)2 − 4( x + h) − (−5 x 2 − 4 x) = h h =

−5 x 2 −10 x(h) − 5h 2 − 4 x 2 − 4h + 5 x 2 + 4 x h

−10 x(h) − 5h 2 − 4h h = −10 x − 5h − 4 =

37.

( g o f )( x) = g [ f ( x) ]

= g [ 3 x + 5] = 2 [ 3 x + 5]

= 6 x +10 − 7 = 6x +3

38.

( g o f )( x) = g [ f ( x) ]

= g [ 2 x − 7]

39.

= 6 x − 21+ 5 = 6 x −16 ( f o g )( x) = f [ g ( x) ]

= f [3 x + 2]

= 2 [3 x + 2] − 7 = 6x + 4−7 = 6x −3

( g o f )( x) = g ⎡⎢ x 2 + 4 x −1⎤⎥ ⎣ ⎦ 2 ⎡ ⎤ = ⎢ x + 4 x −1⎥ + 2 ⎣ ⎦

( g o f )( x) = g ⎡⎢ x 2 −11x ⎤⎥ ⎣ ⎦ 2 = 2 ⎡⎢ x −11x ⎤⎥ + 3 ⎣ ⎦

( g o f )( x) = g [ f ( x) ] = g ⎡⎢ x3 + 2 x ⎤⎥ ⎣ ⎦ 3 ⎡ = −5 ⎢ x + 2 x ⎤⎥ ⎣ ⎦ = −5 x3 −10 x

42.

= 3[ 2 x − 7 ] + 5

= 6 x − 21+ 2 = 6 x −19

= 2 x 2 − 22 x + 3 41.

= f [ 2 x − 7]

= 3[ 2 x − 7 ] + 2

= x 2 + 4 x +1

40.

( f o g )( x) = f [ g ( x)]

( g o f )( x) = g [ f ( x) ] = g ⎡⎢ − x − 7 ⎤⎥ ⎣ ⎦ = ⎡⎢ − x3 − 7 ⎤⎥ +1 ⎣ ⎦ 3

= − x3 − 6

( f o g )( x) = f [ x + 2] 2

= [ x + 2] + 4 [ x + 2]−1 = x 2 + 4 x + 4 + 4 x + 8 −1 = x 2 + 8 x +11 ( f o g )( x) = f [ 2 x + 3] 2

= [ 2 x + 3] −11[ 2 x + 3] = 4 x 2 +12 x + 9 − 22 x − 33 = 4 x 2 −10 x − 24 ( f o g )( x) = f [ g ( x)] = f [ −5 x ] 3

= [ −5 x ] + 2 [ − 5 x ] = −125 x3 −10 x

( f o g )( x) = f [ g ( x)] = f [ x +1]

3

= − [ x +1] − 7 = − x3 − 3 x 2 − 3 x −1− 7 = − x3 − 3 x 2 − 3 x − 8


Section 2.6

43.

149

( g o f )( x) = g [ f ( x)] ⎡ ⎤ =g⎢ 2 ⎥ ⎣ x +1 ⎦ ⎡ ⎤ = 3 ⎢ 2 ⎥ −5 1 x + ⎣ ⎦ 5( x +1) 6 = − x +1 x +1 = 6 −5x −5 x +1 1 − = 5x x +1

44.

( g o f )( x) = g [ f ( x) ] = g ⎡⎣ x + 4 ⎤⎦ 1 x+4 = x+4 x+4 =

( f o g )( x) = f [ g ( x)]

= f [ 3 x − 5] = =

2

[3x − 5]+1 2 3x − 4

( f o g )( x) = f [ g ( x)] =f 1 x = 1 +4 x 1 = + 4x x 1 4x + = x = x 1+ 4 x x 2 = x + 4x x

45.

( g o f )( x) = g [ f ( x)] ⎡ ⎤ =g⎢ 1 ⎥ ⎣ x2 ⎦ =

⎡1 ⎤ ⎢ 2 ⎥ −1 ⎣x ⎦

2 = 1− x 2 x

( f o g )( x) = f [ g ( x)] = f ⎡⎣ x −1 ⎤⎦ 1 = ⎡ x −1 ⎤ ⎣ ⎦ 1 = x −1

2

2 = 1− x | x|

46.

( g o f )( x) = g [ f ( x)]

( f o g )( x) = f [ g ( x)]

⎡ ⎤ =g⎢ 6 ⎥ ⎣ x−2 ⎦ 3 = ⎡ 6 ⎤ 5⎢ ⎣ x − 2 ⎥⎦ = 3 ⎛ 30 ⎞ ⎜ ⎟ ⎝ x−2 ⎠ = 3⋅ x − 2 30 = x−2 10

⎡ ⎤ =f⎢3⎥ ⎣ 5x ⎦ 6 = ⎡3⎤ ⎢⎣ 5 x ⎥⎦ − 2 6 = ⎛ 3-10x ⎞ ⎜ ⎟ ⎝ 5x ⎠ = 30 x 3 −10 x


150

47.


⎡ ( g o f )( x) = g ⎢ 3 ⎢⎣ 5 − x =− 2 ⎡ 3 ⎢ ⎣⎢ 5 − x =

⎤ ⎥ ⎥⎦

⎡ ⎤ ( f o g )( x) = f ⎢ − 2 ⎥ ⎣ x⎦ 3 = ⎡ 2⎤ 5− ⎢− ⎥ ⎣ x⎦ 3 = 5+ 2 x 3 = [5x + 2]

⎤ ⎥ ⎦⎥

−2 5 − x 3

x =

48.

3x 5x + 2

( f o g )( x) = f [ g ( x) ]

( g o f )( x) = g ⎡⎣ 2 x +1 ⎤⎦

= f ⎡⎢3 x 2 −1⎤⎥ ⎣ ⎦ 2 ⎤ ⎡ = 2 ⎢3 x −1⎥ +1 ⎣ ⎦

2

= 3 ⎣⎡ 2 x +1 ⎦⎤ −1 2

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

= 6 x 2 − 2 +1

=12 x 2 +12 x + 3 −1

= 6 x 2 −1

=12 x 2 +12 x + 2 Use the results to work Exercises 49 to 64. 49.

( g o f )( x) = 4 x 2 + 2 x − 6

50.

( g o f )(4) = 4(4)2 + 2(4) − 6 = 64 + 8 − 6 = 66

51.

( f o g )(4) = 2(4)2 −10(4) + 3 = 32 − 40 + 3 = −5

( f o g )( x) = 2 x 2 −10 x + 3

52.

( f o g )(−3) = 2(−3) 2 −10(−3) + 3 =18 + 30 + 3 = 51

53.

57.

= 4−2−6 = −4

( g o h)( x) = 9 x 4 − 9 x 2 − 4

54.

(h o g )( x) = −3x 4 + 30 x3 − 75 x 2 + 4 ( h o g )(0) = −3(0) 4 + 30(0)3 − 75(0)2 + 4 =4

( f o f )( x) = 4 x + 9 ( f o f )(8) = 4(8) + 9 = 41

56.

( h o g )( x) = −3 x 4 + 30 x3 − 75 x 2 + 4 4

( g o f )( x) = 4 x 2 + 2 x − 6 ( g o f )(−1) = 4(−1)2 + 2(−1) − 6

( g o h)(0) = 9(0) 4 − 9(0)2 − 4 = −4

55.

( f o g )( x) = 2 x 2 −10 x + 3

3

58. 2

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ (h o g ) ⎜ 2 ⎟ = −3 ⎜ 2 ⎟ + 30 ⎜ 2 ⎟ − 75 ⎜ 2 ⎟ + 4 ⎝5⎠ ⎝5⎠ ⎝5⎠ ⎝5⎠ 48 240 300 =− + − +4 625 125 25 = −48 +1200 − 7500 + 2500 625 3848 =− 625

( f o f )( x) = 4 x + 9 ( f o f )(−8) = 4(−8) + 9 = −23 ( g o h)( x) = 9 x 4 − 9 x 2 − 4 4

2

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ( g o h) ⎜ − 1 ⎟ = 9 ⎜ − 1 ⎟ − 9 ⎜ − 1 ⎟ − 4 ⎝ 3⎠ ⎝ 3⎠ ⎝ 3⎠ 9 9 = − −4 81 9 = 1 −1− 4 9 = −4 8 or − 44 9 9


Section 2.6

151

( g o f )( x) = 4 x 2 + 2 x − 6

59.

( g o f )( 3) = 4( 3)2 + 2( 3) − 6

61.

( f o g )( x) = 2 x 2 − 10 x + 3

60.

( f o g )( 2) = 2( 2)2 − 10( 2) + 3

= 12 + 2 3 − 6

= 4 − 10 2 + 3

= 6+2 3

= 7 − 10 2

( g o f )( x) = 4 x 2 + 2 x − 6

62.

( f o g )( x) = 2 x 2 −10 x + 3 ( f o g )(3k ) = 2(3k )2 −10(3k ) + 3

( g o f )(2c) = 4(2c)2 + 2(2c) − 6

=18k 2 − 30k + 3

= 16c 2 + 4c − 6 ( g o h)( x) = 9 x 4 − 9 x 2 − 4

63.

( g o h)( k + 1) = 9(k + 1) 4 − 9(k + 1)2 − 4 = 9(k 4 + 4k 3 + 6k 2 + 4k + 1) − 9k 2 − 18k − 9 − 4 = 9k 4 + 36k 3 + 54k 2 + 36k + 9 − 9k 2 − 18k − 13 = 9k 4 + 36k 3 + 45k 2 + 18k − 4

(h o g )( x) = −3x 4 + 30 x3 − 75 x 2 + 4

64.

(h o g )(k −1) = −3(k −1)4 + 30(k −1)3 − 75(k −1)2 + 4 = −3k 4 +12k 3 −18k 2 +12k − 3+ 30k 3 − 90k 2 + 90k − 30 − 75k 2 +150k − 75 + 4 = −3k 4 + 42k 3 −183k 2 + 252k −104 65.

a.

66.

r = 1.5t and A = πr 2

a.

2

so A(t ) = π [ r (t )]

= π (1.5t )2

w = 2 − 0.2t for 0 ≤ t ≤10 = −2 + 0.2t for 10 ≤ t ≤14 or w = 2 − 0.2t

2

A(2) = 2.25π (2) = 9π square feet ≈ 28.27 square feet b.

Note:

r = 1.5t h = 2r = 2(1.5t ) = 3t and 1 V = π r 2h so 3 1 V (t ) = π (1.5t )2 [3t ] 3

l = 3 − 0.5t for 0 ≤ t ≤ 6 = −3 + 0.5t for 6 ≤ t ≤14 or l = 3 − 0.5t

b.

l = lw = 3 − 0.5t 2-0.2t = (3 − 0.5t )(2 − 0.2t )

c.

A is increasing on [6, 8] and on [10, 14] ; and A is decreasing on [0, 6] and on [8, 10] .

d.

The highest point on the graph of A occurs when t = 0 seconds.

= 2.25π t 3 1 1 1 V = π r 2h = (π r 2 ) = hA 3 3 3 1 2 = ( 3t ) (2.25π t ) = 2.25π t 3 3

V (3) = 2.25π (3)3 = 60.75π cubic feet ≈ 190.85 cubic feet

Xmin = 0, Xmax =14, Xscl = 2, Ymin = −1, Ymax = 6, Yscl = 2


152

67.


a.

2 2 2 Since d + 4 = s , 2

68.

2

d = s −16

The perimeter is an increasing function over 0 ≤ t ≤ 14. The graph of P = 2(3 + 0.5t ) + 2 2 − 0.2t is shown below.

d = s 2 −16 d = (48 − t )2 −16 ⋅ s = 48 − t = 2304 − 96t + t 2 −16 = t 2 − 96t + 2288 b.

s (35) = 48 − 35 =13 ft d (35) = 352 − 96(35) + 2288

Xmin = 0, Xmax = 14, Xscl = 5, Ymin = −1, Ymax = 22, Yscl =5

= 153 ≈12.37 ft

69.

(Y o F )( x) = Y ( F ( x)) converts x inches to yards. F takes x inches to feet, and then Y takes feet to yards.

71.

a.

70.

( I o F )( x) = I ( F ( x)) converts x yards to inches. F takes x yards to feet, and then I takes feet to inches.

On [0, 1] , a = 0 Δt = 1 − 0 = 1 C ( a + Δt ) = C (1) = 99.8 C ( a ) = C ( 0) = 0

C (1) − C (0) = 99.8 − 0 = 99.8 1 This is identical to the slope of the line through C (1) − C (0) (0, C(0)) and (1, C(1)) since m = = C (1) − C (0) 1− 0 On [0, 0.5] , a = 0 , Δt = 0.5 Average rate of change =

b.

C (0.5) − C (0) 78.1 − 0 = = 156.2 0.5 0.5 On [1, 2] , a = 1 , Δt = 2 − 1 = 1

Average rate of change = c.

C (2) − C (1) 50.1 − 99.8 = = −49.7 1 1 On [1, 1.5] , a = 1 , Δt = 1.5 − 1 = 0.5

Average rate of change = d.

C (1.5) − C (1) 84.4 − 99.8 −15.4 = = = −30.8 0.5 0.5 0.5 On [1, 1.25] , a = 1 , Δt = 1.25 − 1 = 0.25

Average rate of change = e.

Average rate of change = f.

C (1.25) − C (1) 95.7 − 99.8 −4.1 = = = −16.4 0.25 0.25 0.25

On [1, 1 + Δt ] , Con (1+ Δt ) = 25 (1+ Δt )3 −150 (1+ Δt ) 2 + 225 (1+ Δt ) 3

2

= 25(1+ 3( Δt ) + 3 ( Δt ) ) −150(1+ 2 ( Δt ) + ( Δt ) ) + 225(1+ Δt ) 2

3

2

= 25 + 75( Δt ) + 75 ( Δt ) ) + 25 ( Δt ) −150 − 300 ( Δt ) −150 ( Δt ) + 225 + 225 ( Δt ) 2

=100 − 75 ( Δt ) + 25 ( Δt ) Con (1) =100

3

3

Con (1+ Δt ) − Con (1) 100 − 75( Δt )2 + 25 ( Δt ) −100 = Δt Δt −75( Δt )2 + 25( Δt )3 = Δt = −75( Δt ) + 25( Δt )2

Average rate of change =

As Δt approaches 0, the average rate of change over [1, 1 + Δt ] seems to approach 0. Copyright © Houghton Mifflin Company. All rights reserved.

Section 2.6

72.

a.

153

On [ 2, 3] , a = 2 Δt = 3 − 2 = 1 f (a + Δt ) = f (3) = 6 ⋅ 32 = 54 f (a ) = f (2) = 6 ⋅ 2 2 = 24

f (a + Δt ) − f (a ) f (3) − f (2) = = 54 − 24 = 30 feet per second Δt 1 This is identical to the slope of the line through (2, f (2)) f (3) − f (2) and (3, f (3)) since m = = f (3) − f (2). 3−2 On [ 2, 2.5] , a = 2 ,

Average velocity =

b.

Δt = 2 .5 − 2 = 0 .5 f (a + Δt ) = f (2.5) = 6( 2.5) 2 = 37.5

Average velocity = c.

On [ 2, 2.1] , a = 2

f (2.5) − f (2) 37.5 − 24 13.5 = = = 27 feet per second 0.5 0.5 0.5

Δt = 2 .1 − 2 = 0 .1 f (a + Δt ) = f (2.1) = 6( 2.1) 2 = 26.46

Average velocity = d.

On [ 2, 2.01] , a = 2

f (2.1) − f (2) 26.46 − 24 2.46 = = = 24.6 feet per second 0.1 0.1 0.1

Δt = 2.01 − 2 = 0.01 f (a + Δt ) = f (2.01) = 6( 2.01) 2 = 24.2406

Average velocity = e.

On [ 2, 2.001] , a = 2


Δt = 2.001 − 2 = 0.001 f (a + Δt ) = f ( 2.001) = 6(2.001) 2 = 24.024006

Average velocity = f.


On [ 2, 2 + Δt ] , Con

f (2 +Δt ) − f (2) 6(2 + Δt )2 − 24 6(4 + 4(Δt ) + (Δt )2 − 24 24 + 24(Δt ) + 6(Δt )2 − 24 = = = Δt Δt Δt ( Δt )

24Δt + 6(Δt ) 2 = 24 + 6( Δt ) Δt As Δt approaches 0, the average velocity seems to approach 24 feet per second. =

....................................................... 73.

( g o f )( x) = g [ f ( x) ]

= g [ 2 x + 3]

= 5(2 x + 3) +12 =10 x +15 +12 =10 x + 27 ( g o f )( x) = ( f o g )( x)

( f o g )( x) = f [ g ( x) ]

= f [5 x +12] = 2(5 x +12) + 3 =10 x + 24 + 3 =10 x + 27


Connecting Concepts

154

74.


( g o f )( x) = g [ f ( x) ]

( f o g )( x) = f [ g ( x) ]

= g [ 4 x − 2]

= f [ 7 x − 4] = 4(7 x − 4) − 2 = 28 x −16 − 2 = 28 x −18

= 7(4 x − 2) − 4 = 28 x −14 − 4 = 28 x −18 ( g o f )( x) = ( f o g )( x) 75.

( g o f )( x) = g [ f ( x)]

( f o g )( x) = f [ g ( x)]

⎡ ⎤ = g ⎢ 6x ⎥ ⎣ x −1 ⎦

⎡ ⎤ = f ⎢ 5x ⎥ ⎣ x−2 ⎦ 6 ⎛⎜ 5 x ⎞⎟ = ⎝ x−2 ⎠ 5x −1

⎛ ⎞ 5⎜ 6x ⎟

= ⎝ x−1 ⎠ 6x −2

x−1 30 x = x−1 = 6 x−2 x+2 x−1

x−2

30 x x−1 4 x+ 2 x−1

= 30 x ⋅ x −1 x −1 2(2 x +1) = 15 x 2 x +1

30 x 30 x − x x 2 = = −2 5 x − x+ 2 x−2

4 x+2 x−2

= 30 x ⋅ x − 2 x − 2 2(2 x +1) = 15 x 2 x +1

( g o f )( x) = ( f o g )( x) 76.

( g o f )( x ) = g [ f ( x )]

( f o g )( x ) = f [ g ( x)]

⎡ ⎤ = g ⎢ 5x ⎥ ⎣ x+3 ⎦

⎡ ⎤ = f ⎢− 2 x ⎥ ⎣ x−4 ⎦ 5 ⎛⎜ − 2 x ⎞⎟ = ⎝ x−4 ⎠ 2x − +3

⎛ ⎞ 2 ⎜ 5x ⎟

= − ⎝ x+3 ⎠ 5x −4

x+3 10 x = − x+3 = − 5 x−4 x−12 x+3

= − 10 x ⋅ x + 3 x + 3 x −12 = − 10 x x −12 ( g o f )( x) = ( f o g )( x)

77.

( g o f )( x ) = g [ f ( x )]

= g [ 2 x + 3]

[ 2 x + 3]− 3 = 2

= 2x 2 =x

x−4 −10 x x−4 = = −2 x+3 x−12 x−4

10 x x+3 x−12 x+3

= −10 x ⋅ x − 4 x − 4 x −12 10 = x x −12

−10 x x−4 x−12 x−4

( f o g )( x) = f [ g ( x) ] ⎡ ⎤ = f ⎢ x −3 ⎥ ⎣ 2 ⎦ ⎡ ⎤ = 2 ⎢ x −3 ⎥ + 3 ⎣ 2 ⎦ = x − 3+ 3 =x


Section 2.6

78.

155

( g o f )( x ) = g [ f ( x )]

= g [ 4 x − 5]

[ 4 x − 5] + 5 = 4

= 4x 4 =x

79.

( g o f )( x) = g [ f ( x) ] ⎡ ⎤ =g⎢ 4 ⎥ ⎣ x +1 ⎦ ⎡ ⎤ 4− ⎢ 4 ⎥ 1 + x ⎣ ⎦ = ⎡ 4 ⎤ ⎢⎣ x +1 ⎥⎦ 4x + 4− 4 = x +1 4 x +1 = 4 x ⋅ x +1 x +1 4 =x

80.

( g o f )( x) = g [ f ( x) ] ⎡ ⎤ =g⎢ 2 ⎥ ⎣1− x ⎦ ⎡ 2 ⎤ ⎢ ⎥−2 = ⎣1− x ⎦ ⎡ 2 ⎤ ⎢⎣1− x ⎥⎦ 2− 2+ 2x = 1− x 2 1− x = 2 x ⋅1− x 1− x 2 =x

81.

( g o f )( x) = g [ f ( x) ]

( f o g )( x) = f [ g ( x)] ⎡ ⎤ = f ⎢ x +5 ⎥ ⎣ 4 ⎦ ⎡ ⎤ = 4 ⎢ x +5 ⎥ −5 ⎣ 4 ⎦ = x + 5−5 =x ( f o g )( x) = f [ g ( x) ] ⎡ ⎤ = f ⎢ 4− x ⎥ ⎣ x ⎦ 4 = ⎡ 4− x ⎤ ⎢⎣ x ⎥⎦ +1 4 = 4− x + x x 4 = 4 x = 4⋅ x 4 =x ( f o g )( x ) = f [ g ( x )] ⎡ ⎤ = f ⎢ x−2 ⎥ ⎣ x ⎦ 2 = ⎡ x−2 ⎤ 1− ⎢ ⎣ x ⎥⎦ 2 = x− x+2 x 2 = ⋅ x 1 x− x+2 = 2⋅ x 1 2 =x

( f o g )( x) = f [ g ( x)]

= g ⎡⎢ x3 −1⎤⎥ ⎣ ⎦

= f ⎡ 3 x +1 ⎤ ⎣ ⎦

= 3 ⎡⎢ x3 −1⎤⎥ +1 ⎣ ⎦

= ⎡ 3 x +1 ⎤ −1 ⎣ ⎦ = x +1−1 =x

3

= x3 =x

3


156

82.


( f o g )( x) = f [ g ( x)]

( g o f )( x ) = g [ f ( x )]

= f ⎡3 2 − x ⎤ ⎣ ⎦ = ⎡3 2 − x ⎤ + 2 ⎣ ⎦ = −(2 − x) + 2 = −2 + x + 2 =x

= g ⎡⎢ − x + 2 ⎤⎥ ⎣ ⎦ 3

= 3 2 − ⎡⎢ − x3 + 2 ⎤⎥ ⎣ ⎦ 3

= 2 + x3 − 2 3

= x3 =x

.......................................................


PS1. Slope: − 1 ; y-intercept: (0, 4) 3

PS2. 3x − 4 y =12 y = 3 x −3 4 3 Slope: ; y-intercept: (0, –3) 4

PS3. y = –0.45x + 2.3

PS4.

y + 4 = − 2 ( x − 3) 3 y = − 2 x−2 3

PS6.

f ( x1) − y1 + f ( x2 ) − y2 = (2)2 − 3− (−1) + (4)2 − 3−14

PS5.

f (2) = 3(2)2 + 4(2) −1=12 + 8 −1=19

= 4 − 3+1 + 16 − 3−14 = 2 +1 =3

Section 2.7 1.

The scatter diagram suggests no relationship between x and y.

2.

The scatter diagram suggest a nonlinear relationship between x and y.

3.

The scatter diagram suggests a linear relationship between x and y.

4.

The scatter diagram suggests a linear relationship between x and y.

5.

Figure A better approximates a graph that can be modeled by an equation than does Figure B. Thus Figure A has a coefficient of determination closer to 1.

6.

Figure A better approximates a graph that can be modeled by an equation than does Figure B. Thus Figure A has a coefficient of determination closer to 1.

7.

Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here.

y = 2.00862069x + 0.5603448276 Copyright © Houghton Mifflin Company. All rights reserved.

Section 2.7

8.

157


y = −1.918918919x + 0.4594594595 9.


y = −0.7231182796x + 9.233870968 10.


y =0.6591216216x – 6.658108108 11.


y = 2.222641509x –7.364150943 12.


y = −2.301587302x + 4.813968254 13.


y = 1.095779221x2 – 2.69642857x + 1.136363636


158

14.



y = −0.5714285714x2 + 2.2x + 1.942857143 15.


y = −0.2987274717x2 – 3.20998141x + 3.416463667 16.


y = 1.414285714x2 + 1.954285714x – 2.705714286 17.

18.


a.

y = 23.55706665x – 24.4271215

b.

y = 23.55706665(54) – 24.4271215 ≈ 1248 cm


a.

y = 3.410344828x + 65.09359606

b.

y = 3.410344828x + 65.09359606 ≈ 263 ft


Section 2.7

19.

20.

21.

22.

23.

159


a.

y = 0.194224924x + 0.7978723404

b.

y = 0.194224924(32) + 0.7978723404 ≈ 4.3 m/s


a.

y = 6.357142857x + 90.57142857

b.

y = 6.357142857(7.5) + 90.57142857 = 138.25 or 138,000 bacteria


a.

y = 0.1628623408x – 0.6875682232

b.

y = 0.1628623408(158) – 0.6875682232 ≈ 25


a.

y = −0.6800298805x + 69.05129482

b.

5 feet 8 inches = 68 inches; y = −0.6800298805(68) + 69.05129482 ≈ 23


The value of r is close to 0. Therefore, no, there is not a strong linear relationship between the current and the torque.


160

24.



The value of r is close to –1, so, yes, there is a strong correlation between a man’s age and his average remaining lifetime. 25.


a.

26.

The value of r is close to –1, so, yes, there is a strong linear correlation.

b.

y = −0.9033088235x + 78.62573529

c.

y = −0.9033088235(25) + 78.62573529 ≈ 56 years


y = 0.2711847067 x + 7.91528368 y = 0.2711847067(41) + 7.91528368 ≈19 meters per second

27.


y =113.3111246 x + 21.83605895 a. Positively b. 28.

y =113.3111246(9.5) + 21.83605895 ≈1098 calories


a.

y = −0.0074642857 x 2 + 1.148214286 x + 4.807142857

b.

y = −0.0074642857(65)2 + 1.148214286(65) + 4.807142857 ≈ 47.9 ft


Section 2.7

29.

161


y = −0.6328671329x2 + 33.6160839x – 379.4405594 30.

31.

32.


a.

y = −0.0004093949t2 + 0.2265681259t + 55.57907207

b.

1:00 P.M. is 7 hours, or 420 minutes, after 6:00 A.M. y = −0.0004093949(420)2 + 0.2265681259(420) + 55.57907207 ≈ 78.5°


a.

y = −0.0165034965x2 + 1.366713287x + 5.685314685

b.

y = −0.0165034965(50)2 + 1.366713287(50) + 5.685314685 ≈ 32.8 mpg


a.

y = 0.05208x2 – 3.56026x + 82.32999

b.

− b = − −3.56026 ≈ 34 kilometers per hour 2a 2(0.05208)


162


....................................................... 33.

a.

Enter the data on your calculator. The technique for a TI-83 calculator is illustrated here. 5-lb ball

y = 0.6130952381t2 – 0.0714285714t + 0.1071428571 10-lb ball

y = 0.6091269841t2 – 0.0011904762t – 0.3 15-lb ball

y = 0.5922619048t2 + 0.3571428571t – 1.520833333 b. 34.

All the regression equations are approximately the same. Therefore, there is one equation of motion.


a.

y = 454.1584409x – 40.78364910

b.

y = 454.1584409(1.5) – 40.78364910 ≈ 640 kilometers per second


Connecting Concepts

Chapter Review

163

.......................................................


Graphing Piecewise Functions with a Graphing Calculator 1.

Use Dot mode. Enter the function as Y1=X2*(X 3.067}}

2.

In(3):= NSolve[3x^3-4x^2+x-3==0] Out[3]= {{x -> -0.102814 – 0.799511 I}, {x -> -0.1022814 + 0.799511 I}, {x -> 1.53896}}

3.

In(4):= NSolve[4x^5-3x^3+2x^2-x+2==0] Out[4]= {{x -> -1.25095},, (x -> – 0.156173 – 0.685216 I}, {x -> -0.156173 + 0.685216 I}, {x -> 0.781647 – 0.445283 I}, {x -> 0.781647 + 0.445283 I}}

4.

In(5):= NSolve[-3x^4-6x^3+2x-8==0] Out[5]= {{x -> -1.60199 – 0.623504 I}, {x -> -1.600199 + 0.623504 I}. {x -> 0.601988 – 0.743846 I}, {x -> 0.601988 + 0.7344846 I}}

.......................................................

Assessing Concepts

1.

True

2.

True

3.

False, Descartes’ Rule of Signs indicates that

4.

R( x ) =

x 3 − x 2 + x − 1 has three or one positive zeros. 5.

P( x ) = x − 1 + i is one example.

6.

c

7.

d

8.

f

9.

a

10.

b

11.

e

12.

f

3x 2 is one example. ( x − 2)( x − 5)

....................................................... 1.

3

−11 12 1

4 4

3.

−2

3 3

0 −6 −6

5 3 8 −5 12 7

−2 24 22

4 x2 + x + 8 +

Chapter Review 2.

22 x −3

1

3x 2 − 6 x + 7 + [3.1]

−13 x+2

4.

1 2

−18 5 −13

0 5 5

5

[3.1]

1 −14 −13

5

2 −13 −11

2

7

16

−10

2

1 8

4 20

10 0


5 x 2 + 5 x − 13 +

[3.1]

2 x 2 + 8 x + 20 [3.1]

−11 x −1

Chapter Review

5.

5

229

−10 15 5

3 3

7.

4

1 1

9.

−2

6 6

11.

1

3

1 13.

1

1 1

−5 24 19

2 4 6 0 −12 −12

0 0 0

55 −55 0

1 76 77

−12 24 12 −26 15 −11

2 3 5 −1 1 0

−36 25 −11

−2 0 −2

15.

6.

8.

33 −33 0 1 −2 −1

1 1

P (4) = 77 [3.1] 8 −24 −16

−7

3x 2 + 5 x − 11 [3.1]

1 32 33

6 −14 −8

−65 56 −9

0

−10 −4 −14

8 14 22

−4

−1

4 −4 P (−2) = 33 [3.1]

10.

3

−4

[3.1]

4

−8 15 7

5 5

12.

1 −1 0

9 −7 2

2

14.

1 2

−8 0 −8

8 −8 0

2

−6 69 63

2 21 23

3

−8

3

2

1 4

2 −6

−3 0

16.

x3 + 2 x 2 − 8 x − 9 [3.1]

P (−1) = 22 [3.1]

0 189 189 −31 32 1

2

[3.1]

−63 63 0

−9 567 558 4 −4 0

P (3) = 558 [3.1]

[3.1]

[3.1]

17.

[3.2] [3.2]

[3.2] 18.

19.

20.

[3.2]

[3.2]

[3.2] 21.

p = ±1, ± 2, ± 3, ± 6 [3.3] q

22.

p = ±1, ± 2, ± 3, ± 5, ± 6, ± 10, ± 15, ± 30, ± 1 , ± 3 , ± 5 , ± 15 [3.3] q 2 2 2 2

23.

p = ±1, ± 2, ± 3, ± 4, ± 6, ± 12, ± 1 , ± 2 , ± 4 , ± 1 , ± 2 , ± 3 , ± 4 , ± 6 , ± 12 , ± 1 , ± 2 , ± 4 [3.3] q 3 3 3 5 5 5 5 5 5 15 15 15

24.

p = ±1, ± 2, ± 4, ± 8, ± 16, ± 32, ± 64 [3.3] q

25.

26.

1 1 1 2 p = ±1, ± 2, ± , ± , ± , ± [3.3] q 6 3 2 3

27.

P ( x ) = x 3 + 3x 2 + x + 3 P ( − x ) = − x 3 + 3x 2 − x + 3 no positive and three or one negative real zeros [3.3]

29.

P( x ) = x 4 − x − 1 P( − x ) = x 4 + x − 1 one positive and one negative real zeros [3.3]

28.

P ( x ) = x 4 − 6 x 3 − 5 x 2 + 74 x − 120 P ( − x ) = x 4 + 6 x 3 − 5 x 2 − 74 x − 120 three or one positive and one negative real zeros [3.3]

p = ±1 [3.3] q


230

30.

31.

Chapter 3: Polynomial and Rational Functions

P( x ) = x5 − 4 x 4 + 2 x 3 − x 2 + x − 8 P( − x ) = − x5 − 4 x 4 − 2 x 3 − x 2 − x − 8 five, three or one positive and no negative real zeros [3.3]

1

1

6 1 7

1

−10 10 0

3 7 10

x 2 + 7 x + 10 = 0 ( x + 5)( x + 2) = 0 x = −5 or x = −2

The zeros of x3 + 6 x 2 + 3 x − 10 are 1, − 5, and − 2. [3.3] 32.

2

−10 2 −8

1 1

31 −16 15

−30 30 0

x 2 − 8 x + 15 = 0 ( x − 5)( x − 3) = 0 x = 5 or x = 3

The zeros of x3 − 10 x 2 + 31x − 30 are 2, 5, and 3. [3.3] 33.

−2

6

35 −12 23

6

72 −46 26

60 −52 8

−2

16 −16 0

6

23 −12 11

6

26 −22 4

8 −8 0

The zeros of 6 x 4 + 35 x3 + 72 x 2 + 60 x + 16 are − 2 (multiplicity 2), − 4 /3, and − 1/2. [3.3] 34.

−1

2

2

7

5

7

3

2

−1 6

−3 2

−1 6

−3 0

−3

2

6

2

6

2

−6 0

0 2

−6 0

2x 2 + 2 = 0 2 x 2 = −2 x 2 = −1 x = ± −1 x = ±i

The zeros of 2 x 4 + 7 x3 + 5 x 2 + 7 x + 3 are − 1 / 2, − 3, i, and − i. [3.4] 35.

1

−4 1 −3

1 1

6 −3 3

−4 3 −1

1 −1 0

1

1 1

−3 1 −2

3 −2 1

−1 1 0

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

The zero of x 4 − 4 x3 + 6 x 2 − 4 x + 1 is 1 (multiplicity 4). [3.3] 36.

−1

2

x=

2

−7

22

13

2 x 2 − 8 x + 26 = 0

4 26

−13 0

2( x 2 − 4 x + 13) = 0

2

−1 −8

−(−4) ± (−4)2 − 4(1)(13) 4 ± 16 − 52 4 ± −36 4 ± 6i = = = = 2 ± 3i 2(1) 2 2 2

The zeros of 2 x3 − 7 x 2 + 22 x + 13 are − 1 / 2, 2 + 3 i, and 2 − 3 i. [3.4] 37.

1 – 2i

1 + 2i

1

−4

1

1 – 2i –3 – 2i

6 –7 + 4i –1 + 4i

–4 7 + 6i 3 + 6i

–3 – 2i 1 + 2i 2

–1 + 4i –2 – 4i 3

3 + 6i –3 – 6i 0

1 1

15 15 0

x 2 − 2 x − 3 = ( x − 3)( x + 1) = 0 x = 3, x = −1 The remaining zeros are 1 + 2i, 3, and –1. [3.4] Copyright © Houghton Mifflin Company. All rights reserved.

6x 2 + 11x + 4 = 0 (3x + 4)(2 x + 1) = 0 x = − 4 or x = − 1 3 2

Chapter Review

38.

231

2+i

1 1

2–i

1 1

−1 2+i 1+i

−17 1 + 3i –16 + 3i

1+i 2–i 3

–16 + 3i 6 – 3i −10

55 –35 – 10i 20 – 10i

−50 50 0

20 – 10i –20 + 10i 0

x 2 + 3x − 10 = ( x + 5)( x − 2) = 0 x = −5, x = 2 The remaining zeros are 2 – i, –5, and 2. [3.4] 39.

( x − 4)( x + 3)(2 x −1) = ( x 2 − x −12)(2 x −1) = 2 x3 − x 2 − 2 x 2 + x − 24 x +12 = 2 x3 − 3 x 2 − 23 x +12

40.

[3.4]

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

41.

[3.4]

( x −1)( x − 2)( x − 5i )( x + 5i ) = ( x 2 − 3 x + 2)( x 2 + 25) = x 4 + 25 x 2 − 3 x3 − 75 x + 2 x 2 + 50 = x 4 − 3 x3 + 27 x 2 − 75 x + 50

42.

[3.4]

( x + 2)( x + 2)[ x − (1+ 3i )][ x − (1− 3i )] = ( x 2 + 4 x + 4)( x 2 − 2 x +10) = x 4 − 2 x3 +10 x 2 + 4 x3 − 8 x 2 + 40 x + 4 x 2 − 8 x + 40 = x 4 + 2 x3 + 6 x 2 + 32 x + 40

43.

x2 ⇒ no restrictions on denominator. x +7 The domain is all real numbers. [3.4] F ( x) =

2

[3.4] 44.

45.

x + 2 = 0 ⇒ vertical asymptote : x = −2 3 = 3 ⇒ horizontal asymptote : y = 3 [3.5] 1

47.

48. x + 1 = 0 ⇒ vertical asymptote : x = −1 5 11 −1 2 f ( x) = 2 x + 3+ 8 −2 −3 x +1 ⇒ slant asymptote: y = 2 x + 3 [3.5] 2 3 8

46.

2 2 F ( x ) = 3x2 + 2 x − 5 = 3x + 2 x − 5 ⇒ x ≠ 1 , 4 6 6 x − 25 x + 4 (6 x − 1)( x − 4) 1 The domain is all real numbers except and 4. [3.4] 6

x2 + 2x − 3 = 0 ( x + 3)( x − 1) = 0 ⇒ vertical asymptotes : x = −3, x = 1 2 = 2 ⇒ horizontal asymptote: y = 2 [3.5] 1 2x2 + x + 7 = 0 − 1 ± 12 − 4(1)(7) − 1 ± 1 − 28 − 1 ± − 27 = = 2(1) 2 2 x is not a real number ⇒ vertical asymptote : none x=

6 = 3 ⇒ horizontal asymptote : y = 3 [3.5] 2

49.

50.

51.

52.

[3.5] [3.5]

[3.5]


[3.5]

232


53.

54.

55.

[3.5]

[3.5] 57.

58.

56.

[3.5]

b.

5.75(5000) + 34,200 62,950 = = $12.59 5000 5000 5.75(50,000) + 34,200 321,700 = ≈ $6.43 C (50,000) = 50,000 50,000 y = 5.75. As the number of skateboards that are produced increases, the average cost per skateboard approaches $5.75. [3.5]

a.

F (1) =

60 = 60 = 15o F 12 + 2(1) + 1 4

b.

F (4) =

60 = 60 = 2.4o F 42 + 2(4) + 1 25

c.

F (t ) → 0o F as t → ∞. [3.5]

a.

C (5000) =

59.

a.

As the radius of the blood vessel gets smaller, the resistance gets larger.

b.

As the radius of the blood vessel gets larger, the resistance gets smaller. [3.5]

....................................................... QR1. S (6) = 3.6(6)3 − 36.8(6)2 + 145.2(6) + 8 ≈ 412 3

Quantitative Reasoning QR2.

2

S (8) = 3.6(8) − 36.8(8) + 145.2(8) + 8 ≈ 737.6 There will be about 412,000 hybrid vehicles sold in 2010, and 737,600 hybrid vehicles sold in 2012.

2.5 ⎛⎜ ⎝

96,000 ⎞ ⎛ 96,000 ⎞ + 3600 ⎟ = 2.5 ⎜ ⎟ 18 ⎠ ⎝ 18 + x ⎠ 240,000 240,000 = + 3600 18 18 + x 240,000(18 + x ) = 18(240,000) + 3600(18)(18 + x ) 240,000 x = 1,116,400 + 64,800 x 175,200 x = 1,166,400 x ≈ 6.7 additional mpg

QR3. M (1) = 0.2416(1)3 + 2.3106(1)2 − 1.2373(1) + 25.00 = 26.2879

M (2) = 0.2416(2)3 + 2.3106(2)2 − 1.2373(2) + 25.00 = 33.4846 M (3) = 0.2416(3)3 + 2.3106(3)2 − 1.2373(3) + 25.00 = 47.8777 M (4) = 0.2416(4)3 + 2.3106(4)2 − 1.2373(4) + 25.00 = 70.7548 M (5) = 0.2416(5)3 + 2.3106(5)2 − 1.2373(5) + 25.00 = 103.4035 M (6) = 0.2416(6)3 + 2.3106(6)2 − 1.2373(6) + 25.00 = 147.1114 M (7) = 0.2416(7)3 + 2.3106(7)2 − 1.2373(7) + 25.00 = 203.1661 M (8) = 0.2416(8)3 + 2.3106(8)2 − 1.2373(8) + 25.00 = 272.8552 M (1) + M (2) + M (3) + M (4) + M (5) + M (6) + M (7) + M (8) = 904.9412 ≈ $900 QR4.

[3.5]

3 ⎛⎜ ⎝

96,000 ⎞ ⎛ 96,000 ⎞ + 3600 + 900 ⎟ = 3⎜ ⎟ 18 ⎠ ⎝ 18 + x ⎠ 288,000 288,000 = + 4500 18 18 + x 288,000(18 + x ) = 18(288,000) + 4500(18)(18 + x ) 288,000 x = 1,458,000 + 81,000 x 207,000 x = 1,458,000 x ≈ 7.0 additional mpg


Chapter Test

233

....................................................... 1.

4.

−2

−1 −12 −13

3

5 4 2 −6 3 6 −1 − 13 [3.1] 3x 2 − x + 6 + x+2

−2

Chapter Test

−3

7 6 13 −3 P (−2) = 43 [3.1]

2.

−5 48 43

2 −26 −24

3.

5. P ( x) = −3 x3 + 2 x 2 − 5 x + 2 [3.2] Since An = −3 is negative and n = 3 is odd, the graph of P goes up to the far left and down to the far right.

3x3 + 7 x 2 − 6 x = 0 [3.2]

6.

x(3 x 2 + 7 x − 6) = 0 x(3x − 2)( x + 3) = 0 x = 0, 3 x − 2 = 0, or x + 3 = 0 2 3

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

8.

2

3

P ( x) = ( x − 2) ( x + 2) (2 x − 3)( x + 1) The zeros of P are 2 (multiplicity 2), –2 (multiplicity 2),

−3 −1 1 4 2 2 2 1 1 3 Because P(1) and P(2) have different signs, P must have a real zero between 1 and 2. [3.2]

4

2

5 8 2 13 upper bound: 4 −5

−23 52 29

−38 116 78

5 −23 25 −10 2 2 −5 lower bound: −5 [3.3] 11.

1 2

24 312 336 −38 −10 −48

2

2

−3

−11

6

2

1 −2

−1 −12

−6 0

2

p = ±1, ± 3 q = ±1, ± 2, ± 3, ± 6 1 3 1 1 [3.3] p = ±1, ± 3, ± , ± , ± , ± 2 2 3 6 q

3 (multiplicity 1), and –1 (multiplicity 3). [3.3] 2

9.

[3.1]

P ( x) = 2 x3 − 3 x 2 − x + 1 [3.2] 1 2 1 −3 −1 2 −1 −2 2 −1 −2 −1

2

x = −3

The zeros of 3x 3 + 7 x 2 − 6 x = 0 are 0, 2 , and − 3. [3.2] 3

2

1

x 4 − 4 x3 + 7 x 2 − 6 x + 2.

x=

7.

7 −6 2 −4 1 −3 4 −2 1 −3 4 −2 0 A remainder of 0 implies that x −1 is a factor of 1

10.

P ( x) = x 4 − 3x3 + 2 x 2 − 5 x + 1 P ( − x ) = x 4 + 3 x3 + 2 x 2 + 5 x + 1 four, two, or no positive and no negative real zeros [3.3]

24 240 264

2 x 2 − 2 x − 12 = 0 2( x + 2)( x − 3) = 0 x = −2 or x = 3

12.

2 + 3i

6 6

2 − 3i

6 6

The zeros of 2 x3 − 3x 2 − 11x + 6 are 1/2, − 2, and 3. [3.3]

−5 12 + 18i 7 + 18i 7 + 18i 12 − 18i 19

12 −40 + 57i −28 + 57i −28 + 57i 38 − 57i 10

207 −227 + 30i −20 + 30i −20 + 30i 20 − 30i 0

6 x 2 + 19 x + 10 = 0 (3 x + 2)(2 x + 5) = 0 3x + 2 = 0 2x + 5 = 0 x = −2 / 3

x = −5 / 2 4

The zeros of 6x − 5 x3 +12 x 2 + 207 x +130 are 2 + 3i, 2 − 3i, − 2 / 3, and − 5/ 2. [3.4]


130 −130 0

234

13.


P ( x) = x( x 4 − 6 x3 + 14 x 2 − 14 x + 5) 1 1 5 −6 14 −14 1 −5 9 −5 1 9 0 −5 −5

1

1 1

−5 1 −4

14.

P ( x) = [ x − (1+ i)][ x − (1− i )]( x − 3)( x) [3.4] = ( x 2 − 2 x + 2)( x − 3)( x) = ( x3 − 5 x 2 + 8 x − 6)( x) = x 4 − 5 x3 + 8 x 2 − 6 x

−5 5 0

9 −4 5

x2 − 4 x + 5 = 0 x=

−( −4) ± ( −4)2 − 4(1)(5) 2(1)

x = 4 ± 16 − 20 = 4 ± −4 2 2 4 2 i ± x= =2±i 2 The zeros of x 5 − 6x 4 + 14 x 3 − 14 x 2 + 5 x are 0, 1 (multiplicity 2), 2 + i , and 2 − i. [3.4]

15.

f ( x) =

3x2 − 2 x + 1 x2 − 5x + 6

x2 − 5x + 6 = 0

( x − 3)( x − 2) = 0 x=3 x=2 vertical asymptotes: x = 3, x = 2 [3.5]

17.

16.

f ( x) =

2x2 − 1

horizontal asymptote: y = 18.

[3.5] 19.

3x2 − 2 x + 1

a.

b.

[3.5]

w(t ) = 70t + 120 t + 40 70(1) + 120 70 + 120 190 = = ≈ 5 words per minute w(1) = 1 + 40 41 41 70(10) + 120 700 + 120 820 w(10) = = = ≈ 16 words per minute 10 + 40 50 50 70(20) + 120 1400 + 120 1520 w(20) = = = ≈ 25 words per minute 20 + 40 60 60 As t → ∞, w(t ) → 70 = 70 words per minute [3.5] 1


3 [3.5] 2

Cumulative Review

20.

235

length = 25 – 2(2x) = 25 – 4x width = 18 – 2x height = x volume = length × width × height = (25 − 4 x)(18 − 2 x)( x) = (450 − 122 x + 8 x 2 )( x) = 450 x − 122 x 2 + 8 x3 = 8 x3 − 122 x 2 + 450 x

The value of x (to the nearest 0.001 inch) that will produce a box with the maximum volume is 2.42 inches. The maximum volume (to the nearest 0.1 cubic inch) is 487.9 cubic inches. [3.3]

....................................................... 1.

3.

3 + 4i = 3 + 4i ⋅ 1 + 2i = 3 + 10i + 8i 2 [P.6] 1 − 2i 1 − 2i 1 + 2i 12 − 4i 2 3 + 10i + 8( −1) 3 + 10i − 8 = = 1 − 4( −1) 1+ 4 = −5 + 10i = −1 + 2i 5

Cumulative Review 2.

x=

2x + 5

) = (2 + 2

x −1

)

Check 2: 2(2) + 5 − (2) − 1 = 2 4 + 5 − 2 −1 = 2 9− 1=2 3 −1 = 2 2=2 Yes Check 10: 2(10) + 5 − (10) − 1 = 2 20 + 5 − 10 − 1 = 2 25 − 9 = 2 5−3= 2 2=2 Yes The solutions are x = 2, x = 10. [1.4]

2

2x + 5 = 4 + 4 x − 1 + x − 1 2x + 5 = 3 + 4 x − 1 + x x + 2 = 4 x −1

(

( x + 2) 2 = 4 x − 1

)

2

x 2 + 4 x + 4 = 16( x − 1) x 2 + 4 x + 4 = 16 x − 16 x 2 − 12 x + 20 = 0 ( x − 2)( x − 10) = 0 x = 2, x = 10 4.

x − 3 ≤ 11 −11 ≤ x − 3 ≤ 11 −8 ≤ x ≤ 14 {x | −8 ≤ x ≤ 14}

[1.5]

−( −1) ± ( −1)2 − 4(1)( −1) 1 ± 5 = 2(1) 2

x = 1 − 5 , x = 1 + 5 [1.3] 2 2

2x + 5 − x − 1 = 2 2x + 5 = 2 + x − 1

(

x2 − x − 1 = 0 a = 1, b = −1, c = −1

5.

d = (2 − 7)2 + [5 − ( −11)]2 [2.1] = (2 − 7)2 + (5 + 11)2

6.

Shift the graph of y = x 2 two units to the right and four units up. [2.5]

= ( −5) 2 + (16)2 = 25 + 256 = 281


236


P( x ) = x 2 − 2 x − 3

7.

P( x + h ) − P ( x ) [( x + h )2 − 2( x + h ) − 3] − ( x 2 − 2 x − 3) x 2 + 2 xh + h 2 − 2 x − 2h − 3 − x 2 + 2 x + 3 = = h h h 2 xh h h 2 2 + − = = 2x + h − 2 [2.6] h

8.

f ( x) = 2 x 2 + 5 x − 3 g ( x) = 4 x − 7 ( f o g )( x) = f [ g ( x)] = f (4 x − 7)

[2.6]

9.

( f − g )( x ) = f ( x ) − g ( x )

[2.6]

= x 3 − 2 x + 7 − ( x 2 − 3x − 4) = x 3 − 2 x + 7 − x 2 + 3x + 4 = x 2 − x 2 + x + 11

= 2(4 x − 7)2 + 5(4 x − 7) − 3 = 2(16 x 2 − 56 x + 49) + 5(4 x − 7) − 3 = 32 x 2 − 112 x + 98 + 20 x − 35 − 3 = 32 x 2 − 92 x + 60

10.

12.

−2

−2 −4 16 −28 4 14 −32 3 2 59 4 x − 8 x + 14 x − 32 + [3.1] x+2 4

0 −8 −8

−5 64 59

11.

3

2

2 P(3) = 141 [3.1]

0 6 6

−3 18 15

The leading term has a negative coefficient. The graph of P(x) goes down to the far right. [3.2]

13.

The relative maximum (to the nearest 0.0001) is 0.3997. [3.2] 14.

P( x ) = 3x 4 − 4 x 3 − 11x 2 + 16 x − 4 [3.3] p = ± factors of 4 = ±1, ± 2, ± 4 q = ± factors of 3 = ±1, ± 3 p = ±1, ± 2, ± 4, ± 1 , ± 2 , ± 4 3 3 3 q

15.

P( x ) = x 3 + x 2 + 2 x + 4 has no changes of sign. There are no positive real zeros. P( − x ) = − x 3 + x 2 − 2 x + 4 has three changes of sign. There are three or one negative real zeros. [3.3]


4 45 49

−6 147 141

Cumulative Review

16.

237

P( x ) = x 3 + x + 10 no positive and one negative real zeros p = ±1, ± 2, ± 5, ± 10 q 1 0 1 10 −2 4 −2 −10 1 5 0 −2 x=

17.

If 3 + i is a zero of P(x), then 3 – i is also a zero. P ( x) = [ x − (3 + i )][ x − (3 − i )]( x + 2) = [ x − 3 − i ][ x − 3 + i ]( x + 2) = [( x − 3)2 − i 2 ]( x + 2) = [ x 2 − 6 x + 9 − (−1)]( x + 2) = [ x 2 − 6 x + 9 + 1]( x + 2)

−( −2) ± ( −2)2 − 4(1)(5) 2(1)

= ( x 2 − 6 x + 10)( x + 2) = x 2 ( x + 2) − 6 x( x + 2) + 10( x + 2)

= 2 ± −16 = 2 ± 4i = 1 ± 2i 2 2 The zeros are –2, 1 – 2i, 1 + 2i. [3.4]

18.

P( x ) = x 3 − 2 x 2 + 9 x − 18 three or one positive and no negative real zeros p = ±1, ± 2, ± 3, ± 6, ± 9, ± 18 q 2 1 9 −2 −18 2 0 18 1 0 9 0 x2 + 9 = 0

= x3 + 2 x 2 − 6 x 2 − 12 x + 10 x + 20 = x3 − 4 x 2 − 2 x + 20

19.

F ( x) =

4 x2 x + x−6 2

Vertical asymptotes: x2 + x − 6 = 0 ( x + 3)( x − 2) = 0 x = –3, x = 2 Horizontal asymptote: y = 4 ⇒ y = 4 [3.5] 1

x 2 = −9 x = ± −9 x = ±3i P( x ) = ( x − 2)( x + 3i )( x − 3i ) [3.4]

20.

3 2 F ( x ) = x +24 x + 1 x +4 x+4 +1 x2 + 4 x3 + 4 x2

x3

+ 4x 2

4x − 4x + 1 + 16 4x2 − 4 x − 15 The slant asymptote is y = x + 4. [3.5]


[3.4]

Chapter 4

Exponential and Logarithmic Functions Section 4.1 1.

If f (3) = 7 , then f −1 (7) = 3.

2.

If g ( −3) = 5 , then g −1 (5) = −3.

3.

If h −1 ( −3) = −4 , then h(−4) = −3.

4.

If f −1 (7) = 0 , then f (0) = 7.

5.

If 3 is in the domain of f −1 , then f [ f −1 (3)] = 3.

6.

a.

If f is a one-to-one function and f (0) = 5 , then f −1 (5) = 0.

b.

If f is a one-to-one function and f (1) = 2 , then f −1 (2) = 1.

7.

The domain of the inverse function f −1 is the range of f .

8.

The range of the inverse function f −1 is the domain of f .

9.

10.

11.

12.

Yes, the inverse is a function.

Yes, the inverse is a function. 13.

14.


15.

No, the inverse relation is not a function.

f ( x ) = 4 x; g ( x ) = x 4 x x =4 =x f [ g ( x )] = f 4 4 g [ f ( x )] = g (4 x ) = 4 x = x 4 Yes, f and g are inverses of each other.

() ()


No, the inverse relation is not a function. 18.

Yes, the inverse is a function.

No, the inverse relation is not a function.

f ( x ) = 3 x; g ( x ) = 1 3x 1 f [ g ( x )] = f =3 1 = 1 ≠ x 3x 3x x No, f and g are not inverses of each other.

( ) ( )


Section 4.1

19.

239

f ( x ) = 4 x − 1; g ( x ) = 1 x + 1 4

20.

4

(4 4) = 4( 1 x + 1 ) −1 = x +1−1 4 4

2

2

2

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

g [ f ( x )] = g 1 x − 3 4

4

4

=x Yes, f and g are inverses of each other. f ( x ) = − 1 x − 1 ; g ( x ) = −2 x + 1

=x Yes, f and g are inverses of each other. 22.

2

2

3

3

f [ g ( x )] = f 1 x − 2 3 3

= − 1 ( −2 x + 1) − 1 = x − 1 − 1 2

f ( x ) = 3x + 2; g ( x ) = 1 x − 2

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

f [ g ( x )] = f ( −2 x + 1) 2

2

=x

= 1 (4 x − 1) + 1 = x − 1 + 1

2

2

= 1 (2 x + 3) − 3 = x + 3 − 3

=x g [ f ( x )] = g (4 x − 1)

21.

2

f [ g ( x )] = f (2 x + 3)

f [ g ( x )] = f 1 x + 1

4

f ( x) = 1 x − 3 ; g ( x) = 2 x + 3

2

= x −1 ≠x No, f and g are not inverses of each other.

=x g [ f ( x )] = g (3x + 2) = 1 (3x + 2) − 2 = x + 2 − 2 3

3

3

3

=x Yes, f and g are inverses of each other.

23.

5 ; g ( x) = 5 + 3 x−3 x 5 +3 f [ g ( x )] = f x 5 = = 5 = 5⋅ x 5 + 3−3 5 5 x x =x f ( x) =

24.

( ) ( )

( )

g [ f ( x )] = g

( x 5− 3)

5 + 3= x − 3+ 3 5 x−3 =x Yes, f and g are inverses of each other.

( )

=

25.

f ( x ) = x 3 + 2; g ( x ) = 3 x − 2

f [ g ( x )] = f ( 3 x − 2 ) 3

= (3 x − 2 ) + 2 = x − 2 + 2 =x

g [ f ( x )] = g ( x 3 + 2 ) 3 3

f ( x) = 2 x ; g ( x) = x x −1 x−2 x f [ g ( x )] = f x−2 2x 2x 2 x x−2 = x−2 = x−2 = x − 1 x − ( x − 2) 2 x−2 x−2 x−2 = 2x ⋅ x − 2 x−2 2 =x g [ f ( x )] = g 2 x x −1 2x 2x 2x 1 1 x x x − − = = = −1 2 x − 2 2 x − 2( x − 1) 2 x −1 x −1 x −1 = 2x ⋅ x − 1 x −1 2 =x Yes, f and g are inverses of each other.

26.

f ( x ) = ( x + 5)3 ; g ( x ) = 3 x − 5

f [ g ( x )] = f ( 3 x − 5)

3

= ( 3 x − 5 + 5) = x 3 =x

(

g [ f ( x )] = g ( x + 5)3 3 3

= x +2−2 = x =x Yes, f and g are inverses of each other.

3

)

= 3 ( x + 5)3 − 5 = x + 5 − 5 =x Yes, f and g are inverses of each other.


Chapter 4: Exponential and Logarithmic Functions

240 27.

The inverse of {( −3, 1), ( −2, 2), (1, 5), (4, − 7)} is {(1, − 3), (2, − 2), (5, 1), ( −7, 4)}.

28.

The inverse of {( −5, 4), ( −2, 3), (0, 1), (3, 2), (7, 11)} is {(4, − 5), (3, − 2), (1, 0), (2, 3), (11, 7)}.

29.

The inverse of {(0, 1), (1, 2), (2, 4), (3, 8), (4, 16)} is {(1, 0), (2, 1), (4, 2), (8, 3), (16, 4)}.

30.

The inverse of {(1, 0), (10, 1), (100, 2), (1000, 3), (10000, 4) is {(0, 1), (1, 10), (2, 100), (3, 1000), (4, 10,000). f ( x) = 2 x + 4

31.

32.

f ( x) = 4 x − 8

x = 2y + 4

1x+2= y 4 f −1 ( x ) = 1 x + 2 4

f ( x ) = −3 x − 8 x = −3 y − 8 x + 8 = −3 y

1x+7 = y 3 3 −1 f ( x) = 1 x + 7 3 3

f ( x ) = −2 x + 5 x = −2 y + 5 x − 5 = −2 y

35.

−1 x − 8 = y

f ( x) = − x + 3 x = −y + 3 y = −x + 3

36.

−1 x+ 5 = y 2

3

f −1 ( x ) = − x + 3

2

f −1 ( x ) = − 1 x + 5

( x) = − 1 x − 8

f

x + 7 = 3y

x +8 = 4y

1 x−2= y 2 f −1 ( x ) = 1 x − 2 2

3 3 −1

x = 3y − 7

x = 4y −8

x − 4 = 2y

34.

f ( x ) = 3x − 7

33.

2

3

2

37.

f ( x) = 2 x , x ≠ 1 x −1 2y x= y −1 x ( y − 1) = xy − x = 2 y xy − 2 y = y ( x − 2) = x y= x x−2 −1 f ( x) = x , x ≠ 2 x−2

38.

x , x≠2 x−2 y x= y−2 x ( y − 2) = xy − 2 x = y xy − y = y ( x − 1) = 2 x y = 2x x −1 −1 f ( x) = 2 x , x ≠ 1 x −1

39.

f ( x ) = x − 1 , x ≠ −1 x +1 y −1 x= y +1 x ( y + 1) = xy + x = y − 1 xy − y = − x − 1 y − xy = y (1 − x ) = x + 1 y = x +1 1− x f −1 ( x ) = x + 1 , x ≠ 1 1− x

40.

f ( x ) = 2 x − 1 , x ≠ −3 x+3 2y −1 x= y+3 xy + 3x = 2 y − 1 xy − 2 y = −3x − 1 y = 3x + 1 2− x −1 3 f ( x) = x + 1 , x ≠ 2 2− x

41.

f ( x ) = x 2 + 1, x ≥ 0

42.

f ( x ) = x 2 − 4, x ≥ 0

43.

f ( x) =

x + 4 = y2

x − 1 = y2 x −1 = y

x+4 = y

−1

−1

f ( x ) = x − 1, x ≥ 1 Note: Do not use ± with the radical because the domain of f , and thus the

f ( x ) = x + 4, x ≥ −4 Note: Do not use ± with the radical because the domain of f , and thus the

range of f −1 , is nonnegative.

range of f −1 , is nonnegative.

44.

f ( x ) = x − 2, x ≥ 2 x=

x = y2 − 4

x = y2 + 1

y−2

f ( x) = 4 − x , x ≤ 4 x = 4− y

2

2

x = y−2

x =4− y

2

y = − x2 + 4

x +2= y f −1 ( x ) = x 2 + 2, x ≥ 0 Note: The range of f , is nonnegative, therefore the

f −1 ( x ) = − x 2 + 4, x ≥ 0 Note: The range of f , is non-negative, therefore the

domain of f −1 is also nonnegative.

domain of f −1 is also non-negative.


Section 4.1

241

f ( x ) = x 2 + 4 x , x ≥ −2

45.

f ( x ) = x 2 − 6 x, x ≤ 3

46.

x = y2 + 4 y

x = y2 − 6 y

x + 4 = y2 + 4 y + 4

x + 9 = y2 − 6 y + 9

x + 4 = ( y + 2)2

x + 9 = ( y − 3)2

x+4 = y+2

− x+9 = y−3

y = x+4 −2 f

−1

− x +9 +3= y

f −1 ( x ) = − x + 9 + 3, x ≥ −9

( x ) = x + 4 − 2, x ≥ −4

Note: The range of f , is non-negative, therefore the domain of f

−1

range of f −1 must also be non-positive.

is also non-negative.

f ( x ) = x 2 + 4 x − 1, x ≤ −2

47.

Note: Because the range of f , is non-positive, the

f ( x ) = x 2 − 6 x + 1, x ≥ 3

48.

x = y2 + 4 y − 1

x = y2 − 6 y + 1

x + 1 = y2 + 4 y

x − 1 = y2 − 6 y

x + 1 + 4 = y2 + 4 y + 4

x − 1 + 9 = y2 − 6 y + 9

x + 5 = ( y + 2)2

x + 8 = ( y − 3)2

− x+5 = y+2

x+8 = y −3

− x+5−2= y

x +8 +3= y

−1

49.

f ( x ) = − x + 5 − 2, x ≥ −5 Note: Because the range of f , is non-positive, the

f −1 ( x ) = x + 8 + 3, x ≥ −8 Note: The range of f , is non-negative, therefore the

range of f −1 must also be non-positive.

domain of f −1 is also non-negative.

V ( x) = x3 x= y 3

50.

3

x=y

V −1 ( x ) = 3 x V −1 ( x ) finds the length of a side of a cube given the volume. 51.

f ( x ) = 5 ( x − 32) 9 x = 5 ( y − 32) 9 9 x = y − 32 5

f ( x ) = 12 x x = 12 y x =y 12 −1 f ( x) = x 12 f −1 ( x ) converts x inches into feet.

52.

a. b.

9 x + 32 = y 5 f −1 ( x ) = 9 x + 32 5 f −1 ( x ) is used to convert x degrees Celsius to an equivalent Fahrenheit temperature.

S (96) = 3 (96) + 18 = $162 2 S ( x ) = 3 x + 18 2 x = 3 y + 18 2 x − 18 = 3 y 2 2 x − 12 = y 3 S −1 ( x ) = 2 x − 12 3 −1 2 S (399) = (399) − 12 = $254 3



242 s ( x ) = 2 x + 24 x = 2 y + 24 x − 24 = 2 y

53.

54.

1 x − 12 = y 2 s −1 ( x ) = 1 x − 12 2

K ( x ) = 1.3x − 4.7 x = 1.3 y − 4.7 x + 4.7 = 1.3 y x + 4.7 = y 1.3 K −1 ( x ) = x + 4.7 1.3

55.

E ( s ) = 0.05s + 2500 s = 0.05 y + 2500 s − 2500 = 0.05 y 1 s − 2500 = y 0.05 0.05 20s − 50,000 = y E −1 ( s ) = 20s − 50,000 The executive can use the inverse function to determine the value of the software that must be sold in order to achieve a given monthly income.

56.

No. It is not a one-to-one function. For a given cost, there is more than one weight that can be associated with that cost.

57.

a. b.

p(10) ≈ 0.12 = 12% ; p (30) ≈ 0.71 = 71% The graph of p, for 1 ≤ n ≤ 60 , is an increasing function. Thus p has an inverse that is a function.

c.

p −1 (0.223) represents the number of people required to be in the group for a 22.3% probability that at least two of the people will share a birthday.

a.

D f (13) = 2(13) − 1 = 25 O f (24) = 2(24) − 1 = 47 (space) f (36) = 2(36) − 1 = 71 Y f (34) = 2(34) − 1 = 67 O f (24) = 47 U f (30) = 2(30) − 1 = 59 R f (27) = 2(27) − 1 = 53 (space) f (36) = 71 H f (17) = 2(17) − 1 = 33 O f (24) = 47 M f (22) = 2(22) − 1 = 43 E f (14) = 2(14) − 1 = 27 W f (32) = 2(32) − 1 = 63 O f (24) = 47 R f (27) = 53 K f (20) = 2(20) − 1 = 39 The code is 25 47 71 67 47 59 53 71 33 47 43 27 63 47 53 39.

59.

c. 60.

58.

b.

a. b.

Answers will vary. No. L is not a one-to-one function.

f −1 (49) = 49 + 1 = 25 2 −1 33 f (33) = + 1 = 17 2 −1 47 + 1 = 24 f (47) = 2 f −1 (45) = 45 + 1 = 23 2 −1 27 + 1 = 14 f (27) = 2 f −1 (71) = 71 + 1 = 36 2

P H O N E (space)

f −1 (33) = 17 H f −1 (47) = 24 O f −1 (43) = 43 + 1 = 22 M 2 f −1 (27) = 14 E The message is PHONE HOME.

Answers will vary.

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

g −1 (59) = 59 − 3 = 28 2 −1 31 g (31) = − 3 = 14 2 −1 39 g (39) = − 3 = 18 2 g −1 (73) = 73 − 3 = 35 2

S E I Z

g −1 (31) = 14 E

The message is SEIZE THE DAY.

g −1 (75) = 75 − 3 = 36 (space) 2

g −1 (61) = 61 − 3 = 29 T 2 −1 37 g (37) = − 3 = 17 H 2 g −1 (31) = 14 E g −1 (75) = 36 (space) g −1 (29) = 29 − 3 = 13 D 2 g −1 (23) = 23 − 3 = 10 A 2 g −1 (71) = 71 − 3 = 34 Y 2


Section 4.1

61.

63.

243

f (2) = 7, f (5) = 12, and f (4) = c. Because f is an increasing linear function, and 4 is between 2 and 5, then f (4) is between f (2) and f (5) . Thus, c is between 7 and 12. f is a linear function, therefore f −1 is a linear function.

62.

f (1) = 13, f (4) = 9, and f (3) = c. Because f is a decreasing linear function, and 3 is between 1 and 4, then f (3) is between f (1) and f (4) Thus, c is between 9 and 13.

64.

f is a linear function, therefore f −1 is a linear function.

f (2) = 3 ⇒ f −1 (3) = 2

f (5) = −1 ⇒ f −1 ( −1) = 5

f (5) = 9 ⇒ f −1 (9) = 5

f (9) = −3 ⇒ f −1 ( −3) = 9

Since 6 is between 3 and 9, f −1 (6) is between 2 and 5. 65.

Since −2 is between –1 and –3, f −1 ( −2) is between 5 and 9.

g is a linear function, therefore g −1 is a linear function.

66.

g is a linear function, therefore g −1 is a linear function.

g −1 (3) = 4 ⇒ g (4) = 3

g −1 ( −2) = 5 ⇒ g (5) = −2

g −1 (7) = 8 ⇒ g (8) = 7 Since 5 is between 4 and 8, g (5) is between 3 and 7.

g −1 (0) = −3 ⇒ g ( −3) = 0 Since 0 is between 5 and –3, then g (0) is between –2 and 0.

....................................................... 67.

68.

Connecting Concepts

f ( x ) = mx + b, m ≠ 0 y = mx + b x = my + b x − b = my x−b = y m b 1 −1 f ( x) = x − m m b⎞ 1 ⎛ and the y-intercept is ⎜ 0, − ⎟ . The slope is m⎠ m ⎝

f ( x ) = ax 2 + bx + c, a > 0, x > − b 2a

{

}

2⎫ ⎧ Domain of f is x x > − b , Range of f is ⎨ y y ≥ 4ac − b ⎬ . 4a ⎭ 2a ⎩

{

}

2⎫ ⎧ Domain of f −1 is ⎨ x x ≥ 4ac − b ⎬ , Range of f −1 is y y > − b . 2a 4 a ⎩ ⎭

y = ax 2 + bx + c x = ay 2 + by + c x − c = a y2 + b y a b2 + x − c = ⎛ y 2 + b y + b2 ⎞ Complete the square. ⎜ ⎟ a a 4a 2 4a 2 ⎠ ⎝

(

)

(

b2 + 4ax − 4ac = y + b 2a 4a 2

(

)

2

2 a + b + 4ax2 − 4ac = y + b 2a 4a

)

{

}

Choose the positive root, since the Range of f −1 is y y > − b . 2a

2 Thus f −1 ( x ) = − b + b + 4ax2 − 4ac 2a 4a

2 2 f −1 ( x ) = −b + b + 4ax − 4ac , a ≠ 0, x ≥ 4ac − b 2a 4a



244 69.

The reflection of f across the line given by y=x yields f. Thus f is its own inverse.

70.

The reflection of f across the line given by y=x yields f. Thus f is its own inverse.

71.

There is at most one point where each horizontal line intersects the graph of the function. The function is a one-to-one function.

72.

There is at most one point where each horizontal line intersects the graph of the function. The function is a one-to-one function.

73.

A horizontal line intersects the graph of the function at more than one point. Thus, the function is not a one-to-one function.

74.

A horizontal line intersects the graph of the function at more than one point. Thus, the function is not a one-to-one function.

....................................................... PS2. 3−4 = 1 = 1 = 1 34 3⋅3⋅3⋅3 81

PS1. 23 = 2⋅ 2⋅2 = 8

PS3.

PS5.


1

4+ 1 22 + 2−2 = 4 = 16 +1 = 17 2 2 8 8

PS4. 32 − 3−2 9 − 9 81−1 80 40 = = = = 2 2 18 18 9

f ( x) =10 x

PS6.

f (−1) =10−1 = 1 10 f (0) =100 =1

() f ( −1) = ( ) = 2 f (0) = ( ) =1 f (1) = ( ) = f (2) = ( ) = f ( x) =

1 2

1 2

f (1) =101 =10 f (2) =102 =100

1 2

1 2

x

1 2 0

−1

1

2

1 2

1 4

Section 4.2 1.

f (0) = 30 = 1

2.

f (4) = 34 = 81

4.

g (0) = 40 = 1 g ( −1) = 4 −1 = 1 4

7.

( ) =4 j (4) = ( 1 ) = 1 2 16 j ( −2) = 1 2

4

−2

5.

f (3) = 53 = 125 f ( −2) = 5−2 = 1 25

3.

g (3) = 103 = 1000

( ) = 94 h( −3) = ( 3 ) = 8 2 27 h(2) = 3 2

2

6.

−3

8.

g ( −2) = 10−2 = 1 100

( ) = 25 h(3) = ( 2 ) = 8 5 125 h( −1) = 2 5

3

( ) =4 j (5) = ( 1 ) = 1 4 1024 j ( −1) = 1 4

−1

5


−1

Section 4.2

245

9.

f (3.2) = 23.2 ≈ 9.19

10.

f ( −1.5) = 3−1.5 ≈ 0.19

11.

g (2.2) = e2.2 ≈ 9.03

12.

g ( −1.3) = e −1.3 ≈ 0.27

13.

h( 2) = 5

2

14.

π h(π ) = 5π ≈ 0.11 h(π ) = 5 ≈ 0.11

15.

f ( x ) = 5 x is a basic exponential graph.

19.

f ( x ) = 10 x

22.

f ( x) = 5 2

≈ 9.74

g ( x ) = 1 + 5− x is the graph of f ( x ) reflected across the y-axis and moved up 1 unit. h( x ) = 5x + 3 is the graph of f ( x ) moved to the left 3 units. k ( x ) = 5x + 3 is the graph of f ( x ) moved up 3 units. a. k ( x ) 16.

b. g ( x )

c. h( x )

d. f ( x )

( ) is an exponential function with a base between 0 and 1. g ( x ) = ( 1 ) is the graph of f ( x ) reflected across the y-axis. 4 h( x ) = ( 1 ) is the graph of f ( x ) moved 2 units to the right. 4 k ( x ) = 3 ( 1 ) is the graph of f ( x ) stretched vertically by a factor of 3. 4 x

f ( x) = 1 4

−x

x −2

x

a. k ( x )

b. f ( x )

c. g ( x )

d. h( x )

17.

f ( x ) = 3x

18.

f ( x) = 4 x

20.

f ( x) = 6x

21.

f ( x) = 3 2

23.

f ( x) = 1 3

()

x

()

x

24.

()

f ( x) = 2 3

()

x

x

25.

Shift the graph of f vertically upward 2 units.

26.

Shift the graph of f vertically downward 3 units.

27.

Shift the graph of f horizontally to the right 2 units.

28.

Shift the graph of f horizontally to the left 5 units.

29.

Reflect the graph of f across the y-axis.

30.

Reflect the graph of f across the x-axis.



246 Stretch the graph of f vertically away from the x-axis by a factor of 2.

32.

33.

Reflect the graph of f across the y-axis, and then shift this graph vertically upward 2 units.

34.

Shift the graph of f horizontally 3 units to the right, and then shift this graph vertically upward 1 unit.

35.

Shift the graph of f horizontally 4 units to the right, and then reflect this graph across the x-axis.

36.

Reflect the graph of f across the y-axis, and then reflect this graph across the x-axis.

37.

Reflect the graph of f across the y-axis, and then shift this graph vertically upward 3 units.

38.

Shift the graph of f horizontally 2 units to the left, stretch this graph away from the x-axis by a factor of 3, and then shift this graph vertically downward 1 unit.

31.

39.

x −x f ( x) = 3 + 3 2

40.

x −x f ( x) = e + e 2

43.

f ( x) =

10 , with x ≥ 0 1 + 0.4e −0.5 x

Horizontal asymptote: y = 10 47.

a. b.

2

41.

f ( x ) = − e( x − 4)

44.

46.

f ( x) =

x −x f ( x) = e − e 2

No horizontal asymptote.

Horizontal asymptote: y = 0

No horizontal asymptote 45.

2


No horizontal asymptote 42.

f ( x ) = 4 ⋅ 3− x

Shrink the graph of f vertically towards the x-axis by a factor of 1 .

f ( x ) = 0.5e − x

Horizontal asymptote: y = 0 10 , with x ≥ 0 1 + 1.5e −0.5 x


f ( x ) = 1.353(1.9025) x ⇒ f (9) = 1.353(1.9025)9 ≈ 442 million connections

10 years after January 1, 1998 is in 2008.


Section 4.2

48.

a.

247

A(t ) = 200e −0.014t ⇒ f (45) = 200e −0.014(45) ≈ 107 mg

b.

It will take 99 minutes. 49.

50.

51.

a.

d ( p ) = 25 + 880e −0.18 p ⇒ d (8) = 25 + 880e −0.18(8) ≈ 233 items per month

b.

d ( p ) = 25 + 880e −0.18 p ⇒ d (18) = 25 + 880e −0.18(18) ≈ 59 items per month As p → ∞, d ( p ) → 25. The demand will approach 25 items per month.

a.

I (t ) = 24,000 − 22,000e −0.005t ⇒ I (10) = 24,000 − 22,000e −0.005(10) ≈ $3072.95

b.

I (t ) = 24,000 − 22,000e −0.005t ⇒ I (100) = 24,000 − 22,000e −0.005(100) ≈ $10,656.33 As t → ∞, I (t ) → 24,000. The monthly income will approach $24,000.

a.

P (t ) = 100 ⋅ 22t ⇒ P (3) = 100 ⋅ 22(3) = 100 ⋅ 26 = 100 ⋅ 64 = 6400 bacteria P (t ) = 100 ⋅ 22t ⇒ P (6) = 100 ⋅ 22(6) = 100 ⋅ 212 = 100 ⋅ 4096 = 409,600 bacteria

b.

11.6 hours 52.

a. b.

I ( x ) = 100e −1.5 x ⇒ I (1) = 100e −1.5(1) ≈ 22.3%

5 millimeters 53.

a. b.

P( x ) = (0.9) x ⇒ P(3.5) = (0.9)3.5 ≈ 69.2%

For a transparency of 45%, the UV index is 7.6. 54.

a. b. c.

3600 ⇒ P (0) = 3600 = 3600 = 3600 = 3600 = 3600 = 450 bass 8 1 + 7e −0.05t 1 + 7e −0.05(0) 1 + 7e0 1 + 7(1) 1 + 7 3600 P(t ) = 3600 ⇒ P(12) = ≈ 744 bass 1 + 7e −0.05t 1 + 7e −0.05(12) As t → ∞, P (t ) → 3600. The bass population will increase, approaching 3600. P(t ) =



248 55.

a.

n +1

B(n ) = 3

2

n +1

B(n ) = 3

2

− 3 ⇒ B(5) = 35+1 − 3 = 36 − 3 = 729 − 3 = 726 = 363 beneficiaries 2 2 2 2 − 3 ⇒ B(10) = 310+1 − 3 = 311 − 3 = 177147 − 3 = 177144 = 88,572 beneficiaries 2 2 2 2

b.

13 rounds 56.

a. b.

I ( x ) = 100e −0.95 x ⇒ I (2) = 100e −0.95(2) = 100e −1.9 ≈ 15.0%

about 0.73 foot 57.

a. b.

T (t ) = 65 + 115e −0.042t ⇒ T (10) = 65 + 115e −0.042(10) = 65 + 115e −0.42 ≈ 141o F

28.3 minutes 58.

a. b.

T (t ) = 75 + 95e −0.12t ⇒ T (2) = 75 + 95e −0.12(2) = 75 + 95e −0.24 ≈ 149.7o F

8.3 minutes

59.

c.

As t → ∞, T (t ) → 75. Therefore, room temperature is 75o F.

a. b.

f ( n ) = (27.5)2( n −1) /12 ⇒ f (40) = (27.5)2(40−1) /12 = (27.5)239 /12 = (27.5)23.25 ≈ 261.63 vibrations per second No. The function f ( n ) is not a linear function. Therefore, the graph of f ( n ) does not increase at a constant rate.


Section 4.2

249

....................................................... 60.

x −x cosh( x ) = e + e is an even function. That is, prove 2 cosh( − x ) = cosh(x ).


x −x cosh( x ) = e + e 2

Proof:

x −x sinh( x ) = e − e is an odd function. That is, prove 2 sinh( − x ) = −sinh(x ). x −x sinh( x ) = e − e 2

Proof:

−x x cosh( − x ) = e + e 2

−x x sinh( − x ) = e − e 2

cosh( − x ) =

−x x sinh( − x ) = − e + e 2

( e x − e− x )

2 cosh( − x ) = F ( x )

sinh( − x ) =

( e x − e− x )

2 sinh( − x ) = − F ( x ) 62.

63.

64.

65.

domain: (−∞, ∞)

domain: (−∞, ∞)

66.

67.

domain: 68.

domain:

Let f ( x ) = 2 x and g ( x ) = x 2 + 4 . Then h( x ) = 2

70.

( −∞, 0]

( x2 +4)

= f [ x 2 + 4] = f [ g ( x ) ] = ( f o g ) ( x ) .

69.

[ 0, ∞ )

Let f ( x ) = e x and g ( x ) = 2 x − 5 . Then h( x ) = e

( 2 x −5 )

= f [2 x − 5] = f [ g ( x )] = ( f o g ) ( x ) .

x −x By definition the average of two numbers is their sum divided by 2. The expression e + e shows that f is the average of 2

g ( x ) = e x and h( x ) = e − x .



250

....................................................... PS2. 3− x = 1 = 1 3x 27 1 = 1 3x 33 x=3

PS1. 2 x = 16 x

2 =2 x=4

PS4.


4

f ( x) = 2 x x+3 2y x= y+3 xy + 3x = 2 y 3x = 2 y − xy = y (2 − x ) 3x = y 2− x −1 f ( x ) = 3x 2−x

PS5.

PS3. x 4 = 625 x 4 = 54 x=5

g ( x) = x − 2

PS6. The domain is the set of all positive real numbers.

x−2≥0 x≥2 The domain is {x | x ≥ 2} .

Section 4.3 1.

1 = log10 ⇒ 101 = 10

2.

4 = log10,000 ⇒ 104 = 10,000

3.

2 = log8 64 ⇒ 82 = 64

4.

3 = log 4 64 ⇒ 43 = 64

5.

0 = log7 x ⇒ 70 = x

6.

−4 = log3 1 ⇒ 3−4 = 1 81 81

7.

ln x = 4 ⇒ e4 = x

8.

ln e2 = 2 ⇒ e2 = e2

9.

ln1 = 0 ⇒ e0 = 1

10.

ln x = −3 ⇒ e −3 = x

11.

2 = log(3x + 1) ⇒ 102 = 3x + 1

12.

1 = ln ⎛ x + 1 ⎞ ⇒ e1/ 3 = x + 1 ⎜ 2 ⎟ 3 x2 ⎝ x ⎠

13.

32 = 9 ⇒ log3 9 = 2

14.

53 = 125 ⇒ log5 125 = 3

15.

4 −2 = 1 ⇒ log 4 1 = −2 16 16

16.

100 = 1 ⇒ log1 = 0

17.

b x = y ⇒ logb y = x

18.

2 x = y ⇒ log 2 y = x

19.

y = e x ⇒ ln y = x

20.

51 = 5 ⇒ log5 5 = 1

21.

100 = 102 ⇒ log100 = 2

22.

2 −4 = 1 ⇒ log 2 1 = −4 16 16

23.

e2 = x + 5 ⇒ 2 = ln( x + 5)

24.

3x = 47 ⇒ log3 47 = x

25.

log 4 16 = 2 because 42 = 16

27.

−3

3

= ⎛⎜ 2 ⎞⎟ = 8 27 ⎝ 3⎠

26.

log3/ 2 8 = −3 because ⎛⎜ 3 ⎞⎟ 27 ⎝2⎠

log3 1 = −5 because 3−5 = ⎛⎜ 1 ⎞⎟ = 1 243 243 ⎝ 3⎠

28.

logb 1 = 0 because b0 = 1

29.

ln e3 = 3 because e3 = e3

30.

logb b = 1 because b1 = b

31.

log 1 = −2 because 10−2 = 12 = 1 100 100 10

32.

log1,000,000 = 6 because 106 = 1,000,000

5


Section 4.3

251

()

−4

33.

log0.5 16 = log1/ 2 16 = −4 because 1 2

34.

log0.3 100 = log3/10 100 = −2 because ⎛⎜ 3 ⎞⎟ 9 9 ⎝ 10 ⎠

35.

4log1000 = 12 ⇒ log10004 = 12 because 1012 = (103 ) = (1000 ) 4

37.

= 24 = 16 −2

2

= ⎛⎜ 10 ⎞⎟ = 100 9 ⎝ 3⎠ 36.

log5 1252 = 6 because 56 = 125

38.

3log11 161,051 = 15 ⇒ log11161,0513 = 15

4

2 log7 2401 = 8 ⇒ log7 24012 = 8

because 1115 = (115 ) = (161,051)

2 because 78 = ( 74 ) = ( 2401)

3

2

39.

log3 5 9 = 2 ⇒ log3 91/5 = 2 5 5 because 32/5 = ( 32 )

1/5

41.

y = log 4 x x=4

46.

= (169 )

42.

44.

y

47.

y

45.

2/7

2/7 = ( 343)

y = log12 x x = 12 y

y

y = log1/ 2 x x = (1/ 2)

y = log5 / 2 x x = (5 / 2)

y = log 6 x x=6

1/ 3

= ( 36 )

2log7 7 343 = 6 ⇒ log7 3432/7 = 6 7 7 because 76/7 = ( 73 )

5/ 3

y

y = log8 x x=8

49.

1/3

5log13 3 169 = 10 ⇒ log131695/3 = 10 3 3 5/3

log6 3 36 = 2 ⇒ log6 361/3 = 2 3 3 because 62/3 = ( 62 )

1/ 5

= (9)

because 1310/3 = (132 )

43.

40.

3

48.

y

y = log1/ 4 x x = (1/ 4) y

50.

y = log 7 / 3 x x = (7 / 3) y



252 51.

f ( x ) = log5 ( x − 3)

52.

x −3> 0 x>3 The domain is (3, ∞ ).

54.

H ( x ) = log1/ 4 ( x 2 + 1)

55.

P( x ) = ln( x 2 − 4)

x > −1 True for all real numbers. The domain is ( −∞, ∞ ).

⎛ 2 ⎞ h ( x ) = ln ⎜ x ⎟ ⎝ x −4⎠

58.

x3 − x > 0

60.

( x 2 + 7 x + 10) > 0 ( x + 5)( x + 2) > 0 Critical values are –5 and –2. Product is positive. (−∞, −5) ∪ ( −2, ∞)

62.

4x − 8 = 0 4x = 8 x=2 The domain is ( −∞, 2) ∪ (2, ∞ )

64.

x−4=0 x=4 The domain is ( −∞, 4) ∪ (4, ∞ )

x ( x − 1) > 0 x ( x + 1)( x − 1) > 0 Critical values are 0, −1 and 1. Product is positive. −1 < x < 0 or x > 1 (−1,0) ∪ (1, ∞)

(

)

The domain is 11 , ∞ . 2 3x − 7 > 0 3x > 7 x>7 3

x4 − x2 > 0 x 2 ( x + 1)( x − 1) > 0 Critical values are 0, −1 and 1 Product is positive. x < −1 or x > 1 ( −∞, −1) ∪ (1, ∞)

2

63.

J ( x ) = ln ⎜⎛ x − 3 ⎟⎞ ⎝ x ⎠ x−3 >0 x The critical values are 3 and 0. The quotient is positive. x < 0 or x > 3 The domain is ( −∞, 0) ∪ (3, ∞).

x 2 ( x 2 − 1) > 0

x2 > 0 x−4 The critical values are 0 and 4. The quotient is positive. x>4 The domain is (4, ∞ ).

2 x − 11 > 0 2 x > 11 x > 11 2

56.

( x + 2)( x − 2) > 0 The critical values are –2 and 2. The product is positive. The domain is ( −∞, − 2) ∪ (2, ∞ ).

2

61.

k ( x ) = log 2 / 3 (11 − x ) 11 − x > 0 − x > −11 x < 11 The domain is ( −∞, 11).

x −4>0

x +1> 0

59.

53.

2

2

57.

k ( x ) = log 4 (5 − x ) 5− x > 0 − x > −5 x 7; milk of magnesia is a base 71.

pH = − log[H + ]

72.

9.5 = − log[H + ] 10−9.5 = 10log[H

pH = − log[H + ] 5.6 = − log[H + ]

−9.5 = log[H + ] +

pH = − log[H + ]

log[H + ] = −5.6 10log[H ] = 10−5.6 +

]

[H + ] = 3.16 × 10−10 mole per liter

[H + ] = 2.51 × 10−6 mole per liter



260 73.

dB( I ) = 10log ⎛⎜ I ⎞⎟ ⎝ I0 ⎠ a.

⎛ 1.58 × 108 ⋅ I 0 ⎞ dB(1.58 × 108 ⋅ I 0 ) = 10log ⎜ ⎟⎟ ⎜ I0 ⎝ ⎠

⎛ 10,800 ⋅ I 0 ⎞ dB(10,800 ⋅ I 0 ) = 10log ⎜ ⎟ I0 ⎝ ⎠ = 10log(10,800)

b.

= 10log(1.58 × 108 )

c.

≈ 40.3 decibels

≈ 82.0 decibels ⎛ 3.16 × 1011 ⋅ I 0 ⎞ dB(3.16 × 1011 ⋅ I 0 ) = 10log ⎜ ⎟⎟ ⎜ I0 ⎝ ⎠

⎛ 1.58 × 1015 ⋅ I 0 ⎞ dB(1.58 × 1015 ⋅ I 0 ) = 10log ⎜ ⎟⎟ ⎜ I0 ⎝ ⎠

d.

= 10log(3.16 × 1011 ) ≈ 115.0 decibels

74.

= 10log(1.58 × 1015 ) ≈ 152.0 decibels

⎛ I pain ⎞ 125 = 10log ⎜ ⎟ ⎝ I0 ⎠ ⎛ I pain ⎞ 12.5 = log ⎜ ⎟ ⎝ I0 ⎠ I pain 1012.5 = I0

dB( I ) = 10log ⎛⎜ I ⎞⎟ ⎝ I0 ⎠ ⎛I ⎞ 175 = 10log ⎜ Bronco ⎟ I 0 ⎝ ⎠ ⎛I ⎞ 17.5 = log ⎜ Bronco ⎟ I 0 ⎝ ⎠ I 1017.5 = Bronco I0

I Bronco 1017.5 ⋅ I 0 = 12.5 I pain 10 ⋅ I 0 17.5

= 1012.5 10 = 1017.5−12.5 = 105 = 100,000 times more intense

1012.5 ⋅ I 0 = I pain

1017.5 ⋅ I 0 = I Bronco 75.

⎛I ⎞ 110 = 10log ⎜ 110 ⎟ ⎝ I0 ⎠

⎛ ⎞ dB( I ) = 10log ⎜ I ⎟ ⎝ I0 ⎠

⎛I ⎞ 11 = log ⎜ 110 ⎟ ⎝ I0 ⎠ I 1011 = 110 I0

⎛I ⎞ 120 = 10log ⎜ 120 ⎟ ⎝ I0 ⎠ ⎛I ⎞ 12 = log ⎜ 120 ⎟ ⎝ I0 ⎠ I 1012 = 120 I0

I120 1012 ⋅ I 0 = I110 1011 ⋅ I 0 12

= 1011 = 1012 −11 = 101 10 = 10 times more intense

1011 ⋅ I 0 = I110

1012 ⋅ I 0 = I120 10

10

76.

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ dB(2 I ) − dB ( I ) = 10log ⎜ 2 I ⎟ − 10log ⎜ I ⎟ = log ⎜ 2 I ⎟ I I 0 0 ⎝ ⎠ ⎝ ⎠ ⎝ I0 ⎠

⎛ ⎞ − log ⎜ I ⎟ ⎝ I0 ⎠

⎡ (2)10 ( I )10 ( I )10 ⎡ (2 I )10 ( I 0 )10 ⎤ 0 ⎢ = log ⎢ ⋅ = ⋅ log ⎥ 10 10 10 ⎢ ( I ) ⎦⎥ ( I )10 ⎣⎢ ( I 0 ) ⎣ ( I0 ) 77.

10

⎛ 2I ⎞ ⎜ ⎟ I = log ⎝ 0 ⎠

10

⎛ I ⎞ ⎜ ⎟ ⎝ I0 ⎠

⎡⎛ ⎞10 ⎛ I ⎞10 ⎤ = log ⎢⎜ 2 I ⎟ ⋅ ⎜ 0 ⎟ ⎥ ⎢⎝ I 0 ⎠ ⎝ I ⎠ ⎥ ⎣ ⎦

⎤ ⎥ = log 210 = 10log 2 ≈ 3.0103 decibels ⎥ ⎦

⎛ 100,000 I 0 ⎞ 5 M = log ⎜ ⎟ = log100,000 = log10 = 5 I0 ⎝ ⎠


Section 4.4

78.

261

⎛ ⎞ M = log ⎜ I ⎟ ⎝ I0 ⎠

79.

⎛ 398,107,000 I 0 ⎞ M = log ⎜ ⎟ I0 ⎝ ⎠ = log 398,107,000 ≈ 8.6

81.

⎛ ⎞ log ⎜ I ⎟ = M ⎝ I0 ⎠

80.

⎛ ⎞ log ⎜ I ⎟ = 6.5 ⎝ I0 ⎠ I = 106.5 I0

⎛ ⎞ l og ⎜ I ⎟ = M ⎝ I0 ⎠ ⎛ ⎞ log ⎜ I ⎟ = 9.5 ⎝ I0 ⎠ I = 109.5 I0

I = 106.5 I 0

I = 109.5 I 0

I ≈ 3,162,277.7 I 0

I = 3,162, 277,660 I 0

⎛ ⎞ M = log ⎜ I ⎟ ⎝ I0 ⎠ ⎛I ⎞ ⎛I ⎞ I M 5 = log ⎜ 5 ⎟ ⇒ 5 = log ⎜ 5 ⎟ ⇒ 105 = 5 ⇒ 105 I 0 = I 5 I0 ⎝ I0 ⎠ ⎝ I0 ⎠ ⎛I ⎞ ⎛I ⎞ I M 3 = log ⎜ 3 ⎟ ⇒ 3 = log ⎜ 3 ⎟ ⇒ 103 = 3 ⇒ 103 I 0 = I 3 I I I 0 ⎝ 0⎠ ⎝ 0⎠ I 5 105 I 0 105 = = = 105−3 = 102 = 100 to 1 I 3 103 I 0 103

• short cut: begin with this line

82.

109.5 = 109.5−8.3 = 101.2 ≈ 15.8 times more intense 108.3

83.

108.9 = 108.9 −7.1 = 101.8 to 1 ≈ 63 to 1 1 107.1

84.

108.2 = 108.2 −6.9 = 101.3 to 1 ≈ 20 to 1 1 106.9

85.

M = log A + 3log8t − 2.92 = log18 + 3log[8(31)] − 2.92 ≈ 5.5

86.

M = log A + 3log8t − 2.92 = log 26 + 3log[8(17)] − 2.92 ≈ 4.9

87.

Let r = logb M and s = logb N .

88.

Let x = logb M .

Then M = b r and N = b s . Consider the quotient of M and N M N M N M logb N M logb N

89.

a. c.

Then M = b x . M = bx

r = bs b

(M ) p = (bx )

= br − s

p

M p = b xp logb M p = xp

=r−s

logb M p = p logb M

= logb M − logb N

b. M ≈4 M ≈6 When t = 40, M = log A + 3log8t − 2.92 = log50 + 3log[8(40)] − 2.92 ≈ 6.3 When t = 30, M = log A + 3log8t − 2.92 = log1 + 3log[8(30)] − 2.92 ≈ 4.2 The results from parts a. and b. are close to the magnitudes of 6.3 and 4.2 produced by the amplitude time difference formula.

....................................................... PS1. 36 = 729 ⇒ log3 729 = 6

PS2. log5 625 = 4 ⇒ 54 = 625

Prepare for Section 4.5 PS3. a x + 2 = b ⇒ log a b = x + 2



262 PS4.

4a = 7bx + 2cx 7bx + 2cx = 4a x (7b + 2c) = 4a x = 4a 7b + 2c

PS5.

165 = 300 1 + 12 x 165(1 + 12 x ) = 300

PS6.

165 + 1980 x = 300 1980 x = 135 x = 135 = 3 1980 44

A = 100 + x 100 − x A(100 − x ) = 100 + x 100 A − Ax = 100 + x 100 A − 100 = Ax + x 100( A − 1) = x ( A + 1) 100( A − 1) x= A +1

Section 4.5 1.

2 x = 64 x

2 =2 x=6

5.

2.

6

x

13.

3.

5

3 =3 x=5

25 x + 3 = 1 8

6.

25 x + 3 = 2 −3 5 x + 3 = −3 5 x = −6 x =−6 5 9.

3x = 243

34 x −7 = 1 9

7.

34 x −7 = 3−2 4 x − 7 = −2 4x = 5 x=5 4

5 x = 70

10.

6 x = 50

49 x = 1 343

9x = 1 243

4.

7 2 x = 7 −3 2 x = −3

32 x = 3−5 2 x = −5

x=−3 2

x=−5 2

x

⎛2⎞ = 8 ⎜ ⎟ 125 ⎝ 5⎠ x

⎛2⎞ =⎛2⎞ ⎜ ⎟ ⎜ ⎟ ⎝ 5⎠ ⎝ 5⎠ x=3

8.

x

3

⎛2⎞ = ⎛2⎞ ⎜ ⎟ ⎜ ⎟ ⎝ 5⎠ ⎝ 5⎠ x = −2

3− x = 120

11.

x

⎛ 2 ⎞ = 25 ⎜ ⎟ 4 ⎝ 5⎠ −2

7 − x = 63

12.

log(5 x ) = log 70

log(6 x ) = log 50

log(3− x ) = log120

log(7 − x ) = log 63

x log5 = log 70 log 70 x= log5

x log 6 = log 50 log 50 x= log 6

− x log 3 = log120

− x log 7 = log 63

102 x + 3 = 315 log102 x + 3 = log 315 (2 x + 3) log10 = log 315 2 x + 3 = log 315

x=

14.

106− x = 550 (6 − x ) log10 = log550 6 − x = log550 x = 6 − log550

log120 −x = log 3 log120 x=− log 3

15.

log 63 log 7 log 63 x=− log 7

−x =

e x = 10 ln e x = ln10

log 315 − 3 2


x = ln10

Section 4.5

16.

263

e x +1 = 20

21− x = 3x +1

17.

log 21− x = log 3x +1

ln e x +1 = ln 20 x + 1 = ln 20 x = ln 20 − 1

(1 − x ) log 2 = ( x + 1) log 3 log 2 − x log 2 = x log 3 + log 3 log 2 − x log 2 − x log 3 = log 3 − x log 2 − x log 3 = log 3 − log 2 − x (log 2 + log 3) = log 3 − log 2 (log 3 − log 2) (log 2 + log 3) log 2 − log 3 log 2 − log 3 or x= log 2 + log 3 log 6

x=−

3x − 2 = 42 x +1

18.

log 3x − 2 = log 42 x +1 ( x − 2) log 3 = (2 x + 1) log 4 x log 3 − 2log 3 = 2 x log 4 + log 4 x log 3 − 2log 3 − 2 x log 4 = log 4 x (log 3 − 2 x log 4 = log 4 + 2log 3 log 4 + 2log 3 x= log 3 − 2log 4

20.

53 x = 3x + 4

log 22 x −3 = log5− x −1 (2 x − 3) log 2 = ( − x − 1) log 5 2 x log 2 − 3log 2 = − x log 5 − log 5 2 x log 2 + x log 5 − 3log 2 = − log5 2 x log 2 + x log 5 = 3log 2 − log 5

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

21.

log( x 2 + 19) = 2 x 2 + 19 = 102 x 2 + 19 = 100 x 2 = 81 x = ± 81 x = 9, − 9

3log 2 − log 5 2 log 2 + log 5

log(4 x − 18) = 1 4 x − 18 = 101 4 x − 18 = 10 4 x = 28 x=7

log53 x = log 3x + 4 3x log5 = ( x + 4)log 3 3x log5 = x log 3 + 4log 3 3x log5 − x log 3 = 4log 3 x (3log5 − log 3) = 4log 3 4log 3 x= 3log5 − log 3 22.

22 x −3 = 5− x −1

19.

23.

ln( x 2 − 12) = ln x x 2 − 12 = x x 2 − x − 12 = 0 ( x − 4)( x + 3) = 0 x = 4 or x = −3 (No; not in domain.) x=4



264 24.

25.

log(2 x 2 + 3x) = log(10 x + 30)

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

2 x 2 + 3 x = 10 x + 30

log 2 ( x 2 − 4 x ) = 2

2 x 2 − 7 x − 30 = 0 (2 x + 5)( x − 6) = 0

22 = x 2 − 4 x

x = − 5 or x = 6

0 = x2 − 4 x − 4

−5, 6

x=

2

2

4 ± 16 − 4(1)( −4) 2

x = 4±4 2 2 x =2±2 2

2

2 − 2 2 is not a solution because the logarithm of a negative

number is not defined. The solution is x = 2 + 2 2 . 26.

log3 x + log 3( x + 6) = 3

27.

log(5 x − 1) = 2 + log( x − 2) log(5 x − 1) − log( x − 2) = 2 (5 x − 1) log =2 ( x − 2) (5 x − 1) 102 = ( x − 2) 100( x − 2) = 5 x − 1 100 x − 200 = 5 x − 1 95 x = 199 x = 199 95

29.

ln(1 − x) + ln(3 − x) = ln 8 ln[(1 − x)(3 − x)] = ln 8 (1 − x)(3 − x) = 8

log3 x ( x + 6) = 3 33 = x ( x + 6) 27 = x 2 + 6 x 0 = x 2 + 6 x − 27 0 = ( x + 9)( x − 3) x=3 x = −9 log3 (−9) is not defined. The solution is x = 3.

28.

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

3 − 4 x + x2 = 8 x2 − 4 x − 5 = 0 ( x + 1)( x − 5) = 0 x = −1 or x = 5 (No; not in domain.) The solution is x = −1.

30 x − 10 = (2 x + 1) 28 x = 11

x = 11 28

30.

log(4 − x) = log( x + 8) + log(2 x + 13) log(4 − x) = log[( x + 8)(2 x + 13)] 4 − x = ( x + 8)(2 x + 13) 4 − x = 2 x 2 + 29 x + 104

31.

log x 3 − 17 = 1 2 1 log ( x 3 − 17 ) = 1 2 2 101 = x 3 − 17

2

0 = 2 x + 30 x + 100

27 = x 3

2

0 = 2( x + 15 x + 50) 0 = 2( x + 5)( x + 10) x = −5 or x = −10 (No; not in domain.) The solution is x = −5.

3

27 = x 3 3= x The solution is x = 3. 3


Section 4.5

265

log x 3 = (log x )2

32.

3log x = (log x )

33.

log(log x ) = 1

34.

ln(ln x) = 2 e2 = ln x

1

2

10 = log x 10

2

(log x ) − 3log x = 0 log x (log x − 3) = 0

10

2

=x

ee = x

log x = 0 or log x − 3 = 0 x =1

35.

log x = 3 x = 1000

ln(e3 x ) = 6 3 x ln e = 6 3x(1) = 6

36.

ln x = 1 ln ⎛⎜ 2 x + 5 ⎞⎟ + 1 ln 2 2 ⎝ 2⎠ 2 ln x = 1 ln 2 ⎛⎜ 2 x + 5 ⎞⎟ 2 2⎠ ⎝ 1 ln x = ln(4x + 5) 2

3x = 6 x=2

ln x = ln(4 x + 5)1/ 2 x = 4x + 5 2

x = 4x + 5 0 = x2 − 4 x − 5 0 = ( x − 5)( x + 1) x = 5, − 1

Check:

ln5 = 1 ln ⎛⎜ 10 + 5 ⎞⎟ + 1 ln2 2 ⎝ 2⎠ 2 1.609 = 1.2628 + 0.3465 1.609. = 1.609

ln( − 1) = 1 ln ⎛⎜ −2 + 5 ⎞⎟ + 1 2 ⎝ 2⎠ 2 x = −1 is not a solution because ln(−1) is not defined.

The solution is x = 5. 37.

log7 (5 x ) − log7 3 = log7 (2 x + 1)

( )

38.

39.

eln( x −1) = 4 ln eln( x −1) = ln 4 ln( x − 1)ln e = ln 4 ln( x − 1)(1) = ln 4 ( x − 1) = 4 x=5

log 4 x + log 4 ( x − 2) = log4 15 log 4 x ( x − 2) = log 4 15

log7 5 x = log7 (2 x + 1) 3 5x = 2 x + 1 3 5x = 6 x + 3 −3 = x x = –3 is not a solution because log7 ( −15) is undefined. No solution.

x 2 − 2 x = 15 x 2 − 2 x − 15 = 0 ( x − 5)( x + 3) = 0 x = 5, − 3 x = –3 is not a solution because log 4 ( −3) is undefined. The solution is x = 5. 40.

10log(2 x + 7) = 8 log10log(2 x + 7) = log8 log(2 x + 7) = log8 2x + 7 = 8 2x = 1 x=1 2



266 41.

10 x − 10− x = 20 2

42.

10 x + 10− x = 8 2

10 x (10 x − 10− x ) = 40 (10 x )

10 x (10 x + 10− x ) = (16)10 x

102 x − 1 = 40 (10 x )

102 x + 1 = 16(10 x )

102 x − 40 (10 ) − 1 = 0

102 x − 16 (10 x ) + 1 = 0

Let u = 10 x.

Let u = 10 x.

x

u 2 − 40u − 1 = 0 u=

u=

40 ± 402 − 4 (1)( −1)

2 u = 16 ± 256 − 4 2 u = 16 ± 6 7 2 u =8±3 7

2 = 40 ± 1600 + 4 2 ± 40 1604 = 2 ± 40 2 401 = 2 = 20 ± 401

10 x = 20 + 401

( x = log ( 20 +

16 ± 162 − 4 (1)(1)

10 x = 8 ± 3 7

( ) x = log (8 ± 3 7 )

x log 10 = log 8 ± 3 7

) 401 )

log 10 x = log 20 + 401

43.

10 x + 10− x = 5 10 x − 10− x

44.

10 x + 10− x = 5 (10 x − 10− x )

10 x (10 x + 10− x ) = 5 (10 x − 10− x )10 x 102 x + 1 = 5 (102 x − 1) 4 (102 x ) = 6 2 (102 x ) = 3

(10 x )2 = 3 2

x

10 =

3 2

x log 10 = log 3 2 x = log 3 2

10 x − 10− x = 1 10 x + 10− x 2 10 x (10 x − 10− x ) = 1 (10 x + 10− x )(10 x ) 2 2x 10 − 1 = 1 (102 x + 1) 2 2x 10 − 1 = 1 (102 x ) + 1 2 2 2x 2x ) 3 1 ( 10 − 10 = 2 2 1 (102 x ) = 3 2 2 102 x = 3 2 x log 10 = log 3 2 x = log 3 log 3 x= 2

x = 1 log ⎛⎜ 3 ⎞⎟ 2 ⎝2⎠


Section 4.5

45.

267

e x + e − x = 15 2 e x ( e x + e − x ) = (30)e x

46.

e x ( e x − e − x ) = (30)( e x )

e2 x + 1 = e x (30) e

2x

e2 x − 1 = 30e x

x

− 30e + 1 = 0

e2 x − 30e x − 1 = 0

x

Let u = e .

Let u = e x .

2

u − 30u + 1 = 0

u 2 − 30u − 1 = 0

u = 30 ± 900 − 4 2 ± 30 896 u= 2 u = 30 ± 8 14 2 u = 15 ± 4 14

30 ± 900 − 4( −1) 2 ± 30 904 u= = 30 ± 2 226 2 2 u = 15 ± 226 u=

e x = 15 ± 226

x

e = 15 ± 4 14

x ln e = ln (15 ± 226)

x ln e = ln (15 ± 4 14)

x = ln (15 + 226)

x = ln (15 ± 4 14) 47.

1 =4 e x − e− x 1 = 4( e x − e − x ) x

x

e x − e − x = 15 2

x

1( e ) = 4( e )( e − e

48. −x

e x + e− x = 3 e x − e− x e x ( e x + e − x ) = 3( e x − e − x )e x

)

e2 x + 1 = 3e2 x − 3

e x = 4( e2 x − 1)

4 = 2e 2 x

e x = 4e2 x − 4

2 = e2 x

0 = 4e2 x − e x − 4

ln 2 = 2 x ln e ln 2 = x 2

x

Let u = e . 0 = 4u 2 − u − 4 1 ± 1 − 4(4)( −4) 8 1 65 ± u= 8 + 1 65 x e = 8 ⎛ ⎞ x ln e = ln ⎜ 1 + 65 ⎟ 8 ⎝ ⎠ u=

x = ln (1 + 65) − ln 8 49.

2− x +3 = x + 1

50.

3x − 2 = −2 x − 1

Graph f = 2− x +3 − ( x + 1). Its x-intercept is the solution. x ≈ 1.61

Graph f = 3x − 2 + 2 x + 1. Its x-intercept is the solution. x ≈ −0.53

Xmin = −4, Xmax = 4, Xscl = 1,


Ymin = −4, Ymax = 4, Yscl = 1




268 51.

53.

e3− 2 x − 2 x = 1

52.

Graph f = e3− 2 x − 2 x − 1. Its x-intercept is the solution. x ≈ 0.96

Graph f = 2e x + 2 + 3x − 2. Its x-intercept is the solution. x ≈ −1.05





3 log 2 ( x − 1) = − x + 3

54.

2 log(2 − 3x ) − 2 x + 1. log 3 Its x-intercept is the solution. x ≈ 0.38





Graph f =

ln(2 x + 4) + 1 x = −3 2

56.

Graph f = ln (2 x + 4) + 1 x + 3. 2 Its x-intercept is the solution. x ≈ −1.93

2 x +1 = x 2 − 1

2ln(3 − x) + 3x = 4 Graph f = 2ln (3-x) + 3 x − 4. Its x-intercepts are the solutions. x ≈ 0.81, 2.91



57.

2 log3 (2 − 3 x) = 2 x − 1

3 log(x − 1) + x − 3. log 2 Its x-intercept is the solution. x ≈ 2.20 Graph f =

55.

2e x + 2 + 3x = 2

58.

ln( x ) = − x 2 + 4

Graph f = 2 x +1 − x 2 + 1. Its x-intercept is the solution. x ≈ −1.34

Graph f = ln x + x 2 − 4 . Its x-intercept is the solution. x ≈ 1.84






Section 4.5

59.

a.

269

P (0) = 8500(1.1)0 = 8500(1) = 8500

60.

a.

R (1) = 145e −0.092 ≈ 132 beats per minute

P (2) = 8500(1.1)2 = 10, 285 15,000 = 8500(1.1)t

b.

b.

80 = 145e0.092t ln80 = ln145 − 0.092t ln e ln 80 − ln145 =t −0.092 6=t 6 minutes

a.

⎛1⎞ 1/ 2 A ⎜ ⎟ = 80 ( 0.727 ) ≈ 68 ⎝2⎠ A = 68 mg

ln15,000 = 8500(1.1)t ln 51,000 = ln 8500 + t ln (1.1) ln15,000 − ln 8500 =t ln (1.1) 6≈t The population will reach 15,000 in 6 years.

61.

a.

T (10 ) = 36 + 43e

−0.058(10 )

= 36 + 43e −0.58

62.

T ≈ 60o F 45 = 36 + 43e−0.058t

b.

R ( 0 ) = 145e0 = 145 beats per minute

t

50 = 80 ( 0.727 )

b.

ln ( 45 − 36 ) = ln 43 − 0.058t ln e ln ( 45 − 36 ) − ln 43 −0.058

ln 50 = ln 80 + t ln 0.727 ln 50 − ln 80 =t ln 0.727 t ≈ 1.47 hours ≈ 88 minutes

=t

t ≈ 27 minutes 63.

114 = 198 − (198 − 0.9)e −0.23 x

64.

−84 = −197.1e −0.23 x

−73 = −93.4e −0.21x

84 = e −0.23 x 197.1 ln 84 = −0.23x 197.1 84 197.1 x= ≈ 3.7 years −0.23

73 = e −0.21x 93.4 ln 73 = −0.21x 93.4 73 93.4 x= ≈ 1.2 years −0.21

( )

65.

21 = 94 − (94 − 0.6)e −0.21x

( )

5 + 29ln(t + 1) = 65 Graph f = 29ln( x + 1) − 60. Its x-intercept is the solution. x ≈ 6.9 months

66.


0.37ln x + 0.05 = 2.9 Graph f = 0.37ln x − 2.85. Its x-intercept is the solution. x ≈ 2200 thousand people or 2,200,000 people




270 67.

Consider the first function for time less than 10 seconds. 275 = −2.25 x 2 + 56.26 x − 0.28

68.

a.

0 = −2.25 x 2 + 56.26 x − 275.28 −56.26 ± (56.26)2 − 4( −2.25)( −275.28) 2( −2.25) x = 6.67 or 18.33 18.33 s > 10 s, so it is not a solution. The solution is 6.67 s. x=

363.4 − 88.4ln x = 50 Graph f = 313.4 − 88.4ln x. Its x-intercept is the solution. x ≈ 34.65 2(34.65) = 69.3 m

Consider the second function for time greater than 10 seconds. Xmin = −40, Xmax = 80, Xscl = 10, Ymin = −10, Ymax = 10, Yscl = 1

275 = 8320(0.73) x 275 = (0.73) x 8320 ln 275 = ln 0.73x 8320 ln 275 = x ln 0.73 8320 ln 275 8320 ≈ 10.83 x= ln 0.73 The solutions are 6.67 s and 10.83 s.

( ) ( )

b.

( )

568.2 − 161.5ln x = 125 Graph f = 443.2 − 161.5ln x. Its x-intercept is the solution. x ≈ 15.55 2(15.55) = 31.1 m


69.

a.

b. c. d. 71.

70.

48 hours P = 100 As the number of hours of training increases, the test scores approach 100%.

a.

b. c. d.

b. c. d. 72.

In 27 years or 2026 B = 1000 As the number of years increases, the bison population approaches but never exceeds 1000.

a.

45 weeks P = 100 The more experience a person has, the closer the person’s score is to 100%.

a.

b. c. d.

21 hours Y = 50,000 The number of yeast approaches but never exceeds 50,000.


Section 4.5

73.

a.

c. 74.

271

b.

When T = 100, r ≈ 0.019, or 1.9% t=−

9 ⎛ 24 + v ⎞ ln ⎜ ⎟ 24 ⎝ 24 − v ⎠

1.5 = −

9 ⎛ 24 + v ⎞ ln ⎜ ⎟ 24 ⎝ 24 − v ⎠

a.

24 (1.5 )

75.

a.

e

4

( 24 ) − e

4

− v − e v = 24 − 24e

(

50 = e0.64t − 1 100 e0.64t + 1 0.64 t 0.5 = e0.64t − 1 e +1

0.5( e0.64t + 1) = e0.64t − 1 0.5e01.64t + 0.5 = e0.64t − 1 0.5e0.64t − e0.64t = −1.5

v = 24 + v

4

−0.5e0.64t = −1.5 4

e0.64t = 3

)

v −1 − e4 = 24 v=

⎛ 0.64t ⎞ v = 100 ⎜ e0.64t − 1 ⎟ +1⎠ ⎝e ⎛ 0.64t ⎞ 50 = 100 ⎜ e0.64t − 1 ⎟ +1⎠ ⎝e

⎛ 24 + v ⎞ = ln ⎜ ⎟ 9 ⎝ 24 − v ⎠ ⎛ 24 + v ⎞ 4 = ln ⎜ ⎟ ⎝ 24 − v ⎠ 24 + v e4 = 24 − v − = +v 4 v 24 24 ) e (

0.64t = ln 3 t = ln 3 0.64 t ≈ 1.72 In approximately 1.72 seconds, the velocity will be 50 feet per second.

24 − 24e4

−1 − e4 v ≈ 23.1367 The velocity after 1.5 seconds is approximately 23.14 feet per second.

76.

When r = 3%, or 0.03, T ≈78 years

b.

The vertical asymptote occurs when the denominator of 24 + v is zero, or when v = 24. 24 − v

b.

The horizontal asymptote is the value of ⎡ e0.64t − 1 ⎤ 100 ⎢ ⎥ as t → ∞. Therefore, the ⎢⎣ e0.64t + 1 ⎥⎦ horizontal asymptote is v = 100 feet per second.

c.

The velocity of the object cannot reach or exceed 24 feet per second.

c.

The object cannot fall faster than 100 feet per second.

a.

77.

Graph V = 400,000 − 150,000(1.005) x and V = 100,000. They intersect when x ≈ 138.97. After 138 withdrawals, the account has $101,456.39. After 139 withdrawals, the account has $99,963.67. The designer can make at most 138 withdrawals and still have $100,000.

b.

When s = 100, t ≈ 2.6 seconds.

Xmin = 0,Xmax = 200,Xscl = 25 Ymin = −50000,Ymax = 350000,Yscl = 50000



272 78.

a.

h( x ) = 10( e x / 20 + e − x / 20 ) The lowest height of the cable is in the middle, where x = 0. h(0) = 10( e0 / 20 + e −0 / 20 ) = 10( e0 + e0 ) = 10(1 + 1) = 10(2) = 20 feet

b.

h(10) = 10( e10 / 20 + e −10 / 20 ) = 10( e1/ 2 + e −1/ 2 ) ≈ 22.6 feet 24 = 10( e x / 20 + e − x / 20 ) ⇒ 2.4 = e x / 20 + e − x / 20 ⇒ 2.4e x / 20 = ( e x / 20 + e − x / 20 )e x / 20

c.

2.4e

x / 20

= e2 x / 20 + e0 = ( e x / 20 ) 2 + 1 ⇒ 2.4e x / 20 = (e x / 20 )2 1. Let u = e x / 20 . Then 2.4u = u 2 + 1

0 = u 2 − 2.4u + 1 ⇒ u =

−( −2.4) ± ( −2.4)2 − 4(1)(1) 2.4 ± 1.76 2.4 ± 1.3266 = ≈ 2 2(1) 2

e x / 20 = 2.4 + 1.76 2 2.4 + 1.76 x / 20 = ln 2 x = 20ln 2.4 + 1.76 ≈ 12.4 feet 2

or

e x / 20 = 2.4 − 1.76 2 2.4 − 1.76 x / 20 = ln 2 x = 20ln 2.4 − 1.76 ≈ −12.4 no negative height 2

.......................................................

Connecting Concepts

79.

The second step because log 0.5 < 0. Thus the inequality sign must be reversed.

80.

The third step. log2 (8 + 8) does not equal log 2 8 + log2 8.

81.

log( x + y ) = log x + log y log( x + y ) = log xy Therefore x + y = xy x − xy = − y x(1 − y ) = − y −y x= 1− y y x= y −1

82.

No. The domain of g ( x ) includes negative numbers; the domain of f ( x ) does not. Thus, for any negative value of x, f ( x ) ≠ g ( x ) .

83.

Since e0.336 ≈ 1.4,

84.

2.2 = e− k ln 2.2 = − k ln e

F ( x) = (1.4) x ≈ (e0.336 ) x = e0.336 x = G ( x)

− ln 2.2 = k −0.788 ≈ k

....................................................... 12(2)

PS1.

A = 1000 ⎛⎜ 1 + 0.1 ⎞⎟ 12 ⎠ ⎝

PS3.

0.5 = e14k ln 0.5 = ln e14k ln 0.5 = 14k ln 0.5 = k 14 −0.0495 ≈ k

= 1220.39

Prepare for Section 4.6 PS2.

PS4.

A = 600 ⎛⎜ 1 + 0.04 ⎞⎟ 4 ⎠ ⎝

4(8)

= 824.96

0.85 = 0.5t / 5730 ln 0.85 = ln 0.5t / 5730 t ln 0.5 5730 5730ln 0.85 = t ln 0.5 1340 ≈ t ln 0.85 =


Section 4.6

PS5.

273

70 5 + 9e −12 k 6(5 + 9e −12 k ) = 70 6=

30 + 54e

−12 k

−3

2

4,000,000 = 3n +1 − 3

= 70

3,999,997 = 3n +1

−12 k

= 40 = 20 e 27 −12 k = ln 20 ln e 27 −12k = ln 20 27 1 k = − ln 20 12 27 k ≈ 0.025

54e

n +1

2,000,000 = 3

PS6.

ln 3,999,997 = ln 3n +1 ln 3,999,997 = ( n + 1)ln 3 ln 3,999,997 = n +1 ln 3 ln 3,999,997 −1 = n ln 3 12.8 ≈ n

−12 k

Section 4.6 1.

3.

a.

t = 0 hours, N ( 0 ) = 2200 ( 2 ) = 2200 bacteria

b.

t = 3 hours, N ( 3) = 2200 ( 2 ) = 17,600 bacteria

a.

0

2.

3

N (t ) = N 0 ekt where N 0 = 24600 N (5) = 22,600e

4.

a.

t = 3 years, f ( 3) = 12,400 (1.14 ) ≈ 18,400

b.

t = 4.25 years, f ( 4.25) = 12,400 (1.14 )

a.

k (5)

58,100 = 53,700e4 k 58,100 = e4 k 53,700 58,100 ⎞ 4k ln ⎛⎜ ⎟ = ln ( e ) ⎝ 53,700 ⎠ 58,100 ⎞ ln ⎛⎜ ⎟ = 4k ⎝ 53,700 ⎠ 1 ⎡ln ⎛ 58,100 ⎞ ⎤ = k 4 ⎢⎣ ⎜⎝ 53,700 ⎟⎠ ⎥⎦ 0.01969 ≈ k

0.01368 ≈ k N (t ) = 22,600e0.01368t t = 15 N (15) = 22,600e0.01368(15) = 22,600e0.2052 ≈ 27,700

b.

4.25

≈ 21,600

N (t ) = N 0 ekt where N 0 = 53,700 N (4) = 53,700ek (4)

24,200 = 22,600e5k 24,200 = e 5k 22,600 24,200 ⎞ 5k ln ⎛⎜ ⎟ = ln ( e ) ⎝ 22,600 ⎠ 24,200 ⎞ ln ⎛⎜ ⎟ = 5k ⎝ 22,600 ⎠ 1 ⎡ln 24, 200 ⎤ = k 5 ⎣⎢ 22,600 ⎦⎥

b.

3

N (t ) = 53,700e0.01969t t = 12 N (12) = 53,700e0.01969(12) = 53,700e0.23628 ≈ 68,000



274 N (t ) = N 0 ekt where N 0 = 362,300

5.

N (t ) = N 0 ekt where N 0 = 276,400

6.

N (4) = 362,300ek (4) 379,700 = 362,300e 379,700 = e4k 362,300 379,700 ⎞ 4k ln ⎛⎜ ⎟ = ln ( e ) ⎝ 362,300 ⎠ 379,700 ⎞ ln ⎛⎜ ⎟ = 4k ⎝ 362,300 ⎠ 1 ⎡ln 379,700 ⎤ = k 4 ⎢⎣ 362,300 ⎥⎦ 0.011727 ≈ k

291,800 = 276,400e4 k 291,800 = e4k 276,400 291,800 ⎞ 4k ln ⎛⎜ ⎟ = ln ( e ) ⎝ 276,400 ⎠ 291,800 ⎞ ln ⎛⎜ ⎟ = 4k ⎝ 276,400 ⎠ 1 ⎡ln 291,800 ⎤ = k 4 ⎢⎣ 276,400 ⎥⎦ 0.013555 ≈ k

N (t ) = 362,300e0.011727 t t=9

N (t ) = 276,400e0.013555 t t =8

N (9) = 362,300e0.011727(9)

N (9) = 276,400e0.013555(8)

= 362,300e ≈ 402,600

7.

N (4) = 276,400ek (4)

4k

0.105543

= 276,400e0.10844 ≈ 308,100

a.

c.

Since A = 4 micrograms are present when t = 0, find the time t at which half remains—that is when A = 2.

b.

A(5) = 4e −0.23 ≈ 3.18 micrograms

d.

1 = 4e −0.046t 1 = e −0.046t 4 1 = −0.046t ln 4 ln 1 4 =t −0.046t 30.14 ≈ t The amount of sodium-24 will be 1 microgram after 30.14 hours.

2 = 4e −0.046t

() ()

1 = e −0.046t 2 ln 1 = −0.046t 2 ln 1 2

() ( ) =t

−0.046 15.07 ≈ t The half-life of sodium-24 is about 15.07 hours.

8.

N (t ) = N 0 ekt N (138) = N 0 e138k 138k

0.5N 0 = N 0 e

138k

0.5 = e

138k

ln 0.5 = ln e ln 0.5 = 138k ln e ln 0.5 = 138k ln 0.5 = k 138 −0.005023 ≈ k

9.

N ( t ) = N 0 ( 0.5)

t / 5730

N ( t ) = 0.45 N 0 0.45 N 0 = N 0 ( 0.5) ln ( 0.45) =

t / 5730

t ln 0.5 5730

10.

N ( t ) = N 0 ( 0.5)

t /138

N (730) = N 0 (0.5)730 / 138 ≈ 0.0256 N 0 After 2 years (730 days), only 2.56% of the polonium sample will remain.

5730 ln 0.45 = t ln 0.5 6601 ≈ t The bone is about 6601 years old.

N ( t ) = N 0(0.5)t /138 ≈ N 0 e −0.005023t Copyright © Houghton Mifflin Company. All rights reserved.

Section 4.6

275

N ( t ) = N 0 ( 0.5)

11.

t / 5730

N ( t ) = N 0 ( 0.5)

12.

N ( t ) = 0.75 N 0

N ( t ) = 0.65 N 0 t / 5730

t / 5730

0.75 N 0 = N 0 ( 0.5) ln 0.75 = t ln 0.5 5730 ln 0.75 =t 5730 ln 0.5 2378 ≈ t The Rhind papyrus is about 2378 years old.

13.

a.

0.65 N 0 = N 0 ( 0.5) ln 0.65 = t ln 0.5 5730 ln 0.65 =t 5730 ln 0.5 3561 ≈ t The bone is about 3600 years old.

P = 8000, r = 0.05, t = 4, n = 1

14.

a.

4

⎛ 0.045 ⎞ B = 22,000⎜1 + ⎟ ≈ $24,024.55 1 ⎠ ⎝

7

b.

⎛ 0.05 ⎞ t = 7, B = 8000 ⎜1 + ⎟ ≈ $11, 256.80 1 ⎠ ⎝

a.

P = 38,000, r = 0.065, t = 4, n = 1

10

16.

b.

⎛ 0.045 ⎞ t = 10, B = 22,000⎜1 + ⎟ 1 ⎠ ⎝

a.

P = 12,500, r = 0.08, t = 10, n = 1

4

10

⎛ 0.065 ⎞ B = 38,000⎜1 + ⎟ ≈ $48,885.72 1 ⎠ ⎝

17.

4(365)

b.

⎛ 0.065 ⎞ n = 365, B = 38,000⎜1 + ⎟ 365 ⎠ ⎝

c.

⎛ 0.065 ⎞ n = 8760, B = 38,000⎜1 + ⎟ 8760 ⎠ ⎝

P = 15,000, r = 0.1, t = 5 B = 15,000e

5(0.1)

19.

ln 2 r ln 2 t= 0.0784 t ≈ 8.8 years

21.

B = Pert

t=

⎛ 0.08 ⎞ B = 12,500⎜1 + ⎟ 1 ⎠ ⎝

≈ $49,282.20

4(8760 )

≈ $49,283.30 18.

Let B = 3P

ln 3 r ln 3 t= 0.076 t ≈ 14 years

t=

r = 0.076

⎛ 0.08 ⎞ n = 365, B = 12,500⎜1 + ⎟ 365 ⎠ ⎝

c.

0.08 ⎞ ⎛ n = 8760, B = 12,500⎜1 + ⎟ ⎝ 8760 ⎠

22.

ln 3 r ln 3 t= 0.055 t ≈ 20 years

24.

ln 3 0.055 t = 20 years

t=

t=

t=

≈ $27,816.82 87,600

P = 32,000, r = 0.08, t = 3

ln 2 2 ln 2 t= 0.0588 t ≈ 11.8 years

3 = e rt ln 3 = rt ln e ln 3 t= r 23.

3650

b.

20.

3P = Pert

≈ $34,165.33

≈ $26,986.56

B = 32,000e3(0.08 ) ≈ $40,679.97

≈ $24,730.82

r = 0.0784

P = 22,000, r = 0.045, n = 1, t = 2 2

⎛ 0.05 ⎞ B = 8000⎜1 + ⎟ ≈ $9724.05 1 ⎠ ⎝

15.

t / 5730

r = 0.0588

r = 0.055

r = 0.055


≈ $27,819.16


276 25.

27.

29.

31.

a. b.

1900 0.16

c.

P(0) =

a. b.

157,500 0.04

c.

P (0) =

a. b.

2400 0.12

c.

P(0) =

a=

26.

1900 = 200 1 + 8.5e −0.16(0) 28.

157,500 = 45,000 1 + 2.5e −0.04(0)

30.

2400 = 300 1 + 7e −0.12(0)

c − P0 5500 − 400 = = 12.75 P0 400 c 1 + ae −bt 5500 P (2) = 1 + 12.75e − b(2) 5500 780 = 1 + 12.75e −2b P(t ) =

32.

a. b.

32,550 0.08

c.

P(0) =

a. b.

51 0.03

c.

P(0) =

a. b.

320 0.12

c.

P(0) =

a=

32,550 = 18,600 1 + 0.75e −0.08(0)

51 = 25 1 + 1.04e −0.03(0)

320 = 20 1 + 15e −0.12(0)

c − P0 9500 − 6200 = = 0.53226 P0 6200 c 1 + ae − bt 9500 P (8) = 1 + 0.53226e −b(8) 9500 7100 = 1 + 0.53226e −8b P (t ) =

780(1 + 12.75e −2b ) = 5500

7100(1 + 0.53226e −8b ) = 9500

780 + 9945e −2b = 5500

7100 + 3779e −8b = 9500

9945e −2b = 4720

3779e −8b = 2400

e −2b = 4720 9945 −2 b ln e = ln 4720 9945 4720 −2b = ln 9945 1 b = − ln 4720 2 9945 b ≈ 0.37263 5500 P(t ) = 1 + 12.75e −0.37263 t

e −8b = 2400 3779 −8b ln e = ln 2400 3779 −8b = ln 2400 3779 1 b = − ln 2400 8 3779 b ≈ 0.05675 9500 P(t ) = 1 + 0.53226e −0.05675 t


Section 4.6

33.

a=

277

c − P0 100 − 18 = = 4.55556 P0 18

34.

c 1 + ae − bt 100 P (3) = 1 + 4.55556e − b(3) 100 30 = 1 + 4.55556e −3b 30(1 + 4.55556e −3b ) = 100 P (t ) =

c − P0 P0 c − 3200 a= 3200 3200a + 3200 = c a=

c 1 + ae − bt + 3200 P(22) = 3200a−0.056(22) 1 + ae 3200 5565 = 3200a −+1.232 1 + ae 5565(1 + ae −1.232 ) = 3200a + 3200 P(t ) =

30 + 136.67e −3b = 100 136.67e −3b = 70 e −3b = 70 136.67 ln e −3b = ln 70 136.67 −3b = ln 70 136.67 b = − 1 ln 70 3 136.67 b ≈ 0.22302 100 P(t ) = 1 + 4.55556e −0.22302 t

35.

a.

b.

37.

a.

625,000 1 + 3.1e −0.045 t 625,000 R (1) = ≈ $158,000 1 + 3.1e −0.045(1) 625,000 R (2) = ≈ $163,000 1 + 3.1e −0.045(2) 625,000 R(t ) = , as t → ∞, R(t ) → $625,000 1 + 3.1e −0.045 t R(t ) =

a=

c − P0 1600 − 312 = = 4.12821 312 P0

5565 + 5565ae −1.232 = 3200a + 3200 5565ae −1.232 − 3200a = −2365 a (5565e −1.232 − 3200) = −2365 −2365 5565e −1.232 − 3200 a = 1.5 c = 3200a + 3200 = 3200(1.5) + 3200 = 8000 8000 P(t ) = 1 + 1.5e −0.056 t a=

36.

38.

A(t ) =

b.

A(t ) =

a.

a=

c 1 + ae − bt 1600 P (6) = 1 + 4.12821e −b(6) 1600 416 = 1 + 4.12821e −6b 416(1 + 4.12821e −6b ) = 1600

b.

c − P0 3400 − 240 = = 13.167 240 P0 P(t ) =

416 + 1717.34e −6b = 1600

310 + 4081.6677e − b = 3400

−6 b

= 1184 −6 b = 1184 e 1717.34 ln e −6b = ln 1184 1717.34 −6b = ln 1184 1717.34 1 b = − ln 1184 6 1717.34 b ≈ 0.06198 1600 P(t ) = 1 + 4.12821e −0.06198 t 1600 ≈ 497 wolves P(10) = 1 + 4.12821e −0.06198(10)

1650 , as t → ∞, A(t ) → 1650 cars 1 + 2.4e −0.055 t

c 1 + ae −bt 3400 P(1) = 1 + 13.16667e − b(1) 3400 310 = 1 + 13.16667e − b 310(1 + 13.16667e − b ) = 3400

P(t ) =

1717.34e

1650 1 + 2.4e −0.055 t 1650 A(1) = ≈ 504 cars 1 + 2.4e −0.055(1) 1650 A(2) = ≈ 524 cars 1 + 2.4e −0.055(2)

a.

b.

4081.6677e − b = 3090 e − b = 3090 4081.6677 ln e − b = ln 3090 4081.6677 −b = ln 3090 4081.6677 b = − ln 3090 4081.6677 b ≈ 0.27833 3400 P(t ) = 1 + 13.16667e −0.27833 t 3400 ≈ 1182 groundhogs P(7) = 1 + 13.16667e −0.27833(7)



278 39.

a.

a=

c − P0 8500 − 1500 = = 4.66667 1500 P0

40.

a.

c 1 + ae − bt 8500 P (2) = 1 + 4.66667e − b(2) 8500 1900 = 1 + 4.66667e −2b

c − P0 5500 − 800 = = 5.875 800 P0 c 1 + ae − bt 5500 P (1) = 1 + 5.875e − b(1) 5500 900 = 1 + 5.875e − b

P(t ) =

P (t ) =

1900(1 + 4.66667e −2b ) = 8500

900(1 + 5.875e −b ) = 5500

1900 + 8866.673e −2b = 8500

900 + 5287.5e − b = 5500

8866.673e

−2 b

5287.5e − b = 4600

= 6600

6600 8866.673 ln e −2b = ln 6600 8866.673 −2b = ln 6600 8866.673 b = − 1 ln 6600 2 8866.673 b ≈ 0.14761 8500 P(t ) = 1 + 4.66667e −0.14761t 8500 4000 = 1 + 4.66667e −0.14761t

e − b = 4600 5287.5 ln e − b = ln 4600 5287.5 −b = ln 4600 5287.5 b = − ln 4600 5287.5 b ≈ 0.13929 5500 P(t ) = 1 + 5.875e −0.13929 t

e −2b =

b.

a=

2000 =

b.

5500 1 + 5.875e −0.13929 t

4000(1 + 4.66667e −0.14761t ) = 8500

2000(1 + 5.875e −0.13929 t ) = 5500

1 + 4.66667e −0.14761t = 2.125

1 + 5.875e −0.13929 t = 2.75

4.66667e −0.14761t = 1.125 e −0.14761t = 1.125 4.66667 −0.14761 t = ln 1.125 ln e 4.66667 −0.14761t = ln 1.125 4.66667 1 t=− ln 1.125 0.14761 4.66667 t ≈ 9.6 The population will exceed 4000 in 2007 + 9 = 2016.

5.875e −0.13929 t = 1.75 e −0.13929 t = 1.75 5.875 −0.13929 t = ln 1.75 ln e 5.875 −0.13929t = ln 1.75 5.875 1 t=− ln 1.75 0.13929 5.875 t ≈ 8.7 The population will exceed 2000 in 2003 + 8 = 2011.


Section 4.6

41.

a.

279

A = 34o F, T0 = 75o F, Tt = 65o F, t = 5. Find k .

42.

a.

N ( 2 ) = 100 (1.04 − 0.992 ) N ( 2 ) = 6 wpm

b.

N ( 40 ) = 100 (1.04 − 0.9940 ) N ( 40 ) ≈ 37 wpm

65 = 34 + ( 75 − 34 ) e −5k 31 = 41e −5k 31 = e −5k 41 ⎛ 31 ln ⎜ ⎞⎟ = −5k ⎝ 41 ⎠

c.

0.6 = 1.04 − 0.99t

k = − 1 ln ⎛⎜ 31 ⎞⎟ 5 ⎝ 41 ⎠ k ≈ 0.056

b.

0.44 = 0.99t ln 0.44 = ln 0.99t ln 0.44 = t ln 0.99 82 ≈ t After 82 hours of practice, a student can expect to type 60 wpm.

A = 34o F, k = 0.056, T0 = 75o F, t = 30 Tt = 34 + ( 75 − 34 ) e

−30( 0.056 )

Tt = 34 + (41)e −1.68 Tt ≈ 42o F

c.

60 = 100 (1.04 − 0.99t )

Tt = 36o F, k = 0.056, Tt = 75o F, A = 34o F 36 = 34 + (75 − 34 )e−0.056t 2 = 41e −0.056t t ≈ 54 minutes

43.

a.

44.

10% of 80,000 is 8000.

a.

0.4 = 1 − e −0.03 t

0.1 = 1 − e −0.0005 t −0.9 = − e −0.0005 t

−0.6 = − e −0.03 t

0.9 = e −0.0005 t

0.6 = e −0.03 t

ln 0.9 = −0.0005t ln e

ln ( 0.6 ) = ln e −0.03 t

ln 0.9 = −0.0005t

ln ( 0.6 ) = −0.03t

ln 0.9 = t −0.0005 211 h ≈ t

b.

40% of 1,200,000 is 480,000. 480,000 = 1,200,000 (1 − e −0.03 )

8000 = 80,000 (1 − e −0.0005 t )

ln ( 0.6 ) −0.03

t ≈ 17 days

50% of 80,000 is 40,000. 40,000 = 80,000 (1 − e −0.0005t ) 0.5 = 1 − e −0.0005 t −0.5 = − e −0.0005 t 0.5 = e −0.0005 t ln ( 0.5) = ln ( e −0.0005 t )

=t

b.

960,000 = 1,200,000 (1 − e −0.03t ) 0.8 = 1 − e −0.03t 0.2 = e −0.03t ln 0.2 = t −0.03 t ≈ 54 days

ln ( 0.5) = −0.0005t ln ( 0.5)

=t −0.0005 1386 h ≈ t



280 45.

V (t ) = V0 (1 − r )t

46.

a.

I (0) = 6(1 − e 0 ) = 0 amps

b.

I (0.5) = 6(1 − e −2.5(0.5) ) ≈ 4.28 amps

t

0.5V0 = V0 (1 − 0.20 ) t

0.5 = (1 − 0.20 ) 0.5 = 0.8t

I (t ) − 1 = − e −2.5t 6 I (t ) 1− = e −2.5t 6 I (t ) ln 1 − = −2.5t 6

ln 0.5 = ln 0.8t ln 0.5 = t ln 0.8

(

ln 0.5 =t ln 0.8 3.1 years ≈ t 47.

c. d. 49.

48.

20 = 64(1 − e−t / 2 )

c.

d.

(

I (t )

)

a.

b.

50 = 64(1 − e−t / 2 )

0.625 = 1 − e−t / 2

0.78125 = 1 − e−t / 2

e−t / 2 = 0.375 −t / 2 = ln 0.375 t ≈ 0.98 seconds The horizontal asymptote is v = 32. As time increases, the object’s velocity approaches but never exceeds 32 ft/sec.

e−t / 2 = 0.21875 −t / 2 = ln 0.21875 t ≈ 3.0 seconds The horizontal asymptote is v = 64. As time increases, the object’s velocity approaches but never exceeds 64 ft/sec.

a.

b.

)

t = − 2 ln 1 − 6 5

a.

b.

I (t ) = 6(1 − e −2.56t )

c.

c. d. 50.

The graphs of s = 32t + 32(e −t − 1) and s = 50 intersect when t ≈ 2.5 seconds. The slope m of the secant line containing (1, s(1)) s (2) − s (1) ≈ 24.56 ft/sec and (2, s(2)) is m = 2 −1 The average speed of the object was 24.56 feet per second between t = 1 and t = 2.

51.

a.

b. c.

d.

The graphs of s = 64t + 128(e −t / 2 − 1) and s = 50 intersect when t ≈ 2.1 seconds. The slope m of the secant line containing (1, s(1)) s (2) − s (1) ≈ 33.5 ft/sec and (2, s(2)) is m = 2 −1 The average speed of the object was 33.5 feet per second between t = 1 and t = 2.

52.

Xmin = 0,Xmax = 80,Xscl = 10,

Xmin = 0,Xmax = 24,Xscl = 3,

Ymin = −10,Ymax = 110,Yscl = 15

Ymin = −50,Ymax = 350,Yscl = 50

When P = 75%, t ≈ 45 hours.

When N = 140 circuits, t ≈ 11 weeks. Copyright © Houghton Mifflin Company. All rights reserved.

Section 4.6

281

....................................................... 53.


a.

A(1) = 0.51/ 2 ≈ 0.71 gram

b.

A(4) = 0.54 / 2 + 0.5(4−3) / 2 = 0.52 + 0.51/ 2 ≈ 0.96 gram

Use the TRACE feature and the graph of ⎧0.5x / 2 0≤ x 0 in quadrants I and II. cos θ > 0 in quadrants I and IV. quadrant I

22.

23.

cosθ > 0 in quadrants I and IV. tan θ < 0 in quadrants II and IV. quadrant IV

tan θ < 0 in quadrants II and IV. sin θ < 0 in quadrants III and IV. quadrant IV


Section 5.3

317

24.

sin θ < 0 in quadrants III and IV. cosθ > 0 in quadrants I and IV. quadrant IV

27.

sin θ = −

28.

2 cot θ = −1 = x , x = −1, y = 1 in quadrant II, r = ( −1) + 12 = 2 , cos θ = −1 = − 2 2 y 2

29.

csc θ = 2 = r , r = 2, y = 1, x = ± y

30.

sec θ = 2 3 = r , r = 2 3, x = 3, y = ± x 3

31.

θ is in quadrant IV, sinθ = − 1 =

32.

θ is in quadrant III, tan θ = 1 =

33.

cos θ = 1 , θ is in quadrant I or IV. 2

25.

sin θ < 0 in quadrants III and IV. cos θ < 0 in quadrants II and III. quadrant III

y −y 3 1 y = = , y = −1, r = 2, x = ± 2 2 − (− 1)2 = ± 3 , x = − 3 in quadrant III , tan θ = = x − 3 3 2 r

2

(

2 2 ) − 12 = ±1 , x = −1 in quadrant II , cot θ = −1 = −1 1

( 2 3 ) 2 − 32

= ± 3, y = − 3 in quadrant IV , sin θ = − 3 = − 1 2 2 3

y , y = −1, r = 2, x = 2 2 − 12 = 3 , tan θ = −1 = − 3 3 r 3

y , y = −1, x = −1, r = x

(− 1)2 + (− 1)2 34.

θ is in quadrant I, x = 1, y = 3 , r = 2

35.

r 2 2 3 = = y 3 3

cos θ = − 1 ,θ is in quadrant II or III. 2

sin θ =

37.

= 2 , cosθ =

−1 2

=−

2 2

tan θ = 1 , θ is in quadrant I or III. sin θ = 2 , θ is in quadrant III or IV. 2 θ is in quadrant III, x = 2, y = 2, r = 2 sec θ = r = 2 = 2 x 2

tan θ = 3 , θ is in quadrant I or III.

cscθ =

tan θ < 0 in quadrants II and IV. cosθ < 0 in quadrants II and III. quadrant II

26.

36.

3 , θ is in quadrant I or II. 2

sec θ = 2 3 , θ is in quadrant I or IV. 3 1 sin θ = − , θ is in quadrant II or IV. 2

θ is in quadrant II, x = −1, y = 3 , r = 2

θ is in quadrant IV, x = 3 , y = −1, r = 2

3 x −1 cot θ = = =− y 3 3

cot θ =

θ = 160o Since 90o < θ < 180o ,

θ + θ ′ = 180o θ ′ = 20o

38.

3 x = =− 3 y −1

θ = 255o Since 180o < θ < 270o , θ ′ + 180o = θ θ ′ = 75o


318

39.

Chapter 5: Trigonometric Functions

θ = 351o

40.

Since 270o < θ < 360o ,

41.

θ = 48o

θ = θ ′ = 360o

Since 0o < θ < 48o , θ′ =θ

θ ′ = 9o

θ ′ = 48o

θ=

11π 5

42.

θ is coterminal withα =

α′ =α =θ ′ π θ′ =

π 2

3π < θ < −2π , 2 θ ′ + θ = 2π

Since −

10π , θ > 2π = 5

Since 0 < α
9.1 ( nearest tenth ) .

b.

Xmin = 8, Xm ax = 10, Xscl = 1, Ymin = −0.01, Ymax = 0.01, Yscl = 0.005

39.

a.

The pseudoperiod is 2π = 1. There are 10 complete 2π

Xmin = 22.5, Xma x = 24, Xscl = 1, Ymin = −0.01, Ymax = 0.01, Yscl = 0.005

40.

a.

oscillations of length 1 in 0 ≤ t ≤ 10.

b.

f ( t ) < 0.01 for all t > 6.1( nearest tenth ) .

X min = 5, X max = 7, Xscl = 1, Ymin = −0.01, Ymax = .01, Yscl = 0.005

f ( t ) < 0.01 for all t > 23.0 ( nearest tenth ) .

The pseudoperiod is 2π = 1. There are 10 complete 2π

oscillations of length 1 in 0 ≤ t ≤ 10.

b.

f ( t ) < 0.01 for all t > 4.6 ( nearest tenth ) .

X min = 4, X max = 6, Xscl = 1, Ymin = −0.01, Ymax = 0.01, Yscl = 0.005



351

....................................................... 41.

43.

45.

m 9m , p2 = 2π = 3 p1 k k Increasing the main mass to 9m will triple the period. p1 = 2π

yes


The frequency will double.

44.

no

Xmin = 0, Xma x = 15, Xscl = 1,

Xmin = 0, Xma x = 10, Xscl = 1,

Ymin = −1, Ymax = 1.5, Yscl = 0.25


yes

46.

yes

Xmin = 0, Xmax = 10, Xscl = 1,

Xmin = 0, Xmax = 15, Xscl = 1,


Ymin = −1, Ymax = 1.5, Yscl = 0.25

.......................................................


Sinusoidal Families 1.

All three sine graphs have, a period of 2π , x-intercepts at nπ , and no phase shift, but their amplitudes are 2, 4, and 6 respectively. 2.

All three sine graphs have x-intercepts at n, an amplitude of 1, and no phase shift, but their periods are 2, 1, and 0.5 respectively, and y = sin 2π x and y = 4π x have additional x-intercepts at 0.5n and 0.25n respectively. 3.

All three sine graphs have a period of 2π and an amplitude of 1, but their phase shifts are −π / 4, − π / 6, and − π /12, respectively. Copyright © Houghton Mifflin Company. All rights reserved.

352

4.


Yes, the calculator has displayed all three graphs. All three sine graphs have an amplitude of 1, a period of 2π , and a phase shift of −(2n − 1)π .

.......................................................

Assessing Concepts

1.

True

2.

False; sec2 θ − tan 2 θ = 1 is an identity.

3.

False; 1 rad ≈ 57.3°.

4.

True

5.

π

6.

(0, 1)

7.

The period is

2π = 8 . 3π / 4 3

8.

Shift the graph of y1 to the left π units. 2

9.

All real numbers except multiples of π .

10.

The vertical asymptotes are x = π and x = 3π . 2 2

4

....................................................... 1.

complement: 90° − 65° = 25° [5.1] supplement:180° − 65° = 115°

2.

Chapter Review

θ = 980° = 260° + 2 ⋅ 360° [5.3] θ is coterminal with α = 260° and θ ' = α '. Since 180° < α < 270°, 180° + α ' = α 180° + a′ = 260° α ' = 80° θ = 80°

3.

2 = 2 ⎜⎛ 180° ⎟⎞ [5.1] ⎝ π ⎠ = 114.59°

6.

θ=

s 12 [5.1] = r 40 = 0.3

For exercises 8 to 11, cscθ = 8.

cos θ = x = 5 [5.2] r 3

10.

sin θ =

y 2 [5.2] = r 3

4.

315° = 315° ⎛⎜ π ⎞⎟ [5.1] ⎝ 180° ⎠ 7 π = 4 7.

w=

5.

s = rθ = 3 ( 75° ) ⎛⎜ π ⎞⎟ ⎝ 180° ⎠ [5.1] = 3.93 m

V 50 63360 = ⋅ [5.1] r 16 3600 ≈ 55 rad/sec

3 r = , r = 3, y = 2, and x = 32 − 22 = 5. 2 y 9.

cot θ = x = 5 [5.2] y 2

11.

sec θ = r = 3 = 3 5 [5.2] 5 x 5


Chapter Review

353

12.

2

x = 1, y = −3, r = 12 + ( −3) = 10

13.

15.

3 3 10 =− 10 10

cosθ =

cscθ = −

10 3

secθ = 10

sec 150° =

b.

⎛ 3π ⎞ tan ⎜ − ⎟ = 1 ⎝ 4 ⎠

c.

cot ( −225° ) = −1

d.

cos

a.

b.

16.

a.

b.

17.

a.

b.

(

a.

cos 123° ≈ −0.5446

b.

cot 4.22 ≈ 0.5365

c.

sec612° ≈ −3.2361

d.

tan

2π ≈ 3.0777 [5.3] 5

)

2 3 x = , x = − 3, r = 2, y = − 22 − − 3 = −1 [5.3] 2 r

y 1 =− r 2

tan φ =

y −1 3 = = x − 3 3

3 y = , y = 3, x = −3, r = 3 x

secφ =

( −3 ) 2 + (

3

)

2

= 2 3 [5.3]

r 2 3 2 3 = =− 3 x −3

csc φ =

sin φ = −

14.

2π 1 [5.3] =− 3 2

sin φ =

tan φ = −

tan θ =

2 2 3 =− 3 − 3

a.

cos φ = −

−3 = −3 1 1 cot θ = − 3

1 10 = 10 10

sin θ = −

r 2 3 = =2 y 3

(

)

2 2 , y = − 2, r = 2, x = − 22 − − 2 = 2 [5.3] 2

cos φ =

x 2 = 2 r

cot φ =

x 2 = = −1 y − 2


354

18.


a.

W (π ) = ( −1,0 ) [5.4]

b.

3⎞ ⎛ π ⎞ ⎛1 W ⎜ − ⎟ = ⎜⎜ , − ⎟ 2 ⎟⎠ ⎝ 3⎠ ⎝2

c.

⎛ 5π W⎜ ⎝ 4

19.

= sin x tan x = f ( x)

The function defined by f ( x) = sin( x) tan( x) is an even function. 21.

cos (π + t ) = − x

tan ( −t ) = −

cos t = x

1+

sin 2 φ cos 2 φ

y x

y x tan ( −t ) = − tan t

cos (π + t ) = − cos t

22.

[5.4]

f ( − x ) = sin ( − x ) tan ( − x ) = ( − sin x )( − tan x )

2 2⎞ ⎞ ⎛ ,− ⎟ ⎟ = ⎜⎜ − 2 ⎟⎠ ⎠ ⎝ 2

20.

f ( x ) = sin ( x ) tan ( x )

tan t =

sin φ

2

= 1 + tan φ [5.4]

23.

+1 tan φ + 1 cos φ = cot φ + 1 cos φ + 1

[5.4]

sin φ

= sec2 φ

sin φ + cos φ cos φ = cos φ +sin φ sin φ

=

sin φ ( sin φ + cos φ )

cos φ ( cos φ + sin φ )

= tan φ 24.

cos 2 φ + sin 2 φ 1 [5.4] = csc φ csc φ = sin φ

25.

sin 2 φ (tan 2 φ + 1) = sin 2 φ sec2 φ [5.4] =

sin 2 φ cos2 φ

= tan 2 φ


Chapter Review

26.

1+

1 2

tan φ

355

= =

tan 2 φ + 1 2

tan φ

27.

[5.4]

cos2 φ 2

1 − sin φ

sec2 φ

−1 =

1 − sin 2 φ

1 − sin 2 φ = 1−1

− 1 [5.4]

=0

tan 2 φ

1 cos 2 φ = sin 2 φ cos 2 φ

=

1 sin 2 φ

= csc2 φ 28.

y = 3cos ( 2 x − π ) [5.5]

29.

2π 2π = =π 2 b −π π c = phase shift = − = − 2 2 b

no amplitude; period =

a = 3 = 3; period =

30.

32.

y = 2 tan 3x [5.6]

31.

3π ⎞ ⎛ y = −4sec ⎜ 4 x − ⎟ [5.6] 2 ⎠ ⎝ 2π 2π π = = no amplitude; period = 4 2 b −3π / 2 3π c = phase shift = − = − 4 8 b

=

π 3

33.

2π ⎞ ⎛ y = cos ⎜ 2 x − ⎟ + 2 [5.5] 3 ⎠ ⎝ 2π 2π a = 1 = 1; period = = =π 2 b −2π / 3 π c = phase shift = − = − 2 3 b

π⎞ ⎛ y = 2csc ⎜ x − ⎟ − 3 [5.6] 4⎠ ⎝ 2π 2π = = 2π no amplitude; period = 1 b −π / 4 π c = phase shift = − = − 1 4 b

34.

y = 2 cos π x, p = 2π = 2

35.

y = − sin 2 x , p = 2π = 3π 3 2/3

36.

37.

y = cos ⎛⎜ x − π ⎞⎟ , p = 2π 2⎠ ⎝

38.

y = 1 sin ⎜⎛ 2 x + π ⎟⎞ , p = 2π = π 2 ⎝ 4⎠ 2

39.

phase shift = π 2

b

phase shift = 0

π⎞ ⎛ y = −2sin ⎜ 3 x + ⎟ [5.5] 3⎠ ⎝ 2π 2π a = −2 = 2; period = = 3 b c π /3 π phase shift = − = − =− b 3 9

π

π

phase shift − π 8

y = 2sin 3x , p = 2π = 4π 2 3/ 2 3

y = 3cos3 ( x − π ) , p = 2π 3 phase shift = π


356

40.


y = − tan x , p = π = 2π 2 1/ 2

41.

y = 2 cot 2 x, p = π 2

42.

y = tan ⎛⎜ x − π ⎞⎟ , p = π 2⎠ ⎝

phase shift = π 2

43.

y = − cot ⎛⎜ 2 x + π ⎞⎟ , p = π 4⎠ 2 ⎝

44.

phase shift = − π 8

y = −2 csc ⎜⎛ 2 x − π ⎟⎞ , p = 2π = π 3⎠ 2 ⎝

45.

y = 3 sec ⎛⎜ x + π ⎞⎟ , p = 2π 4⎠ ⎝

phase shift = − π 4

phase shift = π 6

46.

y = 3sin 2 x − 3

47.

y = 2cos3x + 3

48.

y = − cos ⎛⎜ 3x + π ⎞⎟ + 2 2⎠ ⎝

49.

y = 3sin ⎛⎜ 4 x − 2π ⎞⎟ − 3 3 ⎠ ⎝

50.

y = 2 − sin 2 x

51.

y = sin x − 3 cos x

52.

53.

h 1.14 h = 1.14sin 4.5° ≈ 0.089 mi

sin 4.5° =

[5.2]

h 8.55 h = 8.55 tan 55.3 ≈ 12.3 feet [5.2] tan 55.3° =


Quantitative Reasoning

54.

357

Speed for inner ring: v= s t 2π (14.5) = 24 ≈ 3.79609 ft/s

Speed for outer ring: v=s t 2π (21) = 24 ≈ 5.497787 ft/s

55.

The outer swing has a greater speed of 5.497787 − 3.79609 ≈ 1.7 ft/s. [5.1]

56.

y = 2.5sin 50t

80 + x 80 x = + h h h x (2) cot 37° = h x Substitute for in equation (1). h 80 cot18° = + cot 37° h 80 Solve for h. = cot18° − cot 37° h 1 h = 80 cot18° − cot 37° 80 h= ≈ 46 ft cot18° − cot 37° [5.2] (1) cot18° =

57.

[5.8]

amplitude = 0.5 [5.8] 1 f = 2π p =π

amplitude = 2.5 π 2π 2π = = p= 50 25 b 1 25 frequency = = p π

k 1 = m 2π

58.

f ( t ) < 0.01 for all t > 7.2 [5.8]

20 1 = 5 π

⎛1⎞ y = −0.5cos 2π ft = −0.5cos 2π ⎜ ⎟ t ⎝π ⎠ y = −0.5cos 2t

Xmin = 5, Xmax = 10, Xcsl = 1 Ymin = −.01, Ymax = .01, Yscl = .005

....................................................... QR1. a.

c.

e.

QR2.

2π = m ⇒ 3π n

2π n = 3π m ⇒ period = 6π 2π (3) = 3π (2)

π /2 = m ⇒ 2π / 3 n

π n = 2π m

⇒ period = 2π 3 π (4) = 2π (3) 2 3 5/ 2 = m ⇒ 5 n = 3 m ⇒ period = 7.5 3/ 2 n 2 2 5 (3) = 3 (5) 2 2

3 =m ⇒ 2.5 n

2

3n = 2.5m ⇒ period = 15 s 3(5) = 2.5(6)

Quantitative Reasoning b.

d.

f.

2/3 = m ⇒ n 4

2 n = 4m 3

⇒ period = 4

2 (6) = 4(1) 3 3π / 2 = m ⇒ 3π n = 8π m ⇒ period = 24π 8π / 3 n 2 3 3π (16) = 8π (9) 2 3 4π / 5 = m ⇒ 4π n = 4π m ⇒ period = 4π n 4π 5 4π (5) = 4π (1) 5

QR3. 1.25 = m ⇒ 2.25 n

1.25n = 2.25m ⇒ period = 11.25 s 1.25(9) = 2.25(5)


358

QR4.


6n = 4.5m = 27 w ⇒ period = 54 s 6(9) = 4.5(12) = 27(2)

....................................................... 1.

4.

150° = 150° ⎛⎜ π ⎞⎟ ⎝ 180° ⎠ 5 π = 6

[5.1]

rev [5.1] sec rev ⎛ 2π rad ⎞ w=6 ⎜ ⎟ sec ⎝ rev ⎠ w = 12π rad/sec

2.

5.

w=6

π−

11 π π= 12 12

Chapter Test s = rθ

3.

[5.1]

[5.1]

⎛ π ⎞ s = 10 ( 75° ) ⎜ ⎟ ⎝ 180° ⎠ s ≈ 13.1 cm

v = rw [5.1] = 8 ⋅ 10 = 80 cm/sec

6.

r = 7 2 + 32 r = 58

secθ =

7.

9.

csc 67° ≈ 1.0864

[5.2]

11π [5.4] 6 x = cos t y = sin t

t=

=

3 2

=−

8.

10.

tan

π 6

cos

sec2 t − 1 sec2 t

1 2

π 3

π 2

[5.2]

1 1 ⋅ − 1 [5.3] 3 2 1 = −1 2 3 =

=

3 −1 6

=

3 −6 6

1 −1 2t cos = 1 cos 2 t 1−cos 2 t cos 2 t

=

⎛ 3 1⎞ W ( x, y ) = W ⎜⎜ , − ⎟⎟ 2⎠ ⎝ 2

− sin

58 7

[5.4]

1 cos 2 t 2

= 1 − cos t = sin 2 t

11.

period =

π b

=

π 3

[5.6]

12.

a = −3 = 3; period = 2π = 2π = π b 2

phase shift = −

π 4


[5.7]

Chapter Test

13.

359

period = π = 3 [5.7] π /3

14.

y = 3cos 1 x, p = 4π 2

phase shift = − c = − π / 6 = − 1 2 π /3 b

15.

y = −2sec 1 x , 2

p = 4π

16.

Shift the graph [of y = 2sin(2 x)] [5.7] π units to the right and down 1 unit. 4

17.

y = 2 − sin x 2

19.

18.

y = sin x − cos 2 x

20.

p=5 a = 13 y = 13 cos

5=

2π 2π ,b = b 5

2π 2π t or y = 13 sin t 5 5

tan 42.2° = h x h tan 42.2° = h cot 422.2°

x=

tan 37.4o = =

h 5.24 x h 5.24 + h cot 42.2°

Solve for h. h 5.24 + h cot 42.2° tan 37.4°(5.24 + h cot 42.2°) = h 5.24 tan 37.4° + h tan 37.4° cot 42.2° = h h − h tan 37.4° cot 42.2° = 5.24 tan 37.4° h(1 − tan 37.4° cot 42.2°) = 5.24 tan 37.4° 5.24 tan 37.4° h= 1 − tan 37.4° cot 42.2° h ≈ 25.5 meters tan 37.4o =

The height of the tree is approximately 25.5 meters. [5.2]


[5.8]

360


.......................................................

Cumulative Review 3 ÷1 = 3⋅2= 3 2 2 2 1

1.

x 2 − y 2 = ( x + y )( x − y ) [P.4]

2.

3.

A = 1 bh [P.4] 2 = 1 (4)(6) 2 = 12 in 2

4.

−x = − x = − f ( x) [2.5] (− x) 2 + 1 x 2 + 1 Odd function

5.

x [4.1] 2x − 3 y x= 2y − 3 x(2 y − 3) = 2 xy − 3 x = y 2 xy − y = y (2 x − 1) = 3 x y = 3x 2x − 1 f −1 ( x) = 3 x 2x − 1

6.

Domain: (−∞, 4) ∪ (4, ∞) [2.2/3.5]

7.

Range: [0, 2] [2.2]

8.

Shift the graph of y = f (x) horizontally 3 units to the right. [2.5]

9.

Reflect the graph of y = f (x) across the y-axis. [2.5]

10.

300o = 300o ⎛⎜ π o ⎞⎟ = 5π [5.1] ⎝ 180 ⎠ 3

11.

5π = 5π ⎛ 180o ⎞ = 225o [5.1] ⎜ ⎟ 4 4 ⎝ π ⎠

12.

f ⎛⎜ π ⎞⎟ = sin ⎛⎜ π + π ⎞⎟ = sin ⎛⎜ π ⎞⎟ = 1 [5.3] ⎝2⎠ ⎝3⎠ ⎝3 6⎠

13.

f ⎛⎜ π ⎞⎟ = sin ⎛⎜ π ⎞⎟ + sin ⎛⎜ π ⎞⎟ = 3 + 1 = 3 + 1 2 ⎝3⎠ ⎝3⎠ ⎝6⎠ 2 2

14.

⎞ ⎛ ⎞ ⎛ cos2 45o + sin 2 60o + = ⎜ 2 ⎟ + ⎜ 3 ⎟ = 2 + 3 = 5 [5.2] 4 4 4 ⎝ 2 ⎠ ⎝ 2 ⎠

16.

θ = 210o [5.3] Since 180o < θ < 270o , θ ′ + 180o = θ θ ′ = 30o

f ( x) =

15.

negative [5.3]

17.

θ = 2π 3

[5.3]

18.

[5.2]

[P.5]

f (− x) =

2

Domain: ( −∞, ∞ ) [5.4]

19.

Range: [–1, 1]

Since π < θ < π , 2 θ +θ′ = π

θ′ = π

3

20.

tan θ =

opp 3 = adj 4

hypotenuse = 32 + 42 = 9 + 16

sin θ =

opp 3 = hyp 5

[5.2]

= 25 =5


2

[5.4]

Chapter 6

Trigonometric Identities and Equations Section 6.1 1.

tan x csc x cos x =

sin x 1 ⋅ ⋅ cos x = 1 cos x sin x

2.

tan x sec x sin x = tan x ⋅ 1 ⋅ sin x cos x = tan x ⋅ sin x cos x = tan x ⋅ tan x = tan 2 x

3.

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

4.

5.

(sin x − cos x)(sin x + cos x) = sin 2 x − cos 2 x

6.

2

2

= 1 − cos x − cos x

sin 2 x − 2 sin x + 1 (sin x − 1) 2 = = sin x − 1 sin x − 1 sin x − 1

(tan x)(1 − cot x) = tan x − tan x cot x = tan x − 1

= 1 − 2cos 2 x 7.

9.

11.

1 1 cos x sin x − = − sin x cos x sin x cos x sin x cos x cos x − sin x = sin x cos x cos x(1 + sin x) cos x = 1 − sin x (1 − sin x)(1 + sin x) cos x(1 + sin x) = 1 − sin 2 x cos x(1 + sin x) = cos 2 x (1 + sin x) 1 sin x = = + cos x cos x cos x = sec x + tan x

1 − tan 4 x sec2 x

= =

(1 + tan 2 x)(1 − tan 2 x)

8.

10.

12.

sec2 x 2

1 3 cos x 3sin x + = + sin x cos x sin x cos x sin x cos x cos x + 3sin x = sin x cos x sin x(1 + cos x) sin x = 1 − cos x (1 − cos x)(1 + cos x) sin x(1 + cos x) = 1 − cos 2 x sin x(1 + cos x) = sin 2 x 1 + cos x = sin x 1 cos x = + sin x sin x = csc x + cot x sin 4 x − cos 4 x = (sin 2 x + cos 2 x)(sin 2 x − cos2 x) = 1(sin 2 x − cos 2 x)

2

sec x(1 − tan x)

= sin 2 x − cos 2 x

2

sec x

= 1 − tan 2 x 13.

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

14.

(

)

sin x cos x tan x − sin x cos x cos x − sin x = cot x cot x sin x − sin x = cot x =0


362

15.

Chapter 6: Trigonometric Identities and Equations

sin x − 2 + 1

sin x = sin x − 1 sin x

=

sin x − 2 + 1

sin x ⋅ sin x 1 sin x sin x − sin x 2

16.

sin x − 2sin x + 1

sin 2 x − 1 (sin x − 1)(sin x − 1) = (sin x − 1)(sin x + 1) sin x − 1 = sin x + 1

17.

(sin x + cos x)2 = sin 2 x + 2 sin x cos x + cos 2 x

18.

= sin 2 x + cos 2 x + 2 sin x cos x = 1 + 2 sin x cos x 19.

cos x(1 − sin x) cos x = 1 + sin x (1 + sin x)(1 − sin x) cos x(1 − sin x) = 1 − sin 2 x cos x(1 − sin x) = cos 2 x 1 − sin x = cos x 1 sin x = − cos x cos x = sec x − tan x cos x

21.

sin x sin x sin x(1 + cos x) − sin x(1 − cos x) − = 1 − cos x 1 + cos x (1 − cos x)(1 + cos x ) sin x + sin x cos x − sin x + sin x cos x = 1 − cos 2 x 2sin x cos x = sin 2 x 2cos x = sin x = 2cot x

(tan x + 1)2 = tan 2 x + 2 tan x + 1 = 1 + tan 2 x + 2 tan x = sec 2 x + 2 tan x

20.

sin x(1 − cos x) sin x = 1 + cos x (1 + cos x)(1 − cos x) sin x(1 − cos x) = 1 − cos 2 x sin x(1 − cos x) = sin 2 x 1 − cos x = sin x 1 cos x = − sin x sin x = csc x − cot x cos x

sin x

cot x + tan x sin x + cos x = 1 sec x

22.

cos x cos x sin x + sin x cos x = sin x cos x ⋅ 1 sin x cos x cos x 2 2

cos x + sin x sin x 1 = sin x = csc x =

23.

cos x tan x + 2cos x − tan x − 2 cos x(tan x + 2) − (tan x + 2) = tan x + 2 tan x + 2 (tan x + 2)(cos x − 1) = tan x + 2 = cos x − 1

24.

2 sin x cot x + sin x − 4 cot x − 2 sin x(2 cot x + 1) − 2(2 cot x + 1) = 2 cot x + 1 2 cot x + 1 (2 cot x + 1)(sin x − 2) = 2 cot x + 1 = sin x − 2

sin x

+ cot x + tan x sin x cos x = 1 csc x sin x cos x sin x + sin x cos x = sin x cos x ⋅ 1 sin x cos x sin x 2 2

cos x + sin x cos x 1 = cos x = sec x =


Section 6.1

363

25.

1 − sin x 1 sin x = − = sec x − tan x cos x cos x cos x

26.

cos x − 1 cos x 1 = − = cot x − csc x sin x sin x sin x

27.

sin 2 x − cos 2 x = sin 2 x − (1 − sin 2 x)

28.

sin 2 x − cos 2 x = 1 − cos 2 x − cos 2 x

= sin 2 x − 1 + sin 2 x

= 1 − 2cos 2 x

= 2 sin 2 x − 1 29.

1 sin 2 x

+

1 cos 2 x

= =

cos2 x + sin 2 x

30.

tan 2 x

sin 2 x cos 2 x 1 2

1

−

1 cot 2 x

= =

2

sin x cos x

= csc2 x sec2 x

cot 2 x − tan 2 x tan 2 x cot 2 x (csc2 x − 1) − (sec2 x − 1) 1

= csc2 x − 1 − sec2 x + 1 = csc 2 x − sec 2 x

31.

1 − cos x cos x

sec x − cos x = =

32.

=

sin 2 x cos x = sin x tan x 1 1 +1 +1 sin x sin x sin x = ⋅ 1 1 −1 − 1 sin x sin x sin x

34.

1 + sin x 1 − sin x (1 + sin x) (1 + sin x) = ⋅ 1 − sin x 1 + sin x

= =

=

1 + 2 sin x + sin 2 x

=

1 − sin 2 x 1 + 2 sin x + sin 2 x

=

cos 2 x 1 cos 2 x

+

2 sin x cos 2 x

+

1 1 1 1 + + sin x cos x sin x cos x sin x cos x = ⋅ 1 1 1 1 sin x cos x − − sin x cos x sin x cos 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

=

=

sin x cos x + cos x sin x

sin 2 x + cos 2 x sin x cos x 1 = sin x cos x = csc x sec x

1 − cos 2 x cos x

=

33.

tan x + cot x =

cos 2 x − sin 2 x 2

cos x − 2 sin x cos x + sin 2 x cos 2 x − sin 2 x 1 − 2 sin x cos x

sin 2 x cos 2 x

2

= sec x + 2 tan x sec x + tan 2 x 35.

sin 4 x − cos 4 x = (sin 2 x + cos 2 x)(sin 2 x − cos2 x ) = 1(sin 2 x − cos 2 x) = sin 2 x − (1 − sin 2 x) = sin 2 x − 1 + sin 2 x = 2 sin 2 x − 1

36.

sin 6 x + cos6 x = (sin 2 x + cos 2 x)(sin 4 x − sin 2 x cos 2 x + cos 4 x) = sin 4 x − sin 2 x cos 2 x + cos 4 x


364

37.

39.


1 1 1 + cos x = ⋅ 1 − cos x 1 − cos x 1 + cos x 1 + cos x = 1 − cos 2 x 1 + cos x = sin 2 x

38.

sin x cos x sin x − cos x − = 1 − sin x 1 − sin x 1 − sin x

40.

cos 2 x 1 − sin 2 x = 1 − sin x 1 − sin x (1 − sin x)(1 + sin x) = 1 − sin x = 1 + sin x

tan x cot x tan x − cot x − = 1 + tan x 1 + tan x 1 + tan x

sin x cos x − = sin x sin x 1 sin x − sin x sin x

=

41.

43.

45.

tan x cot x − = tan x tan x 1 tan x + tan x tan x 2

1 − cot x csc x − 1

1 − cot x cot x + 1 (1 − cot x)(1 + cot x) = cot x + 1 = 1 − cot x

=

(1 − cos x) − (1 + cos x) 1 1 − = 1 + cos x 1 − cos x (1 + cos x)(1 − cos x) 1 − cos x − 1 − cos x = 1 − cos 2 x −2 cos x = sin 2 x = −2 cot x csc x

1 1 + csc x + csc x sin x sin x sin x = ⋅ 1 1 sin x − sin x − sin x sin x sin x 1+1 = 1 − sin 2 x 2 = cos 2 x cos x 1+ 1 cot x + 1 + csc x = sin x + sin x 1 cos x 1 + csc x cot x 1+ sin x sin x = cos x + sin x + 1 sin x + 1 cos x cos 2 x + (sin x + 1) 2 = cos x(sin x + 1) 2 2 = cos x + sin x + 2sin x + 1 cos x(sin x + 1) = 1 + 2sin x + 1 cos x(sin x + 1) 2(1 + sin x) = cos x(sin x + 1) = 2sec x

42.

44.

1 1 (1 + sin x) − (1 − sin x) − = 1 − sin x 1 + sin x (1 − sin x)(1 + sin x) 1 + sin x − 1 + sin x = 1 − sin 2 x 2sin x = cos 2 x = 2 tan x sec x 2 ⎛ ⎞ ⎟ 2 cot x = tan x ⎜ tan x cot x + tan x tan x ⎜⎜ 1 + tan x ⎟⎟ ⎝ tan x ⎠ 2 = 1 + tan 2 x = 2 sec2 x = 2 cos2 x

46.

1 − 1 cos 2 x sin 2 x 2 2 = sin 2x − cos2 x sin x cos x sin 2 x − cos 2 x = sin x cos 2x sin2 x cos x sin x cos x sin x cos x sin x − cos x = cos x sin x sin x cos x = tan x − cot x sin x cos x

sec2 x − csc 2 x =


Section 6.1

47.

365

1 + sin x 1 + sin x 1 + sin x = ⋅ 1 − sin x 1 − sin x 1 + sin x

48.

(1 + sin x)2

=

1 − sin 2 x (1 + sin x)2

=

cos x ⋅ sin x sin x cot x 2 cos x = cos x / sin x = 2 sin x cos x +

cos x + cot x sin x = cot x

cos 2 x 1 + sin x , cos x > 0 = cos x

49.

sin 3 x + cos3 x (sin x + cos x)(sin 2 x − sin x cos x + cos 2 x) 50. = sin x + cos x sin x + cos x = sin 2 x − sin x cos x + cos2 x = 1 − sin x cos x

1 − sin x 1 + sin x (1 − sin x)2 − (1 − sin x)2 − = 1 + sin x 1 − sin x (1 + sin x)(1 − sin x) =

1 − 2 sin x + sin 2 x − 1 − 2 sin x − sin 2 x

=

1 − sin 2 x −4 sin x

cos 2 x = −4 tan x sec x

51.

sec x − 1 sec x + 1 (sec x − 1)2 − (sec x + 1) 2 − = sec x + 1 sec x − 1 (sec x − 1)(sec x + 1) = =

sec2 x − 2sec x + 1 − sec 2 x − 2sec x − 1 sec2 x − 1 −4sec x

1 cos x 1 + cos x − cos x(1 − cos x) − = 1 − cos x 1 + cos x (1 − cos x)(1 + cos x) = =

tan 2 x 2

−4 cos x ⋅ cos x sin 2 x = −4csc x cot x =

52.

=

1 + cos x − cos x + cos 2 x 1 − cos 2 x 1 + cos 2 x sin 2 x 2 2

sin x

1 + sin x cos x (1 + sin x )(1 − sin x ) − cos x (cos x ) 1 − sin 2 x − cos2 x cos2 x − cos2 x − = = = =0 cos x 1 − sin x cos x (1 − sin x ) cos x (1 − sin x ) cos x (1 − sin x )

54.

( sin x + cos x + 1) 2 = sin 2 x + sin x cos x + sin x + cos x sin x + cos2 x + cos x + sin x + cos x + 1 = 1 + 2sin x cos x + 2sin x + 2 cos x + 1 = 2 ( sin x cos x + cos x + sin x + 1) = 2 ( sin x + 1) ( cos x + 1) sin x 1 sec x + tan x cos x + cos x cos x = ⋅ 1 − sin x cos x sec x − tan x cos x

cos x

1 + sin x = 1 − sin x 1 + sin x 1 + sin x = ⋅ 1 − sin x 1 + sin x = =

(1 + sin x)2 1 − sin 2 x (1 + sin x)2 cos 2 x


1 + 1 − sin 2 x

sin 2 x sin 2 x

= 2csc2 x − 1

53.

55.

−

=

sin 2 x

366

56.


sin 3 x − cos3 x (sin x − cos x )(sin 2 x + sin x cos x + cos2 x ) = sin x + cos x sin x + cos x sin x − cos x (sin x − cos x )(1 + sin x cos x ) = ⋅ sin x − cos x sin x + cos x = = =

=

=

(sin 2 x − 2sin x cos x + cos2 x )(1 + sin x cos x ) sin 2 x − cos2 x (1 − 2sin x cos x )(1 + sin x cos x ) sin 2 x − cos2 x 1 − sin x cos x − 2sin 2 x cos2 x sin 2 x − cos2 x 1 sin 2 x

2 x cos2 x sin x sin 2 x sin 2 x − cos2 x sin 2 x sin 2 x 2

− sin x 2cos x − 2sin

csc2 x − cot x − 2 cos x 1 − cot 2 x

57.

Identity

58.

Identity

59.

Identity

60.

Identity

61.

Identity

62.

Identity

63.

Not an identity

64.

Not an identity

65.

Not an identity. If x = π / 4, the left side is 2 and the right side is 1.

66.

Not an identity. If x = π / 6, the left side is

67.

Not an identity. If x = 0°, the left side is

68.

Not an identity. If x = π , the left side is 1 and the right side is −1.

69.

Not an identity. If x = 0, the left side is –1 and the right side is 1.

70.

Not an identity. If x = π , the left side is 1 and the right side is −1.

3 and the right side is 2 3 / 3.

3 / 2 and the right side is (2 + 3) / 2.


Section 6.1

367

....................................................... 71.

1 − sin x + cos x 1 − sin x + cos x 1 + sin x + cos x = ⋅ 1 + sin x + cos x 1 + sin x + cos x 1 + sin x − cos x

= = =


1 − tan x + sec x (1 − tan x + sec x)(1 + tan x + sec x) = 1 + tan x − sec x (1 + tan x − sec x)(1 + tan x + sec x)

1 − sin 2 x + 2sin x cos x − cos 2 x

=

1 + 2sin x + sin 2 x − cos 2 x 2

2

1 − (sin x + cos x) + 2sin x cos x 2

=

2

2sin 4 x + 2sin 2 x cos2 x − 3sin 2 x − 3cos 2 x 2sin 2 x

= = =

1 + 2 tan x + tan 2 x − sec2 x 1 + 2sec x − (sec2 x − 1) + sec2 x

1 + 2 tan x + tan 2 x − (tan 2 x + 1) 2 + 2sec x = 2 tan x 1 + sec x = tan x

1 + 2sin x + sin x − (1 − sin x) 1 − 1 + 2sin x cos x

1 + 2sin x + sin 2 x − 1 + sin 2 x 2sin x cos x 2sin x cos x = = 2 x(1 + sin x) 2sin 2sin x + 2sin x cos x = 1 + sin x 73.

1 + 2sec x − tan 2 x + sec2 x

2sin 2 x(sin 2 x + cos 2 x) − 3(sin 2 x + cos 2 x) 2sin 2 x (2sin 2 x − 3)(sin 2 x + cos2 x) 2sin 2 x 2sin 2 x − 3 2sin 2 x 2sin 2 x

3 − 2sin 2 x 2sin 2 x 3 = 1 − csc2 x 2 =

74.

4 tan x sec2 x − 4 tan x − sec2 x + 1 4 tan 3 x − tan 2 x

= = =

4 tan x (sec2 x − 1) − (sec2 x − 1) 4 tan 3 x − tan 2 x (4 tan x − 1)(sec2 x − 1) tan 2 x(4 tan x − 1) tan 2 x

tan 2 x =1 75.

sin x cos x cos x sin x − cos x sin x tan x + sin x − 2 sin x = sin x − cos x sin x(tan x − 1) = sin x − cos x sin x(tan x − 1) cos x = sin x cos x − cos x cos x tan x(tan x − 1) = tan x − 1 = tan x

sin x(tan x + 1) − 2 tan x cos x = sin x − cos x

sin x tan x + sin x − 2


368

76.


sin 2 x cos x + cos3 x − sin 3 x cos x − sin x cos3 x 2

1 − sin x

= =

cos x(sin 2 x + cos 2 x) − sin x cos x(sin 2 x + cos 2 x) 1 − sin 2 x cos x − sin x cos x

1 − sin 2 x cos x(1 − sin x) = (1 − sin x)(1 + sin x) cos x = 1 + sin x

77.

sin 4 x + cos 4 x = sin 4 x + 2 sin 2 x cos 2 x + cos4 x − 2 sin 2 x cos 2 x = (sin 2 x + cos 2 x)2 − 2 sin 2 x cos 2 x = 1 − 2 sin 2 x cos 2 x

78.

tan 4 x + sec 4 x = tan 4 x − 2 tan 2 x sec 2 x + sec 4 x + 2 tan 2 x sec 2 x = (tan 2 x − sec 2 x)2 + 2 tan 2 x sec 2 x = 1 + 2 tan 2 x sec2 x

.......................................................

Prepare for Section 6.2 PS2. sin ⎜⎛ π + π ⎟⎞ = sin ⎜⎛ 5π ⎟⎞ = 1 ⎝2 3⎠ ⎝ 6 ⎠ 2

PS1. cos ⎛⎜ π − π ⎞⎟ = cos ⎛⎜ π ⎞⎟ = 1 ⎝2 6⎠ ⎝3⎠ 2 cos ⎛⎜ π ⎞⎟ cos ⎛⎜ π ⎞⎟ + sin ⎛⎜ π ⎞⎟ sin ⎛⎜ π ⎞⎟ = 0 ⋅ 3 + 1 ⋅ 1 = 1 2 2 2 ⎝2⎠ ⎝2⎠ ⎝6⎠ ⎝6⎠ 1 Both functional values equal . 2

sin ⎜⎛ π ⎟⎞ cos ⎜⎛ π ⎟⎞ + cos ⎜⎛ π ⎟⎞ sin ⎜⎛ π ⎟⎞ = 1 ⋅ 1 + 0 ⋅ 3 = 1 2 2 2 ⎝2⎠ ⎝2⎠ ⎝3⎠ ⎝3⎠ 1 Both functional values equal to . 2

PS3. sin(90o − 30o ) = sin(60o ) = 3 = cos(30o ) 2 o o o sin(90 − 45 ) = sin(45 ) = 2 = cos(45o ) 2 sin(90o − 120o ) = sin( −30o ) = − 1 = cos(120o ) 2 For each of the given values of θ , the functional values are equal.

PS4. tan ⎛⎜ π − π ⎞⎟ = tan ⎛⎜ π ⎞⎟ = cot ⎛⎜ π ⎞⎟ ⎝2 6⎠ ⎝3⎠ ⎝6⎠ ⎛ ⎞ ⎛ ⎞ π π π tan ⎜ − ⎟ = tan ⎜ ⎟ = cot ⎛⎜ π ⎞⎟ ⎝2 4⎠ ⎝4⎠ ⎝4⎠ ⎛ ⎞ ⎛ ⎞ ⎛ 4π ⎞ π π π 4 5 tan ⎜ − ⎟ = tan ⎜ − ⎟ = cot ⎜ ⎟ 3 ⎠ ⎝2 ⎝ 6 ⎠ ⎝ 3 ⎠ For each of the given values of θ , the functional values are equal.

PS5. tan ⎛⎜ π − π ⎞⎟ = tan ⎛⎜ π ⎞⎟ = 3 ⎝3 6⎠ ⎝6⎠ 3 ⎛ ⎞ ⎛ ⎞ π π 3 3− 3 2 3 tan ⎜ ⎟ − tan ⎜ ⎟ 3− 3 ⎝3⎠ ⎝6⎠ = 3 = 3 3 = 3 = 3 ⎛ ⎞ ⎛ ⎞ + 1 1 2 3 π π 3 1 + tan ⎜ ⎟ tan ⎜ ⎟ 1 + 3 ⋅ ⎝3⎠ ⎝6⎠ 3

PS6. For k is any integer, the value of (2k + 1) will result in odd integers. Thus sin[(2k + 1)π ] will be 0.

Both functional values equal

3. 3


Section 6.2

369

Section 6.2 1.

sin(45° + 30°) = sin 45° cos30° + cos 45° sin 30°

2.

sin(330° + 45°) = sin 330° cos 45° + cos330° sin 45°

2 3 2 1 = ⋅ + ⋅ 2 2 2 2 6 2 6+ 2 = + = 4 4 4 3.

cos(45° − 30°) = cos 45° cos30° + sin 45° sin 30°

1 2 3 2 =− ⋅ + ⋅ 2 2 2 2 2 6 − 2+ 6 =− + = 4 4 4 4.

cos(120° − 45°) = cos120° cos 45° + sin120° sin 45°

2 3 2 1 ⋅ + ⋅ 2 2 2 2 6 2 6+ 2 = + = 4 4 4

1 2 3 2 =− ⋅ + ⋅ 2 2 2 2 2 6 − 2+ 6 =− + = 4 4 4

=

5.

7.

tan 45° − tan 30° 1 + tan 45° tan 30° 3 3− 3 1− 3− 3 3 = = 3 = = 2− 3 ⎛ 3 ⎞ 3+ 3 3+ 3 1 + 1⎜⎜ ⎟⎟ 3 ⎝ 3 ⎠

tan(45° − 30°) =

5π π 5π π ⎛ 5π π ⎞ − ⎟ = sin cos − cos sin sin ⎜ 4 6 4 6 4 6 ⎝ ⎠ 2 3 ⎛ 2⎞ 1 =− ⋅ − ⎜− ⎟⋅ 2 2 ⎜⎝ 2 ⎟⎠ 2 =−

9.

6.

=

8.

6 2 − 6+ 2 + = 4 4 4

3π π 3π π ⎛ 3π π ⎞ + ⎟ = cos cos − sin sin cos ⎜ 4 6 4 6 ⎝ 4 6⎠

10.

11.

6

4

4π π 4π π ⎛ 4π π ⎞ + ⎟ = sin sin ⎜ cos + cos sin 3 4 3 4 3 4 ⎝ ⎠ =−

3 2 ⎛ 1⎞ 2 ⋅ + ⎜− ⎟⋅ 2 2 ⎝ 2⎠ 2

=−

6 2 6+ 2 − =− 4 4 4

π π π π ⎛π π ⎞ cos ⎜ − ⎟ = cos cos + sin sin 4 3 4 3 ⎝4 3⎠ =

12.

3 +1 3 +3 = 3 = 3 1 − 3 ⋅ 1 3− 3 3 3

( 3 − 1)2 4 − 2 3 = = 2− 3 3 −1 2

2 1 2 3 ⋅ + ⋅ 2 2 2 2 2 6 2+ 6 = + = 4 4 4

2 3 2 1 ⋅ − ⋅ 2 2 2 2 6 2 6+ 2 =− − =− 4 4 4 =−

π π ⎛ π π ⎞ tan 6 + tan 4 tan ⎜ + ⎟ = ⎝ 6 4 ⎠ 1 − tan π tan π

tan 240° − tan 45° 1 + tan 240° tan 45° 3 −1 3 −1 3 −1 = = = 1 + ( 3)(1) 1 + 3 3 +1

tan(240° − 45°) =

11π π tan − tan ⎛ 11π π ⎞ 6 4 tan ⎜ − ⎟= 4 ⎠ 1 + tan 11π tan π ⎝ 6 6 4 =

=

3 +3 3+ 3 9+6 3 +3 ⋅ = 9−3 3− 3 3+ 3

=

=

12 + 6 3 = 2+ 3 6

=

− 3 −1 3

1 + ⎛⎜ − 3 ⎞⎟ (1) ⎝ 3 ⎠

=

− 3 −1 3

1− 3 3

=

( −3 − 3)(3 + 3) (3 − 3)(3 + 3)

−9 − 6 3 − 3 −12 − 6 3 = 9−3 6 = −2 − 3


− 3 −3 3− 3

370


13.

cos 212° cos122° + sin 212° sin122° = cos(212° − 122°) = cos90° = 0

14.

sin167° cos107° − cos167° sin107° = sin(167° − 107°) = sin 60° =

15.

sin

17.

5π π 5π π π 1 ⎛ 5π π ⎞ cos − cos sin = sin ⎜ − ⎟ = sin = 12 4 12 4 6 2 ⎝ 12 4 ⎠

7π π − tan 12 4 = tan ⎛ 7π − π ⎞ = tan π = 3 ⎜ ⎟ 7π π 3 ⎝ 12 4 ⎠ 1 + tan tan 12 4 tan

3 2

16.

18.

cos

π 12

tan

π

cos

π 4

+ tan

− sin

π 12

sin

π

π 1 ⎛π π⎞ = cos ⎜ + ⎟ = cos = 4 3 2 ⎝ 12 4 ⎠

π

3 = tan ⎛ π + π ⎞ = tan π = undefined ⎜ ⎟ π 2 ⎝6 3⎠ 1 − tan tan 6 3 6

π

19.

sin 42o = cos(90o − 42o ) = cos 48o

20.

cos80o = sin(90o − 80o ) = sin10o

21.

tan15o = cot(90o − 15o ) = cot 75o

22.

cot 2o = tan(90o − 2o ) = tan 88o

23.

sec 25o = csc(90o − 25o ) = csc65o

24.

csc84o = sec(90o − 84o ) = sec6o

25.

sin 7 x cos 2 x − cos 7 x sin 2 x = sin(7 x − 2 x) = sin 5 x

26.

sin x cos3x + cos x sin 3 x = sin( x + 3 x) = sin 4 x

27.

cos x cos 2 x + sin x sin 2 x = cos( x − 2 x) = cos(− x) = cos x

28.

cos 4 x cos 2 x − sin 4 x sin 2 x = cos(4 x + 2 x) = cos6 x

29.

sin 7 x cos3 x − cos 7 x sin 3 x = sin(7 x − 3x) = sin 4 x

30.

cos x cos5 x − sin x sin 5 x = cos( x + 5 x) = cos 6 x

31.

cos 4 x cos(−2 x) − sin 4 x sin( −2 x) = cos 4 x cos 2 x + sin 4 x sin 2 x = cos(4 x − 2 x) = cos 2 x

32.

sin(− x)cos3x − cos(− x)sin 3 x = − sin x cos3x − cos x sin 3x = −(sin x cos3x + cos x sin 3 x) = − sin( x + 3 x) = − sin 4 x

33.

x 2x x 2x ⎛ x 2x ⎞ = sin ⎜ + sin cos + cos sin ⎟ = sin x 3 3 3 3 ⎝3 3 ⎠

34.

cos

3x x 3x x x ⎛ 3x x ⎞ cos + sin sin = cos ⎜ − ⎟ = cos 4 4 4 4 2 ⎝ 4 4⎠

35.

tan 3x + tan 4 x = tan(3x + 4 x ) = tan 7 x 1 − tan 3 x tan 4 x

36.

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


Section 6.2

37.

4 4 3 tan α = − , sin α = , cos α = − , 3 5 5 15 15 8 tan β = , sin β = − , cos β = − 8 17 17 a. sin(α − β ) = sin α cos β − cosα sin β

b.

c.

39.

371

⎛ 4 ⎞⎛ 8 ⎞ ⎛ 3 ⎞⎛ 15 ⎞ = ⎜ ⎟⎜ − ⎟ − ⎜ − ⎟⎜ − ⎟ ⎝ 5 ⎠⎝ 17 ⎠ ⎝ 5 ⎠⎝ 17 ⎠ 32 45 77 =− − =− 85 85 85 cos(α + β ) = cos α cos β − sin α sin β

c.

24 24 7 , sin α = , cos α = , 7 25 25 8 15 8 sin β = − , cos β = − , tan β = 17 17 15 a. sin(α + β ) = sin α cos β + cosα sin β tan α =

b.

⎛ 3 ⎞ ⎛ 8 ⎞ ⎛ 4 ⎞ ⎛ 15 ⎞ = ⎜− ⎟⎜− ⎟ − ⎜ ⎟⎜− ⎟ ⎝ 5 ⎠ ⎝ 17 ⎠ ⎝ 5 ⎠ ⎝ 17 ⎠ 24 60 84 = + = 85 85 85 tan α − tan β tan(α − β ) = 1 + tan α tan β

c.

⎛ 24 ⎞⎛ 15 ⎞ ⎛ 7 ⎞⎛ 8 ⎞ = ⎜ ⎟⎜ − ⎟ + ⎜ ⎟⎜ − ⎟ ⎝ 25 ⎠⎝ 17 ⎠ ⎝ 25 ⎠⎝ 17 ⎠ 360 56 416 =− − =− 425 425 425 cos(α + β ) = cos α cos β − sin α sin β

⎛ 7 ⎞ ⎛ 15 ⎞ ⎛ 24 ⎞ ⎛ 8 ⎞ = ⎜ ⎟⎜− ⎟ − ⎜ ⎟⎜− ⎟ ⎝ 25 ⎠ ⎝ 17 ⎠ ⎝ 25 ⎠ ⎝ 17 ⎠ 105 192 87 =− + = 425 425 425 tan α − tan β tan(α − β ) = 1 + tan α tan β

=

− 4 − 15 24 8 = 3 8 ⋅ 1 − 60 24 1 + − 4 15 24 3 8

=

24 − 8 24 − 8 105 7 15 = 7 15 ⋅ 192 8 24 105 1+ 1+ 105 7 15

=

−32 − 45 77 = 24 − 60 36

=

360 − 56 304 = 105 + 192 297

− 4 − 15 3

( )( )

3 4 3 sin α = , cos α = , tan α = , 5 5 4 5 12 12 cos β = − , sin β = , tan β = − 13 13 5 a. sin(α − β ) = sin α cos β − cos α sin β

b.

38.

3 ⎛ 5 ⎞ 4 ⎛ 12 ⎞ = ⎜− ⎟ − ⎜ ⎟ 5 ⎝ 13 ⎠ 5 ⎝ 13 ⎠ 15 48 =− − 65 65 63 =− 65 cos(α + β ) = cos α cos β − sin α sin β 4 ⎛ 5 ⎞ 3 ⎛ 12 ⎞ = ⎜− ⎟ − ⎜ ⎟ 5 ⎝ 13 ⎠ 5 ⎝ 13 ⎠ 20 36 =− − 65 65 56 =− 65 tan α − tan β tan(α − β ) = 1 + tan α tan β

( ) ( )( )

3 − − 12 5 = 4 3 1+ − 12 4 5 3 + 12 20 = 4 5 ⋅ 36 20 1− 20

=

15 + 48 63 =− 20 − 36 16

40.

( )( )

24 7 24 , cos α = − , tan α = − , 25 25 7 4 3 3 cos β = − , sin β = − , tan β = 5 5 4 a. cos( β − α ) = cos β cos α + sin β sin α sin α =

b.

= − 4 ⎛⎜ − 7 ⎞⎟ + ⎛⎜ − 3 ⎞⎟ 24 5 ⎝ 25 ⎠ ⎝ 5 ⎠ 25 28 = − 72 = − 44 125 125 125 sin(α + β ) = sin α cos β + cos α sin β 24 ⎛ 4 ⎞ ⎛ 7 ⎞⎛ 3 ⎞ ⎜ − ⎟ + ⎜ − ⎟⎜ − ⎟ 25 ⎝ 5 ⎠ ⎝ 25 ⎠⎝ 5 ⎠ 96 21 =− + 125 125 75 3 =− =− 125 5 tan α + tan β tan(α + β ) = 1 − tan α tan β =

c.

=

=

− 24 + 3 7

4

( 7 )( 34 )

1 − − 24

− 24 + 3 28 7 4⋅ 1 + 72 28 28

−96 + 21 = 28 + 72 75 3 =− =− 100 4


372

41.


4 3 4 sin α = − , cos α = − , tan α = , 5 5 3 12 5 5 cos β = − , sin β = , tan β = − 13 13 12 a. sin(α − β ) = sin α cos β − cos α sin β

b.

c.

42.

12 ⎞ ⎛ 3 ⎞ 5 = ⎛⎜ − 4 ⎞⎛ ⎟⎜ − ⎟ − ⎜ − ⎟ ⎝ 5 ⎠⎝ 13 ⎠ ⎝ 5 ⎠ 13 = 48 + 15 = 63 65 65 65 cos(α + β ) = cos α cos β − sin α sin β

7 24 7 , cos α = , tan α = − , 25 25 24 8 15 15 cos β = , sin β = − , tan β = − 17 17 8 a. sin(α + β ) = sin α cos β + cos α sin β sin α = −

b.

12 ⎞ ⎛ 4 ⎞ 5 = ⎛⎜ − 3 ⎞⎛ ⎟⎜ − ⎟ − ⎜ − ⎟ ⎝ 5 ⎠⎝ 13 ⎠ ⎝ 5 ⎠ 13 = 36 + 20 = 56 65 65 65 tan α + tan β tan(α + β ) = 1 − tan α tan β

⎛ 7 ⎞ ⎛ 8 ⎞ 24 ⎛ 15 ⎞ = ⎜− ⎟⎜ ⎟ + ⎜− ⎟ ⎝ 25 ⎠ ⎝ 17 ⎠ 25 ⎝ 17 ⎠ 56 360 416 =− − =− 425 425 425 cos(α − β ) = cos α cos β + sin α sin β 24 ⎛ 8 ⎞ ⎛ 7 ⎞ ⎛ 15 ⎞ ⎜ ⎟+⎜− ⎟⎜− ⎟ 25 ⎝ 17 ⎠ ⎝ 25 ⎠ ⎝ 17 ⎠ 192 105 297 = + = 425 425 425 tan α + tan β tan(α + β ) = 1 − tan α tan β =

c.

( ) ( )( )

4+ − 5 12 = 3 4 −5 1− 3 12 4− 5 4− 5 = 3 12 = 3 12 ⋅ 36 1 + 29 1 + 20 36 36 36

= 48 − 15 = 33 36 + 20 56

43.

15 8 8 , sin α = , tan α = , 17 17 15 3 4 3 sin β = − , cos β = − , tan β = 5 5 4 a. sin(α + β ) = sin α cos β + cos α sin β cos α =

8 ⎛ 4 ⎞ 15 ⎛ 3 ⎞ ⎜− ⎟ + ⎜− ⎟ 17 ⎝ 5 ⎠ 17 ⎝ 5 ⎠ 32 45 77 =− − =− 85 85 85 cos(α − β ) = cos α cos β + sin α sin β

=

44.

c.

= 15 ⎛⎜ − 4 ⎞⎟ + 8 ⎛⎜ − 3 ⎞⎟ 17 ⎝ 5 ⎠ 17 ⎝ 5 ⎠ = − 60 − 24 = − 84 85 85 85 tan α − tan β tan(α − β ) = 1 + tan α tan β 8 −3 8 −3 8 −3 60 = 15 4 = 15 4 = 15 4 ⋅ 1 + 24 1 + 24 60 1+ 8 3 60 60 15 4 32 − 45 13 = =− 60 + 24 84

()

24

416 −56 − 360 =− 192 − 105 87

7 24 24 , sin α = , tan α = − , 25 25 7 12 5 12 sin β = − , cos β = , tan β = − 13 13 5 a. sin(α + β ) = sin α cos β + cos α sin β cos α = −

=

b.

( ) ( )( )

− 7 + − 15

8 1 − − 7 − 15 24 8 7 15 − − − 7 − 15 192 = 24 8 = 24 8 ⋅ 1 − 105 1 − 105 192 192 192

=

24 ⎛ 5 ⎞ ⎛ 7 ⎞ ⎛ 12 ⎞ ⎜ ⎟ + ⎜− ⎟⎜− ⎟ 25 ⎝ 12 ⎠ ⎝ 25 ⎠ ⎝ 13 ⎠ 120 84 204 = + = 325 325 325 cos(α + β ) = cos α cos β − sin α sin β =

b.

c.

⎛ 7 ⎞ 5 24 ⎛ 12 ⎞ = ⎜− ⎟ − ⎜− ⎟ ⎝ 25 ⎠ 13 25 ⎝ 13 ⎠ 35 288 253 =− + = 325 325 325 tanα − tan β tan(α − β ) = 1+ tanα tan β =

( 5 ) = − 247 + 125 = − 247 + 125 ⋅ 35 1+ 288 35 1+ ( − 24 )( −12 ) 1+ 288 35 35 7 5 − 24 − −12 7

= −120 + 84 = − 36 35 + 288 323


Section 6.2

45.

373

3 4 4 cos α = − , sin α = − , tan α = , 5 5 3 5 12 5 sin β = , cos β = , tan β = 13 13 12 a. sin(α − β ) = sin α cos β − cos α sin β

b.

c.

46.

5⎞ = ⎛⎜ − 4 ⎞⎟ 12 − ⎛⎜ − 3 ⎞⎛ ⎟⎜ ⎟ ⎝ 5 ⎠ 13 ⎝ 5 ⎠⎝ 13 ⎠ = − 48 + 15 = − 33 65 65 65 cos(α + β ) = cos α cos β − sin α sin β

8 15 15 , sin α = − , tan α = − , 17 17 8 24 7 24 sin β = − , cos β = − , tan β = 25 25 7 a. sin(α − β ) = sin α cos β − cos α sin β cos α =

b.

= ⎛⎜ − 3 ⎞⎟ 12 − ⎛⎜ − 4 ⎞⎟ 5 ⎝ 5 ⎠ 13 ⎝ 5 ⎠ 13 = − 36 + 20 = − 16 65 65 65 tan α + tan β tan(α + β ) = 1 − tan α tan β

⎛ 15 ⎞⎛ 7 ⎞ 8 ⎛ 24 ⎞ = ⎜ − ⎟⎜ − ⎟ − ⎜ − ⎟ ⎝ 17 ⎠⎝ 25 ⎠ 17 ⎝ 25 ⎠ 105 192 297 = + = 425 425 425 cos(α + β ) = cosα cos β − sin α sin β 8 ⎛ 7 ⎞ ⎛ 15 ⎞⎛ 24 ⎞ ⎜ − ⎟ − ⎜ − ⎟⎜ − ⎟ 17 ⎝ 25 ⎠ ⎝ 17 ⎠⎝ 25 ⎠ 56 360 416 =− − =− 425 425 425 tan α + tan β tan(α + β ) = 1 − tan α tan β =

c.

4+ 5 4+ 5 = 3 12 = 3 12 1 − 20 1− 4 5 36 3 12 4+ 5 = 3 12 ⋅ 36 = 48 + 15 = 63 1 − 20 36 36 − 20 16 36

( )

47.

3 4 3 sin α = , cos α = , tan α = , 5 5 4 5 5 12 tan β = , sin β = − , cos β = − 12 13 13 a. sin(α + β ) = sin α cos β + cos α sin β

b.

c.

49.

=

=

36 − 20 16 = 48 + 15 63

48.

15 15 8 , sin α = , cos α = , 8 17 17 7 7 24 tan β = − , sin β = − , cos β = 24 25 25 a. sin(α − β ) = sin α cos β − cos α sin β tan α =

15 ⎛ 24 ⎞ 8 ⎛ 7 ⎞ ⎜ ⎟ − ⎜− ⎟ 17 ⎝ 25 ⎠ 17 ⎝ 25 ⎠ 360 56 416 = + = 425 425 425 cos(α − β ) = cos α cos β + sin α sin β =

b.

c.

8 ⎛ 24 ⎞ 15 ⎛ 7 ⎞ = ⎜ ⎟ + ⎜− ⎟ 7 ⎝ 25 ⎠ 17 ⎝ 25 ⎠ 192 105 87 = − = 425 425 425 tan α + tan β tan(α + β ) = 1 − tan α tan β

( ) ( )

15 + − 7 15 − 7 192 24 = 8 = 8 24 ⋅ 105 192 15 7 + 1 − 1− 192 8 24 360 − 56 304 = = 192 + 105 297

( )( )

π π ⎛π ⎞ cos ⎜ − θ ⎟ = cos cosθ + sin sin θ 2 2 ⎝2 ⎠ = 0 ⋅ cosθ + 1 ⋅ sin θ = sin θ

( )( )

−105 + 192 56 + 360 87 = 416

⎛ 4 ⎞ ⎛ 12 ⎞ ⎛ 3 ⎞ ⎛ 5 ⎞ = ⎜ ⎟⎜− ⎟ + ⎜ ⎟⎜− ⎟ ⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠ 48 15 63 =− − =− 65 65 65 tan α − tan β tan(α − β ) = 1 + tan α tan β 3− 5 3− 5 48 4 12 = 4 12 ⋅ 1 + 15 48 1+ 3 5 48 4 12

− 15 + 24 56 8 7 7 ⋅ = 8 1 + 360 56 1 − − 15 24 56 8 7

=

⎛ 3 ⎞ ⎛ 12 ⎞ ⎛ 4 ⎞ ⎛ 5 ⎞ = ⎜ ⎟⎜− ⎟ + ⎜ ⎟⎜− ⎟ ⎝ 5 ⎠ ⎝ 13 ⎠ ⎝ 5 ⎠ ⎝ 13 ⎠ 36 20 56 =− − =− 65 65 65 cos(α − β ) = cos α cos β + sin α sin β

=

− 15 + 24

50.

cos(θ + π ) = cosθ cos π − sin θ sin π = cosθ (−1) − sin θ (0) = − cosθ


374

51.


sin ⎛⎜ θ + π ⎞⎟ = sin θ cos π + cos θ sin π 2⎠ 2 2 ⎝ = sin θ (0) + cos θ (1)

52.

sin(θ + π ) = sin θ cos π + cosθ sin π = sin θ (−1) + cosθ (0) = − sin θ

54.

tan 2θ = tan(θ + θ )

= cos θ

53.

π π ⎞ tan θ + tan 4 ⎛ tan ⎜θ + ⎟ = 4 ⎠ 1 − tan θ tan π ⎝

tan θ + tan θ 1 − tan θ tan θ 2 tan θ = 1 − tan 2 θ

4

=

tan θ + 1 = 1 − tan θ

55.

3π 3π ⎛ 3π ⎞ cos ⎜ − θ ⎟ = cos cosθ + sin sin θ 2 2 2 ⎝ ⎠ = 0(cosθ ) + (−1)sin θ

56.

3π 3π ⎛ 3π ⎞ sin ⎜ + θ ⎟ = sin cosθ + cos sin θ 2 2 2 ⎝ ⎠ = ( −1) cosθ + (0)sin θ

= − sin θ 57.

= − cosθ

⎛π ⎞ cos (π / 2 − θ ) cot ⎜ − θ ⎟ = ⎝2 ⎠ sin (π / 2 − θ ) =

58.

cos(π + θ ) sin(π + θ ) cos π cosθ − sin π sin θ = sin π cosθ + cos π sin θ (−1)cosθ − (0)sin θ = (0)cosθ + ( −1)sin θ − cosθ = − sin θ = cot θ

cot(π + θ ) =

( cos π2 ) cosθ + (sin π2 ) sinθ (sin π2 ) cosθ − ( cos π2 ) sinθ

(0) cosθ + (1)sin θ (1)cosθ − (0)sin θ sin θ = cosθ = tan θ =

59.

csc(π − θ ) =

1 sin(π − θ )

60.

1 sin π cosθ − cos π sin θ 1 = (0)cosθ − (−1)sin θ 1 = sin θ = cscθ =

61.

sin 6 x cos 2 x − cos 6 x sin 2 x = sin(6 x − 2 x) = sin 4 x = sin(2 x + 2 x) = sin 2 x cos 2 x + cos 2 x sin 2 x = 2sin 2 x cos 2 x

1 sec ⎛⎜ π − θ ⎞⎟ = ⎝2 ⎠ cos π − θ 2

( )

1 cos π cos θ + sin π sin θ 2 2 1 = (0) cos θ + (1)sin θ = 1 sin θ = csc θ =

62.

cos5 x cos3 x + sin 5 x sin 3 x = cos(5 x − 3 x) = cos 2 x = cos( x + x) = cos x cos x − sin x sin x

63.

cos(α + β ) + cos(α − β ) = cosα cos β − sin α sin β + cosα cos β + sin α sin β = 2cosα cos β

64.

cos(α − β ) − cos(α + β ) = cosα cos β + sin α sin β − cosα cos β + sin α sin β = 2sin α sin β

65.

sin(α + β ) + sin(α − β ) = sin α cos β + cosα sin β + sin α cos β − cosα sin β = 2sin α cos β


= cos 2 x − sin 2 x

Section 6.2

66.

67.

69.

71.

375

sin(α − β ) − sin(α + β ) = sin α cos β − cosα sin β − sin α cos β − cosα sin β = −2cos α sin β cos(α − β ) cosα cos β + sin α sin β = sin(α + β ) sin α cos β + cosα sin β cos α cos β sin α sin β + sin α cos β sin α cos β = sin α cos β cosα sin β + sin α cos β sin α cos β cot α + tan β = 1 + cot α tan β sin( x + h) − sin x sin x cos h + cos x sin h − sin x = h h sin x(cos h − 1) cos x sin h = + h h cos h − 1 sin h = sin x + cos x h h

⎛π ⎞ ⎡π ⎤ sin ⎜ + α − β ⎟ = sin ⎢ + (α − β ) ⎥ ⎝2 ⎠ ⎣2 ⎦

π

70.

72.

π

sin(α + β ) sin α cos β + cosα sin β = sin(α − β ) sin α cos β − cosα sin β sin α cos β cosα sin β + sin α cos β sin α cos β = sin α cos β cosα sin β − sin α cos β sin α cos β 1 + cot α tan β = 1 − cot α tan β cos( x + h) − cos x cos x cos h − sin x sin h − cos x = h h cos x(cos h − 1) sin x sin h = − h h cos h − 1 sin h = cos x − sin x h h

⎛π ⎞ ⎡π ⎤ cos ⎜ + α + β ⎟ = cos ⎢ + (α + β ) ⎥ ⎝2 ⎠ ⎣2 ⎦

π

= cos cos(α + β ) − sin sin(α + β ) 2 2 = (0)cos(α + β ) − (1)sin(α + β )

= cos(α − β )

= − sin(α + β )

= cosα cos β + sin α sin β

= −(sin α cos β + cosα sin β )

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

74.

π

cos(α − β ) + cos sin(α − β ) 2 2 = (1) cos(α − β ) + (0)sin(α − β )

= sin

73.

68.

cos3 x = cos(2 x + x) = cos 2 x cos x − sin 2 x sin x = cos( x + x)cos x − sin( x + x)sin x = (cos x cos x − sin x sin x) cos x − (sin x cos x + cos x sin x)sin x = cos3 x − cos x sin 2 x − 2cos x sin 2 x = cos3 x − 3cos x sin 2 x = cos3 x − 3cos x(1 − cos 2 x) = cos3 x − 3cos x + 3cos3 x = 4cos3 x − 3cos x


376


75.

cos(θ + 3π ) = cosθ cos3π − sin θ sin 3π = (cosθ )(−1) − (sin θ )(0) = − cosθ

77.

tan(θ + π ) =

79.

sin(θ + 2kπ ) = sin θ cos(2kπ ) + cosθ sin 2kπ = (sin θ )(1) + (cosθ )(0) = sin θ

80.

We consider two cases, (1) k an odd and (2) k an even integer.

81.

83.

tan θ + tan π 1 − tan θ tan π tan θ + 0 = 1 − (tan θ )(0) = tan θ

(1)

sin(θ − kπ ) = sin θ cos( kπ ) − cos θ sin( kπ ) = (sin θ )(−1) − (cos θ )(0) = − sin θ , provided k is odd

(2)

sin(θ − kπ ) = sin θ cos( kπ ) − cos θ sin( kπ ) = (sin θ )(1) − (cosθ )(0) = sin θ , provided k is even

(

)

y = sin π − x and y = cos x both have the following graph. 2

y = sin 7 x cos 2 x − cos 7 x sin 2 x and y = sin 5 x both have the following graph.

76.

sin(θ + 2π ) = sin θ cos 2π + cosθ sin 2π = (sin θ )(1) + (cosθ )(0) = sin θ

78.

cos[θ + (2k + 1)π ] = cosθ cos[(2k + 1)π ] − sin θ sin[(2k + 1)π ] = (cosθ )(−1) − (sin θ )(0) = − cosθ

82.

y = cos( x + π ) and y = − cos x both have the following graph.

84.

y = sin 3 x and y = 3sin x − 4sin 3 x both have the following graph.

....................................................... 85.

sin( x − y ) ⋅ sin( x + y ) = (sin x cos y − cos x sin y )(sin x cos y + cos x sin y ) = sin 2 x cos 2 y − cos 2 x sin 2 y

86.

sin( x + y + z ) = sin[ x + ( y + z )] = sin x cos( y + z ) + cos x sin( y + z ) = sin x(cos y cos z − sin y sin z ) + cos x (sin y cos z + cos y sin z ) = sin x cos y cos z − sin x sin y sin z + cos x sin y cos z + cos x cos y sin z


Connecting Concepts

Section 6.3

377

87.

cos( x + y + z ) = cos[ x + ( y + z )] = cos x cos( y + z ) − sin x sin( y + z ) = cos x[cos y cos z − sin y sin z ] − sin x[sin y cos z + cos y sin z ] = cos x cos y cos z − cos x sin y sin z − sin x sin y cos z − sin x cos y sin z

88.

sin( x + y ) sin x cos y + cos x sin y = sin x sin y sin x sin y sin x cos y cos x sin y = + sin x sin y sin x sin y = cot y + cot x

90.

ER =

89.

cos( x − y ) cos x cos y + sin x sin y = cos x sin y cos x sin y cos x cos y sin x sin y = + cos x sin y cos x sin y = cot y + tan x

2sin10° + sin 20° 2sin(10° + 20°) 2sin10° + sin 20° = 2sin 30° ER ≈ 0.69

....................................................... PS1. sin 2α = sin(α + α )

PS2. cos 2α = cos(α + α ) = cos α cos α − sin α sin α

= sin α cos α + cos α sin α = 2sin α cos α

= cos 2 α − sin 2 α 3 3 2 = 2 = 3 = 3 1+ cos(60o ) 1+ 1 3 2 2 ⎛ o⎞ sin(90o ) tan ⎜ 90 ⎟ = tan(45o ) =1 = 1 =1 ⎝ 2 ⎠ 1+ cos(90o ) 1+ 0 3 3 o⎞ ⎛ sin(120o ) tan ⎜ 120 ⎟ = tan(60o ) = 3 2 2 = = = 3 ⎝ 2 ⎠ 1+ cos(120o ) 1− 1 1 2 2 For each of the given values of α , the functional values are equal.

⎛ o⎞ PS4. tan ⎜ 60 ⎟ = tan(30o ) = 3 3 ⎝ 2 ⎠

PS3. tan 2α = tan(α + α )

= tan α + tan α 1 − tan α tan α = 2 tan α 1 − tan 2 α

PS5. Let α = 45o ; then the left side of the equation is 1, and

the right side of the equation is


2.

sin(60o )

PS6. Let α = 60o ; then the left side of the equation is 3 , and the right side of the equation is 1 . 2 4

Section 6.3 1.

4.

2sin 2α cos 2α = sin 2(2α ) = sin 4α

2.

2cos 2 2 β − 1 = cos 2(2 β ) = cos 4 β

5.

2sin 3θ cos3θ = sin 2(3θ ) = sin 6θ

3.

cos 2 3α − sin 2 3α = cos 2(3α ) = cos 6α

6.

1 − 2sin 2 5 β = cos 2(5β ) = cos10 β cos 2 6α − sin 2 6α = cos 2(6α ) = cos12α


378

7.

9.


2 tan 3α 1 − tan 2 3α

2 tan 4θ

8.

= tan 2(3α )

1 − tan 2 4θ

= tan 6α

= tan 2(4θ ) = tan 8θ

2

4 3 sin α 3/ 5 3 ⎛4⎞ cos α = − , sin α = 1 − ⎜ ⎟ = , tan α = = =− 5 5 cos α −4 / 5 4 ⎝5⎠ cos 2α = cos2 α − sin 2 α

sin 2α = 2sin α cos α

2

⎛ 3⎞⎛ 4 ⎞ = 2⎜ ⎟⎜ − ⎟ ⎝ 5 ⎠⎝ 5 ⎠ 24 =− 25

⎛ 4⎞ ⎛ 3⎞ = ⎜− ⎟ −⎜ ⎟ ⎝ 5⎠ ⎝ 5⎠ 16 9 = − 25 25 7 = 25

tan 2α =

2

=

=

2 tan α 1 − tan 2 α

( 4 ) = − 64 2 1− 9 1 − (− 3 ) 16 4 2 −3

−6

4 ⋅ 16 = −24 1 − 9 16 16 − 9 16

=−

10.

cos α =

2

24 7 sin α −7 / 25 7 ⎛ 24 ⎞ , sin α = − 1 − ⎜ ⎟ = − , tan α = = =− 25 25 cos α 24 / 25 24 ⎝ 25 ⎠ cos 2α = cos2 α − sin 2 α


2

⎛ 7 ⎞⎛ 24 ⎞ = 2 ⎜ − ⎟⎜ ⎟ ⎝ 25 ⎠⎝ 25 ⎠ 336 =− 625

tan 2α =

⎛ 24 ⎞ ⎛ 7 ⎞ = ⎜ ⎟ − ⎜− ⎟ ⎝ 25 ⎠ ⎝ 25 ⎠ 576 49 = − 625 625 527 = 625

2

=

=


( 24 ) 2 1 − (− 7 ) 24 2 − 7

−7

12 ⋅ 576 1 − 49 576 576

=−

11.

sin α =

24 7

336 336 =− 576 − 49 527

2

8 15 sin α 8 /17 8 ⎛ 8⎞ , cos α = − 1 − ⎜ ⎟ = − , tan α = = =− 17 17 cos α −15/17 15 ⎝ 17 ⎠

sin 2α = 2sin α cos α ⎛ 8 ⎞ ⎛ 15 ⎞ = 2⎜ ⎟⎜ − ⎟ ⎝ 17 ⎠ ⎝ 17 ⎠ 240 =− 289

cos 2α = cos2 α − sin 2 α 2

⎛ 15 ⎞ ⎛ 8⎞ = ⎜− ⎟ − ⎜ ⎟ ⎝ 17 ⎠ ⎝ 17 ⎠ 225 64 = − 289 289 161 = 289

tan 2α = 2

=

=


( 15 ) = − 1615 2 1 − 64 1 − (− 8 ) 225 15 2 −8

− 16

15 ⋅ 225 = −240 1 − 64 225 225 − 64 225

=−

240 161


Section 6.3

12.

379

sin α = −

2

9 40 sin α −9 / 41 9 ⎛ 9⎞ , cos α = − 1 − ⎜ − ⎟ = − , tan α = = = 41 41 cos α −40 / 41 40 ⎝ 41 ⎠ cos 2α = cos2 α − sin 2 α


2

⎛ 9 ⎞ ⎛ 40 ⎞ = 2⎜ − ⎟⎜ − ⎟ ⎝ 41 ⎠ ⎝ 41 ⎠ 720 = 1681

13.

tan α = −

tan 2α =

⎛ 40 ⎞ ⎛ 9⎞ = ⎜− ⎟ − ⎜− ⎟ ⎝ 41 ⎠ ⎝ 41 ⎠ 1600 81 = − 1681 1681 1519 = 1681

2


=

( 40 ) = 209 2 1 − 81 1− ( 9 ) 1600 40

=

9 1600 720 20 ⋅ = 1 − 81 1600 1600 − 81 1600

=

720 1519

2 9

24 24 7 , r = 242 + 72 = 576 + 49 = 625 = 25, sin α = − , cos α = 7 25 25

cos 2α = cos2 α − sin 2 α


2

⎛ 24 ⎞ ⎛ 7 ⎞ = 2⎜ − ⎟⎜ ⎟ ⎝ 25 ⎠ ⎝ 25 ⎠ 336 =− 625

⎛ 7 ⎞ ⎛ 24 ⎞ = ⎜ ⎟ −⎜− ⎟ ⎝ 25 ⎠ ⎝ 25 ⎠ 49 576 = − 625 625 527 =− 625

tan 2α = 2

=

=


( 7) 2 1 − ( − 24 ) 7 2 − 24

− 48

7 ⋅ 49 1 − 576 49 49

336 −336 = = 49 − 576 527 14.

4 4 3 tan α = , r = 42 + 32 = 16 + 9 = 25 = 5, sin α = , cos α = 3 5 5 cos 2α = cos2 α − sin 2 α


2

⎛ 4 ⎞⎛ 3⎞ = 2⎜ ⎟⎜ ⎟ ⎝ 5 ⎠⎝ 5⎠ 24 = 25

⎛ 3⎞ ⎛4⎞ =⎜ ⎟ −⎜ ⎟ ⎝ 5⎠ ⎝5⎠ 9 16 = − 25 25 7 =− 25

2

tan 2α = 2 tan α 1 − tan 2 α 8 2 4 3 = = 3 2 1 − 16 1− 4 9

() (3)

=

8 3 ⋅ 9 = 24 1 − 16 9 9 − 16 9

= − 24 7 15.

2

15 225 289 − 225 64 8 15/17 15 ⎛ 15 ⎞ , cos α = 1 − ⎜ ⎟ = 1 − = = = , tan α = = 17 289 289 289 17 8 /17 8 ⎝ 17 ⎠ 2 tan α tan 2α = sin 2α = 2sin α cos α cos 2α = cos2 α − sin 2 α 1 − tan 2 α 2 2 ⎛ 15 ⎞ ⎛ 8 ⎞ 8 15 ⎛ ⎞ ⎛ ⎞ = 2⎜ ⎟⎜ ⎟ 15 =⎜ ⎟ −⎜ ⎟ 2 15 ⎝ 17 ⎠ ⎝ 17 ⎠ 8 ⎝ 17 ⎠ ⎝ 17 ⎠ = = 4 2 1 − 225 240 64 225 1 − 15 64 = = − 8 289 289 289 15 64 240 161 = 4 ⋅ = =− 225 64 64 − 225 1− 289 sin α =

( ) ( ) 64

240 =− 161


380

16.

Chapter 6: Trigonometric Identities and Equations 2 −3 3 3 9 4 ⎛ 3⎞ sin α = − , cos α = − 1 − ⎜ − ⎟ = − 1 − = − , tan α = 5 = 5 25 5 ⎝ 5⎠ −4 4 5

2


2

cos 2α = cos α − sin α 2

⎛ 3⎞⎛ 4 ⎞ = 2⎜ − ⎟⎜ − ⎟ ⎝ 5⎠⎝ 5 ⎠ 24 = 25

⎛ 4⎞ ⎛ 3⎞ = ⎜− ⎟ − ⎜− ⎟ 5 ⎝ ⎠ ⎝ 5⎠ 16 9 = − 25 25 7 = 25

2

tan 2α = 2 tan α 1 − tan 2 α 3 2 3 4 = = 2 2 1− 9 1− 3 16

() (4)

=

3 2

⋅ 16 = 24 1 − 9 16 16 − 9

= 24 7 17.

cos α =

2

40 1600 9 −9 / 41 9 ⎛ 40 ⎞ , sin α = − 1 − ⎜ ⎟ = − 1 − = − , tan α = =− 41 1681 41 40 / 41 40 ⎝ 41 ⎠ cos 2α = cos2 α − sin 2 α


2

⎛ 9 ⎞ ⎛ 40 ⎞ = 2⎜ − ⎟⎜ ⎟ ⎝ 41 ⎠ ⎝ 41 ⎠ 720 =− 1681

⎛ 40 ⎞ ⎛ 9⎞ = ⎜ ⎟ − ⎜− ⎟ ⎝ 41 ⎠ ⎝ 41 ⎠ 1600 81 = − 1681 1681 1519 = 1681

2

tan 2α = 2 tan α 1 − tan 2 α 2 − 9 − 9 40 20 = = 2 1 − 81 9 1− − 1600 40

( ) ( )

=

− 9

20

1 − 81

1600

=

18.

16

−720 = − 720 1600 − 81 1519

2

4 16 3 −3/ 5 3 ⎛4⎞ cos α = , sin α = − 1 − ⎜ ⎟ = − 1 − = − , tan α = =− 5 5 25 5 4 / 5 4 ⎝ ⎠ cos 2α = cos2 α − sin 2 α


2

⎛ 3⎞⎛ 4 ⎞ = 2⎜ − ⎟⎜ ⎟ ⎝ 5⎠⎝ 5 ⎠ 24 =− 25

⎛4⎞ ⎛ 3⎞ = ⎜ ⎟ −⎜− ⎟ 5 ⎝ ⎠ ⎝ 5⎠ 16 9 = − 25 25 7 = 25

2

tan 2α = 2 tan α 1 − tan 2 α 2 −3 −3 4 2 = = 2 1− 9 3 1− − 16 4

( ) ( )

=

−3

2 ⋅ 16 = −24 1 − 9 16 16 − 9 16

= − 24 7 19.

⋅ 1600 1600

(

)

6cos2 x = 6 1 + cos 2 x = 3 (1 + cos 2 x ) 2


Section 6.3

20.

381

(

)(

)

2

2

sin 4 x cos4 x = 1 − cos 2 x 1 + cos 2 x 2 2 1 = (1 − cos 2 x + cos2 2 x )(1 + cos 2 x + cos2 2 x ) 16 = 1 (1 − 2cos2 2 x + cos4 2 x ) 16 2⎞ ⎛ = 1 ⎜ 1 − 2 1 + cos 4 x + 1 + cos 4 x ⎟ 16 ⎝ 2 2 ⎠

(

) (

)

(

)

= 1 ⎛⎜ 1 − 1 − cos 4 x + 1 + 1 cos 4 x + 1 1 + cos8 x ⎞⎟ 16 ⎝ 4 2 4 2 ⎠ 1 3 1 1 = − cos 4 x + cos8 x 16 8 2 8 = 1 3 − cos 4 x + 1 cos8 x 32 4 4

( (

21.

(

)

)

2

cos4 x = 1 + cos 2 x 2 1 = (1 + 2cos 2 x + cos2 2 x ) 4 = 1 1 + 2cos 2 x + 1 + cos 4 x 4 2 1 = ( 3 + 4 cos 2 x + cos 4 x ) 8

(

23.

(

)(

22.

)

)

2

150° 2 1 − cos150° =+ 2

sin 75° = sin

)

(

)(

24.

)

26.

)

2

sin 2 x cos4 x = 1 − cos 2 x 1 + cos 2 x 2 2 2 1 = (1 − cos 2 x ) (1 + cos 2 x ) 8 = 1 1 − 1 + cos 4 x (1 + cos 2 x ) 8 2 1 1 1 = − cos 4 x (1 + cos 2 x ) 8 2 2 = 1 (1 + cos 2 x − cos 4 x − cos 2 x cos 4 x ) 16

( (

sin 4 x cos2 x = 1 − cos 2 x 1 + cos 2 x 2 2 = 1 (1 − cos2 2 x ) (1 − cos 2 x ) 8 = 1 1 − 1 + cos 4 x (1 − cos 2 x ) 8 2 1 1 1 = − cos 4 x (1 − cos 2 x ) 8 2 2 = 1 (1 − cos 2 x − cos 4 x + cos 4 x cos 2 x ) 16

( (

25.

)

(

)

)

)

3

sin 6 x = 1 − cos 2 x 2 = 1 (1 − 3cos 2 x + 3cos2 2 x − cos3 2 x ) 8 = 1 ⎛⎜ 1 − 3cos 2 x + 3 + 3 cos 4 x − cos 2 x 1 + cos 4 x ⎞⎟ 8⎝ 2 2 2 ⎠ 1 = ( 5 − 7cos 2 x + 3cos 4 x − cos 2 x cos 4 x ) 16

(

210° 2 1 + cos 210° =− 2

cos105° = cos

27.

135° 2 1 − cos135° = sin135° 1 − (− 2 / 2) = 2/2

tan 67.5° = tan

=

1 − (− 3 / 2) 2

=−

1 + (− 3 / 2) 2

=

2+ 3 4

=−

2− 3 4

=

=

2+ 3 2

=−

2− 3 2

=


2+ 2 2 ⋅ 2 2

2 2+2 2 = 2 +1

)

382

28.

31.

34.


330° 2 1 + cos330° =− 2

cos165° = cos

=−

1 + ( 2 / 2) 2

=

1 − ( − 2 / 2) 2

=−

2+ 3 2

=−

2+ 2 4

=

2+ 2 4

=−

2+ 2 2

=

2+ 2 2

45° 2 1 − cos 45° =+ 2

sin

α 2

135° 2 1 + cos135° =+ 2

cos 67.5° = cos

32.

=

1− 2 / 2 2

=

=

2− 2 4

=

=

2− 2 2

=

5π 5π / 4 = cos 8 2 1 + cos5π / 4 =− 2

sin α =

35.

cos

33.

sin

1 + (− 2 / 2)

1− 2 / 2 2

2− 2 4

=

2− 2 4

2− 2 2

=

2− 2 2

36.

sin

=

2− 3 4

=

2+ 2 4

=

2− 3 2

=

2+ 2 2

2− 2 4

=−

2− 2 2 2

5 25 12 ⎛5⎞ =− . , cosα = − 1 − ⎜ ⎟ = − 1 − 13 169 13 ⎝ 13 ⎠

=

=

1 + cos α 2

1 − (−12 /13) 2

=

1 − 12 /13 2

=

=

13 + 12 26

=

13 − 12 26

=

=

25 5 = 26 26

=

1 1 = 26 26

=

26 26

5 26 26

α 2

(

=

=−

cos

3π 3π / 4 = sin 8 2 1 − cos3π / 4 =+ 2

1− 3 / 2 2

=

1 − cosα 2

7π 7π / 4 = sin 8 2 1 − cos 7π / 4 =+ 2 =

2/2

5π 5π / 6 = cos 12 2 1 + cos5π / 6 =+ 2

1− 2 / 2 2

=

=

225° 2 1 − cos 225° =+ 2

sin112.5° = sin

1+ 3 /2 2

=−

37.

30.

=−

sin 22.5° = sin

cos

315° 2 1 + cos315° =− 2

cos157.5° = cos

29.

tan

α 2

=

1 − cos α sin α 1 + 12

13 5 13

13 + 12 5 =5


1− − 2 / 2 2

)

Section 6.3

38.

2

7 49 24 ⎛ 7 ⎞ , cos α = − 1 − ⎜ − ⎟ = − 1 − =− 25 625 25 ⎝ 25 ⎠

sin α = − sin

39.

383

α 2

=

1 − cos α 2

=−

1 + cosα 2

=

1 − (−24 / 25) 2

=−

1 − 24 / 25 2

=

25 + 24 50

=−

25 − 24 50

=

49 50

=−

1 50

=

7 2 10

=−

cos α = −

cos

cos α = sin

41.

α 2

tan

α 2

= =

1 − cos α sin α 1 + 24

25

−7

25

25 + 24 = −7 = −7

2 10

2

cos α = − 1 + cos α 2 2 1 8 /17 − =− 2 = − 17 − 8 34 9 =− 34 3 = − 34 34

tan

α 2

= =

1 − cos α sin α 1 + 18 17

− 15 17

17 + 8 = −15 5 =− 3

2

12 144 5 ⎛ 12 ⎞ , sin α = 1 − ⎜ ⎟ = 1 − = 13 13 169 13 ⎝ ⎠

=

1 − cos α 2

=

1 − 12 /13 2

=

13 − 12 = 26

=

26 26

tan α =

2

8 64 15 ⎛ 8⎞ , sin α = − 1 − ⎜ − ⎟ = − 1 − =− 17 17 289 17 ⎝ ⎠

sin α = 1 − cos α 2 2 1 − (−8 /17) = 2 + 17 8 = 34 25 = 34 5 = 34 34 40.

α

cos

1 26

α 2

=

1 + cos α 2

=

1 + 12 /13 2

=

13 + 12 = 26

=

tan α = 1 − cos α 2 sin α =

25 26

1 − 12

13 5 13

= 13 − 12 5 1 = 5

5 26 26

4 4 3 , r = 32 + 42 = 25 = 5, sin α = , cos = 3 5 5

sin α = 1 − cos α 2 2

cos α = 1 + cos α 2 2

tan

α 2

=

= 1 − 3/ 5 2

= 1 + 3/ 5 2

=

= 5−3 10

= 5+3 10

=

= 1 5

= 4 5

= 5 5

=2 5 5

1 − cos α sin α 1− 3 4 5

5

5−3 4 1 = 2


384

42.


tan α = −

sin

α 2

8 8 15 , r = 82 + 152 = 64 + 225 = 17, sin α = , cos = − 15 17 17

=

1 − cosα 2

=

1 − (−15 /17) 2

cos

α 2

17 + 15 32 = = 34 34

=

43.

cos α = sin

α 2

44.

α 2

= =

2

=

1 17

=

17 17

= =

1 − cosα sin α

( 17 ) = 17 + 15

1 − − 15 8 17

8

=4

α

α

1 − cosα sin α

=−

1 + cos α 2

1 − 24 / 25 2

=−

1 + 24 / 25 2

=

25 − 24 1 = 50 50

=−

25 + 24 49 =− 50 50

=−

cos

2

2 10

=−

tan

2

=

1 − 24

25 = 25 − 24 7 −7 − 25

7 2 10

1 7

2

9 81 40 ⎛ 9⎞ , cosα = 1 − ⎜ − ⎟ = 1 − = 41 41 1681 41 ⎝ ⎠ 1 − cos α 2

cos

α 2

=−

1 + cos α 2

tan

1 − 40 / 41 = 2

1 + 40 / 41 =− 2

41 − 40 1 = = 82 82

41 + 40 81 =− =− 82 82

=

47.

1 + (−15 /17) 2

α

2

1 − cosα 2

=

=

tan

24 576 7 ⎛ 24 ⎞ , sin α = − 1 − ⎜ ⎟ = − 1 − =− 25 625 25 ⎝ 25 ⎠

sin α = − sin

45.

4 17 17

=

=

1 + cosα 2

17 − 15 2 = = 34 34

16 4 = 17 17

=

=

82 82

=−

1 (2sin 3x cos3 x) 2 1 = sin 2(3x) 2 1 = sin 6 x 2

sin 3 x cos3 x =

sin 2 x + cos 2 x = sin 2 x + cos 2 x − sin 2 x = cos 2 x

α 2

= =

1 − 40

41 = 41 − 40

−9

41

=−

9 82 82

46.

1 − cosα sin α

1 9

cos 8 x = cos 2(4 x) = cos2 4 x − sin 2 4 x

48.

cos 2 x = cos 2 x − sin 2 x sin 2 x sin 2 x 2 2 = cos2 x − sin 2 x sin x sin x = cot 2 x − 1


−9

Section 6.3

49.

385

1 + cos 2 x 1 + 2 cos 2 x − 1 = sin 2 x 2 sin x cos x

50.

2 cos 2 x 2 sin x cos x = cot x =

51.

53.

55.

sin 2 x 2

1 − sin x

=

2 sin x cos x

52.

2

cos x = 2 tan x

cos 2 x = cos 2 x − sin 2 x cos 2 x cos 2 x 2 2 = cos 2 x − sin 2 x cos x cos x = 1 − tan 2 x sin 2 x − tan x = 2sin x cos x − =

54.

sin x cos x

56.

1 1 = 1 − cos 2 x 1 − 1 + 2sin 2 x 1 = 2sin 2 x = 1 csc 2 x 2 cos 2 x − sin 2 x cos 2 x = 2 sin x cos x sin 2 x = cot 2 x 2sin x cos x

sin 2 x cos x − sin x cos 2 x = tan 2 x 2

sin 2 x − cot x = 2 sin x cos x −

2sin x cos 2 x − sin x cos x

=

2 sin 2 x cos x − cos x sin x

=

cos 4 x − sin 4 x = (cos 2 x + sin 2 x)(cos 2 x − sin 2 x)

58.

sin 4 x = 2 sin 2 x cos 2 x = 2(2 sin x cos x)(cos 2 x − sin 2 x)

= cos 2 x − sin 2 x = cos 2 x 59.

cos x sin x

cos x(2 sin 2 x − 1) sin x = cot x(− cos 2 x) = − cot x cos 2 x

sin x (2cos 2 x − 1) cos x = tan x cos 2 x =

57.

=

2

= 4 sin x cos3 x − 4 sin 3 x cos x

cos 2 x − 2sin 2 x cos 2 x − sin 2 x + 2sin 4 x = cos2 x (1 − 2sin 2 x ) − sin 2 x(1 − 2sin 2 x) = (1 − 2sin 2 x)(cos 2 x − sin 2 x) = cos 2 x cos 2 x = cos 2 2 x

60.

2cos 4 x − cos2 x − 2sin 2 x cos 2 x + sin 2 x = cos 2 x(2cos 2 x − 1) − sin 2 x(2cos 2 x − 1) = (2cos 2 x − 1)(cos 2 x − sin 2 x) = cos 2 x ⋅ cos 2 x = cos 2 2 x

61.

cos 4 x = cos 2(2 x)

62.

sin 4 x = sin 2(2 x)

= 2cos 2 x − 1

= 2 sin 2 x cos 2 x

= 2(2cos 2 x − 1)2 − 1

= 2(2 sin x cos x)(1 − 2 sin 2 x)

= 2(4cos 4 x − 4cos 2 x + 1) − 1

= 4 sin x cos x − 8 sin 3 x cos x

2

= 8cos 4 x − 8cos 2 x + 1


386

63.


cos3 x − cos x = cos(2 x + x) − cos x = cos 2 x cos x − sin 2 x sin x − cos x = (2cos 2 x − 1) cos x − 2sin x cos x ⋅ sin x − cos x = 2cos3 x − cos x − 2sin 2 x cos x − cos x = 2cos3 x − 2cos x − 2sin 2 x cos x = 2cos3 x − 2cos x − 2(1 − cos 2 x) cos x = 2cos3 x − 2cos x − 2cos x + 2cos3 x = 4cos3 x − 4 cos x

64.

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

65.

sin 3 x + cos3 x = (sin x + cos x)(sin 2 x − sin x cos x + cos 2 x) 2sin x cos x ⎞ ⎛ = (sin x + cos x) ⎜ sin 2 x + cos 2 x − ⎟ 2 ⎝ ⎠ 1 ⎛ ⎞ = (sin x + cos x) ⎜ 1 − sin 2 x ⎟ ⎝ 2 ⎠

66.

cos3 x − sin 3 x = (cos x − sin x)(cos 2 x + sin x cos x + sin 2 x) 2 sin x cos x ⎞ ⎛ = (cos x − sin x) ⎜ cos 2 x + sin 2 x + ⎟ 2 ⎝ ⎠ ⎛ 1 ⎞ = (cos x − sin x) ⎜1 + sin 2 x ⎟ ⎝ 2 ⎠

67.

sin 2

x ⎡ 1 − cos x ⎤ = ⎢± ⎥ 2 ⎣⎢ 2 ⎦⎥

2

68.

cos 2

x ⎡ 1 + cos x ⎤ = ⎢± ⎥ 2 ⎣⎢ 2 ⎦⎥

1 − cos x 2 1 − cos x sec x = ⋅ 2 sec x sec x − 1 = 2 sec x

1 + cos x 2 1 + cos x sec x = ⋅ 2 sec x sec x + 1 = 2 sec x

=

69.

71.

tan

x 1 − cos x = 2 sin x 1 cos x = − sin x sin x = csc x − cot x

x x ⎛ x⎞ 2sin cos = sin 2 ⎜ ⎟ 2 2 ⎝2⎠ = sin x

2

=

70.

72.

tan

x sin x = 2 1 + cos x

cos 2

=

sin x cos x 1 + cos x cos x cos x

=

tan x sec x + 1

x x ⎛ x⎞ − sin 2 = cos 2 ⎜ ⎟ 2 2 ⎝2⎠ = cos x


Section 6.3

73.

387 2

x x⎞ x x x ⎛ 2 x + 2 sin cos + sin 2 ⎜ cos + sin ⎟ = cos 2 2⎠ 2 2 2 2 ⎝ x x ⎛ x⎞ = cos 2 + sin 2 + sin 2 ⎜ ⎟ 2 2 ⎝2⎠ = 1 + sin x

74.

tan 2

x ⎛ 1 − cos x ⎞ =⎜ ⎟ 2 ⎝ sin x ⎠ = =

2

(1 − cos x)2 sin 2 x (1 − cos x)2 1 − cos 2 x

(1 − cos x)2 (1 − cos x)(1 + cos x) 1 − cos x = 1 + cos x

=

1

= cos x

− cos x

cos x 1 + cos x cos x cos x

=

75.

sin 2

2

⎛ 1 − cos x ⎞ x sec x = ⎜⎜ ± ⎟⎟ sec x 2 2 ⎝ ⎠

76.

1 − cos x ⋅ sec x 2 1 = (sec x − 1) 2

2

⎛ 1 + cos x ⎞ x sec x = ⎜⎜ ± ⎟⎟ sec x 2 2 ⎝ ⎠ 1 + cos x ⋅ sec x 2 1 = (sec x + 1) 2

=

=

2

77.

cos 2

sec x − 1 sec x + 1

⎛ ⎞ cos 2 x − cos x = ⎜ ± 1 + cos x ⎟ − cos x 2 2 ⎝ ⎠ = 1 + cos x − cos x 2 = 1 + cos x − 2cos x 2 − x 1 cos = 2 2 x = sin 2

2

78.

79.

sin 2 x − cos 2 x = − ⎛⎜ cos 2 x − sin 2 x ⎞⎟ 2 2 2 2⎠ ⎝ ⎛ ⎞ x = − cos 2 ⎜ ⎟ ⎝2⎠ = − cos x

80.

81.

sin 2 x − cos x = 2sin x cos x − cos x = (cos x)(2sin x − 1)

82.

⎛ ⎞ sin 2 x + cos x = ⎜ ± 1 − cos x ⎟ + cos x 2 2 ⎝ ⎠ = 1 − cos x + cos x 2 = 1 − cos x + 2cos x 2 + x 1 cos = 2 2 x = cos 2

cos 2 x − sin 2 x = cos x 2 2 = 2sin x cos x 2sin x = 1 csc x sin 2 x 2 cos 2 x = 1 − 2sin 2 x sin 2 x sin 2 x 2 = 12 − 2sin2 x sin x sin x = csc 2 x − 2


388

83.

85.


tan 2 x = 2 tan 2x 1 − tan x 2 tan x tan x = 1 − tan 2 x tan x tan x 2 = cot x − tan x

84.

2 2 2cos 2 x = 2 ( cos x − sin x ) sin 2 x 2sin x cos x 2 2 = cos x − sin x sin x cos x sin x cos x = cot x − tan x

sin 2 x + 1 − cos 2 x 1 − cos 2 x + 1 − cos 2 x = sin x(1 + cos x) sin x(1 + cos x)

86.

1 2x 1 = csc 2 2 2sin 2 x

2

2(1 − cos 2 x) = sin x(1 + cos x) 2(1 − cos x)(1 + cos x) = sin x(1 + cos x) 2(1 − cos x) = sin x x = 2 tan 2

1 sin 2 x 1 = 2sin x cos x = 1 csc x sec x 2

87.

csc 2 x =

89.

cos

x ⎛ x⎞ = cos 2 ⎜ ⎟ 5 ⎝ 10 ⎠ = 1 − 2 sin 2

x 10

=

1

(

1− cos x 2

2 ± =

1

2

( 1−cos2 x )

=

)

2

1 1 − cos x

1 ⋅ 1 + cos x 1 − cos x 1 + cos x x = 1 + cos2x = 1 + cos 1 − cos x sin 2 x = 12 + cos2 x sin x sin x = csc2 x + cot x csc x =

1 cos 2 x 1 = 2cos 2 x − 1 2 1 = ⋅ sec 2 x 2 2cos x − 1 sec x 2 = sec x2 2 − sec x

88.

sec 2 x =

90.

sec2

x 1 = 2 cos 2 x

2

=

= =

1 ⎛ ± 1+ cos x ⎞ ⎜ ⎟ 2 ⎝ ⎠ 1

2

1+ cos x 2

2 1 + cos x


Section 6.3

91.

a.

389

1 − cos π 4 M= =1÷ π / 4) 2 ( sin 2 1

=1÷ =1÷ =

b.

M sin sin

1− 2 2 2

α 2

α

2

=1 =

1 M

α

⎛ 1 ⎞ = sin −1 ⎜ ⎟ ⎝M ⎠ −1 ⎛ 1 ⎞ α = 2sin ⎜ ⎟ ⎝M ⎠ 2

2− 2 4 2

2− 2 ≈ 2.61

c.

1 ⎛ 1 ⎞ decreases. So if sin −1 ⎜ ⎟ ⎝M ⎠ M decreases, then α decreases. As M increases

....................................................... 92.

94.

96.

y = sin 2 x + cos 2 x and y = cos 2 x both have the following graph.

x x y = sin cos and y = sin x do not have the same graph. 2 2

sin 3 x + cos3 x (sin x + cos x)(sin 2 x − sin x cos x + cos 2 x) = sin x + cos x sin x + cos x 2sin x cos x 2 2 = sin x + cos x − 2 = 1 − 1 sin 2 x 2


y=

sin 2 x 1 − sin 2 x

and y = 2 tan x both have the following graph.

2

95.

97.

x x⎞ ⎛ y = ⎜ cos + sin ⎟ and y = 1 + sin x both have the 2 2 ⎝ ⎠ following graph.

cos 4 x = cos 2 x ⋅ cos 2 x cos 2 x + 1 cos 2 x + 1 = ⋅ 2 2

(

)

= 1 cos 2 2 x + 2cos 2 x + 1 4

⎛ cos 4 x + 1 ⎞ = 1⎜ + 2cos 2 x + 1⎟ 4⎝ 2 ⎠ = 1 cos 4 x + 1 + 1 cos 2 x + 1 8 8 2 1 = cos 4 x + 1 cos 2 x + 3 8 2 8


4

390

98.


− sin x sin x − sin 2 x sin x − 2sin x cos x sin x − 2sin x cos x sin x(1 − 2cos x ) x = = = = = − tan cos x + cos 2 x cos x + 2cos2 x − 1 2cos2 x + cos x − 1 (2cos x − 1)(cos x + 1) cos x + 1 2

....................................................... PS1.

1 [sin(α + β ) + sin(α − β )] 2 = 1 [sin α cos β + cos α sin β + sin α cos β − cos α sin β ] 2 = sin α cos β


PS3. sin π − sin π = 0 − 1 = − 1 PS4. 6 2 2 ⎛π +π ⎞ ⎛π +π ⎞ ⎜ ⎟ ⎜ 6 ⎟ = 2cos⎛ 7π ⎞cos⎛ 5π ⎞ 2cos⎜ 2 6 ⎟sin ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ 6 ⎠ ⎝ 6 ⎠ ⎝ 1 ⎠ ⎛ ⎛ 7π ⎞ ⎞ ⎛ ⎛ 5π ⎞ ⎞ ⎜ 1+ cos⎜ ⎟ ⎟ ⎜ 1− cos⎜ ⎟ ⎟ ⎝ 6 ⎠ ⎟+⎜ ⎝ 6 ⎠⎟ 2⎜⎜ − ⎟ ⎜ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠

1 [ cos(α + β ) + cos(α − β )] 2 = 1 [ cos α cos β − sin α sin β + cos α cos β − sin α sin β ] 2 = cos α cos β

(

)

()

()

2 sin x + π = 2 ⎡sin x cos π + cos x sin π ⎤ ⎢⎣ 4 4 4 ⎥⎦ ⎡ ⎛ ⎞ ⎛ ⎞⎤ = 2 ⎢ ( sin x ) ⎜ 2 ⎟ + ( cos x ) ⎜ 2 ⎟ ⎥ 2 2 ⎝ ⎠ ⎝ ⎠⎦ ⎣ = sin x + cos x

1− 3 1+ 3 2 2 = −2 2 2 = − 1− 3 = − 1 4 4 1 =− 2 Both functional values equal − 1 . 2 PS5. Answers will vary.

PS6.

(−1)2 + ( 3 ) = 1 + 3 = 4 = 2 2

Section 6.4 1.

1 [sin( x + 2 x) + sin( x − 2 x)] 2 = sin 3 x + sin(− x)

2sin x cos 2 x = 2 ⋅

1 [cos(4 x − 2 x) − cos(4 x + 2 x)] 2 = cos 2 x − cos 6 x

2.

2sin 4 x sin 2 x = 2 ⋅

= sin 3 x − sin x

1 [sin(6 x + 2 x) − sin(6 x − 2 x)] 2 1 = [sin 8 x − sin 4 x ] 2

1 [cos(3x + 5 x) + cos(3x − 5 x)] 2 1 = [ cos8 x + cos(−2 x)] 2 1 = (cos8 x + cos 2 x) 2

3.

cos 6 x sin 2 x =

4.

cos3 x cos5 x =

5.

2sin 5 x cos3 x = sin(5 x + 3 x) + sin(5 x − 3 x) = sin 8 x + sin 2 x

6.

2sin 2 x cos 6 x = sin(2 x + 6 x) + sin(2 x − 6 x) = sin 8 x + sin(−4 x) = sin 8 x − sin 4 x


Section 6.4

391

7.

sin x cos5 x = 1 [ cos( x − 5 x) − cos( x + 5 x) ] 2 = 1 [ cos(−4 x) − cos6 x ] 2 = 1 (cos 4 x − cos 6 x) 2

9.

cos 75° cos15° =

11.

1 [cos(75° + 15°) + cos(75° − 15°)] 2 1 = (cos 90° + cos 60°) 2 1⎛ 1⎞ = ⎜0 + ⎟ 2⎝ 2⎠ 1 = 4

1 [sin(3x + x) − sin(3x − x)] 2 1 = (sin 4 x − sin 2 x) 2

8.

cos 3x sin x =

10.

sin105° cos15° =

1 [sin(105° + 15°) + sin(105° − 15°)] 2 1 = (sin120° + sin 90°) 2 1⎛ 3 ⎞ = ⎜⎜ + 1⎟⎟ 2⎝ 2 ⎠ 3+2 4

=

cos157.5° sin 22.5° = 1 [sin(157.5° + 22.5°) − sin(157.5° − 22.5°)] 2 = 1 (sin180° − sin135°) 2 ⎛ ⎞ 1 = ⎜0 − 2 ⎟ 2⎝ 2 ⎠ =− 2 4

12.

sin195° cos15° = 1 [sin(195° + 15°) + sin(195° − 15°) ] 2 = 1 (sin 210° + sin180°) 2 = 1 ⎛⎜ − 1 + 0 ⎞⎟ 2⎝ 2 ⎠ 1 =− 4

14.

sin 11π sin 7π = 1 ⎡⎢ cos 11π − 7π − cos 11π + 7π ⎤⎥ 12 12 2 ⎣ 12 12 12 12 ⎦

(

) = 1 ( cos π − cos 3π ) 2 3 2 1 1 = ( − 0) 2 2

(

)

13.

15.

=1

⎡ ⎤ sin 13π cos π = 1 ⎢sin ⎛⎜ 13π + π ⎞⎟ + sin ⎛⎜ 13π − π ⎞⎟ ⎥ ⎝ 12 12 ⎠ ⎦ 12 12 2 ⎣ ⎝ 12 12 ⎠ = 1 ⎛⎜ sin 7π + sin π ⎞⎟ 2⎝ 6 ⎠ ⎛ ⎞ 1 1 = ⎜ − + 0⎟ 2⎝ 2 ⎠ 1 =− 4

(

) ( = 1 ⎡⎢sin 2π + sin ( − π ) ⎤⎥ 2⎣ 3 2 ⎦ 2 π π 1 = ( sin − sin ) 2 3 2 = 1 ⎛⎜ 3 − 1⎞⎟ 2⎝ 2 ⎠

4

= 3 −2 4

16.

cos

17π 7π 1 ⎡ ⎛ 17π 7π ⎞ ⎛ 17π 7π ⎞ ⎤ sin = sin ⎜ + − ⎟ − sin ⎜ ⎟⎥ 12 12 2 ⎢⎣ ⎝ 12 12 ⎠ ⎝ 12 12 ⎠ ⎦ 1⎛ 5π ⎞ = ⎜ sin 2π − sin ⎟ 2⎝ 6 ⎠

17.

)

sin π cos 7π = 1 ⎡⎢sin π + 7π + sin π − 7π ⎤⎥ 12 12 2 ⎣ 12 12 12 12 ⎦

4θ + 2θ 4θ − 2θ cos 2 2 = 2sin 3θ cosθ

sin 4θ + sin 2θ = 2sin

1⎛ 1⎞ = ⎜0 − ⎟ 2⎝ 2⎠ 1 =− 4


392


5θ + 3θ 5θ − 3θ sin 2 2 = −2sin 4θ sin θ

18.

cos 5θ − cos 3θ = −2sin

20.

sin 7θ − sin 3θ = 2cos

22.

cos 3θ + cos5θ = 2cos

7θ + 3θ 7θ − 3θ sin 2 2 = 2cos5θ sin 2θ 3θ + 5θ 3θ − 5θ cos 2 2 = 2 cos 4θ cos(−θ )

3θ + θ 3θ − θ cos 2 2 = 2cos 2θ cosθ

19.

cos 3θ + cosθ = 2cos

21.

cos 6θ − cos 2θ = −2 sin

23.

cosθ + cos 7θ = 2 cos

6θ + 2θ 6θ − 2θ sin 2 2 = −2 sin 4θ sin 2θ

θ − 7θ cos 2 2 = 2 cos 4θ cos(−3θ )

= 2 cos 4θ cosθ

24.

= 2 cos 4θ cos3θ

3θ + 7θ 3θ − 7θ cos 2 2 = 2sin 5θ cos(−2θ )


25.



= 2sin 5θ cos 2θ

= 2sin 7θ cos 2θ

5θ + θ 5θ − θ sin 2 2 = −2sin 3θ sin 2θ

26.

cos 5θ − cosθ = −2sin

28.


θ + 7θ


2θ + θ 2θ − θ sin 2 2 3 1 = −2sin θ sin θ 2 2

27.

cos 2θ − cosθ = −2sin

29.

cos θ − cosθ = −2sin 2

θ +θ

2

2

4

32.

3θ + θ 3θ − θ 3θ θ + sin = 2sin 4 2 cos 4 2 4 2 2 2 5 1 = 2sin θ cos θ 8 8

cosθ + cos

θ 2

= 2cos

θ +θ

θ −θ

2

2

2 cos

31.

sin

θ 2

− sin

θ 3

θ +θ

θ −θ

2

2

= 2cos 2 = 2cos

3 sin 2

5 1 θ sin θ 12 12

2

cos (α + β ) + cos (α − β ) = cos α cos β − sin α sin β + cosα cos β + sin α sin β = 2cos α cos β

34.

)

4

3 1 = 2cos θ cos θ 4 4 33.

4

= 2sin 3 θ sin 1 θ 4

sin

(

2

= −2sin 3 θ sin − 1 θ

= 2sin 4θ cos 2θ

30.

θ −θ

sin 2

cos (α − β ) − cos (α + β ) = cosα cos β + sin α sin β − cosα cos β + sin α sin β = 2sin α sin β


3

Section 6.4

35.

393

1 [sin(3x + x) − sin(3x − x)] 2 = sin 4 x − sin 2 x = 2sin 2 x cos 2 x − sin 2 x = sin 2 x(2cos 2 x − 1)

2cos3x sin x = 2 ⋅

1 [sin(5 x + 3x) + sin(5 x − 3x)] 2 1 = (sin8 x + sin 2 x) 2 1 = (2sin 4 x cos 4 x + 2sin x cos x) 2 = sin 4 x cos 4 x + sin x cos x

36.

sin 5 x cos3 x =

38.

sin 3 x cos x =

= 2sin x cos x ⎡⎢ 2(1 − 2sin 2 x) − 1⎤⎥ ⎣ ⎦ = 4sin x cos x − 8sin 3 x cos x − 2sin x cos x = 2sin x cos x − 8cos x sin 3 x 37.

1 [cos(5 x + 7 x) + cos(5 x − 7 x)] 2 = cos12 x + cos(−2 x) = cos12 x + cos 2 x

2 cos 5 x cos 7 x = 2 ⋅

= cos 2 6 x − sin 2 6 x + 2 cos 2 x − 1

1 [sin(3x + x) + sin(3x − x)] 2 1 = (sin 4 x + sin 2 x) 2 1 = (2sin 2 x cos 2 x + sin 2 x) 2 1 = [ (sin 2 x(2cos 2 x + 1)] 2 1 = ⋅ 2sin x cos x ⎡⎢ 2(1 − 2sin 2 x) + 1⎤⎥ ⎣ ⎦ 2 = sin x cos x(2 − 4sin 2 x + 1) = sin x cos x(3 − 4sin 2 x)

39.

= 2(1 − 2sin 2 x)sin x

5 x + 3x 5 x − 3x sin 2 2 = 2sin 4 x sin x = −2(2sin 2 x cos 2 x sin x)

= 2sin x − 4sin 3 x

= −4 ⎡⎢ 2sin x cos x(2cos 2 x − 1)sin x ⎤⎥ ⎣ ⎦

3x + x 3x − x sin 2 2 = 2 cos 2 x sin x

sin 3 x − sin x = 2 cos

40.

cos5 x − cos3 x = −2sin

= −8sin 2 x(2cos3 x − cos x) 41.

2x + 4 2 x − x4 cos 2 2 = 2cos3 x cos(− x) = 2cos3 x cos x

sin 2 x + sin 4 x = 2cos

= 2cos x sin(2 x + x) = 2cos x(sin 2 x cos x + cos 2 x sin x) = 2cos x[(2sin x cos x)cos x + (2cos 2 x − 1)sin x] = 2cos x sin x(4cos 2 x − 1) 42.

3x + x 3x − x cos 2 2 = 2 cos 2 x cos x

cos 3 x + cos x = 2 cos

2

= 2(2 cos x − 1) cos x = 4 cos3 x − 2 cos x

43.

2cos 3 x + x sin 3 x − x sin 3 x − sin x 2 2 = cos3 x − cos x −2sin 3 x + x sin 3 x − x 2

cos 2 x =− sin 2 x = − cot 2 x


2

394

44.

Chapter 6: Trigonometric Identities and Equations 5 x +3 x 5 x −3 x cos5 x − cos3 x −2sin 2 sin 2 = sin 5 x + sin 3x 2sin 5 x +3 x cos 5 x −3 x 2

45.

2

sin 5 x + sin 3 x 4sin x cos3 x − 4sin 3 x cos x

sin 4 x sin x =− sin 4 x cos x = − tan x

46.

=

2sin 5 x +3 x cos 5 x −3 x 2

4x+2 x 4 x−2 x cos 4 x − cos 2 x −2sin 2 sin 2 = sin 2 x − sin 4 x 2cos 2 x + 4 x sin 2 x − 4 x 2

2

− sin 3 x sin x = cos3 x sin(− x) − sin 3 x sin x = − cos3 x sin x = tan 3x 1 [sin( x + y + x − y) + sin( x + y − x + y)] 2 1 = [sin 2 x + sin 2 y ] 2 1 = [ 2sin x cos x + 2sin y cos y ] 2 = sin x cos x + sin y cos y

47.

sin( x + y ) cos( x − y ) =

48.

sin( x + y ) sin( x − y ) =

1 [cos( x + y − x + y) − cos( x + y + x − y )] 2 1 = [ cos 2 y − cos 2 x ] 2 1⎡ = ⎢1 − 2sin 2 y − 1 + 2sin 2 x ⎤⎥ ⎦ 2⎣ = sin 2 x − sin 2 y

49.

a = −1, b = −1, k =

( −1)2 + ( −1)2

α is a third quadrant angle. sin β =

−1

1

= 2 2 β = 45° α = −180° + 45° = −135° y = 2 sin( x − 135°)

= 2,

50.

a = 3, b = −1, k =

(

−3

)

2

α is a fourth quadrant angle. 1 1 = 2 2 β = 30°

sin β = −

α = −30° y = 2sin( x − 30°)


2

4sin x cos x(cos 2 x − sin 2 x) sin 4 x cos x = 2sin x cos x cos 2 x 2sin 2 x cos 2 x cos x = sin 2 x cos 2 x = 2cos x

2

+ ( −1) = 2,

Section 6.4

395 2

51.

2 1 3 3⎞ ⎛1⎞ ⎛ , k = ⎜ ⎟ + ⎜⎜ − a= , b=− ⎟ = 1, 2 2 ⎝ 2 ⎠ ⎝ 2 ⎟⎠ α is a fourth quadrant angle.

2

52.

⎛ 3 ⎞ ⎛ 1 ⎞2 3 1 , b = − , k = ⎜⎜ ⎟⎟ + ⎜ − ⎟ = 1, 2 2 ⎝ 2 ⎠ ⎝ 2⎠ α is a fourth quadrant angle.

a=

−1/ 2 1 = 1 2 β = 30°

3 3 2 sin β = = 1 2 β = 60°

sin β =

−

α = −30°

α = −60°

y = sin( x − 30°)

y = sin( x − 60°) 2

53.

2

1 1 2 ⎛1⎞ ⎛1⎞ a = , b =− , k = ⎜ ⎟ +⎜ ⎟ = , 2 2 2 ⎝2⎠ ⎝2⎠ α is a fourth quadrant angle.

2

54.

−1/ 2 2 = 2 2/2 β = 45° α = −45°

2 ⎛ 3 1 3⎞ ⎛ 1⎞ , b = − , k = ⎜⎜ − ⎟⎟ + ⎜ − ⎟ = 1, 2 2 ⎝ 2 ⎠ ⎝ 2⎠ α is a third quadrant angle.

a=−

−1/ 2 1 = 1 2 β = 30°

sin β =

sin β =

α = 30° − 180° = −150° y = sin( x − 150°)

2 y= sin( x − 45°) 2

55.

2

2

2

a = −3, b = 3, k = (−3) + 3 = 3 2,

56.

α is a second quadrant angle. 3 1 2 = − 2 3 2 2 β = 45°

sin β =

β = 45° α = 45°

y = 3 2 sin( x + 135°)

y = sin( x + 45°)

a = π , b = −π , k = π 2 + (−π )2 = π 2,

58.

a = −0.4, b = 0.4, k = (−0.4)2 + 0.42 = 0.4 2,

α is a second quadrant angle.

α is a fourth quadrant angle. −π 1 2 sin β = = = 2 2 π 2 β = 45°

59.

2/2 2 = 1 2

sin β =

α = 180° − 45° = 135°

57.

0.4 1 2 = = 2 0.4 2 2 β = 45°

sin β =

α = −45°

α = 180° − 45° = 135°

y = π 2 sin( x − 45°)

y = 0.4 2 sin( x + 135°)

a = −1, b = 1, k = (−1) 2 + 12 = 2,

60.

2

a = − 3, b = −1, k = (− 3)2 + ( −1) = 2,


α is a third quadrant angle.

sin β = 1 = 1 = 2 2 2 2

sin β = − 1 = 1 2 2

β =π

4

α = π − π = 3π 4

4 ⎛ y = 2 sin ⎜ x + 3π ⎞⎟ ⎝ 4 ⎠

2

⎛ 2⎞ ⎛ 2⎞ 2 2 , b= , k = ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ = 1, 2 2 ⎝ 2 ⎠ ⎝ 2 ⎠ α is a first quadrant angle. a=

β =π

6

α = π − π = − 5π 6

6 ⎛ ⎞ 5 π y = 2sin ⎜ x − ⎟ 6 ⎠ ⎝


396

Chapter 6: Trigonometric Identities and Equations 2

61.

⎛ 3 ⎞ ⎛ 1 ⎞2 3 1 , b = , k = ⎜⎜ ⎟⎟ + ⎜ ⎟ = 1, 2 2 ⎝ 2 ⎠ ⎝2⎠ α is a first quadrant angle.

62.

a=

α is a first quadrant angle.

β =π

β=

6

y = sin ⎛⎜ x + π ⎞⎟ 6⎠ ⎝

π⎞ ⎛ y = 2sin ⎜ x + ⎟ 3⎠ ⎝

a = −10, b = 10 3, k = (−10)2 + (10 3)2 = 20,

sin β =

β=

64.

β =π

π

3

3

π

α = −π

=

3

y = 6sin ⎛⎜ x − π ⎞⎟ 3⎠ ⎝

a = −5, b = 5, k = (−5)2 + 52 = 5 2,

66.


β=

a = 3, b = −3, k = 32 + (−3)2 = 3 2,

α is a fourth quadrant angle.

5 2 = 2 5 2

sin β =

π 4

a = 3, b = −3 3, k = (3)2 + (−3 3)2 = 6, sin β = −3 3 = 3 6 2

2π 3 3 2π ⎞ ⎛ y = 20sin ⎜ x + ⎟ 3 ⎠ ⎝

sin β =

3

α is a fourth quadrant angle.

10 3 3 = 20 2

α =π −

β=

3π α =π − = 4 4 3π ⎞ ⎛ y = 5 2 sin ⎜ x + ⎟ 4 ⎠ ⎝

π

y = − sin x − 3 cos x

y = sin x + 3 cos x

π⎞ ⎛ y = 2sin ⎜ x + ⎟ 3⎠ ⎝

−3 2 = 2 3 2

π 4

α =−

π 4

π⎞ ⎛ y = 3 2 sin ⎜ x − ⎟ 4⎠ ⎝ 68.

2π ⎞ ⎛ y = 2sin ⎜ x − ⎟ 3 ⎠ ⎝

70.

3

π

α=


67.

π

α =π

6

65.

3 3 = 2 2

sin β =

sin β = 1/ 2 = 1 1 2

63.

a = 1, b = 3, k = ( 3)2 + (1)2 = 2,

y = − 3 sin x + cos x

69.

π⎞ ⎛ y = 2 2 sin ⎜ x + ⎟ 4⎠ ⎝

5π ⎞ ⎛ y = 2sin ⎜ x + ⎟ 6 ⎠ ⎝

71.

y = − 3 sin x − cos x 5π ⎞ ⎛ y = 2sin ⎜ x − ⎟ 6 ⎠ ⎝

y = 2sin x + 2cos x

72.

y = − sin x + cos x 3π ⎞ ⎛ y = 2 sin ⎜ x + ⎟ 4 ⎠ ⎝


Section 6.4

73.

397

74.

y = −5sin x + 5 3 cos x 2π ⎞ ⎛ y = 10sin ⎜ x + ⎟ 3 ⎠ ⎝

75.

3π ⎞ ⎛ y = 2sin ⎜ x + ⎟ 4 ⎠ ⎝

76.

y = 6 3 sin x − 6cos x

π⎞ ⎛ y = 12sin ⎜ x − ⎟ 6⎠ ⎝

77.

a. b.

c. 78.

a. b.

c.

) (

p (t ) = 2sin 2π ⋅ 1336t + 2π ⋅ 770t sin 2π ⋅ 1336t − 2π ⋅ 770t 2 2 = 2sin(2106π t )sin(556π t ) 1336 + 770 = 2106 = 1053 cycles per second 2 2 p (t ) = sin(2π ⋅ 1336t ) + sin(2π ⋅ 852t )

(

y = 5 2 sin x + 5 2 cos x

π⎞ ⎛ y = 10sin ⎜ x + ⎟ 4⎠ ⎝

p (t ) = sin(2π ⋅ 1336t ) + sin(2π ⋅ 770t )

(

y = − 2 sin x + 2 cos x

) (

p (t ) = 2sin 2π ⋅ 1336t + 2π ⋅ 852t sin 2π ⋅ 1336t − 2π ⋅ 852t 2 2 = 2sin(2188π t )sin(484π t ) 1336 + 852 = 2188 = 1094 cycles per second 2 2

)

)

79.

Identity

80.

Identity

81.

Identity

82.

Identity

83.

Identity

84.

Identity


398


....................................................... 85.

Connecting Concepts

Let x = α + β and y = α − β . x + y = α + β + α − β and x − y = α + β − (α − β ) x + y = 2α x − y = 2β x+ y x− y α= 2β = 2 2 cos(α − β ) + cos(α + β ) = 2cosα cos β x+ y x− y ⎡x + y x − y⎤ ⎡x+ y x− y⎤ − cos ⎢ ⎥ + cos ⎢ 2 + 2 ⎥ = 2cos 2 cos 2 2 2 ⎣ ⎦ ⎣ ⎦ x+ y x− y cos y + cos x = 2cos cos 2 2

86.

cos( x − y ) = cos x cos y + sin x sin y cos( x + y ) = cos x cos y − sin x sin y (cos x cos y + sin x sin y ) − (cos x cos y − sin x sin y ) = cos( x − y ) − cos( x + y ) cos x cos y + sin x sin y − cos x cos y + sin x sin y = cos( x − y ) − cos( x + y ) 2sin x sin y = cos( x − y ) − cos( x + y ) sin x sin y = 1 [cos( x − y ) − cos( x + y )] 2

87.

x + y = 180° y = 180° − x sin x + sin y = sin x + sin(180° − x) = sin x + sin 180° cos x − cos180° sin x = sin x + 0(cos x) − ( −1) sin x = 2 sin x

89.

sin 2 x + sin 4 x + sin 6 x = 2sin

90.

sin 4 x − sin 2 x + sin 6 x = 2sin

91.

10 x +8 x 10 x −8 x cos10 x + cos8 x 2cos 2 cos 2 = sin10 x − sin 8 x 2cos 10 x +8 x sin 10 x −8 x

88.

x + y = 360° y = 360° − x cos x + cos y = cos x + cos(360° − x) = cos x + cos360° cos x + sin 360° sin x = cos x + (1)cos x + (0)sin x = 2cos x

2x + 4x 2x − 4x cos + 2sin 3x cos3 x 2 2 = 2sin 3x cos x + 2sin 3 x cos3 x = 2sin 3 x(cos x + cos3 x) = 2sin 3 x ⎛⎜ 2cos x + 3x cos x − 3x ⎞⎟ 2 2 ⎠ ⎝ = 4sin 3 x cos 2 x cos x

4x + 2x 4x − 2x sin + 2sin 3 x cos3 x 2 2 = 2cos3 x sin x + 2sin 3 x cos3 x = 2cos3 x(sin x + sin 3x) = 2cos3 x ⎛⎜ 2sin x + 3x cos x − 3x ⎞⎟ 2 2 ⎠ ⎝ = 2cos3 x(2sin 2 x cos x) = 4cos3 x sin 2 x cos x

2

2

2cos9 x cos x = 2cos9 x sin x = cot x

92.

10 x + 2 x 10 x − 2 x sin10 x + sin 2 x = 2sin 2 cos 2 cos10 x + cos 2 x 2cos 10 x + 2 x cos 10 x − 2 x 2

= sin 6 x cos 4 x cos 6 x cos 4 x = tan 6 x = 2 tan 3 x 1 − tan 2 3 x


2

Section 6.4

93.

94.

399

sin 2 x + sin 4 x + sin 6 x sin 2 x + sin 6 x + sin 4 x = cos 2 x + cos 4 x + cos 6 x cos 2 x + cos6 x + cos 4 x 2x + 6x 2x − 6x + sin 4 x 2sin cos 2 2 = 2x + 6x 2x − 6x + cos 4 x 2cos cos 2 2 2sin 4 x cos 2 x + sin 4 x = 2cos 4 x cos 2 x + cos 4 x sin 4 x(2cos 2 x + 1) = cos 4 x(2cos 2 x + 1) sin 4 x = cos 4 x = tan 4 x 2 x +6 x 2 x −6 x sin 2 x + sin 6 x = 2sin 2 cos 2 cos 6 x − cos 2 x −2sin 6 x + 2 x sin 6 x − 2 x 2

= − 2sin 4 x cos 2 x 2sin 4 x sin 2 x = − cos 2 x sin 2 x = − cot 2 x

2

95.

cos 2 x − sin 2 x = cos x ⋅ cos x − sin x ⋅ sin x = 1 [ cos( x + x) + cos( x − x) ] − 1 [cos( x − x) − cos( x + x)] 2 2 = 1 cos 2 x + 1 cos 0 − 1 cos 0 + 1 cos 2 x 2 2 2 2 = cos 2 x

96.

2sin x cos x = 2 ⋅ 1 [sin( x + x) + sin( x − x)] 2 = sin 2 x + sin 0 = sin 2 x

97.

Let k = a 2 + b 2 , tan α =

a b

a 2 + b 2 ( a sin x + b cos x) a2 + b2 ⎛ ⎞ a b sin x + cos x ⎟ = a2 + b2 ⎜ ⎜ 2 ⎟ 2 2 2 a +b ⎝ a +b ⎠

a sin x + b cos x =

= k (sin α sin x + cos α cos x) because sinα = = k (cos x cos α + sin x sin α ) = k cos( x − α )

98.

Let k = a 2 + b 2 , tan α =

a 2

a +b

2

and cosα =

b 2

a + b2

b a

a 2 + b 2 (a sin cx + b cos cx) a 2 + b2 ⎛ ⎞ a b sin cx + cos cx ⎟ = a 2 + b2 ⎜ ⎜ 2 ⎟ 2 a 2 + b2 ⎝ a +b ⎠

a sin cx + b cos cx =

= k (cos α sin cx + sin α cos cx ) because cosα = = k (sin cx cosα + cos cx sin α ) = k sin(cx + α )

a a 2 + b2

and sinα =

b a2 + b2


400


.......................................................


PS1. A one-to-one function is a function for which each range value (y-value) is paired with one and only one domain value (x-value). PS3.

PS2. If every horizontal line intersects the graph of a function at most once, then the function is a one-to-one function.

f [ g ( x)] = f ⎡⎢ 1 x − 2 ⎤⎥ ⎣2 ⎦ ⎛ ⎞ 1 = 2⎜ x − 2⎟ + 4 ⎝2 ⎠ = x−4+4 =x

PS4.

PS5. The graph of f –1 is the reflection of the graph of f across the line given by y = x.

f [ f −1 ( x)] = x

PS6. No, it does not pass the horizontal line test.

Section 6.5 1.

y = sin −1 1

2.

sin y = 1 with − y=

4.

7.

π

≤ y≤

π 2

3 3

y=

3 3

5.

2 2

2 2

0< y 0 x +1

10.

2a =12 a=6 a 2 = b2 + c2 (6) 2 = b 2 + (4)2 36 −16 = b 2 20 = b 2

The quotient x + 2 is positive. x +1 The critical values are –2 and –1. x+2 x +1

( x − 3)2 ( y + 4) 2 + =1 [8.2] 36 20

(−∞, −2) ∪ (−1, ∞) [1.5] 11.

f ( x + h) − f ( x) ( x + h) 2 − 3( x + h) + 2 − ( x 2 − 3x + 2) = [2.6] h h 2 2 2 = x + 2 xh + h − 3x − 3h + 2 − x + 3 x − 2 h 2 2 3 h xh h + − = h = 2 x − 3+ h

12.

125 x = 1 [4.5] 25 53 x = 5−2 3 x = −2 x=−2 3


760

13.

Chapter 10: Matrices

x−2 x−2 = x − 5 x − 6 ( x + 1)( x − 6) x−2 = A + B ( x + 1)( x − 6) x + 1 x − 6

14.

2

Let u =10 x

x − 2 = A( x − 6) + B( x + 1) x − 2 = Ax − 6 A + Bx + B x − 2 = ( A + B ) x + ( −6 A + B )

u 2 − 2u −1= 0 2 ± 4 − 4(1)(−1) u= 2 = 2± 8 2 =1± 2 10 x =1+ 2 log10 x = log(1+ 2) x = log(1+ 2) ≈ 0.3828

{

1= A+ B (1) −2 = −6 A + B (2)

−1 = − A − B −2 = −6 A + B −3 = −7 A 3 =A 7

[4.5] 10 x −10− x = 2 −x x x x 10 (10 −10 ) = 2(10 ) (10 x ) 2 − 2(10 x ) −1= 0

− 1 times (1) (2)

1= 3 + B 7 4=B 7 4 3 x−2 [9.4] = + x 2 − 5 x − 6 7( x + 1) 7( x − 6)

15.

cos30o sin 30o + sec 45o tan 60o = cos30o sin 30o + 1 o tan 60o sin 45 3 1 2 = ⋅ + ⋅ 3 2 2 2 = 3+2 6 4 2 3 4 6 + = 4

[5.2]

16.

o

sin 292 1 = sin 585 2 2

o

o = − 1 − cos585 2 o = − 1 − cos 225 2

=−

1 − ⎛⎜ − ⎝

2⎞ ⎟ 2 ⎠

2

=− 2+ 2 4 = −1 2+ 2 2 17.

K = 1 bc sin A 2 = 1 (11)(4)sin 65o 2 = 22sin 65o ≈ 20 m 2

[7.2]

18.

[6.3]

cosθ (1 − sin θ ) tan θ + cos θ = tan θ + 1 − sin θ (1 + sin θ )(1 − sin θ ) cosθ (1 − sin θ ) = tan θ + 1 − sin 2 θ cosθ (1 − sin θ ) = tan θ + cos 2 θ − sin 1 sin θ θ = + cos θ cosθ = sin θ + 1 − sin θ cos θ 1 = cos θ = secθ [5.4]


Cumulative Review

19.

761

z = 2 − 2i 3 2

[7.4] 2

r = 2 + ( −2 3) = 4 + 12 = 16 = 4

α = tan −1 b = tan −1 −2 3 = tan −1 − 3 = 60o a

2

z is in the fourth quadrant 270o < θ < 360o ,

θ = 360o − 60o = 300o z = rcisθ = 4(cos300o + i sin 300o )

20.

sin x + cos x = 1 [6.6] 2 1 − cos x + cos x = 1 2 1 − cos x = 1 − cos x 2 1 − cos x = 1 − 2cos x + cos 2 x 2 1 − cos x = 2 − 4cos x + 2cos 2 x 0 = 2cos 2 x − 3cos x + 1 0 = (2cos x − 1)(cos x − 1) 2cos x − 1 = 0 cos x = 1 2

x = π , 5π 3 3


cos x = 1 x=0

Chapter 11

Sequences, Series, and Probability Section 11.1 1.

a1 = 1(1 − 1) = 1 ⋅ 0 = 0

2.

a2 = 2(2 − 1) = 2 ⋅ 1 = 2

a3 = 2 ⋅ 3 = 6

a8 = 8(8 − 1) = 8 ⋅ 7 = 56

a8 = 2 ⋅ 8 = 16

1+1 2 = =2 1 1 2 +1 3 = = 2 2 3 +1 4 = = 3 3 8 +1 9 = = 8 8

a1 = a2 a3 a8

5.

a1 = a2 = a3 = a8 =

7.

10.

a1 =

( −1)2⋅1−1 (−1)2−1 1 = =− 3 ⋅1 3 3

a2 =

3.

a2 = 2 ⋅ 2 = 4

a3 = 3(3 − 1) = 3 ⋅ 2 = 6

4.

a1 = 2 ⋅ 1 = 2

(−1)1+1 12 (−1)2+1 2

2

(−1)3+1 32 (−1)8+1 82

a1 =

(−1)1+1 (−1)2 1 = = 1(1 + 1) 1⋅ 2 2

(−1)3 1 =− 4 4

a2 =

(−1)2+1 (−1)3 1 = =− 2(2 + 1) 2⋅3 6

=

(−1)4 1 = 9 9

a3 =

(−1)3+1 (−1)4 1 = = 3(3 + 1) 3⋅ 4 12

=

(−1)9 1 =− 64 64

a8 =

(−1)8+1 (−1)9 1 = =− 8(8 + 1) 8⋅9 72

=

(−1)2 =1 1

=

a1 =

−1 (−1)1 = = −1 2 ⋅1 − 1 2 − 1

( −1)2⋅2−1 (−1)4−1 1 = =− 3⋅ 2 6 6

a2 =

(−1)2 1 = 2 ⋅ 2 −1 3

a3 =

( −1)2⋅3−1 (−1)6−1 1 = =− 3⋅3 9 9

a3 =

(−1)3 1 =− 2 ⋅ 3 −1 5

a8 =

( −1)2⋅8−1 (−1)16−1 1 = =− 3⋅8 24 24

a8 =

(−1)8 1 = 2 ⋅ 8 − 1 15

8.

1

1 ⎛ −1 ⎞ a1 = ⎜ ⎟ =− 2 ⎝ 2 ⎠

1 a1 = 1 − = 0 1 1 1 a2 = 1 − = 2 2 1 2 a3 = 1 − = 3 3 1 7 a8 = 1 − = 8 8

11.

6.

9.

1

2 ⎛ 2⎞ a1 = ⎜ ⎟ = 3 3 ⎝ ⎠ 2

4 ⎛2⎞ a2 = ⎜ ⎟ = 3 9 ⎝ ⎠ 3

8 ⎛2⎞ a3 = ⎜ ⎟ = 27 ⎝3⎠ 8

256 ⎛2⎞ a8 = ⎜ ⎟ = 3 6561 ⎝ ⎠

a1 = 1 + (−1)1 = 1 + (−1) = 0

2

a2 = 1 + (−1) 2 = 1 + 1 = 2

3

a8 = 1 + (−1)8 = 1 + 1 = 2

1 ⎛ −1 ⎞ a2 = ⎜ ⎟ = 4 ⎝ 2 ⎠

a3 = 1 + (−1)3 = 1 + (−1) = 0

1 ⎛ −1 ⎞ a3 = ⎜ ⎟ = − 2 8 ⎝ ⎠ 8

1 ⎛ −1 ⎞ a8 = ⎜ ⎟ = 256 ⎝ 2 ⎠

12.

a1 = 1 + (−0.1)1 = 1 + (−0.1) = 0.9

13.

a1 = (1.1)1 = 1.1

a2 = 1 + (−0.1)2 = 1 + 0.01 = 1.01

a2 = (1.1)2 = 1.21

a3 = 1 + (−0.1)3 = 1 + (−0.001) = 0.999

a3 = (1.1)3 = 1.331

a8 = 1 + (−0.1)8 = 1 + 0.00000001 = 1.00000001

a8 = (1.1)8 = 2.14358881


Section 11.1

14.

16.

763

1 = 1 =1 1 +1 1+1 2 a2 = 22 = 2 = 2 2 +1 4 +1 5 a3 = 23 = 3 = 3 3 + 1 9 + 1 10 a8 = 28 = 8 = 8 8 + 1 64 + 1 65 a1 =

15.

2

1−1 0 a1 = 3 1 = 3 = 1 2 2 2

17.

a1 =

1! = 1 = 1 = 1 (1 − 1)! 0! 1

19.

a1 = ln 1 = 0

21.

a8 = ln 8 ≈ 2.0794

26.

29.

a3 =

(−1)3+1 (−1)4 1 3 = = = 3 3 3 3

a8 =

(−1)8+1 (−1)9 1 2 = =− =− 4 8 2 2 2 2

a1 = 1!= 1

a1 = log1 = 0

a8 = log 8 ≈ 0.9031

a3 = ln 3 ≈ 1.0986

a1 = 3

(−1) 2+1 ( −1)3 1 2 = =− =− 2 2 2 2

a3 = log 3 ≈ 0.4771

a2 = ln 2 ≈ 0.6931

23.

a2 =

a2 = log 2 ≈ 0.3010

2! = 2 ⋅ 1 = 2 = 2 (2 − 1)! 1! 1 a3 = 3! = 3 ⋅ 2 ⋅ 1 = 3 ⋅ 2 ⋅ 1 = 3 (3 − 1)! 2! 2 ⋅1 a8 = 8! (8 − 1)! 8 = ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 7! ⋅ ⋅ ⋅ ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 = 8 8 7 6 5 = 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1

a2 =

20.

(−1)1+1 ( −1)2 = =1 1 1

a2 = 2! = 2 ⋅ 1 = 2 a3 = 3! = 3 ⋅ 2 ⋅ 1 = 6 a8 = 8! = 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 40,320

2 −1 1 a2 = 3 2 = 3 = 3 4 4 2 3−1 2 3 3 =9 a3 = 3 = 8 8 2 8 −1 7 a8 = 3 8 = 3 = 2187 256 256 2

18.

a1 =

24.

a1 = 1

1 = 0.142857 7

22.

a1 = 0

a2 = 4

a2 = 7

a3 = 2

a3 = 6

a8 = 4

a8 = 7

a1 = −2

25.

1 = 0.076923 13

a1 = 5

a2 = 3

a2 = −2

a2 = 2 ⋅ a1 = 2 ⋅ 5 = 10

a3 = 3

a3 = −2

a3 = 2 ⋅ a2 = 2 ⋅ 10 = 20

a8 = 3

a8 = −2

a1 = 2

27.

a1 = 2

28.

a2 = 3 ⋅ a1 = 3 ⋅ 2 = 6

a2 = 2 ⋅ a1 = 2 ⋅ 2 = 4

a3 = 3 ⋅ a2 = 3 ⋅ 6 = 18

a3 = 3 ⋅ a2 = 3 ⋅ 4 = 12

a1 = 2

a2 = (a1 )2 = (2 )2 = 4 a3 = (a2 )2 = (4)2 = 16

30.

a1 = 4

a2 = 22 ⋅ a1 = 4 ⋅ 1 = 4 a3 = 32 ⋅ a2 = 9 ⋅ 4 = 36 31.

a2 = 1 = 1 a1 4 a3 = 1 = 11 = 4 a2

a1 = 1

a1 = 2

a2 = 2 ⋅ 2 ⋅ a1 = 4 ⋅ 2 = 8 a3 = 2 ⋅ 3 ⋅ a2 = 6 ⋅ 8 = 48

4


764

32.

Chapter 11: Sequences, Series, and Probability

a1 = 2

33.

a2 = (−3) ⋅ 2 ⋅ a1 = −6 ⋅ 2 = −12

a2 = ( a1 )1 2 = (3)1 2 = 3

a3 = (−3) ⋅ 3 ⋅ a2 = −9 ⋅ (−12) = 108

34.

a1 = 2 2

a1 = 3 a3 = ( a2 )1 3 = ( 31 2 )

13

= 31 6 = 6 3

35.

a1 = 1 a2 = 3 a3 = 1 ( a2 + a1 ) = 1 (3 + 1) = 1 (4) = 2 2 2 2 a 4 = 1 ( a3 + a 2 ) = 1 ( 2 + 3 ) = 1 ( 5 ) = 5 2 2 2 2 1 1 5 1 9 ⎛ ⎞ ⎛ ⎞ a5 = ( a4 + a3 ) = ⎜ + 2 ⎟ = ⎜ ⎟ = 9 2 2⎝2 ⎠ 2⎝2⎠ 4

38.

668 + 866 1534 Sorting 1534 gives 1345

2

a2 = ( a1) = (2) = 4 a3 = ( a2 )3 = (4)3 = 64

36.

a1 = 1 a2 = 4 a3 = a2 ⋅ a1 = 4 ⋅ 1 = 4 a4 = a3 ⋅ a2 = 4 ⋅ 4 = 16 a5 = a4 ⋅ a3 = 16 ⋅ 4 = 64

37.

an a3 a4 a5

39.

7!− 6! = 7 ⋅ 6!− 6! = 6!( 7 − 1) = 6!⋅ 6 = 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 ⋅ 6 = 4320

40.

(4!)2 = (4 ⋅ 3 ⋅ 2 ⋅ 1)2 = (24)2 = 576

41.

9! = 9 ⋅ 8 ⋅ 7! = 72 7! 7!

42.

10! = 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5! = 30, 240 5! 5!

43.

8! = 8 ⋅ 7 ⋅ 6 ⋅ 5! = 56 3!5! 3 ⋅ 2 ⋅ 1 ⋅ 5!

44.

12! = 12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8! = 495 4!8! 4 ⋅ 3 ⋅ 2 ⋅ 1 ⋅ 8!

45.

100! = 100 ⋅ 99! = 100 99! 99!

46.

100! = 100 ⋅ 99 ⋅ 98! = 4950 98!2! 98!⋅ 2 ⋅ 1

47.

∑i = 1 + 2 + 3 + 4 + 5 = 15

48.

= an −1 + an − 2 = 3+1 = 4 =4+3=7 = 7 + 4 = 11

4

∑i

2

5

i =1

= 12 + 22 + 32 + 42 = 1 + 4 + 9 + 16 = 30

i =1 5

49.

∑i (i − 1) = 1(1 − 1) + 2 ( 2 − 1) + 3(3 − 1) + 4 ( 4 − 1) + 5(5 − 1) i =1

= 1⋅ 0 + 2 ⋅1 + 3 ⋅ 2 + 4 ⋅ 3 + 5 ⋅ 4 = 0 + 2 + 6 + 12 + 20 = 40 7

50.

∑i ( 2i + 1) = ( 2 ⋅1 + 1) + ( 2 ⋅ 2 + 1) + 2 ( 2 ⋅ 3 + 1) + ( 2 ⋅ 4 + 11) + ( 2 ⋅ 5 + 1) + ( 2 ⋅ 6 + 1) + ( 2 ⋅ 7 + 1) i =1

= ( 2 + 1) + ( 4 + 1) + ( 6 + 1) + ( 8 − 1) + (10 + 1) + (12 + 1) + (14 + 1) = 3 + 5 + 7 + 9 + 11 + 13 + 15 = 63


1345 + 5431 6776 Sorting 6776 gives 6677

Section 11.1

765

4

51.

∑ 1k = 11 + 12 + 13 + 14 = 1212 + 126 + 124 + 123 = 1225 k =1 6

52.

∑ k ( k1+ 1) = 1(11+ 1) + 2 ( 21+ 1) + 3(31+ 1) + 4 ( 41+ 1) + 5(51+ 1) + 6 ( 61+ 1) k =1

1 1 1 1 1 1 + + + + + 1⋅ 2 2 ⋅ 3 3 ⋅ 4 4 ⋅ 5 5 ⋅ 6 6 ⋅ 7 210 70 35 21 14 10 360 6 = + + + + + = = 420 420 420 420 420 420 420 7 =

8

53.

∑

8

∑ j =2 (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = 2 (36) = 72

2j =2

j =1

6

54.

∑

j =1

( 2i + 1)( 2i − 1) =

i =1

6

∑(

6

6

i =1

i =1

) ∑ i2 − ∑ (1) = 4 (12 + 22 + 32 + 42 + 52 + 62 ) − (1 + 1 + 1 + 1 + 1 + 1)

4i 2 − 1 = 4

i =1

= 4(1 + 4 + 9 + 16 + 25 + 36 ) − 6 = 4(91) − 6 = 358 5

55.

( −1)i 2i = (−1)3 23 + (−1) 4 24 + (−1)5 25 = (−1)8 + (1)16 + (−1)32 = −8 + 16 − 32 = −24 ∑ i=3 5

56.

i

(−1) ∑ i i=3 2

=

(−1)3 3

2

+

( −1)4 2

4

+

(−1)5 5

2

=

−1 1 − 1 − 4 2 − 1 3 + + = + + =− 8 16 32 32 32 32 32

7

57.

⎛ n +1⎞ ⎛1+1⎞ ⎛ 2 +1⎞ ⎛ 3 +1⎞ ⎛ 4 + 1⎞ ⎛ 5 +1⎞ ⎛ 6 +1⎞ ⎛ 7 +1⎞ log ⎜ ⎟ = log ⎜ ⎟ + log ⎜ ⎟ + log ⎜ ⎟ + log ⎜ ⎟ + log ⎜ ⎟ + log ⎜ ⎟ + log ⎜ ⎟ ∑ ⎝ n ⎠ ⎝ 1 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 4 ⎠ ⎝ 5 ⎠ ⎝ 6 ⎠ ⎝ 7 ⎠ n =1 ⎛ 3 4 5 6 7 8⎞ = log ⎜ 2 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⎟ = log8 = 3log 2 ≈ 0.9031 ⎝ 2 3 4 5 6 7⎠ 8

58.

2 n 2 3 5 6 7 8 ⎛2 3 4 5 6 7 8⎞ ln = ln + ln + ln + ln + ln + ln = ln⎜ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⎟ = ln ≈ −1.5041 ∑ 3 4 6 7 8 9 n +1 3 4 5 6 7 8 9⎠ 9 ⎝ n=2 8

59.

∑0 k !(88!− k )! = 0!(88!− 0)! + 1!(88!− 1)! + 2!(88!− 2)! + 3!(88!− 3)! + 4!(88!− 4)! + 5!(88!− 5)! + 6!(88!− 6)! + 7!(88!− 7)! + 8!(88!− 8)! k=

8! 8! 8! 8! 8! 8! 8! 8! 8! + + + + + + + + 0!8! 1!7! 2!6! 3!5! 4!4! 5!3! 6!2! 7!1! 8!0! 8⋅7 8⋅7 ⋅6 8⋅7⋅6⋅5 8⋅7⋅6 8⋅7 =1+ 8 + + + + + + 8 +1 4⋅3⋅ 2 3 ⋅ 2 ⋅1 2 2 3⋅ 2 = 1 + 8 + 28 + 56 + 70 + 56 + 28 + 8 + 1 = 256 =


766

Chapter 11: Sequences, Series, and Probability 7

60.

1 ∑0 k1! = 0!1 + 1!1 + 2!1 + 3!1 + 4!1 + 5!1 + 6!1 + 7!1 = 11 + 11 + 12 + 16 + 241 + 1201 + 7201 + 5040 k=

5040 5040 2520 840 210 42 7 1 + + + + + + + 5040 5040 5040 5040 5040 5040 5040 5040 13700 1370 685 = = = 5040 504 252 =

61.

62.

1 1 1 1 1 1 1 1 1 1 1 1 + + + + + = + + + + + = 1 4 9 16 25 36 12 22 32 42 52 62

6

∑1 i12 i=

2 + 4 + 6 + 8 + 10 + 12 + 14 = 2(1) + 2(2) + 2(3) + 2(4 ) + 2(5) + 2(6 ) + 2(7 ) =

7

∑1 2i i=

63.

2 − 4 + 8 −16 + 32 − 64 +128 = 21( −1)1+1 + 22 (−1)2+1 + 23 (−1)3+1 + 24 (−1) 4+1 + 25 (−1)5+1 + 26 (−1)6+1 + 27 (−1)7 +1 =

7

∑1 2 (−1) i

i +1

i=

64.

1 − 8 + 27 − 64 + 125 = 13 − 23 + 33 − 43 + 53 = 13 (−1)1+1 + 23 (−1)2 +1 + 33 (−1)3+1 + 43 (−1)4 +1 +53 (−1)5+1 =

5

∑1 i3(−1)

i +1

i=

65.

7 + 10 + 13 + 16 + 19 = 7 + (7 + 3) + (7 + 3 ⋅ 2 ) + (7 + 3 ⋅ 3) + (7 + 3 ⋅ 4 ) =

4

∑0 (7 + 3i) i=

66.

30 + 26 + 22 + 18 + 14 + 10 = 30 + ( 30 − 4 ) + ( 30 − 4 ⋅ 2 ) + ( 30 − 4 ⋅ 3) + ( 30 − 4 ⋅ 4 ) + ( 30 − 4 ⋅ 5) =

5

∑(30 − 4i ) i =0

67.

68.

1 1 1 1 1 1 1 1 + + + = + + + = 2 4 8 16 2 22 23 24

1−

4

∑1 21i i=

5 2 3 4 5 i 5 ⎛ 2 ⎞i 2 4 8 16 32 2 ⎛2⎞ ⎛2⎞ ⎛2⎞ ⎛2⎞ ⎛ 2⎞ + − + − = 1 − + ⎜ ⎟ − ⎜ ⎟ + ⎜ ⎟ − ⎜ ⎟ = ∑ ⎜ ⎟ (− 1)i = ⎜− ⎟ 3 9 27 81 243 3 ⎝3⎠ ⎝3⎠ ⎝3⎠ ⎝3⎠ ⎝ 3⎠ i = 0⎝ 3 ⎠ i =0

∑

....................................................... 69.

Let N = 7. a1 = 7 = 3.5 2 1 a2 = ⎛⎜ 3.5 + 7 ⎞⎟ = 2.75 2⎝ 3.5 ⎠


Let N = 10. a1 = 10 = 5 2 1 a2 = ⎛⎜ 5 + 10 ⎞⎟ = 3.5 2⎝ 5⎠

a3 = 1 ⎛⎜ 2.75 + 7 ⎞⎟ ≈ 2.6477273 2⎝ 2.75 ⎠

a3 = 1 ⎛⎜ 3.5 + 10 ⎞⎟ ≈ 3.1785714 2⎝ 3.5 ⎠

7 ⎞ ≈ 2.6457520 a4 ≈ 1 ⎛⎜ 2.6477273 + ⎟ 2⎝ 2.6477273 ⎠

⎞ ≈ 3.1623194 10 a4 = 1 ⎛⎜ 3.1785714 + ⎟ 2⎝ 3.1785714 ⎠ ⎞ ≈ 3.1622777 10 a5 = 1 ⎛⎜ 3.1623194 + ⎟ 2⎝ 3.1623194 ⎠


Section 11.1

71.

767

a20 ≈ 1.0000037

72.

a1 = i, a2 = −1, a3 = −i, a4 = 1 a5 = i, a6 = −1, a7 = −i, a8 = 1

a100 ≈ 1

Notice that the sequence repeats itself in groups of 4. To find a237 divide 237 by 4. a237 = i r where r is the remainder after

the division. Thus a237 = i1 = i. 73.

1+1 = 2 1+1+ 2 = 4 1+1+ 2 + 3 = 7 The sum of the first n terms of a Fibonacci sequence equals the (n + 2) term – 1. Therefore the sum of the first ten terms is (10+2) term – 1, i.e. 12th term – 1. 144 1 = 143 10

75.

⎛1+ 5 ⎞ ⎜ ⎟ 2 ⎠ ⎝ F10 =

10

⎛ ⎞ − ⎜1− 5 ⎟ ⎝ 2 ⎠ 5

15

74.

76.

= 55

⎛n⎞ 2π n ⎜ ⎟ ⎝e⎠

n

10

⎛ 10 ⎞ When n = 10 we have 2π (10) ⎜ ⎟ ⎝ e ⎠

15

⎛1+ 5 ⎞ ⎛1− 5 ⎞ ⎜ ⎟ −⎜ ⎟ 2 2 ⎠ ⎝ ⎠ ⎝ F15 = 5

81 = 5 + 21 + 55

= 610

⎛ 20 ⎞ When n = 20 we have 2π (20) ⎜ ⎟ ⎝ e ⎠ ⎛ 30 ⎞ When n = 30 we have 2π (30) ⎜ ⎟ ⎝ e ⎠

≈ 3.5986956 × 106.

20

30

≈ 2.4227869 × 1018. ≈ 2.6451710 × 1032.

n

77.

cai = ca1 + ca2 + ca3 + ⋅ ⋅ ⋅ + can ∑ i =1

= c ( a1 + a2 + a3 + ⋅ ⋅ ⋅ + an ) n

=c

∑a

i

i =1

....................................................... PS1.

−3 = 25 + (15 − 1)d

Prepare for Section 11.2 PS2. 13 = 3 + (5 − 1)d 13 = 3 + 4d 10 = 4d 5 =d 2

−3 = 25 + 14d −28 = 14d −2 = d 50 ⎡ 2(2) + (50 − 1) 5 ⎤ 50 ⎡ 4 + 245 ⎤ 4 ⎦⎥ = ⎣⎢ 4 ⎦⎥ ⎣⎢ 2 2 50 ⎡ 261 ⎤ ⎢ ⎥ = ⎣ 4 ⎦ = 6525 2 4

PS3. S =

PS4. a5 = 5 + (5 − 1)4 = 21

PS5. a20 = 52 + (20 − 1)( −3) = −5

PS6. 5 – 2 = 3 8–5=3 Yes


768


Section 11.2 1.

d = 10 − 6 = 4

2.

an = 6 + ( n − 1)4 = 6 + 4n − 4 = 4n + 2 a9 = 4 ⋅ 9 + 2 = 36 + 2 = 38 a24 = 4 ⋅ 24 + 2 = 98 3.

d = 4 − 6 = −2

a n = 7 + ( n − 1)5 = 5n + 2 a 9 = 5 ⋅ 9 + 2 = 45 + 2 = 47 a 24 = 5 ⋅ 24 + 2 = 120 + 2 = 122 4.

a n = 6 + ( n − 1)( −2) = 6 − 2n + 2 = 8 − 2n a 9 = 8 − 2 ⋅ 9 = 8 − 18 = −10 a 24 = 8 − 2 ⋅ 24 = 8 − 48 = −40

5.

d = −5 − (−8) = 3

d = 12 − 7 = 5

d = 4 − 11 = −7 a n = 11 + ( n − 1)( −7) = 11 − 7n + 7 = 18 − 7n a 9 = 18 − 7 ⋅ 9 = 18 − 63 = −45 a 24 = 18 − 7 ⋅ 24 = 18 − 168 = −150

6.

a n = −8 + ( n − 1)3 = −8 + 3n − 3 = 3n − 11 a 9 = 3 ⋅ 9 − 11 = 27 − 11 = 16 a 24 = 3 ⋅ 24 − 11 = 72 − 11 = 61

d = −9 − (−15) = 6 a n = −15 + ( n − 1)6 = −15 + 6n − 6 = 6n − 21 a 9 = 6 ⋅ 9 − 21 = 33 a 24 = 6 ⋅ 24 − 21 = 123

7.

d = 4 −1 = 3 a n = 1 + ( n − 1)3 = 1 + 3n − 3 = 3n − 2 a 9 = 3 ⋅ 9 − 2 = 27 − 2 = 25 a 24 = 3 ⋅ 24 − 2 = 72 − 2 = 70

8.

d = 1 − (−4 ) = 5 a n = −4 + ( n − 1)5 = −4 + 5n − 5 = 5n − 9 a 9 = 5 ⋅ 9 − 9 = 45 − 9 = 36 a 24 = 5 ⋅ 24 − 9 = 120 − 9 = 111

9.

d = (a + 2 ) − a = 2

10.

d = (a + 1) − (a − 3) = a + 1 − a + 3 = 4 a n = a − 3 + ( n − 1)4 = a − 3 + 4n − 4 = a + 4n − 7 a 9 = a + 4 ⋅ 9 − 7 = a + 36 − 7 = a + 29 a 24 = a + 4 ⋅ 24 − 7 = a + 96 − 7 = a + 89

a n = a + ( n − 1)2 = a + 2n − 2 a 9 = a + 2 ⋅ 9 − 2 = a + 18 − 2 = a + 16 a 24 = a + 2 ⋅ 24 − 2 = a + 48 − 2 = a + 46

11.

13.

d = log14 − log 7 = log 14 = log 2 7 a n = log 7 + ( n − 1) log 2 a 9 = log 7 + 8log 2 a 24 = log 7 + 23log 2

2

d = log a 2 − log a = log a = log a a a n = log a + ( n − 1) log a = (1 + n − 1)log a = n log a a 9 = 9log a a 24 = 24log a

12.

d = ln16 − ln 4 = ln 16 = ln 4 4

a n = ln 4 + ( n − 1) ln 4 = ln 4 + ln 4 ( n −1) = ln ( 4 ⋅ 4 ( n − 1) ) = ln 4 n = n ln 4 a 9 = 9ln 4 a 24 = 24ln 4 14.

d = log 2 5a − log 2 5 = log 2 5a = log 2 a 5 a n = log 2 5 + ( n − 1) log 2 a a 9 = log 2 5 + 8log 2 a a 24 = log 2 5 + 23log 2 a


Section 11.2

15.

769

d = 15 − 13 = 2 a 4 = a1 + ( 4 − 1) 2 = 13 a1 + 6 = 13 a1 = 7 a 20 = 7 + ( 20 − 1) 2 = 7 + (19 ) 2 = 7 + 38 = 45

16.

a 6 = −14 = a1 + ( 6 − 1) d

a8 = a1 + ( 8 − 1) d = −20 a1 + 7d 14 − 5d + 7d −14 + 2d 2d d

−14 = a1 + 5d −14 − 5d = a1

= −20 = −20 = −20 = −6 = −3

a1 = −14 − 5( −3) = −14 + 15 = 1 a15 = 1 + (15 − 1)( −3) = 1 + 14( −3) = 1 − 42 = −41

17.

a5 = −19 = a1 + ( 5 − 1) d

a 7 = −29 = a1 + ( 7 − 1) d

18.

a1 = 3 (1) + 2 = 5

3d = 12 d =4

22 = a1 + 3d a1 = 22 − 3d a1 = 22 − 3 ⋅ 4 = 10 a 23 = 10 + ( 23 − 1) 4 = 10 + ( 22 ) 4 = 10 + 88 = 98

a10 = 3 (10 ) + 2 = 32

a1 = 4(1) − 3 = 1 a12 = 4(12) − 3 = 45

S10 = 10 ( a1 + a10 ) = 5(5 + 32) = 185

S12 = 12 ( a1 + a12 ) = 6(1 + 45) = 276

20.

2

2

21.

23.

a1 = 3 − 5 (1) = −2

22.

a 20 = 1 − 2 ( 20 ) = −39

S15 = 15 ( a1 + a15 ) = 15 ( −2 + ( −72 ) ) = 15 ( −74 ) = −555 2 2 2

S 20 = 20 ( a1 + a 20 ) = 10 ( −1 + ( −39 ) ) = 10 ( −40 ) = −400 2

a1 = 6 (1) = 6 a12 = 6 (12 ) = 72

24.

a1 = 1 + 8 = 9 a25 = 25 + 8 = 33

S14 = 14 ( a1 + a14 ) = 7 ( 7 + 98 ) = 735 2

26.

a1 = −1 a30 = −30

28.

a1 = 1 + x a12 = 12 + x

a1 = 4 − 1 = 3 a40 = 4 − 40 = −36

S40 = 40 ( a1 + a40 ) = 20 ( 3 + ( −36 ) ) = 20 ( −33) = −660 2 30.

S12 = 12 ( a1 + a12 ) = 6 (1 + x + 12 + x ) = 78 + 12 x 2

a1 = 1 − 4 = −3 a 25 = 25 − 4 = 21 S25 = 25 ( a1 + a25 ) = 25 ( −3 + 21) = 25 (18 ) = 225 2 2 2

S30 = 30 ( a1 + a30 ) = 15 ( −1 + ( −30 ) ) = 15 ( −31) = −465 2

29.

a1 = 7 (1) = 7

a14 = 7 (14 ) = 98

S25 = 25 ( a1 + a25 ) = 25 ( 9 + 33) = 25 ( 42 ) = 525 2 2 2

27.

a1 = 1 − 2 (1) = −1

a15 = 3 − 5 (15) = −72

S12 = 12 ( a1 + a12 ) = 6 ( 6 + 72 ) = 6 ( 78 ) = 468 2

25.

34 = a1 + 6d 34 = 22 − 3d + 6d 34 = 22 + 3d

a1 = −4( −5) − 19 = 1 a17 = 1 + (17 − 1)( −5) = 1 + 16( −5) = 1 − 80 = −79

19.

a 4 = 22 = a1 + ( 4 − 1) d a 7 = 34 = a1 + ( 7 − 1) d

−29 = a1 + 6d −29 = ( −19 − 4d ) + 6d −29 = −19 + 2d −10 = 2d d = −5

−19 = a1 + 4d −19 − 4d = a1

a1 = 2 (1) − x = 2 − x a15 = 2 (15) − x = 30 − x S15 = 15 ( a1 + a15 ) = 15 ( 2 − x + 30 − x ) = 15 ( 32 − 2 x ) 2 2 2 = 240 − 15 x


770

31.


a1 = (1) x = x a20 = ( 20 ) x = 20 x

32.

S20 = 20 ( a1 + a20 ) = 10 ( x + 20 x ) = 210 x 2

33.

a1 = − (1) x = − x

a14 = − (14 ) x = −14 x

− 1, c2 , c3 , c4 , c5 , c6 , 23

S14 = 14 ( a1 + a14 ) = 7 ⎣⎡ − x + ( −14 x )⎦⎤ = 7 ( −15 x ) = −105 x 2

34.

7, c2 , c3 , c4 , c5 , c6 , 19

a1 = 7 a7 = a1 + ( n − 1) d 19 = 7 + ( 7 − 1) d 12 = 6d d =2 c2 = 7 + ( 2 − 1) 2 = 7 + 2 = 9 c3 = 7 + ( 3 − 1) 2 = 7 + ( 2 ) 2 = 11 c4 = 7 + ( 4 − 1) 2 = 7 + ( 3) 2 = 13 c5 = 7 + ( 5 − 1) 2 = 7 + ( 4 ) 2 = 15 c6 = 7 + ( 6 − 1) 2 = 7 + ( 5) 2 = 17

a1 = −1 a7 = a1 + ( n − 1) d 23 = −1 + ( 7 − 1) d 23 = −1 + 6d 24 = 6d d =4 c2 = −1 + ( 2 − 1) 4 = −1 + 4 = 3 c3 = −1 + ( 3 − 1) 4 = −1 + ( 2 ) 4 = 7 c4 = −1 + ( 4 − 1) 4 = −1 + ( 3) 4 = 11 c5 = −1 + ( 5 − 1) 4 = −1 + ( 4 ) 4 = 15 c6 = −1 + ( 6 − 1) 4 = −1 + ( 5) 4 = 19 35.

3, c2 , c3 , c4 , c5 , 1 2 a1 = 3 a6 = a1 + ( n − 1)d 1 = 3 + (6 − 1)d 2 − 5 = 5d 2 d =−1 2 c2 = 3 + (2 − 1) ⎛⎜ − 1 ⎞⎟ = 3 − 1 = 5 2 2 ⎝ 2⎠ 1 ⎛ ⎞ c3 = 3 + (3 − 1) ⎜ − ⎟ = 3 − 1 = 2 ⎝ 2⎠ c4 = 3 + (4 − 1) ⎛⎜ − 1 ⎞⎟ = 3 − 3 = 3 2 2 ⎝ 2⎠ 1 ⎛ ⎞ c5 = 3 + (5 − 1) ⎜ − ⎟ = 3 − 2 = 1 ⎝ 2⎠

36.

11 , c , c , c , c , 6 2 3 4 5 3 a1 = 11 3 a6 = a1 + ( n − 1)d 6 = 11 + (6 − 1)d 3 7 = 5d 3 d= 7 15 c2 = 11 + (2 − 1) 7 = 11 + 7 = 62 3 15 3 15 15 c3 = 11 + (3 − 1) 7 = 11 + 14 = 69 3 15 3 15 15 11 c4 = + (4 − 1) 7 = 11 + 21 = 76 3 15 3 15 15 11 c5 = + (5 − 1) 7 = 11 + 28 = 83 3 15 3 15 15

37.

a1 = 1, d = 2

38.

a1 = 2 an = 2 n

n[2(1) + (n − 1)2] n Sn = = [2n] = n 2 2 2

39.

a1 = 25, d = –1 a6 = 25 + (6 − 1)(−1) = 25 − 5 = 20 S6 = 6 (25 + 20) = 3(45) = 135 2 20 logs stacked on sixth row, 135 logs in the six rows

Sn = n ( an + a1 ) = n [ 2n + 2] = n 2 + n 2 2

40.

a1 = 27, d = 2 a10 = 27 + (10 − 1)(2) = 27 + 18 = 45 S10 = 10 (27 + 45) = 5(72) = 360 2 45 seats in the tenth row, 360 seats in the ten rows


Section 11.2

41.

771

a1 = 5000, d = −250

42.

a15 = 5000 + (15 − 1)(−250) = 5000 − 3500 = 1500

The fifteenth prize is $1500. 15 15 ( 5000 + 1500 ) = ( 6500 ) = 48,750 2 2 The total amount of money distributed is $48,750. S15 =

43.

a1 = 16, d = 32

a1 = 15, d = 5 an = a1 + ( n − 1) d 60 = 15 + ( n − 1) 5 60 = 15 + 5n − 5 50 = 5n n = 10 In 10 weeks, a person will be walking 60 minutes a day.

44.

S7 = 7 ⎡⎣ 2 (16 ) + ( 7 − 1) 32 ⎤⎦ = 7 [32 + 192] = 7 [ 224] = 784 2 2 2

an = 2n – 1 a10 = 2 (10 ) − 1 = 20 − 1 = 19 a1 = 2 (1) − 1 = 1 S10 = 10 (1 + 19 ) = 100 2

The total distance the object falls is 784 ft.

The distance the ball rolls in the tenth second is 19 ft. The total distance is 100 ft. 45.

a n = a1 + ( n − 1)d 46.9 = 3.5 + ( n − 1)1.4 43.4 = ( n − 1)1.4 31 = n − 1 32 = n

46.

S n = n ( a1 + a n ) 2 S 32 = 32 ( 3.5 + 46.9 ) 2 S 32 = 806.4 mm

....................................................... 47.

If f(x) is a linear function, then f(x) = mx + b. To show that f(n), where n is a positive integer, is an arithmetic sequence, we must show that f ( n + 1) − f ( n ) is a constant. We have f (n + 1) − f (n) = (m(n + 1) + b) − (m(n) + b) = mn + m + b − mn − b =m

Thus, the difference between any two successive terms is m, the slope of the linear function. 49.


a1 = 3, an = an–1 + 5 Rewriting an = an −1 + 5 as an − an −1 = 5, the difference between successive terms is the same constant 5. Thus the sequence is an arithmetic sequence with a1 = 3 and d = 5. an = a1 + (n – 1)d Substituting, an + 3 + (n – 1)5 = 5n – 2

a1 = 4, an = an −1 − 3

Rewriting an = an −1 − 3 as an − an −1 = −3, we find that the difference between successive terms is the same constant –3. Thus the sequence is an arithmetic sequence with a1 = 4 and d = −3. an = a1 + ( n − 1)d Substituting, an = 4 + ( n − 1)( −3) = 4 − 3n + 3 = 7 − 3n


772

50.


a1 = 4, an = bn −1 + 5; b1 = 2, bn = an −1 + 1

To show that an is an arithmetic sequence, we must show that the difference between successive terms is some constant d. We begin by finding a relationship between an and an − 2 . an = bn −1 + 5 = an −1−1 + 1 + 5 = an − 2 + 6

a1 = 4 a2 = b1 + 5 = 2 + 5 = 7 a3 = a1 + 6 a4 = a2 + 6 a5 = a3 + 6 = ( a1 + 6 ) + 6 = a1 + 2 ( 6 )

a6 = a4 + 6 = ( a2 + 6 ) + 6 = a2 + 2 ( 6 ) a7 = a5 + 6 = ( a3 + 6 ) + 6 = a1 + 3 ( 6 )

a8 = a6 + 6 = ( a4 + 6 ) + 6 = a2 + 3 ( 6 ) Thus we have an = an − 2 + 6. This establishes a relationship between alternate successive terms. We now examine the terms of an . Now consider two cases. First, n is an even integer, n = 2 k . From the list of terms, we have a2 k = a2 + ( k − 1) 6

k≥2

Now consider the case when n is an odd integer, n = 2 k − 1. a2 k −1 = a1 + ( k − 1) 6

k≥2

Thus a2 k − a2 k −1 = ( a2 + ( k − 1) 6 ) − ( a1 + ( k − 1) 6 ) = a2 − a1 = 7 − 4 = 3 Therefore, the difference between successive terms is the constant 3. To find a100 , use an = a1 + ( n − 1) d . a100 = 4 + (100 − 1)( 3 ) = 4 + ( 99 )( 3 ) = 301

To show that bn is also an arithmetic sequence, we have

bn − bn −1 = ( an −1 + 1) − ( an − 2 + 1) = an −1 − an − 2

Because an is an arithmetic sequence, an −1 − an − 2 is a constant. Thus bn is an arithmetic sequence.


Section 11.2

51.

773

a1 = 1, an = bn −1 + 7; b1 = −2, bn = an −1 + 1 To show that an is an arithmetic sequence, we must show that an − an −1 = d , where d is a constant. We begin by finding a relationship between an and an − 2 . an = bn−1 + 7 = an−2 + 1 + 7

an = an−2 + 8

This establishes a relationship between alternate successive terms. We now examine some terms of an . a1 = 1 a2 = b1 + 7 = −2 + 7 = 5 a3 = a1 + 8 ( an = a n − 2 + 8 ) a4 = a2 + 8 a5 = a3 + 8 = ( a1 + 8 ) + 8 = a1 + 2 (8 ) a6 = a4 + 8 = ( a2 + 8 ) + 8 = a2 + 2 (8 )

a7 = a5 + 8 = ( a1 + 2 (8 ) ) + 8 = a1 + 3 (8 ) a8 = a6 + 8 = ( a2 + 2 (8 ) ) = a2 + 3 ( 8 )

Now consider two cases. First, n is an even integer, n = 2 k . a2 k = a2 + ( k − 1) 8 k≥2 When n is an odd integer, n = 2 k − 1. a2 k −1 = a1 + ( k − 1) 8

k≥2

Thus a2 k − a2 k −1 = ( a2 + ( k − 1) 8 ) − ( a1 + ( k − 1) 8 ) = a2 − a1 = 5 − 1 = 4 Therefore, the difference between successive terms is the constant. To find a50 , use an = a1 + ( n − 1) d . a50 = 1 + ( 49 )( 4 ) = 197 To show that bn is an arithmetic sequence, we have

bn − bn −1 = ( an −1 + 1) − ( an − 2 + 1) = an −1 − an − 2 Because an is an arithmetic sequence, an −1 − an − 2 is a constant. Thus bn is an arithmetic sequence.

....................................................... PS1.

4 = 2, 8 = 2 2 4 The ratio is 2.

5 PS3. S = 3(1 − ( −2) ) = 33 1 − ( −2)


4

∑ 2n1−1 = 210 + 211 + 212 + 213 = 158

n =1

PS4.

S − rS = a − ar 2 S (1− r ) = a (1− r 2 ) a (1+ r )(1− r ) S= (1− r ) S = a (1+ r )

1

PS5.

⎛ ⎞ a1 = 3 ⎜ − 1 ⎟ = − 3 2 ⎝ 2⎠ 2

⎛ ⎞ a2 = 3 ⎜ − 1 ⎟ = 3 ⎝ 2⎠ 4 3 ⎛ ⎞ a3 = 3 ⎜ − 1 ⎟ = − 3 ⎝ 2⎠ 8

PS6. S1 = 2

S2 = 2 + 4 = 6 S3 = 2 + 4 + 8 = 14


774


Section 11.3 1.

3.

5.

4i +1 = 4, geometric, r = 4 4i 1 i + 1 = i , not geometric 1 i +1 i 2(i +1) x 2ix

=

2ix + x 2ix

7.

3(2(i +1) −1)

9.

x 2(i +1)

3(2i −1)

x 11.

2.

2i

=

=

x

(− 1)(i +1)−1e(i +1)x = (− 1)i eix + x = (− 1)e x (− 1)i −1eix (− 1)i −1eix geometric, r = (− 1)e x

6.

= 2, geometric, r = 2

8.

5(− 2 )(i +1)−1

10.

3(i + x )xi +1

2i −1

2i

1 i 2i +1 = 2 = 1 , geometric, r = 1 i + 1 2 2 1 2 i 2

= 2 x , geometric, r = 2 x

2i

x 2i + 2

4.

6i +1 = 6, geometric, r = 6 6i

= x 2 , geometric, r = x 2

13.

r = 8 = 4, 2 2 n −1 an = 2 ⋅ 4n −1 = 2 ⋅ 2 ( ) = 22 n −1

15.

r = 12 = −3, an = −4 ( −3) −4

18.

r = 6 = 3 , an = 8 ⎛⎜ 3 ⎞⎟ 8 4 ⎝4⎠

21.

r = −3 = − 1 , an = 9 ⎛⎜ − 1 ⎞⎟ 9 3 ⎝ 3⎠

23.

n −1 n −1 r = − x = − x, an = 1( − x ) = ( − x ) 1

n −1

( )

n −1

n −1

12.

= ⎛⎜ − 1 ⎞⎟ ⎝ 3⎠

r=

=

2i −1

n −1 r = 6 = −2, an = −3 ( −2 ) −3

19.

r = − 5 , an = −6 ⎛⎜ − 5 ⎞⎟ 6 ⎝ 6⎠

n −1

n −3

(− 2)i = −2, geometric, r = −2 (− 2)i −1

(i + 1)x , not geometric

=

i

2i log x i −1

2

log x

=

2i i −1

22

17.

r = 4 = 2 , an = 6 ⎛⎜ 2 ⎞⎟ 6 3 ⎝3⎠

20.

r=−

−4 3

= −1 , an = 8 ⎛⎜ − 1 ⎞⎟ 8 6 ⎝ 6⎠

= − 4 = − 2 , an = −2 ⎛⎜ − 2 ⎞⎟ −2 6 3 ⎝ 3⎠

n −1

r=

24.

n −1 r = 2a = a, an = 2 ( a ) = 2a n −1 2

26.

n −1 4 n −1 n r = x 2 = − x 2 , an = − x 2 ( − x 2 ) = −1 ⋅ x 2 ⎡ ( −1) ⋅ x 2 n − 2 ⎤ = ( −1) x 2 n ⎣ ⎦ −x

27.

r=

29.

n −1 n r = 0.05 = 0.1, an = 0.5 ( 0.1) = 5 ( 0.1) 0.5

n −1

4 3

22.

5 r = c 2 = c 3 , an = c 2 c 3 c

= 2, geometric, r = 2

5 = 5, an = 1(5)n −1 = 5n −1 1

16.

25.

3 10,000 3 100

i

log x 2(i +1)−1 log x

14.

n −1

5(− 2 )i −1

3ix

ln 5(i + 1) ⎛ ln 5 ⎞⎛ ln(i + 1) ⎞ =⎜ ⎟⎜ ⎟, not geometric ln 5i ⎝ ln 5i ⎠⎝ ln 5i ⎠

=

= c 2c3n −3 = c3n −1

= 1 , an = 3 ⎛⎜ 1 ⎞⎟ 100 100 ⎝ 100 ⎠

n −1

= 3 ⎛⎜ 1 ⎞⎟ ⎝ 100 ⎠

n

7 10,000 7 10

=

1 ,a = 7 ⎛ 1 ⎞ ⎜ ⎟ n 1000 100 ⎝ 1000 ⎠

n −1

= 7 ⎛⎜ 1 ⎞⎟ ⎝ 10 ⎠

3n − 2

28.

r=

30.

n −1 2 n −1 r = 0.004 = 0.01, an = 0.4 ( 0.01) = 4 ( 0.1) 0.4


n −1

Section 11.3

775

31.

n −1 n r = 0.0045 = 0.01, an = 0.45 ( 0.01) = 45 ( 0.01) 0.45

33.

a1 = 2,

a5 = 162,

an = a1r n −1

34.

162 = 2r 5−1

32.

a3 = 1,

a8 =

a3 = 2(3)3−1 = 2 ⋅ 9 = 18

a4 = 4 ⎜⎛ 1 ⎟⎞ ⎝2⎠

a4 = 8 9 64 243

8 , 9

=

a7 =

64 243

37.

a1r 4 −1

2 a1 = , 3

a1 = 3,

S5 =

7 −1

a1r 27 = 1 8 r3 r3 = 8 27 r=2 3 4 −1 8 = a ⎛2⎞ 1⎜ ⎟ 9 ⎝3⎠ 8 a1 = ⋅ 27 = 3 9 8 5 −1 a5 = 3 ⎛⎜ 2 ⎞⎟ = 16 27 ⎝3⎠

39.

1 32

4 2 9 r= = 2 3 3

4 a2 = , 9

41.

a3 =

4 −1

4 3 − 32 81

a2 = 9,

( )

40.

38.

a1 = 4 , a2 = 16 , r = 3 9 14 ⎤ ⎡ 4 1− ⎛ 4 ⎞ ⎢ ⎥ ⎜ ⎟ 3 ⎝3⎠

a6 = −

32 81

3−1

=

a1 ( r )

6 −1

a1 = 2,

S7 =

16 9 4 3

(

a2 = 4,

)

=4 3

1 1 a1 = 1, a2 = − , r = − 3 3

⎡ ⎛ 1 ⎞8 ⎤ 1⎢1 − ⎜ − ⎟ ⎥ 6560 ⎢ ⎝ 3⎠ ⎥ ⎦ = 6561 = 1640 S8 = ⎣ 4 ⎛ 1⎞ 2187 1− ⎜− ⎟ 3 ⎝ 3⎠


r=

4 =2 2

2 1 − 27 2(− 127 ) = = 254 1− 2 −1

4 ⎛ −263,652,487 ⎞ ⎢ ⎥⎦ 3 ⎝⎜ 4,782,969 ⎠⎟ S14 = ⎣ = −1 1− 4 3 3 1,054,609,948 = ≈ 220.49 4,782,969 42.

9⎤ ⎡ 1 ⎢1 − ⎛⎜ − 2 ⎞⎟ ⎥ 1,953,637 ⎢ ⎝ 5 ⎠ ⎦⎥ 1,953,125 279,091 = = S9 = ⎣ 7 390,625 1 − ⎛⎜ − 2 ⎞⎟ 5 ⎝ 5⎠

9 =3 3

3 1 − 35 3(− 242) = = 363 1− 3 −2

3

2 2 a1 = 1, a2 = − , r = − 5 5

r=

4 , 3

a1 ( r ) −27 = 1 8 r3 3 r =− 8 27 r = −2 3 3−1 4 = a ⎛− 2⎞ ⎟ 1⎜ 3 ⎝ 3⎠ ⎛ 4 a1 = ⎜ 9 ⎞⎟ = 3 3⎝ 4⎠ a2 = 3 ⎛⎜ −2 ⎞⎟ = −2 ⎝ 3 ⎠

= 4 ⎜⎛ 1 ⎟⎞ = 1 ⎝8⎠ 2

⎡ ⎛ 2 6 ⎞⎤ 2 ⎢1 − ⎜ ⎟ ⎥ 2 ⎛ 665 ⎞ 3 ⎢ ⎜ 3 ⎟⎥ ⎟ ⎜ ⎝ ⎠ ⎦ 3 ⎝ 729 ⎠ 1330 S6 = ⎣ = = 2 1 729 1− 3

35.

2 1 = a1r 1 7 a1r 32 32 = 15 r 5 r = 1 32 r=1 2 2 1 = a1 ⎛⎜ 1 ⎞⎟ ⎝2⎠ a1 = 4

r 4 = 81 r =3

36.

n −1 n r = 0.000234 = 0.001, an = 0.234 ( 0.001) = 234 ( 0.001) 0.234

776

43.


a1 = 1, a2 = −2, r = −2

S7 =

46.

1[1 − (−2) ] = −341 1 − ( −2) 47.

2[1 − (−4)11] = 1,677,722 1 − (−4)

a1 =

2[1 − 5 ] = 195,312 1− 5

S10 =

1 1 1 a1 = , a2 = , r = 3 3 9

a1 =

48.

50.

2 3

3 5

()

a1 =

51.

53.

−

a1 = 0.1, r = 0.1

a1 = 0.4, r = 0.5

54.

0.1 0.1 1 S= = = 1 − 0.1 0.9 9

7 7 7 10 10 S= = = 7 3 3 1− 10 10

S=

55.

a1 = 1, r = −0.4 1 S= = 1 =5 1 − ( −0.4 ) 1.4 7

56.

a1 = 1, r = −0.8 1 = 1 =5 S= 1 − ( −0.8 ) 1.8 9

57.

0.3 = 3 + 3 + 3 + ⋅ ⋅ ⋅ 10 100 1000

58.

0.5 = 5 + 5 + 5 + ⋅ ⋅ ⋅ 10 100 1000

3

a1 =

0.3 =

59.

=

=1 3

0.5 =

45 0.45 = 45 + 45 + + ⋅⋅⋅ 100 10,000 1,000,000 a1 = 0.45 =

61.

a1 =

10

3 10 9 10

60.

45 10,000 = 1 45 100 100 100 45 45 100 = 100 = 5 99 11 1− 1 100 100 45

,r =

123 1000 1− 1 1000

= 123 = 41 999 333

= 1 , r = 100 5 10 10 5

5 10 1− 1 10

0.63 =

62.

10 5 = 10 9 10

=5 9

63 0.63 = 63 + 63 + + ⋅⋅⋅ 100 10,000 1,000,000 a1 =

123 + + ⋅⋅⋅ 0.123 = 123 + 123 100 1,000,000 1,000,000,000 a1 = 123 , r = 1 1000 1000 0.123 =

0.5 0.5 = =1 1 − 0.5 0.5

5

= 1 , r = 100 3 10 10 3

3 10 1− 1 10

9 9 ,r = 100 100

9 9 9 100 100 S= = = 9 91 91 1− 100 100

3 3 5 S= = =− 8 3⎞ 8 ⎛ 1− ⎜− ⎟ 5 ⎝ 5⎠

7 7 ,r = 10 10

3 9 3 , a2 = , r = 4 4 16

3 S= = 4 =3 1 3 1− 4 4

3 3 a1 = − , r = − 5 5 −

5[1 − 310 ] = 147,620 1− 3

3 4

1 1 S= = 3 = 2 1 2 1− 3 3

2 2 S= =− 3 =− 5 2⎞ 5 ⎛ 1− ⎜− ⎟ 3 ⎝ 3⎠

52.

a1 = 5, a2 = 15, r = 3

45.

8

1 3

2 2 a1 = − , r = − 3 3 −

a1 = 2, a2 = 10, r = 5

S8 =

a1 = 2, a2 = −8, r = −4

S11 =

49.

44.

10

63 100

,r =

63 100 1− 1 100

63 10,000 63 100

= 1 100

= 63 = 7 99 11

⎡ ⎤ 95 0.395 = 3 + ⎢ 95 + 95 + + ⋅ ⋅ ⋅⎥ 10 ⎣1000 100,000 10,000,000 ⎦ 1 95 a1 = ,r = 100 1000 95

0.395 = 3 + 10001 = 3 + 95 = 392 = 196 10 1 − 10 990 990 495 100


Section 11.3

63.

422 + 422 + ⋅⋅⋅ 0.422 = 422 + 1000 1,000,000 1,000,000,000 a1 = 422 , r = 1 1000 1000 0.422 =

65.

777

422 1000 1− 1 1000

64.

= 422 999

0.355 =

⎤ 25 ⎡ 4 4 4 +⎢ + + + ⋅ ⋅ ⋅⎥ 100 ⎣1,000 10,000 100,000 ⎦ 1 4 ,r = a1 = 10 1000

0.25 4 =

66.

⎡ ⎤ 84 84 1.2084 = 1 + 2 + ⎢ 84 + + + ⋅ ⋅ ⋅⎥ 10 ⎣10,000 1,000,000 100,000,000 ⎦ a1 = 84 , r = 1 10, 000 100

68.

⎡ ⎤ 90 90 + ⋅⋅⋅ 02.2590 = 225 + ⎢ 90 + 100 ⎣10,000 1,000,000 100,000,000 ⎥⎦ a1 = 90 , r = 1 10, 000 100 90

10,000 = 9 + 1 = 497 2.2590 = 225 + 100 1 − 1 4 110 220 100

100

P = 100

⎡ ⎤ 2 0.372 = 37 + ⎢ 2 + 2 + ⋅⋅⋅ 100 ⎣1,000 10,000 100,000 ⎥⎦ a1 = 2 , r = 1 10 1000

10

84

A = 100, i = 0.09, n = 2, r = 8, r =

= 355 999

2

10,000 12 = + 7 = 1994 = 997 1.2084 = 1 + 2 + 10 1− 1 10 825 1650 825

69.

355 1000 1− 1 1000

0.372 = 37 + 10001 = 37 + 2 = 335 = 67 100 1 − 100 900 900 180

4 25 1000 25 4 229 0.25 4 = + = + = 100 1− 1 100 900 900 10

67.

355 355 + + ⋅⋅⋅ 0.355 = 355 + 1000 1,000,000 1,000,000,000 a1 = 355 , r = 1 1000 1000

i 0.09 = = 0.049, m = nt = 2 ⋅ 8 = 16 n 2

[(1+0.045) −1] ≈ 2271.93367 16

0.045

The future value is $2271.93.

70.

A = 250, i = 0.08, n = 12, t = 4, r = 8 = 2 , m = 12 ⋅ 4 = 48 12 3 P = 250

71.

⎡⎛ 0.08 ⎞48 ⎤ ⎢⎜1+ ⎟ −1⎥ ⎣⎢⎝ 12 ⎠ ⎦⎥ 0.08 12

≈ $14,087.48

When a name was removed from the top of the list, the letter had been sent to

72.

A = 0.5, n = 4, k = −0.876, t = 4

74.

( 12 ) S12 = 5 1 − 5 = 305,175,780 1− 5 For a population of 127,000,000, the entire population would receive the letter on the 12th level.

5 ( 55 ) = 15,625 people who sent 10 cents each for a total of 0.10(15,625) = $1562.50 for each recipient..

73.

A + Aekt + Ae2 kt Ae3kt = 0.5 + 0.5e −0.867(4) + 0.5e2( −0.867)(4) + 0.5e3( −0.867)(4) ≈ 0.52 mg

( 11 ) S11 = 5 1 − 5 = 61,035,155 1− 5

A + Aekt + L + Ae( n −1) kt + L = 2 A (1 + ekt + L + e( n −1) kt + L) = 2 A =2 1 − ekt

Or A = 0.5, r = ekt , n = 4, k = −0.876, t = 4

( −.867(4)(3) ) ≈ 0.52 mg S4 = 0.5 1 − e−.867(4) 1− e


A = 2 (1 − ekt ) A = 2 (1 − e −0.25(12) ) A ≈ 1.90 mg

778

75.


D (1 + g ) i−g 1.87(1 + 0.15) = 0.20 − 0.15 = $43.01

Stock value =

76. D = 1.87, g = 0.15, i = 0.20

D (1 + g ) i−g 1.32(1 + g ) 67 = Solve for g . 0.20 − g 67(0.20 − g ) = 1.32(1 + g ) 13.4 − 67 g = 1.32 + 1.32 g 12.08 = 68.32 g 0.1768 = g The dividend growth rather is 17.68%. Stock value =

77.

Stock value (no dividend growth) = D = 2.94 = $19.60 i 0.15

78.

Stock value (no dividend growth) = D i 3.24 16 = i 3.24 i= 16 i = 0.2025 The expected rate of return is 20.25%.

79.

If g was not less than i in the Gordon model of stock valuation, the common ratio of the geometric sequence would be greater than 1 and the sum of the infinite geometric series would not be defined.

80.

Using the multiplier effect, 50 = 500 1 − 0.90 The net effect of $50 million is $500 million.

81.

Using the multiplier effect, 25 = 100 1 − 0.75 The net effect of $25 million is $100 million.

82.

Using the multiplier effect, 500,000 ≈ $833,000 1 − 0.40 About $833,000 is used before it is removed.

....................................................... 83.

Connecting Concepts

Let an be a geometric sequence. Thus

an = a1r n −1, a1 ≠ 0,

r≠0

and log an = log a1r n −1 = log a1 + log r n −1

= log a1 + (n − 1)log r Since r is a constant, log r is a constant. Conjecture: The sequence log an is an arithmetic sequence. To prove this conjecture, we must show that log an − log an −1 is a constant.

log an − log an −1 = ( log a1 + ( n − 1) log r ) − ( log a1 + ( n − 2 ) log r )

= log r Since log r is a constant, the sequence log an is an arithmetic sequence.


Section 11.3

84.

779

Let an be a geometric sequence. Thus

an = a1 + (n − 1)d where d is a constant.

( )

n −1 2an = 2a1 + (n −1)d = 2a1 ⋅ 2(n −1)d = 2a1 ⋅ 2d

Conjecture : The sequence 2an is a geometric sequence. To prove this conjecture let bn = 2an and show that bn + 1 = r where r is a constant. bn

( )n ( )

2a1 ⋅ 2d bn +1 = = 2d . bn a1 d n −1 2 ⋅2 Because d is a constant (see above), 2d is a constant. Thus 2an is a geometric sequence. 85.

Yes. Because x ≠ 0, the first term is 1 and the common ratio is x. If x < 1, the geometric series converges to

1 . If x ≥ 1, the 1− x

geometric series does not converge. 86.

To find the area of the n th inscribed square, begin by finding the area of the first few squares. The area of the first square is 12 = 1. Thus a1 = 1 The length of a side of the second square can be found by using the Pythagorean Theorem. 2

2

S 2 = ⎛⎜ 1 ⎞⎟ + ⎛⎜ 1 ⎞⎟ = 1 + 1 = 1 4 4 2 ⎝2⎠ ⎝2⎠ Notice here that S 2 is the area of the second square. Thus a2 = 1 . 2 The length of the side of the third square is 2

2

⎛ ⎞ ⎛ ⎞ S2 = ⎜ 1 ⎟ + ⎜ 1 ⎟ = 1 + 1 = 2 = 1 8 8 8 4 ⎝2 2⎠ ⎝2 2⎠ Again S 2 is the area of the square. Thus a3 = 1 . 4 From here we conjecture that the area of the n th square is an = n1−1 . 2

87.

If an is a geometric sequence, then an = a1r n −1

Since an = ar n −1 , then a1 = a. Pn = a1 ⋅ a2 ⋅ a3 ⋅ ⋅ ⋅ ⋅ ⋅ an = a ⋅ a2 ⋅ a3 ⋅ ⋅ ⋅ ⋅ ⋅ an

= a ⋅ ar ⋅ ar 2 ⋅ ⋅ ⋅ ⋅ ⋅ ar n −1 = a n r [( n −1) n ]/ 2

The exponent on r is found by using the sum of an arithmetic series formula. Note that

a ⋅ ar ⋅ ar 2 ⋅ ar 3 ⋅ ⋅ ⋅ ⋅ ⋅ ar n −1 = a n ⋅ r 1 + 2 + 3 + ⋅⋅⋅+ n − 1 and 1 + 2 + 3 + ⋅ ⋅ ⋅ + n − 1 = 88.

( n − 1) n .

Let an = f (n ) = ab n . Thus an +1 ab n +1 = =b an ab n Since b is a constant, an is a geometric sequence.


2

780

89.


The distance the ball travels each bounce is given by a1 = 5 a2 = 2 ( 0.8 ) 5 Multiply by 2 for the distance up and down. a3 = 2 ( 0.8 )( 0.8 )( 5 ) = 2 ( 0.8 ) 5 2

a4 = 2 ( 0.8 )( 0.8 ) 5 = 2 ( 0.8 ) 5 2

M

3

n −1

an = 2 ( 0.8 ) ⋅ 5 This is a geometric sequence (after a1 ). The sum of this sequence is the total distance travelled by the ball. n

2 ( 0.8 ) ( 5 ) an +1 = = 0.8 n +1 an 2 ( 0.8 ) ( 5 ) Because 0.8 < 1, the geometric series converges. S = 5 + 8 = 5 + 8 = 5 + 40 = 45 0.2 1 − 0.8 The distance traveled is 45 feet. Note from our calculation that the geometric series begins with a2 . The first term (a1 = 5) is added to the series. 90.

The distance of each swing of the bob is a term of a sequence 91. a1 = 30 a2 = ( 0.9 )( 30 ) 2

a3 = ( 0.9 )( 0.9 )( 30 ) = ( 0.9 ) ( 30 ) M an = ( 0.9 )

n −1

( 30 ) a ( 0.9 )n ( 30 ) = 0.9, a constant, the sequence is Because n +1 = an ( 0.9 )n −1 ( 30 ) a geometric sequence. Since 0.9 < 1, the infinite geometric series converges. The sum of this series is the total distance traveled by the bob. 30 30 S= = = 300 1 − 0.9 0.1 The bob traveled 300 inches.

The n th generation has an = 2n grandparents. Since this is a geometric sequence, the sum can be found by a formula. Sn =

a1 1 − r n

(

)

1− r 2 1 − 210

)

(

2 (1 − 1024 ) = = 2046 1− 2 −1 When n = 1, an = 2 and there are no grandparents. Therefore there are 2046 − 2 = 2044 grandparents by the 10 th generation. S10 =

....................................................... PS1.

4

∑ i(i 1+ 1) = 1(11+ 1) + 2(21+ 1) + 3(31+ 1) + 4(41+ 1)

Prepare for Section 11.4 PS2. k ( k + 1)(2k + 1) + 6( k + 1)2 = ( k + 1)[k (2k + 1) + 6( k + 1)] = ( k + 1)[2k 2 + k + 6k + 6]

i =1

=1+1+ 1 + 1 =4= 4 2 6 12 20 5 4 + 1

PS3.

k + 1 1 = k +2⋅ k + k + 1 ( k + 1)( k + 2) k + 2 k + 1 ( k + 1)( k + 2) 2 ( k + 1)( k + 1) k + 1 = k + 2k + 1 = = ( k + 1)( k + 2) ( k + 1)( k + 2) k + 2

PS5. Sn + an +1 = n( n + 1) + n + 1 2 2 2 = n + n + 2n + 2 = n + 3n + 2 2 2 2 ( n + 1)( n + 2) = 2

= ( k + 1)[2k 2 + 7k + 6] = ( k + 1)( k + 2)(2k + 3)

PS4. 12 > 2(1) + 1 = 3

22 > 2(2) + 1 = 5 32 > 2(3) + 1 = 7 3 PS6. Sn + an +1 = 2n +1 − 2 + 2n +1

= 2n + 2 − 2 = 2(2n +1 − 1)


Section 11.4

781

Section 11.4 1.

1. 2.

1(3 ⋅ 1 − 1) 2 Assume the statement is true for some positive integer k. k (3k − 1) (Induction Hypothesis) Sk = 1 + 4 + 7 + ⋅ ⋅ ⋅ + 3k − 2 = 2 Verify that the statement is true when n = k + 1 (k + 1)(3k + 2) . Sk +1 = 2 a k = 3k − 2, a k +1 = 3k + 1

Let n = 1. S1 = 3 ⋅ − 2 = 1 =

Sk +1 = Sk + ak +1 =

3k 2 − k 6k + 2 k (3k − 1) + 3k + 1 = + 2 2 2

3k 2 + 5k + 2 (k + 1)(3k + 2) = 2 2 By the Principle of Mathematical Induction, the statement is true for all positive integers. =

2.

1.

Let n =1. S1 = 2 ⋅ 1 = 2 = 1(1 + 1)

2.

Assume Sk = 2 + 4 + 6 + ⋅ ⋅ ⋅ + 2k = k (k + 1) is true for some positive integer k (Induction Hypothesis). Verify that Sk +1 = ( k + 1)( k + 2) is true when n = k + 1 ak = 2k ,

ak +1 = 2(k + 1)

Sk +1 = Sk + ak +1 = k (k + 1) + 2(k + 1) = (k + 1)(k + 2) By the Principle of Mathematical Induction, the statement is true for all positive integers.

3.

12 (1 + 1)

2

1.

Let n = 1. S1 = 13 = 1 =

2.

Assume Sk = 1 + 8 + 27 + ⋅ ⋅ ⋅ + k 3 = Verify that Sk +1 =

4 k 2 (k + 1)2 is true for some positive integer k.(Induction Hypothesis). 4

( k + 1)2 (k + 2)2 . 4

ak = k 3 , ak +1 = (k + 1)3 Sk +1 = Sk + ak +1 =

2 2 3 k 2 (k + 1)2 3 k ( k + 1) + 4( k + 1) + ( k + 1) = 4 4

( k + 1) 2 (k 2 + 4k + 4) (k + 1)2 (k + 2)2 = 4 4 By the Principle of Mathematical Induction, the statement is true for all positive integers. =

4.

1.

Let n = 1 S1 = 21 = 2 = 2(21−1)

2.

Assume that Sk = 2 + 4 + 8 + ⋅ ⋅ ⋅ + 2k = 2(2k − 1) is true for some positive integer k (Induction Hypothesis).

)

(

Verify that S k +1 = 2 2 k +1 − 1 . k

a k = 2 , a k +1 = 2

k +1

(

)

S k +1 = S k + a k +1 = 2 2 k − 1 + 2 k +1 = 2 k +1 − 2 + 2 k +1

(

)

2 ⋅ 2 k +1 − 2 = 2 2 k +1 − 1 By the Principle of Mathematical Induction, the statement is true for all positive integers. Copyright © Houghton Mifflin Company. All rights reserved.

782

5.


1.

Let n = 1. S1 = 4 ⋅ 1 − 1 = 3 = 1(2 ⋅ 1 + 1)

2.

Assume that Sk = 3 + 7 + 11 + ⋅ ⋅ ⋅ + 4k − 1 = k (2k + 1) is true for some positive integer k (Induction Hypothesis). Verify that Sk +1 + (k + 1)(2k + 3) . a k = 4k − 1,

a k +1 = 4k + 3

Sk +1 = Sk + ak +1 = k (2k + 1) + 4k + 3 = 2k 2 + 5k + 3 = (k + 1)(2k + 3) By the Principle of Mathematical Induction, the statement is true for all positive integers.

6.

3(31 − 1) =3 2

1.

Let n = 1. S1 = 31 =

2.

Assume that S k = 3 + 9 + 27 + ⋅ ⋅ ⋅ + 3 k = Verify that S k +1 = a k = 3k ,

3(3k − 1) is true for some positive integer k (Induction Hypothesis). 2

3(3k + 1 − 1) . 2

a k +1 = 3 k +1

Sk +1 = Sk + ak +1 =

3(3k − 1) + 3k +1 2

3k +1 − 3 + 2 ⋅ 3k +1 3 ⋅ 3k +1 − 3 3(3k +1 − 1) = = 2 2 2 By the Principle of Mathematical Induction, the statement is true for all positive integers. =

7.

1.

Let n = 1. S1 = (2 ⋅ 1 − 1)3 = 1 = 12 (2 ⋅ 12 − 1)

2.

Assume that Sk = 1 + 27 + 125 + ⋅ ⋅ ⋅ + (2k − 1)3 = k 2 (2k 2 − 1) is true for some positive integer k (Induction Hypothesis). Verify that Sk +1 = (k + 1)2 (2k 2 + 4k + 1) . ak = (2k − 1)3 ,

ak +1 = (2k + 1)3

Sk +1 = Sk + ak +1 = k 2 (2k 2 − 1) + (2k + 1)3 = 2k 4 − k 2 + 8k 3 + 12k 2 + 6k + 1 = 2k 4 + 8k 3 + 11k 2 + 6k + 1 = (k + 1)(2k 3 + 6k 2 + 5k + 1) = (k + 1)2 (2k 2 + 4k + 1) By the Principle of Mathematical Induction, the statement is true for all positive integers. 8.

1(1 + 1)(1 + 2) 3

1.

Let n = 1. S1 = 1(1 + 1) = 2 =

2.

Assume that Sk = 2 + 6 + 12 + ⋅ ⋅ ⋅ + k ( k + 1) =

k (k + 1)(k + 2) is true for some positive integer k (Induction Hypothesis). 3

(k + 1)(k + 2)(k + 3) . 3 ak +1 = (k + 1)(k + 2)

Verify that Sk +1 = ak = k (k + 1),

k (k + 1)(k + 2) k (k + 1)(k + 2) + 3(k + 1)(k + 2) + (k + 1)(k + 2) = 3 3 (k + 1)( k + 2)( k + 3) = 3 By the Principle of Mathematical Induction, the statement is true for all positive integers. Sk +1 =


Section 11.4

9.

783

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

1.

Let n = 1. S1 =

2.

Assume that Sk =

1 1 1 1 k + + + ⋅⋅⋅ + = for some positive integer k (Induction Hypothesis). 1⋅ 3 3 ⋅ 5 5 ⋅ 7 (2k − 1)(2k + 1) 2k + 1

Verify that S k +1 = ak =

k +1 . 2k + 3

1 , (2k − 1)(2k + 1)

ak +1 =

1 (2k + 1)(2k + 3)

k 1 2k 2 + 3k + 1 (2k + 1)(k + 1) k +1 + = = = 2k + 1 (2k + 1)(2k + 3) (2k + 1)(2k + 3) (2k + 1)(2k + 3) 2k + 3 By the Principle of Mathematical Induction, the statement is true for all positive integers. Sk +1 =

10.

1 1 1 = = 2 ⋅ 1(2 ⋅ 1 + 2) 8 4(1 + 1)

1.

Let n = 1. S1 =

2.

Assume that Sk =

1 1 1 1 k + + + ⋅⋅⋅ + = for some positive integer k (Induction Hypothesis). 2⋅4 4⋅6 6⋅7 2k (2k + 2) 4(k + 1)

Verify that Sk +1 =

k +1 . 4(k + 2)

1 1 , ak +1 = 2k (2k + 2) (2k + 2)(2k + 4) 2 k ( k + 2) + 1 1 1 k k S k +1 = + = + = = k + 2k + 1 4(k + 1) (2k + 2)(2k + 4) 4(k + 1) 4(k + 1)(k + 2) 4(k + 1)(k + 2) 4(k + 1)(k + 2) (k + 1) 2 = = k +1 4(k + 1)(k + 2) 4(k + 2) By the Principle of Mathematical Induction, the statement is true for all positive integers. ak =

11.

1(1 + 1)(2 ⋅ 1 + 1)(3 ⋅ 12 + 3 ⋅ 1 − 1) 2 ⋅ 3 ⋅ 5 = =1 30 30

1.

Let n = 1. S1 = 14 = 1 =

2.

Assume that Sk = 1 + 16 + 81 + ⋅ ⋅ ⋅ + k 4 = Verify that Sk +1 = ak = k 4 ,

k (k + 1)(2k + 1)(3k 2 + 3k − 1) for some positive integer k (Induction Hypothesis). 30

( k + 1)(k + 2)(2k + 3)(3k 2 + 9k + 5) . 30

ak +1 = (k + 14 )

k (k + 1)(2k + 1)(3k 2 + 3k − 1) + (k + 1) 4 30 (k + 1)[k (2k + 1)(3k 2 + 3k − 1) + 30( k + 1)3 ] = 30 (k + 1)[6k 4 + 39k 3 + 91k 2 + 89k + 30] (k + 1)(k + 2)(6k 3 + 27 k 2 + 37 k + 15) = = 30 30 2 (k + 1)(k + 2)(2k + 3)(3k + 9k + 5) = 30 By the Principle of Mathematical Induction, the statement is true for all positive integers. S k +1 =


784

12.


1. 2.

1 ⎞ 1 1 ⎛ Let n = 1. P1 = ⎜1 − ⎟= = 1 1 2 1 + 1 + ⎠ ⎝ 1⎞ ⎛ Assume Pk = ⎜1 − ⎟ 2⎠ ⎝ 1 Verify Pk +1 = . k+2

⎛ 1⎞ ⎜1 − ⎟ 3⎠ ⎝

1⎞ ⎛ 1 ⎞ 1 ⎛ is true for some positive integer k (Induction Hypothesis). ⎟= ⎜1 − ⎟ ⋅ ⋅ ⋅ ⎜ 1 − +1 4 1 + k k ⎠ ⎠ ⎝ ⎝

Pk +1 = ⎛⎜1 − 1 ⎞⎟ ⎛⎜ 1 − 1 ⎞⎟ ⎛⎜ 1 − 1 ⎞⎟ ⋅ ⋅ ⋅ ⎛⎜ 1 − 1 ⎞⎟ ⎛⎜1 − 1 ⎞⎟ ⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 4 ⎠ ⎝ k +1⎠ ⎝ k + 2 ⎠ = Pk ⎛⎜1 − 1 ⎞⎟ ⎝ k +2⎠ ⎛1 − 1 ⎞ 1 = ⎜ ⎟ k +1 ⎝ k + 2 ⎠ k +1 1 = 1 − = k + 2 −1 = = 1 k + 1 (k + 1)(k + 1) (k + 1)(k + 2) ( k + 1)( k + 2) k + 2 By the Principle of mathematical Induction, the statement is true for all positive integers.

13.

4

1.

1 81 ⎛3⎞ Let n = 4. Then ⎜ ⎟ = = 5 ;4 +1= 5 16 16 ⎝2⎠ n

⎛3⎞ Thus, ⎜ ⎟ > n + 1 for n = 4. ⎝2⎠ k

2.

⎛3⎞ Assume ⎜ ⎟ > k + 1 is true for some positive integer k ≥ 4 (Induction Hypothesis). ⎝2⎠ ⎛3⎞ Verify that ⎜ ⎟ ⎝2⎠ ⎛3⎞ ⎜ ⎟ ⎝2⎠

k +1

> k + 2.

k

1 1 ⎛ 3⎞ ⎛3⎞ ⎛ 3⎞ 1 = ⎜ ⎟ ⎜ ⎟ > ( k + 1) ⎜ ⎟ = ( 3k + 3) = ( 2k + k + 3) > ( 2k + 1 + 3) = k + 2 2 2 ⎝ 2⎠ ⎝2⎠ ⎝ 2⎠ 2

⎛3⎞ Thus ⎜ ⎟ ⎝2⎠ 14.

k +1

k +1

n

⎛3⎞ > k + 2. By the Principle of Mathematical Induction, ⎜ ⎟ > n + 1 for all n ≥ 4. ⎝2⎠ 7

1.

16384 1075 ⎛4⎞ Let n = 7. ⎜ ⎟ = =7 >7 3 2187 2187 ⎝ ⎠ n

⎛4⎞ Thus, ⎜ ⎟ > n for n = 7. ⎝3⎠ k

2.

⎛4⎞ Assume ⎜ ⎟ > k is true for some positive integer k ≥ 7 (Induction Hypothesis). ⎝3⎠ ⎛4⎞ Verify that ⎜ ⎟ ⎝3⎠ ⎛4⎞ ⎜ ⎟ ⎝3⎠

k +1

k +1

> k + 1.

k

1 1 ⎛4⎞ ⎛ 4⎞ ⎛ 4⎞ 1 = ⎜ ⎟ ⎜ ⎟ > k ⎜ ⎟ = ( 4k ) = ( 3k + k ) > ( 3k + 3) = k + 1 3 3 3 3 3 3 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛4⎞ Thus ⎜ ⎟ ⎝3⎠

k +1

n

⎛ 4⎞ > k + 1. By the Principle of Mathematical Induction, ⎜ ⎟ > n for all n ≥ 7. ⎝ 3⎠


Section 11.4

15.

1.

785

Let n = 1. 0 < a 1 ⋅ a1 a1+1 > a1 Thus, if a > 1, then a n +1 > a n for n = 1.

2.

Assume that if a > 1, then a k +1 > a k for some positive integer k (Induction Hypothesis). Verify a k + 2 > a k +1. a >1 a ⋅ a k +1 > 1 ⋅ a k +1 a k + 2 > a k +1 By the Principle of Mathematical Induction, if a > 1, then a n + 2 > a n +1 for all positive integers.

17.

1.

Let n = 4. 1 ⋅ 2 ⋅ 3 ⋅ 4 = 24, 24 = 16 Thus, 1 ⋅ 2 ⋅ 3 ⋅ 4 > 2n for n = 4.

2.

Assume 1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ ⋅ ⋅ ⋅ ⋅ k > 2k is true for some positive integer k ≥ 4 (Induction Hypothesis). Verify 1 ⋅ 2 ⋅ 3 ⋅ ⋅ ⋅ ⋅ ⋅ k ⋅ (k + 1) > 2k +1. 1 ⋅ 2 ⋅ 3 ⋅ ⋅ ⋅ ⋅ ⋅ k ⋅ (k + 1) > 2k ( k + 1) > 2k ⋅ 2 = 2k +1

Thus, 1 ⋅ 2 ⋅ 3 ⋅ ⋅ ⋅ ⋅ ⋅ k ⋅ (k + 1) > 2k +1. By the Principle of Mathematical Induction, 1 ⋅ 2 ⋅ 3 ⋅ ⋅ ⋅ ⋅ ⋅ n > 2n for all n ≥ 4. 18.

1.

1

Let n = 1. Thus

1 1

+ 1

Assume

1 1

=1= 1

2

+ 1

+

2

1

1

2

+

1 3

+ ⋅⋅⋅ + 1

+

3

1

1 n

+ ⋅⋅⋅ +

3

+ ⋅⋅⋅ +

1

≥ n for n = 1. 1 k ≥

≥ k is true for some positive integer k (Induction Hypothesis). 1

> k + 1. n k +1 1 1 1 1 1 1 + + + ⋅⋅⋅ + + ≥ k+ 1 2 3 k k +1 k +1

Verify

+

1 1

k k +1 1 k k 1 + > + k +1 k +1 k +1 k +1 k +1 = = k +1 k +1 1 + 1 + 1 + ⋅ ⋅ ⋅ + 1 + 1 > k + 1. By the Principle of Mathematical Induction, 1 + 1 + 1 + ⋅ ⋅ ⋅ + 1 ≥ n for Thus, 1 2 3 k k +1 1 2 3 n all positive integers n. =


786

19.


1.

Let n = 1 and a > 0.

(1 + a )1 = 1 + a = 1 + 1 ⋅ a

Thus (1 + a )n ≥ 1 + na for n = 1. 2.

Assume (1 + a) k ≥ 1 + ka is true for some positive integer k (Induction Hypotheses). Verify (1 + a )k +1 > 1 + (k + 1)a. (1 + a )k +1 = (1 + a )k (1 + a) ≥ (1 + ka)(1 + a) = 1 + (k + 1) a + ka 2 > 1 + ( k + 1) a

Thus (1 + a )k +1 > 1 + (k + 1)a. By the Principle of Mathematical Induction, (1 + a) n > 1 + na for all positive integers n. 20.

1.

Let n = 1. log10 1 = 0 < 1

2.

Assume log10 k < k is true for some positive integer k (Induction Hypothesis).

Thus log10 n < n for n = 1. Verify log10 (k + 1) < k + 1.

log10 (k + 1) ≤ log10 (k + k ) = log10 2k = log10 2 + log10 k < 1 + k because log10 2 < 1 and log10 k < k .

Thus log10 (k + 1) < k + 1. By the Principle of Mathematical Induction, log10 n < n for all positive integers n. 21.

1.

Let n = 1. 12 + 1 = 2,

2 = 2 ⋅1

Thus 2 is a factor of n 2 + n for n = 1. 2.

Assume 2 is a factor of k 2 + k for some positive integer k (Induction Hypothesis). Verify 2 is a factor of (k + 1)2 + k + 1.

(k + 1)2 + k + 1 = (k + 1)(k + 1 + 1) = (k + 1)(k + 2) Since k 2 + k = k (k + 1) , 2 is a factor of k or k + 1. If 2 is a factor of k + 1, then 2 is a factor of (k + 1)(k + 2). If 2 is a factor of k, then 2 is a factor of k + 2. Thus, 2 is a factor of (k + 1)2 + k + 1. By the Principle of Mathematical Induction, 2 is a factor of n 2 + n for all positive integers. 22.

1.

Let n = 1. 13 − 1 = 0,

0 = 0⋅3

Thus, 3 is a factor of n 3 − n for n = 1. 2.

Assume 3 is a factor of k 3 − k for some positive integer k (Induction Hypothesis). Verify 3 is a factor of (k + 1)3 − (k + 1) when n = k + 1. (k + 1)3 − ( k + 1) = (k + 1)[(k + 1)2 − 1] = (k + 1)(k 2 + 2k ) = (k + 1)(k + 2) k = k (k + 1)(k + 2) Since k 3 − k = k (k + 1)( k − 1) , then 3 is a factor of k, k + 1, or k − 1 . If 3 is a factor of k, then 3 is a factor of k (k + 1)(k + 2). If 3 is a factor of k + 1, then 3 is a factor of k (k + 1)(k + 2). If 3 is a factor of k − 1, then 3 is a factor of k + 2. Since k + 2 = k − 1 + 3, the sum of two multiples of 3 is a multiple of 3, so 3 is a factor of k (k + 1)(k + 2).

Thus, 3 is a factor of (k + 1)3 − (k + 1). By the Principle of Mathematical Induction, 3 is a factor of n 3 − n for all positive integers. 23.

1.

Let n = 1. 51 − 1 = 4,

4 = 4 ⋅1

Thus, 4 is a factor of 5 n − 1 for n = 1. 2.

Assume 4 is a factor of 5 k − 1 for some positive integer k (Induction Hypothesis). Verify 4 is a factor of 5 k +1 − 1. Now 5k +1 − 1 = 5 ⋅ 5k − 5 + 4 = 5 (5k − 1) + 4 which is the sum of two multiples of 4.

Thus, 4 is a factor of 5 k +1 − 1. By the Principle of Mathematical Induction, 4 is a factor of 5 n − 1 for all positive integers. Copyright © Houghton Mifflin Company. All rights reserved.

Section 11.4

24.

1.

787

Let n = 1. 61 − 1 = 5,

5 = 5 ⋅1 n

Thus, 5 is a factor of 6 − 1 for n = 1. 2.

Assume 5 is a factor of 6 k − 1 for some positive integer k (Induction Hypothesis). Verify 5 is a factor of 6 k +1 − 1. Now 6k +1 − 1 = 6 ⋅ 6k − 6 + 5 = 6(6k − 1) + 5 which is the sum of two multiples of 5. Thus, 5 is a factor of 6 k +1 − 1. By the Principle of Mathematical Induction, 5 is a factor of 6 n − 1 for all positive integers.

25.

1.

Let n = 1. ( xy )1 = xy = x1 y1 Thus, ( xy )n = x n y n for n = 1.

2.

Assume ( xy )k = x k y k is true for some positive integer k (Induction Hypothesis). Verify ( xy )k +1 = x k +1 y k +1. ( xy )k +1 = ( xy )k ( xy )1 = x k y k ⋅ xy = x k +1 y k +1

Thus ( xy )k +1 = x k +1 y k +1. By the Principle of Mathematical Induction, ( xy )n = x n y n for all positive integers. 1

26.

1.

1 ⎛ ⎞ Let n = 1. ⎜ x ⎟ = x = x1 . y y ⎝ y⎠ n

n ⎛ ⎞ Thus, ⎜ x ⎟ = x n for n = 1. y y ⎝ ⎠ k

2.

k ⎛ ⎞ Assume ⎜ x ⎟ = x k is true for some positive integer k (Induction Hypothesis). y ⎝ y⎠

Verify ⎛⎜ x ⎞⎟ ⎝ y⎠ ⎛x⎞ ⎜ y⎟ ⎝ ⎠

k +1

1.

k +1 = x k +1 . y

k

1

k k +1 = ⎛⎜ x ⎞⎟ ⎛⎜ x ⎞⎟ = x k ⋅ x = x k +1 y y y ⎝ y⎠ ⎝ y⎠

Thus, ⎛⎜ x ⎞⎟ ⎝ y⎠ 27.

k +1

k +1

n

k +1 n = x k +1 . By the Principle of Mathematical Induction, ⎛⎜ x ⎞⎟ = x n for all positive integers. y y ⎝ y⎠

Let n = 1. a1 − b1 = a − b Thus a − b is a factor of a n − b n for n = 1.

2.

Assume a − b is a factor of a k − b k for some positive integer k (Induction Hypothesis). Verify a − b is a factor of a k +1 − b k +1. a k +1 − b k +1 = (a ⋅ a k − ab k ) + (ab k − b ⋅ b k ) = a(a k − b k ) + b k (a − b) The sum of two multiples of a − b is a multiple of a − b. Thus, a − b is a factor of a k +1 − b k +1 . By the Principle of Mathematical Induction, a − b is a factor of a n − b n for all positive integers.

28.

1.

Let n = 1. a 2⋅1+1 + b 2⋅1+1 = a3 + b3 = (a + b)(a 2 − ab + b 2 ) Thus, a + b is a factor of a 2n+1 + b 2n+1 for n = 1.

2.

Assume a + b is a factor of a 2k +1 + b 2k +1 for some positive integer k (Induction Hypothesis). Verify a + b is a factor of a 2k +3 + b 2k +3 . a 2k + 3 + b 2k + 3 = ( a 2k + 2 + b 2k + 2 )(a + b) − ab(a 2k +1 + b 2k +1 ) The sum of two multiples of a + b is a multiple of a + b. Thus, a + b is a factor of a 2 k + 3 + b 2 k + 3. By the Principle of Mathematical Induction, a + b is a factor of a 2 n +1 + b 2 n +1 for all positive integers.


788

29.


1.

a(1 − r1) 1− r Thus, the statement is true for n = 1. Let n = 1. ar1−1 = a ⋅ 1 = a = j

2.

Assume

−

= j +1

Verify

∑ k 1

ar k −1 =

=

j +1

∑

j

a (1 − r ) is true for some positive integer j. ar k 1 = ∑ 1− r k 1 a(1 − r j +1) is true for n = j + 1. 1− r

j

ar k −1 =

k =1

∑ (ar k =1

k +1

− 1) =

a (1 − r j ) + ar j 1− r

a (1 − r j ) + ar j (1 − r ) a[1 − r j + r j − r j +1 ] = = 1− r 1− r j +1 a (1 − r ) = 1− r n

By the Principle of Mathematical Induction,

ar k ∑ k

−1

=

=1

a (1 − r n ) 1− r

1

30.

1[(1 + 1)a + 2b] = a+b ( ak + b ) = a + b and ∑ 2 k =1

1.

Let n = 1.

2.

Therefore, the statement is true for n = 1. Assume the statement is true for n = i. i

i[(i + 1)a + 2b] ( ak + b ) and ∑ 2 k =1 i +1

Prove the statement is true for n = i +1. That is, prove

(i + 1)[(i + 2)a + 2b] ( ak + b ) = ∑ 2 k =1

i +1

i

( ak + b ) = ∑ ( ak + b ) + a ( i + 1) + b ∑ n k =1

=1

i[(i + 1)a + 2b] i[(i + 1)a + 2b] + [2a(i + 1) + 2b] + a (i + 1) + b = 2 2 i (i + 1)a + 2bi + 2a (i + 1) + 2b i (i + 1)a + 2b(i + 1) + 2a (i + 1) = = 2 2 (i + 1)[ai + 2a + 2b] (i + 1)[a (i + 2) + 2b] = = 2 2 Therefore, the statement is true for all positive integers n. =

....................................................... 31.

1. 2.

If N = 25, then log25! ≈ 25.19 > 25. Assume logk! > k for k > 25 (Induction Hypothesis). Prove log(k + 1)! > k + 1. log(k + 1)! = log[(k + 1)k!] = log(k + 1) + logk! > log(k + 1) + k Because k > 25, log(k + 1) > 1. Thus, log(k + 1)! > k + 1. Therefore, log n! > n for all n > 25.


Connecting Concepts

Section 11.4

32.

1.

789

Let

a n+1 < r for n ≥ N . We are to prove that a N + k < a N r k for each positive integer k . an

a When k = 1, we have N +1 < r. Thus a N +1 < a N r which is true by the condition of the sequence. aN Thus, the statement is true for k = 1. 2.

Assume that for some integer k, a N + k < a N r k (Induction Hypothesis). Prove a N + k +1 < a N r k +1. a N + k +1 < r by the Induction Hypothesis, we have a N =k

Since

a N + k +1 < ra N + k < r (a N r k ) = a N r k +1 Thus, the statement is true for all positive integers n. 33.

1.

When n = 1, we have ( x m )1 = x m and x m ⋅ 1 = x m . Therefore, the statement is true for n = 1.

2.

Assume the statement is true for n = k. That is, assume ( x m ) k = x mk (Induction Hypothesis). Prove the statement is true for n = k + 1. x m( k +1) = x mk + m = x mk ⋅ x m = ( x m ) k ⋅ x m = ( x m )k +1 Thus, the statement is true for all positive integers n and m.

34.

1.

When n = 1, 1

1 1 1 = + =1+1= 2 ∑ i ! 0 ! 1! i =0

1 3 − = 3 −1= 2 1 Thus, the statement is true for n = 1. k

2.

Assume the statement is true for n = k. That is, assume

1 1 ≤ 3 − (Induction Hypothesis). ∑ i! k i =0

Now prove the statement is true for n = k + 1. k +1

∑ i =0

1 = i!

Because

k

1 1 1 1 + ≤ 3− + ∑ i ! (k + 1)! k (k + 1)! i =0

1 1 1 1 1 1 1 ≤ ,3 − + ≤ 3− + = 3− . (k + 1)! k (k + 1) k (k + 1)! k k (k + 1) (k + 1)

k +1

Thus,

1 1 ≤3− . ∑ i! k +1 i =0

The statement is true for all positive integers n.


790

35.


1.

3

64 ⎛ 3 +1⎞ ⎛ 4 ⎞ When n = 3, we have ⎜ < 3. ⎟ =⎜ ⎟ = 27 ⎝ 3 ⎠ ⎝3⎠ Thus the statement is true for n = 3. k

2.

⎛ k +1⎞ Assume the statement is true for n = k. That is, assume ⎜ ⎟ < k (Induction Hypothesis). ⎝ k ⎠ ⎛k +2⎞ Prove the statement is true for n = k + 1. That is, prove ⎜ ⎟ ⎝ k +1 ⎠

k +1

< k + 1.

⎛ k + 2 ⎞ k +1 . Therefore We begin by noting that ⎜ ⎟< k ⎝ k +1 ⎠ ⎛k +2⎞ ⎜ ⎟ ⎝ k +1 ⎠

k +1

⎛ k +1⎞ (k + 1)k k By Induction Hypothesis k k ≥ k !. We have (k + 1)k k ≥ (k + 1)k ! = (k + 1)! Therefore (k + 1)k +1 ≥ (k + 1)! Therefore the statement is true for all integers n ≥ 1. 62.

1. 2.

For n = 9, 9! = 362,880, 49 = 262,144 [11.4] Since 362,880 > 262,144, the statement is true for n = 9. Assume the statement is true for n = k .

k ! > 4k Induction Hypothesis Prove the statement is true for n = k + 1. That is, prove (k + 1)! > 4k +1 (k + 1)! = (k + 1) k ! > (k + 1)4k By Induction Hypothesis Since k ≥ 9, k + 1 ≥ 4. Thus (k + 1)4k > 4 ⋅ 4k = 4k +1 Thus (k + 1)! > 4k +1 The statement is true for all integers n ≥ 9.


Chapter Review

63.

1. 2.

809

When n = 1, we have 13 + 2(1) = 3. Since 3 is a factor of 3, the statement is true for n = 1. [11.4] Assume the statement is true for n = k .

3 is a factor of k 3 + 2k Induction Hypothesis Prove the statement is true for n = k + 1. That is, prove 3 is a factor of (k + 1)3 + 2(k + 1).

(k + 1)3 + 2(k + 1) = k 3 + 3k 2 + 3k + 1 + 2k + 2 = ( k 3 + 2k ) + 3(k 2 + k + 1) By Induction Hypothesis, 3 is a factor of k 3 + 2k . Three is also a factor of 3(k 2 + k + 1). Thus 3 is a factor of (k + 1)3 + 2(k + 1). The statement is true for all positive integers n. 64.

1.

When n = 1, a1 = 2 < 2. The statement is true for n = 1. [11.4]

2.

Assume the statement is true for some integer k . ak < 2 Induction Hypothesis Prove the statement is true for n = k + 1. That is, prove ak +1 < 2. By the Induction Hypothesis, ak < 2. Thus ( 2) ak < ( 2)2 = 2 But ( 2)ak = ak +1. Thus ak +1 < 2 The statement is true for all positive integers n. 5

65.

(4a − b)5 =

∑ ⎛⎜⎝ 5i ⎞⎟⎠ (4a)

5−i

( −b )i

[11.5]

i =0

⎛5⎞ ⎛ 5⎞ ⎛5⎞ ⎛ 5⎞ ⎛5⎞ ⎛ 5⎞ = ⎜ ⎟ (4a )5 + ⎜ ⎟ (4a )4 (−b)1 + ⎜ ⎟ (4a )3 (−b)2 + ⎜ ⎟ (4a ) 2 (−b)3 + ⎜ ⎟ (4a )(−b)4 + ⎜ ⎟ (−b)5 ⎝0⎠ ⎝1 ⎠ ⎝ 2⎠ ⎝ 3⎠ ⎝ 4⎠ ⎝ 5⎠ = 1(1024a 5 ) + 5(−256a 4b) + 10(64a3b 2 ) + 10(−16a 2b3 ) + 5(4ab 4 ) + (−b5 ) = 1024a 5 − 1280a 4b + 640a3b 2 − 160a 2b3 + 20ab 4 − b5

66.

6

( x + 3 y )6 = ∑ ⎛⎜ i6 ⎞⎟ x6−i ( 3 y )i i=0 ⎝

[11.5]

⎠

⎛ 6⎞ ⎛ 6⎞ ⎛6⎞ ⎛ 6⎞ ⎛ 6⎞ ⎛ 6⎞ ⎛ 6⎞ = ⎜ ⎟ x6 + ⎜ ⎟ x5 (3 y ) + ⎜ ⎟ x 4 (3 y ) 2 + ⎜ ⎟ x3 (3 y )3 + ⎜ ⎟ x 2 (3 y ) 4 + ⎜ ⎟ x(3 y )5 + ⎜ ⎟ (3 y )6 ⎝ 0⎠ ⎝1 ⎠ ⎝ 2⎠ ⎝3⎠ ⎝ 4⎠ ⎝5⎠ ⎝ 6⎠ = x6 + 6 x5 ( 3 y ) +15 x 4 (9 y 2 ) + 20 x3 (27 y 3 ) +15 x 2 (81 y 4 ) + 6 x(243 y 5 ) +1(729 y 6 ) = x6 +18 x5 y +135 x 4 y 2 + 540 x3 y 3 +1215 x 2 y 4 +1458 xy 5 + 729 y 6

67.

8

( a + 2 b ) = ∑ ⎛⎜⎜ 8i ⎞⎟⎟( a )8−i (2 b )i ⎝ ⎠ 8

[11.5]

i =0

⎛8 ⎞ ⎛8⎞ ⎛8 ⎞ ⎛8⎞ ⎛8 ⎞ ⎛8⎞ = ⎜ ⎟ ( a )8 + ⎜ ⎟ ( a )7 (2 b ) + ⎜ ⎟ ( a )6 (2 b )2 + ⎜ ⎟ ( a )5 (2 b )3 + ⎜ ⎟ ( a )4 (2 b )4 + ⎜ ⎟ ( a )3 (2 b )5 ⎝0⎠ ⎝1 ⎠ ⎝ 2⎠ ⎝ 3⎠ ⎝ 4⎠ ⎝ 5⎠ ⎛8 ⎞ ⎛8 ⎞ ⎛8⎞ + ⎜ ⎟ ( a )2 (2 b )6 + ⎜ ⎟ ( a )(2 b )7 + ⎜ ⎟ (2 b )8 ⎝ 6⎠ ⎝7⎠ ⎝8⎠ = 1(a 4 ) + 8a 7 / 2 (2b1/ 2 ) + 28a3 (4b) + 56a5 / 2 (8b3 / 2 ) + 70a 2 (16b 2 ) + 56a3 / 2 (32b5 / 2 ) + 28a(64b3 ) + 8a1/ 2 (128b7 / 2 ) + 1(256b 4 ) = a 4 + 16a 7 / 2b1/ 2 + 112a3b + 448a5 / 2b3 / 2 + 1120a 2b 2 + 1792a3 / 2b5 / 2 + 1792ab3 + 1024a1/ 2b7 / 2 + 256b 4


810

68.


1 ⎞ ⎛ ⎟ = ⎜ 2x − 2x ⎠ ⎝

7

∑ i =0

i

⎛7⎞ ⎛ 1 ⎞ ⎜⎜ ⎟⎟(2 x )7 −i ⎜ − ⎟ [11.5] i ⎝ 2x ⎠ ⎝ ⎠ 2

3

4

⎛7⎞ ⎛7⎞ ⎛7⎞ ⎛7⎞ ⎛7⎞ ⎛7⎞ = ⎜ ⎟ (2 x)7 + ⎜ ⎟ (2 x)6 ⎛⎜ − 1 ⎞⎟ + ⎜ ⎟ (2 x)5 ⎛⎜ − 1 ⎞⎟ + ⎜ ⎟ (2 x) 4 ⎛⎜ − 1 ⎞⎟ + ⎜ ⎟ (2 x)3 ⎛⎜ − 1 ⎞⎟ + ⎜ ⎟ (2 x) 2 ⎛⎜ − 1 ⎞⎟ ⎝ 2x ⎠ ⎝ 2 ⎠ ⎝ 2x ⎠ ⎝ 3 ⎠ ⎝ 2x ⎠ ⎝ 4⎠ ⎝ 2x ⎠ ⎝ 5 ⎠ ⎝ 2x ⎠ ⎝0⎠ ⎝1 ⎠ 6 7 7 7 ⎛ ⎞ ⎛ ⎞ + ⎜ ⎟ (2 x) ⎛⎜ − 1 ⎞⎟ + ⎜ ⎟ ⎛⎜ − 1 ⎞⎟ ⎝ 2x ⎠ ⎝ 7 ⎠ ⎝ 2x ⎠ ⎝6⎠ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = 1(128 x7 ) − 7(64 x6 ) ⎛⎜ 1 ⎞⎟ + 21(32 x5 ) ⎜ 1 2 ⎟ − 35(16 x 4 ) ⎜ 13 ⎟ + 35(8 x3 ) ⎜ 1 4 ⎟ − 21(4 x 2 ) ⎜ 1 5 ⎟ ⎝ 2x ⎠ ⎝ 4x ⎠ ⎝ 8x ⎠ ⎝ 16 x ⎠ ⎝ 32 x ⎠ ⎛ 1 ⎞ ⎛ 1 ⎞ + 7(2 x) ⎜ ⎟ − 1⎜ ⎟ ⎝ 64 x6 ⎠ ⎝ 128 x7 ⎠ 7 5 3 = 128 x − 224 x + 168 x − 70 x + 35 − 213 + 7 5 − 1 7 2x 8x 32 x 128 x 69.

The fifth term of (3 x − 4 y )7 is

⎛7⎞ 3 4 ⎜⎜ 4 ⎟⎟ (3 x) ( −4 y ) ⎝ ⎠

70.

The eighth term of (1 − 3x)9 is

⎛9⎞ 2 7 ⎜⎜ ⎟⎟ (1) (−3 x ) ⎝7⎠

71.

There are 26 choices for each letter. By the Fundamental Counting Principle, there are 268 possible passwords. [11.6]

72.

Using the Fundamental Counting Principle, we have 106 ⋅ 26 possible serial numbers. [11.6]

73.

This is a permutation with n = 15 and r = 3. [11.6] 15! 15! P (15,3) = = = 2730 (15 − 3)! 12!

74.

There are ⎛⎜

4⎞ ⎟ ⎝1 ⎠

⎛ 4⎞ ⎜⎜ ⎟⎟ ⎝1 ⎠

75.

⎛12 ⎞ ⎜⎜ ⎟⎟ ⎝3 ⎠

5

= 35(27 x3 )(256 y 4 ) = 241,920 x3 y 4 . [11.5]

= 36 ⋅ 1 ⋅ (−2187 x7 ) = −78,732 x7 . [11.5]

ways to choose a supervisor and

⎛12 ⎞ ⎜ ⎟ ⎝3 ⎠

ways to choose 3 regular employees. Thus, there are ⎛⎜

4 ⎞ ⎛12 ⎞ ⎟ ⎜ ⎟ ways ⎝1 ⎠ ⎝ 3 ⎠

to do both.

= 4 ⋅ 220 = 880 shifts have 1 supervisor. [11.6]

This problem is solved in stages. First, there are ⎛⎜

10 ⎞ ⎟ ⎝5 ⎠

together. Second, there are ⎛⎜

10 ⎞⎛ 2 ⎞ ⎟⎜ ⎟ ⎝ 4 ⎠⎝1 ⎠

Altogether there are

⎛10 ⎞ ⎛10 ⎞⎛ 2 ⎞ ⎜ ⎟ + ⎜ ⎟⎜ ⎟ ⎝ 5 ⎠ ⎝ 4 ⎠⎝1 ⎠

ways to choose a committee excluding both people who refuse to serve

ways to choose a committee that includes one person but not the other.

ways to choose the committee.

⎛10 ⎞ ⎛10 ⎞⎛ 2 ⎞ ⎜ ⎟ + ⎜ ⎟⎜ ⎟ = 252 + 210 ( 2 ) = 672 possible committees [11.6] ⎝ 5 ⎠ ⎝ 4 ⎠⎝1 ⎠

76.

⎛10 ⎞ ⎟⎟ ⎝4 ⎠

There are ⎜⎜

= 210 ways of choosing 4 calculators from 10. If the inspector is to choose 1 defective calculator, then 3 nondefective ⎛ 2⎞ ⎛ 8⎞ ⎟⎟ ⎜⎜ ⎟⎟ ⎝1 ⎠ ⎝ 3 ⎠

calculators must also be chosen. There are ⎜⎜ 112 210

77.

=

8 . 15

= 2 ( 56 ) = 112 ways to accomplish that. Therefore, the probability of the event is

[11.7]

The probability is 1 ⋅ 1 ⋅ 1 = 1 . 2 2 2 8 The probability of one tail and therefore two heads is 3 ⎜⎛ 1 ⋅ 1 ⋅ 1 ⎟⎞ = 3 . [11.4] ⎝2 2 2⎠ 8


Chapter Review

78.

811 ⎛10 ⎞ ⎟⎟ ⎝4 ⎠

There are ⎜⎜

⎛5⎞⎛5⎞ ⎟⎟ ⎜⎜ ⎟⎟ ⎝ 2⎠ ⎝ 2⎠

ways to draw 4 cards from 10. There are ⎜⎜

ways to draw 2 red and 2 black cards The probability of drawing 2 red

⎛ 5 ⎞⎛ 5 ⎞ ⎜ ⎟⎜ ⎟ 2 2 10 ⋅ 10 10 = . [11.7] and 2 black cards is ⎝ ⎠⎝ ⎠ = 210 21 ⎛10 ⎞ ⎜ ⎟ 4 ⎝ ⎠ 79.

We look at the possibility for each case. If the middle digit is zero, there are If the middle digit is one, there is If the middle digit is two, there are If the middle digit is three, there are If the middle digit is four, there are If the middle digit is five, there are If the middle digit is six, there are If the middle digit is seven, there are If the middle digit is eight, there are If the middle digit is nine, there are Total The probability is

80.

0 numbers 1 number 4 numbers 9 numbers 16 numbers 25 numbers 36 numbers 49 numbers 64 numbers 81 numbers 285 numbers

285 = 0.285. [11.7] 1000

The probability that the sum of two numbers is 9 when the numbers are selected with replacement from 1, 2, 3, 4, 5, 6 is The probability that the sum is 7 is

6 36

= 1 . Therefore the probability that it is not 7 and not 9 is 6

1 − ⎜⎛ 1 + 1 ⎟⎞ = 1 − 10 = 13 . 36 18 ⎝9 6⎠ First selection, sum is 9. Second selection, probability of 9 is 1 .

9 13 ⋅ 1 . 18 9 probability is 13 ⋅ 13 ⋅ 1 . 18 18 9

Third selection, probability is Fourth selection,

M

The total probability is the sum of this infinite process. 2

1 + 13 ⋅ 1 + ⎛ 13 ⎞ ⎛ 1 ⎞ + ⋅ ⋅ ⋅ ⎜ ⎟ ⎜ ⎟ 9 18 9 ⎝ 18 ⎠ ⎝ 9 ⎠ This is a geometric series with a1 = 1 , r = 13 . S=

1 9

1 − 13 18

=

1 9 5 18

9

=2 5

The probability is 2 . 5

81.

18

[11.7]

The probability of drawing an ace and a 10 card from one regular deck of playing cards is ⎛ 4 ⎞ ⎛ 16 ⎞ ⎜1 ⎟ ⎜1 ⎟ ⎝ ⎠ ⎝ ⎠ = 4 ⋅ 16 ≈ 0.0483. 52 ⋅ 51 ⎛ 52 ⎞ ⎜2 ⎟ 2 ⎝ ⎠ The probability of drawing an ace and a 10 card from two regular decks of playing cards is ⎛ 8 ⎞ ⎛ 32 ⎞ ⎜1 ⎟ ⎜1 ⎟ ⎝ ⎠ ⎝ ⎠ = 8 ⋅ 32 ≈ 0.0478. 104 ⋅ 103 ⎛ 104 ⎞ ⎜2 ⎟ 2 ⎝ ⎠ Drawing an ace and a 10 card from one deck has the greater probability. [11.7]


4 36

= 1. 9

812

82.

83.

84.


Probability = (probability of 2)(probability of 1) [11.7] + (probability of 3)(probability of 1 or 2) + (probability of 4)(probability of 1 or 2 or 3) + (probability of 5) 1 = ⋅1 +1⋅2+1⋅3+1⋅4 5 4 5 4 5 4 5 4 = 1 + 2 + 3 + 4 = 10 = 1 20 20 20 20 20 2 ⎛12 ⎞ ⎛11⎞ There are ⎜ ⎟ ways of choosing 3 people from 12. There are ⎜ ⎟ ⋅ 1 ways of choosing 2 people and the person with [11.7] 3 ⎝ ⎠ ⎝2 ⎠ badge number 6. ⎛11⎞ 11 ⋅ 10 ⎜⎜ ⎟⎟ ⋅ 1 2⎠ 1 ⎝ 2 Probability = = = 12 ⋅ 11 ⋅ 10 4 ⎛12 ⎞ ⎜⎜ ⎟⎟ 3⋅ 2 ⎝3 ⎠ Stock value =

D (1 + g ) 1.27(1 + 0.03) = ≈ $14.53 [11.3] 0.12 − 0.03 i−g

85.

Using the multiplier effect, [11.3] 15 = 75 1 − 0.80 The net effect of $15 million is $75 million.

....................................................... i ⎤, i = r QR1. PMT = L ⎡ ⎢ 1 − (1 + i ) − nt ⎥ n ⎣ ⎦ 0.09 ⎡ ⎤ ⎢ ⎥ 12 = 12,000 ⎢ −12(5) ⎥ 0.09 ⎢1 − 1 + ⎥ 12 ⎣ ⎦ ≈ $249.10

(

)

Quantitative Reasoning i ⎤, i = r QR2. PMT = L ⎡ ⎢ 1 − (1 + i ) − nt ⎥ n ⎣ ⎦ 0.06 ⎡ ⎤ ⎢ ⎥ 12 = 18,000 ⎢ −12(4) ⎥ 0.06 ⎢1 − 1 + ⎥ 12 ⎣ ⎦ ≈ $422.73

(

)

i ⎤, i = r QR3. PMT = L ⎡ ⎢ 1 − (1 + i ) − nt ⎥ n ⎣ ⎦ 0.085 ⎡ ⎤ ⎢ ⎥ 12 = 15,000 ⎢ −12(5) ⎥ 0.085 ⎢1 − 1 + ⎥ 12 ⎣ ⎦ ≈ $307.75 Total payments made over 12 years: 307.75(12)(5) = 18,465 Total interest paid 18,465 − 15,000 = $3465 .

(

)

....................................................... 1.

3 a3 = 2 = 8 = 4 3! 6 3 5 a5 = 2 = 32 = 4 5! 120 15

[11.1]

Chapter Test 2.

a2 = 2 ⋅ a1 = 2 ⋅ 3 = 6 [11.1]

a3 = 2 ⋅ a2 = 12 • a4 = 2 ⋅ a3 = 24 a5 = 2 ⋅ a4 = 48 • Copyright © Houghton Mifflin Company. All rights reserved.

Chapter Test

3.

813

an +1 − an = [−2(n + 1) + 3] − (−2n + 3) [11.3]

an +1 − an = 2(n + 1)2 − 2n 2

4.

= −2n − 2 + 3 + 2n − 3 = −2 = constant

= 4n + 2 ≠ constant an +1 2(n + 1)2 = an 2n 2 2 1 = 1+ + ≠ constant n n2 neither

arithmetic

5.

( −1)n +1−1 n +1 an +1 −1 = 3 = = constant [11.3] an 3 ( −1)n −1 3n

geometric 6.

6

120 + 60 + 40 + 30 + 24 + 20 ∑ 1i = 1 + 12 + 13 + 14 + 15 + 16 = 120 120 120 120 120 120

[11.1]

i =1

= 294 = 49 120 20

7.

8.

()

1 ⎛1 − 1 ⎜⎜ 2 1 = 1 + 1 + 1 + 1 +L 1 = 2 ⎝ ∑ 2 j 2 4 8 16 1024 1− 1 j =1 2 1 1023 =1− = 1024 1024 10

10 ⎞

⎟⎟ 10 ⎠ =1− ⎛ 1 ⎞ ⎜ ⎟ ⎝ 2 ⎠ [11.3]

20

(1 + 58) = 10 ( 59 ) ∑ ( 3k − 2 ) = 1 + 4 + 7 + 10 + L + 58 = 20 2

k =1

[11.2]

= 590

9.

a3 = a1 + ( 3 − 1) d = 7,

a1 + 2d = 7

a8 = a1 + ( 8 − 1) d = 22

a1 + 7d = 22 − 5d = −15 d =3

10.

3 3 k ⎛ 3 ⎞ = a1 = 8 = 8 = 3 [11.3] ∑⎜ ⎟ 1− r 5 5 k =1 ⎝ 8 ⎠ 1 − ⎛⎜ 3 ⎞⎟ ⎝8⎠ 8

11.

0.15 = 0.15 + 0.0015 + 0.000015 + ... =

a1 = a3 − 2 ( 3) =7−6 =1

a20 = a1 + ( 20 − 1) d [11.2] = 1 + (19 )( 3) = 58

∞

0.15 0.15 15 5 [11.3] = = = 1 − 0.01 0.99 99 33


[11.1]

814

12.


1.

1(1 − 3(1)) = −1 [11.4] 2 Thus the statement is true for n = 1. Let n = 1. 2 − 3(1) = −1 k

2.

Assume

∑( 2 − 3i ) =

k (1 − 3k ) 2

i =1 k +1

Verify

is true for some positive number k .

∑ (2 − 3i) = (k + 1)[1 −2 3(k + 1)] = (k + 1)(12− 3k − 3) = (k + 1)(−23k − 2) = − (k + 1)(32 k + 2) i =1

k (1 − 3k ) k (1 − 3k ) + [2 − 3(k + 1)] = + (−3k − 1) 2 2 =

k − 3k 2 − 6k − 2 2

(3k 2 + 5k + 2) 2 (k + 1)(3k + 2) =− 2 Thus the formula has been established by the extended principle of mathematical induction. =−

13.

1.

Let n = 7 [11.4] 7!= 50,407 37 = 2187

Thus n!> 3n for n = 7. 2.

Assume k!> 3k Verify (k + 1)! > 3k +1 k ! > 3k k +1 > 3 (k + 1)k ! > 3 ⋅ 3k (k + 1)! > 3k +1 Thus the formula has been established by the extended principle of mathematical induction.

14.

( x − 2 y )5 = x5 − 5( x)4 (2 y ) + 10( x)3 (2 y )2 − 10( x)2 (2 y )3 + 5( x)(2 y )4 − (2 y )5 = x5 − 10 x 4 y + 40 x3 y 2 − 80 x 2 y3 + 80 xy 4 − 32 y5

15.

(a + b)6 = a 6 + 6a5b + 15a 4b 2 + 20a3b3 + 15a 2b 4 + 6ab5 + b6 [11.5] 6

2

1⎞ ⎛ 6 5⎛ 1 ⎞ 4⎛1⎞ ⎜ x + ⎟ = x + 6( x) ⎜ ⎟ + 15( x) ⎜ ⎟ x x ⎝ ⎠ ⎝ ⎠ ⎝ x⎠ 15 = x6 + 6 x 4 + 15 x 2 + 20 + + x2 16.

18.

[11.5]

3

4

5

⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ + 20( x)3 ⎜ ⎟ + 15( x)2 ⎜ ⎟ + 6 x ⎜ ⎟ + ⎜ ⎟ x x ⎝ ⎠ ⎝ ⎠ ⎝ x⎠ ⎝ x⎠ 6 1 + x 4 x6

⎛ 8 ⎞ 3 6 −1 [11.5] 6th term of (3 x + 2 y )8 = ⎜ ⎟ (3 x) (2 y ) ⎝ 6 − 1⎠ ⎛8⎞ = ⎜ ⎟ (3x)3 (2 y )5 ⎝ 5⎠ = 56 ⋅ 27 x3 ⋅ 32 y 5 = 48,384 x3 y 5 26 ⋅ 25 ⋅ 24 ⋅ 9 ⋅ 8 ⋅ 23 ⋅ 22 = 568,339,200 [11.6]

6

17.

52 ⋅ 51 ⋅ 50 = 132,600 [11.6]

19.

C (8, 3)C (10, 2) 56 ⋅ 45 5 = = ≈ 0.294118 [11.7] C (18, 5) 8568 17


Cumulative Review

20.

Stock value =

815

D (1 + g ) 0.86(1 + 0.06) = ≈ $10.13 [11.3] i−g 0.15 − 0.06

....................................................... 1.

Cumulative Review 2.

5+

x = −3 [2.3] 3 x = −8 3 x = −24

y = 1.7x + 3.6 [2.7] 3.

2 x 2 − 3 x = 4 [1.3] 2 x − 3x − 4 = 0

4.

⎛ xy 2 ⎞ logb ⎜ 3 ⎟ = logb x + logb y 2 − logb z 3 [4.4] ⎜ z ⎟ ⎝ ⎠ = logb x + 2logb y − 3logb z

6.

(1) ⎧2 x − 3 y = 8 ⎨ ⎩ x + 4 y = −7 (2)

2

3 ± (−3) 2 − 4(2)(−4) 3 ± 9 + 32 = 2(2) 4 ± 3 41 = 4

x=

5.

16 x 2 + 25 y 2 − 96 x + 100 y − 156 = 0 16( x 2 − 6 x) + 25( y 2 + 4 y ) = 156 16( x − 3) 2 + 25( y + 2) 2 = 156 + 144 + 100 16( x − 3) 2 + 25( y + 2) 2 = 400 16( x − 3) 2 25( y + 2) 2 + =1 400 400 ( x − 3)2 ( y + 2) 2 + =1 25 16

Solve (2) for x and substitute into (1). x = −4 y − 7 2(−4 y − 7) − 3 y = 8 −8 y − 14 − 3 y = 8 −11y = 22 y = −2 x = −4(−2) − 7 = 1 The solution is (1, –2). [9.1]

a 2 = 25, b 2 = 16 c 2 = a 2 − b 2 = 25 − 16 = 9 c=3 e = c = 3 [5.2] a 5

7.

⎡ −1 3 A − 2 B = 3 ⎢⎢ 5 ⎢⎣ 0 ⎡ −3 = ⎢⎢15 ⎣⎢ 0

9.

y = 0 [3.5]

2⎤ ⎡7 −3⎤ ⎥ 3 ⎥ − 2 ⎢⎢ 6 5 ⎥⎥ [10.2] ⎢⎣1 −2 ⎥⎦ 3 ⎥⎦ 6 ⎤ ⎡14 −6 ⎤ ⎡ −17 12 ⎤ 9 ⎥⎥ − ⎢⎢12 10 ⎥⎥ = ⎢⎢ 3 −1⎥⎥ 9 ⎦⎥ ⎣⎢ 2 −4 ⎦⎥ ⎣⎢ −2 13 ⎥⎦

8.

⎛h⎞ h(−3) −3 − 2 = = −5 [2.6] ⎜ ⎟ (−3) = g (−3) (−3)2 − (−3) + 4 16 ⎝g⎠

10.

log1/ 2 64 = −6 [4.3]


816

11.


42 x +1 = 3x − 2 [4.5] ln 42 x +1 = ln 3x − 2 (2 x + 1)ln 4 = ( x − 2)ln 3 2 x ln 4 + ln 4 = x ln 3 − 2ln 3 x ln 42 − x ln 3 = −2ln 3 − ln 4 x (ln16 − ln 3) = −2ln 3 − ln 4 x = −2ln 3 − ln 4 ≈ −2.1 ln16 − ln 3

12.

⎧⎪ x 2 + y 2 + xy = 10 (1) ⎨ x − y = 1 (2) ⎪⎩ Solve (2) for x and substitute into (1). x = y +1 ( y + 1)2 + y 2 + ( y + 1) y = 10 y + 2 y + 1 + y 2 + y 2 + y = 10 3y2 + 3y − 9 = 0 3( y 2 + y − 3) = 0 2

y=

−1 ± 12 − 4(1)( −3) −1 ± 1 + 12 −1 ± 13 = = 2(1) 2 2

x = −1 ± 13 + 1 = 1 ± 13 2 2 ⎛ 1 + 13 −1 + 13 ⎞ The solutions are ⎜ , ⎟ and 2 ⎝ 2 ⎠

⎛ 1 − 13 −1 − 13 ⎞ , ⎜ ⎟ . [9.3] 2 ⎝ 2 ⎠ 13.

⎡3 2⎤ ⎡ 2 9 11 −3⎤ ⎢ −2 1 ⎥ ⎡ 2 3 1 1 ⎤ = ⎢ −6 −6 2 −5⎥ [10.2] ⎢ ⎥ ⎢ −2 0 4 −3⎥ ⎢ ⎥ ⎦ ⎢10 3 −15 13 ⎥ ⎢⎣ 1 −4 ⎥⎦ ⎣ ⎣ ⎦

14.

Let t = 5, [4.5] 5 = − 175 ln ⎛⎜1 − v ⎞⎟ 32 ⎝ 175 ⎠ ⎛ ⎞ ⎛ v ⎞ 32 5⎜ − ⎟ = ln ⎜ 1 − ⎟ ⎝ 175 ⎠ ⎝ 175 ⎠ −32 / 35 v e =1− 175 −175 e −32 / 35 − 1 = v

(

15.

17.

)

v ≈ 105 mph

opp −1 = [5.3] hyp 2 hyp = 2 secθ = 1 = 2 3 = cos θ 3 adj 3 adj = 3 =− 3 cot θ = 1 = tan θ opp −1

16.

sin x + 1 + cos x = sin x(1 − cos x) + 1 + cos x 1 + cos x sin x sin x 1 − cos 2 x sin x(1 − cos x) 1 + cos x = + sin x sin 2 x 1 cos 1 cos − x x = + + sin x sin x sin x = 1 + 1 sin x sin x = 2csc x

C = 180o − 40o − 65o = 75o [7.1] a = c sin A sin C o a = 20sin 40 ≈ 13 cm o sin 75 b = c sin B sin C o b = 20sin 65 ≈ 19 cm o sin 75

18.

A = 9, B = 4, C = 6 [8.4]

sin θ =

cot 2α = A − C = 9 − 6 = 3 B 4 4 2α ≈ 54o

α ≈ 27o


[6.1]

Cumulative Review

19.

(

sin 1 cos −1 4 2 5 Let

817

)

[6.5]

θ = cos −1 4 cosθ = 4 5

20.

2 v − 3w = 2(2i + 5 j) − 3(3i − 6 j) = 4i + 10 j − 9i + 18 j = −5i + 28 j

5

1− 4 5 sin 1 θ = 1 − cos θ = 2 2 2 = 5−4 = 1 10 10 10 = 10


[7.3]

Responses to Projects in the Text

819

Chapter P

College Algebra and Trigonometry

Responses to Projects

Preliminary Concepts CHAPTER P of College Algebra and Trigonometry P.1 The Real Number System 1. Number Puzzle n is 1 less than a multiple of 6 and also 1 less than a multiple of 5 and also 1 less than a multiple of 4 and also 1 less than a multiple of 3 and also 1 less than a multiple of 2. The smallest value of n is 1 less than the least common multiple of 6, 5, 4, 3, and 2. Thus n is 1 less than 60, or n = 59. 2. Operations on Intervals a. (−4)2 = 16, 02 = 0, 22 = 4 . The square of every number in the interval (−4, 2) is (−4, 2)2 = [0, 16) . b. |4| = 4, |0| = 0, |5| = 5. The absolute value of every number in the interval (−4, 5) is ABS(−4, 5) = [0, 5). c. 0 = 0, 9 = 3 . The square root of every number in the interval (0, 9) is (0,9) = (0,3) d. The reciprocal of every number in the interval (0, 1) is (1,∞ ) . 3.

Factors of a Number Whenever you square a natural number, there is a repeated factor, so in listing the pairs of factors, that one would repeat, therefore the number of factors is an odd number. For example, squaring 6 gives us 36. The factors are 1 × 36, 2 × 18, 3 × 12,4 × 9, and 6 × 6. We do not count the repeated 6 twice, so there are an odd number of factors.

P.2 Integer and Rational Number Exponents ⎡ ⎤ ⎡ ⎤ 1 1 2 ⎢ ⎥ − 1 mv 2 ⎥ − 1 v2 1 1 mc 2 ⎢ − c − 2 2 ⎢ 1 − v2 ⎥ 2 ⎢ 1 − v2 ⎥ 2 c c ⎣ ⎦ ⎣ ⎦ × 100 = × 100 1. Relativity Theory % error = ⎡ ⎤ ⎡ ⎤ 1 1 − 1⎥ − 1⎥ mc 2 ⎢ c2 ⎢ ⎢ 1 − v22 ⎥ ⎢ 1 − v22 ⎥ c c ⎣ ⎦ ⎣ ⎦ a. For v = 30 meters per second, % error = 0.000303176 b. For v = 240 meters per second, % error = 1.75483 × 10−6 c. For v = 3 × 107 meters per second, % error = 0.750628 d. For v = 1.5 × 108 meters per second, % error = 19.1987 e. For v = 2.7 × 108 meters per second, % error = 68.0755 f. The percent errors is very small for everyday speeds. |m −m| m0 × 100 where m = and v = 0.99 c . % change is approximately 609%. g. % change = 0 1− v 2 m0 2 c

h. As the speed of the object approaches the speed of light, the denominator of kinetic energy equation approaches 0, which implies that the kinetic energy is approaching infinity. Thus it would require an infinite amount of energy to move a particle at the speed of light. P.3 Polynomials 1. Odd Numbers An even number is a number that is a multiple of 2. Let m and n be natural numbers. Then 2m and 2n are even natural numbers. The product ( 2m )( 2n ) = 4mn = 2 ( 2mn ) , which is an even number (it is a multiple of 2). Therefore, an even number times an even number is an even number. An odd number is a number that is not a multiple of 2. Let m and n be natural numbers. Then 2m + 1 and 2n + 1 are odd numbers because they are not multiples of 2. The product ( 2m + 1)( 2n + 1) = 4mn + 2 ( m + n ) + 1 is not a multiple of 2 (2 is not a common factor) and is therefore an odd number. Therefore, the product of two odd numbers is an odd number. Consider the product 2m ( 2n + 1) = 4mn + 2m = 2 ( 2mn + m ) . Because the product is a multiple of 2, it is an even number. Thus the product of an odd number and an even number is an even number.

820


Chapter P 2.



Prime Numbers a. The contrapositive of “If A, then B” is “If not B, then not A.” The contrapositive of “If two triangles are congruent, then they are similar” is “If two triangles are not similar, then they are not congruent.” b. Fermat’s Little Theorem does not say anything about n when n is not a prime number. c. The converse of “If A, then B” is “If B, then A.” The converse of Fermat’s Little Theorem is “If a is any natural number and a n − a is divisible by n, then n is a prime number.” d. No. In fact, 561 is not a prime number. The converse of Fermat’s Little Theorem is false. e. No. See part (d). f. A Carmichael number n is a number for which a n − a is divisible by every natural number a and n is not prime. The first three Carmichael numbers are 561 , 1105 , and 1729 . Any Carmichael number can be used to show that the converse of Fermat’s Little Theorem is false. Carmichael numbers are sometimes called pseudoprimes.

P.4 Factoring 1. Geometry a. I + II + III b. II + III + V c. Because the area of I is the same as the area of V, the sum of the areas of region I, II, and III equals the sum of the areas of regions II, III, and V. 2. Geometry

( x + y) 3.

2

= x 2 + 2 xy + y 2

Geometry x 3 − y 3 = x ( x − y ) + xy ( x − y ) + xy ( x − y ) + y 2 ( x − y ) 2

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

P.5 Rational Expressions 1. Continued Fractions 1 1 1 = = = 0.6 a. C2 = 1 + 1+11 1 + 12 1.5 b.

c.

C3 =

C5 =

1 1+

1 1 + 1+11

1 1 + 12

1+

=

1 1 3 = 5 = = 0.6 2 1+ 3 3 5 1

=

1 1 1 1+ 1+1

1

1+ 1+

1

=

1

1+

1 1+

2.

1+

1 1+

d.

1

=

1 1 + 12

1

1+ 1+

1 1 + 23

=

1 1 1+ 1 + 35

=

1 8 = 5 1 + 8 13

8 −1 + 5 ≈ 0.615 ≈ 0.618 , C5 = 13 2

Representation of π An excellent source for π and its history can be found in an article by Dario Castellanos in Mathematics Magazine, vol. 61, no. 2 (April 1988). Here are two results (both attributable to Euler) from that article. 2 ⎛ ⎞ ⎜1 + ⎟ 1⋅ 3 12 + 3 ⎜ ⎟ = and π 2 π = 3+ 3⋅5 ⎟ 32 ⎜ + 4 6+ 2 7 ⎟ ⎜ 4 + 45+⋅L 6 + 65+L ⎝ ⎠ Copyright © Houghton Mifflin Company. All rights reserved.

821

Chapter P



P.6 Complex Numbers 1. – 8.

9.

2 + 5i = 22 + 52 = 4 + 25 = 29

10. 4 − 3i = 42 + (−3)2 = 16 + 9 = 25 =5

11. −2 + 6i = (−2)2 + 62 = 4 + 36 = 40 = 4 ⋅10 = 2 10

12. −3 − 5i = (−3)2 + (−5)2 = 9 + 25 = 34 13.

a + bi = (a )2 + (b)2 = a 2 + b 2

−a − bi = (− a) 2 + (−b)2 = a 2 + b 2 14. A complex number and its additive inverse are the same distance from the origin in the complex plane. The real parts of a complex number and its additive inverse are the same distance from the imaginary axis but on opposite sides of the imaginary axis. The imaginary parts of a complex number and its additive inverse are the same distance from the real axis but on opposite sides of the real axis.

822


Chapter 1



Equations and Inequalities CHAPTER 1 of College Algebra and Trigonometry 1.1 Linear Equations 1. Perfect Games a. The probability that one of the teams will get a perfect game is 0.7 27 . Because every game involves exactly two teams, the probability of a perfect game in any one game is 2 ( 0.7 27 ) . Therefore, the number of perfect games we can expect in x games is p = 2 ( 0.7 27 ) x.

b. A baseball almanac shows that during the years 1962 to 2002, there were 10 perfect games. It is difficult to determine the exact number of games played in this period, but it is about 71,000. c. From part (a), we should expect the number of perfect games during the period 1962 to 2002 to be p = 2 ( 0.7 27 ) 71, 000 ≈ 9.3 . This result is very close to the actual result found in part (b). 1.2 Formulas and Applications 1. A Work Problem and Its Extensions a.

1 . A 1 If a pump can fill a pool in B hours, then the part of the pool it fills every hour is . B Let T be the total time it takes the pumps to fill the pool when they both work together. T ( 1A ) = the part of the pool filled by pump A. T ( B1 ) = the part of the pool filled by pump B. Because

If a pump can fill a pool in A hours, then the part of the pool it fills every hour is

T ⎛⎜ 1 ⎞⎟ + T ⎛⎜ 1 ⎞⎟ = 1 ⎝ A⎠ ⎝B⎠ ⎛ ⎞ ⎛ ⎞ 1 1 T ⎜ ⎟ AB + T ⎜ ⎟ AB = 1 ⋅ AB ⎝ A⎠ ⎝B⎠ TB + TA = AB T ( B + A ) = AB T = AB A+ B

we fill exactly 1 pool, we have

b. Using the procedure in part (a) yields T =

c.

2.

ABC AB + AC + BC

Observe that T is the product of the individual times divided by the sum of the products of the times taken two at a time. A1 A2 A3 L An T= L + L A A A A A A A A ( 2 3 4 n ) ( 1 3 4 n ) + ( A1 A2 A4 L An )L + ( A1 A2 A3 L An −1 )

That is, T is given by the product of the A’s divided by the sum of products of the A’s taken (n – 1) at a time. d. One method is as follows: Use the alignment chart to determine that together the pump that can fill it in 6 hours working with the pump that can fill it in 12 hours would take a total of 4 hours. Think of these two pumps as one 4-hour pump. Now use the alignment chart again, using 4 hours and 6 hours as the individual times for two pumps to produce the final answer of 2.7 hours (nearest 0.1 hour). Resistance of Parallel Circuits In electronics it can be shown that if two resistors (one with resistance R1 ohms and the other with resistance R2 ohms) are placed in parallel, the total resistance R provided by the two resistors is R=

R1R2 R1 + R2

This formula has the same form as the formula give in part 1(a). Although the setting is different, the mathematics needed to solve a parallel resistance problem is exactly the same as that used to solve a combined work problem. The parallel resistance problem can also be extended to consider more than two resistors, in a manner analogous to the work problems in 1(b) and 1(c).


823

Chapter 1



1.3 Quadratic Equations 1. The Sum and Product of the Roots Theorem The quadratic equation ax 2 + bx + c = 0 has roots r1 =

−b + b 2 − 4ac −b − b 2 − 4ac and r2 = 2a 2a

The sum of the roots is r1 + r2 =

b −b + b 2 − 4ac −b − b 2 − 4ac −2b + = =− 2a 2z 2a a

The product of the roots is ⎛ −b + b 2 − 4ac ⎞ ⎛ −b − b 2 − 4ac ⎞ b 2 − ( b 2 − 4ac ) b 2 − b 2 + 4ac c ⎟⎜ ⎟= = = r1r2 = ⎜ ⎜ ⎟⎜ ⎟ 2a 2a 4a 2 4a 2 a ⎝ ⎠⎝ ⎠ Visual Insight The reasons for the steps in President Garfield’s proof of the Pythagorean Theorem are as follows: i) The area of the large triangle plus the area of the two small triangles is equal to the total area of the region, which is a trapezoid. ii) The height of the trapezoid is a + c. The sum of the bases of the trapezoid is also a + b. iii) Expand the right side and subtract ab from each side of the equation. iv) Multiply each side of the equation by 2. 2

2.

1.4 Other Types of Equations 1. The Reduced Cubic a.

Given the reduced cubic x 3 + mx + n = 0. Let x =

m − z. 3z

3

⎛m ⎞ ⎛m ⎞ ⎜ − z ⎟ + m⎜ − z ⎟ + n = 0 ⎝ 3z ⎠ ⎝ 3z ⎠ 3 2 2 4 6 2 2 m − 9m z + 27 mz − 27 z m − 3mz + +n =0 3 27 z 3z m3 − 9m 2 z 2 + 27mz 4 − 27 z 6 + 9m 2 z 2 − 27 mz 4 + 27 nz 3 =0 27 z 3 m3 − 9m 2 z 2 + 27mz 4 − 27 z 6 + 9m 2 z 2 − 27 mz 4 + 27 nz 3 = 0 −27 z 6 + 27 z 3 n + m3 = 0 z 6 − nz 3 −

m3 =0 27

m3 =0. 27 At this point Francois Vieta knew he could solve the original reduced cubic in part (a), because he knew he could use the quadratic formula, which had been around for centuries, to solve the foregoing quadratic. The work follows in part (c).

b. Let u = z 3 and u 2 = z 6 . Then the last equation in part (a) can be written as u 2 − nu −

n ± n 2 + 427m

3

c.

z =u= 3

2

=

n 1 2 4m 3 n n 2 m3 ± n + = ± + 2 2 27 2 4 27

If we use the positive root in this equation, we can show that z =

824

3

n n 2 m3 + + . 2 4 27


Chapter 1



d. Rewrite the reduced cubic equation x 3 + 3 x = 14 as x 3 + 3 x − 14 = 0. Thus we have a reduced cubic equation with m = 3 and n = −14. Substituting for z in the equation from part (c) gives z=

3

−14 + 2

It is easy to verify that

3

( −14 )

+

33 = 3 −7 + 49 + 1 = 3 −7 + 5 2 = 2 − 1 27

−7 + 5 2 = 2 − 1. Just show that

more difficult to determine that 3

4

2

−7 + 5 2 can be expressed as

3

(

)

3

2 − 1 = −7 + 5 2. It is somewhat

−7 + 5 2 = 2 − 1. The motivating idea is first to suspect that 2 + k for some real constant k and then solve

for k. Hence the real solution of x + 3 x = 14 is 3 m x= −z= − 3z 3 2 −1

(

2+k

)

3

= −7 + 5 2

3

(

2.

)

(

)

2 −1 =

1 2 −1

−

(

)

2 −1 = 2 .

Fermat’s Last Theorem The History of Fermat’s Last Theorem The conjecture that x n + y n = z n is impossible for all integers n > 2 is known as Fermat’s Last Theorem. The actual statement of the conjecture was given by Pierre de Fermat in 1637. Fermat wrote the conjecture as a paragraph in the margin of the text The Arithmetic of Diophantus.

It is impossible to write a cubic as the sum of two cubes, a fourth power as the sum of two fourth powers, and in general any power beyond the second as the sum of two similar powers. For this I have discovered a truly wonderful proof but the margin of this book is too small to contain it. Many mathematicians have tried to find either a proof of Fermat’s Last Theorem or a counterexample to disprove the result. In 1780 Leonhard Euler proved the theorem for n = 3. Other mathematicians proved it true for n = 5, n = 7, and n = 13. Before June of 1993, the theorem had been established for 2 < n < 25, 000 . The Relationship Between Fermat’s Last Theorem and the Pythagorean Theorem If we remove the restriction that n must be larger than 2 and consider the equation x 2 + y 2 = z 2 , then we can find an infinite number of solutions. For example, we know that x = 3, y = 4, and z = 5 are solutions

of this equation, as are 3k, 4k, and 5k for any positive integer k. Because x 2 + y 2 = z 2 is in the form of the Pythagorean Theorem, positive-integer solutions of this equation are called Pythagorean triples. Dr. Andrew Wile’s Proof of Fermat’s Last Theorem In June of 1993, Andrew Wiles of Princeton University announced that he had produced a proof of Fermat’s Last Theorem. At first it appeared that he had, in fact, written a proof of Fermat’s Last Theorem. However, an error was soon discovered. At this time Andrew Wiles was extremely disappointed. He had spent over 7 years working on his proof. He had even given a presentation at Cambridge University in which he outlined his proof to his peers. At first Wiles could not find a way to repair his proof, but eventually, after an additional year of work, and with the assistance of the mathematician Richard Taylor, a valid proof of Fermat’s Last Theorem was achieved. Additional information about Wiles’ proof of Fermat’s Last Theorem can be found on the NOVA ONLINE internet site Solving Fermat: Andrew Wiles at: http://www.pbs.org/wgbh/nova/proof/wiles.html

Andrew Wiles’ proof consists of a 150-page document. It makes use of 20th century mathematics that was not available to Fermat. Thus we are certain that Wiles’ proof is not the same as the proof that Fermat indicated he had produced. It is interesting to note that on the NOVA ONLINE internet site mentioned above, Andrew Wiles states “I don’t believe that Fermat had a proof.”


825

Chapter 1



1.5 Inequalities 1. Triangles a. x + x + 5 > x + 9, ⇒ x > 4 b. x + x 2 + x > 2 x 2 + x − x2 + x > 0 − x ( x − 1) > 0 − x ( x − 1) | + + + | − − − − − x 0 1 x is between 0 and 1.

c.

1 1 1 + > x + 2 x +1 x 1 1 1 + − >0 x + 2 x +1 x x ( x + 1) + x ( x + 2 ) − ( x + 2 )( x + 1) >0 x ( x + 1)( x + 2 ) x 2 + x + x 2 + 2 x − x 2 − 3x − 2 >0 x ( x + 1)( x + 2 ) x2 − 2 >0 x ( x + 1)( x + 2 ) x2 − 2 > 0 and x > 0 x ( x + 1)( x + 2 )

|− − − − − −| + + x 0 2 The solution of the above inequalities is given by x > 2 .

2.

Fair Coins t − 500 ≤ 2.33 if and only if a. 15.81 t − 500 ≤ 2.33 15.81 −36.8373 ≤ t − 500 ≤ 36.8373 −2.33 ≤

463.1627 ≤

t

≤ 536.8373

Because t must be a nonnegative integer, we have 464 ≤ t ≤ 536. Therefore, according to this definition, a coin will be considered a fair coin if, in 1000 flips of the coin, the number of tails is greater than or equal to 464, but less than or equal to 536. b. Answers will vary.

826


Chapter 1



1.6 Variation and Applications 1.

A Direct Variation Formula

Given f ( x) = k x , prove f ( x2 ) = f ( x1 )

x1 . x2

Proof: f ( x1 ) = k x1 and f ( x2 ) = k x2 , so f ( x2 ) f ( x1 )

f ( x2 ) f ( x1 )

=

k x2 k x1

=

x2 x1

f ( x2 ) = f ( x1 )

x2 x1

Let f1 = 17 kg , f 2 = 22 kg , and d ( f1 ) = 8.5 centimeters. Then d ( f 2 ) = d ( f1 )

f2 f1

22 17 d ( 22 ) = 11 centimeters d ( 22 ) = 8.5 ⋅

2.

An Inverse Variation Formula x k Given f ( x ) = , prove f ( x2 ) = f ( x1 ) 1 . x x2

Proof: f ( x ) = The ratio

k k k , so f ( x2 ) = and f ( x1 ) = . x x2 x1

f ( x2 ) f ( x1 )

is given by f ( x2 ) f ( x1 )

f ( x2 ) = f ( x1 )

Thus f ( x2 ) = f ( x1 )

k x2 k x1

=

k x2 k x1

= f ( x1 )

k x1 ⋅ x2 k

x1 . x2

Let x1 = 280 , x2 = 330 , and V ( 280 ) = 2.4 . V ( x2 ) = V ( x1 )

x1 x2

V ( 330 ) = V ( 280 )

280 280 = ( 2.4 ) ≈ 2.0 liters 330 330


827

Chapter 2



Functions and Graphs CHAPTER 2 of College Algebra and Trigonometry 2.1 A Two-Dimensional Coordinate System and Graphs 1. Verify a Geometric Theorem Consider any right triangle ABC. Place the triangle in a coordinate system with the vertex point of its right angle at the origin and its legs on the x-axis and the y-axis as shown on page 177 of the text. B ( 0, b )

By the midpoint formula, the coordinates of point M are ⎛ a+0 0+b⎞ ⎛ a b⎞ , ⎜ ⎟=⎜ , ⎟ 2 ⎠ ⎝ 2 2⎠ ⎝ 2

M C ( 0, 0 )

A ( a, 0 ) 2

2

2

2

2

2

2

2

2

2

a2 + b2 2

2

a 2 + b2 2

d ( A, M ) = ⎛⎜ a − a ⎞⎟ + ⎛⎜ 0 − b ⎞⎟ = ⎛⎜ a ⎞⎟ + ⎛⎜ − b ⎞⎟ = ⎝ ⎝ 2⎠ ⎝ 2⎠ 2⎠ ⎝ 2⎠ d ( B, M ) = ⎛⎜ 0 − a ⎞⎟ + ⎛⎜ b − b ⎞⎟ = ⎛⎜ − a ⎞⎟ + ⎛⎜ b ⎞⎟ = ⎝ ⎝ 2⎠ ⎝2⎠ 2⎠ ⎝ 2⎠

2.

2

2 2 d ( C , M ) = ⎛⎜ 0 − a ⎞⎟ + ⎛⎜ 0 − b ⎞⎟ = ⎛⎜ − a ⎞⎟ + ⎛⎜ − b ⎞⎟ = a + b ⎝ ⎝ 2⎠ ⎝ 2⎠ 2⎠ ⎝ 2⎠ 2 Thus the midpoint of the hypotenuse of any right triangle is equidistant from each of the vertices of the triangle. Solve a Quadratic Equation Geometrically a. See the following figure.

Q

T a

b

E

a

A

a

S

B

P b

C

b. Using the figure from part (a), we see that d ( Q, B ) = a + d ( A, B ) = a + a 2 + b 2

Let x = a + a 2 + b 2 in the equation x 2 = 2ax + b 2 to show that both sides equal the expression 2a 2 + 2a a 2 + b 2 + b 2 . Thus d ( Q, B ) is a solution of x 2 = 2ax + b 2 .

c.

Using the figure from part (a), we see that d ( P, B ) = d ( A, B ) − d ( A, P ) = a 2 + b 2 − a

Let x = a 2 + b 2 − a in the equation x 2 = −2ax + b 2 to show that both sides equal the expression 2a 2 − 2a a 2 + b 2 + b 2 . Thus d ( P, B ) is a solution of x 2 = −2ax + b 2 .

828


Chapter 2



d. Because d (T , E ) = a 2 − b 2 and d ( E , B ) = a, we know that d (T , B ) = a + a 2 − b 2 . Let x = a + a 2 − b 2 in x 2 = 2ax − b 2 to show that each side simplifies to 2a 2 + 2a a 2 − b 2 − b 2 . Thus d (T , B ) is a solution of x 2 = 2ax − b 2 .

Also note that d ( S , B ) = a − d ( E , T ) = a − a 2 − b 2 . Let x = a − a 2 − b 2 in x 2 = 2ax − b 2 to show that each side simplifies to 2a 2 − 2a a 2 − b 2 − b 2 . Thus d ( S , B ) is a solution of x 2 = 2ax − b 2 . 2.2 Introduction to Functions 1. Day of Week a.

Let m = 10, d = 7, c = 19, and y = 41. Then 13m − 1 y c z = 5 + 4 + 4 + d + y − 2c 13 10 1 19 ⋅ − + 41+ + 7 + 41 − 2 ⋅19 = 5 4 4 = 25 + 10 + 4 + 7 + 41 − 38

= 49 The remainder of 49 divided by 7 is 0. Thus December 7, 1941, was a Sunday.

b. This one is tricky. Because we are finding a date in the month of January, we must use 11 for the month and we must use the previous year, which is 2009. Thus we let m = 11, d = 1, c = 19, and y = 109. Then 13m − 1 y c z= 5 + 4 + 4 + d + y − 2c 13 ⋅11 − 1 109 19 + = 5 4 + 4 + 1 + 109 − 2 ⋅19 = 28 + 27 + 4 + 1 + 109 − 38 = 131 The remainder of 131 divided by 7 is 5. Thus January 1, 2010, will be a Friday. c.

Let m = 5, d = 4, c = 17, and y = 76. Then 13m − 1 y c z = 5 + 4 + 4 + d + y − 2c 13 ⋅ 5 − 1 17 + 76 + + 4 + 76 − 2 ⋅ 17 = 5 4 4 = 12 + 19 + 4 + 4 + 76 − 34 = 81 The remainder of 81 divided by 7 is 4. Thus July 4, 1776 was a Thursday.

d. Answers will vary.


829

Chapter 2



2.3 Linear Functions 1. Visual Insight y G

x = x1

1 + m2

1 E C ( x1 , mx1 + b )

B

m

F

y = mx + b d

A ( x1 , y1 ) D

x

The distance d between the point A ( x1 , y1 ) and the line y = mx + b is d =

mx1 + b − y1 1 + m2

.

Triangle ABC is similar to triangle EFG. Thus d ( A, B )

d ( A, C )

=

d ( E, F )

• Corresponding sides of similar triangles are proportional.

d ( E, G )

d 1 = mx1 + b − y1 1 + m2

d=

2.

• Substitute for d ( A, B ) , d ( A, C ) , and d ( E , G ) .

mx1 + b − y1

• Solve for d, which is the distance from point A to the line.

1 + m2

Verify Geometric Theorems Place an arbitrary triangle in a coordinate system and label its vertices as shown. y

( b, c ) ( − a, 0 )

( a, 0 )

x

The coordinates of the endpoints of the line segment that connects the midpoints of two of the sides of the triangle are given by ⎛ −a + b 0 + c ⎞ ⎛ a+b 0+c⎞ and , , ⎜ ⎟ ⎜ ⎟ 2 2 2 ⎠ ⎝ ⎠ ⎝ 2 The slope of the line segment that connects the midpoints of two sides of the triangle is c c − 0 0 2 2 = = =0 a + b − a + b 22a a − 2 2 The slope of the third side of the triangle is also 0. Thus the two line segments are parallel.

830


Chapter 2 3.



Verify Geometric Theorems Place an arbitrary square in a coordinate system and label its vertices as shown. y

( a, a )

( 0, a )

( 0, 0 )

x

( a, 0 )

The slope of the diagonal through the origin is a / a = 1. The slope of the other diagonal is a / ( −a ) = −1.

Applying the Parallel and Perpendicular Lines Theorem from Section 2.3 enables us to state that the diagonals are perpendicular.

) = ( , ) . The midpoint ) = ( , ) . Thus the midpoint of each diagonal is M = ( , ) .

The midpoint of the diagonal through the origin is the point of the other diagonal is the point

(

0+a,a+0 2 2

(

a+0,0+a 2 2

a a 2 2

a a 2 2

a a 2 2

Use the distance formula to determine that the distance from the midpoint M to each of the vertices is 2

2

2

2

⎛ a − 0⎞ + ⎛ a − a ⎞ = ⎛ a ⎞ + ⎛ − a ⎞ = a2 = a = 2 a ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ a 2 ⎝2 ⎠ ⎝2 ⎠ ⎝2⎠ ⎝ 2⎠ 2 Thus the diagonals of any square are perpendicular bisectors of each other.

2.4 Quadratic Functions 1. The Cubic Formula In the fifteenth century, the Italian mathematician Luca Pacioli (ca. 1445-1509) expressed his belief that cubic equations of the form x 3 + bx 2 + cx + d = 0 could not be solved, in general, by algebraic procedures involving radicals. His assertion challenged mathematicians to search for a “solution.” The mathematician Scipione del Ferro (1465-1526) was not able to find such a solution, however, he did discover a formula that solved “depressed cubic” equations of the form x 3 + mx = n. At that time del Ferro decided to keep his solution a secret. The reason for this decision was based on the fact that mathematicians of that time period occasionally faced challenges from other mathematicians. If someone challenged him with a set of problems to be solved, then he could challenge the person to solve a set of reduced cubic equations. Even if he did poorly on his problems, he was confident that the challenger would not be able to solve any of the depressed cubic equations. In such a situation, del Ferro would be considered the winner of the challenge. This is in sharp contrast to the present time where a mathematician’s reputation is based on his or her published works. Just prior to his death, del Ferro shared his solution with his student Antonio Fior (ca. 1506-?). In 1535 Fior challenged the scholar Niccolo Fontana (1499-1557), (also known as Tartaglia – the Stammerer) to solve a set of depressed cubic equations. The challenge problems that Fior received from Tartaglia concerned several different mathematical topics. Thus Tartaglia was in a difficult situation. If he could find the solution to depressed cubic equations, then he would be able to solve all of the challenge problems, but if he could not discover the solution to depressed cubics, then he would probably not be able to solve any of the challenge problems. Tartaglia began a desperate assault on the problem of finding the solution for depressed cubics. Many days passed and Tartaglia had not found the solution. However, Tartaglia was a talented mathematician and with continued effort he did discover the solution. With the solution in hand, it was easy for Tartaglia to solve all of the challenge problems. Fior was unable to solve all of the challenge problems he had received from Tartaglia. Thus Tartaglia was the undisputed winner of the challenge. Tartaglia had discovered the solution for the depressed cubic equation, but the solution of the general cubic equations x 3 + bx 2 + cx + d = 0 was still unknown. It was at this point that the Italian mathematician Gerolamo Cardano (1501-1576) contacted Tartaglia to learn about his wonderful solution for depressed cubics. Tartaglia was not eager to share his solution with Cardano, but Cardano was Copyright © Houghton Mifflin Company. All rights reserved.

831

Chapter 2

2.



persistent and after a period of about 4 years, Tartaglia met with Cardano. Before Tartaglia divulged the solution, he made Cardano take an oath to never publish the solution. For many years Cardano kept Tartaglia’s solution a secret. Cardano even determined how to solve the general cubic equation x 3 + bx 2 + cx + d = 0, but the procedure consisted of using a substitution to write the general cubic equation in the form of a depressed cubic equation. Cardano was stymied. He could not publish his wondrous new solution because of the promise he had made to Tartaglia. Ludovico Ferrari, a protégé of Cardano’s even discovered how to solve general fourth degree polynomial equations, but his solution also depended upon writing the polynomial in the form of the depressed cubic equation and then applying Tartaglia’s solution. Then in 1543, Cardano and Ferrari happened to examine the mathematical papers of Scipione del Ferro. To their surprise, they found that del Ferro had in fact been the first to discover the solution of the depressed cubic equation and that he had left a written copy of his discovery. Cardano felt that he no longer needed to keep his oath to Tartaglia, because the original solution had been done by del Ferro and only rediscovered by Tartaglia. In 1545 Cardano published a mathematical manuscript that included the solution to the general cubic equation x 3 + bx 2 + cx + d = 0. Note that the above account is only a brief discussion of the events surrounding the development of the solution of the general cubic equation. The mathematical techniques used in solving a cubic equation are given in the response to Project 1, of Section 3.4. Simpson’s Rule The equation of the Parabola in the following figure is y = Ax 2 + Bx + C. y

P0 ( − h, y0 )

−h

P1 ( 0, y1 ) P2 ( h, y2 )

h

x

2 y0 = A ( −h ) + B ( − h ) + C = Ah 2 − Bh + C 2 y1 = A ( 0 ) + B ( 0 ) + C = C 2 y2 = A ( h ) + B ( h ) + C = Ah 2 + Bh + C

Thus y0 + 4 y1 + y2 = Ah 2 − Bh + C + 4C + Ah 2 + Bh + C = 2 Ah 2 + 6C. 2.5 Properties of Graphs 1. Dirichlet Function a. The domain of the Dirichlet function is the set of all real numbers. b. The range of the Dirichlet function is {0,1}. c. The Dirichlet function has an x-intercept at every point ( a, 0 ) where a is a rational number. d. The Dirichlet function has a y-intercept of ( 0, 0 ) .

832

e. f.

The Dirichlet function is an even function. Graphing calculators use only rational numbers. Thus a graphing calculator will fail to show any of the points ( b,1) where b is an irrational number.

g.

The Dirichlet function is said to be discontinuous at every point because between any two rational numbers we can find an irrational number, and between any two irrational numbers we can find a rational number. The graph of the Dirichlet function looks like to horizontal lines (one at y = 0 and one at y = 1), except that the lines are “full of holes.” Some mathematicians have called the Dirichlet function a shotgun function because the graph can be thought of as two horizontal lines in which we use a shotgun to blast holes. The holes on the line y = 1 occur at (a, 1) for all rational numbers a. The holes on the line y = 0 occur at (b, 0) for all irrational numbers b. Copyright © Houghton Mifflin Company. All rights reserved.

Chapter 2 2.

4.


Isolated Point

The point (1,1) is a solution of y =

3.


( x − 1) ( x − 2 ) + 1 because 1 = ( (1) − 1) ( (1) − 2 ) + 1 . 2

2

Graphing utilities use only a finite number of domain values in the construction of a graph. Your graphing utility may have graphed the function for some values of x < 1 and for some values of x > 1, but not for x = 1. A Line with a Hole Graphing utilities use only a finite number of domain values in the construction of a graph. Thus your graphing utility may fail to show the proper result exactly at x = 2. Finding a Complete Graph To produce the part of the graph to the left of the y-axis, it may be necessary to enter the function as y = −3 x

5/3

−6 x

4/3

+ 2 for x ≤ 0

This procedure may be necessary because some graphing utilities do not evaluate fractional powers of negative numbers. You need y1 as shown in text and y2 as above with a condition at the end:

(

y2 = −3 x

5/3

−6 x

4/3

)

+ 2 ( x ≤ 0 ) (using TI-83)

2.6 The Algebra of Functions 1. A Graphing Utility Project a.

Let x = 5 and y = 9 . Then Maximum ( 5,9 ) =

b. Let x = 201 and y = 80 .

5 + 9 5 − 9 14 4 + = + = 7+2 = 9 . 2 2 2 2

201 + 80 201 − 80 281 121 + = + = 140.5 + 60.5 = 201 . 2 2 2 2 For each real number x ≥ 0, the graph of y3 is a graph whose range value is the maximum of y1 ( x )

Then Maximum ( 201,80 ) = c.

and y2 ( x ) . Thus for each x ≥ 0, y3 can be graphed by plotting points from the graph of y1 if its graph is higher than the graph of y2 , and by plotting points from the graph of y2 if it is the higher graph. d. The domain of y1 :{x x ∈ ℜ} (that is, all real numbers). The domain of y2 :{x x ≥ 0}. The domain of y3 :{x x > 0}. The domain of y3 is the intersection of the domain of y1 and the domain of y2 .

e. 2.

Minimum ( f ( x ) , g ( x ) ) =

f ( x) + g ( x) 2

−

f ( x) − g ( x) 2

The Never-Negative Function a.

b.

6000

0 −1000

3

6000

0

3

−1000

The graph of M + is the graph of M provided that M ≥ 0. For each x such that M < 0, the graph of M + is the point (x, 0). b. The maximum mosquito population is 4850 mosquitoes per acre. This maximum occurs at 1 3 5 t = , t = and t = (the middle of each month). 2 2 2 a.


833

Chapter 2 c.



Assume the months have 30 days. The best time to fish if you wish to minimize your exposure to mosquitoes is during the first 3 or the last 3 days of each month. During this period, the mosquito population will be zero. During the period from the 4th day to the 26th day of the month, you will be exposed to mosquitoes.

2.7 Modeling Data Using Regression The Median-Median Line

834

1.

y ≈ 24.8558 x − 41.9712

2.

y ≈ 3.5429 x + 58.0952

3.

a. y = 2.0000 x + 1.0000 b. y = 2.0000 x + 1.0000 c. If a least-squares regression line provides an exact fit for a set of data, then the median-median line for the data will be exactly the same as the least-squares regression line.

4.

a. median-median line: y ≈ 1.3571x − 0.3333; regression line: y ≈ 1.2636 x + 0.5182 b. median-median line: y ≈ 1.2143 x + 0.3333; regression line: y ≈ 0.6727 x + 3.4727 c. A median for a set of data is generally not changed by increasing one of the larger data values or by decreasing one of the smaller data values. For instance, consider the data 3, 5, 7, 11, 20, which has a median of 7. Now change the 20 to a larger number, say 30. This new set of data, 3, 5, 7, 11, 30, still has a median of 7. In general, a median-median line is less sensitive to a single change in one value than is a least-squares regression line. In fact, any single change in a set of data will cause a change in the least-squares regression line for the data, but often this same change will not change the medianmedian line for the data. Some mathematical texts refer to the median-median line of a set of data as the “resistant line” for the data.


Chapter 3



Polynomial and Rational Functions CHAPTER 3 of College Algebra and Trigonometry 3.1 The Remainder Theorem and the Factor Theorem 1. Horner’s Polynomial Form P ( x ) = 3 x5 − 4 x 4 + 5 x 3 − 2 x 2 + 3 x − 8

a.

P (6) = 3(6)5 − 4(6) 4 + 5(6)3 − 2(6) 2 + 3(6) − 8 = 3(7776) − 4(1296) + 5(216) − 2(36) + 3(6) − 8 = 23328 − 5184 + 1080 − 72 + 18 − 8 = 18144 + 1080 − 72 + 18 − 8 = 19224 − 72 + 18 − 8 = 19152 + 18 − 8 = 19170 − 8 = 19162 P ( x ) = 3 x5 − 4 x 4 + 5 x3 − 2 x 2 + 3 x − 8

b.

= (3x 4 − 4 x3 + 5 x 2 − 2 x + 3) x − 8 = ((3 x3 − 4 x 2 + 5 x − 2) x + 3) x − 8 = (((3 x 2 − 4 x + 5) x − 2) x + 3) x − 8 = ((((3x − 4) x + 5) x − 2) x + 3) x − 8 P (6) = ((((3 ⋅ 6 − 4)(6) + 5)(6) − 2)(6) + 3)(6) − 8 = (((18 − 4)(6) + 5)(6) − 2)(6) + 3)(6) − 8 = (((14)(6) + 5)(6) − 2)(6) + 3)(6) − 8 = (((84 + 5)(6) − 2)(6) + 3)(6) − 8 = (((89)(6) − 2)(6) + 3)(6) − 8 = ((534 − 2)(6) + 3)(6) − 8 = ((532)(6) + 3)(6) − 8 = (3192 + 3)(6) − 8 = (3195)(6) − 8 = 19170 − 8 = 19162 The evaluation in part b. involves easier calculations.

3.2

Polynomial Functions of Higher Degree 1.

3.3 1.

The student is correct. The polynomial P (n) = n3 − n can be written in factored form as P (n) = n(n − 1)(n + 1) . In this form it is easy to see that P(n) is the product of three consecutive natural numbers, one of which must be an even number and one of which must be a multiple of three. Thus P (n) must be a multiple of 6 for any natural number n. Zeros of Polynomial Functions Relationships Between Zeros and Coefficients a. r1 = 1, r2 = 2, r3 = 3.

P ( x) = x3 + C1 x 2 + C2 x + C3 = x3 − 6 x 2 + 11x − 6 ⇒ C1 = −6, C2 = 11, C3 = −6 1 + 2 + 3 = 6 = −(−6) ⇒ r1 + r2 + r3 = −C1 1(2) + 1(3) + 2(3) = 2 + 3 + 6 = 11 ⇒ r1r2 + r1r3 + r2 r3 = C2 1(2)(3) = 6 = −(−6) ⇒ r1r2 r3 = −C3

1(2)(3) = 6 = (−1)3 (−6) ⇒ r1r2 r3 = (−1) n Cn Responses will vary.

b. 3.4

The Fundamental Theorem of Algebra 1. Investigate the Roots of Cubic Equation a.

Given x 3 + bx 2 + cx + d = 0. Let x = y −

b and substitute into the equation. Then simplify. 3 x 3 + bx 2 + cx + d = 0

3

⎛ ⎛ ⎛ b⎞ b⎞ b⎞ ⎜ y − ⎟ + b⎜ y − ⎟2 + c⎜ y − ⎟ + d = 0 3⎠ 3⎠ 3⎠ ⎝ ⎝ ⎝ 2 3 2 yb by 2 ⎛ ⎞ − b + b ⎜ y2 − + b ⎟ + c ⎛⎜ y − b ⎞⎟ + d = 0 y 3 − by 2 + 3 27 3 9 ⎠ ⎝ 3⎠ ⎝ 2 2 ⎛ ⎞ ⎛ 3 b b bc 2 − + d ⎟⎞ = 0 y + ⎜c − ⎟ y + ⎜ 3⎠ ⎝ ⎝ 27 3 ⎠ 2 3 Now let m = c − b and n = − ⎛⎜ 2b − bc + d ⎞⎟ . Then y 3 + my = n . 3 ⎝ 27 3 ⎠


835

Chapter 3



b. To solve the equation x3 − 6 x 2 + 20 x − 33 = 0, let b = −6, c = 20 and d = −33. Then

m = 20 −

( −6 ) 3

2

⎛ 2 ( −6 )3 ( −6 )( 20 ) ⎞ = 8 and n = − ⎜ − − 33 ⎟ = 9 ⎜ 27 ⎟ 3 ⎝ ⎠

A solution is given by y=

3

n n 2 m3 3 n n 2 m3 + + − − + + 2 4 27 2 4 27

=

3

9 92 83 3 9 92 83 + + − − + + 2 4 27 2 4 27

=1 b Substituting y and b into x = y − gives x = 3 is a solution of x3 − 6 x 2 + 20 x − 33 = 0. To find the 3 remaining solutions, use synthetic division to determine the reduced equation, which is 3 + i 35 3 − i 35 x 2 − 3 x + 11 = 0. The solutions of this equation are and . 2 2

3.5

Graphs of Rational Functions and Their Applications 1.

Parabolic Asymptotes a.

Yes. As x → ∞ and x → −∞, the graph of F approaches the graph of the parabola. b. Divide R(x) by S(x) to find the quotient Q(x). The equation y = Q(x) is the equation of the parabolic asymptote. c.

x2 + 2 x2 − 1 x4 + x2 + 2 x4 − x2 2x2 + 2 2x2 − 2 4

Yes d. Answers will vary.

836


Chapter 4



Exponential and Logarithmic Functions CHAPTER 4 of College Algebra and Trigonometry 4.1 Inverse Functions 1. Intersection Points for the Graphs of f and f–1 i. f ( x) = 2 x − 4 x = 2y − 4 x + 4 = 2y −9.1 1 x+2= y 2 f −1 ( x) = 12 x + 2 f ( x) = − x + 2 x = −y + 2 y = −x + 2

ii.

f −1 ( x) = − x + 2

6

9.1

−6 6

−9.1

9.1

−6

iii.

6

f ( x ) = x3 + 1 x = y3 + 1 x − 1 = y3 3

f

−9.1

x −1 = y

−1

( x) = 3 x − 1

−6

f ( x) = x − 3

iv.

9.1

6

x = y −3 x+3= y f

−1

( x) = x + 3

−9.1

9.1

−6 f ( x ) = −3x + 2

v.

6

x = −3 y + 2 3y = −x + 2 f

−1

( x ) = − 13 x + 32

−9.1

9.1

−6

vi.

f ( x ) = 1x x = 1y y = 1x

a. No

4 −4

f −1 ( x ) = 1x b. They are equal.

4

c. Yes. Consider the function f ( x) = x .


837

Chapter 4



4.2 Exponential Functions and Their Applications 1. The Saint Louis Gateway Arch a.

b. Use the TRACE feature. Choose a value of x and press ENTER for each successive value of x.

h(0) ≈ 625.1 feet

h(100) ≈ 587.5 feet

h(200) ≈ 433.5 feet

h(299) ≈ 1.6 feet

c.

Width of the catenary at ground level =|−299.2261 – 299.22611| = |−598.45221| ≈ 598.5 feet Maximum height of the catenary = h(0) ≈ 625.1 feet d. 625.1 − 598.5 = 26.6 feet 2.

An Exponential Reward a. From the chart, create a pattern from the total number of grains of wheat on squares 1 to n For square n = 1, 1 = 21 – 1 For square n = 2, 3 = 22 – 1 For square n = 3, 7 = 23 – 1 For square n = 4, 15 = 24 – 1 For square n = 5, 31 = 25 – 1 For square n = 6, 63 = 26 – 1 So for square n = 64, 264 – 1 = 1.8446744 × 1019 grains is the total. b. To find the total weight, multiply the number of grains by the weight for each one (1.8446744 × 1019 )(0.000008) = 1.4757395 × 1014 kilograms c. To find how long it would take, convert the weight to metric tons and divide by the amount per year 1year 1.4757395 × 1014 kg ⋅ 1 metric ton ⋅ ≈ 227 years 1000 kg 6.5 × 108 metric tons

838


Chapter 4



4.3 Logarithmic Functions and Their Applications 1.

a.

d

1 2 3 4 5 6 7 8 9 b. c.

⎛ 1⎞ P (d ) = log ⎜1 + ⎟ ⎝ d⎠ 0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.046

(

)

P (6) = log 1 + 16 = log

( 76 ) ≈ 0.067 or 6.7%

( ) P (9) = log (1 + 19 ) = log ( 10 ≈ 0.046 or 4.6% 9) P (1) = log 1 + 11 = log ( 2 ) ≈ 0.301 or 30.1%

P (1) 0.301 = ≈ 6.54 times as many P (9) 0.046

d.

Most high school students are teenagers.

4.4 Logarithms and Logarithmic Scales 1. Logarithmic Scales Mass (g) ln(Mass) a. Animal Rotifer 0.000000006 −8.22 Dwarf Goby 0.30 −0.52 Lobster 15,900 4.20 Leatherback Turtle 851,000 5.93 Giant Squid 1,820,000 6.26 Whale Shark 44,700,000 7.65 Blue Whale 120,000,000 8.08

b.

The logarithmic number line is more helpful when comparing different animals.

c.

101 = 10 times heavier 102 = 100 times heavier


839

Chapter 4



2. Logarithmic Scales a. b.

Planet

Distance (million km) 58 108 150 228 778 1427 2871 4497 5913

Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto

ln(Distance) 1.76 2.03 2.18 2.36 2.89 3.15 3.46 3.65 3.77

c. Answers will vary. d. 103 = 1000 . One distance is 1000 times greater than the other. 3.

Biological Diversity a.

D = − ( 15 log 2 15 + 15 log 2 15 + 15 log 2 15 + 15 log 2 15 + 15 log 2

b.

D = − ( 18 log 2 18 + 83 log 2 83 + 161 log 2 161 + 18 log 2 18 + 165 log 2 165 )

1 5

log 1

) = − log 2 15 = − log 25 ≈ 2.322

⎛ 1 log 81 + 83 log 83 + 161 log 161 + 18 log 81 + 165 log 165 ⎞ = −⎜ 8 ⎟ ≈ 2.055 log 2 ⎝ ⎠ This system has less diversity than the one given in Table 1. D = − ( 14 log 2 14 + 34 log 2 34 )

c.

⎛ 0.25log 0.25 + 0.75log 0.75 ⎞ = −⎜ ⎟ ≈ 0.811 log 2 ⎝ ⎠ This system has less diversity than the one given in Table 2. d. D = − (1 log1) = 0

This value means that the system has no variety of species. 4.5 Exponential and Logarithmic Equations 1.

Navigating a.

If v = w, then the equation becomes y = 1 − ( 2x ) . The graph of this equation is shown below. The 2

boat reaches the other shore 1 mile from point O. y 1.0 0.8 0.6 0.4 0.2 O

840

0.5

1.0

1.5

2.0

x


Chapter 4



b. Now suppose w > v . For instance, let w = 1.1v. Then the equation becomes y = ( 2x )

−0.1

− ( 2x ) . The 2.1

graph of this equation is shown below. In this case, the y-axis is a vertical asymptote and the boat never reaches the shore. y

1.5 1.0 0.5 O

c.

0.5

1.0

1.5

2.0

x

Now suppose w < v . For instance, let w = 0.5v . Then the equation becomes y = ( 2x )

0.5

− ( 2x ) . The 1.5

graph of this equation is shown below. In this case, the boat reaches the shore at point O. y

0.3 0.2 0.1 O

0.5

1.0

1.5

2.0

x

4.6 Exponential Growth and Decay 1.

2.

3.

A Declining Fish Population 1000 1000 = = 1500 fish a. P0 = 1 + (−0.3333) 0.6667 b. As t → ∞, P(t ) → 1000 fish. A Declining Deer Population 1800 1800 = = 2300 deer a. P2 = 1 + (−0.25)(0.869) 0.782 b. As t → ∞, P(t ) → 1800 deer. Modeling World Record Times in the Men’s Mile Race 199.13 = 219.41 s = 3 min, 39.41 s a. WR (107) = 1 + (−0.21726)(0.42536) WR (137) =

b.

199.13 = 214.75 s = 3 min, 34.75 s 1 + (−0.21726)(0.33471)

As t → ∞, WR (t ) → 199.13 s = 3 min, 19.13 s.

4.7 Modeling Data with Exponential and Logarithmic Functions 1.

A Modeling Project a. Answers will vary b. Answers will vary c. Answers will vary d. Answers will vary e. Answers will vary f. Answers will vary


841

Chapter 5



Trigonometric Functions CHAPTER 5 of College Algebra and Trigonometry 5.1 Angles and Arcs 1. Conversion of Units 14 tollars ⎛ 4 nollars ⎞ ⎛ 3 mollars ⎞ 14 ⋅ 4 ⋅ 3 24 4 a. 14 tollars = mollars = mollars =4 mollars ⎜ ⎟ ⎜ ⎟= 1 7 tollars 5 nollars 7 ⋅ 5 5 5 ⎝ ⎠ ⎝ ⎠ 4 4 4 mollars ⎛ 5 lollars ⎞ b. 4 mollars = 4 mollars + mollars = 4 mollars + ⎜ ⎟ = 4 mollars + 4 lollars 5 5 5 ⎝ 1 mollar ⎠ c. Start with the given quantity. Multiply by unit fractions that eliminate the given units and yield results in terms of the desired units. We knew we wanted to eliminate the tollars unit in the numerator. Thus ⎛ 4 nollars ⎞ we need to multiply by the unit fraction: ⎜ ⎟ ⎝ 7 tollars ⎠ ⎛ π ⎞ d. ⎜ ⎟ ⎝ 180° ⎠ 2.

Space Shuttle Let d be the portion of a revolution that the Galápagos Islands rotates as the shuttle revolves 1 + d revolutions. The time required for the shuttle will be ⎛ (1 + d ) revolutions ⎞ ⎛ 2.231 hours ⎞ t1 = ⎜ ⎟ ⎜ ⎟ = (1+d )( 2.231) hours 1 ⎝ ⎠ ⎝ 1 revolution ⎠ The time required for the earth to rotate d revolutions is ⎛ d revolutions ⎞ ⎛ 23.933 hours ⎞ t2 = ⎜ ⎟ ⎜ ⎟ = 23.933d hours 1 ⎝ ⎠ ⎝ 1 revolution ⎠ Setting t1 = t2 yields 2.231 + 2.231d = 23.933d . Solving this equation gives d ≈ 0.1028 revolution. The time required for the shuttle to complete 1 + d revolutions is 2.231 hours ⎞ (1 + 0.1028) revolutions ⎛⎜ ⎟ = 2.460 hours (to the nearest 0.001 hour) ⎝ 1 revolution ⎠

5.2 Trigonometric Functions of Acute Angles 1. Perimeter of a Regular n-gon a.

P = 2 xn

x

sin 180° n

x

r =1

b.

842

180° x = n r 180° 180° = sin n n 180° P = 2n sin n x = r sin

180° ≈ 6.18034 10 P50 = 2 ( 50 ) sin 180° ≈ 6.27905 50 P100 = 2 (100 ) sin 180° ≈ 6.282152 100 P1000 = 2 (1000 ) sin 180° ≈ 6.283175 1000 P10,000 = 2 (10, 000 ) sin 180° ≈ 6.283185 10, 000 Pn approaches 2π as n increases because the perimeter approaches the circumference of the circle.

P10 = 2 (10 ) sin


Chapter 5 2.



Area of a Regular n-gon a. 1 hr ⋅ n 2 n 360° = r ⋅ r sin n 2 n 360° = sin n 2

A= r

h 360° n

r

b.

sin

360° h = n r h = r sin

360° n

r =1

10 360° sin ≈ 2.938926261 2 10 50 360° ≈ 3.133330839 A50 = sin 2 50 100 360° ≈ 3.139525976 sin A100 = 2 100 1000 360° ≈ 3.141571983 sin A1000 = 2 1000 10, 000 360° ≈ 3.141592447 A10,000 = sin 2 10, 000 An approaches π as n increases because the area of the polygon approaches the area of the circle, A20 =

which is π ⋅12 = π . 5.3 Trigonometric Functions of Any Angle 1.

Find Sums or Products a. 0 The sum is 0 because cos n° = − cos (180° − n° ) for all integers n such that 0° ≤ n ≤ 90° . b.

0 The sum is 0 because sin n° = − sin ( 360° − n° ) for all integers n such that 0° ≤ n ≤ 180° .

c.

0 The sum is 0 because cot n° = − cot (180° − n° ) for all integers n such that 1° ≤ n ≤ 89° .

d. e.

0 The product is 0 because cos 90° = 0 and the product of 0 and any real number is 0. 179 This is the sum of 359 numbers. However, the sum can be regrouped so that it consists of 179 pairs of the form ( cos x ) + ( cos 90° − x ) = ( cos x ) + ( sin x ) = 1 . The term that is not paired up 2

2

2

2

with another number in the list is cos 90° = 0 . Thus the sum is 179. 5.4 Trigonometric Functions of Real Numbers 1.

y

Visual Insight E

C

Unit circle OD = cos φ DC = sin φ

a.

φ

F

B

φ O

D

A (1, 0)

x

Consider the triangle ABC. By definition, opp d ( A, B ) d ( A, B ) = = = the length of line segment AB. tan φ = adj d ( O, A ) 1


843

Chapter 5

2.



b.

Triangle ODC is similar to triangle FEO. Therefore, d ( O, D ) d ( F , E ) d ( F , E ) = = = the length of line segment EF. cot φ = d ( D, C ) d ( E , O ) 1

c.

Consider triangle AOB. By definition, hyp d ( O, B ) d ( O, B ) = = = the length of line segment OB. sec γ = adj d ( O, A ) 1

d.

Triangle OEF is similar to triangle CDO. Therefore, d ( O, C ) d ( F , O ) d ( F , O ) csc γ = = = = the length of line segment OF. d ( C , D ) d ( O, E ) 1

Functions Defined by a Square a. ssin 3.2 = 0.8 because traveling 3.2 units counterclockwise around the square from (1, 0 ) places you

at the point ( −1, 0.8 ) . The y-value of this ordered pair is 0.8. b.

scos 4.4 = −1 because traveling 4.4 units counterclockwise around the square from (1, 0 ) places you at

the point ( −1, −0.4 ) . The x-value of this ordered pair is –1. c. d. e. f.

−1 1 ssin 5.5 = −1 and scos 5.5 = − , thus, stan 5.5 = 1 = 2 . 2 (− 2 )

ssin 11.2 = 0.8 (Note that the function y = ssin x is periodic with a period of 8. Thus ssin 11.2 = ssin 3.2 which equals 0.8 from part (a).) scos − 5.2 = −0.8 . ssin ( −6.5 ) 1 stan − 6.5 = = =2 scos ( −6.5 ) 0.5

5.5 Graphs of the Sine and Cosine Functions 1. Cepheid Variable Stars and the Period-Luminosity Relationship a. A Cepheid is usually a population I giant yellow star, pulsing regularly by expanding and contracting, resulting in a regular oscillation of its luminosity. Named for the prototype of this class found in the constellation Cepheus, classical Cepheids have periods from about 1.5 days to over 50 days and are Population I stars. The longer the period of such a star, the greater its natural brightness; this relationship was discovered in 1912 by the American astronomer Henrietta Leavitt (b. 1868-d. 1921). The relationship between a Cepheid variable's luminosity and variability period is quite precise, and has been used as a standard candle for almost a century. b. Because of this correlation (discovered by Henrietta Leavitt in 1912), a Cepheid variable can be used as a standard candle to determine the distance to its host cluster or galaxy. Since the period-luminosity relation can be calibrated with great precision using the nearest Cepheid stars, the distances found with this method are among the most accurate available. 5.6 Graphs of the Other Trigonometric Functions 1. A Technology Question The function f ( x ) = tan x is undefined at x = 32 π , which is between the domain values 4.7123 and

4.7124. Because y = tan x approaches ∞ as x approaches 32 π from the left and y = tan x approaches −∞ as x approaches 32 π from the right, it is possible to produce large changes in your range values with small changes in your domain values as they change from slightly less than 32 π to slightly greater than 32 π .

844


Chapter 5 2.



Solutions of a Trigonometric Equation The equation tan ( 1x ) = 0 has an infinite number of solutions on the interval −1 ≤ x ≤ 1 . This can be

determined by observing that as x varies from –1 to 1, the fraction 1/x takes on all values less than or equal to –1 and all values greater than or equal to 1. Because the tangent function is periodic with a period of π , the expression tan ( 1x ) will equal 0 whenever 1/x is a multiple of π for – for instance, when x = π1 , 21π , 31π ,K . Thus there are an infinite number of solutions.

5.7 Graphing Techniques 1. Predator-Prey Relationships The graphs are shown below. Because the assumption is that the wolves prey on the rabbits, as the rabbit population increases, there is more food for the wolves, which in turn allows the wolf population to increase. However, as the wolf population increases, the demand for rabbits increases, and the rabbit population starts to decline. This effects a decline in the wolf population. But as the wolf population declines, there is less danger to the rabbits, and their population starts to rise. The process then repeats itself. Rabbit

Population

1000 800 600 400 200

Wolf 4

8

12

16

20

24

t

A possible equation for the graph given in the text of the population model for rabbits is r ( t ) = 503 t + 200sin ( π6t ) + 400 . This equation is based on the assumption that there is a linear increase in the sine function. Other answers may be given.


845

Chapter 5



5.8 Harmonic Motion—An Application of the Sine and Cosine Functions 1. Three Types of Damped Harmonic Motion The displacement of f ( t ) starts at its maximum when t = 0 and then descends toward its equilibrium

point, getting closer and closer as t increases without bound. See the following figure. 1

15

0

−1

The displacement of g ( t ) starts at its maximum when t = 0 . It next falls to a minimum displacement below the equilibrium position and then rises to approach equilibrium as t increases without bound. See the following figure. 1

15

0

−1

The displacement of h ( t ) starts at its maximum displacement when t = 0 . As t increases, it oscillates about its equilibrium point. As it oscillates, the maximum displacement of each cycle tends to 0. See the following figure. 4.2

15

0

−4.2

2.

Logarithmic Decrement a. γ ≈ 3.51 b.

846

Δ ≈ 1.26; ln γ = Δ


Chapter 6



Trigonometric Equations and Identities CHAPTER 6 of College Algebra and Trigonometry 6.1 Verification of Trigonometric Identities 1. Grading a Quiz 1,2, and 4 are correct. 6.2 Sum, Difference, and Cofunction Identities 1. Intersecting Lines a.

In the following figure, we see that if a line intersects the x-axis at an angle of θ , then y −y tan θ = 2 1 = m , which is the slope of the line. x2 − x1 y

P2 ( x2 , y2 )

θ P1 ( x1 , y1 ) x

Let γ be the smallest positive angle from l1 to l 2 , as shown in the following figure. y

l2

l1

γ

α

β x

It is possible to show that β = α + γ . (Use the exterior angle theorem and the theorem on vertical angles.) Thus γ = β − α , and tan γ = tan ( β − α ) =

m − m1 tan β − tan α = 2 . 1 + tan β tan α 1 + m1m2

b. The line l1 given by y = x + 5 has a slope of m1 = 1 . The slope of the line l 2 given by y = 3x − 4 has a slope of m2 = 3 . From part (a), we know that the tangent of the angle γ (the acute angle between the lines) is m − m1 3 −1 2 1 = = = tan γ = 2 1 + m1m2 1 + (1)( 3) 4 2

A graph of y = tan γ and y =

1 (on the interval 0 < γ < 90° ) shows that γ ≈ 26.6° . 2


847

Chapter 6

c.



Let m2 be the slope of the second line. From part (a), m2 − 0.5 tan 60° = 1 + ( 0.5 )( m2 ) 3=

m2 − 0.5 1 + ( 0.5 )( m2 )

3 + 0.5 3m2 = m2 − 0.5 3 + 0.5 = m2 − 0.5 3m2

(

3 + 0.5 = m2 1 − 0.5 3

)

3 + 0.5

= m2 1 − 0.5 3 m2 ≈ 16.66

The second line passes through the point (1,5) with a slope of about 16.66, so y − 5 ≈ 16.66 ( x − 1)

• The equation of line l 2 in slope-intercept form

y ≈ 16.66 x − 11.66

6.3 Double- and Half-Angle Identities 1. Visual Insight The measure of a central angle is equal to the measure of its intercepted arc. The measure of an inscribed angle is one-half the measure of its intercepted arc. Therefore, the measure of the small marked angle at the right must be half the measure of the central angle θ . Thus the measure of the small marked angle is θ / 2. The side opposite θ in the small triangle is sin θ . The side adjacent to θ in the small triangle is cos θ . θ opp sin θ . By definition, tan = = 2 adj 1 + cos θ 6.4 Identities Involving the Sum of Trigonometric Functions 1. Beats a.

Y3

Y3, Y4, Y5 The graph of Y3 is bounded by the graphs of Y4 and Y5. b. Y3 = 2sin 2π 442 + 440 x ⋅ cos 2π 442 − 440 x 2 2 = 2sin ( 2π ⋅ 441x ) cos ( 2π x ) c. 568 − 564 = 4 beats per second d. 2 beats per second

(

848

)

(

)


Chapter 6



6.5 Inverse Trigonometric Functions 1. Visual Insight a. Verify the identity tan −1 13 + tan −1 12 = π4 . 1

α

γ

β

1

1

1

1 3

and tanβ = 12 .

Let α = tan −1 13 and β =tan -1 12 , as in the figure. Thus tan α = tan −1 1 + tan −1 1 = α + β 3 2 = tan −1 ⎣⎡ tan (α + β )⎦⎤ ⎛ tan α + tan β ⎞ = tan −1 ⎜ ⎟ ⎝ 1 − tan β tan β ⎠ ⎡ ⎤ = tan −1 ⎢ 1/ 3 + 1/ 2 ⎥ ⎣ 1 − (1/ 3)(1/ 2 ) ⎦

= tan −1 1 = π 4

Hence tan −1 13 + tan −1 12 = π4 . b. The figure in part (a) shows that α = tan −1 13 , β = tan −1 12 , and γ = π4 . Substituting in the result from

part (a) produces α + β = γ . 6.6 Trigonometric Equations 1. The Moons of Saturn a. Titan completes one cycle in about 15.95 days. Thus for Titan,

y 1 ≈ 1.0000sin(0.3963 x − 2.5696) + 0.0026

b. Rhea completes one cycle in about 4.52 days. Thus for Rhea,

y 2 ≈ 0.4002sin( −1.3963x − 2.0944) − 0.0000


849

Chapter 6 c.



This solution can be found by using the regression formula directly, or using the graphing calculator graph at value x = 40 as shown below. At 11:59PM on May 10, 2000, x = 40, Using the regression formulas from parts a and b, y 1 (40) ≈ 1.0000sin(0.3963(40) − 2.5696) + 0.0026 ≈ 0.65888 y 2 (40) ≈ 0.4002 sin( −1.3963(40) − 2.0944) − 0.0000 ≈ −0.13689 Using the graphing calculator to graph both regression formulas.

They were on opposite sides of Saturn.

850


Chapter 7



Applications of Trigonometry CHAPTER 7 of College Algebra and Trigonometry 7.1 The Law of Sines 1. Fermat’s Principle and Snell’s Law Fermat’s Principle: Light traveling from one point to another will follow a path such that, compared with nearby paths, the time required is either a minimum or maximum or will remain unchanged. Snell’s Law, which follows, is derived by using Fermat’s Principle. sin θ Snell’s Law: sin θ12 = n21 , where n21 is a constant called the index of refraction of medium 2 with respect to medium 1. See the diagram below. (Note: The index of refraction depends on the wavelength. As the wavelength increases, the index of refraction decreases.)

θ1 Because a diamond has a higher index of refraction than glass, light entering the diamond is refracted at a greater angle than the same light entering a piece of glass. The result is a narrower “rainbow” as the light leaves the diamond than as it leaves the glass.

θ2

7.2 The Law of Cosines and Area 1. Visual Insight d e

b

c

C

a

a a

Center of circle

bd = ( a + c ) e

• If two chords of a circle intersect, then the product of the lengths of the segments on one chord equals the product of the lengths of the segments on the other chord.

bd = ( a + c )( a − c )

• e = a−c

b ( 2a cos C − b ) = ( a + c )( a − c )

• In the right triangle cos C = ( d + b ) / ( 2a ) . Solving for d gives us d = 2a cos C − b .

2ab cos C − b = a − c 2

2

2

c 2 = a 2 + b 2 − 2ab cos C

• Simplify. • Solve for c 2 .


851

Chapter 7



7.3 Vectors 1. Same Direction or Opposite Directions

If c > 0 and v = cw , then v and w have the same direction, v⋅w = cos 0° = 1 . and the angle between the vectors is θ = 0° . Thus v w If c < 0 and v = cw , then v and w are vectors that have opposite directions, and the angle between the v⋅w = cos180° = −1 . vectors is θ = 180° . Thus v w

2.

The Law of Cosines and Vectors v−w

2

• Because v ⋅ w = v w cos α , we see that this equation

= ( v − w )( v − w ) = v⋅v − v⋅w − w⋅v + w⋅w 2

is a restatement of the Law of Cosines in vector form.

2

= v + w − 2v ⋅ w

3.

Projection Relationships Let v and w be two nonzero vectors. Let α be the measure of the angle between the vectors. By definition, projw v = v cos α . a.

If projw v = 0 , then v cos α = 0 , and cos α = 0 , which implies α = 90° . Hence v and w are

perpendicular (orthogonal). b. If projw v = v , then v cos α = v , and cos α = 1 . Thus, α = 0° . Hence v and w have the same direction. 7.4 Trigonometric Form of Complex Numbers 1. A Geometrical Interpretation Multiplication of a real number a by i produces the product ai . Note that in an Argand diagram, the numbers a and ai are both placed the same distance from the origin. Also note that the position of ai can be determined by rotating the position of the real umber a 90° counterclockwise about the origin. Multiplication of a complex number a + bi by i also produces a number that is located in an Argand diagram the same distance from the origin as the original number a + bi . Once again, the position of the product −b + ai can be determined by rotating the position of the original number a + bi 90° counterclockwise about the origin. It is also worth noting that multiplicity a number by −1 = i 2 can be thought of geometrically as a 180° counterclockwise rotation of the original number about the origin. 7.5 De Moivre’s Theorem 1. Verify Identities

z = ( cos θ + i sin θ )

z 2 = cos 2θ + i sin 2θ by De Moivre's Theorem. z 2 = cos 2 θ + 2i sin θ cos θ + i 2 sin 2 θ by squaring both sides.

Because z 2 = z 2 , we have ( cos 2θ + i sin 2θ ) = ( cos 2 θ + 2i sin θ cosθ + i 2 sin 2 θ )

( cos 2θ + i sin 2θ ) = ( cos 2 θ + 2i sin θ cosθ − sin 2 θ ) Equating the real parts, we have cos 2θ = cos 2 θ − sin 2 θ

Equating the imaginary parts, we have sin 2θ = 2sin θ cosθ

852


Chapter 7 2.



Discover Identities x = ( cos θ + i sin θ ) z 4 = cos 4θ + i sin 4θ by De Moivre's Theorem. z 4 = cos 4 θ + 4i cos3 θ sin θ + 6i 2 cos 2 θ sin 2 θ + 4i 3 cosθ sin 3 θ + i 4 sin θ by taking z to the fourth power.

Because z 4 = z 4 , we have

( cos 4θ + i sin 4θ ) = ( cos 4 θ + 4i cos3 θ sin θ + 6i 2 cos 2 θ sin 2 θ + 4i 3 cosθ sin 3 θ + i 4 sin θ ) = cos 4 θ + 4i cos3 θ sin θ − 6 cos 2 θ sin 2 θ − 4i cosθ sin 3 θ + sin 4 θ

Equating the real parts, we have cos 4θ = cos 4 θ − 6 cos 2 θ sin 2 θ + sin 4 θ = cos 4 θ − 2 cos 2 θ sin 2 θ + sin 4 θ − 4 cos 2 θ sin 2 θ = ( cos 2 θ − sin 2 θ ) − 4 cos 2 θ sin 2 θ 2

= cos 2 2θ − 2sin 2 2θ

Equating the imaginary parts, we have

sin 4θ = 4 cos3 θ sin θ − 4 cosθ sin 3 θ


853

Chapter 8



Topics in Analytic Geometry CHAPTER 8 of College Algebra and Trigonometry 8.1 Parabolas 1.

3-D Optical Illusion Responses will vary. For example, http://www.exploratorium.edu/snacks/parabolas.html gives the following explanation. You are seeing an image formed by two concave mirrors facing one another. The object is placed at the center of the bottom mirror. The curvature of the mirrors is such that the object is at the focal point of the top mirror. When light from a point on the object hits the top mirror, it reflects in parallel rays. These parallel rays hit the bottom mirror and reflect so that they reassemble to form a point located at one focal length from the bottom mirror. The mirrors are placed so that the focal point of the bottom mirror is located at the hole in the top of the device. The end result is that light from every point on the object is assembled into an image in the hole. The ray diagram may help explain this effect. The image produced by this apparatus is known as a real image, because the light that forms it actually passes through the location of the image. However, if you place a piece of wax paper or onionskin paper at the location of the real image, the image will not appear on the paper. The outside regions of the mirrors that do not reflect light to your eyes do reflect light to the paper. The edges of the mirrors have large aberrations and create an image so blurred that it cannot be seen.

8.2 Ellipses 1.

2.

854

Kepler’s Laws Kepler was born Dec. 17, 1571, and died Nov. 15, 1630. He was among the first strong supporters of the heliocentric theory. In 1596 Kepler published Cosmographic Mystery, in which he defended the Copernican theory. Tycho Brahe, mathematician at the court of Emperor Rudolph II, was impressed with the work of Kepler and invited him to Prague as his assistant. When Brahe died the following year (1601), Kepler was appointed to the position held by Brahe. Between 1609 and 1619, Kepler published his three laws of motion: 1. Each planet moves about the sun in an orbit that is an ellipse, with the sun at one focus of the ellipse. 2. The straight line joining a planet and the sun sweeps out equal areas in space in equal intervals of time. 3. The squares of the sidereal periods of the planets are in direct proportion to the cubes of the semimajor axes of their orbits. That is, P 2 = ka 3 . The value of k depends on the units of measurement. If astronomical units are used, then k = 1 . Kepler also made contributions to optics and telescope lenses and gave a physical explanation of nova. His text Introduction to Copernican Astronomy was one of the most widely read treatises on astronomy. a. A planet’s velocity is greatest when it is at perihelion. This follows from Kepler’s second law. b. The period of Mars is 1.87 years. This follows from the third law. Neptune Neptune was discovered as a result of mathematical prediction. The perturbative effects of Jupiter and Saturn on Uranus alone did not allow for observed discrepancies in Uranus’s orbit. Using Newtonian gravitational theory, John Couch Adams produced mathematical evidence that an unknown planet could account for the irregularities in Uranus’s orbit. Adams sent his calculations, along with the region of the sky in which to search for the new planet, to Sir George Airy, the Royal Astronomer, and asked him to begin a search for the planet. But Airy had no faith in Adams’ calculations and did not look for Neptune. Copyright © Houghton Mifflin Company. All rights reserved.

Chapter 8

3.



In June of 1846, Leverrier, a French mathematician, independently reproduced the work of Adams. His calculations also were sent to Airy. This time the Royal Astronomer suggested that Cambridge University begin a search for the new planet. The work of Challis, director of Cambridge Observatory, was sloppy and negligent. Consequently, he did not find the planet. In September 1846, Leverrier suggested that Galle of the Berlin Observatory search the Aquarius region of the sky for the planet. Galle found the planet during his first observation. The discovery of Neptune was a great achievement for the time and was a major triumph of gravitational theory. Graph the Colosseum a. A graph of the Colosseum produced by using the Maple commands given in the text.

z

200

y

100

200

−300

−100 0 100 x

300

0 −200

b. The following image appears to be constructed with ellipses because the window is not a “square” 9

-4.7

4.7 -4

window. That is, one unit on the x-axis is not the same length as one unit on the y-axis. The graphs of the equations on a “square” window would appear as semicircles, but on this non-square window they are distorted and appear to be elliptical. 8.3 Hyperbolas a. A Hyperbolic Paraboloid The general equation is

y2 b2

− ax 2 = cz . The graph for a = 3, b = 4, and c = 1 is shown below. The 2

graph is symmetric with respect to the planes x = 0 and y = 0 . z Hyperbola

y

x

The section in the plane x = 0 is y 2 =

Parabola b2 c

z , which is a parabola that opens upward and has its vertex at

the origin. 2 The section in the plane y = 0 is x 2 = − ac z , which is a parabola that opens downward and has its vertex at the origin. Copyright © Houghton Mifflin Company. All rights reserved.

855

Chapter 8



If the surface is cut by the plane z = k > 0 , then the section is the hyperbola axis is parallel to the y-axis and whose vertices are on the parabola y 2 = If k < 0 , then the section is the hyperbola

whose vertices lie on the parabola x 2 = −

x2 a2

2

− by2 =

k c

b2 c

y2 b2

− ax 2 = kc , whose focal 2

z.

, whose focal axis is parallel to the x-axis and

a2 z. c

b. A Hyperboloid of One Sheet x2 y 2 z 2 The general equation is 2 + 2 + 2 = 1 . The graph for a = 3, b = 4 , and c = 5 is shown below. The a b c graph is symmetric with respect to each of the coordinate planes. z Ellipse

y

x Hyperbola

The sections cut by the coordinate planes are x = 0:

y2 z2 − =1 b2 c2

y = 0:

x2 z 2 − =1 a2 c2

The plane z = k cuts the surface in an ellipse with equation is on the z-axis, and its vertices fall on the hyperbolas

y2 b2

x2 y 2 + =1 a2 b2

z = 0: x2 a2

=

− cz 2 = 1 and 2

y2 b2

x2 a2

= 1 + kc2 . The center of the ellipse 2

− cz 2 = 1 . 2

Cooling towers are built in the shape of a hyperboloid of one sheet, because such structures are strong, they have a large carrying capacity, and they can be constructed of many straight, narrow boards. 8.4 Rotation of Axes 1.

Use the Invariant Theorems We are given 10 x 2 + 24 xy + 17 y 2 − 26 = 0 . Thus A = 10, B = 24, C = 17, D = 0, E = 0, and F = 26 . We seek the equation of the form A′ ( x′ ) + C ′ ( y ′ ) − F = 0 . 2

2

The invariant theorems in Exercises 41 and 41 of Section 8.4 indicate that 2 A′ + C ′ = A + C and ( B′ ) − 4 A′C ′ = B 2 − 4 AC . A′ + C ′ = A + C implies A′ + C ′ = 10 + 17 = 27. Hence C ′ = 27 − A′.

( B′ )

856

2

− 4 A′C ′ = B 2 − 4 AC = 242 − 4 (10 )(17 ) = −104 . In the x′y ′ -coordinate system B ′ = 0 , so we have


Chapter 8


−4 A′C ′ = −104


• Substitute 27 − A′ for C ′ .

−4 A′ ( 27 − A′ ) = −104 4 ( A′ ) − 108 A′ + 104 = 0 2

( A′ ) − 27 A′ + 26 = 0 ( A′ − 1)( A′ − 26 ) = 0 2

Hence A′ = 1 or A′ = 26. If A′ = 1, then C ′ = 26. If A′ = 26, then C ′ = 1. Thus we conclude that the conic given by 10 x 2 + 24 xy + 17 y 2 − 26 = 0 is also represented by 26 ( x′ ) + 1( y ′ ) − 26 = 0 or 1( x′ ) + 26 ( y ′ ) − 26 = 0 2

2

2

in the x′y ′ -coordinate system. An examination of the following graph shows that we should pick

( x′ )

2

+ 26 ( y ′ ) − 26 = 0 if we plan to obtain the x′ -axis by a 53° counterclockwise rotation of the x-axis 2

and that we should pick 26 ( x′ ) + ( y ′ ) − 26 = 0 if we plan to obtain the x′ -axis by a 143° 2

2

counterclockwise rotation of the x-axis. 5

8

-8

-5

8.5 Introduction to Polar Coordinates 1.

A Polar Distance Formula a. Let P1 , the origin, and P2 form a triangle ΔPOP 1 2. If we let a = the distance from P1 to the origin, b = the distance from the origin to P2 , and c = the

distance from P1 to P1 , the Law of Cosines defines c 2 = a 2 + b 2 − 2ab cos C , where C is the angle formed at the origin. Substituting r1 for the distance to P1 from the origin ( r1 = a ) and r2 for the distance from the origin to P2 ( r2 = b ) , we find that the distance squared between the two points P1 and P2 is now d [ P1 , P2 ] = r12 + r22 − 2r1r2 cos C . 2

Because the angle C is simply the difference between the angles formed by the two points and the polar axis, we can substitute (θ 2 − θ1 ) for C and arrive at the desired result: d ( P1 , P2 ) = r12 + r22 − 2r1r2 cos (θ 2 − θ1 ) .

b.

d ( ( 3, 60° ) , ( 5, 170° ) ) = 32 + 52 − 2 ( 3)( 5 ) cos (170° − 60° ) ≈ 6.65

c.

Because (θ 2 − θ1 ) = − (θ1 − θ 2 ) and because cos x = cos ( − x ) , this distance formula can also be written as d ( P1 , P2 ) = r12 + r22 − 2r1r2 cos (θ1 − θ 2 ) .


857

Chapter 8 2.


Another Polar Form for a Circle a.


r = a sin θ + b cos θ

r 2 = ar sin θ + br cos θ x 2 + y 2 = ay + bx x 2 − bx + y 2 − ay = 0 x 2 − bx +

b2 a 2 a 2 b2 + y 2 − ay + = + 4 4 4 4 2

2

b⎞ ⎛ a⎞ a 2 + b2 ⎛ − + − = x y ⎜ ⎟ ⎜ ⎟ 2⎠ ⎝ 2⎠ 4 ⎝ 2 2 b⎞ ⎛ a ⎞ ⎛ a2 + b2 ⎛ ⎜ x − ⎟ + ⎜ y − ⎟ = ⎜⎜ 2⎠ ⎝ 2⎠ ⎝ 2 ⎝

Thus the center of the circle is

( b2 , a2 )

⎞ ⎟ ⎟ ⎠

2

and the radius is r =

a 2 + b2 2

.

8.6 Polar Equations of the Conics 1.

Polar Equation of a Line From the figure k = d sin θ p , h = d cos θ p and the slope of the line segment perpendicular to the line is m = tan θ p . Therefore, the equation of the line is given by

⎛ 1 y − d sin θ p = ⎜ − ⎜ tan θ p ⎝

⎞ ⎟⎟ ( x − d cos θ p ) ⎠

Switching to polar, we have ⎛ 1 ⎞ r sin θ − d sin θ p = ⎜ − r cos θ − d cos θ p ) ⎜ tan θ ⎟⎟ ( p ⎠ ⎝ − r sin θ tan θ p + d sin θ p tan θ p = r cosθ − d cosθ p −r sin θ

sin θ p cos θ p

+ d sin θ p

sin θ p cosθ p

= r cos θ − d cosθ p

−r sin θ sin θ p + d sin 2 θ p = r cosθ cos θ p − d cos 2 θ p d sin 2 θ p + d cos 2 θ p = r cosθ cos θ p + r sin θ sin θ p d ( sin 2 θ p + cos 2 θ p ) = r ( cos θ cosθ p + sin θ sin θ p ) d = r cos (θ − θ p ) r=

858

d

cos (θ − θ p )


Chapter 8 3.



Polar Equation of a Circle That Passes Through the Pole From the figure, we have the center of the circle ( h, k ) with h = a cos θ C , k = a sin θ C , and radius a.

( x − h) + ( y − k ) = a2 2 2 ( x − a cosθ C ) + ( y − a sin θ C ) = a 2 2

2

Changing to polar yields

(r

( r cosθ − a cosθ C )

2

2

+ ( r sin θ − a sin θ C ) = a 2 2

cos 2 θ − 2ra cosθ cos θ C + a 2 cos 2 θ C ) + ( r 2 sin 2 θ − 2ra sin θ sin θ C + a 2 sin 2 θ C1 ) = a 2

r 2 ( cos 2 θ + sin 2 θ ) − 2ra ( cosθ cosθ C + sin θ sin θ C ) + a 2 ( cos 2 θ C + sin 2 θ C ) = a 2 r 2 − 2ra ( cos θ cos θ C + sin θ sin θ C ) + a 2 = a 2 r 2 − 2ra ⎡⎣ cos (θ − θ C ) ⎤⎦ = 0 r − 2a ⎣⎡ cos (θ − θ C ) ⎦⎤ = 0 r = 2a ⎡⎣ cos (θ − θ C ) ⎤⎦

8.7 Parametric Equations 1.

Parametric Equations in an xyz-Coordinate System a. Graph of x = 3 cos t , y = 3 sin t , z = 0.5t , for 0 ≤ t ≤ 12 . z 6

4

y

2 2

-2

2

-2

x

b. Graph of x = 3 cos t , y = 6sin t , z = 0.5t , for 0 ≤ t ≤ 12 . z 6 4

y

2 2

-2 -2

2

x

c. The graph in part (a) is “circular,” whereas the graph in part (b) is “elliptical.” d. Each of the curves is a helix.


859

Chapter 9



Systems of Equations CHAPTER 9 of College Algebra and Trigonometry 9.1 Systems of Linear Equations in Two Variables 1. Independent and Dependent Conditions a.

⎧ x + y = 30 . Independent. The system of equations is ⎨ ⎩ x − y = 10 The solution of this system of equations is ( 20, 10 ) .

b. Dependent. The equation is 20 x + 30 y = 1000 . Because three are two variables and only one equation, there are an infinite number of possible solutions. The ordered pairs ( 50, 0 ) and ( 35, 10 ) are two possible solutions.

⎧ x + y = 20 The system of equations is ⎨ . The system of equations has no solution. This means it ⎩ 2 x = 10 − 2 y is impossible to satisfy the two conditions of the problem at the same time. d. Answers will vary. The answer should contain a word problem with two independent conditions. e. Answers will vary. The answer should contain a word problem with two dependent conditions. c.

9.2 Nonlinear Systems of Equations 1. Concept of Dimension In Flatland, people are two-dimensional polygons. A person’s station in life is determined by the number of sides of the polygon. People walk by sliding along the plane. The Flatlanders visit the people from the world of one dimension. These people live on a line. When the Flatlanders tell the ruler of the one-dimensional people that they can change position with their neighbors, the one-dimensional people cannot comprehend how such a movement would take place. The Flatlanders are dutifully smug about their ability to move in the plane and think they are superior to the one-dimensional people. Then strange phenomena begin when a three-dimensional person enters the world of the Flatlanders. For example, the Flatlanders cannot understand how the three-dimensional people enter their homes even though all the doors and windows are latched. 2.

Abilities of a Four-Dimensional Human Some of the best accounts have appeared in Scientific American. Descriptions of the capabilities of a fourdimensional person range from turning a tennis ball inside out without tearing it, to reaching into a closed safe and removing the contents.

9.3 Nonlinear Systems of Equations 1. Proving a Geometry Theorem a.

Substitute y = mx into ( x − a ) + y 2 = a 2 and solve for x. 2

( x − a)

2

+ m2 x 2 = a 2 x 2 − 2ax + a 2 + m 2 x 2 = a 2 (1 + m2 ) x 2 − 2ax = 0

(

)

x (1 + m 2 ) x − 2a = 0

Thus x = 0 or (1 + m 2 ) x − 2a = 0 .

2a 1 + m2 If x = 0 , then y = m ( 0 ) = 0 . One intersection is ( 0, 0 ) . x=

If x = 1+2ma 2 , then y = m

860

( )= 2a 1+ m 2

2 am 1+ m2

. A second intersection is

(

2a 1+ m2

)

, 12+am . m2


Chapter 9


b. The slope of the line through P and ( 2a, 0 ) is: Slope =

2.

2am 1+ m 2 2a 1+ m 2

−0 − 2a

=

2 am 1+ m2 −2 am2 1+ m2

=−


1 . m

c.

The line segment OP is on the line given by y = mx . Thus the slope of OP is m.

d.

OP is perpendicular to PQ because their slopes are negative reciprocals of each other.

e. Since OP is perpendicular to PQ , angle OPQ is a right angle and triangle OPQ is a right triangle. Finding Zeros of a Polynomial Let a, b, and c be zeros of P(x) = x3 + 2x2 + Cx – 6 such that a = b + c . Because these are the zeros of P, the polynomial must factor (by the Factor Theorem) as (x – a)(x – b)(x – c). Now multiply this out and equate the coefficients to those of P. The results enables us to form the system of equations • Equating coefficients of x. a + b + c = −2 abc = −6

• Equating the constant term.

a =b+c

Solving this system yields a = −1, b =

• This is a condition of the problem. −1+ i 23 2

, and c =

−1− i 23 2

.

To find C, solve P ( −1) = 0 = ( −1) + 2 ( −1) + C ( −1) − 6, or C = −9 . 3

9.4 Partial Fractions 1. Using a Computer Algebra System The answer to this question will depend on the rational functions for which the student attempted a partial fraction decomposition using some type of computer algebra system. 9.5 Inequalities in Two Variables and Systems of Inequalities 1.

A Parallelogram Coordinate System Coordinate lines are drawn parallel to the coordinate axes and form a parallelogram. See the accompanying illustration. A point is located in much the same manner as in a rectangular coordinate system. However, displacement is along the edge of a parallelogram rather than of a rectangle.

The graph of 3x + 4 y = 12 is a straight line. In fact, all linear equations in two variables have a graph that is a straight line.

y

x

( 0, 0 ) 3 x + 4 y = 12

Here are some other observations a student may include: Transformation equations between rectangular coordinates and parallelogram coordinates are given by x′ = x and y ′ = y − x , where x′ and y ′ are the coordinates in the parallelogram system. One way to define the distance d between P1 ( x1 , y1 ) and P2 ( x2 , y2 ) is d = x1 − x2 + y1 − y2 .


861

Chapter 9



9.6 Linear Programming 1. History of Linear Programming Linear programming is a program, or method, of allocating resources. It is used in planning food distribution, building rockets, supplying military units with essential equipment, and farming, as well as in many other applications. George Dantzig developed the “simplex method” of solving linear programming problems in the late 1940s. The simplex method is basically a matrix method, analogous to row reduction. The result is the best method of allocating resources. In the early 1980s, Narendra Karmarkar of AT&T suggested an improvement on the simplex method that greatly reduced the time required for a computer to determine the optimal solution of a linear programming problem. In the mid-1980s, L. G. Khachian introduced a method that was supposed to revolutionize the technique of solving linear programming problems. Although his method has some theoretical importance, its practical applications have not been demonstrated.

862


Chapter 10



Matrices CHAPTER 10 of College Algebra and Trigonometry 10.1 Gaussian Elimination Method 1. Echelon Form by Using a Graphing Calculator 1 4⎤ ⎡ 2 −3 ⎢ a. ⎢ 1 2 −2 −2 ⎥⎥ ⎢⎣ 3 1 −3 4 ⎥⎦ b.

R1 ↔ R2

⎡ 1 2 −2 −2 ⎤ ⎢ 2 −3 1 4 ⎥⎥ ⎢ ⎢⎣ 3 1 −3 4 ⎥⎦

−3R1 + R3 → R3

5R2 + R3 → R3

−2R1 + R2 → R2

⎡ 1 2 −2 −2 ⎤ ⎢ 0 −7 5 8 ⎥⎥ ⎢ ⎢⎣ 0 −5 3 10 ⎥⎦ ⎡ 1 ⎢ ⎢ 0 ⎢⎣ 0

2

−2

1

−

0

−

5 7 4 7

⎛ 1⎞ ⎜ − ⎟ R2 ⎝ 7⎠

−2 ⎤ ⎥ − 87 ⎥ − 307 ⎥⎦

⎛ 7⎞ ⎜ − ⎟ R3 ⎝ 4⎠

⎡ 1 2 −2 −2 ⎤ ⎢ 0 −7 5 8 ⎥⎥ ⎢ ⎢⎣ 3 1 −3 4 ⎥⎦ 2 −2 −2 ⎤ ⎡ 1 ⎢ 0 1 − 75 − 78 ⎥⎥ ⎢ ⎢⎣ 0 −5 3 10 ⎥⎦ ⎡ 1 ⎢ ⎢ 0 ⎢⎣ 0

2 1 0

−2

−2 ⎤ ⎥ − − 87 ⎥ 1 − 152 ⎥⎦ 5 7

10.2 The Algebra of Matrices 1. Transformations ⎡ t ⎤ ⎡ 0 −1 0⎤ ⎡ t ⎤ ⎡ −t − 2⎤ a. R90 ⋅ ⎢ t + 2 ⎥ = ⎢1 0 0⎥ ⎢ t + 2 ⎥ = ⎢ t ⎥ ⇒ ( −t − 2, t ) ⇒ x = −t − 2, y = t ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎣ 1 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 1 ⎥⎦ ⎢⎣ 1 ⎥⎦ Using substitution, x = − y − 2 y = −x − 2 b.

c.

⎡ t ⎤ ⎡ −1 0 0 ⎤ ⎡ t ⎤ ⎡ − t ⎤ R y ⋅ ⎢ 3t − 1⎥ = ⎢ 0 1 0⎥ ⎢ 3t − 1⎥ = ⎢ 3t − 1⎥ ⇒ ( −t , 3t − 1) ⇒ x = − t , y = 3t − 1 ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎣ 1 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 1 ⎥⎦ ⎢⎣ 1 ⎥⎦ t = −x Using substitution, y = 3( − x ) − 1 y = −3 x − 1 ⎡ t ⎤ ⎡1 0 −1⎤ ⎡ t ⎤ ⎡ t − 1 ⎤ ⎢1⎥ ⎢ ⎢ ⎥ ⎢ ⎥ T−1,−1 ⋅ ⎢ ⎥ = 0 1 −1⎥ ⎢ 1 ⎥ = ⎢ 1 − 1⎥ ⎢ ⎥ ⎢ t ⎥ ⎢0 0 1 ⎥ ⎢ t ⎥ ⎢ t ⎥ ⎣ ⎦ ⎢1⎥ ⎢ 1 ⎥ ⎣⎢ 1 ⎦⎥ ⎣ ⎦ ⎣ ⎦ ⎡ t − 1 ⎤ ⎡ −1 0 0 ⎤ ⎡ t − 1 ⎤ ⎡ − t + 1 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ R180 ⋅ ⎢ 1 − 1⎥ = ⎢ 0 −1 0⎥ ⎢ 1 − 1⎥ = ⎢ − 1 + 1⎥ ⎢ ⎥ ⎢ t ⎥ ⎢ 0 0 1⎥ ⎢ t ⎥ ⎢ t ⎥ ⎣ ⎦⎢ 1 ⎥ ⎢ 1 ⎥ ⎣⎢ 1 ⎦⎥ ⎣ ⎦ ⎣ ⎦

⎡ −t + 1 ⎤ ⎡1 0 1⎤ ⎡ −t + 1 ⎤ ⎡ −t + 2 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ T1, 1 ⋅ ⎢ − 1 + 1⎥ = ⎢ 0 1 1⎥ ⎢ − 1 + 1⎥ = ⎢ − 1 + 2⎥ ⇒ ( −t + 2, − 1 + 2) ⇒ x = −t + 2, y = − 1 + 2 ⎢ ⎥ t t ⎢ t ⎥ ⎢ 0 0 1⎥ ⎢ t ⎥ ⎢ t ⎥ t = −x + 2 ⎣ ⎦ ⎢⎣ 1 ⎦⎥ ⎣⎢ 1 ⎦⎥ ⎣⎢ 1 ⎦⎥ Using substitution, y = − 1 + 2 = −1 + −2 x + 4 = −2 x + 3 = 2 x − 3 −x + 2 −x + 2 −x + 2 −x + 2 x−2


863

Chapter 10



⎡ t ⎤ ⎡1 0 2 ⎤ ⎡ t ⎤ ⎡ t + 2 ⎤ d. T2,−1 ⋅ ⎢ t 2 ⎥ = ⎢0 1 −1⎥ ⎢t 2 ⎥ = ⎢ t 2 − 1⎥ ⇒ (t + 2, t 2 − 1) ⇒ x = t + 2, y = t 2 − 1 ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ t = x−2 ⎣⎢ 1 ⎦⎥ ⎣⎢0 0 1 ⎦⎥ ⎢⎣ 1 ⎦⎥ ⎣⎢ 1 ⎦⎥ Using substitution, y = ( x − 2) 2 − 1 y = x2 − 4 x + 3

e.

f.

2 ⎡ t ⎤ ⎡ 0 1 0⎤ ⎡ t ⎤ ⎡t ⎤ ⎢ ⎥ R270 ⋅ ⎢ t 2 ⎥ = ⎢ −1 0 0⎥ ⎢ t 2 ⎥ = ⎢ t ⎥ ⇒ (t 2 , t ) ⇒ x = t 2 , y = t ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣⎢ 1 ⎦⎥ ⎣⎢ 0 0 1⎦⎥ ⎢⎣ 1 ⎦⎥ ⎣⎢ 1 ⎦⎥ Using substitution, x = y2

2 ⎡ t ⎤ ⎡ 0 −1 0 ⎤ ⎡ t ⎤ ⎡ − t ⎤ ⎢ ⎥ R90 ⋅ ⎢ t 2 ⎥ = ⎢1 0 0⎥ ⎢ t 2 ⎥ = ⎢ t ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢⎣ 1 ⎥⎦ ⎢⎣ 0 0 1⎥⎦ ⎢⎣ 1 ⎥⎦ ⎣⎢ 1 ⎦⎥

⎡ − t 2 ⎤ ⎡ 1 0 −2 ⎤ ⎡ − t 2 ⎤ ⎡ − t 2 − 2 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ T−2, 1 ⋅ ⎢ t ⎥ = ⎢0 1 1 ⎥ ⎢ t ⎥ = ⎢ t − 1 ⎥ ⇒ ( −t 2 − 2, t − 1) ⇒ x = − t 2 − 2, y = t − 1 ⎢ ⎥ ⎢⎣ 1 ⎥⎦ ⎣⎢0 0 1 ⎦⎥ ⎣⎢ 1 ⎦⎥ ⎣⎢ 1 ⎦⎥ t = y +1 Using substitution, x = −( y + 1) 2 − 2

2.

x = − y2 − 2 y − 3 Translations This project appears on our Internet site at college.hmco.com/info/aufmannCAT.

10.3 The Inverse of a Matrix 1. Multiple Regression Models ⎡ 90,500 ⎤ ⎢ 73,750 ⎥ ⎢ ⎥ B = ⎢117, 200⎥ ⎢ ⎥ ⎢ 59,500 ⎥ ⎢⎣ 74,800 ⎥⎦

⎡ 20 6 1⎤ ⎢16 5 1⎥ ⎢ ⎥ AT = ⎢19 12 1⎥ ⎢ ⎥ ⎢13 4 1⎥ ⎢⎣13 7 1⎥⎦

a.

⎡ 20 16 19 13 13⎤ A = ⎢ 6 5 12 4 7 ⎥ ⎢ ⎥ ⎢⎣ 1 1 1 1 1 ⎥⎦

d.

⎛ ⎡ 20 6 1⎤ ⎞ ⎛ ⎡ 90,500 ⎤ ⎞ ⎜ ⎢16 5 1⎥ ⎟ ⎜ 20 16 19 13 13 ⎢ 73,750 ⎥ ⎟ L 20 16 19 13 13 ⎡ ⎤ ⎜⎡ ⎤ ⎢ ⎤ ⎢ ⎥⎟ ⎜ ⎡ ⎥⎟ ⎢ M ⎥ = ⎜ ⎢ 6 5 12 4 7 ⎥ ⋅ ⎢19 12 1⎥ ⎟ ⋅ ⎜ ⎢ 6 5 12 4 7 ⎥ ⋅ ⎢117, 200⎥ ⎟ ⎢ ⎥ ⎜⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎟ ⎜ ⎢ ⎥⎟ ⎢⎣ N ⎥⎦ ⎜ ⎢⎣ 1 1 1 1 1 ⎥⎦ ⎢13 4 1⎥ ⎟ ⎜ ⎢⎣ 1 1 1 1 1 ⎥⎦ ⎢ 59,500 ⎥ ⎟ ⎜ ⎢⎣13 7 1⎥⎦ ⎟⎠ ⎜⎝ ⎢⎣ 74,800 ⎥⎦ ⎟⎠ ⎝

b.

c.

−1

−1

⎡1355 571 81⎤ ⎡6,962,700⎤ ⎡ 0.03098 −0.01613 −0.39214 ⎤ ⎡6,962,700 ⎤ = ⎢ 571 270 34 ⎥ ⋅ ⎢ 3,079,750 ⎥ ≈ ⎢ −0.01613 0.03417 0.02890 ⎥ ⋅ ⎢ 3,079,750 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 81 34 5 ⎥⎦ ⎢⎣ 415,750 ⎥⎦ ⎢⎣ −0.39214 0.02890 6.35622 ⎥⎦ ⎢⎣ 415,750 ⎥⎦ ⎡ 2974.14 ⎤ ≈ ⎢ 4963.20⎥ ⎢ ⎥ ⎣⎢ 1219.11 ⎦⎥

864


Chapter 10 e. f. 2.



S = 2974.14 x + 4963.20 y + 1219.11 S (10, 8) = 2974.14(10) + 4963.20(8) + 1219.11

= $70,666.10 Cryptography a. The ASCII (American Standard Code for Information Interchange) is a method by which each letter, punctuation mark, and numeral is given a two-number code. This system is used by all computer systems to exchange information. b. Answers will vary. The students should have an m × n matrix, W, in which m represents the length of a code packet and n represents the number of characters in the message. c. Answers will vary. The student should construct an m × m matrix that has an inverse. We will call this matrix E. d. Answers will vary depending on parts (b) and (c) above. However, the student should exhibit the product EW = M . e. Answers will vary. The student should show that E −1 M = W . That is, the product of the inverse of E and the encoded message should be the original message.

10.4 Determinants 1. Determinants, Matrices, and Area ⎡ 2 1 ⎤ ⎡ 4 4 10 10 ⎤ ⎡ 12 16 28 24 ⎤ a. AM = ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎣ 3 2 ⎦ ⎣ 4 8 8 4 ⎦ ⎣ 20 28 46 38⎦ The new figure is a parallelogram. To find the area of the parallelogram, we will use the distance formula to find the length of the base and the formula for the distance between a point and a line to find the height. d=

(12 − 24 )

2

+ ( 20 − 38 ) = 468 = 6 13 2

To find the height, first determine the equation of the line through (12, 20) and (24, 38). 38 − 20 3 m= = 24 − 12 2 3 y − 20 = ( x − 12 ) 2 3 y = x+2 2 3 ( −2 4 13 2 16 ) + 2 − 28 h= = = 2 13 13 4 ( 32 ) + 1 ⎛ 4 13 ⎞ Area of parallelogram = 6 13 ⎜⎜ ⎟⎟ = 24 ⎝ 13 ⎠ Area of rectangle = ( 6 )( 4 ) = 24

(

)

The areas are the same, and det ( A ) = 4 − 3 = 1 .


865

Chapter 10

b.



⎡3 1⎤ ⎡ 4 4 10 10 ⎤ ⎡ 16 20 38 34⎤ AM = ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎣1 1⎦ ⎣ 4 8 8 4 ⎦ ⎣ 8 12 18 14 ⎦ The new figure is a parallelogram. To find the area of the parallelogram, we will use the distance formula to find the length of the base and the formula for the distance between a point and a line to find the height.

d=

(16 − 34 )

2

+ ( 8 − 14 ) = 360 = 6 10 2

To find the height, first determine the equation of the line through (12, 20) and (24, 38). 14 − 8 1 = m= 34 − 16 3 1 ( x − 34 ) 3 1 8 y = x+ 3 3

y − 14 =

h=

( 13 ) 20 + 83 − 12 2 ( − 13 ) + 1

=

− 83 10 9

=

4 10 5

⎛ 4 10 ⎞ Area of parallelogram = 6 10 ⎜⎜ ⎟⎟ = 48 ⎝ 5 ⎠ Area of rectangle = ( 6 )( 4 ) = 24

(

)

The area of the new figure is twice that of the rectangle, and det ( A ) = 3 − 1 = 2 . c.

⎡ 1 2 ⎤ ⎡ 4 4 10 10 ⎤ ⎡ 12 20 26 18⎤ AM = ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎣0.5 1 ⎦ ⎣ 4 8 8 4 ⎦ ⎣ 6 10 13 9 ⎦ The new figure is a line segment and therefore has no area. The student should verify this by showing that all the points lie on the same line. The area is 0; det ( A) = 1 − 1 = 0 .

d. The absolute value of the determinant of A times the original area equals the area of the figure produced by AM. e. Some students may not have realized that the absolute value of the determinant is required for their conjecture (in part d).

866


Chapter 10


10.5 Cramer’s Rule 1. Cramer’s Rule Prove Cramer’s Rule for a 3 by 3 system of linear equations. and z are variables. ax + by + cz = d ex + fy + gz = h jx + ky + mz = n


Consider the following system, where x, y, (I) (II) (III)

Multiply Equation (I) by g, and Equation (II) by (–c), and add. agx + bgy + cgz = dg −cex − cfy − cgz = −ch (ag − ce) x + (bg − cf ) y = dg − ch

(IV)

Multiply Equation (I) by m, and Equation (III) by (–c), and add. amx + bmy + cmz = dm −cjx − cky − cmz = −cn (am − cj ) x + (bm − ck ) y = dm − cn

(V)

Multiply Equation (IV) by ( bm − ck ) , and Equation (V) by ( −bg + cf ) , and add.

( bm − ck )( ag − ce ) x + ( bm − ck )( bg − cf ) y = ( bm − ck )( dg − ch ) ( −bg + cf )( am − cj ) x + ( −bg + cf )( bm − ck ) y = ( −bg + cf )( dm − cn ) ⎡⎣( bm − ck )( ag − ce ) + ( −bg + cf )( am − cj ) ⎤⎦ x = ( bm − ck )( dg − ch ) + ( −bg + cf )( dm − cn ) Thus x= = =

=

The results y =

( bm − ck )( dg − ch ) + ( −bh + cf )( dm − cn ) ( bm − ck )( ag − ce ) + ( −bg + cf )( am − ej ) bdgm − bchm − cdgk + c 2 hk − bdgm + bcgn + cdfm − c 2 fn abgm − bcem − acgk + c 2 ek − abgm + begj + acfm − cefj c {( dfm + bgn + chk ) + ( −cfn − bhm − dgk )} c {( afm + bgj + cek ) + ( −efj − bem − akg )} d b c h f g n k m a b c e f g j k m

a d c e h g j n m a b c e f g j k m

, z=

a b d e f h j k n a b c e f g j k m

can by established by using a similar approach.


867

Chapter 11



Sequences, Series, and Probability CHAPTER 11 of College Algebra and Trigonometry 11.1 Infinite Sequences and Summation Notation 1. Formulas for Infinite Sequences a. The idea is to write a formula that gives 2n for n = 1, 2, 3, 4 and 43 when n = 5. The formula contains two parts. The first part is zero for n = 1, 2, 3, 4. The second part is 2n. ⎡ n ( n − 1)( n − 2 )( n − 3)( n − 4 ) ⎤ an = ( 43 − 2n ) ⎢ ⎥ + 2n n! ⎣ ⎦ The first five terms of this sequence are 2, 4, 6, 8, 43. b. The idea is to write a formula that gives 2n for n = 1, 2, 3, 4 and x when n = 5. The formula contains two parts. The first part is zero for n = 1, 2, 3, 4. The second part is 2n. ⎡ n ( n − 1)( n − 2 )( n − 3)( n − 4 ) ⎤ an = ( x − 2n ) ⎢ ⎥ + 2n n! ⎣ ⎦ The first five terms of this sequence are 2, 4, 6, 8, x. 11.2 Arithmetic Sequences and Series 1. Angles of a Triangle a. 360° b. 540° c. 720° d. ( n − 3)180° 2.

Prove a Formula We know that S n =

( a1 + an )

n 2

and that an = a1 + ( n − 1) d . Substitute for an in the formula for S n and

simplify. Sn =

n n ⎡⎣ a1 + a1 + ( n − 1) d ⎤⎦ = ⎡⎣ 2a1 + ( n − 1) d ⎤⎦ 2 2

11.3 Geometric Sequences and Series 1. Fractals a.

The perimeter after completing the process n times is Pn = 4 ( 53 )

b. As n → ∞, Pn = 4 (

868

)

5 n −1 3

n −1

.

approaches infinity. Therefore, the perimeter is infinite.

k −1

c.

An = 1 + 4Σ nk =1 532 k

d.

25 An = 1 + 4Σ nk =1 532 k = 1 + 4 ( 19 + 815 + 729 +L) k −1


Chapter 11

Beginning with

5 81


, the series is an infinite geometric series with a =

5 81


and r = 95 .

25 ⎛1 5 ⎞ ⎛ 1 5 / 81 ⎞ 1+ 4⎜ + + + L⎟ = 1 + 4 ⎜ + ⎟ ⎝ 9 81 729 ⎠ ⎝ 9 1− 5 / 9 ⎠ ⎛1 9 ⎞ = 1+ 4⎜ + ⎟ ⎝ 9 36 ⎠ ⎛1⎞ = 1+ 4⎜ ⎟ = 2 ⎝4⎠

The area is 2 square units. 11.4 Mathematical Induction 1. Steps in a Mathematical Induction Proof a. Assume 2 + 4 + 8 + L + 2 k = 2 k +1 + 1 . Prove the formula is true for n = k + 1 . That is, prove that 2 + 4 + 8 + L + 2k + 2k +1 = 2k + 2 + 1 . S k +1 = 2 + 4 + 8 + L + 2k + 2k +1 = S k + 2k +1 = 2k +1 + 1 + 2k +1 = 2 ⋅ 2k +1 + 1 = 2k + 2 + 1 Thus the statement is true for n = k + 1 . b. Let n = 1 . 21 ≠ 2 2 + 1 = 5 . This statement is not true for n = 1 .

c.

2 + 4 + 8 + L + 2k = 2 (1 + 2 + 4 + 8 + L + 2k −1 )

Let N = 1 + 2 + 4 + 8 + L + 2k −1 . Thus 2 + 4 + 8 + L + 2k = 2 ( N ) = even number 2k+1 + 1 = 2 ⋅ 2k + 1 = even number +1 = odd number Therefore, the left side is always an even number and the right side is always an odd number. Thus the two values can never be equal. d. The Principle of Mathematical Induction requires that we first establish that there is at least one element in the set S. We did not do that, and consequently, we apparently “proved” a statement that is always false. 2.

The Tower of Hanoi a. The proof is by induction. If there is one disk ( n = 1) , then 21 − 1 = 2 − 1 = 1 and the game is

completed in one move. Assume that for k disks, the game can be completed in 2k − 1 moves. Prove that for n = k + 1 disks, the game is completed in 2k +1 − 1 moves. Consider one peg in which k + 1 disks are arranged. By the Induction Hypothesis, 2k − 1 moves are required to move the first k disks to another peg. Now move the k + 1 disk to the unoccupied peg. The total number of moves is now 2k − 1 + 1 . Now move the k disks back to the disk containing the k + 1 disk. This requires 2k − 1 moves (Induction Hypothesis). The total number of moves is 2k − 1 + 1 + 2k − 1 = 2 ⋅ 2k − 1 = 2k +1 − 1 Thus the statement is true for n = k + 1 , and the formula is established. b. From Exercise 2(a), it will take 264 − 1 seconds to complete the transfer. 264 − 1 seconds ≈ 5.85 × 1011 years = 585 billion years Thus the legend predicts that the universe will exist for approximately 580 billion more years.


869

Chapter 11



11.5 The Binomial Theorem 1. Pascal’s Triangle Two references a student can check are The History of Mathematics: An Introduction by David M. Burton, Dubuque, Iowa: William C. Brown, publishers, 1988 and The History of Mathematics: An Introduction by Victor J. Katz, New York: Harper Collins, 1993. 2.

Some Other Factorial Functions a. b.

For integers, Pochammer ( m, n ) =

( m + n − 2)! ( n −1)!

.

n !! = n ( n − 2 )( n − 4 )K 2 when n is an even integer and n !! = n ( n − 2 )( n − 4 )K1 when n is an odd

integer. 8.6 Permutations and Combinations 1. Explain Permutations and Combinations The student should prepare a lesson to explain permutations and combinations to a classmate. The lesson should contain at least five examples of permutations and five examples of combinations. 2.

Application of Counting a. n = 5, k = 3 ⎛ 3 ⎞ 5 ⎛ 3⎞ ⎛3⎞ 5 5 5 5 ⎜ ⎟ 3 − ⎜ ⎟ ( 3 − 1) + ⎜ ⎟ ( 3 − 2 ) = 3 − 3 ( 2 ) + 3 = 150 0 1 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

b.

n = 10, k = 4 ⎛ 4 ⎞ 10 ⎛ 4 ⎞ ⎛ 4⎞ ⎛ 4⎞ 10 10 10 10 10 10 ⎜ ⎟ 4 − ⎜ ⎟ ( 4 − 1) + ⎜ ⎟ ( 4 − 2 ) − ⎜ ⎟ ( 4 − 3) = 4 − 4 ( 3 ) + 6 ( 2 ) − 4 = 818,520 ⎝0⎠ ⎝1 ⎠ ⎝ 2⎠ ⎝3⎠

11.7 Introduction to Probability 1. Monte Hall Problem If the contestant stays with his or her first choice, then the probability that the contestant will win the grand prize is 1/3. The only other possibility is to go with one of the other curtains when given the opportunity to switch to a different curtain. This event of switching to another curtain is the compliment of staying with the first choice. Thus the probability of winning the grand prize by switching to a different curtain is 1 2 1 − = . This analysis shows that a contestant will double their chance of winning the grand prize by 3 3 switching rather than staying with their first choice. Many Internet sites discuss this famous problem. Here are two recommended sites: http://www.cut-the-knot.com/hall.html http://www.math.rice.edu/~ddonavan/montyurl.html

870

2.

Probability and Automatic Garage Door Openers The probability of at least two having the same garage door opener sequence is 1 – (the probability of none having the same sequence). 0 500 ⎛ 500 ⎞ ⎛ 1 ⎞ ⎛ 63 ⎞ P = 1− ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ≈ 1 − 0.00038 = 0.99962 ⎝ 0 ⎠ ⎝ 64 ⎠ ⎝ 64 ⎠ There is more than a 99.9% chance that at least two people will choose the same code sequence.

3.

Overbooking by Airlines This project appears on our Internet site at http://www.college.hmco.com.



Additional College Trigonometry, 6e Solutions Section 1.1 1.

2 x + 10 = 40 2 x = 40 − 10 2 x = 30 x = 15

2.

−3 y + 20 = 2 −3 y = 2 − 20 −3 y = −18 y=6

3.

5 x + 2 = 2 x − 10 5 x − 2 x = −10 − 2 3x = −12 x = −4

4.

4 x − 11 = 7 x + 20 4 x − 7 x = 20 + 11 −3x = 31 x = − 31 3

5.

2( x − 3) − 5 = 4( x − 5)

6.

6(5s − 11) − 12(2 s + 5) = 0 30s − 66 − 24 s − 60 = 0 6s − 126 = 0 6s = 126 s = 21

2 x − 6 − 5 = 4 x − 20 2 x − 11 = 4 x − 20 2 x − 4 x = −20 + 11 −2 x = −9 x=

9 2

7.

3x+1 = 2 4 2 3 ⎛ 3 1 12 ⎜ x + ⎞⎟ = 12 ⎛⎜ 2 ⎞⎟ 2⎠ ⎝4 ⎝ 3⎠ 9x + 6 = 8 9x = 8 − 6 9x = 2 x=2 9

8.

x −5= 1 4 2 4 ⎛⎜ x − 5 ⎞⎟ = 4 ⎛⎜ 1 ⎞⎟ ⎝4 ⎠ ⎝2⎠ x − 20 = 2 x = 2 + 20 x = 22

9.

2 x −5= 1 x −3 3 2 ⎛ ⎞ 2 6 ⎜ x − 5 ⎟ = 6 ⎛⎜ 1 x − 3 ⎞⎟ ⎝2 ⎠ ⎝3 ⎠ 4 x − 30 = 3x − 18 4 x − 3x = −18 + 30 x = 12

10.

1 x + 7 − 1 x = 19 2 4 2 4 ⎛⎜ 1 x + 7 − 1 x ⎞⎟ = 4 ⎛⎜ 19 ⎞⎟ 4 ⎠ ⎝2 ⎝ 2⎠ 2 x + 28 − x = 38 x = 38 − 28 x = 10

11.

0.2 x + 0.4 = 3.6 0.2 x = 3.2 x = 16

12.

0.04 x − 0.2 = 0.07 0.04 x = 0.27 x = 6.75

13.

3 ( n + 5) − 3 ( n − 11) = 0 5 4 3 3 ⎡ 20 n + 5) − ( n − 11)⎤ = 20 − 0 ⎢⎣ 5 ( ⎥⎦ 4 12 ( n + 5) − 15 ( n − 11) = 0 12n + 60 − 15n + 165 = 0 −3n = −225 n = 75

15.

3( x + 5)( x − 1) = (3 x + 4)( x − 2) 2

16.

2

18.

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

3x + 12 x − 15 = 3 x − 2 x − 8 14 x = 7 x=

− 5 ( p + 11) + 2 ( 2 p − 5) = 0 7 5 ⎡ 5 2 35 − ( p + 11) + ( 2 p − 5) ⎤ = 35 − 0 ⎢⎣ 7 ⎥⎦ 5 −25 ( p + 11) + 14 ( 2 p − 5) = 0 −25 p − 275 + 28 p − 70 = 0 3 p = 345 p = 115

14.

17.

0.08 x + 0.12(4000 − x) = 432 0.08 x + 480 − 0.12 x = 432 −0.04 x = −48 x = 1200

20.

3x − 5 y = 15 −5 y = −3x + 15 y = 3x−3 5

2

5 x − 80 = 5 x − 11x − 12 11x = 68

1 2

x=

0.075 y + 0.06(10,000 − y ) = 727.50 0.075 y + 600 − 0.06 y = 727.50 0.015 y = 127.5 y = 8500

19.

68 11

x + 2y = 8 2 y = −x + 8 y =−1x+4 2


874


2 x + 5 y = 10 2 x = −5 y + 10 x = −5 y +5 2

22.

24.

ax + by = c by = − ax + c y = − ax + c b

25.

27.

x 2 − 2 x − 15 = 0 a = 1 b = −2 c = −15

28.

21.

x 2 − 5 x − 24 = 0 a = 1 b = −5 c = −24

ay − by = c y ( a − b) = c y=

26.

29.

c a−b

2y − 3 y −1 x ( y − 1) = 2 y − 3 xy − x = 2 y − 3 xy − 2 y = x − 3 y ( x − 2) = x − 3 y = x−3 x−2 x=

x2 + x − 1 = 0 a = 1 b = 1 c = −1

− (−2) ± ( −2) 2 − 4(1)(−15) 2(1)

x=

− (−5) ± (−5) 2 − 4(1)(−24) 2(1)

x=

x=

2 ± 4 + 60 2

x=

5 ± 25 + 96 2

x=

−1± 1+ 4 2

x=

−1± 5 2

2 ± 64 2 ± 8 = 2 2 2 + 8 10 = = 5 or x= 2 2 2−8 −6 = = −3 x= 2 2

5 ± 121 5 ± 11 = 2 2 5 + 11 16 x= = = 8 or 2 2 5 − 11 − 6 x= = = −3 2 2 x=

x2 + x − 2 = 0 a = 1 b = 1 c = −2 x=

31.

− 1 ± 12 − 4(1)(−2) 2(1)

−1± 1+ 8 1± 9 −1± 3 = = 2 2 2 −1+ 3 2 x= = = 1 or 2 2 −1− 3 − 4 x= = = −2 2 2 x=

33.

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

23.

x=

x=

30.

5 x − 4 y = 10 5 x = 4 y + 10 x= 4 y+2 5

3x 2 − 5 x − 3 = 0 a = 3 b = −5 c = −3

34.

2 x2 + 4 x + 1 = 0 a = 2 b = 4 c =1

32.

− 1 ± 12 − 4(1)(−1) 2

2 x2 + 4 x − 1 = 0 a = 2 b = 4 c = −1

x=

− 4 ± 42 − 4( 2)(1) 2(2)

x=

− 4 ± 42 − 4(2)(−1) 2( 2)

x=

− 4 ± 16 − 8 4

x=

− 4 ± 16 + 8 4

x=

−4± 8 −4±2 2 = 4 4

x=

− 4 ± 24 − 4 ± 2 6 = 4 4

x=

−2± 2 2

x=

−2± 6 2

3x 2 − 5 x − 4 = 0 a = 3 b = −5 c = −4

x=

− (−5) ± (−5) 2 − 4(3)(−3) 2(3)

x=

− (−5) ± (−5) 2 − 4(3)(−4) 2(3)

x=

5 ± 25 + 36 6

x=

5 ± 25 + 48 6

x=

5 ± 61 6

x=

5 ± 73 6



35.

1 2 3 x + x −1 = 0 2 4 1 3 a= b= c = −1 2 4

x=

x=

−3 ± 4

−3 ± 4

()

875

36.

()

2

3 2 − 4 1 (−1) 4 2 2 1 2

()

2 2 1 x − 5x + = 0 3 2 2 1 a= b = −5 c = 3 2

x=

9 +2 16

x=

1 3 41 x=− ± 4 16

x=

3 41 x=− ± 4 4 −3 ± 41 x= 4

x=

a= 2

( 3 ) ( 12 )

−( −5) ± (−5) − 4 2

(3)

2 2

x=

−3 ± 32 − 4 ⋅ 2 ⋅ 2 2 2

x=

−3 ± 9 − 8 2 2

x=

3

4 3 71 3

5±

b=3 c= 2

−3 + 1 = 2 2 −3 − 1 = x= 2 2

25 − 4

5±

2 x2 + 3x + 2 = 0

37.

−2 2 =− 2 2 2 −4 =− 2 2 2

or

4 3

(5 ± 213 ) ⎛ 3 ⎞ 3 ⎜ ⎟ 4 ⎝3⎠ 3

15 ± 213 x= 4 38.

2 x2 + 5 x − 3 = 0

a=2 b= 5

39.

x 2 − 3x − 5 = 0 a =1 b = −3

c = −3

− 5 ± ( 5 )2 − 4(2)(−3) x= 2⋅2 x=

− 5 ± 5 + 24 4

− 5 ± 29 x= 4

41.

44.

47.

x 2 − 2 x − 15 = 0 ( x + 3)( x − 5) = 0 or x − 5 = 0 x+3= 0 x = −3 x=5

42.

12w2 − 41w + 24 = 0 (4 w − 3)(3w − 8) = 0 4 w − 3 = 0 or 3w − 8 = 0 3 8 w= w= 4 3

45.

( x − 5) 2 − 9 = 0 [( x − 5) − 3][( x − 5) + 3] = 0

48.

x − 8 = 0 or x − 2 = 0 x=8 x=2

x 2 = 3x + 5

40.

− x2 = 7 x − 1 − x2 − 7 x + 1 = 0 a = −1 b = −7

c = −5

c =1

x=

− (−3) ± (−3)2 − 4(1)(5) 2(1)

x=

− (−7) ± (−7)2 − 4(−1)(1) 2( −1)

x=

3 ± 9 + 20 2

x=

7 ± 49 + 4 −2

x=

3 ± 29 2

x=

7 ± 53 −2

x=

− 7 ± 53 2

y 2 + 3 y − 10 = 0 ( y + 5)( y − 2) = 0 y+5 = 0 or y − 2 = 0 y = −5 y=2

43.

3x 2 − 7 x = 0 x (3 x − 7 ) = 0 x = 0 or 3x − 7 = 0 7 x= 3

46.

(3 x + 4)2 − 16 = 0 [(3x + 4) − 4)][(3 x + 4) + 4] = 0 3x = 0 or 3x + 8 = 0

49.

x=0

x=−

8 y 2 + 189 y − 72 = 0 (8 y − 3)( y + 24) = 0

8y − 3 = 0 3 y= 8

or

y + 24 = 0 y = −24

5 x 2 = −8 x 2

5x + 8x = 0 x(5 x + 8) = 0 x = 0 or 5 x + 8 = 0 x = −8 5 2 x + 3 < 11 2x < 8 x 16 3 x > 21 x>7

51.

x + 4 > 3x + 16 −2 x > 12 x < −6

52.

5x + 6 < 2x + 1 3x < −5 5 x 2( x − 4) −12 x + 20 > 2 x − 8 − 14 x > −28 x 0 x( x + 7) > 0

60.

The product is positive. The critical values are 0 and –7.

The product is negative or zero. The critical values are 0 and 5.

x( x − 7)

x( x − 5)

(−∞, − 7) ∪ (0, ∞) 61.

[0, 5]

x 2 + 7 x + 10 < 0

62.

x2 + 5x + 6 < 0

( x + 5)( x + 2) < 0

( x + 3)( x + 2) < 0

The product is negative. The critical values are –5 and –2.

The product is negative. The critical values are –3 and –2.

( x + 5)( x + 2)

( x + 3)( x + 2)

(−3, − 2)

(−5, − 2) 63.

x2 − 5x ≤ 0 x( x − 5) ≤ 0

x 2 − 3x ≥ 28 x − 3x − 28 ≥ 0 ( x + 4)( x − 7) ≥ 0

x 2 + x − 30 < 0 ( x − 5)( x + 6) < 0

The product is positive or zero. The critical values are –4 and 7. ( x + 4)( x − 7)

(−∞, − 4] ∪ [7, ∞)

x 2 < − x + 30

64.

2

The product is negative. The critical values are 5 and –6. ( x − 5)( x + 6)

(−6, 5)



877

6 x2 − 4 ≤ 5x

65.

12 x 2 + 8 x ≥ 15

66.

6 x2 − 5x − 4 ≤ 0 (3x − 4)(2 x + 1) ≤ 0

12 x 2 + 8 x − 15 ≥ 0 (6 x − 5)(2 x + 3) ≥ 0

The product is negative or zero.

The product is positive or zero.

The critical values are 4 and − 1 . 3 2

The critical values are 5 and − 3 . 6

(6 x − 5)(2 x + 3)

(3 x − 4)(2 x + 1)

⎛ 3⎤ ⎡5 ⎞ ⎜⎜ − ∞,− ⎥ ∪ ⎢ , ∞ ⎟⎟ 2⎦ ⎣6 ⎠ ⎝

⎡ 1 4⎤ ⎢− 2 , 3 ⎥ ⎣ ⎦ 67.

70.

68.

x 2

x −1 < 9

−4 < x < 4

x < −2 or x > 2

−9 < x − 1 < 9 − 8 < x < 10

(−4, 4)

(−∞, − 2) ∪ (2, ∞)

(−8, 10)

71.

x − 3 < 10

72.

x + 3 > 30 x + 3 < −30 or x + 3 > 30

−10 < x − 3 < 10 − 7 < x < 13

x < −33

74.

2x − 1 > 4 2 x − 1 < −4 or 2 x − 1 > 4 2 x < −3 2x > 5 3 5 x 2 2

2x − 9 < 7

x+4 < 2 −2 < x + 4 < 2

x > 27

− 6 < x < −2

(−∞, − 33) ∪ (27, ∞)

(−7, 13) 73.

2

(−6, − 2) 75.

−7 < 2 x − 9 < 7

x+3 ≥ 5 x + 3 ≤ −5 or x + 3 ≥ 5

2 < 2 x < 16 1< x < 8

x ≤ −8

x≥2

(−∞, − 8] ∪ [2, ∞)

(1, 8)

3⎞ ⎛ ⎛5 ⎞ ⎜ − ∞, − ⎟ ∪ ⎜ , ∞ ⎟ 2⎠ ⎝ ⎝2 ⎠ 76.

77.

x − 10 ≥ 2

78.

−14 ≤ 3 x − 10 ≤ 14 − 4 ≤ 3 x ≤ 24 − 4 ≤ x≤8 3

x − 10 ≤ −2 or x − 10 ≥ 2 x≤8

3 x − 10 ≤ 14

x ≥ 12

(−∞, 8) ∪ (12, ∞)

2 x − 5 ≤ −1 2x ≤ 4 x≤2

80.

4 − 5 x ≥ 24 4 − 5 x ≤ 24 −5 x ≤ −28 x ≥ 28 5

or

4 − 5 x ≥ 24 −5 x ≥ 20 x ≤ −4

⎡ (−∞, − 4] ∪ ⎢ 28 , ∞ ⎟⎞ ⎠ ⎣5

3 − 2x ≤ 5 −5 ≤ 3 − 2 x ≤ 5 − 8 ≤ −2 x ≤ 2 4 ≥ x ≥ −1

[−1, 4]

or 2 x − 5 ≥ 1 or

2x ≥ 6 x≥3

(−∞, 2] ∪ [3, ∞)

⎡ 4 ⎤ ⎢ − 3 , 8⎥ ⎣ ⎦ 79.

2x − 5 ≥ 1

81.

x −5 ≥ 0

Because an absolute value is always nonnegative, the inequality is always true. The solution set consists of all real numbers. (−∞, ∞)


878

82.


83.

x−7 ≥ 0

Because an absolute value is always nonnegative, the inequality is always true. The solution set consists of all real numbers.

84.

x−4 ≤ 0

2x + 7 ≤ 0

Because an absolute value is always nonnegative, the inequality x − 4 < 0

2x + 7 = 0 2 x = −7 7 x=− 2

has no solution. Thus the only solution of the inequality x − 4 ≤ 0 is the solution of the equation x – 4 = 0.

(−∞, ∞)

x=4 85.

A = 35

86.

LW = 60

LW = 35 L=

A = 60 A = LW

A = LW 35 W

W =

P = 34 P = 2 L + 2W 2 L + 2W = 34 L + W = 17 60 L + =17 L

P = 27 P = 2 L + 2W 2 L + 2W = 27

⎛ 35 ⎞ 2 ⎜ ⎟ + 2W = 27 ⎝W ⎠ 70 + 2W 2 = 27W

L2 − 17 L + 60 = 0 ( L − 12)( L − 5) = 0

2W 2 − 27W + 70 = 0 (2W − 7)(W − 10) = 0 W =

7 2

L=5

L = 12

or W = 10

W =

7 L 2 70 = 7 L 10 = L 35 =

60 L

35 = LW 35 = 10 L 3.5 = L

60 =5 12

W =

60 = 12 5

The rectangle measures 5 ft by 12 ft.

The rectangle measures 3.5 cm by 10 cm. 87.

A = 1500 = lw P = 600 − 2l + 3w 15000 l= w

88.

2l + 3w = 600 ⎛ 15000 ⎞ 2⎜ ⎟ + 3w = 600 ⎝ w ⎠

P = 4 w + 2l = 400 2 w + l = 200 A = 4800 = lw 4800 l= w

2w +

4800 = 200 w

2 w2 + 4800 = 200w

30,000 + 3w2 = 600w

w2 − 100w + 2400 = 0 ( w − 60)( w − 40) = 0

3w2 − 600 w + 30,000 = 0 3( w2 − 200w + 10,000) = 0 3( w − 100)( w − 100) = 0

w = 60 4800 = 80 l= 60

w = 100 ft 15000 l= = 150 ft 100

The dimensions are 100 feet by 150 feet.

w = 40 4800 = 120 l= 40

There are two solutions: 60 yd × 80 yd or 40 yd × 120 yd.



89.

91.

879

Plan A: 5 + 0.01x

90.

Company A: 19 + 0.12M

Plan B: 1 + 0.08x

Company B: 12 + 0.21M

5 + 0.01x < 1 + 0.08x 4 < .07x 57.1 < x

19 + 0.12M < 12 + 0.21M 7 < 0.09M 77.7 < M

Plan A is less expensive if you use at least 58 checks.

Company A is less expensive if you drive at least 78 miles.

Plan A: 100 + 8x

92.

Plan A: 15 + 1.49x

93.

Plan B: 250 + 3.5x

Plan B: 1.99x

100 + 8x > 250 + 3.5x 4.5x > 150 x > 33.3

1.99x < 15 + 1.49x 0.50x < 15 x < 30

Plan A pays better if at least 34 sales are made.

If fewer than 30 videos are rented, Plan B is less expensive.

68 ≤ F ≤ 104 9 68 ≤ C + 32 ≤ 104 5 9 36 ≤ C ≤ 72 5 20 ≤ C ≤ 40

....................................................... 94.

a.

l w = w l−w


l (l − w) = w2 l 2 − lw = w2

So 1 + 2 + 3 + L + 21 + 22 = 253 .

0 = w2 + lw − l 2 w=

− l ± l 2 − 4(−l 2 ) 2

Since w > 0, w =

b.

96.

⎛ −1+ 5 ⎞ −l +l 5 ⎟ = l⎜ ⎟ ⎜ 2 2 ⎠ ⎝

⎛ −l + 5 ⎞ ⎟ ≈ 62.4 ft w = 101 ⎜ ⎟ ⎜ 2 ⎠ ⎝ 464 =

a.

n (n − 3) 2

n 2 − 3n − 928 = 0

R = 420 x − 2 x 2

2 x(210 − x) > 0 The product is positive. The critical values are 0 and 210.

The polygon has 32 sides. 12 =

97.

420 x − 2 x 2 > 0

(n − 32)(n + 29) = 0 n = 32

b.

253 = 1 n( n + 1) 2 n 2 + n − 506 = 0 ( n + 23)( n − 22) = 0 n = 22

2 x(210 − x)

n (n − 3) 2

24 = n 2 − 3 0 = n 2 − 3 − 24

(0, 210)

n 2 − 3n − 24 is not factorable over the integers. Thus, the polygon in a. cannot have 12 diagonals.


880

98.


99.

x −3 < 8

a.

s = −16t 2 + v0t + s0

v0 = 64,

s0 = 0

−16t + 64t > 48

x− j < k

b.

s > 48,

2

− 16t + 64t − 48 > 0 − 16(t 2 − 4t + 3) > 0 − 16(t − 3)(t − 1) > 0 The product is positive. The critical values are t = 3 and t = 1. −16(t − 3)(t − 1)

1 second < t < 3 seconds The ball is higher than 48 ft between 1 and 3 seconds. 100.

v0 = 80,

s0 = 32

101. a.

s − 4.25 ≤ 0.01

b.

s − 4.25 = 0.01,

− 16t 2 + 80t + 32 > 96 − 16t 2 + 80t − 64 > 0

or

s = 4.26

− 16(t 2 − 5t + 4) > 0 − 16(t − 4)(t − 1) > 0

s − 4.25 = −0.01 s = 4.24 critical values

4.24 ≤ s ≤ 4.26

The product is positive. The critical values are t = 4 and t = 1. −16(t − 4)(t − 1)

1 second < t < 4 seconds The ball is higher than 96 ft between 1 and 4 seconds.

.......................................................


PS1.

4 + (−7) −3 = 2 2

PS2.

50 = 25 2 = 5 2

PS3.

y = 3x − 2

PS4.

y = (−3) 2 − 3( −3) − 2 y =9+9−2 y = 16

?

5 = 3(−1) − 2 No, the equation is not true. 5 ≠ −5

PS5.

−3 − ( −1) = −3 + 1 = −2 = 2

PS6.

(−3)2 − 4(−2)(2) = 9 + 16 = 25 = 5



881

Section 1.2

....................................................... PS1.

y = 3x + 12 0 = 3x + 12 −12 = 3x −4 = x

PS3.

y2 = x a2 =9 a = −3 or 3


y = x 2 − 4x + 3 0 = x 2 − 4x + 3 0 = ( x − 1)( x − 3) x −1 = 0 x − 3 = 0 x =1 x=3 1 and 3

PS4. d = (3 − (−4))2 + (−2 −1)2 = 49 + 9 = 58

y2 = x b2 = 9 b = −3 or 3

PS5. –4

PS6. Decrease

Section 1.3

....................................................... PS1.


f (3) =

3(3)4 (3) 2 + 1

f ( −3) =

= 243 = 24.3 10

3( −3)4

( −3)2 + 1 f (3) = f ( −3)

= 243 = 24.3 10

The graph of g is one unit above the graph of f. PS3.

f ( −2) = 2( −2)3 − 5( −2) = −16 + 10 = −6

PS4.

− f (2) = −[2(2)3 − 5(2)] = −[16 − 10] = −6 f ( −2) = − f (2)

f ( −2) − g ( −2) = ( −2) 2 − [ −2 + 3] = 4 − 1 = 3 f ( −1) − g ( −1) = ( −1)2 − [ −1 + 3] = 1 − 2 = −1 f (0) − g (0) = (0) 2 − [0 + 3] = 0 − 3 = −3 f (1) − g (1) = (1)2 − [1 + 3] = 1 − 4 = −3 f (2) − g (2) = (2)2 − [2 + 3] = 4 − 5 = −1

PS5.

− a + a = 0, b + b = b 2 2 midpoint is (0, b)

PS6.

− a + a = 0, −b + b = 0 2 2 midpoint is (0, 0)

Section 1.4

....................................................... PS1.

f (3) − g (3) = ( 3 2 + 3(3) + 1) − ( 4(3) + 5) = 19 − 17 =2


f ( −2) ⋅ g ( −2) = ( 3( −2) 2 − ( −2) − 4 ) ⋅ ( 2( −2) − 5) = (12 + 2 − 4 ) ⋅ ( −9 ) = 10 ⋅ ( −9) = −90


882

PS3.


f (3a ) = 2(3a )2 − 5(3a ) + 2

f (2 + h ) = 2(2 + h )2 − 5(2 + h ) + 2

PS4.

= 2h 2 + 8h + 8 − 5h − 10 + 2

= 18a 2 − 15a + 2

= 2h 2 + 3h PS6. 2 x − 8 = 0 x=4 Domain: x > 4 or [4, ∞)

PS5. Domain: all real numbers except x = 1

Section 1.5

.......................................................


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

PS1. 2 x + 5 y = 15 5 y = −2 x + 15 y =−2x+3 5

PS2.

PS4. (3, 7)

PS5. All real numbers.

PS3.

f (−1) =

2(−1) 2 = 2 = −1 (−1) − 1 −2

PS6. x + 2 ≥ 0 x ≥ −2 { x x ≥ −2}

Section 1.6

.......................................................


See CAT Prepare for Section 2.7 solutions on page 156.

....................................................... 1.

a, c, d, e

2.

Chapter 1 Assessing Concepts

f [ g ( x )] = 3(2 x + 4) + 8 = 6 x + 12 + 8 = 6 x + 20

3.

f (2) = 3 f ( x ) = f ( x + 4) f (2) = f (2 + 4) = f (6) f (6) = f (6 + 4) = f (10) f (10) = f (10 + 4) = f (14) f (14) = f (14 + 4) = f (18) Thus, f (18) = f (2) = 3.

6.

(3, –2)

g [ f ( x )] = 2(3x + 8) + 4 = 6 x + 16 + 4 = 6 x + 20 Thus f [ g ( x )] = 6 x + 20 = g [ f ( x ) ] .

To be inverse functions, f [ g ( x )] = x = g [ f ( x )] . No. They are not inverse functions. 4.

7.

x+2 5 − 2(3a − 4) 7 a > 5 − 6a + 8 13a > 13 a >1

11.

−3±

s=

− 3 ± 9 + 32 4

s=

− 3 ± 41 4

x 2 − x − 12 ≥ 0 ( x − 4)( x + 3) ≥ 0

2

Critical values are 4 and − 3.

Critical values are − −

14.

2x − 5 > 3 2 x − 5 > 3 or 2x > 8 x>4

2x < 2 x 6 − 5x > 2 x < −2

x>2

3

( −∞,

−2 5

)

5

∪ (2, ∞)



6.

d = ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2

893

7.

d = [4 − ( −2)] 2 + ( −2 − 5) 2 d = 36 + 49 d = 85

4−2 =1 2 −1 + 3 =1 ym = 2

xm =

length = ( x2 − x1)2 + ( y2 − y1)2

midpoint = (1, 1)

= [4 − (−2)]2 + (−1 − 3)2 = 62 + ( −4)2 = 36 + 16 = 52 = 2 13

8.

x = 2 y2 − 4 x=0

9.

2 y2 − 4 = 0 2 y2 = 4 y2 = 2 y=± 2

If y = 0,

x = −4

intercepts (0, − 2), (0,

10.

x 2 − 4x + y 2 + 2 y − 4 = 0 ( x 2 − 4 x + 4) + ( y 2 + 2 y + 1) = 4 + 4 + 1 ( x − 2) 2 + ( y + 1) 2 = 9

center: (2, − 1) 12.

2), (−4, 0)

11.

f ( −3) = − 25 − ( −3) 2 f ( −3) = − 16 f ( −3) = −4

radius: 3

x 2 − 16 ≥ 0 ( x − 4)( x + 4) ≥ 0

f ( x ) = − 25 − x 2

13.

Critical values 4 and −4. Domain {x | x ≥ 4 or x ≤ −4}

f is increasing on (−∞, 2] f is decreasing on [2, ∞ )



894

14.

First shift the graph of f(x) horizontally 2 units to the left. Next, reflect the graph across the x-axis. Finally, shift the graph vertically down 1 unit.

15.

a.

f (− x) = (− x)4 − ( − x) 2

= x4 − x2 = f ( x) The function of f ( x) = x 4 − x 2 is an even function. b.

f ( − x ) = ( − x )3 − ( − x ) = − x3 + x = −( x3 − x ) = − f ( x)

The function f ( x) = x3 − x is an odd function. c.

f (− x) = − x − 1 The function f ( x) = x − 1 is neither an even nor an odd function.

Thus, only b defines an odd function. 16.

( f + g )( x ) = ( x 2 − 1) + ( x − 2) = x2 + x−3 ⎛ f ⎞ x 2 −1, x ≠ 2 ⎜ ⎟ ( x) = g x−2 ⎝ ⎠

17.

18.

( f o g )( x ) = ( x − 2 ) 2 − 2 x − 2 + 1 = x − 2 − 2 x − 2 +1 = x − 2 x − 2 −1

19.

2 2 f ( x + h ) − f ( x ) ⎡⎣ ( x + h ) + 1⎤⎦ − ( x + 1) = h h 2 2 2 = x + 2 xh + h + 1 − x − 1 h 2 = 2 xh + h h = 2x + h

x x +1 Interchange x and y. Then solve for y. y x= y +1 x( y + 1) = y y=

xy + x = y xy − y = − x y ( x − 1) = − x x −x y= = x −1 1− x x −1 f ( x) = 1− x 20.

a.

Enter the data on your calculator. The technique for a TI-83 is illustrated here.

y = −7.98245614 x + 767.122807 b.

y = −7.98245614(89) + 767.122807 ≈ 57 Calories



895

....................................................... 1.

d = (4 − (−3))2 + (1 − 2) 2 = 49 + 1 = 50

[1.1]

Chapter 2 Cumulative Review 2.

c2 = a 2 + b2

[1.1]

2

1 = ⎛⎜ 1 ⎞⎟ + b 2 ⎝2⎠ 3 = b2 4 3 =b 2 2

3.

Intercepts: (–3, 0), (3, 0), (–9, 0) [1.2]

4.

−x = − x = − f ( x ) [1.4] (− x) 2 + 1 x 2 + 1 Odd function

5.

f ( x) =

x [1.5] 2x − 3 y x= 2y − 3 x(2 y − 3) = 2 xy − 3 x = y 2 xy − y = y (2 x − 1) = 3 x y = 3x 2x − 1 f −1 ( x) = 3 x 2x − 1

6.

Domain: (−∞, 4) ∪ (4, ∞) [1.3]

7.

x2 + x − 6 = 0 ( x + 3)( x − 2) = 0

8.

Shift the graph of y = f (x) horizontally 3 units to the right. [1.4]

f (− x) =

x+3=0 x−2=0 x = −3 x=2 The solutions are –3 and 2. [1.1] 9.

Reflect the graph of y = f (x) across the y-axis. [1.4[

10.

⎛ ⎞ 300o = 300o ⎜ π o ⎟ = 5π [2.1] ⎝ 180 ⎠ 3

11.

5π = 5π ⎛ 180o ⎞ = 225o [2.1] ⎜ ⎟ 4 4 ⎝ π ⎠

12.

f ⎜⎛ π ⎟⎞ = sin ⎜⎛ π + π ⎟⎞ = sin ⎜⎛ π ⎟⎞ = 1 [2.3] ⎝2⎠ ⎝3⎠ ⎝3 6⎠

13.

f ⎛⎜ π ⎞⎟ = sin ⎛⎜ π ⎞⎟ + sin ⎛⎜ π ⎞⎟ = 3 + 1 = 3 + 1 2 ⎝3⎠ ⎝3⎠ ⎝6⎠ 2 2

14.

⎞ ⎛ ⎞ ⎛ cos 2 45o + sin 2 60o + = ⎜ 2 ⎟ + ⎜ 3 ⎟ = 2 + 3 = 5 [2.2] 4 4 4 ⎝ 2 ⎠ ⎝ 2 ⎠

16.

θ = 210o [2.3] Since 180o < θ < 270o , θ ′ + 180o = θ θ ′ = 30o

15.

negative [2.3]

[2.2]

2


2


896

17.

θ = 2π 3

[2.3]

18.

Domain: ( −∞, ∞ ) [2.4]

19.

Range: [–1, 1]

[2.4]

Since π < θ < π , 2 θ +θ′ = π

θ′ = π

3

20.

tan θ =

opp 3 = adj 4

hypotenuse = 32 + 42

sin θ =

= 9 + 16

opp 3 = hyp 5

[2.2]

= 25 =5

.......................................................

Chapter 3 Cumulative Review

1.

−2 x + 1 < 7 −2 x < 6 x > −3

2.

Shift the graph of y = f (x) horizontally 1 unit to the left and up 2 units.

3.

Reflect the graph of y = f (x) across the x-axis.

4.

f (− x) = − x − sin(− x) = − x + sin x = −( x − sin x) = − f ( x) odd function

5.

f ( x) = 5 x x −1 y = 5x x −1 5y x= y −1 x( y − 1) = 5 y xy − 5 y = x y ( x − 5) = x y= x x−5 f −1 ( x) = x x−5

6.

8.

sin π = 3 3 2

9.

⎛ ⎞ 240o = 240o ⎜ π o ⎟ = 4π ⎝ 180 ⎠ 3

csc60o =

[2.1]

1 = 2 =2 3 3 3 sin 60o

7.

10.

5π = 5π ⎛ 180o ⎞ = 300o ⎜ ⎟ 3 3 ⎝ π ⎠

sin θ =

opp 2 = hyp 3

adjacent side = 32 − 22 = 9−4 = 5 opp tan θ = = 2 =2 5 adj 5 5


[2.1]


11.

cot θ > 0 in quadrant III Positive [2.3]

897

12.

θ = 310o

13.

[2.3] o

θ = 5π

[2.3]

3

o

Since 270 < θ < 360 ,

Since 3π < θ < 2π , 2 θ = θ ′ = 2π

θ = θ ′ = 360o θ ′ = 50o

θ′ = π

3

14.

t=π 3

15.

[2.4]

y = sin t = sin π = 3 3 2 x = cos t = cos π = 1 3 2 The point on the unit circle corresponding to t = π is ⎛⎜ 1 , 3 ⎞⎟ . 3 ⎝2 2 ⎠

17.

cos −1 ( −0.8 ) = 2.498

19.

Range: − π , π 2 2

(

)

16.

y = 0.43cos ⎛⎜ 2 x − π ⎞⎟ [2.7] 6⎠ ⎝ amplitude: 0.43

y = sin −1 sin y =

0 ≤ 2 x − π ≤ 2π

6 π ≤ 2 x ≤ 13π 6 6 π ≤ x ≤ 13π 12 12

y=

1 2

π

1 2 −

[3.5]

π 2

≤y≤

π 2

6

period = π , phase shift = π 12

[3.5]

[3.5]

18.

Domain: [–1, 1].

[3.5]

20.

2cos 2 x − 1 = − sin x 1 − 2sin 2 x = − sin x 0 = 2sin 2 x − sin x − 1 0 = (2sin x + 1)(sin x − 1)

2sin x + 1 = 0 sin x = − 1 2 π 7 , 11π x= 6 6 The solutions are

sin x − 1 = 0 sin x = 1 x=

π 2

π 7π 11π

, , . 2 6 6

.......................................................


1.

An oblique triangle that does not contain a right angle.

2.

The Law of Cosines

3.

SSA

4.

The variable s represents the semiperimeter of the triangle.

5.

A scalar

6.

A scalar

7.

True

8.

False

9.

True

10.

True


898


....................................................... 1.

Chapter 4 Test 2.

B = 180° − 70° − 16° B = 94°

b a = sin B sin A

c a = sin C sin A 14 sin 70° a= sin 16° a ≈ 48

⎛ b sin A ⎞ B = sin −1 ⎜ ⎟ [4.1] ⎝ a ⎠ ⎛ 13 sin 140° ⎞ B = sin −1 ⎜ ⎟ 45 ⎝ ⎠ B ≈ 11°

c b [4.1] = sin C sin B 14 sin 94° b= sin16° b ≈ 51

3.

4.

a 2 + c2 − b2 2ac ⎛ 322 + 182 − 242 ⎞ ⎟ [4.2] B = cos −1 ⎜ ⎜ ⎟ 2(32)(18) ⎝ ⎠ B ≈ 48°

c 2 = a 2 + b 2 − 2ab cos C

cos B =

c 2 = 202 + 122 − 2(20)(12) cos 42° [4.2] c ≈ 14

5.

1 ab sin C 2 1 K = (7)(12)(sin110°) 2 K ≈ 39 square units [4.2] K=

7.

A = 180° − 42° − 75° A = 63°

6.

2 K = b sin A sin C 2sin B 2 12 sin 63° sin 75° K= 2sin 42° K ≈ 93 square units [4.2]

8.

v = (−2)2 + (3) 2 = 13

s = 1 (a + b + c) 2 s = 1 (17 + 55 + 42 ) = 57 2 K = s ( s − a )( s − b)( s − c) K = 57(57 − 17)(57 − 55)(57 − 42) K ≈ 260 square units [4.2]


[4.3]


899

9.

a1 = 12cos 220° ≈ −9.2 [4.3] a2 = 12sin 220° ≈ −7.7 v = a1i + a2 j v = −9.2i − 7.7 j

10.

3u − 5v = 3(2i − 3 j) − 5(5i + 4 j) [4.3] = (6i − 9 j) − (25i + 20 j) = (6 − 25)i + (−9 − 20) j = −19i − 29 j

11.

u ⋅ v = (−2i + 3 j) ⋅ (5i + 3 j) [4.3]

12.

cosθ =

= (−2 ⋅ 5) + (3 ⋅ 3) = −10 + 9 = −1

3,5 ⋅ −6, 2 u⋅v = 2 u v 3 + 52 (−6) 2 + 22

[4.3]

−18 + 10 −8 = 34 40 34 40 θ ≈ 103°

cosθ =

13.

14.

A = 142° − 65° = 77°

A = 82° − 32° = 50°

R 2 = 242 + 182 − 2(24)(18) cos 77° R ≈ 27 miles

C = 180° − 50° − 32° − 72° = 26° a = 12 sin 50° sin 26° a = 12sin 50° sin 26° a ≈ 21 miles [4.3]

[4.3]

15.

1 (112 + 165 + 140 ) = 208.5 2 K = 208.5(208.5 − 112)(208.5 − 165)(208.5 − 140) K ≈ 7743 cost ≈ 8.50(7743) cost ≈ $66,000 [4.2] S=

....................................................... 1.

d = (−3 − 4)2 + (4 − (−1)) 2 = 49 + 25 = 74

3.

( f o g )( x) = f [ g ( x)] [1.5] = f [cos x] = sec(cos x))

[1.2]


f ( x) + g ( x) = sin x + cos x

4.

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

[1.5]

[1.6]

f −1 ( x ) = 2 x + 6

5.

Shifted 2 units to the right and 3 units up. [1.4]

6.

sin 27o = 15 a a = 15 o ≈ 33 cm sin 27


[2.2]


900

7.

y = 3sin(2π x)

8.

y = 1 tan(2 x) 4

[2.6]

[2.5]

9.

( )

y = 2sin π x + 1 2

10.

y = sin x + cos x Amplitude:

13.

15.

12.

[2.7]

( )

4

[3.1]

tan ⎛⎜ cos −1 12 ⎞⎟ [3.5] 13 ⎠ ⎝ Let θ = cos −1 12 and find y = tan θ . 13 12 Then cosθ = , and 0 ≤ θ ≤ π . 13 132 − 122 = 25 = 5 Thus tanθ = 5 . 12 y= 5 12

[2.7]

c 2 = a 2 + b 2 − 2ab cos C

[4.2]

c 2 = (10)2 + (12)2 − 2(10)(12) cos50o

2 , period: 2π , phase shift: π

1 − cos x = 1 − cos 2 x cos x cos x 2 x sin = cos x = sin x tan x

)

amplitude = 3, period = 6π , phase shift = 3π 2

[2.7]

11.

(

y = 3sin 1 x − π 3 2 1 π 0 ≤ x − ≤ 2π 3 2 π ≤ 1 x ≤ 5π 2 3 2 3π ≤ x ≤ 15π 2 2

= 244 − 240cos50o ≈ 9.473 c ≈ 9.5 cm

14.

16.

( )

⎛ ⎞ sin −1 sin 2π = sin −1 ⎜ 3 ⎟ = π 3 ⎝ 2 ⎠ 3

sin x tan x − 1 tan x = 0 [3.6] 2 tan x sin x − 1 = 0 2 sin x − 1 = 0 tan x = 0 2 x = 0, π sin x = 1 2

(

[3.5]

)

x = π , 5π 6 6



17.

v = 42 + (−3)2 α = tan −1 v = 16 + 9

α θ θ θ

v =5

901

−3 = tan −1 3 4 4

≈ 36.9° = 360° − α ≈ 360° − 36.9° ≈ 323.1°

19.

α = sin −1

1 mph 1 mph

20.

α

18.

[4.3]

cos θ =

v⋅w v w

[4.3]

1, 2 ⋅ −2, 3 2

1 + (2)2 ( −2)2 + 32

1(−2) + (2)(3) 5 13 4 cos θ = ≈ 0.49613 5 13 θ = 60.3° cos θ =

heading = θ = 270° + α θ ≈ 270° + 19.5° θ ≈ 289.5°

1 1 = sin −1 3 3

α ≈ 19.5°

3 mph

cos θ =

[4.3]

θ

AB = 515(cos36i + sin 36 j) ≈ 416.6i + 302.7 j [4.3] AD = 150[cos(−30°)i + sin(−30°) j] ≈ 129.9i − 75 j AC = AB + AD AC = 416.6i + 302.7 j + 129.9i − 75 j AC ≈ 546.5i + 227.7 j AC = 546.52 + (227.7)2 AC ≈ 592 mph

( 546.5 )

α = 90o − θ = 90o − tan −1 227.7 ≈ 67.4o

Section 5.1 See CAT Section P.6 solutions on page 26 for exercises 1 – 62. 63.

65.

z 1 = ( −2 + i ) + 4 + 3i = 2 + 4i z 2 = (2 + 4i ) + 4 + 3i = 6 + 7i z 3 = (6 + 7i ) + 4 + 3i = 10 + 10i z 4 = (10 + 10i ) + 4 + 3i = 14 + 13i z 5 = (14 + 13i ) + 4 + 3i = 18 + 16i

64.

z 1 = i (1 − i ) = i − i 2 = 1 + i z 2 = i (1 + i ) = i + i 2 = −1 + i z 3 = i ( −1 + i ) = −i + i 2 = −1 − i

66.

z4 z5 z6 z7 z8

= i ( −1 − i ) = −i − i 2 = 1 − i = z1 = z2 = z3 = z4

z 1 = 2i (1 + 3i ) = 2i + 6i 2 = −6 + 2i z 2 = 2i ( −6 + 2i ) = −12i + 4i 2 = −4 − 12i z 3 = 2i ( −4 − 12i ) = −8i − 24i 2 = 24 − 8i

z 4 = 2i (24 − 8i ) = 48i − 16i 2 = 16 + 48i z 5 = 2i (16 + 48i ) = 32i + 96i 2 = −96 + 32i z 1 = (0.5i ) 2 = 0.25i 2 = −0.25 z 2 = ( −0.25) 2 = 0.0625 z 3 = (0.0625) 2 = 0.00390625



902

67.

Use a = 3, b = −3, c = 3.

68.

Use a = 2, b = 4, c = 4. 2 −b + b2 − 4ac = −(4) + (4) − 4(2)(4) 2a 2(2)

2

−b + b − 4ac = −( −3) + ( −3) − 4(3)(3) 2a 2(3) 2

= −4 + 16 − 32 = −4 + −16 4 4 i 4 16 − + i 4 4 − + = = 4 4 i 4 4 − = + = −1 + i 4 4

= 3 + 9 − 36 = 3 + −27 6 6 i i + + 3 27 3 3 3 = = 6 6 = 3+ 3 3i = 1 + 3i 6 6 2 2 69.

Use a = 2 b = 6, c = 6.

70. 2

Use a =2, b = 1, c = 3. 2 −b + b2 − 4ac = −(1) + (1) − 4(2)(3) 2a 2(2)

−b + b − 4ac = −(6) + (6) − 4(2)(6) 2a 2(2) 2

= −1 + 1 − 24 4 = −1 + −23 = −1 + i 23 4 4 23 1 =− + i 4 4

= −6 + 36 − 48 = −6 + −12 4 4 6 2 3 − + i 6 12 − + i = = 4 4 2 3 i 6 3 − = + =− + 3i 4 4 2 2 71.

Use a = 4, b = −4, c = 2.

72. 2

Use a = 3, b = −2, c = 4. 2 −b + b2 − 4ac = −( −2) + ( −2) − 4(3)(4) 2a 2(3)

−b + b − 4ac = −( −4) + ( −4) − 4(4)(2) 2a 2(4) 2

= 2 + 4 − 48 = 2 + −44 6 6 = 2 + i 44 = 2 + 2i 11 6 6 2 11 i 2 1 = + = + 11 i 6 6 3 3

= 4 + 16 − 32 = 4 + −16 8 8 = 4 + i 16 = 4 + 4i 8 8 4 4 1 1 i = + = + i 8 8 2 2

.......................................................

Connecting Concepts

73.

x 2 + 16 = x 2 + 42 = ( x + 4i )( x − 4i )

74.

x 2 + 9 = x 2 + 32 = ( x + 3i )( x − 3i )

75.

z 2 + 25 = z 2 + 52 = ( z + 5i )( z − 5i )

76.

z 2 + 64 = z 2 + 82 = ( z + 8i )( z − 8i )

77.

4 x 2 + 81 = (2 x )2 + 92 = (2 x + 9i )(2 x − 9i )

78.

9 x 2 + 1 = (3x )2 + 12 = (3x + i )(3x − i )

79.

If x = 1 + 2i, then x 2 − 2 x + 5 = (1 + 2i )2 − 2(1 + 2i ) + 5 = 1 + 4i + 4i 2 − 2 − 4i + 5 = 1 + 4i + 4(−1) − 2 − 4i + 5 = 1 + 4i − 4 − 2 − 4i + 5 = (1 − 4 − 2 + 5) + (4i − 4i ) = 0

80.

If x = 1 − 2i, then x 2 − 2 x + 5 = (1 − 2i )2 − 2(1 − 2i ) + 5 = 1 − 4i + 4i 2 − 2 + 4i + 5 = 1 − 4i + 4( −1) − 2 + 4i + 5 = 1 − 4i − 4 − 2 + 4i + 5 = (1 − 4 − 2 + 5) + ( −4i + 4i ) = 0



81.

903

Verify that ( −1 + i 3)3 = 8 . ( −1 + i 3)3 = ( −1 + i 3)( −1 + i 3)2 = ( −1 + i 3)[( −1)2 + 2( −1)(i 3) + (i 3)2 ] = ( −1 + i 3)[1 − 2i 3 + 3i 2 ] = ( −1 + i 3)[1 − 2i 3 + 3( −1)] = ( −1 + i 3)[1 − 2i 3 − 3] = ( −1 + i 3)( −2 − 2i 3) = −1( −2 − 2i 3) + i 3( −2 − 2i 3) = 2 + 2i 3 − 2i 3 − 2i 2 ( 3)2 = 2 + 2i 3 − 2i 3 − 2( −1)(3) = 2 + 2i 3 − 2i 3 + 6 = (2 + 6) + (2i 3 − 2i 3) =8

Verify that ( −1 − i 3)3 = 8 . ( −1 − i 3)3 = ( −1 − i 3)( −1 − i 3)2 = ( −1 − i 3)[( −1)2 + 2( −1)( −i 3) + ( −i 3)2 ] = ( −1 − i 3)[1 + 2i 3 + 3i 2 ] = ( −1 − i 3)[1 + 2i 3 + 3( −1)] = ( −1 − i 3)[1 + 2i 3 − 3] = ( −1 − i 3)( −2 + 2i 3) = −1( −2 + 2i 3) − i 3( −2 + 2i 3) = 2 − 2i 3 + 2i 3 − 2i 2 ( 3)2 = 2 − 2i 3 + 2i 3 − 2( −1)(3) = 2 − 2i 3 + 2i 3 + 6 = (2 + 6) + ( −2i 3 + 2i 3) =8 82.

⎡

⎤

2

Verify that ⎢ 2 (1 + i ) ⎥ = i . ⎣ 2 ⎦ 2

2

⎡ 2 ⎤ 2 2 2 2 1 1 1 ⎢ 2 (1 + i ) ⎥ = 2 (1 + i ) = 4 (1 + 2i + i ) = 2 [1 + 2i + ( −1)] = 2 (1 + 2i − 1) = 2 (2i ) = i 2 ⎣ ⎦ 83.

i + i 2 + i 3 + i 4 + ... + i 28 = 7(i + i 2 + i 3 + i 4 ) = 7(i + ( −1) + ( −i ) + 1) = 7(0) = 0

84.

i + i 2 + i 3 + i 4 + ... + i100 = 25(i + i 2 + i 3 + i 4 ) = 25(i + ( −1) + ( −i ) + 1) = 25(0) = 0

.......................................................


See CAT Prepare for Section 7.4 solutions on page 462.

.......................................................


The Mandelbrot Iteration Procedure 1.

z 5 ≈ 0.4001878333

z 10 ≈ 0.4353725328

z 100 ≈ 0.4906925008

z 200 ≈ 0.4951947016

Answers will vary. 2.

z0 = i z 1 = −1 + i z 2 = −1 z 3 = −1 + i z 4 = −i The iterates continue to cycles back and forth between –1 + i and –i.



904

.......................................................


1.

True

True

5.

The four roots are equally spaced around a circle with center 6. (0, 0) and radius 1.

1

7.

5

–3 + 5i

2.

No

8.

True

3.

9.

4.

False

10.

2

....................................................... 1.

4.

3 − −64 = 3 − 8i 3 + 8i

2.

[5.1]

5.

−5 + −27 = −5 + 3i 3 [5.1] −5 − 3i 3

Chapter 5 Review

−4 + 6 = 6 + 2i 6 – 2i

3.

[5.1]

[5.1]

−2 − i 5

(

−4 )( −4 ) = (2i )(2i ) = 4i 2 = −4 [5.1]

6.

9.

(3 + 7i) + (2 − 5i) = 5 + 2i [5.1]

8.

(3 − 4i ) + (−6 + 8i ) = −3 + 4i

10.

(−3 − 5i ) − (2 + 10i ) = −5 − 15i [5.1]

11.

(5 + 3i )(2 − 5i) = 10 − 19i − 15i 2 [5.1] 12. = 25 − 19i

13.

−2i ⋅ 3 + 4i = −6i − 8i 2 = 8 − 6 i 3 − 4i 3 + 4i 9 + 16 25 25

[5.1]

15.

i(2i ) − (1 + i ) 2 = 2i 2 − 1 − 2i − i 2 = −2 − 2i

[5.1]

(−

−27 )( −3 ) = ( −3i 3 )( i 3 ) [5.1] = −9i 2 = 9

7.

[5.1]

−2 + −5 = −2 + i 5

(6 − 8i ) − (9 − 11i ) = −3 + 3i [5.1] (−2 − 3i )(−4 + 7i ) = 8 − 2i − 21i 2 = 29 − 2i

14.

4 + i ⋅ 7 + 2i = 28 + 15i + 2i 2 = 26 + 15 i 7 − 2i 7 + 2i 49 + 4 53 53

16.

(2 − i )3 = (4 − 4i + i 2 )(2 − i) [5.1] = (3 − 4i )(2 − i )

[5.1]

[5.1]

= 6 − 11i + 4i 2 = 2 − 11i 17.

(3 + −4) − (−3 − −16) = 3 + 2i + 3 + 4i = 6 + 6i

19.

(2 − −3)(2 + −3) = 4 − (−3) = 7 [5.1]

[5.1]

18.

(−2 + −9) + ( −3 − −81) = −2 + 3i − 3 − 9i = −5 − 6i

20.

(3 − −5)(2 + −5) = 6 + i 5 − (−5)

[5.1]

[5.1]

= 11 + i 5 21.

25.

22.

i 27 = i 3 = −i [5.1]

( 0 ) 2 + ( −8 ) 2

−8i =

i105 = i 26.

[5.2]

2 − 3i =

=8 28.

−1 − i =

23.

[5.1]

24.

i = 1 = 1 = 1 [5.1] i17 i16 1

( 2 )2 + ( −3)2

[5.2]

27.

−4 + 5i =

= 4 + 9 = 13

( −1)2 + ( −1)2

= 1+1 = 2

[5.2]

29.

r=

( 2 ) 2 + ( −2 ) 2

r=2 2

α = tan −1

( −4 ) 2 + ( 5 ) 2

= 16 + 25 = 41

30.

[5.2]

i 62 = i 2 = −1 [5.1]

r=

(− 3)

2

+ (1)

2

r=2 −2 2

1 3 −1 1 = tan = 30o 3

α = tan −1

= tan −1 1 = 45o

α = 360o − 45o = 315o o

z = 2 2 cis 315

α = 180o − 30o = 150o z = 2 cis 150o


[5.2]

[5.2]


31.

r=

( −3)2 + ( 2 )2

905

32.

[5.2]

r = 13

( 4 )2 + ( −1)2

[5.2]

r = 17 −1 4 1 = tan −1 = 14.04o 4

2 −3 2 = tan −1 = 33.7o 3

33.

r=

α = tan −1

α = tan −1

α = 180o − 33.7o = 146.3o

α = 360o − 14.04o = 345.96o

z = 13 cis 146.3o

z = 17 cis 345.96o

z = 5(cos 315o + i sin 315o ) [5.2]

z = 2 ( cos 2 + i sin 2 ) [5.2] z ≈ 2 ( −0.4161 + 0.9093i ) z ≈ −0.832 + 1.819i

37.

z1z2 z1z2 z1z2 z1z2 z1z2

= 3(cos 225° + i sin 225°) ⋅ 10(cos 45° + i sin 45°) = 30[cos(225° + 45°) + i sin(225° + 45°)] = 30(cos 270° + i sin 270°) = 30(0 − 1i ) = −30i

39.


= 3(cos 12° + i sin 12°) ⋅ 4(cos 126° + i sin 126°) = 12[cos(12° + 126°) + i sin(12° + 126°)] = 12(cos 138° + i sin 138°) ≈ 12( −0.74314 + 0.66913i ) ≈ −8.918 + 8.030i

36.

z = 3(cos 115o + i sin 115o ) [5.2] z ≈ 3 ( −0.4226 + 0.9063i ) z ≈ −1.27 + 2.72i

41.


= 3(cos 1.8 + i sin 1.8) ⋅ 5(cos 2.5 + i sin 2.5) [5.2] = 15[cos(1.8 + 2.5) + i sin(1.8 + 2.5)] = 15(cos 4.3 + i sin 4.3) ≈ 15(−0.4008 − 0.9162i ) ≈ −6.012 − 13.743i

43.

z1 6(cos 50o + i sin 50o ) [5.2] = z2 2(cos 150o + i sin 150o ) z1 = 3 [cos(50o − 150o ) + i sin(50o − 150o )] z2 z1 = 3(cos − 100o + i sin − 100o ) = 3cis (−100o ) z2

[5.2] 38.

[5.2]

)

z = 6 cos 4π + i sin 4π 3 3 ⎛ 1 ⎞ z = 6⎜ − − 3 i ⎟ ⎝ 2 2 ⎠ z = −3 − 3 3i

⎛ ⎞ z = 5⎜ 2 − 2 i ⎟ 2 ⎠ ⎝ 2 5 2 5 z= − 2i 2 2 ≈ 3.536 − 3.536i

35.

(

34.

[5.2]

z1z2 = 5(cos 162° + i sin 162°) ⋅ 2(cos 63° + i sin 63°) [5.2] z1z2 = 10[cos(162° + 63°) + i sin(162° + 63°)] z1z2 = 10(cos 225° + i sin 225°) ⎛ 2 2 ⎞ − z1z2 = 10 ⎜ − i⎟ ⎝ 2 2 ⎠ z1z2 = −5 2 − 5i 2

40.


= (cos 23° + i sin 23°) ⋅ 4(cos 233° + i sin 233°) [5.2] = 4[cos(23° + 233°) + i sin(23° + 233°)] = 4(cos 256° + i sin 256°) ≈ 4 ( −0.24192 − 0.97030i ) ≈ −0.968 − 3.881i

42.


= 6(cos 3.1 + i sin 1.8) ⋅ 5(cos 4.3 + i sin 4.3) = 30[cos(3.1 + 4.3) + i sin(3.1 + 4.3)] = 30(cos 7.4 + i sin 7.4) ≈ 30(0.439 + 0.899i ) ≈ 13.2 + 27.0i

44.

z1 30(cos 165o + i sin 165o ) [5.2] = z2 10(cos 55o + i sin 55o ) z1 = 3 [cos(165o − 55o ) + i sin(165o − 55o )] z2 z1 = 3(cos 110o + i sin 110o ) = 3cis (110o ) z2


[5.2]

906


45.


46.


47.

z1 10(cos 3.7 + i sin 3.7) = z2 6(cos 1.8 + i sin 1.8)

48.

z1 4(cos 1.2 + i sin 1.2) = z2 8(cos 5.2 + i sin 5.2)

[5.2]

z1 5 = [cos(3.7 − 1.8) + i sin(3.7 − 1.8)] z2 3 z1 5 = (cos 1.9 + i sin 1.9) = 5 cis (1.9) z2 3 3 49.

z1 1 = [cos(1.2 − 5.2) + i sin(1.2 − 5.2)] z2 2 z1 1 = (cos − 4 + i sin − 4) = 1 cis (−4) z2 2 2

[3(cos 45° + i sin 45°)]5 = 35[cos(5 ⋅ 45°) + i sin(5 ⋅ 45°)] [5.3] o

z =1− i 3

[5.3]

r = 12 + ( − 3 )

2

(

52.

z = −2 − 2i [5.3]

8

α = tan −1

r = (−2)2 + (−2)2

1

r=2 2

θ = 300o

)

z = 2 2(cos 225° + i sin 225°) 7

10

(1 − i 3) = [2(cos300° + i sin 300°)]

(−2 − 2i )

= [2 2(cos 225° + i sin 225°)]10

= 128[cos(7 ⋅ 300°) + i sin(7 ⋅ 300°)] = 128(cos 2100° + i sin 2100°)

= 32,768[cos(10 ⋅ 225°) + i sin(10 ⋅ 225°)]

= 128(cos300° + i sin 300°)

= 32,768(cos90° + i sin 90°)

= 64 − 64i 3

= 0 + 32,768i

= 32,768(cos 2250° + i sin 2250°)

≈ 64 − 110.851i

53.

z = 2 −i 2 2

54.

[5.3]

r = ( 2 ) + (− 2 ) r=2

2

α = tan −1 − 2 = 45o 2

o

θ = 315

z = 2(cos315° + i sin 315°) ( 2 − i 2)5 = [2(cos315° + i sin 315°)]5 = 32[cos(5 ⋅ 315°) + i sin(5 ⋅ 315°)] = 32(cos1575° + i sin1575°) = 32(cos135° + i sin135°) = −16 2 + 16i 2

z = 3 − 4i

[5.3]

r = 32 + ( −4)2 r =5

α = tan −1

−4 = 53.13o 3

θ = 306.87o z = 5(cos306.87° + i sin 306.87°) 5

(3 − 4i ) = [5(cos306.87° + i sin 306.87°)]5 = 3125[cos(5 ⋅ 306.87°) + i sin(5 ⋅ 306.87°)] = 3125(cos1534.35° + i sin1534.35°) = 3125(cos94.35° + i sin 94.35°) = −237 + 3116i

≈ −22.627 + 22.627i


[5.3]

−2 = 45o −2

θ = 225o

z = 2(cos300° + i sin 300°) 7

)

⎡cos 11π + i sin 11π ⎤ = cos 8 ⋅ 11π + i sin 8 ⋅ 11π ⎢⎣ 8 8 ⎥⎦ 8 8 = cos (11π ) + i sin (11π ) = cos π + i sin π = −1 + 0i = −1

α = tan −1 − 3 = 60o

r=2

(

50.

o

= 243[cos 225 + i sin 225 ] ⎛ ⎞ = 243 ⎜ − 2 − 2 i ⎟ ⎝ 2 2 ⎠ = − 243 2 − 243 2 i 2 2 ≈ −171.827 − 171.827i 51.

[5.2]


55.

(

27i = 27 cos90o + i sin 90o

)

907

[5.3]

wk = 271/ 3 ⎛⎜ cos 90° + 360°k + i sin 90° + 360°k ⎞⎟ 3 3 ⎝ ⎠

(

w0 = 3 cos 90° + i sin 90° 3 3 w0 = 3(cos30° + i sin 30°) w0 = 3 cis 30° 56.

(

8i = 8 cos90o + i sin 90o

)

)

w1 = 3 ⎜⎛ cos 90° + 360° + i sin 90° + 360° ⎟⎞ 3 3 ⎝ ⎠ w1 = 3(cos150° + i sin150°) w1 = 3 cis 150°

(

)

w0 = 4 8 cis 22.5°

w1 = 4 8 cis 112.5°

w2 = 4 8 ⎜⎛ cos 90° + 360° ⋅ 2 + i sin 90° + 360° ⋅ 2 ⎟⎞ 4 4 ⎝ ⎠ 4 w2 = 8(cos 202.5° + i sin 202.5°) w2 = 4 8 cis 202.5°

(

k = 0, 1, 2, 3

w1 = 4 8 ⎜⎛ cos 90° + 360° + i sin 90° + 360° ⎟⎞ 4 4 ⎝ ⎠ 4 w1 = 8(cos112.5° + i sin112.5°)

w0 = 4 8 cos 90° + i sin 90° 4 4 4 w0 = 8(cos 22.5° + i sin 22.5°)

256 = 256 cos0o + i sin 0o

w2 = 3 ⎜⎛ cos 90° + 360° ⋅ 2 + i sin 90° + 360° ⋅ 2 ⎟⎞ 3 3 ⎝ ⎠ w2 = 3(cos 270° + i sin 270°) w2 = 3 cis 270°

[5.3]

wk = 81/ 4 ⎛⎜ cos 90° + 360°k + i sin 9° + 360°k ⎞⎟ 4 4 ⎝ ⎠

57.

k = 0, 1, 2

w3 = 4 8 ⎛⎜ cos 90° + 360° ⋅ 3 + i sin 90° + 360° ⋅ 3 ⎞⎟ ⎝ ⎠ 4 4 4 w3 = 8(cos 292.5° + i sin 292.5°) w3 = 4 8 cis 292.5°

)

[5.3]

wk = 811/ 4 ⎛⎜ cos 0° + 360°k + i sin 0° + 360°k ⎞⎟ 4 4 ⎝ ⎠

k = 0, 1, 2, 3

w0 = 3 cis 0°

w1 = 3 ⎜⎛ cos 0° + 360° + i sin 0° + 360° ⎟⎞ 4 4 ⎝ ⎠ w1 = 3(cos90° + i sin 90°) w1 = 3 cis 90°

w2 = 3 ⎜⎛ cos 0° + 360° ⋅ 2 + i sin 0° + 360° ⋅ 2 ⎟⎞ 4 4 ⎝ ⎠ w2 = 3(cos180° + i sin180°) w2 = 3 cis 180°

w3 = 3 ⎛⎜ cos 0° + 360° ⋅ 3 + i sin 0° + 360° ⋅ 3 ⎞⎟ 4 4 ⎝ ⎠ w3 = 3(cos 270° + i sin 270°) w3 = 3 cis 270°

w0 = 3(cos 0° + i sin 0°)


908

58.


−16 2 − 16i 2 = 32(cos 225° + i sin 225°)

[5.3]

wk = 321/ 5 ⎜⎛ cos 225° + 360°k + i sin 225° + 360°k ⎟⎞ 5 5 ⎝ ⎠

k = 0, 1, 2, 3, 4

w0 = 2 ⎜⎛ cos 225° + i sin 225° ⎟⎞ 5 5 ⎠ ⎝ w0 = 2(cos 45° + i sin 45°) w0 = 2 cis 45°

o o o o⎞ ⎛ w1 = 2 ⎜ cos 225 + 360 + i sin 225 + 360 ⎟ 5 5 ⎝ ⎠ w1 = 2(cos117° + i sin117°) w1 = 2 cis 117°

w2 = 2 ⎛⎜ cos 225° + 360° ⋅ 2 + i sin 225° + 360° ⋅ 2 ⎞⎟ 5 5 ⎝ ⎠ w2 = 2(cos189° + i sin189°) w2 = 2 cis 189°


w4 = 2 ⎜⎛ cos 225° + 360° ⋅ 4 + i sin 225° + 360° ⋅ 4 ⎟⎞ 5 5 ⎝ ⎠ w4 = 2(cos333° + i sin 333°) w4 = 2 cis 333° 59.

(

81 = 81 cos 0o + i sin 0o

)

[5.3]

wk = 2561/ 4 ⎜⎛ cos 0° + 360°k + i sin 0° + 360°k ⎟⎞ 4 4 ⎝ ⎠

k = 0, 1, 2, 3

w0 = 4 cis 0°

w1 = 4 ⎛⎜ cos 0° + 360° + i sin 0° + 360° ⎞⎟ 4 4 ⎝ ⎠ w1 = 4(cos90° + i sin 90°) w1 = 4 cis 90°

w2 = 4 ⎜⎛ cos 0° + 360° ⋅ 2 + i sin 0° + 360° ⋅ 2 ⎟⎞ ⎝ 4 4 ⎠ w2 = 4(cos180° + i sin180°) w2 = 4 cis 180°

w3 = 4 ⎛⎜ cos 0° + 360° ⋅ 3 + i sin 0° + 360° ⋅ 3 ⎞⎟ 4 4 ⎝ ⎠ w3 = 4(cos 270° + i sin 270°) w3 = 4 cis 270°

w0 = 4(cos 0° + i sin 0°)

60.

(

−125 = 125 cos180o + i sin180o

)

[5.3]

wk = 1251/ 3 ⎛⎜ cos 180° + 360°k + i sin 180° + 360°k ⎞⎟ 3 3 ⎝ ⎠

(

w0 = 5 cos 180° + i sin 180° 3 3 w0 = 5(cos 60° + i sin 60°) w0 = 5 cis 60°

)

k = 0, 1, 2

w1 = 5 ⎛⎜ cos 180° + 360° + i sin 180° + 360° ⎞⎟ 3 3 ⎝ ⎠ w1 = 5(cos180° + i sin180°) w1 = 5 cis 180° = −5

....................................................... QR1.

w2 = 5 ⎛⎜ cos 180° + 360° ⋅ 2 + i sin 180° + 360° ⋅ 2 ⎞⎟ 3 3 ⎝ ⎠ w2 = 5(cos300° + i sin 300°) w2 = 5 cis 300°

Chapter 5 Quantitative Reasoning QR2. –0.25 + 0.25i, –1 + 0.1i, and 0.1 + 0.2i



909

QR4. The first iterate of 2i is z 1 = −4 + 2i . 2i is not an element of

QR3. –2 is an element of the Mendelbrot set because all of its iterates equal 2. QR5.

the Mandelbrot set because − 4 + 2i > 3 .

(Z 2 + S ) → Z

QR6. Answers will vary.

....................................................... 1.

6 + −9 = 6 + 3i

2.

4.

(−1 + −25) − (8 − −16) = −1 + 5i − 8 + 4i = −9 + 9i

[5.1]

6.

i 263 = i 3 = −i

9.

(3 − 5i )(−3 + 5i ) = −9 + 30i − 25i 2 [5.1] = 16 + 30i

11.

2 − 7i ⋅ 4 − 3i = 8 − 34i + 21i 2 = − 13 − 34 i 4 + 3i 4 − 3i 16 + 9 25 25

13.

z =

7.

[5.1]

( 3)2 + ( −5)2

14.

[5.2]

−18 = 3i 2

Chapter 5 Test

(

5.

[5.1]

3.

[5.1]

−12 ) ( −3 ) = ( 2i 3 )( i 3 ) [5.1] = 6i 2 = −6

(3 + 7i) − (−2 − 9i ) = 5 + 16i

[5.1]

r=

[5.1]

8.

4 − 5i ⋅ −i = −4i + 5i 2 = −5 − 4i −i i −i 2

12.

6 + 2i ⋅ 1 + i = 6 + 8i + 2i 2 = 4 + 8 i = 2 + 4i 1− i 1+ i 1+1 2 2

( 3)2 + ( −3)2

α = tan −1

(−6 − 9i )(4 + 3i ) = −24 − 54i − 27i 2 = 3 − 54i [5.1]

10.

15.

[5.2]

r =3 2

= 9 + 25 = 34

(3 + −4) + (7 − −9) = 3 + 2i + 7 − 3i = 10 − i [5.1]

−3 3

[5.1]

2

r = ( 0 ) + ( −6 ) r =6

α = tan −1

2

[5.1]

[5.2]

−6 = 90o 0

α = 360o − 90o = 270o

= tan −1 1 = 45o

α = 360o − 45o = 315o

z = 6 cis 270o

z = 3 2 cis 315o 16.

17.

z = 4(cos 120o + i sin 120o ) [5.2] ⎛ ⎞ z = 4⎜ − 1 + 3 i ⎟ ⎝ 2 2 ⎠ z = −2 + 2i 3

18.

z = 5(cos 225o + i sin 225o )

[5.2]

⎛ ⎞ z = 5⎜ − 2 − 2 i ⎟ 2 ⎠ ⎝ 2 z = −5 2 − 5 2 i 2 2

z1z2 = 3(cos 28° + i sin 28°) ⋅ 4(cos 17° + i sin 17°) z1z2 = 12[cos(28° + 17°) + i sin(28° + 17°)] z1z2 = 12(cos 45° + i sin 45°) ⎛ 2 2 ⎞ + z1z2 = 12 ⎜ i⎟ ⎝ 2 2 ⎠ z1z2 = 6 2 + 6i 2

[5.2]

19.


= 5(cos 115° + i sin 115°) ⋅ 4(cos 10° + i sin 10°) = 20[cos(115° + 10°) + i sin(115° + 10°)] = 20(cos 125° + i sin 125°) ≈ 20(−0.5736 + 0.81915i ) ≈ −11.472 + 16.383i


[5.2]

910

20.


21.

z1 24(cos 258o + i sin 258o ) = [5.2] z2 6(cos 78o + i sin 78o ) z1 = 4 [cos(258o − 78o ) + i sin(258o − 78o )] z2 z1 = 4(cos 180o + i sin 180o ) = −4 + 0i = −4 z2

22.

z1 18(cos 50o + i sin 50o ) = [5.2] z2 3(cos 140o + i sin 140o ) z1 = 6 [cos(50o − 140o ) + i sin(50o − 140o )] z2 z1 = 6(cos − 90o + i sin − 90o ) = 0 − 6i = −6i z2

z = 2 − 2i 3 [5.3]

r = 22 + ( −2 3 ) r=4

2

α = tan −1 −2 3 = 60o 2

θ = 300o

z = 4(cos300° + i sin 300°) 12

(2 − 3i 3)

23.

= [4(cos300° + i sin 300°)]12 = 16,777, 216[cos(12 ⋅ 300°) + i sin(10 ⋅ 300°)] = 16,777, 216(cos3600° + i sin 3600°) = 16,777, 216(cos 0° + i sin 0°) = 16,777, 216 + 0i

64 = 64(cos 0° + i sin 0°) [5.3] 0° + 360°k 0° + 360°k ⎞ ⎛ + i sin wk = 641/ 6 ⎜ cos ⎟ 6 6 ⎝ ⎠

k = 0, 1, 2, 3, 4, 5

w0 = 2 cis 0°

w1 = 2 ⎛⎜ cos 0 + 360° + i sin 0 + 360° ⎞⎟ 6 6 ⎝ ⎠ w1 = 2(cos 60° + i sin 60°) w1 = 2 cis 60°

w2 = 2 ⎜⎛ cos 0° + 360° ⋅ 2 + i sin 0° + 360° ⋅ 2 ⎟⎞ 6 6 ⎝ ⎠ w2 = 2(cos120° + i sin120°) w2 = 2 cis 120°

w3 = 2 ⎛⎜ cos 0° + 360° ⋅ 3 + i sin 0° + 360° ⋅ 3 ⎞⎟ 6 6 ⎝ ⎠ w3 = 2(cos180° + i sin180°) w3 = 2 cis 180°


w5 = 2 ⎛⎜ cos 0° + 360° ⋅ 5 + i sin 0° + 360° ⋅ 5 ⎞⎟ 6 6 ⎝ ⎠ w5 = 2(cos300° + i sin 300°) w5 = 2 cis 300°

w0 = 2(cos 0° + i sin 0°)

24.

−1 + 3i = 2(cos120° + i sin120°) [5.3] wk = 21/ 3 ⎛⎜ cos 120° + 360°k + i sin 120° + 360°k ⎞⎟ 3 3 ⎝ ⎠ w0 = 3 2 ⎛⎜ cos 120° + i sin 120° ⎞⎟ 3 3 ⎠ ⎝ w0 = 3 2(cos40°+ i sin 40°) w0 = 3 2 cis 40°

k = 0, 1, 2

w1 = 3 2 ⎛⎜ cos 120° + 360° + i sin 120°+ 360° ⎞⎟ 3 3 ⎝ ⎠ w1 = 3 2(cos160°+ i sin160°) w1 = 3 2 cis 160°

w2 = 3 2 ⎛⎜ cos 120° + 360°⋅ 2 + i sin 120° + 360°⋅ 2 ⎞⎟ 3 3 ⎝ ⎠ w2 = 3 2(cos280° + i sin 280°) w2 = 3 2 cis 280°



25.

x5 + 32 = 0

911

[5.3]

5

x = −32

Find the five fifth roots of –32. −32 = 32(cos180° + i sin180°) wk = 321/ 5 ⎛⎜ cos 180° + 360°k + i sin 180° + 360°k ⎞⎟ 5 5 ⎝ ⎠

k = 0, 1, 2, 3, 4 o o w2 = 2 cis 180 + 360 ⋅ 2 5 w2 = 2 cis 180°

w0 = 2 cis 180° 5 w0 = 2 cis 36°

w1 = 2 cis 180° + 360° 5 w1 = 2 cis 108°

o o w3 = 2 cis 180 + 360 ⋅ 3 5 w3 = 2 cis 252°

o o w4 = 2 cis 180 + 360 ⋅ 4 5 w4 = 2 cis 324°

....................................................... 1.


x 2 − x − 6 ≤ 0 [1.1 ( x − 3)( x + 2) ≤ 0 The product is negative or zero. The critical values are –2 and 3.

x2 − 4 = 0 ( x − 2)( x + 2) = 0 x = −2, 2 Domain: all real numbers except –2 and 2.

( x − 3)( x + 2) [–2, 3] 3.

f (c ) = 2

[1.3]

4.

( f o g )( x) = f [ g ( x)] ⎡ 2 ⎤ = f ⎢ x − 1⎥ ⎣ 3 ⎦ 2 ⎛ ⎞ = sin ⎜ 3 ⋅ x − 1 ⎟ 3 ⎠ ⎝ = sin( x 2 − 1)

7.

[2.2] cos38o = 20 c c = 20 o ≈ 25.4 cm cos38

2= c c +1 2(c + 1) = c 2c + 2 = c c = −2

6.

3π ⎛ 180o ⎞ = 270o [2.1] ⎜ ⎟ 2 ⎝ π ⎠

9.

y = 3sin π x

[2.5]

10.

[1.5]

y = 1 tan π x 2 4

5.

[1.6] f ( x) = x x −1 y= x x −1 y x= y −1 x( y − 1) = y xy − x = y xy − y = x y ( x − 1) = x y= x x −1 f −1 ( x) = x x −1 f −1 (3) = 3 = 3 3 −1 2

8.

−1 ≤ sin t ≤ 1 [2.4] a = –1, b = 1

[2.6]


912


11.

cos x = cos x ⋅ 1 − sin x 1 + sin x 1 + sin x 1 − sin x = cos x − sin 2x cos x 1 − sin x x cos x = cos x − sin cos 2 x = 1 − sin x cos x cos x = sec x − tan x

13.

[3.2] sin α = 4 , cos α = 3 5 5 cos β = 12 , sin β = − 5 13 13 cos(α + β ) = cos α cos β − sin α sin β

[3.1]

12.

[3.2] sin 2 x cos3 x − cos 2 x sin 3 x = sin(2 x − 3 x) = sin( − x) or − sin x

14.

⎡ ⎛3⎞ ⎛ 5 ⎞⎤ y = sin ⎢sin −1 ⎜ ⎟ + cos−1 ⎜ − ⎟ ⎥ ⎝5⎠ ⎝ 13 ⎠ ⎦ ⎣

2

Let α = sin −1 3 , sin α = 3 , cosα = 1− ⎛⎜ 3 ⎞⎟ = 4 . 5 5 5 ⎝5⎠

( 13)

⎛ 3 ⎞⎛ 12 ⎞ ⎛ 4 ⎞⎛ 5 ⎞ = ⎜ ⎟⎜ ⎟ − ⎜ ⎟⎜ − ⎟ ⎝ 5 ⎠⎝ 13 ⎠ ⎝ 5 ⎠⎝ 13 ⎠ 48 15 56 =− + = 65 65 65

15.

16.

20.

(

−27 = 27 cos180o + i sin180o

[4.3]

(

)

c 2 = a 2 + b 2 − 2ab cos C

[4.2]

= 36,500 − 36, 400cos 78o ≈ 28,932 c ≈ 170 cm

18.

W = F ⋅s W= F

[4.3] s cos α

W = (100)(15)(cos15°) W ≈ 1449 foot-pounds

19.

r=

( 2 ) 2 + ( 2 )2

[5.2]

r=2 2

α = tan −1

2 2

= tan −1 1 = 45o z = 2 2 cis 45o

)

[5.3]

wk = 271/ 3 ⎛⎜ cos 180° + 360°k + i sin 180° + 360°k ⎞⎟ 3 3 ⎝ ⎠ w0 = 3 cos 180° + i sin 180° 3 3 w0 = 3(cos 60° + i sin 60°) ⎛ ⎞ w0 = 3 ⎜ 1 + 3 i ⎟ = 3 + 3 3 2 ⎝2 2 ⎠ 2

13

= (140)2 + (130)2 − 2(140)(130) cos 78o

2cos x − 3 = 0

cos θ = v ⋅ w v w (3i + 2 j) ⋅ (5i − 3 j) cos θ = 32 + 22 52 + (−3) 2 3(5) + (2)(−3) cos θ = 13 34 9 cos θ = =0 442 o θ = 64.7

⎝ 13 ⎠

13

y = sin(α + β ) = sin α cos β + cos α sin β = 3 ⎛⎜ − 5 ⎞⎟ + 4 ⎛⎜ 12 ⎞⎟ 5 ⎝ 13 ⎠ 5 ⎝ 13 ⎠ = − 15 + 48 = 33 65 65 65

cos x = 3 2 x = π , 11π 6 6 π 11 π The solutions are 0, , π , . 6 6

17.

2

β = cos −1 − 5 , cosβ = − 5 , sinβ = 1− ⎛⎜ − 5 ⎞⎟ = 12 .

sin 2 x = 3 sin x [3.6] 2sin x cos x − 3 sin x = 0 sin x (2cos x − 3) = 0

sin x = 0 x = 0, π

[3.5]

k = 0, 1, 2

w1 = 3⎛⎜ cos 180°+ 360° + i sin 180°+ 360° ⎞⎟ 3 3 ⎝ ⎠ w1 = 3(cos180° + i sin180°) w1 = 3( −1+ 0i ) = −3

w2 = 3⎛⎜ cos 180°+ 360°⋅ 2 + i sin 180° + 360°⋅ 2 ⎞⎟ 3 3 ⎝ ⎠ w2 = 3(cos300° + i sin300°) ⎛ ⎞ w2 = 3⎜ 1 − 3 i ⎟ = 3 − 3 3 i ⎝2 2 ⎠ 2 2



913

Section 6.7 See CAT Section 8.7 solutions on page 577 for exercises 1 – 18. 19.

y = 3 + 2sin t y − 3 = 2sin t y−3 = sin t 2

x = 2 + 3cos t x − 2 = 3cos t x − 2 = cos t 3

( ) ( x −3 2 ) x−2 3

2

2

2

⎛ y − 3⎞ 2 2 +⎜ ⎟ = cos t + sin t ⎝ 2 ⎠ 2

⎛ y − 3⎞ +⎜ ⎟ =1 ⎝ 2 ⎠

At t = 0 , x = 2 + 3cos0 = 5 At t = π , x = 2 + 3cos π = −1

y = 3 + 2sin 0 = 3 y = 3 + 2sin π = 3

The point traces the top half of the ellipse

( )

2

2 x − 2 + ⎛ y − 3 ⎞ = 1 , as shown in the figure. The point ⎜ ⎟ 3 ⎝ 2 ⎠ starts at (5, 3) and moves counterclockwise along the ellipse until it reaches the point (–1, 3) at time t = π .

20.

x = sin t

y = − cos t

x 2 + y 2 = sin 2 t + cos2 t x2 + y2 = 1

At t = 0 , x = sin 0 = 0 3 At t = π , 2 x = sin 3π = −1 2

y = − cos0 = −1

y = − cos 3π = 0 2

The point traces a portion of the circle x 2 + y 2 = 1 , as shown in the figure. The point starts at (0, –1) and moves counter clockwise along the circle until it reaches the point (–1, 0) at time t = 3π . 2



914

21.

y = t +1 ⇒ t = y −1 x = 2t − 1 x = 2( y − 1) − 1

22.

y= t y = x − 1 or y 2 = x − 1

x = 2y − 3 x + 3 = 2y y = 1 x+3 2

At t = 0 ,

At t = 0 , x = 2(0) − 1 = −1 At t = 3 , x = 2(3) − 1 = 5

y = 0 +1=1

( )

x = tan π − t 4

y = 3+1 = 4

y2 − x2

( ) = sec ( π − t ) − tan ( π − t ) 4 4

y2 − x2 = 1 At t = 0 ,

y = sec π − t 4

2

At t = π , 2

24.

x = 4 +1= 5

y= 4 =2

x =1− t ⇒ t = 1− x 2

y = (1 − x ) or y = ( x − 1)

( )

y = sec π − 0 = 2 4

x =1− 0 =1 At t = 2 , x = 1 − 2 = −1

y = (0)2 = 0 y = (2)2 = 4

The point traces a portion of the parabola given by

(

)

x = tan π − π = −1 4 2

(

)

y = sec π − π = 2 4 2

The point traces a portion of the top branch of the

y = ( x − 1)2 , as shown in the figure. The point starts at (1, 0) and moves along the parabola until it reaches (–1, 4) at time t = 2.

hyperbola y 2 − x 2 = 1 , as shown in the figure. The point starts at (1,

2

At t = 0 ,

Since 1 + tan 2 θ = sec2 θ

( )

y= 0 =0

y = t2

2

x = tan π − 0 = 1 4

x = 0 +1=1 At t = 4 ,

The point traces a portion of the parabola y 2 = x − 1 , as shown in the figure. The point starts at (1, 0) and moves along the parabola until it reaches the point (5, 2) at time t = 4.

The point traces a line segment, as shown in the figure. The point starts at (–1, 1) and moves along the line segment until it reaches the point (5, 4) at time t = 3.

23.

x = t +1 ⇒ t = x −1

2 ) and moves along the hyperbola until it

reaches the point ( −1,

2 ) at time t = π . 2



25.

915

26.

C1 : x = 2 + t 2 y = 1 − 2t 2

C1 : x = sec2 t y = tan 2 t

x = 2 + t2 → t2 = x − 2 y = 1 − 2( x − 2) y = −2 x + 5 x ≥ 2, y ≤ 1 C2 : x = 2 + t

y = 1− 2t x = 2+t →t = x−2 y = 1 − 2( x − 2) y = −2 x + 5 x ∈ R, y ∈ R The graph of C1 is a ray beginning at (2, 1) with slope −2. The graph of C 2 is a line passing through (2, 1) with slope –2.

tan 2 t + 1 = sec 2 t y +1 = x y = x −1

Because 0 ≤ t < π , 2 1 ≤ sec2 t < ∞ and 0 ≤ tan 2 t < ∞ 1 ≤ x < ∞ and 0 ≤ y 0, y ∈ R, 0 ≤ t < 2π x = h + a sin t → sin t = x − h a y−k y = k + b cos t → cos t = b

x = h + a sin t y = k + b cos t

Let α = ∠COP. Because the smaller circle does not slip, b bθ = aα or α = θ . a

From the figure, OR = a cosθ and QP = aθ sin θ . Thus, x = a cosθ + aθ sin θ TR = a sin θ and TQ = aθ cosθ . Thus, y = a sin θ − aθ cosθ The parametric equations are x = a cosθ + aθ sin θ y = a sin θ − aθ cosθ

The coordinates of P(x, y) are given by x = BC + DP y = OB − OD

π⎞ ⎛ a+b Thus, x = (a + b) cosθ + a sin ⎜ θ− ⎟ 2⎠ ⎝ a ⎛a+b ⎞ = (a + b)cosθ − a cos ⎜ θ⎟ ⎝ a ⎠ π⎞ ⎛a+b θ− ⎟ y = (a + b)sin θ − a cos ⎜ 2⎠ ⎝ a ⎛a+b ⎞ = (a + b)sin θ − a cos ⎜ θ⎟ ⎝ a ⎠ The parametric equations are ⎛ a+b ⎞ θ⎟ x = (a + b) cosθ − a cos⎜ ⎠ ⎝ a ⎛ a+b ⎞ θ⎟ y = (a + b) sin θ − a sin ⎜ ⎝ a ⎠



918

45.

Because the circle moves without slipping, bθ = aα . bθ Therefore, α = . Let P( x, y ) be the coordinates of the moving point. a π ⎛b−a⎞ Angle φ = − ⎜ ⎟θ 2 ⎝ a ⎠ ⎡π ⎛ b − a ⎞ ⎤ Thus, x = (b − a ) cosθ + a sin ⎢ − ⎜ ⎟θ ⎥ ⎣2 ⎝ a ⎠ ⎦ ⎡π ⎛ b − a ⎞ ⎤ y = (b − a )sin θ − a cos ⎢ − ⎜ ⎟θ⎥ ⎣2 ⎝ a ⎠ ⎦

Simplifying, we have ⎛b−a ⎞ θ⎟ x = (b − a) cosθ + a cos⎜ ⎠ ⎝ a ⎛b−a ⎞ θ⎟ y = (b − a ) sin θ − a sin ⎜ ⎝ a ⎠

....................................................... 1.

log5 25 = x [7.2]

2.

x

5.

log3 81 = x [7.2]

Chapter 6 Chapter Review 3.

x

5 = 25

3 = 81

5 x = 52 x=2

3x = 34 x=4

32 x + 7 = 27 [7.4]

6.

x

5 x − 4 = 625 [7.4]

7.

2x + 7 = 3 2 x = −4 x = −2

2x = 1 8

ln eπ = x

[7.2]

π

x

3

e =e x=3

5 x − 4 = 54 x−4=4 x =8

32 x + 7 = 33

4.

ln e3 = x [7.2]

e =e x =π

8.

[7.4]

27 ( 3x ) = 3−1 [7.4] 27 ( 3x ) = 1 3 x 3 = 1 81

2 x = 2 −3 x = −3

3x = 3−4 x = −4

9.

log x 2 = 6 106 = x 2 1,000,000 = x 2 ± 1,000,000 = x ±1000 = x

13.

[7.4]

10.

1 log x = 5 2 log x = 10 1010 = x

[7.4]

11.

10log 2 x = 14 [7.4] 2 x = 14 x=7

x = ±1010

14.

15.


12.

2

eln x = 64 [7.4] x 2 = 64 x = ±8


919

16.

17.

19.

18.

20.

21.

23.

25.

22.

24.

log 4 64 = 3

[7.2]

log1/ 2 8 = −3 [7.2]

26.

3

4 = 64

29.

53 = 125 [7.2] log5 125 = 3

33.

log b

35.

ln xy 3 = ln x + 3 ln y [7.3]

⎛1⎞ ⎜ ⎟ ⎝2⎠

−3

x2 y3 = 2 log b x + 3 log b y − log b z [7.3] z

log

(

=8

210 = 1024 [7.2] log2 1024 = 10

30.

27.

2

4 = 4 [7.2]

28.

31.

100 = 1 [7.2] log10 1 = 0

34.

logb

36.

ln

ln1 = 0 [7.2] e0 = 1

2) = 4 4

32.

81/ 2 = 2 2 [7.2] log8 2 2 = 1 2

x = 1 log x − 2log y + log z [7.3] ( b b b ) y2 z 2 = 1 logb x − 2logb y − logb z 2 xy

z

4

= 1 ( ln x + ln y ) − 4 ln z [7.3] 2 1 = ln x + 1 ln y − 4 ln z 2 2

37.

2 log x + 1 log ( x + 1) = log ( x 2 3 x + 1 ) [7.3] 3

38.

5log x − 2log( x + 5) = log

39.

1 ln 2 xy − 3ln z = ln 2 xy [7.3] 2 z3

40.

ln x − (ln y − ln z ) = ln

41.

log 5 101 =

log 101 ≈ 2.86754 [7.3] log 5

42.

log 3 40 =

log 40 ≈ 3.35776 [7.3] log 3

43.

log 4 0.85 =

log 0.85 ≈ −0.117233 [7.3] log 4

44.

log 8 0.3 =

log 0.3 ≈ −0.578989 [7.3] log 8

x5 [7.3] ( x + 5)2

x xz = ln [7.3] y z y



920

45.

4 x = 30

46.

[7.4]

x

log 4 = log 30 x log 4 = log 30 log 30 x= log 4

48.

ln ( 3x ) + ln 2 = ln1 [7.4]

49.

ln ( 3x ⋅ 2 ) = 1 ln ( 6 x ) = 1 1

e = 6x e=x 6

5 x +1 = 41 [7.4] ( x + 1) log5 = log 41 log 41 x +1= log5 log 41 x= −1 log5

47.

ln x + 2 e ( ) = 6 [7.4]

50.

3x = 4 x − 4 4=x log 2 x +1) 10 ( = 31 [7.4]

2 x + 1 = 31

x+2=6 x=4

2 x = 30 x = 15

5 x + 5− x = 8 2

52.

5 x ( 5 x + 5− x ) = 16 ( 5x )

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

52 x + 1 = 16 ( 5x )

42 x + 1 = 2 ( 42 x − 1)

52 x − 16 ( 5 x ) + 1 = 0

42 x + 1 = 2 ( 42 x − 1)

Let 5 x = u

42 x − 2 ⋅ 42 x + 3 = 0

u 2 − 16u + 1 = 0

42 x = 3 2 x ln 4 = ln 3 x = ln 3 2 ln 4

[7.4]

ln 3x = ln 4 x −1 3x = 4 x −1 3x = 4 ( x − 1)

( x + 2) = 6

4 x + 4− x = 2 4 x − 4− x

51.

ln ( 3x ) − ln ( x − 1) = ln 4

u=

16 ±

162 − 4 (1)(1)

2 ± 16 252 u= 2 ± 16 6 7 u= 2 u =8±3 7

[7.4]

5x = 8 ± 3 7 x= 53.

log ( log x ) = 3

[7.4]

54.

ln ( ln x ) = 2

=x

log x − 5 = 3

e = ln x

10 = log x 10

55.

2

3

(10 3 )

[7.4]

ln (8 ± 3 7 ) [7.4] ln 5

e

( e2 )

=x

3

10 = x − 5 106 = x − 5 106 + 5 = x x = 1,000,005

101000 = x


[7.4]


56.

log x + log ( x − 15) = 1

921

57.

log x ( x − 15) = 1

3 =x 81 = x

1 = log5 x 2

[7.4]

5 = x2

15 ± 152 − 4 (1)( −10 )

± 5=x

[7.4]

2

x = 15 ± 265 2 + 15 265 x= 2

[7.4]

log5 x 3 = log5 16 x [7.4]

60.

25 = 16log4 x [7.4] 25 = 4

2

25 = 4log4 x

x = 16 x=4

⎛ I ⎞ m = log ⎜ ⎟ [7.3] ⎝ I0 ⎠ ⎛ 51,782,000 I 0 ⎞ = log ⎜ ⎟ I0 ⎝ ⎠ = log 51,782,000 ≈ 7.7

61.

2 log 4 x

3

x = 16 x

62.

70 = log5 x 2

4

0 = x 2 − 15 x − 10

59.

log7 (log5 x 2 ) = 0

4 = log3 x

10 = x 2 − 15 x

x=

58.

log 4 (log3 x ) = 1

2

25 = x 2 ±5 = x 5= x

M = log A + 3log8t − 2.92

[7.3]

63.

= log18 + 3log8(21) − 2.92 = log18 + 3log168 − 2.92 ≈ 5.0

⎛I ⎞ log ⎜ 1 ⎟ = 7.2 ⎝ I0 ⎠ I1 = 107.2 I0

⎛I ⎞ log ⎜ 2 ⎟ = 3.7 ⎝ I0 ⎠ I2 = 103.7 I0

and

I1 = 107.2 I 0

[7.3]

I 2 = 103.7 I 0

I1 107.2 I 0 103.5 3162 = = ≈ I 2 103.7 I 0 1 1 3162 to 1

64.

I1 = 600 = 10 x [7.3] I2

65.

log 600 = log10 x log 600 = x 2.8 ≈ x

[7.3] pH = − log ⎡⎢ H3O + ⎤⎥ ⎣ ⎦ = − log ⎡⎢6.28 × 10−5 ⎤⎥ ⎣ ⎦ ≈ 4.2

5.4 = − log ⎡⎢ H3O + ⎤⎥ [7.3] ⎣ ⎦

66.

−5.4 = log ⎡⎢ H3O + ⎤⎥ ⎣ ⎦ 10−5.4 = H3O + 0.00000398 ≈ H3O + H3O + ≈ 3.98 × 10−6

67.

P = 16,000, r = 0.08, t = 3 [7.5] a. b.

0.08 ⎞ ⎛ B = 16,000⎜1 + ⎟ 12 ⎠ ⎝

36

68.

≈ $20,323.79

B = 16,000e 0.08(3)

P = 19,000, r = 0.06, t = 5 [7.5] 1825

a.

0.06 ⎞ ⎛ B = 19,000⎜1 + ⎟ 365 ⎠ ⎝

b.

B = 19,000e 0.3 ≈ $25,647.32

B = 16,000e 0.24 ≈ $20,339.99


≈ $25,646.69


922

69.

S (n ) = P (1 − r )n , P = 12,400, r = 0.29, t = 3 [7.5]

70.

N ( t ) = N 0 e −0.12t

a.

S (n ) = 12,400(1 − 0.29 )3 ≈ $4438.10

N (10 ) = N 0 e N (10 ) N0

[7.5]

−0.12(10 )

= e −1.2 = .301

N (10 ) N0

= 30.1% healed

100% − 30.1% = 69.9% healed b.

N (t ) N0

= 0.5

0.5 = e −0.12t ln 0.5 = −0.12t ln 0.5 = t −0.12 t ≈ 6 days c.

N (t ) N0

= 0.1

0.1 = e −0.12t ln 0.1 = −0.12t ln 0.1 = t −0.12 t ≈ 19 days 71.

N (2) = 5

N ( 0) = 1 k 0 1 = N 0e ( )

72.

5 = e2 k ln 5 = 2k k = ln 5 ≈ 0.8047 2

1 = N0

N (0) = N 0 = 2 and N (3) = N 0 e 3k = 2e 3k = 11 e3k = 11 2 3k 11 e = 2 3k = ln ⎛⎜ 11 ⎞⎟ ⎝2⎠

Thus N ( t ) = e0.8047 t [7.5]

k = 1 ln ⎛⎜ 11 ⎞⎟ 3 ⎝2⎠ ≈ 0.5682

Thus N (t ) = 2e 0.5682t [7.5] 73.

4 = N (1) = N 0e k and thus

4 = e k . Now, we also N0 5

⎛ 4 ⎞ 1024 ⎟ = . have N (5) = 5 = N 0e5k = N 0 ⎜⎜ ⎟ N 04 ⎝ N0 ⎠ N0 = 4

1024 ≈ 3.783 5

74.

1 = N (0 ) = N 0 and 2 = N (− 1) = N 0e− k .

Since N 0 = 1 , we have 2 = 1 ⋅ e− k . ln 2 = − k k ≈ −0.6931 Thus N ( t ) = e−0.6931 t . [7.5]

Thus 4 = 3.783ek . ⎛ 4 ⎞ ln⎜ ⎟=k ⎝ 3.783 ⎠ k ≈ 0.0558 Thus N 0 = 3.783e0.0558 t . [7.5] Copyright © Houghton Mifflin Company. All rights reserved.


75.

a.

923

[7.5] N (1) = 25, 200ek (1) = 26,800 k 26,800 e = 25, 200 ⎛ 26,800 ⎞ k ln e = ln ⎜ ⎟ ⎝ 25, 200 ⎠ k ≈ 0.061557893

76.

P(t) = 0.5 t / 5730 = 0.96 [7.5]

t log 0.5 = log 0.96 5730 t log 0.96 = 5730 log 0.5

N (t ) = 25, 200e0.061557893 t

b.

)

(

log 0.5 t / 5730 = log 0.96

⎛ log 0.96 ⎞ t = 5730 ⎜ ⎟ ≈ 340 years ⎝ log 0.5 ⎠

N (7) = 25, 200e0.061557893(7) = 25, 200e0.430905251 ≈ 38,800

77.

Answers will vary.

78.

a.

b. c. 79.

a.

exponential: R ≈ 179.949 ( 0.968094t ) , r ≈ −0.99277 logarithmic: R ≈ 171.19665 − 35.71341ln t , r ≈ −0.98574 The exponential equation provides a better fit to the data.

R ≈ 179.949 ( 0.968094108 ) ≈ 5.4 per 1000 live births [7.6] P(t ) =

mP0 P0 + ( m − P0 )e

P(3) = 360 =

− kt

80.

128 = 128 = 128 = 128 = 21 1 3 6 1 + 5e −0.27(0) 1 + 5e0 1 + 5

a.

P(0) =

b.

As t → ∞, e−0.27t → 0. P(t ) → 128 = 128 = 128 [7.5] 1 + 5(0) 1

1400(210)

210 + (1400 − 210)e − k (3) 294000 360 = 210 + 1190e −3k

360 ( 210 + 1190e −3k ) = 294000

210 + 1190e −3k = 294000 360 −3k 29400 = − 210 1190e 36 e −3k = 29400 / 36 − 210 1190 −3k 29400 / 36 − 210 ⎞ ⎛ ln e = ln ⎜ ⎟ 1190 ⎝ ⎠ −3k = ln ⎛⎜ 29400 / 36 − 210 ⎞⎟ 1190 ⎝ ⎠

b.

k = − 1 ln ⎛⎜ 29400 / 36 − 210 ⎞⎟ 3 ⎝ 1190 ⎠ k ≈ 0.2245763649 294000 1400 P(t ) = = 210 + 1190e −0.22458t 1 + 17 e −0.22458t 3 294000 P(13) = 210 + 1190e −0.22458(13) 294000 = 210 + 1190e −2.919492744 ≈ 1070 coyotes [7.5]



924

....................................................... 1.

a.

logb (5 x − 3) = c [7.2]

Chapter 6 Chapter Test b.

bc = 5 x − 3

2.

logb

z2 y

3

x

3x / 2 = y x 2

log3 y =

= logb z 2 − logb y 3 − logb x1/ 2 [7.3]

3.

3 log10 ( 2 x + 3) − 3log10 ( x − 2 ) = log10 ( 2 x + 3) − log10 ( x − 2 )

= log10 2 x + 33 [7.3] ( x − 2)

1 = 2logb z − 3logb y − logb x 2

4.

log12 [7.3] log 4 ≈ 1.7925

5.

log 4 12 =

45− x = 7 x

8.

5− x

6.

[7.4] x

ln 4 = ln 7 (5 − x ) ln 4 = x ln 7 5ln 4 − x ln 4 = x ln 7 5ln 4 = x ln 7 + x ln 4 5ln 4 = x (ln 7 + ln 4) 5ln 4 = x ln 28

11.

a.

(

) 12(5) = 20,000 (1 + 0.078 ) 12

nt A = P 1+ r n

= 20,000(1.0065) = $29,502.36 b.

60

A = Pert = 20,000e0.078(5) = 20,000e0.39 = $29,539.62

[7.5]

9.

7.

log( x + 99) − log(3x − 2) = 2 [7.4] x + 99 log =2 3x − 2 x + 99 = 102 3x − 2 x + 99 = 100(3x − 2) x + 99 = 300 x − 200 −299 x = −299 x =1

12.

10 − 7 x + x 2 = 37 − x x 2 − 6 x − 27 = 0 ( x − 9)( x + 3) = 0 x = 9 (not in domain) or x = −3 x = −3 [7.4]

(

(

nt

[7.5]

)

0.04 12t 12 0.04 12t

2P = P 1 + 2 = 1+

ln(2 − x) + ln(5 − x) = ln(37 − x) ln(2 − x)(5 − x) = ln(37 − x) (2 − x)(5 − x) = (37 − x)

10.

r⎞ ⎛ A = P ⎜1 + ⎟ ⎝ n⎠

(

12

)

)

0.04 12t 12 0.04 ln 2 = 12t ln 1 + 12

ln 2 = ln 1 +

12t = t=

(

(

ln 2

ln 1 +

0.04 12

)

)

1 ln 2 ⋅ 12 ln 1 + 0.04

(

5 x = 22 [7.4] x log5 = log 22 log 22 x= log5 x ≈ 1.9206

t ≈ 17.36 years

12

)



13.

a.

925

⎛ I ⎞ [7.3] M = log ⎜ ⎟ ⎝ I0 ⎠ ⎛ 42,304,000 I 0 ⎞ = log ⎜ ⎟ I0 ⎝ ⎠ = log 42,304,000 ≈ 7.6

14.

a.

N (3) = 34600ek (3) = 39800 34600e3k = 39800 39800 e3k = 34600 ⎛ 398 ⎞ ln e3k = ln ⎜ ⎟ ⎝ 346 ⎠ ⎛ 398 ⎞ 3k = ln ⎜ ⎟ ⎝ 346 ⎠ ⎛ 398 ⎞ k = 1 ln ⎜ 3 ⎝ 346 ⎟⎠ k ≈ 0.0466710767 N (t ) = 34600e0.0466710767 t [7.5]

b.

⎛I ⎞ log ⎜ 1 ⎟ = 6.3 ⎝ I0 ⎠ I1 = 106.3 I0

and

⎛I ⎞ log ⎜ 2 ⎟ = 4.5 ⎝ I0 ⎠ I2 = 104.5 I0

I1 = 106.3 I 0

b.

N (10) = 34600e0.0466710767(10) = 34600e0.466710767 ≈ 55,000

I 2 = 104.5 I 0

I1 106.3 I 0 101.8 63 = = ≈ 1 1 I 2 104.5 I 0 Therefore the ratio is 63 to 1.

P (t ) = 0.5 t / 5730 = 0.92 [7.5]

15.

log 0.5 t / 5730 = log 0.92 t log 0.5 = log 0.92 5730 log 0.92 t = 5730 log 0.5 ⎛ log 0.92 ⎞ t = 5730 ⎜ ⎟ ⎝ log 0.5 ⎠ t ≈ 690 years

16.

a.

y = 1.671991998(2.471878247) x

b.

y = 1.671991998(2.471878247)7.8 ≈ 1945

17.

a.

Logarithmic: d ≈ 67.35501 + 2.54015ln t ; Logistic: d ≈

b.

Logarithmic: d ≈ 67.35501 + 2.54015ln(12) ≈ 73.67 m ; 72.03783 Logistic: d ≈ ≈ 72.03 m [7.6] 1 + 0.15279e −0.67752 (12)

72.03783 1 + 0.15279e −0.67752 t


[7.6]

926

18.


a.

a=

c − P0 1100 − 160 = = 5.875 P0 160

b.

P(t ) =

1100 ≈ 457 raccoons 1 + 5.875e −0.20429(7)

c 1 + ae − bt 1100 P(1) = 1 + 5.875e − b(1) 1100 190 = 1 + 5.875e − b P (t ) =

190(1 + 5.875e − b ) = 1100 190 + 1116.25e − b = 1100 1116.25e − b = 910 e − b = 910 1116.25 −b ln e = ln 910 1116.25 −b = ln 910 1116.25 b = − ln 910 1116.25 b ≈ 0.20429 1100 P(t ) = [7.5] 1 + 5.875e −0.20429 t

....................................................... 1.

x 2 + 4 x − 6 = 0 [1.1]

2.

y = x3 − 4 x [1.4]


( g o f )( x) = g[ f ( x)] = g [sin x ] = 3sin x − 2

6.

opp sin t = − 3 = 2 hyp

2

−4 ± 4 − 4(1)(−6) −4 ± 40 = 2(1) 2 4 2 10 − ± = = −2 ± 10 2 The solutions are 2 ± 10. x=

[1.5]


4.

240o = 240o ⎛⎜ π o ⎞⎟ = 4π ⎝ 180 ⎠ 3

[2.1]

5.

[2.1] v = ωr ⋅ ⋅ ⋅ ⋅ 3 2π 60 60 10 = 12 ⋅ 5280 ≈ 11 mph

[2.4]

adj = 22 − ( −3 ) = 1 = 1 2

tan t =


opp − 3 = =− 3 adj 1


7.

9.

11.

13.

927

tan 43o = a [2.4] 20 o a = 20 tan 43 a = 19 cm

( )

y = 2 tan π x 3 period: 3

8.

10.

[3.2] sin α = 3 , cos α = − 4 5 5 cos β = − 5 , sin β = 12 13 13 sin(α + β ) = sin α cos β + cos α sin β 5 ⎞ ⎛ 4 ⎞⎛ 12 ⎞ = ⎛⎜ 3 ⎞⎛ ⎟⎜ − ⎟ + ⎜ − ⎟⎜ ⎟ ⎝ 5 ⎠⎝ 13 ⎠ ⎝ 5 ⎠⎝ 13 ⎠ = − 15 − 48 = − 63 65 65 65

12.

( )

a d c b C = 180o − A − B = 180o − 71o − 80o = 29o b = c sin B sin C o b = 155sin 80 ≈ 314.9 ft o sin 29 a = c sin A sin C o a = 155sin 71 ≈ 302.3 ft o sin 29 sin 71o = d 314.9 d = 314.9sin 71o [4.1] ≈ 298 ft

[4.2] 2

2

cos B = a + c − b = 4 + 3.6 − 2.5 = 22.7 2ac 2(4)(3.6) 28.8 B ≈ 38o

C A

b 2 = a 2 + c 2 − 2ac cos B 2

52 − (1) opp = =2 6 hyp 5 5

B

14.

[3.6]

( )

2

[3.5]

⎡ ⎛ 1 ⎞⎤ 2 6 y = sin ⎢cos −1 ⎜ ⎟ ⎥ = 5 ⎝ 5 ⎠⎦ ⎣

( )

2

[3.1]

2

2 2 adj = 13 −12 = 5 hyp 13 13

2

⎡ ⎛ 1 ⎞⎤ y = sin ⎢ cos −1 ⎜ ⎟ ⎥ ⎝ 5 ⎠⎦ ⎣

sin α =

x = cos ⎡sin −1 12 ⎤ = 5 ⎢⎣ 13 ⎥⎦ 13

15.

sin x = sin x ⋅ 1 + cos x 1 − cos x 1 − cos x 1 + cos x sin x (1 + cos x) = 1 − cos 2 x sin x (1 + cos x) = sin 2 x 1 = + cos x sin x sin x = csc x + cot x

adj Let α = cos −1 1 , cosα = 1 = 5 5 hyp

opp Let α = sin −1 12 , sin α = 12 = 13 13 hyp cosα =

[2.5]

period: 6, amplitude: 1 2

[2.6]

cos −1 x = sin −1 12 13 ⎡ x = cos sin −1 12 ⎤ ⎢⎣ 13 ⎥⎦

( )

y = 1 cos π x 2 3

16.

a1 = 30cos145o ≈ −24.6 [4.3] a2 = 30sin145o ≈ 17.2

v = −24.6i + 17.2 j


928

17.


v ⋅ w = (3i + 2 j) ⋅ (5i − 7 j) [4.3] = 3(5) + (2)( −7) =1≠ 0 No. the vectors are not orthogonal.

18.

z = −2 + 2i 3

[5.2]

2

r = (−2) + (2 3) 2 = 16 = 4

α = tan −1 2 3 = tan −1 3 = π −2

3

θ = π − π = 2π 3

3

( )

z = 4 cis 2π 3

19.

r = 3sin(2θ ) [6.5]

20.

x = 2t − 1 [6.7] x + 1 = 2t t = x +1 2 y = 4t 2 + 1

( )

2 ⎛ 2 ⎞ y = 4 x + 1 + 1 = 4⎜ x + 2x + 1 ⎟ + 1 2 4 ⎝ ⎠ y = x2 + 2 x + 2

....................................................... 1.

( f o g )( x) = f [ g ( x)]

[1.5]

2.

f ( x) = 2 x + 8 y = 2x + 8 x = 2y + 8 x − 8 = 2y x −8 = y 2 f −1 ( x) = 1 x − 4 2

5.

y = 3sin 1 x − π 3 2

2

= f ( x + 1) = cos( x 2 + 1)

4.

[2.2] cos 26o = 15 a a = 15 o ≈ 16.7 cm cos 26

(

Chapter 7 Cumulative Review

)

[1.6]

3.

c = 32 + 42 = 25 = 5 sin θ = 3 , cosθ = 4 , tan θ = 3 5 5 4

[2.5]

6.

y = sin x + cos x

0 ≤ 2 x − π ≤ 2π 2 π ≤ 2 x ≤ 5π 2 2 π ≤ x ≤ 5π 4 4 amplitude = 4, period = π ,

Amplitude:

phase shift = π 4


[3.4]

2 , period: 2π

[2.2]


7.

929

f ( − x) = sin(− x) = − sin x = − f ( x) odd function [2.4]

8.

1 − sin x = 1 − sin 2 x sin x sin x 2 x cos = sin x = cos x cos x sin x = cos x cot x

9.

[3.1]

( )

tan ⎛⎜ sin −1 12 ⎞⎟ 13 ⎠ ⎝

[3.5]

opp Let α = sin −1 12 , sin α = 12 = 13 13 hyp opp 12 tanα = = = 12 adj 132 −122 5

( )

tan ⎡sin −1 12 ⎤ = 12 ⎢⎣ 13 ⎥⎦ 5

10.

2cos 2 x + sin x − 1 = 0

[3.6]

11.

2

2(1 − sin x) + sin x − 1 = 0 2sin 2 x − sin x − 1 = 0 (2sin x + 1)(sin x − 1) = 0 2sin x + 1 = 0 sin x = − 1 2 x = 7π , 11π 6 6

12.

14.

v = (−3)2 + 42

α = tan −1

v = 9 + 16

α θ θ θ

v =5 sin x − 1 = 0 sin x = 1 x=π 2

cos θ = v ⋅ w v w (2i − 3 j) ⋅ ( −3i + 4 j) cos θ = 22 + (−3)2 (−3) 2 + 42 2( −3) + (−3)(4) cos θ = 13 25 cos θ = −18 = −0.9985 325 θ = 176.8o

13

[4.3]

a = b sin A sin B o sin A = a sin B = 42sin 32 ≈ 0.445 b 50 A ≈ 26o

AB = 400(cos 42i + sin 42 j) ≈ 297.3i + 267.7 j [4.3] AD = 55[cos(−25°)i + sin(−25°) j] ≈ 49.8i − 23.2 j AC = AB + AD AC = 297.3i + 267.7 j + 49.8i − 23.2 j AC ≈ 347.1i + 244.5 j AC = 347.12 + (244.5)2 AC ≈ 425 mph

( 347.1 )

α = 90o − θ = 90o − tan −1 244.5 ≈ 55o 15.

[4.1]

z =1− i

[5.3]

r = 12 + (−1)2 r= 2

α = tan −1 θ = 315o

(1 − i)8 = [ 2(cos315° + i sin 315°)]8 = 16[cos(8 ⋅ 315°) + i sin(8 ⋅ 315°)] = 16(cos 2520° + i sin 2520°) = 16(cos 0° + i sin 0°) = 16 + 0i = 16 i = 1( cos90o + i sin 90o )

[4.3]

≈ 53.1° = 180° − α ≈ 180° − 53.1° ≈ 126.9°

z = 2(cos315° + i sin 315°)

16.

4 = tan −1 4 −3 3

[5.3]

wk = 11/ 2 ⎛⎜ cos 90° + 360°k + i sin 90° + 360°k ⎞⎟ 2 2 ⎝ ⎠

k = 0, 1

w0 = cos 90° + i sin 90° 2 2 w0 = cos 45° + i sin 45°

w1 = cos 90° + 360° + i sin 90° + 360° 2 2 w1 = cos225°+ i sin 225°

w0 = 2 + 2 i 2 2

w1 = − 2 − 2 i 2 2


−1 = 45o 1


930

17.

y x = tan −1 1 1 o

θ = tan −1

r = x2 + y 2

[6.5]

18.

r = 2 − 2cosθ

[6.5]

()

= 12 + 12 = 2

= 45

The polar coordinates of the point are ( 2, 45o ).

19.

5 x = 10 x

log 5 = log10 x log 5 = 1 x = 1 ≈ 1.43 log 5

[7.4]

20.

N (t ) = N 0ekt N (138) = N 0e138k 0.5 N 0 = N 0e138k 0.5 = e138k ln 0.5 = ln e138k ln 0.5 = 138k ln e ln 0.5 = 138k ln 0.5 =k 138 −0.005023 ≈ k N ( t ) = N 0(0.5)t /138 ≈ N 0e−0.005023t N (100 ) = 3 ( 0.5 )

100 /138

≈ 1.8 mg


[7.5]

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