Solution Manual for Trigonometry, 5th Edition

INSTRUCTOR’S SOLUTIONS MANUAL EDGAR REYES Southeastern Louisiana University T RIGONOMETRY FIFTH EDITION Mark Dugopolski Southeastern Louisiana University The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Reproduced by Pearson from electronic files supplied by the author. Copyright © 2020, 2015, 2011 by Pearson Education, Inc. 221 River Street, Hoboken, NJ 07030. All rights reserved. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. ISBN-13: 978-0-13-520744-4 ISBN-10: 0-13-520744-4 Table of Contents Chapter P …………………………………………………………………………………….1 Chapter 1 ……………………………………………………………………………………32 Chapter 2 ……………………………………………………………………………………83 Chapter 3 ………………………………………………………………………………….137 Chapter 4 ………………………………………………………………………………….195 Chapter 5 ………………………………………………………………………………….245 Chapter 6 ………………………………………………………………………………….302 P.1 The Cartesian Coordinate System 1 For Thought 18. (1, 5), Quadrant I 1. False, the point (2, −3) is in Quadrant IV. 19. c = 2. False, the point (4, 0) does not belong to any quadrant. 20. Since a2 + a2 = Then a = 1. 3. False, since the distance is (a − c)2 + (b − d)2 . √ ( 3)2 + 12 = √ 4=2 √ 2 2 , we get 2a2 = 2 or a2 = 1. 4. False, since Ax + By = C is a linear equation. 21. Since b2 + √ 22 = 32 , we get b2 + 4 = 9 or b2 = 5. Then b = 5. 5. True 22. Since b2 + 6. False, since √ 72 + 9 2 = √ 130 ≈ 11.4 7. True 2 1 1 = 12 , we get b2 + = 1 2 4 √ 3 3 or b2 = . Thus, b = . 4 2 P.1 Exercises 23. Since a2 + 32 = 52 , we get a2 + 9 = 25 or a2 = 16. Then a = 4. √ √ √ 24. c = 32 + 22 = 9 + 4 = 13 √ √ 25. 4 · 7 = 2 7 √ √ 26. 25 · 2 = 5 2 1. ordered 27. √59 = 2. Cartesian 3 28. √16 = 8. True 9. True 10. False, since the radius is 3. 3. x-axis 4. origin √ √ √ √ 5 3 3 4 √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ 29. √23 · √33 = 6 3 5. Pythagorean theorem 30. √35 · √55 = 6. circle 31. 2√53 · √55 = 2 515 7. linear equation 32. √53 · √33 = 5 3 3 8. y-intercept 33. √13 · √33 = 9. (4, 1), Quadrant I 10. (−3, 2), Quadrant II 15 5 3 3 34. √32 · √22 = 3 2 2 11. (1, 0), x-axis 35. √23 · √33 = 12. (−1, −5), Quadrant III 36. √52 · √22 = 13. (5, −1), Quadrant IV 37. Distance is (4 − 1)2 + (7 − 3)2 = √ 25 = 5, midpoint is (2.5, 5) √ 38. Distance is 144 + 25 = 13, midpoint is (3, 0.5) 14. (0, −3), y-axis 15. (−4, −2), Quadrant III 16. (−2, 0), x-axis 17. (−2, 4), Quadrant II 6 3 10 2 √ 9 + 16 = 39. Distance is √ (−1 − 1)2 + (−2 − 0)2 = √ 4 + 4 = 2 2, midpoint is (0, −1) c 2020 Pearson Education, Inc. Copyright 2 Chapter P 40. Distance is √ √ 4 + 4 = 2 2, midpoint is (0, 1) √ 2 2 −0 + 41. Distance is 2 √ 2 −0 2 2 = 2 2 √ + = 1 = 1, midpoint is 4 4 √ √ √ √ 2/2 + 0 2/2 + 0 2 2 = , , 2 2 4 4 √ 2 42. Distance is 3 − 0 + (1 − 0)2 = √ 3 + 1 = 2, midpoint is √ √ 3+0 1+0 3 1 = , , 2 2 2 2 √ 2 √ √ √ 2 43. Distance is 18 − 8 + 12 − 27 = √ √ √ √ (3 2 − 2 2)2 + (2 3 − 3 3)2 = √ √ √ ( 2)2 + (− 3)2 = 5, √ √ √ √ 18 + 8 12 + 27 = midpoint is , 2 2 √ √ √ √ √ √ 3 2+2 2 2 3+3 3 5 2 5 3 = , , 2 2 2 2 √ √ √ √ 2 2 44. Distance is 72 − 50 + 45 − 20 = √ √ √ √ (6 2 − 5 2)2 + (3 5 − 2 5)2 = √ √ √ ( 2)2 + ( 5)2 = 7, √ √ √ √ 72 + 50 45 + 20 = midpoint is , 2 2 √ √ √ √ √ √ 6 2+5 2 3 5+2 5 11 2 5 5 = , , 2 2 2 2 √ Algebraic Prerequisites π2 + 4 , midpoint is 2 50. Distance is π , midpoint is 2 51. Distance is √ π 2 25 + = 36 36 + (1 − 1)2 = π + π/2 1 + 1 , 2 2 2 π π− 2 π π − 2 3 2 = π2 = 4 3π ,1 4 1 1 + − − 3 2 2 = π 2 + 25 , 6 ⎛ ⎞ π π 1 1 ⎜3 + 2 2 − 3⎟ 5π 1 ⎜ ⎟ = midpoint is ⎝ , , 2 2 ⎠ 12 12 π− 52. Distance is π2 1 + = 9 4 √ 2π 3 2 + −1 + 1 2 2 = 4π 2 + 9 , 6 ⎛ ⎞ 2π 1 ⎜ 3 + π −2 − 1⎟ 5π 3 ⎜ ⎟ = midpoint is ⎝ , ,− 2 2 ⎠ 6 4 53. Center(0, 0), radius 4 y 5 3 3 5 x 45. Distance is (1.2 + 3.8)2 + (4.4 + 2.2)2 = √ √ 25 + 49 = 74, midpoint is (−1.3, 1.3) √ √ 46. Distance is 49 + 81 = 130, midpoint is (1.2, −3) √ π2 + 4 3π 1 47. Distance is , midpoint is , 2 4 2 π 1 , 4 2 √ 49. Distance is (2π − π)2 + (0 − 0)2 = π 2 = π, 2π + π 0 + 0 3π = midpoint is , ,0 2 2 2 48. Distance is 54. Center (0, 0), radius 1 y 2 -2 2 -2 c 2020 Pearson Education, Inc. Copyright x P.1 The Cartesian Coordinate System √ 66. Note, the distance between ( 3/2, 1/2) and the origin is 1. Thus, the radius is 1. The standard equation is x2 + y 2 = 1. 55. Center (−6, 0), radius 6 y 6 -14 √ 67. The radius is (−1 − 0)2 + (2 − 0)2 = 5. The standard equation is (x+1)2 +(y−2)2 = 5. x -10 3 -6 68. Since the center is (0, 0) and the radius is 2, the standard equation is x2 + y 2 = 4. 56. Center (0, 3), radius 3 69. Note, the center is (1, 3) and the radius is 2. The standard equation is (x−1)2 +(y−3)2 = 4. √ 70. The radius is (2 − 0)2 + (2 − 0)2 = 8. The standard equation is (x − 2)2 + (y − 2)2 = 8. y 7 3 -3 3 x 71. We solve for a. √ 57. Center (2, −2), radius 2 2 a2 + y 2 5 2 3 = 1 5 a2 = 1 − x 9 25 16 25 4 a = ± 5 a2 = -2 -5 √ 58. Center (4, −2), radius 2 5 72. We solve for a. y 4 a2 + − x 10 2 = 1 1 4 a2 = 1 − -4 -6 59. x2 + y 2 = 7 1 2 a2 = a = ± √ 60. x2 + y 2 = 12 since (2 3)2 = 12 61. (x + 2)2 + (y − 5)2 = 1/4 73. We solve for a. 62. (x + 1)2 + (y + 6)2 = 1/9 2 − 5 63. The √ distance between (3, 5) and the origin is 34 which is the radius. The standard equation is (x − 3)2 + (y − 5)2 = 34. 3 4√ 64. The √ distance between (−3, 9) and the origin is 90 which is the radius. The standard equation is (x + 3)2 + (y − 9)2 = 90. √ √ 65. Note, the distance between ( 2/2, 2/2) and the origin is 1. Thus, the radius is 1. The standard equation is x2 + y 2 = 1. c 2020 Pearson Education, Inc. Copyright 2 3 2 + a2 = 1 4 25 a2 = 1 − a2 = 21 25√ a = ± 21 5 4 Chapter P 74. Solve for a: Algebraic Prerequisites 79. x + y = 80 goes through (0, 80), (80, 0). 2 2 3 y +a 2 = 1 80 4 a2 = 1 − 9 5 2 a = 9√ 5 a = ± 3 75. y = 3x − 4 goes through (0, −4), 20 40 4 ,0 . 3 80 x 80. 2x + y = −100 goes through (0, −100), (−50, 0). y y x 20 40 2 5 x 50 -3 -4 76. y = 5x − 5 goes through (0, −5), (1, 0). 100 y 5 81. x = 3y − 90 goes through (0, 30), (−90, 0). 2 4 y x -5 30 77. 3x − y = 6 goes through (0, −6), (2, 0). 15 y -90 3 1 3 x -45 x 82. x = 80 − 2y goes through (0, 40), (80, 0). y -6 78. 5x − 2y = 10 goes through (0, −5), (2, 0). y 40 5/2 1 -4 -5 3 x 20 40 c 2020 Pearson Education, Inc. Copyright 80 x P.1 The Cartesian Coordinate System 83. 1 1 x − y = 600 goes through (0, −1800), 2 3 (1200, 0). 5 87. x = 5 y 5 y x 1200 600 7 3 x -5 -900 88. y = −2 y -1800 84. 2 1 y − x = 400 goes through (0, 600), 3 2 (−800, 0). -5 5 -1 x -3 y 89. y = 4 y 600 5 3 300 -800 -4 x -400 4 x 85. Intercepts are (0, 0.0025), (0.005, 0). 90. x = −3 y y 3 0.0025 x x 0.005 -4 -2 -3 86. Intercepts are (0, −0.3), (0.5, 0). 91. x = −4 y y 4 -5 0.3 -0.5 -0.3 0.5 x x -3 -4 c 2020 Pearson Education, Inc. Copyright 6 Chapter P Algebraic Prerequisites 96. y = 999x − 100 goes through (0, −100), (100/999, 0). 92. y = 5 y 6 y 4 -5 x 5 100 0.5 93. Solving for y, we have y = 1. 1 x 100 y 2 -1 97. y = 3000 − 500x goes through (0, 3000), (6, 0). x 1 y -1 3000 94. Solving for x, we get x = 1 . y 1000 4 6 x 1 -1 x 2 1 (200x − 1) goes through (0, −1/300), 300 (1/200, 0). 98. y = -1 y 0.01 95. y = x − 20 goes through (0, −20), (20, 0). y 20 20 20 x 0.01 0.01 20 x 0.01 √ √ 99. The hypotenuse is 62 + 82 = 100 = 10. √ √ √ 100. The other leg is 102 − 42 = 84 = 2 21 ft. 101. a) Let r be the radius of the smaller circle. Consider the right triangle with vertices at the origin, another vertex at the center of a smaller circle, and a third vertex at the center of the circle of radius 1. By the Pythagorean Theorem, we obtain 1 + (2 − r)2 = (1 + r)2 5 − 4r = 1 + 2r c 2020 Pearson Education, Inc. Copyright P.1 The Cartesian Coordinate System 7 4 = 6r 2 r = . 3 The diameter of the smaller circle is 2r = 43 . b) The smallest circles are centered at (±r, 0) or (±4/3, 0). The equations of the circles are 4 2 4 x− + y2 = 3 9 and 4 2 4 x+ + y2 = 3 9 102. Draw a right triangle with vertices at the centers of the circles, and another vertex at a point of intersection of the two circles. The legs of the right triangle are 5 and 12. By the √ Pythagorean theorem, the hypotenuse is 52 + 122 = 13. 103. Let C(h, k) and r be the center and radius of the smallest circle, respectively. Then k = −r. We consider two right triangles each of which has a vertex at C. The right triangles have sides that are perpendicular to the coordinate axes. Also, one side of each right triangle passes through the center of a larger circle. Applying the Pythagorean Theorem, we list a system of equations (r + 1)2 = h2 + (1 − r)2 (2 − r)2 = h2 + r2 . The solutions are r = 1/2, h = −r = −1/2. √ 105. The midpoint of (0, 20.8) and (48, 27.4) is 0 + 48 20.8 + 27.4 , 2 2 = (24, 24.1). In 1994 (= 1970 + 24), the median age at first marriage was 24.1 years. 106. a) If h = 0, then 0 = 0.229n + 5.203. Then n = −5.203/0.229 ≈ −22.72. The n-intercept is (−22.72, 0). There were no unmarried couples in 1977 (≈ 2000 − 22.7). Nonsense. b) If n = 0, then h = 0.229(0) + 5.203 = 5.203. The h-intercept is (0, 5.203). In 2000, there where 5,203,000 unmarriedcouple households. 107. The distance between (10, 0) and (0, 0) is 10. The √ distance between (1, 3) and the origin is 10. If two points have integer coordinates, then them is of the form √ the distance 2between 2 2 2 s + t where s , t ∈ {0, 1, 22 , 32 , 42 , …} = {0, 1, 4, 9, 16, …}. Note, there are no numbers s2 and t2 in {0, 1, 4, 9, 16, …} satisfying s2 + t2 = 19. Thus, one cannot find two points with integer coordinates whose distance between them is √ 19. 108. One can assume the vertices √ of the right triangle are A(0, 0), B(1, 3), and C(1, 0). 2, and k = The equation of the smallest circle is √ (x − 2)2 + (y + 1/2)2 = 1/4. 104. We apply symmetry to the centers of the remaining three circles. From the answer or equation in Exercise 103, the equations of the remaining circles are √ (x − 2)2 + (y − 1/2)2 = 1/4 √ (x + 2)2 + (y − 1/2)2 = 1/4 √ (x + 2)2 + (y + 1/2)2 = 1/4. The midpoint of the hypotenuse AB is √ 1 3 . The distance between the midpoint , 2 2 and C is 1 2 −1 2 √ + 3 2 −0 2 = 1, which is also the distance from the midpoint to A, and the distance from the midpoint to B. 111. On day 1, break off a 1-dollar piece and pay the gardener. On day 2, break of a 2-dollar and pay the gardener. The gardener will give you back your change which is a 1-dollar piece. c 2020 Pearson Education, Inc. Copyright 8 Chapter P On day 3, you pay the gardener with the 1dollar piece you received as change from the previous day. On day 4, pay the gardener with the 4-dollar bar. The gardener will give you back your change which will consist of a 1-dollar piece and a 2-dollar piece. On day 5, you pay the gardener with the 1dollar piece you received as change from the previous day. On day 6, pay the gardener with the 2-dollar piece you received as change from day 4. The gardener will give you back your change which is a 1-dollar piece. On day 7, pay the gardener with the 1-dollar piece you received as change from day 6. 112. Let ABC be a right triangle with vertices at A(2, 7), B(0, −3), and C(6, 1). Notice, the midpoints of the sides of ABC are (3, −1), (4, 4) and (1, 2). The area of ABC is 1 AC × BC = 2 = 1 2 4 + 6 2 62 + 4 2 2 1 (52) = 26. 2 P.1 Pop Quiz 1. The distance is √ 16 + 4 = √ √ 20 = 2 5. Algebraic Prerequisites For Thought 1 True, since the number of gallons purchased is 20 divided by the price per gallon. 2. False, since a student’s exam grade is a function of the student’s preparation. If two classmates had the same IQ and only one prepared then the one who prepared will most likely achieve a higher grade. 3. False, since {(1, 2), (1, 3)} is not a function. 4. True 5. True 6. True 7. False, the domain is the set of all real numbers. 8. True 9. True, since f (0) = 0−2 = −1. 0+2 10. True, since if a − 5 = 0 then a = 5. P.2 Exercises 1. function 2. independent, dependent 2. Center (3, −5), radius 9 3. domain, range 3. Completing the square, we find 4. parabola (x2 + 4x + 4) + (y 2 − 10y + 25) = −28 + 4 + 25 5. function (x + 2)2 + (y − 5)2 = 1. 6. function The center is (−2, 5) and the radius is 1. b . Then a 7. Note, b = 2πa is equivalent to a = 2π 4. The distance between (3, 4) and the origin is is a function of b, and b is a function of a. 5, which is the radius. The circle is given by (x − 3)2 + (y − 4)2 = 25. 5. By setting x = 0 and y = 0 in 2x − 3y = 12 we find −3y = 12 and 2x = 12, respectively. Since y = −4 and x = 6 are the solutions of the two equations, the intercepts are (0, −4) and (6, 0). 6. (5, −1) b − 10 . 2 Then a is a function of b, and b is a function of a. 8. Note, b = 2(5 + a) is equivalent to a = 9. a is a function of b since a given denomination has a unique length. Since a dollar bill and a five-dollar bill have the same length, then b is not a function of a. c 2020 Pearson Education, Inc. Copyright P.2 Functions 9 10. Since different U.S. coins have different diameters, then a is a function of b, and b is a function of a. 11. Since an item has only one price, b is a function of a. Since two items may have the same price, a is not a function of b. √ 24. d = s 2 27. A = P 2 /16 y 2 1 b is not a function of a since it is possible for two adults with the same age to have different shoe sizes. 15. Since 1 in ≈ 2.54 cm, a is a function of b and b is a function of a. 16. Since there is only one cost for mailing a first class letter, then a is a function of b. Since two letters with different weights each under 1/2-ounce cost 47 cents to mail first class, b is not a function of a. √ 17. Since b = a3 and a = 3 b, we get that b is a function of a, and a is a function of b. √ 18. Since b = a4 and a = ± 4 b, we get that b is a function of a, but a is not a function of b. 19. Since b = |a|, we get b is a function of a. Since a = ±b, we find a is not a function of b. √ 20. Note, b = a since a ≥ 0, and a = b2 . Thus, b is a function of a, and a is a function of b. √ √ 2d 2 22. s = A 23. s = 21. A = s 2 x 1 3 30. y = −x + 3 has domain (−∞, ∞) and range (−∞, ∞), some points are (0, 3) and (3, 0) b is not a function of a since it is possible that two different students can spend the same time studying but obtain different final exam scores. 14. a is not a function of b since it is possible that two adult males can have the same shoe size but have different ages. s = P/4 29. y = 2x − 1 has domain (−∞, ∞) and range (−∞, ∞), some points are (0, −1) and (1, 1) 12. a is not a function of b since there may be two students with the same semester grades but different final exams scores. b is not a function of a since there may be identical final exam scores with different semester grades. 13. a is not a function of b since it is possible that two different students can obtain the same final exam score but the times spent on studying are different. P = 4s 26. √ 28. d = 2A 25. y 3 2 x 3 2 31. y = 5 has domain (−∞, ∞) and range {5}, some points are (0, 5) and (1, 5) y 6 4 2 2 x 32. y = −4 has domain (−∞, ∞) and range {−4}, some points are (0, −4) and (1, −4) y 3 3 3 5 c 2020 Pearson Education, Inc. Copyright x 10 Chapter P 33. y = x2 − 20 has domain (−∞, ∞) and range [−20, ∞), some points are (0, −20) and (6, 16) Algebraic Prerequisites 37. y = x3 has domain (−∞, ∞) and range (−∞, ∞), some points are (0, 0) and (2, 8) y y 8 30 10 6 6 1 x 30 2 x 8 34. y = x2 + 50 has domain (−∞, ∞) and range [50, ∞), some points are (0, 50) and (5, 75) 38. y = −x3 has domain (−∞, ∞) and range (−∞, ∞), some points are (0, 0) and (0, −8) y 150 y 8 100 1 x 10 5 2 x 8 35. y = 40 − x2 has domain (−∞, ∞) and range (−∞, 40], some points are (0, 40) and (6, 4) √ 39. y = x − 10 has domain [10, ∞) and range [0, ∞), some points are (10, 0) and (14, 2) y y 50 4 30 6 6 10 x x 40 10 2 36. y = −10 − x2 has domain (−∞, ∞) and range (−∞, −10], some points are (0, −10) and (4, −26) √ 40. y = x + 30 has domain [−30, ∞) and range [0, ∞), some points are (−30, 0) and (−26, 2) y 6 y 10 4 4 20 x x 10 30 30 2 c 2020 Pearson Education, Inc. Copyright P.2 Functions 11 √ 41. y = x + 30 has domain [0, ∞) and range [30, ∞), some points are (0, 30) and (400, 50) 45. y = |x − 20| has domain (−∞, ∞) and range [0, ∞), some points are (0, 20) and (20, 0) y y 40 60 20 30 10 x 50 20 x 600 200 √ 42. y = x − 50 has domain [0, ∞) and range [−50, ∞), some points are (0, −50) and (900, −20) 46. y = |x + 30| has domain (−∞, ∞) and range [0, ∞), some points are (0, 30) and (−30, 0) y y 10 30 x 2000 1000 10 60 x 30 50 43. y = |x| − 40 has domain (−∞, ∞) and range [−40, ∞), some points are (0, −40) and (40, 0) y 10 48. 3(16) + 4 = 52 49. −4 − 2 = −6 50. −8 − 2 = −10 51. |8| = 8 | − 1| = 1 52. 53. 4 + (−6) = −2 x 60 20 47. 3 · 4 − 2 = 10 55. 80 − 2 = 78 57. 3a2 − a 58. 54. 56. 24 · 6 = 144 2/2 = 1 4b − 2 59. f (−x) = 3(−x)2 − (−x) = 3×2 + x 40 60. g(−x) = 4(−x) − 2 = −4x − 2 44. y = 2|x| has domain (−∞, ∞) and range [0, ∞), some points are (0, 0) and (1, 2) 61. Factoring, we get x(3x − 1) + 0. So x = 0, 1/3. 62. Since 4x − 2 = 3, we get x = 5/4. y 4 63. Since |a + 3| = 4 is equivalent to a + 3 = 4 or a + 3 = −4, we have a = 1, −7. 2 64. Since 3t2 − t − 10 = (t − 2)(3t + 5) = 0, we find t = 2, −5/3. 1 1 2 x 65. C = 353n 66. P = 580n c 2020 Pearson Education, Inc. Copyright 12 Chapter P 67. C = 35n + 50 Algebraic Prerequisites 73. Solving for B, 68. C = 2.50 + 0.50n D = 69. We find 4D πS · N 4B 4(12 + 11/12) √ C= √ = ≈ 1.822 3 3 22, 800 D 4B and a sketch of the graph of C = √ 3 22, 800 is given below. C π 2 B ·S·N 4 = B2 B = 2 D . πS · N 74. Solving for V , 3 CR = 1 + 2 1 5 B 20 10 CR − 1 = πB 2 · S 4V 1 CR − 1 = 4V πB 2 · S V = πB 2 · S . 4(CR − 1) 70. Solving for B, we get √ 3 4B 22, 800 < 2 √ 3 22, 800 B < 2 B < 14 ft, 2 in. Then the maximum displacement is 14 ft, 2 in. With D fixed, we get that C becomes larger as the beam B becomes larger. Thus, a boat is more likely to capsize as the beam gets larger. 71. Let N = 2, B = 3.498, and S = 4.250. Then π 2 D = B ·S·N 4 = 81.686 in. 3 Then D ≈ 81.7 in.3 . 75. Pythagorean, legs, hypotenuse 76. circle, radius, center 77 . (2 + 3)2 + (−4 + 6)2 = 78. 4 − 6 −8 + 16 , 2 2 Using the unrounded answer to Exercise 71, the difference in the displacement is √ 29 = (−1, 4) 79. If we replace x = 0 in 4x − 6y = 40, then −6y = 40 or y = −20/3. The y-intercept is (0, − 20 3 ). If we replace y = 0 in 4x − 6y = 40, then 4x = 40 or x = 10. The x-intercept is (10, 0). 80. The diagonal is 72. Let N = 2, B = 3.518, and S = 4.250. Then π 2 D = B ·S·N 4 = 82.622 in.3 πB 2 · S 4V √ 32 + 72 = √ 58 ft. 81. Rewriting the equation, we find 1 1 1 x · 3100 · 4 · 32x = ·3 3 3 3 3 1 1 · 3100 · 4 · 32x = 3x−1 3 3 3 32x+93 = 3x−1 3 82.622 − 81.686 ≈ 0.94 in. . c 2020 Pearson Education, Inc. Copyright 2x + 93 = x − 1 x = −94. P.3 Families of Functions, Transformations, and Symmetry 82. First, 9 = (a + b)2 = (a2 + b2 ) + 2ab = 89 + 2ab. Then ab = −40. Thus, 13 P.3 Exercises 1. rigid a3 + b3 = (a + b)(a2 + b2 − ab) = 3(89 + 40) 2. nonrigid = 387. 3. reflection 4. upward translation, downward translation P.2 Pop Quiz 5. right, left 1. Yes, since r ≥ 0 and r = 6. stretching, shrinking A/π 2. Since A = s2 and s ≥ 0, we obtain s = √ A. 7. odd 3. No, since b = ±a. 8. even 4. [1, ∞) 9. transformation 5. [2, ∞) 10. family 6. f (3) = 3(3) + 6 = 15 11. f (x) = √ √ x, g(x) = − x y 7. We find 2 2a − 4 = 10 2a = 14 2 x 4 a = 7. -2 For Thought 12. f (x) = x2 + 1, g(x) = −x2 − 1 y 1. False, it is a reflection in the y-axis. 4 2. False, the graph of y = x2 − 4 is shifted down 4 units from the graph of y = x2 . -2 3. False, rather it is a left translation. x 2 -4 4. True 5. True 13. y = x, y = −x y 6. False, the down shift should come after the reflection. 4 7. True -4 4 8. False, since the domains are different. -4 9. True 10. True, since f (−x) = −f (x) where f (x) = x3 . c 2020 Pearson Education, Inc. Copyright x 14 14. y = Chapter P √ √ 4 − x2 , y = − 4 − x2 Algebraic Prerequisites 19. y = x2 , y = (x − 3)2 y y 3 1 1 3 x 4 1 15. f (x) = |x|, g(x) = |x| − 4 y -2 x 3 20. y = |x|, y = |x + 2| 4 y -4 4 x 8 -4 16. f (x) = √ x, g(x) = √ 4 x+3 y -2 8 x 21. f (x) = x3 , g(x) = (x + 1)3 3 y 8 1 1 x 4 17. f (x) = x, g(x) = x + 3 y 4 3 1 2 x 4 3 -4 4 8 x 22. f (x) = -4 √ x, g(x) = √ x−3 y 3 18. f (x) = x2 , g(x) = x2 − 5 y 2 4 -3 1 3 x 3 -5 6 1 c 2020 Pearson Education, Inc. Copyright x 12 P.3 Families of Functions, Transformations, and Symmetry 23. y = √ √ x, y = 3 x 15 41. y = y √ x − 1 + 2; right by 1, up by 2 y 4 2 3 2 1 x 4 1 1 24. y = |x|, y = |x| 3 3 5 x 2 y 42. y = √ x + 5 − 4; left by 5, down by 4 y 2 4 1 -3 3 x 5 10 15 x 1 1 25. y = x2 , y = x2 4 y 4 43. y = |x − 1| + 3; right by 1, up by 3 4 y 1 -2 2 x 5 26. y = x2 , y = 4 − x2 3 y 10 1 6 1 x 44. y = |x + 3| − 4; left by 3, down by 4 y 2 3 3 1 3 x 2 6 27. g 28. h 29. b 30. d 31. c 32. a 33. f 34. e x 1 -3 -4 45. y = 3x − 40 y 35. y = (x − 10)2 + 4 √ 36. y = x + 5 − 12 40 37. y = −3|x − 7| + 9 38. y = −2(x + 6) − 8 or y = −2x − 20 √ √ 39. y = −(3 x + 5) or y = −3 x − 5 20 -40 40. y = − (x − 13)2 − 6 or y = −(x − 13)2 + 6 c 2020 Pearson Education, Inc. Copyright 40 x 16 Chapter P Algebraic Prerequisites 1 51. y = − |x + 4|, left by 4, 2 reflect about x-axis, shrink by 1/2 46. y = −4x + 200 y y 200 100 1 x 400 -3 -5 -1 1 x -3 -400 1 47. y = x − 20 2 y 52. y = 3|x − 2|, right by 2, stretch by 3 y 20 -40 x 40 -20 6 3 1 48. y = − x + 40 2 y 1 80 x 2 53. y = −(x − 3)2 + 1; right by 3, reflect about x-axis, up by 1 40 -80 x 80 y 1 2 3 8 x 1 49. y = − |x| + 40, shrink by 1/2, 2 reflect about x-axis, up by 40 y 8 50 -80 80 x 54. y = −(x + 2)2 − 4; left by 2, reflect about x-axis, down by 4 -40 y 2 50. y = 3|x| − 200, stretch by 3, down by 200 6 2 2 y 600 4 8 -100 100 x -200 c 2020 Pearson Education, Inc. Copyright x P.3 Families of Functions, Transformations, and Symmetry 55. y = −2(x + 3)2 − 4; left by 3, stretch by 2, reflect about x-axis, down by 4 17 62. Symmetric about the origin, odd function since f (−x) = −f (x) y 6 3 2 63. Neither symmetry, neither even nor odd since f (−x) = f (x) and f (−x) = −f (x) x 4 64. Neither symmetry, neither even nor odd since f (−x) = f (x) and f (−x) = −f (x) 65. No symmetry, not an even or odd function since f (−x) = −f (x) and f (−x) = −f (x) 24 66. Symmetric about the y-axis, even function since f (−x) = f (x) 56. y = 3(x + 1)2 − 5; left by 1, stretch by 3, down by 5 67. Symmetric about the origin, odd function since f (−x) = −f (x) y 2 3 x 2 68. Symmetric about the origin, odd function since f (−x) = −f (x) 5 69. No symmetry, not an even or odd function since f (−x) = f (x) and f (−x) = −f (x) 1 1 √ 57. y = −2 x + 3 + 2, left by 3, stretch by 2, reflect about x-axis, up by 2 y 2 -3 6 71. No symmetry, not an even or odd function since f (−x) = f (x) and f (−x) = −f (x) 72. Symmetric about the y-axis, even function since f (−x) = f (x) x -4 1√ 58. y = − x + 2 + 4, left by 2, shrink by 1/2, 2 reflect about x-axis, up by 4 y 73. Neither symmetry, not an even or odd function since f (−x) = f (x) and f (−x) = −f (x) 74. Symmetric about the y-axis, even function since f (−x) = f (x) 75. Symmetric about the y-axis, even function since f (−x) = f (x) 4 2 -2 70. No symmetry, not an even or odd function since f (−x) = f (x) and f (−x) = −f (x) 2 76. Symmetric about the y-axis, even function since f (−x) = f (x) x 59. Symmetric about y-axis, even function since f (−x) = f (x) 60. Symmetric about y-axis, even function since f (−x) = f (x) 61. No symmetry, neither even nor odd since f (−x) = f (x) and f (−x) = −f (x) since f (−x) = f (x) since f (−x) = f (x) 77. e 78. a 79. g 80. h 81. b 82. d 83. c 84. f 85. N (x) = x + 2000 c 2020 Pearson Education, Inc. Copyright 18 Chapter P 86. N (x) = 1.05x+3000. Yes, if the merit increase is followed by the cost of living raise then the new salary becomes higher and is N (x) = 1.05(x + 3000) = 1.05x + 3150. (d) The second graph is obtained by translating the first one to the right by 2 units and 3 units up. y 87. If inflation rate is less than 50%, then √ 1 1 √ 1 − x < . This simplifies to < x. After 2 2 1 squaring we have 25%. 4 3 -2 88. If production is at least 28 windows, √ then 1.75 x ≥ 28. They need at least x= 28 1.75 2 Algebraic Prerequisites x 2 -8 90. The graph of y = x3 + 6×2 + 12x + 8 or equivalently y = (x + 2)3 can be obtained by shifting the graph of y = x3 to the left by 2 units. = 256 hrs. 89. y (a) Both functions are symmetric about the y-axis, and the graphs are identical. 8 y 4 -2 2 x -4 -2 x 2 -1 (b) One graph is a reflection of the other about the y-axis, and both are symmetric about the y-axis. y 91. x2 + y 2 = 1 92. y = 5 93. x = 4 1 -2 2 x 94. If A is the area and s is the length of a side, then A = s2 . 95. f (6) = 2(36) − 3(6) = 72 − 18 = 54 -1 f (−x) = 2(−x)2 − 3(−x) = 2×2 + 3x (c) The second graph is obtained by translating the first one to the left by 1 unit. 96. Solve for a: y 2a2 + 1 = 9 2a2 = 8 a2 = 4 2 -2 a = ±2. 2 x -1 c 2020 Pearson Education, Inc. Copyright P.4 Compositions and Inverses 19 97. Draw the numbers in {1, 2, 3, …, 999} that can be written in the form (2n + 1) · 2 2m for n, m ≥ 0. These means that you will have drawn 666 numbers. Then the 667th number you will draw has the form (2n + 1) · 22m+1 which must be twice one of the first 666 numbers. x2 − 1 3 =− . 3 − x2 3 √ Solving, x2 = 2 + 10. Since the distance AB is the same as the distance between (1, −1) and (3, 5), we obtain √ √ 40 = 2 10 x2 − x1 = √ x1 = x2 − 2 10 √ √ x1 = (2 + 10) − 2 10 √ x1 = 2 − 10. √ Then the other vertices are (2 ± 10, 2). 2. False, rather y = (x − 1)2 = x2 − 2x + 1. √ 3. False, rather (f ◦ g)(x) = x − 2. 4. True 5. False, since (h ◦ g) (x) = x2 − 9. 1 1 6. False, rather f −1 (x) = x − . 2 2 8. False; g −1 does not exist since the graph of g which is a parabola fails the horizontal line test. 9. False, since y = x2 is a function which does not have an inverse function. 10. True P.4 Exercises 1. composition 2. one-to-one 3. invertible 4. inverse 5. switch-and-solve 6. symmetric 7. y = 2(3x + 1) − 3 = 6x − 1 P.3 Pop Quiz 1. y = 1 True, since A = P 2 /16. 7. True 98. Let A(x1 , 2) and B(x2 , 2) be the other two opposite vertices where x1 b > 0 and n > 0, then 6. Interchange x and y then solve for y. 3 b b a a + > + . n n+1 n n+1 y + 1 = (x + 4)3 y = (x + 4)3 − 1 Thus, the arrangement with the largest sum is 2025 2024 2 1 + + … + + . 1 2 2024 2025 94. If f (x) = mx + b, then (f ◦ f ◦ f )(x) = m3 x + b(m2 + m + 1). Since (f ◦ f ◦ f )(x) = 27x + 26 we obtain m3 = 27 or m = 3. Also, b(m2 + m + 1) = 26. If we substitute m = 3, then 13b = 26 or m = 2. Thus, f (x) = 3x + 2. The y-intercept is (0, 2). y+1−4 = x g −1 (x) = (x + 4)3 − 1 Chapter P Review Exercises √ 49 · 2 = 7 2 √ √ 2. 100 · 2 = 10 2 √ √ 5 3 3 5 3. √ · √ = 5 5 5 √ √ 13 4 4 13 4. √ · √ = 13 13 13 √ √ √ 5 6 30 5. √ · √ = 6 6 6 √ √ √ 3 2 6 6. √ · √ = 4 2 2 2 √ √ √ 2 8 8 2 =4 2 7. √ · √ = 2 2 2 √ √ √ 9 3 9 3 8. √ · √ = =3 3 3 3 3 1. √ c 2020 Pearson Education, Inc. Copyright Chapter P Review Exercises 25 (−3 − 2)2 + (5 − (−6))2 = √ √ (−5)2 + 112 = 25 + 121 = 146. −3 + 2 5 − 6 The midpoint is = , 2 2 1 1 − ,− . 2 2 9. The distance is 15. Circle with center at (0, 1) and radius 1 y 3 1 10. The distance is (−1 − (−2))2 + (1 − (−3))2 = √ √ 12 + 42 = 17. The midpoint is 3 −1 − 2 1 − 3 = − , −1 . , 2 2 2 π− 11. The distance is π 2 π 2 2 + (1 − 1)2 = 1 16. Circle with center at (2, −1) and radius 1 y 1 π . The midpoint is 2 2 π/2 + π 1 + 1 3π = , ,1 . 2 2 4 +0= 12. The distance is 2 π π − 2 3 2 x 1 1 2 3 x 2 + (2 − 2)2 = π π + 0 = . The midpoint is 6 6 π/2 + π/3 2 + 2 5π = , ,2 . 2 2 12 17. The line through the points (10, 0) and (0, −4). y 2 5 10 x 13. Circle with center at (0, 0) and radius 3 4 y 4 2 2 4 18. The line through the points (25, 0) and (0, −50). x y 15 14. Circle with center at (0, 0) and radius 5 25 y 6 50 4 4 6 x c 2020 Pearson Education, Inc. Copyright 25 x 26 Chapter P 19. The vertical line through the point (5, 0). y Algebraic Prerequisites 23. Domain (−∞, ∞) and range {4}. The horizontal line through the point (0, 4). y 3 6 2 4 6 x 2 3 4 4 x 20. The horizontal line through the point (0, 3). 24. Domain (−∞, ∞) and range {−2} y 4 y 2 2 x 2 1 3 3 x 3 21. Domain (−∞, ∞) and range (−∞, ∞) 25. Domain (−∞, ∞) and range [−3, ∞) y 2 y 2 4 x 3 4 1 22. Domain (−∞, ∞) and range (−∞, ∞) x 2 4 26. Domain (−∞, ∞) and range (−∞, 6] y y 7 4 5 2 3 2 4 x 2 c 2020 Pearson Education, Inc. Copyright 3 x