Algebra and Trigonometry, 9th Edition Solution Manual

NOT FOR SALE C H A P T E R 2 Functions and Their Graphs Section 2.1 Linear Equations in Two Variables ……………………………………………165 Section 2.2 Functions…………………………………………………………………………………178 Section 2.3 Analyzing Graphs of Functions …………………………………………………186 Section 2.4 A Library of Parent Functions …………………………………………………..197 Section 2.5 Transformations of Functions ……………………………………………………201 Section 2.6 Combinations of Functions: Composite Functions………………………212 Section 2.7 Inverse Functions……………………………………………………………………..221 Review Exercises …………………………………………………………………………………………234 Problem Solving …………………………………………………………………………………………243 Practice Test ……………………………………………………………………………………………….248 INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE C H A P T E R 2 Functions and Their Graphs Section 2.1 Linear Equations in Two Variables 14. The line appears to go through (0, 7) and (7, 0). 1. linear 2. slope Slope = 3. point-slope y2 − y1 0 −7 = = −1 x2 − x1 7 −0 15. y = 5 x + 3 4. parallel Slope: m = 5 5. perpendicular y-intercept: (0, 3) 6. rate or rate of change 7. linear extrapolation 8. general 9. (a) m = 23 . Because the slope is positive, the line rises. Matches L2 . (b) m is undefined. The line is vertical. Matches L3. (c) m = −2. The line falls. Matches L1. 10. (a) m = 0. The line is horizontal. Matches L2 . (b) 16. Slope: m = −1 y -intercept: (0, −10) m = − 34 . Because the slope is negative, the line falls. Matches L1. (c) m = 1. Because the slope is positive, the line rises. Matches L3 . 11. 17. y = − 12 x + 4 Slope: m = − 12 y-intercept: (0, 4) 12. 13. Two points on the line: (0, 0) and ( 4, 6) y2 − y1 6 3 = = x2 − x1 4 2 INSTRUCTOR USE ONLY Slope p = © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 165 166 NOT FOR SALE Chapter 2 Functions ctions and Their Graphs 18. Slope: m = 3 2 y -intercept: (0, 6) 22. 3 y + 5 = 0 3 y = −5 y = − 53 Slope: m = 0 ( y-intercept: 0, − 53 ) 19. y − 3 = 0 y = 3, horizontal line Slope: m = 0 y-intercept: (0, 3) 23. 7 x − 6 y = 30 − 6 y = − 7 x + 30 y = 76 x − 5 Slope: m = 76 y -intercept: ( 0, − 5) 20. x + 5 = 0 x = −5 Slope: undefined (vertical line) No y-intercept 24. 2 x + 3 y = 9 3 y = −2 x + 9 y = − 23 x + 3 Slope: m = − 23 21. 5 x − 2 = 0 y-intercept: (0, 3) x = 52 , vertical line Slope: undefined No y-intercept INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.1 Linear Equations in Tw Two Variables T 25. m = 0−9 −9 3 = = − 6−0 6 2 29. m = 26. m = 8 2 −8 − 0 = = 0 − 12 12 3 30. m = 27. m = 6 − ( −2) 1 − ( −3) = 8 = 2 4 31. m = −7 − ( −7) 8−5 = 167 0 = 0 3 −5 − 1 = 3 − 4 − ( −2) 4 − ( −1) −6 − ( −6) = 5 0 m is undefined. 28. m = −4 − 4 = −4 4−2 32. m = 0 − ( −10) −4 − 0 = − 5 2 INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 168 Chapter 2 33. m = NOT FOR SALE Functions ctions and Their Graphs 41. Point: ( −1, − 6), Slope: m = − 12 1.6 − 3.1 −1.5 = = 0.15 −5.2 − 4.8 −10 Because m = − 12 , y decreases by 1 unit for every two unit increase in x. Three additional points are (1, − 7), (3, − 8), and (−13, 0). 42. Point: (7, − 2), Slope: m = 12 Because m = 12 , y increases by 1 unit for every two − 34. m = unit increase in x. Three additional points are (9, −1), 1 § 4· − ¨− ¸ 1 3 © 3¹ = − 3 11 7 − − 2 2 (11, 0), and (13, 1). 43. Point: (0, − 2); m = 3 y + 2 = 3( x − 0) y = 3x − 2 35. Point: ( 2, 1), Slope: m = 0 Because m = 0, y does not change. Three points are (0, 1), (3, 1), and (−1, 1). 44. Point: (0, 10); m = −1 36. Point: (3, − 2), Slope: m = 0 Because m = 0, y does not change. Three other points are ( − 4, − 2), (0, − 2), and (5, − 2). y − 10 = −1( x − 0) y − 10 = − x y = − x + 10 37. Point: ( − 8, 1), Slope is undefined. Because m is undefined, x does not change. Three points are ( − 8, 0), ( − 8, 2), and ( −8, 3). 38. Point: (1, 5), Slope is undefined. Because m is undefined, x does not change. Three other points are (1, − 3), (1, 1), and (1, 7). 45. Point: ( −3, 6); m = −2 39. Point: ( − 5, 4), Slope: m = 2 Because m = 2 = 12 , y increases by 2 for every one unit increase in x. Three additional points are ( − 4, 6), y − 6 = −2( x + 3) y = −2 x (−3, 8), and (−2, 10). 40. Point: (0, − 9), Slope: m = −2 Because m = −2, y decreases by 2 for every one unit increase in x. Three other points are ( − 2, − 5), INSTRUCTOR USE ONLY (1, −11), and (3, −15). © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. Section 2.1 Linear Equations in Two Tw T Variables 46. Point: (0, 0); m = 4 50. Point: ( −2, − 5); m = 34 y − 0 = 4( x − 0) y + 5 = 34 ( x + 2) y = 4x 169 4 y + 20 = 3x + 6 4 y = 3x − 14 y = 34 x − 72 51. Point: (6, −1); m is undefined. Because the slope is undefined, the line is a vertical line. x = 6 47. Point: ( 4, 0); m = − 13 y − 0 = − 13 ( x − 4) y = − 13 x + 43 52. Point: ( −10, 4); m is undefined. Because the slope is undefined, the line is a vertical line. x = −10 48. Point: (8, 2); m = 14 y − 2 = 14 ( x − 8) y − 2 = 14 x − 2 y = 14 x ( ) 53. Point: 4, 52 ; m = 0 y − 52 = 0( x − 4) y − 52 = 0 y = 52 49. Point: ( 2, − 3); m = − 1 2 1 ( x − 2) 2 1 y +3 = − x +1 2 1 y = − x − 2 2 y − ( −3) = − INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 170 Chapter 2 NOT FOR SALE Functions ctions and Their Graphs 58. ( −1, 4), (6, 4) 54. Point: ( −5.1, 1.8); m = 5 y − 1.8 = 5( x − ( −5.1)) y − 4 = y = 5 x + 27.3 4− 4 ( x + 1) 6 − ( −1) y − 4 = 0( x + 1) y − 4 = 0 y = 4 § 1· §1 5· 59. ¨ 2, ¸, ¨ , ¸ © 2¹ © 2 4¹ 5 1 − 1 2 ( x − 2) y − = 4 1 2 − 2 2 1 1 y = − ( x − 2) + 2 2 1 3 y = − x + 2 2 55. (5, −1), ( −5, 5) 5+1 ( x − 5) −5 − 5 3 y = − ( x − 5) − 1 5 3 y = − x + 2 5 y +1 = 2· § 60. (1, 1), ¨ 6, − ¸ 3¹ © 2 −1 y −1 = 3 ( x − 1) 6 −1 1 y − 1 = − ( x − 1) 3 1 1 y −1 = − x + 3 3 1 4 y = − x + 3 3 56. ( 4, 3), ( −4, − 4) − −4 − 3 ( x − 4) −4 − 4 7 y − 3 = ( x − 4) 8 7 7 y −3 = x − 8 2 7 1 y = x − 8 2 y −3 = 61. (1, 0.6), ( −2, − 0.6) 57. ( −8, 1), ( −8, 7) Because both points have x = −8, the slope is undefined, and the line is vertical. x = −8 −0.6 − 0.6 ( x − 1) −2 − 1 y = 0.4( x − 1) + 0.6 y − 0.6 = y = 0.4 x + 0.2 INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.1 62. ( −8, 0.6), ( 2, − 2.4) y − 0.6 = m1 = 12 L2 : y = − 12 x + 1 3 y − 0.6 = − ( x + 8) 10 10 y − 6 = −3( x + 8) m2 = − 12 The lines are neither parallel nor perpendicular. 10 y − 6 = −3 x − 24 68. L1 : y = − 54 x − 5 10 y = −3 x − 18 3 9 x − 10 5 or y = −0.3x − 1.8 m2 = 54 The lines are perpendicular. −1 − ( −1) y +1 = ( x − 2) 1 − 2 3 y +1= 0 69. L1 : (0, −1), (5, 9) m1 = y = −1 9+1 = 2 5−0 L2 : (0, 3), ( 4, 1) The line is horizontal. m2 = §7 · §7 · 64. ¨ , − 8¸, ¨ , 1¸ ©3 ¹ ©3 ¹ 1−3 1 = − 4−0 2 The lines are perpendicular. 1 − ( −8) 9 and is undefined. = 7 7 0 − 3 3 7 x = 3 The line is vertical. m1 = − 54 L2 : y = 54 x + 1 §1 · 63. ( 2, −1), ¨ , −1¸ ©3 ¹ m = 171 67. L1 : y = 12 x − 3 −2.4 − 0.6 ( x + 8) 2 − ( −8) y = − Linear Equations in Two Tw T Variables 70. L1 : ( −2, −1), (1, 5) m1 = 5 − ( −1) 6 = = 2 1 − ( −2) 3 L2 : (1, 3), (5, − 5) m2 = −5 − 3 −8 = = −2 5 −1 4 The lines are neither parallel nor perpendicular. 71. L1 : (3, 6), ( −6, 0) m1 = 65. L1 : y = 13 x − 2 m1 = 13 L2 : y = 13 x + 3 0−6 2 = −6 − 3 3 § 7· L2 : (0, −1), ¨ 5, ¸ © 3¹ 7 +1 2 m2 = 3 = 5−0 3 The lines are parallel. m2 = 13 The lines are parallel. 66. L1 : y = 4 x − 1 m1 = 4 L2 : y = 4 x + 7 m2 = 4 INSTRUCTOR USE ONLY The lines are parallel. © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 172 Chapter 2 NOT FOR SALE Functions ctions and Their Graphs 72. L1 : ( 4, 8), ( − 4, 2) 76. 5 x + 3 y = 0 3 y = −5 x 2−8 3 −6 = = 4 −4 − 4 −8 m1 = 1· § L2 : (3, − 5), ¨ −1, ¸ 3¹ © 1 16 − ( −5) 4 3 m2 = = 3 = − 3 −1 − 3 −4 y = − 53 x Slope: m = − 53 ( ) (a) m = − 53 , 78 , 34 ( The lines are perpendicular. (2, 1), m = 2 24 y = − 40 x + 53 y = − 53 x + 53 24 ( ) (b) m = 53 , 87 , 43 y − 1 = 2( x − 2) ( y − 34 = 53 x − 78 y = 2x − 3 40 y = 24 x + 9 9 y = 53 x + 40 = − 12 x + 2 77. y + 3 = 0 74. x + y = 7 y = −3 y = −x + 7 Slope: m = −1 Slope: m = 0 (a) m = −1, ( −3, 2) (a) y − 2 = −1( x + 3) y − 2 = −x − 3 (b) m = 1, ( −3, 2) y − 2 = 1( x + 3) y = x +5 (b) ( −1, 0), m is undefined. x = −1 78. x − 4 = 0 x = 4 Slope: m is undefined. (a) 75. 3x + 4 y = 7 y = − 34 x + 74 Slope: m = − 34 (−1, 0), m = 0 y = 0 y = −x − 1 ( ) 40 y − 30 = 24 x − 21 y − 1 = − 12 ( x − 2) (a) ( ) 40 y − 30 = 24 x − 78 (b) ( 2, 1), m = − 12 y ) 24 y − 18 = − 40 x + 35 = 2 x − 32 Slope: m = 2 (a) ( 24 y − 18 = − 40 x − 78 73. 4 x − 2 y = 3 y ) y − 34 = − 53 x − 78 (3, − 2), m is undefined. x = 3 (b) (3, − 2), m = 0 y = −2 ) − 23 , 78 , m = − 34 ( ( )) y − 78 = − 34 x − − 23 y = − 34 x + 83 (b) (− 23 , 78 ), m = 43 ( ( )) y − 78 = 43 x − − 23 y = 43 x + 127 72 INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.1 79. x − y = 4 y = x − 4 x y + =1 2 3 −2 (2.5, 6.8), m = 1 3x y − =1 2 2 3x − y − 2 = 0 y − 6.8 = 1( x − 2.5) y = x + 4.3 (b) ( 2.5, 6.8), m = −1 y − 6.8 = ( −1)( x − 2.5) y = − x + 9.3 2 y = −6 x + 9 y = −3x + 92 Slope: m = −3 (−3.9, −1.4), m = −3 x y + = 1, c ≠ 0 c c x + y = c 3 = c x + y = 3 x + y −3 = 0 86. ( d , 0), (0, d ), ( −3, 4) y + 1.4 = −3 x − 11.7 x y + =1 d d x + y = d y = −3x − 13.1 −3 + 4 = d y − ( −1.4) = −3( x − ( −3.9)) (b) ( −3.9, −1.4), m = 13 y − ( −1.4) = 13 ( x − ( −3.9)) y + 1.4 = 13 x + 1.3 y = 13 x − 0.1 x y 81. + =1 2 3 3x + 2 y − 6 = 0 82. ( −3, 0), (0, 4) x y + =1 −3 4 x y (−12) + (−12) = ( −12) ⋅ 1 −3 4 4 x − 3 y + 12 = 0 83. 85. 1+ 2 = c 80. 6 x + 2 y = 9 (a) 173 §2 · 84. ¨ , 0 ¸, (0, − 2) ©3 ¹ Slope: m = 1 (a) Linear Equations in Two Tw T Variables 1= d x + y =1 x + y −1 = 0 87. (a) m = 135. The sales are increasing 135 units per year. (b) m = 0. There is no change in sales during the year. (c) m = − 40. The sales are decreasing 40 units per year. 88. (a) greatest increase = largest slope (9, 36.54), (10, 65.23) m1 = So, the sales increased the greatest between the years 2009 and 2010. least increase = smallest slope (8, 32.48), (9, 36.54) x y + =1 −1 6 −2 3 3 6 x + y = −1 2 12 x + 3 y + 2 = 0 65.23 − 36.54 = 28.69 10 − 9 m2 = 36.54 − 32.48 = 4.06 9 −8 So, the sales increased the least between the years 2008 and 2009. (b) ( 4, 8.28), (10, 65.23) m = 65.23 − 8.28 56.95 = ≈ 9.49 10 − 4 6 The slope of the line is about 9.49. INSTRUCTOR USE ONLY (c) The sales increased $9.49 billion each year between the he years 2004 and 2010. © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 174 Chapter 2 NOT FOR SALE Functions ctions and Their Graphs 6 89. y = 100 x 6 y = 100 (200) = 12 feet 90. (a) and (b) x 300 600 900 1200 1500 1800 2100 y –25 –50 –75 –100 –125 –150 –175 (c) m = −50 − ( −25) 600 − 300 = 1 −25 = − 300 12 1 ( x − 600) 12 1 y + 50 = − x + 50 12 1 y = − x 12 y − ( −50) = − 1 , for every change in the horizontal measurement of 12 feet, the vertical 12 measurement decreases by 1 foot. (d) Because m = − (e) 1 ≈ 0.083 = 8.3% grade 12 91. (10, 2540), m = −125 95. Using the points (0, 875) and (5, 0), where the first coordinate represents the year t and the second coordinate represents the value V, you have V – 2540 = −125(t − 10) V − 2540 = −125t + 1250 V = −125t + 3790, 5 ≤ t ≤ 10 92. (10, 156), m = 4.50 96. Using the points (0, 24,000) and (10, 2000), where the V – 156 = 4.50(t − 10) V − 156 = 4.50t − 45 V = 4.5t + 111, 5 ≤ t ≤ 10 93. The C-intercept measures the fixed costs of manufacturing when zero bags are produced. The slope measures the cost to produce one laptop bag. 94. W = 0.07 S + 2500 0 − 875 = −175 5−0 V = −175t + 875, 0 ≤ t ≤ 5. m = first coordinate represents the year t and the second coordinate represents the value V, you have m = 2,000 − 24,000 − 22,000 = = − 2200. 10 − 0 10 Since the point (0, 24,000) is the V -intercept, b = 24,000, the equation is V = − 2200t + 24,000, 0 ≤ t ≤ 10. INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.1 97. Using the points (0, 32) and (100, 212), where the first coordinate represents a temperature in degrees Celsius and the second coordinate represents a temperature in degrees Fahrenheit, you have m = Linear Equations in Two Tw T Variables The equipment must be used 1750 hours to yield a profit of 0 dollars. 100. (a) 212 − 32 180 9 = = . 100 − 0 100 5 10 m Since the point (0, 32) is the F- intercept, b = 32, the x 15 m 9 equation is F = C + 32. 5 x (b) y = 2(15 + 2 x) + 2(10 + 2 x) = 8 x + 50 98. (a) Using the points (1, 970) and (3, 1270), you have m = 175 (c) 1270 − 970 300 = = 150. 3−1 2 Using the point-slope form with m = 150 and the point (1, 970), you have (d) Because m = 8, each 1-meter increase in x will increase y by 8 meters. y − y1 = m(t − t1 ) y − 970 = 150(t − 1) 101. False. The slope with the greatest magnitude corresponds to the steepest line. y − 970 = 150t − 150 y = 150t + 820. (b) The slope is m = 150. The slope tells you the amount of increase in the weight of average male child’s brain each year. (c) Let t = 2: y = 150( 2) + 820 y = 300 + 820 y = 1120 The average brain weight at age 2 is 1120 grams. 102. False. The lines are not parallel. (−8, 2) and (−1, 4): m1 = 4− 2 2 = −1 − ( −8) 7 (0, − 4) and (−7, 7): m2 = 7 − ( −4) 11 = −7 − 0 −7 103. Find the slope of the line segments between the points A and B, and B and C. (d) Answers will vary. (e) Answers will vary. Sample Answer: No. The brain stops growing after reaching a certain age. 99. (a) Total Cost = cost for cost fuel and + for maintainance purchase + cost operator C = 9.5t + 11.5t + 42,000 C = 21.0t + 42,000 m AB = 7 −5 2 1 = = 3 − ( −1) 4 2 mBC = −4 3−7 = = −2 5−3 2 (b) Revenue = Rate per hour ⋅ Hours R = 45t (c) P = R − C P = 45t − ( 21t + 42,000) P = 24t − 42,000 (d) Let P = 0, and solve for t. 0 = 24t − 42,000 42,000 = 24t 1750 = t Since the slopes are negative reciprocals, the line segments are perpendicular and therefore intersect to form a right angle. So, the triangle is a right triangle. 104. On a vertical line, all the points have the same x-value, y − y1 , you would have so when you evaluate m = 2 x2 − x1 a zero in the denominator, and division by zero is undefined. INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 176 Chapter 2 NOT FOR SALE Functions ctions and Their Graphs 105. No. The slope cannot be determined without knowing the scale on the y-axis. The slopes will be the same if the scale on the y-axis of (a) is 2 12 and the scale on the y-axis of (b) is 1. Then the slope of both is 54 . 106. d1 = ( x2 − x1 )2 + ( y2 − y1 )2 ( x2 − x1 )2 + ( y2 − y1 )2 d2 = = (1 − 0)2 + ( m1 − 0)2 = (1 − 0)2 + ( m2 − 0)2 = 1 + (m 1 ) = 1 + (m 2 ) 2 2 Using the Pythagorean Theorem: (d1 ) + (d 2 ) 2 2 2 = (distance between (1, m1 ), and (1, m2 )) 2 § 1 + m 2· + § 1 + m 2· = § ( 1) ¸ ¨ ( 2) ¸ ¨ ¨ © ¹ © ¹ © (1 − 1) + (m2 − m1 ) ·¸ 2 2 2 ( m 1 ) + (m 2 ) + 2 = (m 2 ) − 2m1m2 + (m 1 ) 2 2 2 ¹ 1 + ( m 1 ) + 1 + ( m 2 ) = ( m2 − m1 ) 2 2 2 2 2 2 = −2m1m2 − 1 = m1 m2 107. No, the slopes of two perpendicular lines have opposite signs. (Assume that neither line is vertical or horizontal.) 108. Because −4 > 52 , the steeper line is the one with a slope of – 4. The slope with the greatest magnitude corresponds to the steepest line. 109. The line y = 4 x rises most quickly. 110. (a) Matches graph (ii). The slope is –20, which represents the decrease in the amount of the loan each week. The y-intercept is (0, 200), which represents the original amount of the loan. (b) Matches graph (iii). The slope is 2, which represents the increase in the hourly wage for each unit produced. The y-intercept is (0, 12.5), which represents the hourly rate if the employee produces no units. (c) Matches graph (i). The slope is 0.32, which represents the increase in travel cost for each mile driven. The y-intercept is (0, 32), which represents the fixed cost of $30 per The line y = −4 x falls most quickly. day for meals. This amount does not depend on the number of miles driven. (d) Matches graph (iv). The slope is –100, which represents the amount by which the computer depreciates each year. The yintercept is (0, 750), which represents the original purchase price. The greater the magnitude of the slope (the absolute value of the slope), the faster the line rises or falls. INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.1 Linear Equations in T Two Tw Variables 177 111. Set the distance between ( 4, −1) and ( x, y ) equal to the distance between ( −2, 3) and ( x, y ). ( x − 4) + ¬ª y − (−1)¼º 2 2 2 ¬ª x − ( −2)¼º + ( y − 3) = 2 ( x − 4)2 + ( y + 1)2 = ( x + 2)2 + ( y − 3)2 x 2 − 8 x + 16 + y 2 + 2 y + 1 = x 2 + 4 x + 4 + y 2 − 6 y + 9 −8 x + 2 y + 17 = 4 x − 6 y + 13 0 = 12 x − 8 y − 4 0 = 4(3 x − 2 y − 1) 0 = 3x − 2 y − 1 This line is the perpendicular bisector of the line segment connecting ( 4, −1) and ( −2, 3). 112. Set the distance between (6, 5) and ( x, y ) equal to the distance between (1, − 8) and ( x, y ). ( x − 1)2 + ( y − (−8)) ( x − 6)2 + ( y − 5)2 = ( x − 6) + ( y − 5) 2 = ( x − 1) + ( y + 8) 2 2 2 2 x 2 − 12 x + 36 + y 2 − 10 y + 25 = x 2 − 2 x + 1 + y 2 + 16 y + 64 x 2 + y 2 − 12 x − 10 y + 61 = x 2 + y 2 − 2 x + 16 y + 65 −12 x − 10 y + 61 = −2 x + 16 y + 65 −10 x − 26 y − 4 = 0 −2(5 x + 13 y + 2) = 0 5 x + 13 y + 2 = 0 ( ) 113. Set the distance between 3, 52 and ( x, y ) equal to the distance between ( −7, 1) and ( x, y ). ( x − 3) + ( y − 52 ) 2 2 ( x − 3)2 + ( y − 52 ) 2 2 ª¬ x − ( −7)¼º + ( y − 1) = = ( x + 7) + ( y − 1) 2 2 2 x 2 − 6 x + 9 + y 2 − 5 y + 25 = x 2 + 14 x + 49 + y 2 − 2 y + 1 4 −6 x − 5 y + 61 = 14 x − 2 y + 50 4 −24 x − 20 y + 61 = 56 x − 8 y + 200 80 x + 12 y + 139 = 0 ( ) This line is the perpendicular bisector of the line segment connecting 3, 52 and ( −7, 1). ( ) 114. Set the distance between − 12 , − 4 and ( x, y ) equal to the distance between ( x − (− )) + ( y − (−4)) = ( x − ) + ( y − ) 1 2 2 2 7 2 2 ( x + 12 ) + ( y + 4) = ( x − 72 ) + ( y − 54 ) 2 2 2 5 4 ( 72 , 54 ) and ( x, y). 2 2 25 x 2 + x + 14 + y 2 + 8 y + 16 = x 2 − 7 x + 49 + y 2 − 52 y + 16 4 x 2 + y 2 + x + 8 y + 65 = x 2 + y 2 − 7 x − 52 y + 221 4 16 x + 8 y + 65 = −7 x − 52 y + 221 4 16 39 8 x + 21 y + 16 = 0 2 128 x + 168 y + 39 = 0 INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 178 Chapter 2 NOT FOR SALE Functions ctions and Their Graphs Section 2.2 Functions 1. domain; range; function 16. y = x +5 Yes, y is a function of x. 2. independent; dependent 17. y = 4 − x 3. implied domain Yes, y is a function of x. 4. difference quotient 5. Yes, the relationship is a function. Each domain value is matched with exactly one range value. 6. No, the relationship is not a function. The domain value of –1 is matched with two output values. 18. y = 4− x Ÿ y = 4− x or y = −( 4 − x ) No, y is not a function of x. 19. y = −75 y = −75 + 0 x or Yes, y is a function of x. 7. No, it does not represent a function. The input values of 10 and 7 are each matched with two output values. 8. Yes, the table does represent a function. Each input value is matched with exactly one output value. 9. (a) Each element of A is matched with exactly one element of B, so it does represent a function. 20. x − 1 = 0 x =1 No, this is not a function of x. 21. f ( x) = 2 x − 3 (a) f (1) = 2(1) − 3 = −1 (b) The element 1 in A is matched with two elements, –2 and 1 of B, so it does not represent a function. (b) f ( −3) = 2( −3) − 3 = −9 (c) Each element of A is matched with exactly one element of B, so it does represent a function. (c) f ( x − 1) = 2( x − 1) − 3 = 2 x − 5 (d) The element 2 in A is not matched with an element of B, so the relation does not represent a function. 10. (a) The element c in A is matched with two elements, 2 and 3 of B, so it is not a function. (b) Each element of A is matched with exactly one element of B, so it does represent a function. (c) This is not a function from A to B (it represents a function from B to A instead). (d) Each element of A is matched with exactly one element of B, so it does represent a function. 3 ( ) π r3 (c) V ( 2r ) = 43 π ( 2r ) = 43 π 8r 3 = 32 3 3 23. g (t ) = 4t 2 − 3t + 5 (a) g ( 2) = 4( 2) − 3( 2) + 5 2 (c) g (t ) − g ( 2) = 4t 2 − 3t + 5 − 15 = 4t 2 − 3t − 10 ) 24. h(t ) = t 2 − 2t Yes, y is a function of x. (a) h( 2) = 22 − 2( 2) = 0 14. ( x − 2) + y 2 = 4 2 (b) h(1.5) = (1.5) − 2(1.5) = −0.75 2 4 − ( x − 2) No, y is not a function of x. 15. y = ( ) = 34π ( 278 ) = 92π = 4t 2 − 19t + 27 Yes, y is a function of x. y = ± () (b) V 32 = 34 π 32 2 12. x 2 + y = 4 Ÿ y = 4 − x 2 ( 3 (b) g (t − 2) = 4(t − 2) − 3(t − 2) + 5 No, y is not a function of x. 13. 2 x + 3 y = 4 Ÿ y = (a) V (3) = 43 π (3) = 43 π ( 27) = 36π = 15 11. x 2 + y 2 = 4 Ÿ y = ± 4 − x 2 1 4 − 2x 3 22. V ( r ) = 43 π r 3 2 (c) h( x + 2) = ( x + 2) − 2( x + 2) = x 2 + 2 x 2 16 − x 2 INSTRUCTOR USE ONLY Yes, y is a function of x. x. © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. Section 2.2 2. 25. f ( y ) = 3 − 4 =1 (a) f ( −1) = 2( −1) + 1 = −1 (b) f (0.25) = 3 − 0.25 = 2.5 (c) f ( 4 x 2 ) = 3 − 4 x2 = 3 − 2 x 26. f ( x ) = (c) f ( 2) = 2( 2) + 2 = 6 ­4 − 5 x, x ≤ −2 ° −2 2 ¯ (−8) + 8 + 2 = 2 (b) f (1) = (1) + 8 + 2 = 5 (c) f ( x − 8) = 27. q( x) = (b) f (0) = 2(0) + 2 = 2 x +8 + 2 (a) f ( −8) = ( x − 8) + 8 + 2 = x + 2 1 x2 − 9 (a) q(0) = 1 1 = − 0 −9 9 2 f ( −2) = ( −2) − 3 = 1 2 1 ( y + 3) − 9 2 = f ( −1) = ( −1) − 3 = −2 2 1 y2 + 6 y 2t 2 + 3 t2 (b) q(0) = (b) f ( 4) = ( 4) + 1 = 17 33. f ( x) = x 2 − 3 2 (c) q( y + 3) = (a) q( 2) = (a) f ( −3) = 4 − 5( −3) = 19 (c) f ( −1) = 0 1 (b) q(3) = 2 is undefined. 3 −9 28. q(t ) = f (0) = (0) − 3 = −3 2 f (1) = (1) − 3 = −2 2 f ( 2) = ( 2) − 3 = 1 2 2( 2) + 3 2 ( 2)2 = 8+3 11 = 4 4 2(0) + 3 x –2 −1 0 1 2 f ( x) 1 –2 –3 –2 1 2 34. h(t ) = 12 t + 3 ( 0) 2 Division by zero is undefined. h( −5) = 12 −5 + 3 = 1 2( − x) + 3 h( −4) = 12 −4 + 3 = 12 (c) q( − x) = 29. f ( x) = 2 ( − x)2 = 2×2 + 3 x2 h( −1) = 12 −1 + 3 = 1 x (b) f ( −2) = h( −3) = 12 −3 + 3 = 0 h( −2) = 12 −2 + 3 = 12 x (a) f ( 2) = 179 ­2 x + 1, x < 0 31. f ( x) = ® ¯2 x + 2, x ≥ 0 y (a) f ( 4) = 3 − Functions 2 =1 t −2 h(t ) 2 −2 (c) f ( x − 1) = = −1 –5 –4 1 1 2 –3 –2 –1 0 1 2 1 ­−1, if x 1 ¯1, x −1 30. f ( x) = x + 4 (a) f ( 2) = 2 + 4 = 6 (b) f ( −2) = −2 + 4 = 6 (c) f ( x 2 ) = x 2 + 4 = x 2 + 4 INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 180 Chapter 2 NOT FOR SALE Functions ctions and Their Graphs 35. f ( x) = ­°− 12 x + 4, x ≤ 0 ® 2 °̄( x − 2) , x > 0 41. x 2 − 9 = 0 x2 = 9 x = ±3 f ( −2) = − 12 ( −2) + 4 = 5 f ( −1) = − 12 ( −1) + 4 = 4 12 = 92 x 2 − 8 x + 15 = 0 f (0) = − 12 (0) + 4 = 4 ( x − 5)( x − 3) = 0 f (1) = (1 − 2) = 1 2 x −5 = 0 Ÿ x = 5 f ( 2) = ( 2 − 2) = 0 2 x f ( x) x −3 = 0 Ÿ x = 3 –2 –1 0 1 2 5 9 2 4 1 0 x( x 2 − 1) = 0 x( x + 1)( x − 1) = 0 x = 0, x = −1, or x = 1 f ( x) = x3 − x 2 − 4 x + 4 44. f (1) = 9 − (1) = 8 2 x3 − x 2 − 4 x + 4 = 0 f ( 2) = 9 − ( 2) = 5 2 x 2 ( x − 1) − 4( x − 1) = 0 f (3) = (3) − 3 = 0 ( x − 1)( x 2 − 4) = 0 f ( 4) = ( 4) − 3 = 1 x −1 = 0 Ÿ x =1 f (5) = (5) − 3 = 2 x 2 − 4 = 0 Ÿ x = ±2 x 1 2 3 4 5 f ( x) 8 5 0 1 2 f ( x ) = g ( x) 45. x2 = x + 2 x2 − x − 2 = 0 37. 15 − 3 x = 0 3 x = 15 ( x − 2)( x + 1) = 0 x = 5 x − 2 = 0 x +1 = 0 x = 2 f ( x) = 5 x + 1 5x + 1 = 0 2 x + 2x + 1 = 7 x − 5 x2 − 5x + 6 = 0 3x − 4 = 0 39. 5 3x − 4 = 0 ( x − 3)( x − 2) = 0 x −3 = 0 x − 2 = 0 x = 3 4 x = 3 12 − x 2 f ( x) = 5 12 − x 2 = 0 5 x 2 = 12 x = ± 12 = ±2 3 x = −1 f ( x ) = g ( x) 46. x = − 15 40. x3 − x = 0 43. 36. f ( x) = °­9 − x 2 , x −6 or (−6, ∞). Domain: y − 10 ≥ 0 y ≥ 10 The domain is all real numbers y such that y ≥ 10. 54. f (t ) = 3 t + 4 Because f (t ) is a cube root, the domain is all real numbers t. s −1 s − 4 59. f ( x) = x − 4 x The domain is all real numbers x such that x > 0 or (0, ∞). 60. f ( x) = x + 2 x − 10 x − 10 > 0 x > 10 The domain is all real numbers x such that x > 10. INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 182 Chapter 2 61. (a) NOT FOR SALE Functions ctions and Their Graphs 63. A = s 2 and P = 4s Ÿ Height, x Volume, V 1 484 2 800 P2 §P· A = ¨ ¸ = 16 ©4¹ 3 972 64. A = π r 2 , C = 2π r 4 1024 5 980 6 864 P = s 4 2 r = C 2π 2 C2 §C · A = π¨ ¸ = 4π © 2π ¹ The volume is maximum when x = 4 and V = 1024 cubic centimeters. (b) 1 2 y = − 10 x + 3x + 6 65. 1 y(30) = − 10 (30) + 3(30) + 6 = 6 feet 2 If the child holds a glove at a height of 5 feet, then the ball will be over the child’s head because it will be at a height of 6 feet. 66. (a) V = l ⋅ w ⋅ h = x ⋅ y ⋅ x = x 2 y where 4 x + y = 108. So, y = 108 − 4 x and V = x 2 (108 − 4 x) = 108 x 2 − 4 x3. Domain: 0 < x < 27 V is a function of x. (c) V = x( 24 − 2 x) (b) 2 Domain: 0 < x 100 = (90 − 0.15 x + 15) x − 60 x = (105 − 0.15 x) x − 60 x = 105 x − 0.15 x 2 − 60 x = 45 x − 0.15 x 2 , x > 100 1− y 0 −1 = x − 2 2 −0 1− y −1 = x − 2 2 2 y = +1 x − 2 x y = x − 2 So, A = 1 § x · x2 x¨ . ¸ = 2 © x − 2¹ 2( x − 2) The domain of A includes x-values such that x 2 ª¬2( x − 2)¼º > 0. By solving this inequality, the INSTRUCTOR USE ONLY omain is x > 2. domain © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.2 2. 68. A = l ⋅ w = ( 2 x) y = 2 xy Functions 183 69. For 2004 through 2007, use p(t ) = 4.57t + 27.3. But y = 36 − x 2 , so A = 2 x 36 − x 2 . The domain is 0 < x 1 as it is possible to find the square root of 0. However, 1 cannot be included in the domain of g ( x) as it causes f (8) = 82 3 + 1 = 5 f ( x) − f (8) x −1 The value 1 may be included in the domain of f ( x) f ( x) = x 2 3 + 1 84. 93. f ( x) = x2 3 + 1 − 5 x2 3 − 4 = ,x ≠ 8 x −8 x −8 85. By plotting the points, we have a parabola, so g ( x) = cx 2 . Because ( −4, − 32) is on the graph, you have −32 = c( −4) Ÿ c = −2. So, g ( x) = −2 x 2 . 2 86. By plotting the data, you can see that they represent a ( ) line, or f ( x) = cx. Because (0, 0) and 1, 14 are on the line, the slope is 14 . So, f ( x) = 14 x. 87. Because the function is undefined at 0, we have r ( x) = c x. Because ( −4, − 8) is on the graph, you a zero to occur in the denominator which results in the function being undefined. 94. Because f ( x) is a function of an even root, the radicand cannot be negative. g ( x) is an odd root, therefore the radicand can be any real number. So, the domain of g is all real numbers x and the domain of f is all real numbers x such that x ≥ 2. 95. No; x is the independent variable, f is the name of the function. 96. (a) The height h is function of t because for each value of t there is a corresponding value of h for 0 ≤ t ≤ 2.6. have −8 = c −4 Ÿ c = 32. So, r ( x) = 32 x. (b) Using the graph when t = 0.5, h ≈ 20 feet and 88. By plotting the data, you can see that they represent (c) The domain of h is approximately 0 ≤ t ≤ 2.6. h( x ) = c x . Because −4 = 2 and −1 = 1, and the corresponding y-values are 6 and 3, c = 3 and h( x ) = 3 x. 89. False. The equation y 2 = x 2 + 4 is a relation between x and y. However, y = ± a function. x 2 + 4 does not represent 90. True. A function is a relation by definition. 91. False. The range is [−1, ∞). 92. True. The set represents a function. Each x-value is mapped to exactly one y-value. when t = 1.25, h ≈ 28 feet. (d) No, the time t is not a function of the height h because some values of h correspond to more than one value of t. 97. (a) Yes. The amount that you pay in sales tax will increase as the price of the item purchased increases. (b) No. The length of time that you study the night before an exam does not necessarily determine your score on the exam. 98. (a) No. During the course of a year, for example, your salary may remain constant while your savings account balance may vary. That is, there may be two or more outputs (savings account balances) for one input (salary). (b) Yes. The greater the height from which the ball is dropped, the greater the speed with which the ball will strike the ground. Section 2.3 Analyzing Graphs of Functions 1. Vertical Line Test 7. Domain: ( −∞, ∞); Range: [−4, ∞) 2. zeros (a) f ( −2) = 0 3. decreasing (b) f ( −1) = −1 4. maximum (c) f 12 = 0 5. average rate of change; secant (d) f (1) = − 2 () 6. odd INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.3 8. Domain: ( −∞, ∞); Range: ( −∞, ∞) Analyzing Graphs of Functions f ( x) = 3 x 2 + 22 x − 16 16. (a) f ( −1) = 4 3x 2 + 22 x − 16 = 0 (3x − 2)( x + 8) = 0 (b) f ( 2) = 4 3x − 2 = 0 Ÿ x = 23 (c) f (0) = 2 x + 8 = 0 Ÿ x = −8 (d) f (1) = 0 17. 9. Domain: ( −∞, ∞); Range: ( −2, ∞) f ( x) = x 9×2 − 4 x = 0 9×2 − 4 x = 0 (a) f ( 2) = 0 (b) f (1) = 1 (c) f (3) = 2 x 2 − 9 x + 14 = 0 4x ( x − 7)( x − 2) = 0 10. Domain: ( −∞, ∞); Range: ( – ∞, 1] (a) f ( −2) = −3 x −7 = 0 Ÿ x = 7 (b) f (1) = 0 x − 2 = 0 Ÿ x = 2 (c) f (0) = 1 19. (d) f ( 2) = −3 f ( x) = 12 x3 − x 1 x3 − x = 0 2 x 3 − 2 x = 2(0) 11. y = 14 x3 x ( x 2 − 2) = 0 A vertical line intersects the graph at most once, so y is a function of x. 12. x − y 2 = 1 Ÿ y = ± x = 0 or x2 − 2 = 0 x2 = 2 x −1 x = ± 2 y is not a function of x. Some vertical lines intersect the graph twice. x3 − 4 x 2 − 9 x + 36 = 0 A vertical line intersects the graph more than once, so y is not a function of x. x 2 ( x − 4) − 9( x − 4) = 0 ( x − 4)( x 2 − 9) = 0 14. x 2 = 2 xy − 1 x − 4 = 0 Ÿ x = 4 A vertical line intersects the graph at most once, so y is a function of x. f ( x) = 2 x 2 − 7 x − 30 x = − 52 f ( x) = 4 x3 − 24 x 2 − x + 6 21. 4 x 2 ( x − 6) − 1( x − 6) = 0 (2 x + 5)( x − 6) = 0 or x 2 − 9 = 0 Ÿ x = ±3 4 x3 − 24 x 2 − x + 6 = 0 2 x 2 − 7 x − 30 = 0 2x + 5 = 0 f ( x) = x3 − 4 x 2 − 9 x + 36 20. 2 13. x + y = 25 15. x 2 − 9 x + 14 4x f ( x) = 18. (d) f ( −1) = 3 2 187 x −6 = 0 x = 6 ( x − 6)(4 x 2 − 1) = 0 ( x − 6)(2 x + 1)(2 x − 1) = 0 x − 6 = 0 or 2 x + 1 = 0 or 2 x − 1 = 0 x = − 12 x = 12 x = 6 INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 188 Chapter 2 NOT FOR SALE Functions ctions and Their Graphs f ( x) = 9 x 4 − 25 x 2 22. 27. (a) 9 x 4 − 25 x 2 = 0 x 2 (9 x 2 − 25) = 0 x2 = 0 Ÿ x = 0 9 x 2 − 25 = 0 Ÿ x = ± 53 f ( x) = 23. 2x − 1 Zero: x = − 11 2 (b) f ( x) = 2 x + 11 2x − 1 = 0 2 x + 11 = 0 2x = 1 2 x + 11 = 0 x = − 11 2 2x = 1 x = 12 f ( x) = 24. 28. (a) 3x + 2 3x + 2 = 0 3x + 2 = 0 − 23 = x Zero: x = 26 f ( x) = (b) 25. (a) 3x − 14 − 8 3 x − 14 − 8 = 0 3 x − 14 = 8 3 x − 14 = 64 x = 26 5 Zero: x = − 3 (b) 29. (a) f ( x) = 3 + 5x 3 + 5x = 0 3x + 5 = 0 x = − 53 26. (a) Zero: x = (b) f ( x) = 1 3 3x − 1 x −6 3x − 1 = 0 x −6 3x − 1 = 0 Zeros: x = 0, x = 7 (b) x = 1 3 f ( x ) = x( x − 7) x( x − 7) = 0 x = 0 x −7 = 0 Ÿ x = 7 INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.3 30. (a) Analyzing Graphs of Functions 189 ­2 x + 1, x ≤ −1 38. f ( x) = ® 2 ¯x − 2, x > −1 The function is decreasing on ( −1, 0) and increasing on (−∞, −1) and (0, ∞). Zeros: x = ±2.1213 39. f ( x) = 3 2 x2 − 9 f ( x) = 3− x (b) (a) 2 x2 − 9 = 0 3− x 2 x2 − 9 = 0 Ÿ x = ± 31. f ( x) = 3 2 = ±2.1213 2 3 x 2 Constant on ( −∞, ∞) (b) The function is increasing on ( −∞, ∞). 32. f ( x) = x 2 − 4 x The function is decreasing on ( −∞, 2) and increasing on (2, ∞). x –2 −1 0 1 2 f ( x) 3 3 3 3 3 40. g ( x) = x (a) 33. f ( x) = x3 − 3 x 2 + 2 The function is increasing on ( −∞, 0) and ( 2, ∞) and decreasing on (0, 2). 34. f ( x) = Increasing on ( −∞, ∞) (b) x2 − 1 The function is decreasing on ( −∞, −1) and increasing on (1, ∞). x –2 −1 0 1 2 g ( x) –2 –1 0 1 2 41. g ( s ) = 35. f ( x) = x + 1 + x − 1 The function is increasing on (1, ∞). s2 4 (a) The function is constant on ( −1, 1). The function is decreasing on ( −∞, −1). 36. The function is decreasing on ( −2, −1) and ( −1, 0) and increasing on ( −∞, − 2) and (0, ∞). ­x + 3, x ≤ 0 ° 37. f ( x) = ®3, 0 2 ¯ Decreasing on ( −∞, 0); Increasing on (0, ∞) (b) s –4 −2 0 2 4 g ( s) 4 1 0 1 4 The function is increasing on ( −∞, 0) and ( 2, ∞). The function is constant on (0, 2). INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 190 Chapter 2 NOT FOR SALE Functions ctions and Their Graphs 42. f ( x) = 3x 4 − 6 x 2 46. f ( x) = x 2 3 (a) (a) Increasing on ( −1, 0), (1, ∞); Decreasing on (−∞, −1), (0, 1) (b) Decreasing on ( – ∞, 0); Increasing on (0, ∞) (b) x –2 −1 0 1 2 f ( x) 24 –3 0 –3 24 x –2 –1 0 1 2 f ( x) 1.59 1 0 1 1.59 47. f ( x) = 3x 2 − 2 x − 5 43. f ( x) = 1− x (a) Relative minimum: Decreasing on ( −∞, 1) (b) x –3 f ( x) 2 44. f ( x) = x −2 –1 3 2 0 1 1 0 ( 13 , − 163 ) or (0.33, − 5.33) 48. f ( x) = − x 2 + 3 x − 2 x +3 (a) Relative maximum: (1.5, 0.25) 49. f ( x) = −2 x 2 + 9 x Increasing on ( −2, ∞); Decreasing on ( −3, − 2) (b) x –3 −2 –1 0 1 f ( x) 0 –2 –1.414 0 2 45. f ( x) = x Relative maximum: ( 2.25, 10.125) 32 50. f ( x) = x( x − 2)( x + 3) (a) Increasing on (0, ∞) (b) x 0 1 2 Relative minimum: (1.12, – 4.06) 3 4 Relative maximum: ( −1.79, 8.21) INSTRUCTOR NST TRUC USE ONLY f ( x) 0 1 2.8 5.2 8 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.3 51. f ( x) = x3 − 3 x 2 − x + 1 Analyzing Graphs of Functions 191 56. f ( x) = 4 x + 2 f ( x) ≥ 0 on ª¬− 12 , ∞ ) 4x + 2 ≥ 0 4 x ≥ −2 x ≥ − 12 Relative maximum: ( −0.15, 1.08) Relative minimum: ( 2.15, − 5.08) 52. h( x) = x3 − 6 x 2 + 15 ª− 1 , ∞ ¬ 2 ) 57. f ( x) = 9 − x 2 f ( x) ≥ 0 on [−3, 3] Relative minimum: ( 4, −17) Relative maximum: (0, 15) 53. h( x) = ( x − 1) x 58. f ( x) = x 2 − 4 x f ( x) ≥ 0 on ( −∞, 0] and [4, ∞) x2 − 4 x ≥ 0 x ( x − 4) ≥ 0 Relative minimum: (0.33, − 0.38) (−∞, 0], [4, ∞) 54. g ( x) = x 4 − x 59. f ( x) = Relative maximum: ( 2.67, 3.08) 55. f ( x) = 4 − x f ( x) ≥ 0 on ( −∞, 4] x −1 f ( x) ≥ 0 on [1, ∞) x −1 ≥ 0 x −1 ≥ 0 x ≥1 [1, ∞) 60. f ( x ) = −(1 + x ) f ( x) is never greater than 0. ( f ( x) < 0 for all x.) INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 192 NOT FOR SALE Chapter 2 Functions ctions and Their Graphs 61. f ( x) = −2 x + 15 f (3) − f (0) 3−0 = 67. s0 = 6, v0 = 64 (a) s = −16t 2 + 64t + 6 9 − 15 = −2 3 (b) The average rate of change from x1 = 0 to x2 = 3 is −2. 62. f ( x) = x 2 − 2 x + 8 f (5) − f (1) 5−1 = 23 − 7 16 = = 4 4 4 The average rate of change from x1 = 1 to x2 = 5 is 4. 3−1 = s(3) − s(0) 3−0 = 54 − 6 = 16 3 (d) The slope of the secant line is positive. (e) s(0) = 6, m = 16 63. f ( x ) = x3 − 3 x 2 − x f (3) − f (1) (c) −3 − ( −3) 2 Secant line: y − 6 = 16(t − 0) = 0 y = 16t + 6 The average rate of change from x1 = 1 to x2 = 3 is 0. (f ) 64. f ( x ) = − x 3 + 6 x 2 + x f (6) − f (1) 6−6 0 = = 0 6 −1 5 5 The average rate of change from x1 = 1 to x2 = 6 is 0. = 65. (a) 68. (a) s = −16t 2 + 72t + 6.5 (b) (b) To find the average rate of change of the amount the U.S. Department of Energy spent for research and development from 2005 to 2010, find the average rate of change from (5, f (5)) to (10, f (10)). f (10) − f (5) 10 − 5 = 10,925 − 8501.25 = 484.75 5 The amount the U.S. Department of Energy spent for research and development increased by about $484.75 million each year from 2005 to 2010. 66. Average rate of change = = s(t2 ) − s(t1 ) (c) The average rate of change from t = 0 to t = 4: s ( 4) − s ( 0) 4 −0 second = 38.5 − 6.5 32 = = 8 feet per 4 4 (d) The slope of the secant line through (0, s(0)) and (4, s(4)) is positive. (e) The equation of the secant line: m = 8, y = 8t + 6.5 (f ) t2 − t1 s(9) − s(0) 9−0 540 − 0 = 9 −0 = 60 feet per second. As the time traveled increases, the distance increases rapidly, causing the average speed to increase with each time increment. From t = 0 to t = 4, the average speed is less than from t = 4 to t = 9. Therefore, the overall average from t = 0 to t = 9 falls below the average found in part (b). INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. Section 2.3 69. v0 = 120, s0 = 0 71. Analyzing Graphs of Functions 193 f ( x) = x6 − 2 x 2 + 3 f ( − x) = ( − x) − 2( − x) + 3 6 (a) s = −16t 2 + 120t (b) 2 = x6 − 2 x 2 + 3 = f ( x) The function is even. y-axis symmetry. (c) The average rate of change from t = 3 to t = 5: s(5) − s(3) 5−3 second = 72. g ( − x) = ( − x) − 5( − x) 3 200 − 216 16 = − = − 8 feet per 2 2 = − x3 + 5 x = − g ( x) The function is odd. Origin symmetry. (d) The slope of the secant line through (3, s(3)) and (5, s(5)) is negative. 73. h( x ) = x x +5 h( − x ) = ( − x ) − x + 5 (e) The equation of the secant line: m = − 8 Using (5, s(5)) = (5, 200) we have = −x 5 − x ≠ h( x ) y − 200 = − 8(t − 5) ≠ − h( x ) y = − 8t + 240. (f ) g ( x) = x3 − 5 x The function is neither odd nor even. No symmetry. 270 74. 0 f ( x) = x 1 − x 2 f ( − x ) = − x 1 − ( − x) 8 2 = − x 1 − x2 0 = − f ( x) 2 70. (a) s = −16t + 80 (b) The function is odd. Origin symmetry. 75. f ( s) = 4 s 3 2 = 4( − s ) ≠ f ( s) (c) The average rate of change from t = 1 to t = 2: s( 2) − s(1) 2 −1 per second ≠ − f ( s) 16 − 64 48 = = − = −48 feet 1 1 The function is neither odd nor even. No symmetry. 76. (d) The slope of the secant line through (1, s(1)) and (2, s(2)) is negative. 32 g ( s ) = 4s 2 3 g ( − s ) = 4( − s ) 23 = 4s 2 3 (e) The equation of the secant line: m = −48 Using (1, s(1)) = (1, 64) we have = g ( s) The function is even. y-axis symmetry. y − 64 = −48(t − 1) y = −48t + 112. (f ) INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 194 Chapter 2 Functions ctions and Their Graphs 80. h( x) = x 2 − 4 77. The graph of f ( x) = −9 is symmetric to the y-axis, which implies f ( x) is even. The graph displays y-axis symmetry, which implies h( x) is even. f ( − x) = −9 h( − x ) = ( − x ) − 4 = x 2 − 4 = h( x ) 2 = f ( x) The function is even. The function is even. 81. f ( x ) = 78. f ( x) = 5 − 3 x The graph displays no symmetry, which implies f ( x) is 1− x neither odd nor even. The graph displays no symmetry, which implies f ( x) is neither odd nor even. f ( − x ) = 5 − 3( − x) f ( − x) = = 5 + 3x = ≠ f ( x) 1 − ( − x) 1+ x ≠ f ( x) ≠ − f ( x) ≠ − f ( x) The function is neither even nor odd. 79. f ( x) = − x − 5 The function is neither even nor odd. 82. g (t ) = 3 t − 1 The graph displays no symmetry, which implies f ( x) is neither odd nor even. The graph displays no symmetry, which implies g (t ) is neither odd nor even. f ( x ) = − ( − x) − 5 g ( −t ) = 3 ( −t ) − 1 = − −x − 5 = 3 −t − 1 ≠ f ( x) ≠ g (t ) ≠ − f ( x) ≠ − g (t ) INSTRUCTOR USE ONLY The function is neither even nor odd. odd The function is neither even nor odd. © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.3 83. h = top − bottom Analyzing Graphs of Functions 195 (c) When x = 4, the resulting figure is a square. = 3 − (4 x − x ) 2 By = 3 − 4x + x2 84. h = top − bottom = (4 x − x 2 ) − 2 x = 2x − x2 85. L = right − left By the Pythagorean Theorem, 42 + 42 = s 2 Ÿ s = 32 = 4 2 meters. = 2 − 3 2y 86. L = right − left 89. (a) For the average salaries of college professors, a scale of $10,000 would be appropriate. 2 = −0 y = (b) For the population of the United States, use a scale of 10,000,000. 2 y 87. L = −0.294 x 2 + 97.744 x − 664.875, 20 ≤ x ≤ 90 (a) (c) For the percent of the civilian workforce that is unemployed, use a scale of 1%. 90. (a) 70 0 24 0 (b) L = 2000 when x ≈ 29.9645 ≈ 30 watts. 88. (b) The model is an excellent fit. (c) The temperature is increasing from 6 A.M. until noon ( x = 0 to x = 6). Then it decreases until 2 A.M. ( x = 6 to x = 20). Then the temperature increases until 6 A.M. ( x = 20 to x = 24). (d) The maximum temperature according to the model is about 63.93°F. According to the data, it is 64°F. The minimum temperature according to the model is about 33.98°F. According to the data, it is 34°F. () (a) A = (8)(8) − 4 12 ( x)( x) = 64 − 2 x 2 Domain: 0 ≤ x ≤ 4 (e) Answers may vary. Temperatures will depend upon the weather patterns, which usually change from day to day. (b) Range: 32 ≤ A ≤ 64 INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 196 Chapter 2 NOT FOR SALE Functions ctions and Their Graphs 91. (a) y = x (b) y = x2 (c) y = x3 (d) y = x 4 (e) y = x5 (f ) y = x 6 All the graphs pass through the origin. The graphs of the odd powers of x are symmetric with respect to the origin and the graphs of the even powers are symmetric with respect to the y-axis. As the powers increase, the graphs become flatter in the interval −1 < x −4 = −7ced5 12 fhg + 6 = −7(5) + 6 = −29 INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 200 Chapter 2 NOT FOR SALE Functions ctions and Their Graphs °­ 4 + x , x 2 °̄ ( ) 41. s( x) = 2 14 x − cde 14 xfgh (a) (b) Domain: ( −∞, ∞) ; Range: [0, 2) ( ) 42. k ( x) = 4 12 x − ced 12 xfhg 2 (a) (b) Domain: ( −∞, ∞) ; Range: [0, 4) 43. (a) W (30) = 14(30) = 420 W ( 40) = 14( 40) = 560 ­°x 2 + 5, x ≤1 38. f ( x) = ® 2 °̄− x + 4 x + 3, x > 1 W ( 45) = 21( 45 − 40) + 560 = 665 W (50) = 21(50 − 40) + 560 = 770 0 45 44. (a) ­4 − x 2 , x < −2 ° 39. h( x) = ®3 + x, −2 ≤ x < 0 °x 2 + 1, x ≥ 0 ¯ The domain of f ( x) = −1.97 x + 26.3 is 6 < x ≤ 12. One way to see this is to notice that this is the equation of a line with negative slope, so the function values are decreasing as x increases, which matches the data for the corresponding part of the table. The domain of f ( x ) = 0.505 x 2 − 1.47 x + 6.3 is then 1 ≤ x ≤ 6. ­2 x + 1, x ≤ −1 ° 40. k ( x) = ®2 x 2 − 1, −1 1 ¯ (b) f (5) = 0.505(5) − 1.47(5) + 6.3 2 = 0.505( 25) − 7.35 + 6.3 = 11.575 f (11) = −1.97(11) + 26.3 = 4.63 These values represent the revenue in thousands of dollars for the months of May and November, respectively. (c) These values are quite close to the actual data values. INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.5 45. Answers will vary. Sample answer: Interval Input Pipe Drain Pipe 1 Drain Pipe 2 [0, 5] Open Closed Closed [5, 10] Open Open Closed [10, 20] Closed Closed Closed [20, 30] Closed Closed Open [30, 40] Open Open Open [40, 45] Open Closed Open [45, 50] Open Open Open [50, 60] Open Open Closed Transformations of Functions 201 47. For the first two hours the slope is 1. For the next six hours, the slope is 2. For the final hour, the slope is 12 . ­t , 0 ≤ t ≤ 2 ° f (t ) = ®2t − 2, 2 < t ≤ 8 ° 1 t + 10, 8 < t ≤ 9 ¯2 To find f (t ) = 2t − 2, use m = 2 and ( 2, 2). 46. (a) Cost = Flat fee + fee per pound y − 2 = 2(t − 2) Ÿ y = 2t − 2 C ( x) = 26.10 + 4.35a xb To find f (t ) = 12 t + 10, use m = 12 and (8, 14). (b) y − 14 = 12 (t − 8) Ÿ y = 12 t + 10 Total accumulation = 14.5 inches 48. f ( x) = x 2 f ( x) = x3 (a) Domain: ( −∞, ∞) (a) Domain: ( −∞, ∞) Range: [0, ∞) Range: ( −∞, ∞) (b) x-intercept: (0, 0) (b) x-intercept: (0, 0) y-intercept: (0, 0) y-intercept: (0, 0) (c) Increasing: (0, ∞) (c) Increasing: ( −∞, ∞) Decreasing: ( −∞, 0) (d) Odd; the graph has origin symmetry. (d) Even; the graph has y-axis symmetry. 49. False. A piecewise-defined function is a function that is defined by two or more equations over a specified domain. That domain may or may not include x- and y-intercepts. 50. False. The vertical line x = 2 has an x-intercept at the point ( 2, 0) but does not have a y-intercept. The horizontal line y = 3 has a y-intercept at the point (0, 3) but does not have an x-intercept. Section 2.5 Transformations of Functions 1. rigid 3. vertical stretch; vertical shrink 2. − f ( x); f ( − x) 4. (a) iv (b) ii (c) iii INSTRUCTOR USE ONLY (d) d) i © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 202 Chapter 2 NOT FOR SALE Functions ctions and Their Graphs 5. (a) f ( x) = x + c Vertical shifts c = −1: f ( x) = x − 1 1 unit down c = 1: f ( x) = x + 1 1 unit up c = 3: f ( x) = x + 3 3 units up (b) f ( x) = x − c Horizontal shifts c = −1: f ( x) = x + 1 1 unit left c = 1: f ( x) = x − 1 1 unit right c = 3: f ( x) = x − 3 3 units right 6. (a) f ( x ) = x + c Vertical shifts c = −3: f ( x ) = x −3 3 units down c = −1: f ( x) = x −1 1 unit down c = 1: f ( x ) = x +1 1 unit up c = 3: f ( x) = x +3 3 units up (b) f ( x ) = x −c Horizontal shifts c = −3: f ( x ) = x+3 3 units left c = −1: f ( x) = x +1 1 unit left c = 1: f ( x ) = x −1 1 unit right c = 3: f ( x) = x −3 3 units right 7. (a) f ( x) = a xb + c Vertical shifts c = −2: f ( x) = a xb − 2 2 units down c = 0: f ( x) = a xb Parent function c = 2: f ( x) = a xb + 2 2 units up (b) f ( x) = a x + cb Horizontal shifts c = −2: f ( x) = a x − 2b 2 units right c = 0: f ( x) = a xb Parent function c = 2: f ( x) = a x + 2b 2 units left INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.5 ­°x 2 + c, x < 0 8. (a) f ( x) = ® 2 °̄− x + c, x ≥ 0 (b) ­°( x + c)2 , x < 0 f ( x) = ® 2 °̄−( x + c) , x ≥ 0 y = f ( − x) (b) y = f ( x) + 4 9. (a) Reflection in the y-axis (d) y = − f ( x − 4) Reflection in the x-axis and a horizontal shift 4 units to the right Transformations of Functions (e) y = f ( x) − 3 Vertical shift 3 units downward y = 2 f ( x) (c) Vertical shift 4 units upward 203 Vertical stretch (each y-value is multiplied by 2) y = − f ( x) − 1 (f ) Reflection in the x-axis and a vertical shift 1 unit downward (g) y = f ( 2 x) Horizontal shrink (each x-value is divided by 2) INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 204 NOT FOR SALE Chapter 2 Functions ctions and Their Graphs 10. (a) y = f ( x − 5) (b) y = − f ( x) + 3 Horizontal shift 5 units Reflection in the x-axis and a to the right vertical shift 3 units upward (d) y = − f ( x + 1) (e) Reflection in the x-axis and a horizontal shift 1 unit to the left y = 13 f ( x) (c) Vertical shrink (each y-value is multiplied by 13 ) y = f ( − x) y = f ( x) − 10 (f ) Reflection in the y-axis Vertical shift 10 units downward ( ) (g) y = f 13 x Horizontal stretch (each x-value is multiplied by 3) 13. Parent function: f ( x) = x 11. Parent function: f ( x) = x 2 (a) Vertical shift 1 unit downward g ( x) = x − 1 2 (b) Reflection in the x-axis, horizontal shift 1 unit to the left, and a vertical shift 1 unit upward g ( x) = −( x + 1) + 1 2 12. Parent function: f ( x ) = x (a) Reflected in the x-axis and shifted upward 1 unit g ( x) = − x3 + 1 = 1 − x3 (b) Shifted to the left 3 units and down 1 unit 3 g ( x) = − x + 3 (b) Horizontal shift 2 units to the right and a vertical shift 4 units downward g ( x) = x − 2 − 4 3 g ( x) = − ( x + 3) − 1 (a) Reflection in the x-axis and a horizontal shift 3 units to the left 14. Parent function: f ( x ) = x (a) Shifted downward 7 units and to the left 1 unit g ( x) = x +1 −7 (d) Reflected about the x- and y-axis and shifted to the right 3 units and downward 4 units INSTRUCTOR USE ONLY g ( x) = − −x + 3 − 4 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.5 15. Parent function: f ( x ) = x 3 22. g ( x) = ( x − 8) Horizontal shift 2 units to the right y = ( x − 2) Transformations of Functions 3 205 2 (a) Parent function: f ( x ) = y = x 2 (b) Horizontal shift of 8 units to the right (c) 16. Parent function: y = x Vertical shrink y = 12 x 17. Parent function: f ( x) = x 2 Reflection in the x-axis y = − x2 (d) g ( x) = f ( x − 8) 18. Parent function: y = a xb Vertical shift 23. g ( x) = x 3 + 7 y = a xb + 4 19. Parent function: f ( x ) = (a) Parent function: f ( x ) = x 3 x (b) Vertical shift 7 units upward Reflection in the x-axis and a vertical shift 1 unit upward y = − (c) x +1 20. Parent function: y = x Horizontal shift y = x + 2 21. g ( x) = 12 − x 2 (d) g ( x) = f ( x) + 7 (a) Parent function: f ( x) = x 2 (b) Reflection in the x-axis and a vertical shift 12 units upward (c) 24. g ( x) = − x3 − 1 (a) Parent function: f ( x ) = x 3 (b) Reflection in the x-axis, vertical shift of 1 unit downward (c) (d) g ( x) = 12 − f ( x) (d) g ( x) = − f ( x) − 1 INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 206 Chapter 2 NOT FOR SALE Functions ctions and Their Graphs 25. g ( x) = 23 x 2 + 4 28. g ( x) = − 14 ( x + 2) − 2 2 (a) Parent function: f ( x) = x 2 (a) Parent function: f ( x) = x 2 (b) Vertical shrink of two-thirds, and a vertical shift 4 units upward (b) Horizontal shift 2 units to the left, vertical shrink, reflection in the x-axis, vertical shift 2 units downward (c) (c) (d) g ( x) = 23 f ( x) + 4 26. g ( x) = 2( x − 7) (d) g ( x) = − 14 f ( x + 2) − 2 2 29. g ( x) = (a) Parent function: f ( x) = x 2 (b) Vertical stretch of 2 and a horizontal shift 7 units to the right of f ( x) = x 2 3x (a) Parent function: f ( x ) = (b) Horizontal shrink by 13 (c) (c) (d) g ( x) = 2 f ( x − 7) (d) g ( x) = f (3x) 27. g ( x) = 2 − ( x + 5) 2 30. g ( x) = 1x 4 (a) Parent function: f ( x) = x 2 (a) Parent function: f ( x) = (b) Reflection in the x-axis, horizontal shift 5 units to the left, and a vertical shift 2 units upward (b) Horizontal stretch of 4 (c) (d) g ( x) = 2 − f ( x + 5) x x (c) ( ) (d) g ( x) = f 14 x INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. Section 2.5 Transformations of Functions 34. g ( x) = − 12 ( x + 1) 31. g ( x) = ( x − 1) + 2 3 207 3 (a) Parent function: f ( x) = x3 (a) Parent function: f ( x) = x3 (b) Horizontal shift 1 unit to the right and a vertical shift 2 units upward (b) Horizontal shift one unit to the right, vertical shrink (each y-value is multiplied by 12 ), reflection in the (c) x-axis. (c) (d) g ( x) = f ( x − 1) + 2 (d) g ( x) = − 12 f ( x + 1) 32. g ( x) = ( x + 3) − 10 3 (a) Parent function: f ( x) = x3 35. g ( x) = − x − 2 (b) Horizontal shift of 3 units to the left, vertical shift of 10 units downward (a) Parent function: f ( x) = x (c) (b) Reflection in the x-axis, vertical shift 2 units downward (c) (d) g ( x) = f ( x + 3) − 10 (d) g ( x) = − f ( x) − 2 33. g ( x) = 3( x − 2) 3 36. g ( x) = 6 − x + 5 (a) Parent function: f ( x) = x3 (b) Horizontal shift 2 units to the right, vertical stretch (each y-value is multiplied by 3) (c) (a) Parent function: f ( x) = x (b) Reflection in the x-axis, horizontal shift of 5 units to the left, vertical shift of 6 units upward (c) (d) g ( x) = 3 f ( x − 2) (d) g ( x) = 6 − f ( x + 5) INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 208 Chapter 2 Functions ctions and Their Graphs 37. g ( x) = − x + 4 + 8 40. g ( x) = 12 x − 2 − 3 (a) Parent function: f ( x) = x (a) Parent function: f ( x) = x (b) Reflection in the x-axis, horizontal shift 4 units to the left, and a vertical shift 8 units upward (b) Horizontal shift 2 units to the right, vertical shrink, vertical shift 3 units downward (c) (c) (d) g ( x) = − f ( x + 4) + 8 (d) g ( x) = 12 f ( x − 2) − 3 38. g ( x) = − x + 3 + 9 41. g ( x) = 3 − a xb (a) Parent function: f ( x) = x (a) Parent function: f ( x) = a xb (b) Reflection in the y-axis, horizontal shift of 3 units to the right, vertical shift of 9 units upward (b) Reflection in the x-axis and a vertical shift 3 units upward (c) (c) (d) g ( x) = f ( − ( x − 3)) + 9 (d) g ( x) = 3 − f ( x) 39. g ( x) = − 2 x − 1 − 4 42. g ( x) = 2a x + 5b (a) Parent function: f ( x) = x (a) Parent function: f ( x) = a xb (b) Horizontal shift one unit to the right, vertical stretch, reflection in the x-axis, vertical shift four units downward (b) Horizontal shift of 5 units to the left, vertical stretch (each y-value is multiplied by 2) (c) (d) g ( x) = −2 f ( x − 1) − 4 (c) (d) g ( x) = 2 f ( x + 5) INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE 43. g ( x) = x −9 (a) Parent function: f ( x) = Section 2.5 Transformations of Functions 46. g ( x) = 3x + 1 (a) Parent function: f ( x) = x (b) Horizontal shift 9 units to the right (c) 209 x ( ) (b) Horizontal shrink each x-value is multiplied by 13 , vertical shift of 1 unit upward (c) (d) g ( x) = f ( x − 9) 44. g ( x) = (d) g ( x) = f (3x) + 1 x + 4 +8 (a) Parent function: f ( x) = 47. g ( x) = ( x − 3) − 7 2 x (b) Horizontal shift of 4 units to the left, vertical shift of 8 units upward (c) 48. g ( x) = − ( x + 2) + 9 2 49. f ( x) = x3 moved 13 units to the right g ( x) = ( x − 13) 3 50. f ( x) = x3 moved 6 units to the left, 6 units downward, and reflected in the y-axis (in that order) g ( x) = ( − x + 6) − 6 3 (d) g ( x) = f ( x + 4) + 8 45. g ( x) = 51. g ( x) = − x + 12 7 − x − 2 or g ( x) = (a) Parent function: f ( x) = − ( x − 7) − 2 x 53. f ( x) = (b) Reflection in the y-axis, horizontal shift 7 units to the right, and a vertical shift 2 units downward (c) 52. g ( x) = x + 4 − 8 x moved 6 units to the left and reflected in both the x- and y-axes g ( x) = − −x + 6 54. f ( x) = x moved 9 units downward and reflected in both the x-axis and the y-axis g ( x) = − ( − x − 9) 55. f ( x) = x 2 (d) g ( x) = f (7 − x) − 2 (a) Reflection in the x-axis and a vertical stretch (each y-value is multiplied by 3) g ( x) = − 3 x 2 (b) Vertical shift 3 units upward and a vertical stretch (each y-value is multiplied by 4) g ( x) = 4 x 2 + 3 INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 210 Chapter 2 NOT FOR SALE Functions ctions and Their Graphs 56. f ( x) = x3 63. Parent function: f ( x) = ( (a) Vertical shrink each y -value is multiplied by 14 ) g ( x) = 14 x3 (b) Reflection in the x-axis and a vertical stretch (each y -value is multiplied by 2) g ( x) = − 2 x 3 (a) Reflection in the x-axis and a vertical shrink each y -value is multiplied by 12 ) g ( x) = − 12 x (b) Vertical stretch (each y-value is multiplied by 3) and a vertical shift 3 units downward g ( x) = 3 x − 3 58. f ( x) = (each y-value is multiplied by 12 ) g ( x) = 12 −x 64. Parent function: f ( x) = x g ( x) = − 2 x − 2 65. Parent function: f ( x) = x3 Reflection in the x-axis, horizontal shift 2 units to the right and a vertical shift 2 units upward g ( x) = − ( x − 2) + 2 3 66. Parent function: f ( x) = x x (a) Vertical stretch (each y-value is multiplied by 8) g ( x) = 8 Reflection in the y-axis, vertical shrink Reflection in the x-axis, vertical shift of 2 units downward, vertical stretch (each y-value is multiplied by 2) 57. f ( x) = x ( x x Horizontal shift of 4 units to the left and a vertical shift of 2 units downward g ( x) = x + 4 − 2 (b) Reflection in the x-axis and a vertical shrink (each y-value is multiplied by 14 ) g ( x) = − 14 x Reflection in the x-axis and a vertical shift 3 units downward x g ( x) = − 59. Parent function: f ( x) = x3 Vertical stretch (each y-value is multiplied by 2) g ( x) = 2 x 67. Parent function: f ( x) = 3 x −3 68. Parent function: f ( x) = x 2 Horizontal shift of 2 units to the right and a vertical shift of 4 units upward 60. Parent function: f ( x) = x g ( x) = ( x − 2) + 4 2 Vertical stretch (each y-value is multiplied by 6) g ( x) = 6 x 69. (a) 61. Parent function: f ( x) = x 2 Reflection in the x-axis, vertical shrink (each y-value is multiplied by 12 ) (b) g ( x) = − 12 x 2 2 § x · § x · § x · H ¨ ¸ = 0.002¨ ¸ + 0.005¨ ¸ − 0.029 1.6 1.6 © ¹ © ¹ © 1.6 ¹ 62. Parent function: y = a xb Horizontal stretch (each x-value is multiplied by 2) g ( x) = cde 12 xfgh H ( x) = 0.002 x 2 + 0.005 x − 0.029 § x2 · § x · = 0.002¨ ¸ + 0.005¨ ¸ − 0.029 2.56 © 1.6 ¹ © ¹ = 0.00078125 x 2 + 0.003125 x − 0.029 § x · The graph of H ¨ ¸ is a horizontal stretch of the © 1.6 ¹ graph of H ( x). INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.5 70. (a) The graph of N ( x) = − 0.068( x − 13.68) + 119 is 2 a reflection in the x-axis, a vertical shrink of a factor of 0.068, a horizontal shift of 13.68 units to the right and a vertical shift of 119 units upward of the graph f ( x) = x 2 . Transformations of Functions 211 76. (a) Answers will vary. Sample Answer: To graph f ( x) = 3x 2 − 4 x + 1 use the point-plotting method since it is not written in a form that is easily identified by a sequence of translations of the parent function y = x 2 . (b) Answers will vary. Sample Answer: To graph f ( x) = 2( x − 1) − 6 use the method of translating 2 the parent function y = x 2 since it is written in a form such that a sequence of translations is easily identified. (b) The average rate of change from t = 3 to t = 10 is given by the following. N (10) − N (3) 118.079 − 111.244 ≈ 10 − 3 7 6.835 = 7 ≈ 0.976 77. (a) (b) Each year, the number of households in the United States increases by an average of 976,000 households. (c) Let t = 18: N (18) = − 0.068(18 − 13.68) + 119 2 (c) ≈ 117.7 In 2018, the number of households in the United States will be about 117.7 million households. Answers will vary. Sample answer: No, because the number of households has been increasing on average. 71. False. y = f ( − x) is a reflection in the y-axis. 72. False. y = − f ( x) is a reflection in the x-axis. 73. True. Because x = − x , the graphs of f ( x) = x + 6 and f ( x) = − x + 6 are identical. 74. False. The point ( − 2, − 61) lies on the transformation. 75. y = f ( x + 2) − 1 78. (a) Increasing on the interval ( − 2, 1) and decreasing on the intervals ( − ∞, − 2) and (1, ∞) (b) Increasing on the interval ( −1, 2) and decreasing on the intervals ( − ∞, −1) and ( 2, ∞) (c) Increasing on the intervals ( − ∞, −1) and ( 2, ∞) and decreasing on the interval ( −1, 2) (d) Increasing on the interval (0, 3) and decreasing on the intervals ( − ∞, 0) and (3, ∞) (e) Increasing on the intervals ( − ∞, 1) and ( 4, ∞) and Horizontal shift 2 units to the left and a vertical shift 1 unit downward (0, 1) → (0 − 2, 1 − 1) = (− 2, 0) (1, 2) → (1 − 2, 2 − 1) = (−1, 1) (2, 3) → (2 − 2, 3 − 1) = (0, 2) decreasing on the interval (1, 4) 79. (a) The profits were only 34 as large as expected: g (t ) = 34 f (t ) (b) The profits were $10,000 greater than predicted: g (t ) = f (t ) + 10,000 (c) There was a two-year delay: g (t ) = f (t − 2) INSTRUCTOR USE ONLY 80. No. g ( x) = − x 4 − 2. Yes. h( x) = − ( x − 3) 4 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 212 NOT FOR SALE Chapter 2 Functions ctions and Their Graphs Section 2.6 Combinations of Functions: Composite Functions 1. addition; subtraction; multiplication; division 2. composition 3. (a) x 0 1 2 3 f 2 3 1 2 g –1 0 1 2 0 3 3 2 2 f + g 6. f ( x) = 2 x − 5, g ( x) = 2 − x 1 ( f + g )( x) = 2 x − 5 + 2 − x = x − 3 (b) ( f − g )( x) = 2 x − 5 − ( 2 − x) = 2x − 5 − 2 + x = 3x − 7 (c) ( fg )( x) = (2 x − 5)(2 − x) = 4 x − 2 x 2 − 10 + 5 x = −2 x 2 + 9 x − 10 §f· 2x − 5 (d) ¨ ¸( x) = 2 − x ©g¹ Domain: all real numbers x except x = 2 7. f ( x ) = x 2 , g ( x) = 4 x − 5 (a) 4. ( f + g )( x) = f ( x) + g ( x) = x 2 + ( 4 x − 5) x –2 0 1 2 4 = x2 + 4 x − 5 f 2 0 1 2 4 (b) ( f − g )( x) = f ( x) − g ( x) g 4 2 1 0 2 f + g 6 2 2 2 6 = x 2 − ( 4 x − 5) = x2 − 4x + 5 (c) ( fg )( x) = f ( x) ⋅ g ( x) = x 2 ( 4 x − 5) = 4 x3 − 5 x 2 f ( x) §f· (d) ¨ ¸( x) = g ( x) ©g¹ = Domain: all real numbers x except x = 5. f ( x) = x + 2, g ( x) = x − 2 (a) x2 4x − 5 5 4 ( f + g )( x) = f ( x) + g ( x) = ( x + 2) + ( x − 2) = 2x (b) ( f − g )( x) = f ( x) − g ( x) = ( x + 2) − ( x − 2) = 4 (c) ( fg )( x) = f ( x) ⋅ g ( x) = ( x + 2)( x − 2) = x2 − 4 f ( x) x + 2 §f· (d) ¨ ¸( x) = = x g x x − 2 ( ) © ¹ INSTRUCTOR USE ONLY Domain: all real numbers x except xcept x = 2 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section on 2.6 8. f ( x) = 3x + 1, g ( x) = 5 x − 4 (a) 11. f ( x) = ( f + g )( x) = f ( x) + g ( x) = 8x − 3 (b) ( f − g )( x) = f ( x) − g ( x) = 3 x + 1 − (5 x − 4) ( fg )( x) = f ( x) ⋅ g ( x) = (3x + 1)(5 x − 4) (b) ( f − g )( x) = f ( x) − g ( x) = 1 1 x −1 − 2 = x x x2 12. f ( x) = Domain: all real numbers x except x = 4 5 ( f + g )( x) = f ( x) + g ( x) = x 2 + 6 + (b) ( f − g )( x) = f ( x) − g ( x) = x + 6 − ( fg )( x) = f ( x) ⋅ g ( x) = ( x 2 + 6) 1 − x ( x 2 + 6) 1 − x f ( x) §f· x2 + 6 (d) ¨ ¸( x) = = = g ( x) 1− x 1− x ©g¹ Domain: x 6, g ( x) contributes most to the magnitude. x , g ( x) = 2 x ( f + g )( x) = + 2 28. f ( x) = x x = (52 + 1)(5 − 4) + ( 42 + 1) = 26 ⋅ 1 + 17 = 43 25. f ( x) = 12 x, g ( x) = x − 1 ( f + g )( x) = 32 x − 1 g ( x) contributes most to the magnitude of the sum for 0 ≤ x ≤ 2. f ( x) contributes most to the magnitude of the sum for x > 6. 29. f ( x) = 3x + 2, g ( x) = − ( f + g ) x = 3x − x +5 x +5 + 2 For 0 ≤ x ≤ 2, f ( x) contributes most to the magnitude. For x > 6, f ( x) contributes most to the magnitude. INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. Section on 2.6 30. f ( x) = x 2 − 12 , g ( x) = −3x 2 − 1 ( f + g )( x) = −2 x 2 − 32 1 x 34. f ( x) = x3 , g ( x) = (a) For 0 ≤ x ≤ 2, g ( x) contributes most to the magnitude. 1 © x¹ ( f D g )( x) = f ( g ( x)) = f ( x − 1) = ( x − 1)2 (b) ( g D f )( x) = g ( f ( x)) = g ( x 2 ) = x 2 − 1 (c) ( g D g )( x) = g ( g ( x)) = g ( x − 1) = x − 2 1 = 20 − 3x (b) ( g D f )( x) = g ( f ( x)) Domain: all real numbers x ( f D g )( x) = f ( g ( x)) = f ( x 2 ) = = g 2 Domain: all real numbers x g ( x) = x3 + 1 Domain: all real numbers x (a) ( f D g )( x) = f ( g ( x)) = f ( x3 + 1) = 3 x3 + 1 − 5 ( g D g )( x) = g ( g ( x)) = g (5 − x) = x = 3 x3 − 4 Domain: all real numbers x 33. f ( x) = 3 x − 1, g ( x) = x3 + 1 (b) ( g D f )( x) = g ( f ( x)) ( f D g )( x) = f ( g ( x)) ( = f ( x3 + 1) = g 3 x −5 = 3 ( x3 + 1) − 1 = ( 37. f ( x) = x 2 + 1 ) g ( x) = ( x − 1) + 1 3 3 ( x − 5) + 1 3 Domain: all real numbers x = g 3 x −1 = ) 3 = x −5+1= x − 4 = 3 x3 = x = ( x − 1) + 1 = x (c) ( x + 4) = ( x + 4) = x + 4 36. f ( x) = 3 x − 5 = −3 x (b) ( g D f )( x) = g ( f ( x)) x2 + 4 (b) ( g D f )( x) = g ( f ( x)) = g (3 x + 5) = 5 − (3x + 5) (a) Domain: x ≥ −4 x + 4 Domain: x ≥ − 4 ( f D g )( x) = f ( g ( x)) = f (5 − x) = 3(5 − x) + 5 (c) 1 x3 Domain: all real numbers x 32. f ( x) = 3 x + 5, g ( x) = 5 − x (a) 1 x3 © x¹ 35. f ( x) = (a) 1 © x¹ ( g D g )( x) = g ( g ( x)) = g §¨ ·¸ = x g ( x) = x 2 31. f ( x) = x 2 , g ( x) = x − 1 (a) (c) 3 ( f D g )( x) = f ( g ( x)) = f §¨ ·¸ = §¨ ·¸ = (b) ( g D f )( x) = g ( f ( x)) = g ( x3 ) = For x > 6, g ( x) contributes most to the magnitude. 215 Combinations of Functions: Composite Compos Composi Functions ( g D g )( x) = g ( g ( x)) (a) Domain: all real numbers x Domain: x ≥ 0 x ( f D g )( x) = f ( g ( x)) ( x) = ( x) + 1 = f 2 = g ( x3 + 1) = x +1 = ( x3 + 1) + 1 3 = x9 + 3 x 6 + 3×3 + 2 Domain: x ≥ 0 (b) ( g D f )( x) = g ( f ( x)) = g ( x 2 + 1) = x2 + 1 INSTRUCTOR USE ONLY Domain: all real numbers x © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 216 Chapter 2 Functions ctions and Their Graphs 38. f ( x) = x 2 3 Domain: all real numbers x g ( x) = x 6 Domain: all real numbers x (a) ( f D g )( x) = f ( g ( x)) = f ( x6 ) = ( x 6 ) 39. f ( x) = x g ( x) = x + 6 23 (a) = x4 Domain: all real numbers x ( f D g )( x) = f ( g ( x)) = f ( x + 6) = x + 6 Domain: all real numbers x Domain: all real numbers x (b) ( g D f )( x ) = g ( f ( x)) = g ( x ) = x + 6 (b) ( g D f )( x) = g ( f ( x)) = g ( x 2 3 ) = ( x 2 3 ) = x 4 6 Domain: all real numbers x Domain: all real numbers x 40. f ( x) = x − 4 Domain: all real numbers x g ( x) = 3 − x Domain: all real numbers x (a) Domain: all real numbers x ( f D g )( x) = f ( g ( x)) = f (3 − x) = (3 − x) − 4 = − x − 1 Domain: all real numbers x (b) ( g D f )( x) = g ( f ( x)) = g ( x − 4 ) = 3 − ( x − 4 ) = 3 − x − 4 Domain: all real numbers x 41. f ( x ) = 1 x g ( x) = x + 3 (a) Domain: all real numbers x except x = 0 Domain: all real numbers x ( f D g )( x) = f ( g ( x)) = f ( x + 3) = 1 x +3 Domain: all real numbers x except x = −3 1 §1· +3 (b) ( g D f )( x) = g ( f ( x)) = g ¨ ¸ = x © x¹ Domain: all real numbers x except x = 0 42. f ( x ) = 3 x2 − 1 g ( x) = x + 1 (a) Domain: all real numbers x except x = ±1 Domain: all real numbers x ( f D g )( x) = f ( g ( x)) = f ( x + 1) = 3 ( x + 1) − 1 2 = 3 3 = 2 x2 + 2 x + 1 − 1 x + 2x Domain: all real numbers x except x = 0 and x = − 2 3 3 + x2 − 1 x2 + 2 § 3 · +1= = 2 (b) ( g D f )( x) = g ( f ( x)) = g ¨ 2 ¸ = 2 2 x −1 x −1 x −1 © x − 1¹ Domain: all real numbers x except x = ±1 43. (a) ( f + g )(3) = f (3) + g (3) = 2 + 1 = 3 f ( 2) §f· 0 (b) ¨ ¸( 2) = = = 0 g ( 2) 2 ©g¹ 44. (a) 45. (a) ( f D g )(2) = f ( g (2)) = f (2) = 0 (b) ( g D f )( 2) = g ( f ( 2)) = g (0) = 4 ( f − g )(1) = f (1) − g (1) = 2 − 3 = −1 (b) ( fg )( 4) = f ( 4) ⋅ g ( 4) = 4 ⋅ 0 = 0 46. (a) ( f D g )(1) = f ( g (1)) = f (3) = 2 (b) ( g D f )(3) = g ( f (3)) = g ( 2) = 2 INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section on 2.6 47. h( x) = ( 2 x 2 + 1) 2 then ( f D g )( x) = h( x). (5 x + 2) 2 One possibility: Let g ( x) = 5 x + 2 and f ( x ) = 4 , x2 then ( f D g )( x) = h( x). 3 One possibility: Let g ( x) = 1 − x and f ( x ) = x 3 , then ( f D g )( x) = h( x). One possibility: Let f ( x ) = − x2 + 3 4 − x2 53. h( x) = One possibility: Let f ( x ) = 49. h( x) = 3 x 2 − 4 3 x and g ( x) = x − 4, then ( f D g )( x) = h( x). x +3 and g ( x) = − x 2 , 4 + x 2 then ( f D g )( x) = h( x). 50. h( x) = 217 4 52. h( x) = One possibility: Let f ( x) = x 2 and g ( x) = 2 x + 1, 48. h( x) = (1 − x) Combinations of Functions: Composite Compos Composi Functions 54. h( x) = 27 x 3 + 6 x 10 − 27 x 3 One possibility: Let g ( x) = x3 and 9− x One possibility: Let g ( x) = 9 − x and f ( x ) = x, then ( f D g )( x) = h( x). f ( x) = 27 x + 6 3 x , then ( f D g )( x) = h( x). 10 − 27 x 1 x2 55. (a) T ( x) = R( x) + B( x) = 34 x + 15 1 51. h( x) = x + 2 (b) One possibility: Let f ( x) = 1 x and g ( x) = x + 2, then ( f D g )( x) = h( x). (c) B( x); As x increases, B( x) increases at a faster rate. 56. (a) c(t ) = b (t ) − d (t ) p (t ) × 100 (b) c(5) represents the percent change in the population due to births and deaths in the year 2005. 57. (a) p(t ) = d (t ) + c(t ) (b) p(5) represents the number of dogs and cats in 2005. (c) h(t ) = p (t ) n (t ) = d (t ) + c ( t ) n (t ) h(t ) represents the number of dogs and cats at time t compared to the population at time t or the number of dogs and cats per capita. 58. (a) T is a function of t since for each time t there corresponds one and only one temperature T. (b) T ( 4) ≈ 60°; T (15) ≈ 72° (c) H (t ) = T (t − 1); All the temperature changes would be one hour later. (d) H (t ) = T (t ) − 1; The temperature would be decreased by one degree. INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 218 NOT FOR SALE Chapter 2 Functions ctions and Their Graphs (e) The points at the endpoints of the individual functions that form each “piece” appear to be (0, 60), (6, 60), (7, 72), (20, 72), ( 21, 60), and (24, 60). Note that the value t = 24 is chosen for the last ordered pair because that is when the day ends and the cycle starts over. From t = 0 to t = 6: This is the constant function T (t ) = 60. From t = 6 to t = 7: Use the points (6, 60) and (7, 72). 72 − 60 = 12 7 −6 y − 60 = 12( x − 6) Ÿ y = 12 x − 12, or m = T (t ) = 12t − 12 From t = 7 to t = 20: This is the constant function T (t ) = 72. From t = 20 to t = 21: Use the points ( 20, 72) and ( 21, 60). 72 − 60 = −12 20 − 21 y − 60 = −12( x − 21) Ÿ y = −12 x + 312, or T (t ) = −12t + 312 m = From t = 21 to t = 24: This is the constant function T (t ) = 60. ­60, ° °12t − 12, ° A piecewise-defined function is T (t ) = ®72, °− ° 12t + 312, °¯60, 59. (a) r ( x ) = 6 < t < 7 7 ≤ t ≤ 20 20 < t < 21 21 ≤ t ≤ 24 61. (a) f ( g ( x)) = f (0.03 x) = 0.03x − 500,000 x 2 (b) g ( f ( x)) = g ( x − 500,000) = 0.03( x − 500,000) (b) A( r ) = π r 2 (c) 0 ≤ t ≤ 6 ( A D r )( x) = A(r ( x)) = A§¨ ·¸ = 𠧨 ·¸ x © 2¹ x © 2¹ g ( f ( x)) represents your bonus of 3% of an amount 2 ( A D r )( x) represents the area of the circular base of the tank on the square foundation with side length x. over $500,000. 62. (a) R( p ) = p − 2000 the cost of the car after the factory rebate. (b) S ( p ) = 0.9 p the cost of the car with the dealership 60. (a) N (T (t )) = N (3t + 2) = 10(3t + 2) − 20(3t + 2) + 600 discount. 2 = 10(9t 2 + 12t + 4) − 60t − 40 + 600 = 90t 2 + 60t + 600 = 30(3t 2 + 2t + 20), 0 ≤ t ≤ 6 This represents the number of bacteria in the food as a function of time. (b) Use t = 0.5. ( ) N (T (0.5)) = 30 3(0.5) + 2(0.5) + 20 = 652.5 2 After half an hour, there will be about 653 bacteria. (c) 30(3t 2 + 2t + 20) = 1500 3t 2 + 2t + 20 = 50 3t 2 + 2t − 30 = 0 By the Quadratic Formula, t ≈ −3.513 or 2.846. (c) ( R D S )( p) = R(0.9 p) = 0.9 p − 2000 ( S D R)( p) = S ( p − 2000) = 0.9( p − 2000) = 0.9 p − 1800 ( R D S )( p) represents the factory rebate after the dealership discount. ( S D R)( p) represents the dealership discount after the factory rebate. (d) ( R D S )( p) = ( R D S )( 20,500) = 0.9( 20,500) − 2000 = $16,450 ( S D R)( p) = ( S D R)(20,500) = 0.9( 20,500) − 1800 = $16,650 ( R D S )(20,500) yields the lower cost because 10% of the price of the car is more than $2000. INSTRUCTOR USE ONLY Choosing the positive value for t, you have t ≈ 2.846 hours. © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section on 2.6 Combinations of Functions: Composite Compos Composi Functions 219 63. Let O = oldest sibling, M = middle sibling, Y = youngest sibling. Then the ages of each sibling can be found using the equations: O = 2M M = 12 Y + 6 ( ) (a) O( M (Y )) = 2 12 (Y ) + 6 = 12 + Y ; Answers will vary. (b) Oldest sibling is 16: O = 16 Middle sibling: O = 2M 16 = 2 M M = 8 years old Youngest sibling: M = 12 Y + 6 8 = 12 Y + 6 2 = 12 Y Y = 4 years old ( ) 64. (a) Y ( M (O )) = 2 12 O − 12 = O − 12; Answers will vary. (b) Youngest sibling is 2 → Y = 2 Middle sibling: M = 12 Y + 6 M = 12 ( 2) + 6 M = 7 years old Oldest sibling: O = 2M O = 2(7) O = 14 years old 65. False. ( f D g )( x) = 6 x + 1 and ( g D f )( x) = 6 x + 6 66. True. The range of g must be a subset of the domain of f for ( f D g )( x) to be defined. 68. (a) f ( p ): matches L 2 ; For example, an original price of p = $15.00 corresponds to a sale price of S = $7.50. (b) g ( p ): matches L1; For example an original price of p = $20.00 corresponds to a sale price of S = $15.00. 67. Let f ( x) and g ( x) be two odd functions and define h( x) = f ( x) g ( x). Then (c) h( − x ) = f ( − x ) g ( − x ) = ª− ºª− g ( x)¼º ¬ f ( x)¼¬ because f and g are odd = f ( x) g ( x) So, h( x) is even. Let f ( x) and g ( x) be two even functions and define h( x) = f ( x) g ( x). Then So, h( x) is even. (d) ( f D g ) ( p) matches L 3 ; This function represents 69. Let f ( x) be an odd function, g ( x) be an even function, and define h( x) = f ( x) g ( x). Then h( − x ) = f ( − x ) g ( − x ) h( − x ) = f ( − x ) g ( − x ) = h( x). applying a 50% discount to the original price p, then subtracting a $5 discount. subtracting a $5 discount from the original price p, then applying a 50% discount. = h( x). = f ( x) g ( x ) ( g D f ) ( p): matches L 4 ; This function represents because f and g are even = ª− ¬ f ( x)¼º g ( x) because f is odd and g is even = − f ( x ) g ( x) = − h( x). So, h is odd and the product of an odd function and an even ven function is odd. INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 220 Chapter 2 Functions ctions and Their Graphs 1 ª f ( x) + f ( − x)¼º 2¬ 70. (a) g ( x) = (c) To determine if g ( x) is even, show g ( − x) = g ( x). 1 ª f ( − x) + f ( − ( − x))º¼ 2¬ 1 = ª¬ f ( − x) + f ( x)º¼ 2 1 = ª¬ f ( x) + f ( − x)º¼ 2 g ( − x) = = g ( x) 9 1 ª f ( x) − f ( − x)¼º 2¬ h( x ) = To determine if h( x) is odd show h( − x) = − h( x). 1 ª f ( − x) − f ( − ( − x))º ¼ 2¬ 1 = ª¬ f ( − x) − f ( x)º¼ 2 1 = − ª¬ f ( x) − f ( − x)¼º 2 h( − x ) = = − h( x ) 9 (b) Let f ( x) = a function f ( x) = x 2 − 2 x + 1 f ( x ) = g ( x ) + h( x ) 1 ª f ( x) + f ( − x)¼º 2¬ 1 2 = ª x 2 − 2 x + 1 + ( − x) − 2( − x) + 1º ¼ 2¬ 1 2 2 = ª¬ x − 2 x + 1 + x + 2 x + 1º¼ 2 1 = ª¬2 x 2 + 2º¼ = x 2 + 1 2 1 h( x) = ¬ª f ( x) − f ( − x)¼º 2 1ª 2 2 = « x − 2 x + 1 − ( − x) − 2( − x) + 1 º» ¼ 2¬ 1 2 2 = ª¬ x − 2 x + 1 − x − 2 x − 1º¼ 2 1 = [− 4 x] = −2 x 2 g ( x) = ( f ( x) = ( x 2 + 1) + ( − 2 x) 1 x +1 k ( x ) = g ( x ) + h( x ) k ( x) = Using the result from part (a) g ( x) is an even 1 ªk ( x) + k ( − x)¼º 2¬ 1ª 1 1 º = « + 2 ¬ x + 1 − x + 1»¼ function and h( x) is an odd function. = 1 ª1 − x + x + 1º 2 «¬ ( x + 1)(1 − x) »¼ 1 1 ª f ( x) + f ( − x)¼º + ¬ª f ( x) − f ( − x)¼º 2¬ 2 1 1 1 1 = f ( x ) + f ( − x) + f ( x) − f ( − x) 2 2 2 2 = 1ª 2 º 2 «¬ ( x + 1)(1 − x) »¼ = f ( x) 9 = f ( x) = even function + odd function. g ( x) = f ( x ) = g ( x ) + h( x ) = ) = 1 ( x + 1)(1 − x) −1 ( x + 1)( x − 1) 1 ªk ( x) − k ( − x)º¼ 2¬ 1ª 1 1 º = « − 2 ¬ x + 1 1 − x »¼ 1 ª1 − x − ( x + 1) º = « » 2 «¬ ( x + 1)(1 − x) »¼ º − 2x 1ª = « » 2 ¬« ( x + 1)(1 − x) ¼» −x = ( x + 1)(1 − x) h( x ) = = x ( x + 1)( x − 1) § · § · −1 x + k ( x) = ¨ ¨ ( x + 1)( x − 1) ¸¸ ¨¨ ( x + 1)( x − 1) ¸¸ © ¹ © ¹ INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.7 Inver Inverse Functions 9. f ( x) = 3x + 1 Section 2.7 Inverse Functions f −1 ( x) = 1. inverse x −1 3 § x − 1· § x − 1· f ( f −1 ( x)) = f ¨ ¸ = 3¨ ¸ +1 = x © 3 ¹ © 3 ¹ 2. f −1 3. range; domain f −1 ( f ( x)) = f −1 (3 x + 1) = 4. y = x 10. f ( x) = 5. one-to-one f −1 ( x) = 5 x + 1 7. f ( x) = 6 x f ( f −1 ( x)) = f (5 x + 1) = x 1 = x 6 6 § x· § x· −1 f ( f ( x)) = f ¨ ¸ = 6¨ ¸ = x ©6¹ ©6¹ f −1 ( x) = f −1 ( f ( x)) = f −1 (6 x) = 3 5x + 1 − 1 5x = = x 5 5 § x − 1· § x − 1· f −1 ( f ( x)) = f −1 ¨ ¸ = 5¨ ¸ +1 5 © ¹ © 5 ¹ = x −1+1 = x 6x = x 6 11. f ( x) = 3 x f −1 ( x) = x3 8. f ( x) = 13 x f ( f −1 ( x)) = f ( x3 ) = 3 x3 = x f −1 ( x) = 3 x −1 (3x + 1) − 1 = x x −1 5 6. Horizontal f(f 221 ( ) ( x) = x f −1 ( f ( x)) = f −1 3 x = ( x)) = f (3x) = ( ) = x 1 3x 3 ( ) ( ) 3 3 12. f ( x) = x5 f −1 ( f ( x)) = f −1 13 x = 3 13 x = x f −1 ( x) = 5 x f ( f −1 ( x)) = f ( x) = ( x) = x 5 5 5 f −1 ( f ( x)) = f −1 ( x5 ) = 5 x5 = x 7§ 2x + 6 · § 2x + 6 · 13. ( f D g )( x) = f ( g ( x)) = f ¨ − ¸ = − ¨− ¸ −3 = x +3−3 = x 7 ¹ 2© 7 ¹ © § 7 · 2¨ − x − 3¸ + 6 − ( − 7 x) 7 2 § · © ¹ = = x ( g D f )( x) = g ( f ( x)) = g ¨ − x − 3¸ = − 7 7 © 2 ¹ 4x + 9 − 9 4x = = x 4 4 x − 9· § x − 9· ( g D f )( x) = g ( f ( x)) = g §¨ ¸ = 4¨ ¸ +9 = x −9+9 = x © 4 ¹ © 4 ¹ 14. ( f D g )( x) = f ( g ( x)) = f ( 4 x + 9) = 15. ( f D g )( x) = f ( g ( x)) = f ( x − 5) = ( x − 5) + 5 = x − 5 + 5 = x 3 3 3 ( g D f )( x) = g ( f ( x)) = g ( x3 + 5) = 3 x3 + 5 − 5 = 3 x3 = x 16. ( f D g )( x) = f ( g ( x)) = f ( 3 ) 2x = ( 2x ) = 2x = x 3 3 2 2 §x · §x · 3 3 ¸ = 3 2¨ ¸ = x = x ©2¹ ©2¹ ( g D f )( x) = g ( f ( x)) = g ¨ 3 3 INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 222 Chapter 2 17. y NOT FOR SALE Functions ctions and Their Graphs 22. f ( x) = x − 5, g ( x) = x + 5 (a) f ( g ( x)) = f ( x + 5) = ( x + 5) − 5 = x 3 2 g ( f ( x)) = g ( x − 5) = ( x − 5) + 5 = x 1 −3 −1 (b) x 1 2 3 −1 −2 −3 18. y 7 6 5 4 3 23. f ( x) = 7 x + 1, g ( x) = 2 1 x −1 −1 1 19. 2 3 4 5 6 x −1 7 § x − 1· § x − 1· (a) f ( g ( x)) = f ¨ ¸ = 7¨ ¸ +1= x 7 © ¹ © 7 ¹ 7 g ( f ( x)) = g (7 x + 1) = y 4 (7 x + 1) − 1 = x 7 (b) 3 2 1 x −1 1 2 3 4 −1 20. y 3 24. f ( x) = 3 − 4 x, g ( x) = 2 1 −3 −2 §3 − x· §3 − x· (a) f ( g ( x)) = f ¨ ¸ = 3 − 4¨ ¸ © 4 ¹ © 4 ¹ x −1 1 2 3 = 3 − (3 − x ) = x −2 −3 21. f ( x) = 2 x, g ( x) = 3− x 4 g ( f ( x )) = g (3 − 4 x ) = x 2 3 − (3 − 4 x ) 4 = 4x = x 4 (b) § x· § x· (a) f ( g ( x)) = f ¨ ¸ = 2¨ ¸ = x © 2¹ © 2¹ 2x = x g ( f ( x )) = g ( 2 x ) = 2 (b) INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.7 25. f ( x) = 27. f ( x) = x3 , g ( x) = 3 8 x 8 (a) f ( g ( x)) = f ( 8x ) = 3 ( 8x ) = 8x = x 8 223 x − 4, g ( x) = x 2 + 4, x ≥ 0 (a) f ( g ( x)) = f ( x 2 + 4), x ≥ 0 3 3 Inver Inverse Functions = 8 ( x 2 + 4) − 4 = x ( x − 4) = ( x − 4) + 4 = x g ( f ( x)) = g § x3 · § x3 · g ( f ( x)) = g ¨ ¸ = 3 8¨ ¸ = 3 x3 = x ©8¹ ©8¹ 2 (b) (b) 26. f ( x) = 1 1 , g ( x) = x x 28. f ( x) = 1 − x3 , g ( x) = 3 1 − x x 1 1 §1· (a) f ( g ( x)) = f ¨ ¸ = =1÷ = 1⋅ = x 1 1 x x x © ¹ (a) f ( g ( x)) = f ( 1 − x) = 1 − ( 1 − x) 3 3 3 = 1 − (1 − x) = x 1 1 x §1· g ( f ( x )) = g ¨ ¸ = =1÷ = 1⋅ = x 1 1 x x x © ¹ g ( f ( x)) = g (1 − x3 ) = 3 1 − (1 − x3 ) (b) = 3 x3 = x (b) 29. f ( x) = 9 − x 2 , x ≥ 0; g ( x) = (a) f ( g ( x)) = f 9 − x, x ≤ 9 ( 9 − x ), x ≤ 9 = 9 − ( 9 − x ) = x 2 g ( f ( x)) = g (9 − x 2 ), x ≥ 0 = 9 − (9 − x 2 ) = x (b) INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 224 Chapter 2 30. f ( x) = NOT FOR SALE Functions ctions and Their Graphs 1 1− x , x ≥ 0; g ( x) = ,0 < x ≤ 1 1+ x x 1 1 1 §1 − x · = = = x (a) f ( g ( x)) = f ¨ ¸ = 1 x 1− x §1 − x · © x ¹ + 1+¨ ¸ x x x © x ¹ § 1 · g ( f ( x )) = g ¨ ¸ = ©1 + x ¹ § 1 · 1+ x 1 x 1−¨ − ¸ ©1 + x ¹ = 1 + x 1 + x = 1 + x = x ⋅ x + 1 = x 1 1 1+ x 1 § 1 · ¨ ¸ x x + + 1 1 ©1 + x ¹ (b) 31. f ( x) = x −1 5x + 1 , g ( x) = − x +5 x −1 § 5x + 1 · − 1¸ ¨− − (5 x + 1) − ( x − 1) x −1 − 6x x −1 § 5x + 1· © ¹ (a) f ( g ( x)) = f ¨ − ⋅ = = = x ¸ = − (5 x + 1) + 5( x − 1) −6 § 5x + 1 · x −1 © x −1¹ 5 − + ¨ ¸ © x −1 ¹ ª § x − 1· º «5¨ x + 5 ¸ + 1» x + 5 5( x − 1) + ( x + 5) 6x § x − 1· © ¹ ¬ ¼ ⋅ g ( f ( x )) = g ¨ = − = − = x ¸ = − x −1 5 5 1 5 6 x x x x + + − − + − ª º ( ) ( ) © ¹ 1 − «¬ x + 5 »¼ (b) INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.7 32. f ( x) = Inver Inverse Functions 225 x +3 2x + 3 , g ( x) = x − 2 x −1 2x + 3 2 x + 3 + 3x − 3 +3 5x § 2x + 3 · x x −1 1 − (a) f ( g ( x)) = f ¨ = = = x ¸ = 2x + 3 2x + 3 − 2x + 2 5 © x −1¹ − 2 x −1 x −1 § x + 3· 2 6 + 3x − 6 x + 2¨ ¸ +3 5x x − 2¹ § x + 3· © x − 2 g ( f ( x)) = g ¨ = = = x ¸ = x +3 3 2 + − + x x 2 5 − x © ¹ −1 x − 2 x − 2 (b) 33. No, {( −2, −1), (1, 0), ( 2, 1), (1, 2), ( −2, 3), ( −6, 4)} does not represent a function. −2 and 1 are paired with two different values. 42. f ( x) = 18 ( x + 2) − 1 2 34. Yes, {(10, − 3), (6, − 2), ( 4, −1), (1, 0), ( −3, 2), (10, 2)} does represent a function. 35. 36. f does not pass the Horizontal Line Test, so f does not have an inverse. x –2 0 2 4 6 8 f −1 ( x) –2 –1 0 1 2 3 x –10 –7 –4 –1 2 5 –3 –2 –1 0 1 2 f −1 ( x) 43. f ( x) = −2 x 16 − x 2 37. Yes, because no horizontal line crosses the graph of f at more than one point, f has an inverse. 38. No, because some horizontal lines intersect the graph of f twice, f does not have an inverse. f does not pass the Horizontal Line Test, so f does not have an inverse. 44. h( x) = x + 4 − x − 4 39. No, because some horizontal lines cross the graph of f twice, f does not have an inverse. 40. Yes, because no horizontal lines intersect the graph, of f at more than one point, f has an inverse. 41. g ( x) = ( x + 5) h does not pass the Horizontal Line Test, so h does not have an inverse. 3 g passes the Horizontal Line Test, so g has an inverse. INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 226 Chapter 2 NOT FOR SALE Functions ctions and Their Graphs f ( x) = 2 x − 3 45. (a) (b) f ( x) = 49. (a) y = 2x − 3 x = 2y − 3 x +3 y = 2 x +3 −1 f ( x) = 2 4 − x2 x = 4 − y2 y2 = 4 − x2 y = (d) The domains and ranges of f and f −1 are all real numbers. f ( x) = 3 x + 1 y = x2 = 4 − y2 (c) The graph of f −1 is the reflection of the graph of f in the line y = x. 46. (a) 4 − x2 , 0 ≤ x ≤ 2 f −1 ( x) = 4 − x2 4 − x2 , 0 ≤ x ≤ 2 (b) (b) y = 3x + 1 x = 3y + 1 x −1 = y 3 x −1 f −1 ( x) = 3 (c) The graph of f −1 is the same as the graph of f. (c) The graph of f −1 is the reflection of f in the line y = x. (d) The domains and ranges of f and f numbers. f ( x) = x − 2 5 47. (a) −1 are all real (d) The domains and ranges of f and f −1 are all real numbers x such that 0 ≤ x ≤ 2. f ( x) = x 2 − 2, x ≤ 0 50. (a) y = x2 − 2 x = y2 − 2 (b) ± y = x5 − 2 x + 2 = y f −1 ( x) = − x = y5 − 2 x + 2 (b) y = 5 x + 2 f −1 ( x) = 5 x + 2 (c) The graph of f −1 is the reflection of the graph of f in the line y = x. (d) The domains and ranges of f and f −1 are all real numbers. f ( x) = x + 1 3 48. (a) (b) (d) [− 2, ∞) is the range of f and domain of f −1. 3 y = x +1 x = y3 + 1 x −1 = y 3 (c) The graph of f −1 is the reflection of f in the line y = x. (− ∞, 0] is the domain of f and the range of f −1. 3 x −1 = y f −1 ( x) = 3 x − 1 (c) The graph of f −1 is the reflection of f in the line y = x. (d) The domains and ranges of f and f −1 are all real numbers. INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.7 51. (a) 4 x 4 y = x 4 x = y f ( x) = x −3 x + 2 x −3 y = x + 2 y −3 x = y + 2 xy + 2 x − y + 3 = 0 xy = 4 y ( x − 1) = − 2 x − 3 4 y = x 4 −1 f ( x) = x − 2x − 3 x −1 − 2x − 3 −1 f ( x) = x −1 y = (c) The graph of f −1 is the same as the graph of f. (d) The domains and ranges of f and f −1 are all real numbers except for 0. 52. (a) 2 x 2 y = − x 2 x = − y f ( x) = − (b) (b) (c) The graph of f −1 is the reflection of the graph of f in the line y = x. 2 x 2 f −1 ( x) = − x y = − (d) The domain of f and the range of f −1 is all real numbers except x = − 2. (c) The graphs are the same. The range of f and the domain of f −1 is all real numbers x except x = 1. (d) The domains and ranges of f and f −1 are all real numbers except for 0. 53. (a) x +1 x − 2 x +1 y = x − 2 y +1 x = y − 2 f ( x) = 227 f ( x) = 54. (a) (b) Inver Inverse Functions 55. (a) (b) f ( x) = 3 x − 1 y = 3 (b) x −1 x = 3 y −1 x3 = y − 1 y = x3 + 1 f −1 ( x) = x3 + 1 x ( y − 2) = y + 1 xy − 2 x = y + 1 (c) The graph of f −1 is the reflection of the graph of f in the line y = x. xy − y = 2 x + 1 y ( x − 1) = 2 x + 1 (d) The domains and ranges of f and f −1 are all real numbers. 2x + 1 y = x −1 2x + 1 −1 f ( x) = x −1 (c) The graph of f −1 is the reflection of graph of f in the line y = x. (d) The domain of f and the range of f −1 is all real numbers except 2. The range of f and the domain of f −1 is all real numbers except 1. INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 228 NOT FOR SALE Chapter 2 56. (a) Functions ctions and Their Graphs f ( x) = x3 5 y = x 60. (b) y = 3x + 5 x = 3y + 5 x = y3 5 x5 3 = ( y 3 5 ) x − 5 = 3y x −5 = y 3 53 x5 3 = y This is a function of x, so f has an inverse. f −1 ( x) = x5 3 (c) The graph of f −1 is the reflection of the graph of f in the line y = x. (d) The domains and ranges of f and f −1 are all real numbers. f −1 ( x) = x −5 3 61. p( x) = − 4 y = −4 Because y = − 4 for all x, the graph is a horizontal line 57. f ( x) = x 4 and fails the Horizontal Line Test. p does not have an inverse. y = x4 x = y4 62. y = ±4 x This does not represent y as a function of x. f does not have an inverse. 1 x2 1 y = 2 x 1 x = 2 y 1 y2 = x 58. f ( x) = y = ± f ( x) = 3 x + 5 35 3x + 4 5 3x + 4 y = 5 3y + 4 x = 5 5x = 3 y + 4 5x − 4 = 3 y 5x − 4 = y 3 f ( x) = This is a function of x, so f has an inverse. f −1 ( x) = 1 x 5x − 4 3 63. f ( x) = ( x + 3) , x ≥ − 3 Ÿ y ≥ 0 2 This does not represent y as a function of x. f does not have an inverse. y = ( x + 3) , x ≥ − 3, y ≥ 0 2 x = ( y + 3) , y ≥ − 3, x ≥ 0 2 x 59. g ( x) = 8 x y = 8 y x = 8 y = 8x x = y + 3, y ≥ − 3, x ≥ 0 y = x − 3, x ≥ 0, y ≥ − 3 This is a function of x, so f has an inverse. f −1 ( x) = This is a function of x, so g has an inverse. 64. x − 3, x ≥ 0 q( x) = ( x − 5) 2 y = ( x − 5) 2 g −1 ( x) = 8 x x = ( y − 5) ± 5± 2 x = y −5 x = y This does not represent y as a function of x, so q does not have an inverse. INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Section 2.7 Inver Inverse Functions 229 3 2x + 3 Ÿ x ≥ − , y ≥ 0 2 3 y = 2 x + 3, x ≥ − , y ≥ 0 2 3 x = 2 y + 3, y ≥ − , x ≥ 0 2 3 x 2 = 2 y + 3, x ≥ 0, y ≥ − 2 2 x −3 3 y = , x ≥ 0, y ≥ − 2 2 ­x + 3, x 0 f −1 ( x) = x2 − 3 ,x ≥ 0 2 f ( x) = 70. x − 2 Ÿ x ≥ 2, y ≥ 0 y = x − 2, x ≥ 2, y ≥ 0 x = y − 2, y ≥ 2, x ≥ 0 2 x = y − 2, x ≥ 0, y ≥ 2 2 x + 2 = y, x ≥ 0, y ≥ 2 This is a function of x, so f has an inverse. f −1 ( x) = x 2 + 2, x ≥ 0 The graph fails the Horizontal Line Test, so f does not have an inverse. 67. h( x) = − 4 x2 y 71. 6x + 4 4x + 5 6x + 4 y = 4x + 5 6y + 4 x = 4y + 5 f ( x) = x( 4 y + 5) = 6 y + 4 2 4 xy + 5 x = 6 y + 4 4 xy − 6 y = − 5 x + 4 x −4 4 y( 4 x − 6) = − 5 x + 4 −2 − 5x + 4 4x − 6 5x − 4 = 6 − 4x y = The graph fails the Horizontal Line Test so h does not have an inverse. 68. f ( x) = x − 2 , x ≤ 2 Ÿ y ≥ 0 This is a function of x, so f has an inverse. f −1 ( x) = 5x − 4 6 − 4x y = x − 2 , x ≤ 2, y ≥ 0 x = y − 2 , y ≤ 2, x ≥ 0 x = y − 2 or 2+ x = y −x = y − 2 or 2 − x = y The portion that satisfies the conditions y ≤ 2 and x ≥ 0 is 2 − x = y. This is a function of x, so f has an inverse. INSTRUCTOR USE ONLY f −1 ( x) = 2 − x, x ≥ 0 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 230 Chapter 2 NOT FOR SALE Functions ctions and Their Graphs 72. The graph of f passes the Horizontal Line Test. So, you know f is one-to-one and has an inverse function. 5x − 3 2x + 5 5x − 3 y = 2x + 5 5y − 3 x = 2y + 5 f ( x) = 75. f ( x) = x + 2 domain of f : x ≥ − 2, range of f : y ≥ 0 f ( x) = x + 2 y = x + 2 x = y + 2 x − 2 = y x( 2 y + 5) = 5 y − 3 So, f −1 ( x) = x − 2. 2 xy + 5 x = 5 y − 3 domain of f −1 : x ≥ 0, range of f −1 : y ≥ − 2 2 xy − 5 y = − 5 x − 3 76. f ( x) = x − 5 y ( 2 x − 5) = − (5 x + 3) domain of f : x ≥ 5, range of f : y ≥ 0 5x + 3 2x − 5 5x + 3 f −1 ( x) = − 2x − 5 y = − 73. f ( x) = ( x − 2) f ( x) = x − 5 y = x −5 x = y −5 2 x +5 = y domain of f : x ≥ 2, range of f : y ≥ 0 f ( x ) = ( x − 2) 2 y = ( x − 2) 2 x = ( y − 2) domain f −1 : x ≥ 0, range of f −1 : y ≥ 5 77. f ( x) = ( x + 6) 2 x + 2 = y x + 2. domain of f −1 : x ≥ 0, range of f −1 : x ≥ 2 74. f ( x) = 1 − x 4 f ( x ) = ( x + 6) 2 y = ( x + 6) 2 x = ( y + 6) 2 x = y + 6 x −6 = y domain of f : x ≥ 0, range of f : y ≤ 1 f ( x) = 1 − x 4 So, f −1 ( x) = x − 6. domain of f −1 : x ≥ 0, range of f −1 : y ≥ − 6 y = 1 − x4 78. f ( x) = ( x − 4) x = 1 − y4 2 domain of f : x ≥ 4, range of f : y ≥ 0 x − 1 = − y4 4 2 domain of f : x ≥ − 6, range of f : y ≥ 0 x = y − 2 So, f −1 ( x) = So, f −1 ( x) = x + 5. 1− x = y f ( x ) = ( x − 4) 2 So, f −1 ( x) = 4 1 − x . y = ( x − 4) 2 domain of f −1 : x ≤ 1, range of f −1 : y ≥ 0 x = ( y − 4) 2 x = y − 4 x + 4 = y So, f −1 ( x) = x + 4. domain of f −1 : x ≥ 0, range of f −1 : y ≥ 4 INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. Section 2.7 domain of f : x ≥ 1, range of f : y ≤ − 2 domain of f : x ≥ 0, range of f : y ≤ 5 f ( x) = − 2 x 2 + 5 f ( x) = − x − 1 − 2 y = − 2×2 + 5 y = − x −1 − 2 x = −( y − 1) − 2 x = − 2 y2 + 5 x − 5 = −2y x = −y − 1 2 5 − x = 2 y2 −x − 1 = y 5− x = y 2 So, f −1 ( x) = − x − 1. 5− x ⋅ 2 2 − 2( x − 5) 2 −1 In Exercises 83– 88, f ( x ) = 18 x – 3, f –1 ( x ) = 8( x + 3), g ( x ) = x 3 , g –1 ( x ) = 3 x . = y So, f −1 ( x) = domain of f domain of f −1 : x ≤ − 2, range of f −1 : y ≥ 1 2 = y 2 2(5 − x) 83. ( f −1 D g −1 )(1) = f −1 ( g −1 (1)) . ( x): x ≤ 5, range of f ( x): y ≥ 0 domain of f : x ≥ 0, range of f : y ≥ −1 ( 84. ( g −1 D f −1 )( −3) = g −1 ( f −1 ( − 3)) = g −1 (8( − 3 + 3)) = g −1 (0) = 3 0 = 0 85. ( f −1 D f −1 )(6) = f −1 ( f −1 (6)) x = 12 y 2 − 1 = f −1 (8[6 + 3]) x + 1 = 12 y 2 = 8¬ª8(6 + 3) + 3¼º = 600 2x + 2 = y2 2x + 2 = y So, f ( x) = domain of f −1 ) = 8 3 1 + 3 = 32 1 2 x −1 2 y = 12 x 2 − 1 −1 ( ) = f −1 3 1 −1 1 x2 − 1 2 f ( x) = 231 82. f ( x) = − x − 1 − 2 79. f ( x) = − 2 x 2 + 5 80. f ( x) = Inver Inverse Functions 86. ( g −1 D g −1 )( − 4) = g −1 ( g −1 ( − 4)) 2 x + 2. ( = g −1 3 − 4 : x ≥ −1, range of f −1 :y ≥ 0 81. f ( x) = x − 4 + 1 domain of f : x ≥ 4, range of f : y ≥ 1 ) = 3 3 −4 = 9 −4 87. ( f D g )( x) = f ( g ( x)) = f ( x3 ) = 18 x3 − 3 y = 18 x3 − 3 f ( x) = x − 4 + 1 x = 18 y 3 − 3 y = x −3 x + 3 = 18 y 3 x = y −3 x +3 = y 8( x + 3) = y 3 So, f −1 ( x) = x + 3. ( ) = y 3 8 x + 3 domain of f −1 : x ≥ 1, range of f −1 : y ≥ 4 −1 ( f D g ) ( x) = 2 3 x + 3 INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 232 Chapter 2 NOT FOR SALE Functions ctions and Their Graphs 88. g −1 D f −1 = g −1 ( f −1 ( x)) x = 10 + 0.75 y = g −1 (8( x + 3)) x − 10 = 0.75 y = 3 8( x + 3) x − 10 = y 0.75 = 23 x + 3 So, f −1 ( x) = In Exercises 89–92, f ( x ) = x + 4, f –1 ( x ) = x – 4, g ( x ) = 2 x – 5, g –1 ( x ) = 89. ( g −1 D f −1 x +5 . 2 (b) y = −1 24.25 − 10 = 19 0.75 So, 19 units are produced. = g −1 ( x − 4) = ( x − 4) + 5 y = 0.03x 2 + 245.50, 0 < x < 100 94. (a) 2 x +1 = 2 90. ( f −1 D g −1 Ÿ 245.50 < y < 545.50 x = 0.03 y 2 + 245.50 x − 245.50 = 0.03 y 2 )( x) = f ( g ( x)) −1 x − 10 . 0.75 x = hourly wage, y = number of units produced )( x) = g ( f ( x)) −1 y = 10 + 0.75 x 93. (a) −1 x − 245.50 = y2 0.03 § x + 5· = f −1 ¨ ¸ © 2 ¹ x +5 = − 4 2 x +5−8 = 2 x −3 = 2 x − 245.50 = y, 245.50 < x < 545.50 0.03 f −1 ( x) = x − 245.50 0.03 x = temperature in degrees Fahrenheit y = percent load for a diesel engine (b) 91. ( f D g )( x) = f ( g ( x)) = f ( 2 x − 5) = ( 2 x − 5) + 4 = 2x − 1 ( f D g) −1 ( x) = (c) 0.03 x 2 + 245.50 ≤ 500 x +1 2 0.03 x 2 ≤ 254.50 x 2 ≤ 8483.33 Note: Comparing Exercises 89 and 91, ( f D g ) ( x) = ( g −1 D f −1 )( x). −1 92. ( g D f )( x) = g ( f ( x)) = g ( x + 4) = 2( x + 4) − 5 = 2x + 8 − 5 = 2x + 3 y = 2x + 3 x ≤ 92.10 Thus, 0 −1 22. h( x) = (a) h( − 2) = 2( − 2) + 1 = − 3 (b) h( −1) = 2( −1) + 1 = −1 2 Domain: All real numbers x except x = − 2, 3 (c) h(0) = 02 + 2 = 2 (d) h( 2) = 22 + 2 = 6 21. f ( x) = 25 − x 2 25 − x 2 ≥ 0 Domain: (5 + x)(5 − x) ≥ 0 Critical numbers: x = ± 5 23. v(t ) = − 32t + 48 Test intervals: ( − ∞, − 5), ( − 5, 5), (5, ∞) v(1) = 16 feet per second 2 Test: Is 25 − x ≥ 0? Solution set: − 5 ≤ x ≤ 5 24. 0 = − 32t + 48 Domain: all real numbers x such that − 5 ≤ x ≤ 5, or [− 5, 5] 48 t = 32 = 1.5 seconds 25. f ( x) = 2 x 2 + 3 x − 1 f ( x + h) − f ( x) h ª2( x + h) + 3( x + h) − 1º − ( 2 x 2 + 3 x − 1) ¼ = ¬ h 2 = = 2 x 2 + 4 xh + 2h 2 + 3 x + 3h − 1 − 2 x 2 − 3 x + 1 h h( 4 x + 2h + 3) h = 4 x + 2h + 3, h ≠ 0 f ( x) = x3 − 5 x 2 + x 26. f ( x + h) = ( x + h) − 5( x + h) + ( x + h) 3 2 = x3 + 3x 2 h + 3 xh 2 + h3 − 5 x 2 − 10 xh − 5h 2 + x + h f ( x + h) − f ( x) h = x3 + 3x 2 h + 3 xh 2 + h3 − 5 x 2 − 10 xh − 5h 2 + x + h − x3 + 5 x 2 − x h = 3 x 2 h + 3 xh 2 + h3 − 10 xh − 5h 2 + h h = h(3 x 2 + 3 xh + h 2 − 10 x − 5h + 1) h INSTRUCTOR USE ONLY 2 2 = 3 x + 3 xh + h − 10 x − 5h + 1, h ≠ 0 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Review Exercises for f Chapter 2 27. y = ( x − 3) 34. f ( x ) = ( x 2 − 4) 2 237 2 A vertical line intersects the graph no more than once, so y is a function of x. 28. x = − 4 − y A vertical line intersects the graph more than once, so y is not a function of x. f is increasing on ( − 2, 0) and ( 2, ∞). 29. f ( x ) = 5 x 2 + 4 x − 1 f is decreasing on ( − ∞, − 2) and (0, 2). 5×2 + 4 x − 1 = 0 (5 x − 1)( x + 1) = 0 35. f ( x ) = − x 2 + 2 x + 1 Relative maximum: (1, 2) 5 x − 1 = 0 Ÿ x = 15 x + 1 = 0 Ÿ x = −1 30. f ( x ) = 8x + 3 11 − x 36. f ( x) = x3 − 4 x 2 − 1 8x + 3 = 0 11 − x 8x + 3 = 0 Relative minimum: (2.67, −10.48) x = − 83 31. f ( x) = Relative maximum: (0, −1) 2x + 1 37. f ( x) = − x 2 + 8 x − 4 2x + 1 = 0 f ( 4) − f (0) 2x + 1 = 0 4−0 2 x = −1 38. f ( x) = 2 − 32. f ( x ) = x 3 − x 2 x3 − x 2 = 0 f (7) − f (3) x 2 ( x − 1) = 0 7 −3 x = 0 12 − ( − 4) 4 = 4 The average rate of change of f from x1 = 0 to x2 = 4 is 4. x = 12 x 2 = 0 or = = (2 − ) 8 − ( 2 − 2) 4 2− 2 2 1− 2 = = 4 2 x −1 = 0 x =1 The average rate of change of f from x1 = 3 to x2 = 7 ( is 1 − 33. f ( x) = x + x + 1 f is increasing on (0, ∞). x +1 39. f is decreasing on ( − ∞, −1). ) 2 2. f ( x) = x5 + 4 x − 7 f ( − x) = ( − x) + 4( − x) − 7 5 f is constant on ( −1, 0). = − x5 − 4 x − 7 ≠ f ( x) ≠ − f ( x) Neither even nor odd 40. f ( x ) = x 4 − 20 x 2 f ( − x ) = ( − x) − 20( − x) = x 4 − 20 x 2 = f ( x) 4 2 INSTRUCTOR USE ONLY The functionn is i even. © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE 238 Chapter 2 Functions ctions and Their Graphs 41. f ( x) = 2 x 45. f ( x) = x 2 + 5 x2 + 3 f ( − x) = 2( − x) = − 2x ( − x) + 3 2 x2 + 3 = − f ( x) The function is odd. 42. f ( x) = 5 6 x 2 f ( − x ) = 5 6( − x ) = 5 6 x 2 = f ( x) 2 46. g ( x) = − 3 x3 The function is even. 43. (a) f ( 2) = − 6, f ( −1) = 3 Points: ( 2, − 6), ( −1, 3) m = 3 − ( − 6) −1 − 2 = 9 = −3 −3 y − ( − 6) = − 3( x − 2) y + 6 = − 3x + 6 y = − 3x 47. f ( x ) = f ( x) = − 3 x x +1 (b) 44. (a) f (0) = − 5, f ( 4) = − 8 (0, − 5), (4, − 8) − 8 − ( − 5) 3 = − m = 4−0 48. g ( x) = 1 x +5 4 3 ( x − 0) 4 3 y = − x −5 4 3 f ( x) = − x − 5 4 y − ( − 5) = − (b) INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Review Exercises ffor Chapter 2 49. g ( x) = a x + 4b 53. (a) f ( x ) = (b) h( x) = − 239 x x + 4 Vertical shift 4 units upward, reflection in the x-axis (c) x ≥ −1 ­5 x − 3, 50. f ( x) = ® ¯− 4 x + 5, x < −1 (d) h( x) = − f ( x) + 4 54. (a) f ( x) = x (b) h( x) = x + 3 − 5 Horizontal shift 3 units to the left; vertical shift 5 units downward (c) 51. (a) f ( x) = x 2 (b) h( x) = x 2 − 9 Vertical shift 9 units downward (c) (d) h( x) = f ( x + 3) − 5 55. (a) f ( x) = x 2 (b) h( x) = − ( x + 2) + 3 2 Horizontal shift two units to the left, vertical shift 3 units upward, reflection in the x-axis. (d) h( x) = f ( x) − 9 (c) 52. (a) f ( x ) = x 3 (b) h( x) = ( x − 2) + 2 3 Horizontal shift 2 units to the right; vertical shift 2 units upward (c) (d) h( x) = − f ( x + 2) + 3 INSTRUCTOR USE ONLY (d) h( x) = f ( x − 2) + 2 (d © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 240 Chapter 2 NOT FOR SALE Functions ctions and Their Graphs 59. (a) f ( x) = a xb 56. (a) f ( x) = x 2 (b) h( x) = 5a x − 9b (b) h( x) = 12 ( x − 1) − 2 2 Horizontal shift 9 units to the right and a vertical stretch (each y-value is multiplied by 5) Horizontal shift one unit to the right, vertical shrink, vertical shift 2 units downward (c) (c) (d) h( x) = 5 f ( x − 9) (d) h( x) = 12 f ( x − 1) − 2 60. (a) f ( x ) = x3 57. (a) f ( x) = a xb (b) h( x) = − 13 x3 (b) h( x) = −a xb + 6 Reflection in the x-axis; vertical shrink (each y-value is multiplied by 13 ) Reflection in the x-axis and a vertical shift 6 units upward (c) (c) (d) h( x) = − f ( x) + 6 58. (a) f ( x ) = (b) h( x) = − (d) h( x) = − 13 f ( x) x 61. f ( x) = x 2 + 3, g ( x) = 2 x − 1 x +1 +9 Reflection in the x-axis, a horizontal shift 1 unit to the left, and a vertical shift 9 units upward (c) (a) ( f + g )( x) = ( x 2 + 3) + (2 x − 1) = x 2 + 2 x + 2 (b) ( f − g )( x) = ( x 2 + 3) − ( 2 x − 1) = x 2 − 2 x + 4 (c) ( fg )( x) = ( x 2 + 3)( 2 x − 1) = 2 x3 − x 2 + 6 x − 3 §f· x2 + 3 1 (d) ¨ ¸( x) = , Domain: x ≠ 2x − 1 2 ©g¹ 62. f ( x ) = x 2 − 4, g ( x) = ( f + g )( x) = f ( x) + g ( x) = x 2 − 4 + 3− x (b) ( f − g )( x) = f ( x) − g ( x) = x 2 − 4 − 3− x (a) (d) h( x) = − f ( x + 1) + 9 3− x (c) ( fg )( x) = f ( x) g ( x) = ( x 2 − 4)( 3− x ) f ( x) §f· x2 − 4 = , Domain: x 4 and y > 0. 2 y = 2( x − 4) 2 x = 2( y − 4) , x > 0, y > 4 2 x 2 = ( y − 4) 2 71. (a) x = y − 4 2 f ( x) = 12 x − 3 (b) x + 4 = y 2 y = 12 x − 3 x = 12 y − 3 f −1 ( x) = x + 3 = 12 y 2( x + 3) = y x + 4, x > 0 2 74. f ( x) = x − 2 is increasing on ( 2, ∞). f −1 ( x) = 2 x + 6 (c) The graph of f −1 is the reflection of the graph of f in the line y = x. (d) The domains and ranges of f and f −1 are the set of all real numbers. Let f ( x) = x − 2, x > 2, y > 0. y = x − 2 x = y − 2, x > 0, y > 2 x + 2 = y, x > 0, y > 2 f −1 ( x) = x + 2, x > 0 INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Problem Solving ffor Chapter 2 75. False. The graph is reflected in the x-axis, shifted 9 units to the left, then shifted 13 units down ward. 243 76. True. If f ( x) = x3 and g ( x) = 3 x , then the domain of g is all real numbers, which is equal to the range of f and vice versa. Problem Solving for Chapter 2 1. (a) W1 = 0.07 S + 2000 (b) W2 = 0.05S + 2300 (c) Point of intersection: (15,000, 3050) Both jobs pay the same, $3050, if you sell $15,000 per month. (d) No. If you think you can sell $20,000 per month, keep your current job with the higher commission rate. For sales over $15,000 it pays more than the other job. 2. Mapping numbers onto letters is not a function. Each number between 2 and 9 is mapped to more than one letter. {(2, A), (2, B), (2, C ), (3, D), (3, E ), (3, F ), (4, G), (4, H ), (4, I ), (5, J ), (5, K ), (5, L), (6, M ), (6, N ), (6, O), (7, P), (7, Q), (7, R), (7, S ), (8, T ), (8, U ), (8, V ), (9, W ), (9, X ), (9, Y ), (9, Z )} Mapping letters onto numbers is a function. Each letter is only mapped to one number. {( A, 2), ( B, 2), (C , 2), ( D, 3), ( E , 3), ( F , 3), (G, 4), ( H , 4), ( I , 4), ( J , 5), ( K , 5), ( L, 5), ( M , 6), ( N , 6), (O, 6), ( P, 7), (Q, 7), ( R, 7), ( S , 7), (T , 8), (U , 8), (V , 8), (W , 9), ( X , 9), (Y , 9), ( Z , 9)} 3. (a) Let f ( x) and g ( x) be two even functions. (b) Let f ( x) and g ( x) be two odd functions. Then define h( x) = f ( x) ± g ( x). Then define h( x) = f ( x) ± g ( x). h( − x ) = f ( − x ) ± g ( − x ) h( − x ) = f ( − x ) ± g ( − x ) = f ( x) ± g ( x) because f and g are even = − f ( x) ± g ( x) because f and g are odd = h( x ) = − h( x ) So, h( x) is also odd. ( If f ( x) ≠ g ( x)) So, h( x) is also even. (c) Let f ( x) be odd and g ( x) be even. Then define h( x) = f ( x) ± g ( x). h( − x ) = f ( − x ) ± g ( − x ) = − f ( x) ± g ( x) because f is odd and g is even ≠ h( x ) ≠ − h( x ) So, h( x) is neither odd nor even. INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 244 Chapter 2 NOT FOR SALE Functions ctions and Their Graphs 4. f ( x) = x g ( x) = − x ( f D f )( x) = x and ( g D g )( x) = x These are the only two linear functions that are their own inverse functions since m has to equal 1 m for this to be true. General formula: y = − x + c 5. f ( x ) = a2 n x 2 n + a2 n − 2 x 2 n − 2 + ” + a2 x 2 + a0 f ( − x ) = a2 n ( − x ) 2n + a2 n − 2 ( − x ) 2n − 2 + ” + a2 ( − x) + a0 = a2 n x 2 n + a2 n − 2 x 2 n − 2 + ” + a 2 x 2 + a0 = f ( x) 2 So, f ( x) is even. 6. It appears, from the drawing, that the triangles are equal; thus ( x, y ) = (6, 8). The line between ( 2.5, 2) and (6, 8) 7. (a) April 11: 10 hours April 12: 24 hours April 13: 24 hours is y = 12 x − 16 . 7 7 x + 128 . The line between (9.5, 2) and (6, 8) is y = − 12 7 7 2 April 14: 23 hours 3 The path of the ball is: Total: ­°12 x − 16 , 2.5 ≤ x ≤ 6 7 f ( x) = ® 7 128 12 °̄− 7 x + 7 , 6 < x ≤ 9.5 (b) Speed = 2 81 hours 3 distance 2100 180 5 = = = 25 mph 2 time 7 7 81 3 180 t + 3400 7 1190 Domain: 0 ≤ t ≤ 9 Range: 0 ≤ D ≤ 3400 (c) D = − (d) INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. Problem Solving ffor Chapter 2 8. (a) f ( x2 ) − f ( x1 ) (b) f ( x2 ) − f ( x1 ) (c) f ( x2 ) − f ( x1 ) (d) f ( x2 ) − f ( x1 ) (e) f ( x2 ) − f ( x1 ) x2 − x1 x2 − x1 x2 − x1 x2 − x1 x2 − x1 = = = = = f ( 2) − f (1) 2 −1 = f (1.5) − f (1) 1.5 − 1 245 1−0 =1 1 = 0.75 − 0 = 1.5 0.5 = 0.4375 − 0 = 1.75 0.25 = 0.234375 − 0 = 1.875 0.125 = 0.12109375 − 0 = 1.9375 0.625 f (1.25) − f (1) 1.25 − 1 f (1.125) − f (1) 1.125 − 1 f (1.0625) − f (1) 1.0625 − 1 (f ) Yes, the average rate of change appears to be approaching 2. (g) a. (1, 0), ( 2, 1), m = 1, y = x − 1 b. (1, 0), (1.5, 0.75), m = 0.75 = 1.5, y = 1.5 x − 1.5 0.5 c. (1, 0), (1.25, 0.4375), m = 0.4375 = 1.75, y = 1.75 x − 1.75 0.25 d. (1, 0), (1.125, 0.234375), m = 0.234375 = 1.875, y = 1.875 x − 1.875 0.125 e. (1, 0), (1.0625, 0.12109375), m = 0.12109375 = 1.9375, y = 1.9375 x − 1.9375 0.0625 (h) (1, f (1)) = (1, 0), m → 2, y = 2( x − 1), y = 2 x − 2 9. (a)–(d) Use f ( x) = 4 x and g ( x) = x + 6. (a) ( f D g )( x) = f ( x + 6) = 4( x + 6) = 4 x + 24 (b) ( f D g ) −1 ( x) = the length of the trip over land is 1 + (3 − x ) . 2 The total time is x − 24 1 = x −6 4 4 T ( x) = 1 x 4 g −1 ( x ) = x − 6 (c) f −1 ( x ) = 4 + x2 + 2 1 = 2 1 §1 · (d) ( g −1 D f −1 )( x) = g −1 ¨ x ¸ = x − 6 4 4 © ¹ 22 + x 2 , and 10. (a) The length of the trip in the water is 1 + (3 − x ) 1 4+ x + 4 2 2 4 2 x − 6 x + 10. (b) Domain of T ( x): 0 ≤ x ≤ 3 (c) (e) f ( x) = x + 1 and g ( x) = 2 x 3 ( f D g )( x) = f (2 x) = (2 x)3 + 1 = 8 x3 + 1 −1 ( f D g ) ( x) = 3 x −1 1 3 x −1 = 8 2 f −1 ( x) = 3 x − 1 g −1 ( x) = 1 x 2 ( g −1 D f −1 )( x) = g −1( 3 x − 1) = 12 3 x − 1 (d) T ( x) is a minimum when x = 1. (e) Answers will vary. Sample answer: To reach point Q in the shortest amount of time, you should row to a point one mile down the coast, and then walk the rest of the way. (f ) Answers will vary. (g) Conjecture: ( f D g ) −1 ( x) = ( g −1 D f −1 )( x) INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 246 Chapter 2 Functions ctions and Their Graphs ­1, x ≥ 0 11. H ( x) = ® ¯0, x < 0 (a) H ( x) − 2 (b) H ( x − 2) (c) − H ( x) (d) H ( − x) (e) 1 H 2 ( x) (f ) − H ( x − 2) + 2 y 3 2 −3 −2 −1 x 1 2 3 −1 −2 −3 1 1− x (a) Domain: all real numbers x except x = 1 1 1 § x − 1· (c) f f ( f ( x)) = f ¨ = = x ¸ = 1 © x ¹ 1 − § x − 1· ¨ ¸ x © x ¹ Range: all real numbers y except y = 0 The graph is not a line. It has holes at (0, 0) and 12. f ( x) = y = ( § 1 · (b) f ( f ( x)) = f ¨ ¸ ©1 − x ¹ 1 1 = = 1− x −1 § 1 · 1−¨ ¸ 1− x ©1 − x ¹ x −1 1− x = = −x x ) (1, 1). Domain: all real numbers x except x = 0 and x =1 INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Problem Solving ffor Chapter 2 247 ( ) (( f D g ) D h)( x) = ( f D g )(h( x)) = f ( g (h( x))) = ( f D g D h)( x) 13. ( f D ( g D h))( x) = f (( g D h)( x)) = f g ( h( x)) = ( f D g D h)( x) 14. (a) f ( x + 1) (b) f ( x) + 1 (c) 2 f ( x) (d) f ( − x) (e) − f ( x) (f ) f ( x) (g) f ( x ) 15. (a) f ( f −1 ( x)) x f ( x) f −1 ( x) –4 — 2 –4 –3 4 1 –2 –2 1 0 0 –1 0 — 4 f ( f −1 ( 4)) = f ( −3) = 4 0 –2 –1 1 –3 –2 x ( f ⋅ f −1 )( x) 2 –4 — –3 3 — — 4 — –3 (c) x (b) f ( f −1 ( −4)) = f ( 2) = −4 f ( f −1 ( −2)) = f (0) = −2 f ( f −1 (0)) = f ( −1) = 0 (d) x ( f + f −1 )( x) –3 f ( −3) + f −1 ( −3) = 4 + 1 = 5 –2 f ( −2) + f −1 ( −2) = 1 + 0 = 1 0 f (0) + f −1 (0) = −2 + ( −1) = −3 1 f (1) + f −1 (1) = −3 + ( −2) = −5 x f −1 ( x) f ( −3) f −1 ( −3) = ( 4)(1) = 4 –4 f −1 ( −4) = 2 = 2 –2 f ( −2) f −1 ( −2) = (1)(0) = 0 –3 f −1 ( −3) = 1 = 1 0 f (0) f −1 (0) = ( −2)( −1) = 2 0 f −1 (0) = −1 = 1 1 f (1) f −1 (1) = ( −3)( −2) = 6 4 f −1 ( 4) = −3 = 3 INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. 248 Chapter 2 Functions ctions and Their Graphs Practice Test for Chapter 2 1. Find the equation of the line through ( 2, 4) and (3, −1). 2. Find the equation of the line with slope m = 4 3 and y-intercept b = −3. 3. Find the equation of the line through ( 4, 1) perpendicular to the line 2 x + 3 y = 0. 4. If it costs a company $32 to produce 5 units of a product and $44 to produce 9 units, how much does it cost to produce 20 units? (Assume that the cost function is linear.) 5. Given f ( x) = x 2 − 2 x + 1, find f ( x − 3). 6. Given f ( x) = 4 x − 11, find f ( x) − f (3) x −3 7. Find the domain and range of f ( x) = 36 − x 2 . 8. Which equations determine y as a function of x? (a) 6 x − 5 y + 4 = 0 (b) x 2 + y 2 = 9 (c) y 3 = x 2 + 6 9. Sketch the graph of f ( x) = x 2 − 5. 10. Sketch the graph of f ( x) = x + 3 . ­2 x + 1, if x ≥ 0, 11. Sketch the graph of f ( x) = ® 2 ¯x − x, if x < 0. 12. Use the graph of f ( x) = x to graph the following: (a) f ( x + 2) (b) − f ( x) + 2 13. Given f ( x) = 3x + 7 and g ( x) = 2 x 2 − 5, find the following: (a) ( g − f )( x) (b) ( fg )( x) 14. Given f ( x) = x 2 − 2 x + 16 and g ( x) = 2 x + 3, find f ( g ( x)). 15. Given f ( x) = x3 + 7, find f −1 ( x). 16. Which of the following functions have inverses? (a) f ( x) = x − 6 (b) f ( x) = ax + b, a ≠ 0 (c) f ( x) = x3 − 19 INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved. NOT FOR SALE Practice Test ffor Chapter 2 17. Given f ( x) = 249 3− x , 0 < x ≤ 3, find f −1 ( x). x Exercises 18–20, true or false? 18. y = 3 x + 7 and y = 13 x − 4 are perpendicular. 19. ( f D g ) −1 = g −1 D f −1 20. If a function has an inverse, then it must pass both the Vertical Line Test and the Horizontal Line Test. INSTRUCTOR USE ONLY © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. © Cengage Learning. All Rights Reserved.