Math Chapter 2 Worked Out Answer

Chapter 2 Review Exercises 5. y = (2x ¡ 1)(x + 1) = 2x2 + x ¡ 1 x y ¡3 14 ¡2 5 ¡1 0 0 ¡1 1 2 2 9 3 20 Pairs: (¡3; 14); (¡2; 5); (¡1; 0); (0; ¡1); (1; 2); (2; 9); (3; 20) Range: f¡1; 0; 2; 5; 9; 14; 20g 6. y = x y x x2 + 1 ¡3 3 ¡ 10 Pairs: ¡2 ¡ 25 ¡1 ¡ 12 0 0 1 2 3 1 2 2 5 3 10 ¡ ¢ ¡ ¢ ¡ ¢ 3 ¡3; ¡ 10 ; ¡2; ¡ 25 ; ¡1; ¡ 12 ; ¢ ¡ ¢ ¡ 3¢ ¡ (0; 0); 1; 12 ; 2; 25 ; 3; 10 Chapter 2 Review Exercises Range: © 3 3 2 1 ¡ 12 ; ¡ 25 ; ¡ 10 ; 0; 10 ; 5; 2 ª 8. f(x) = 2x2 + 5 and g(x) = 3x2 + 4x ¡ 1 (a) f (¡3) = 2(¡3)2 + 5 = 23 (b) g(2) = 3(2)2 + 4(2) ¡ 1 = 19 (c) f (3m) = 2(3m)2 + 5 = 18m2 + 5 (d) g(¡k) = 3(¡k)2 + 4(¡k) ¡ 1 = 3k2 ¡ 4k ¡ 1 (e) f (x + h) = 2(x + h)2 + 5 = 2(x2 + 2xh + h2 ) + 5 = 2x2 + 4xh + 2h2 + 5 7. f (x) = 5x2 ¡ 3 and g(x) = ¡x2 + 4x + 1 (a) f(¡2) = 5(¡2)2 ¡ 3 = 17 (b) g(3) = ¡(3)2 + 4(3) + 1 = 4 (c) f(¡k) = 5(¡k)2 ¡ 3 = 5k2 ¡ 3 (f ) g(x + h) = 3(x + h)2 + 4(x + h) ¡ 1 = 3(x2 + 2xh + h2 ) + 4x + 4h ¡ 1 = 3x2 + 6xh + 3h2 + 4x + 4h ¡ 1 (g) (d) g(3m) = ¡(3m)2 + 4(3m) + 1 = ¡9m2 + 12m + 1 (e) f(x + h) = 5(x + h)2 ¡ 3 = 5(x2 + 2xh + h2 ) ¡ 3 = 5x2 + 10xh + 5h2 ¡ 3 (f ) g(x + h) = ¡(x + h)2 + 4(x + h) + 1 = ¡(x2 + 2xh + h2 ) + 4x + 4h + 1 = ¡x2 ¡ 2xh ¡ h2 + 4x + 4h + 1 (g) f(x + h) ¡ f (x) h 5(x + h)2 ¡ 3 ¡ (5x2 ¡ 3) = h 5(x2 + 2hx + h2 ) ¡ 3 ¡ 5x2 + 3 = h 2 5x + 10hx + 5h2 ¡ 5x2 = h 10hx + 5h2 = h = 10x + 5h (h) g(x + h) ¡ g(x) h ¡(x + h)2 + 4(x + h) + 1 ¡ (¡x2 + 4x + 1) = h ¡(x2 + 2xh + h2 ) + 4x + 4h + 1 + x2 ¡ 4x ¡ 1 = h ¡x2 ¡ 2xh ¡ h2 + 4h + x2 = h 2 ¡2xh ¡ h + 4h = h = ¡2x ¡ h + 4 (h) f (x + h) ¡ f (x) h 2(x + h)2 + 5 ¡ (2x2 + 5) = h 2 2(x + 2xh + h2 ) + 5 ¡ 2x2 ¡ 5 = h 2x2 + 4xh + 2h2 + 5 ¡ 2x2 ¡ 5 = h 4xh + 2h2 = h = 4x + 2h g(x + h) ¡ g(x) h 3(x + h)2 + 4(x + h) ¡ 1 ¡ (3x2 + 4x ¡ 1) = h 2 2 3(x + 2xh + h ) + 4x + 4h ¡ 1 ¡ 3x2 ¡ 4x + 1 = h 3x2 + 6xh + 3h2 + 4x + 4h ¡ 1 ¡ 3x2 ¡ 4x + 1 = h 6xh + 3h2 + 4h = h = 6x + 3h + 4 9. y = ln(x + 7) x+7>0 x > ¡7 Domain: (¡7; 1): 10. y = ln(x2 ¡ 16) x2 ¡ 16 > 0 x2 > 16 x > 4 or x < ¡4 Domain: (¡1; ¡4) [ (4; 1) Chapter 2 3x ¡ 4 x x 6= 0 1 14. y = ¡ x2 + x + 2 4 The graph is a parabola. Let y = 0: 1 0 = ¡ x2 + x + 2 4 11. y = Domain: (¡1; 0) [ (0; 1) p x¡2 12. y = 2x + 3 Multiply by 4. x ¡ 2 ¸ 0 and 2x + 3 6= 0 x¸2 2x 6= ¡3 x 6= ¡ 3 2 Domain: [2; 1) 13. y = 2x2 + 3x ¡ 1 The graph is a parabola. Let y = 0: 0 = 2x2 + 3x ¡ 1 p ¡3 § 32 ¡ 4(2)(¡1) x= 2(2) p ¡3 § 9 + 8 = 4 p ¡3 § 17 = 4 The x-intercepts are p ¡3¡ 17 ¼ ¡1:48: 4 p ¡3+ 17 4 NONLINEAR FUNCTIONS 0 = ¡x2 + 4x + 8 p ¡4 § 42 ¡ 4(¡1)(8) x= 2(¡1) p ¡4 § 48 = ¡2 p =2§2 3 p The x-intercepts are 2 + 2 3 ¼ 5:46 p and 2 ¡ 2 3 ¼ ¡1:46: Let x = 0: 1 y = ¡ (0)2 + 0 + 2 4 y = 2 is the y-intercept. Vertex: x = ¼ 0:28 and 1 y = ¡ (2)2 + 2 + 2 4 = ¡1 + 4 =3 Let x = 0: y = 2(0)2 + 3(0) ¡ 1 ¡1 is the y-intercept. ¡3 3 ¡b = =¡ 2a 2(2) 4 μ ¶ μ ¶2 3 3 +3 ¡ y=2 ¡ ¡1 4 4 9 9 = ¡ ¡1 8 4 17 =¡ 8 ¡ ¢ The vertex is ¡ 34 ; ¡ 17 8 : ¡1 ¡b = ¡ 1¢ = 2 2a 2 ¡4 The vertex is (2; 3): Vertex: x = 15. y = ¡x2 + 4x + 2 Let y = 0: 0 = ¡x2 + 4x + 2 p ¡4 § 42 ¡ 4(¡1)(2) x= 2(¡1) p ¡4 § 24 = ¡2 p =2§ 6 Chapter 2 Review Exercises p Thepx-intercepts are 2 + 6 ¼ 4:45 and 2 ¡ 6 ¼ ¡0:45: Let x = 0: y = ¡02 + 4(0) + 2 2 is the y-intercept. Vertex: x = 18. f (x) = 1 ¡ x4 = ¡x4 + 1 Re‡ect the graph of y = x4 vertically then translate 1 unit upward. ¡4 ¡4 ¡b = = =2 2a 2(¡1) ¡2 y = ¡22 + 4(2) + 2 = 6 The vertex is (2; 6): 19. y = ¡(x ¡ 1)4 + 4 Translate the graph of y = x4 1 unit to the right and re‡ect vertically. Translate 4 units upward. 16. y = 3x2 ¡ 9x + 2 x-intercepts: 0:24 and 2:76 y-intercept: ¡ 2 ¢ Vertex: 32 ; ¡ 19 4 20. y = ¡(x + 2)3 ¡ 2 17. f (x) = x3 ¡ 3 Translate the graph of f (x) = x3 3 units down. Translate the graph of y = x3 2 units to the left and re‡ect vertically. Translate 2 units downward. Chapter 2 8 x Vertical asymptote: x = 0 x y 21. f (x) = ¡3 1:75 ¡2 2 ¡1 3 NONLINEAR FUNCTIONS 0 ¡2 1 0:5 2 0:86 Horizontal asymptote: 8 x approaches zero as x gets larger. y = 0 is an asymptote. x y ¡4 ¡2 ¡3 ¡2:7 ¡2 ¡4 ¡1 ¡8 1 8 2 4 3 2:7 4 2 6x x+2 Vertical asymptote: x = ¡2 Horizontal asymptote: y = 6 24. f(x) = x y ¡5 10 ¡4 12 ¡3 18 ¡2 ¡1 1 16 1 4 0 1 ¡1 ¡6 0 0 2 3x ¡ 6 Vertical asymptote: 3x ¡ 6 = 0 or x = 2 2 Horizontal asymptote: y = 0; since 3x¡6 approaches zero as x gets larger. 22. f (x) = x y 0 ¡ 13 1 ¡ 23 3 4 5 2 3 1 3 2 9 25. y = 4x x y 4x ¡ 2 3x + 1 Vertical asymptote: 1 4 2 16 23. f (x) = 26. y = 4¡x + 3 3x + 1 = 0 1 x=¡ 3 Horizontal asymptote: As x gets larger, 4x 4 4x ¡ 2 ¼ = : 3x ¡ 1 3x 3 y= 4 3 is an asymptote. x y ¡2 19 ¡1 7 0 1 2 4 13 4 49 16 1 2 2 3 3 1 Chapter 2 Review Exercises μ ¶2x¡3 1 27. y = 5 x y 28. y = x y 0 125 1 5 30. 2 1 5 μ ¶x¡1 1 2 ¡2 8 31. ¡1 4 0 2 29. y = log2 (x ¡ 1) 2y = x ¡ 1 x = 1 + 2y x y 2 0 3 1 5 2 1 1 2 1 2 y = 1 + log3 x y ¡ 1 = log3 x 3y¡1 = x x 1 9 1 3 1 3 9 y ¡1 0 1 2 3 y = ¡ ln (x + 3) ¡y = ln (x + 3) e¡y = x + 3 e¡y ¡ 3 = x x y ¡2:63 1 ¡2 0 ¡0:28 ¡1 4:39 ¡2 32. y = 2 ¡ ln x2 9 3 x y ¡4 ¡0:8 x y 1 2 2 0:6 ¡3 ¡0:2 3 ¡0:2 ¡2 0:6 ¡1 2 4 ¡0:8 Chapter 2 1 8 1 = 3 2 40. 101:07918 = 12 33. 2x+2 = 2x+2 The equation in logarithmic form is log10 12 = 1:07918: 2x+2 = 2¡3 x + 2 = ¡3 x = ¡5 μ 34. 41. log2 32 = 5 The equation in exponential form is ¶x 3 9 = 16 4 μ ¶2x μ ¶1 3 3 = 4 4 25 = 32: 1 2 The equation in exponential form is 42. log9 3 = 2x = 1 91=2 = 3: 1 x= 2 35. 2y+3 43. ln 82:9 = 4:41763 = 27 9 (32 )2y+3 = (33 )y 34y+6 = 33y 4y + 6 = 3y y = ¡6 y The equation in exponential form is e4:41763 = 82:9: 44. log 3:21 = 0:50651 μ ¶1=4 b 1 = 36. 2 4 μ ¶4 b 1 = 2 4 μ ¶4 1 4 =b 2 μ ¶ 1 4 =b 16 1 =b 4 37. 35 = 243 The equation in logarithmic form is log3 243 = 5: 38. 51=2 = NONLINEAR FUNCTIONS p 5 The equation in logarithmic form is p 1 log5 5 = : 2 39. e0:8 = 2:22554 The equation in logarithmic form is ln 2:22554 = 0:8: The equation in exponential form is 100:50651 = 3:21: Recall that log x means log10 x: 45. log3 81 = x 3x = 81 3 x = 34 x=4 46. log32 16 = x 32x = 16 25x = 24 5x = 4 x= 4 5 47. log4 8 = x 4x = 8 2 x (2 ) = 23 2x = 3 x= 3 2 48. log100 1000 = x 100x = 1000 (102 )x = 103 2x = 3 x= 3 2 Chapter 2 Review Exercises 49. log5 3k + log5 7k3 = log5 3k(7k3 ) = log5 (21k4 ) 56. k=¡ 50. log3 2y3 ¡ log3 8y 2 = log3 = log3 2y 3 8y 2 y 4 57. 52. 3 log4 r2 ¡ 2 log4 r = log4 (r2 )3 ¡ log4 r2 μ 6¶ r = log4 r2 ¼ 1:581 54. = 11 3 ln 3z¡2 = ln 11 (z ¡ 2) ln 3 = ln 11 z¡2 ln 11 z¡2= ln 3 ln 11 z= +2 ln 3 ¼ 4:183 55. 21¡m = 7 ln 21¡m = ln 7 (1 ¡ m) ln 2 = ln 7 ln ln ln ¡m = ln 7 2 7 ¡1 2 ln 7 m=1¡ ln 2 1¡m= ¼ ¡1:807 5 5 5 5+5 ln 5 + 5 ¡2 ¼ ¡3:305 58. e3x¡1 = 14 ln (e3x¡1 ) = ln 14 3x ¡ 1 = ln 14 3x = 1 + ln 14 1 + ln 14 3 ¼ 1:213 x= 6p = 17 ln 6p = ln 17 p ln 6 = ln 17 ln 17 ln 6 e¡5¡2x = 5 ln e¡5¡2x = ln (¡5 ¡ 2x) ln e = ln (¡5 ¡ 2x) ¢ 1 = ln ¡2x = ln x= = log4 (r4 ) p= ln 9 ln 12 ¼ ¡0:884 51. 4 log3 y ¡ 2 log3 x = log3 y 4 ¡ log3 x2 μ 4¶ y = log3 x2 53. 12¡k = 9 ln 12¡k = ln 9 ¡k ln 12 = ln 9 59. ³ m ´5 1+ 3 ·³ ¸1=5 ´ 5 m 1+ 3 m 1+ 3 m 3 = 15 = 151=5 = 151=5 = 151=5 ¡ 1 m = 3(151=5 ¡ 1) ¼ 2:156 μ ¶2 2p =3 60. 1 + 5 p 2p =§ 3 1+ 5 p 5 + 2p = §5 3 p 2p = ¡5 § 5 3 p ¡5 § 5 3 p= 2 p ¡5 + 5 3 ¼ 1:830 p= 2 p ¡5 ¡ 5 3 or p = ¼ ¡6:830 2 Chapter 2 NONLINEAR FUNCTIONS (f ) f is increasing if a > 1: 61. logk 64 = 6 k6 = 64 k 6 = 26 k=2 (g) f is decreasing if 0 < a < 1: 62. log3 (2x + 5) = 5 35 = 2x + 5 243 = 2x + 5 238 = 2x x = 119 63. log(4p + 1) + log p = log 3 log[p(4p + 1)] = log 3 log(4p2 + p) = log 3 4p2 + p = 3 2 4p + p ¡ 3 = 0 (4p ¡ 3)(p + 1) = 0 4p ¡ 3 = 0 or p + 1 = 0 3 p = ¡1 4 p cannot be negative, so p = 34 . p= 64. log2 (5m ¡ 2) ¡ log2 (m + 3) = 2 log2 68. (a) For x in the interval 0 < x · 1; the renter is charged the …xed cost of $60 and 1 day’s rent of $60 so ¡ ¢ C 34 = $40 + $60(1) = $40 + $60 = $100: (b) C 65. f (x) = ax ; a > 0; a 6= 1 9 10 ¢ = $40 + $60(1) = $40 + $60 = $100: (c) C(1) = $40 + $60(1) = $40 + $60 = $100: (d) For x in the interval 1 < x · 2; the renter is charged the …xed cost of $40 and 2 days rent of $120, so ¡ ¢ C 1 58 = $40 + $60(2) = $40 + $120 = $160: 5m ¡ 2 =2 m+3 5m ¡ 2 = 22 m+3 5m ¡ 2 = 4(m + 3) 5m ¡ 2 = 4m + 12 m = 14 ¡ (e) For x in the interval 2 < x · 3 the renter is charged the …xed cost of $40 and 3 days rent of $180. So ¡ ¢ C 2 19 = $40 + $60(3) = $40 + $180 = $220: (a) The domain is (¡1; 1): (b) The range is (0; 1): (c) The y-intercept is 1. (d) The graph has no discontinuities. (f ) (e) The x-axis, y = 0; is a horizontal asymptote. (f ) The function is increasing if a > 1: (g) The function is decreasing if 0 < a < 1: 66. f (x) = loga x; a > 0; a 6= 1 (a) The domain is (0; 1): (b) The range is (¡1; 1): (c) The x-intercept is 1. (d) There are no discontinuities. (g) The independent variable is the number of days, or x: (e) The y-axis, x = 0; is a vertical asymptote. (h) The dependent variable is the cost, or C(x): Chapter 2 Review Exercises 69. y = 7x 100 ¡ x (a) y = 560 7(80) = = 28 100 ¡ 80 20 The cost is $28,000. (b) y = 72. $1000 deposited at 6% compounded semiannually. ³ r ´tm A=P 1+ m To double: μ ¶t¢2 0:06 2(1000) = 1000 1 + 2 350 7(50) = =7 100 ¡ 50 50 2 = 1:032t ln 2 = 2t ln 1:03 The cost is $7000. (c) 630 7(90) = = 63 100 ¡ 90 10 t= The cost is $63,000. (d) Plot the points (80; 28); (50; 7); and (90; 63): ln 2 2 ln 1:03 ¼ 12 years To triple: μ ¶t¢2 0:06 3(1000) = 1000 1 + 2 3 = 1:032t ln 3 = 2t ln 1:03 ln 3 2 ln 1:03 ¼ 19 years t= (e) No, because all of the pollutant would be removed when x = 100; at which point the denominator of the function would be zero. 70. P = $6902; r = 6%; t = 8; m = 2 ³ r ´tm A=P 1+ m μ ¶8(2) 0:06 A = 6902 1 + 2 73. $2100 deposited at 4% compounded quarterly. ³ r ´tm A=P 1+ m To double: μ ¶t¢4 0:04 2(2100) = 2100 1 + 4 2 = 1:014t ln 2 = 4t ln 1:01 = 6902(1:03)16 = $11; 075:68 Interest = A ¡ P = $11; 075:68 ¡ $6902 = $4173:68 71. P = $2781:36; r = 4:8%; t = 6; m = 4 ³ r ´tm A=P 1+ m μ ¶(6)(4) 0:048 A = 2781:36 1 + 4 = 2781:36(1:012)24 = $3703:31 Interest = $3703:31 ¡ $2781:36 = $921:95 ln 2 4 ln 1:01 ¼ 17:4 t= Because interest is compounded quarterly, round the result up to the nearest quarter, which is 17.5 years or 70 quarters. To triple: ¶t¢4 μ 0:04 3(2100) = 2100 1 + 4 3 = 1:014t ln 3 = 4t ln 1:01 t= ln 3 4 ln 1:01 ¼ 27:6 Chapter 2 Because interest is compounded quarterly, round the result up to the nearest quarter, which is 27.75 years or 111 quarters. 74. P = $12,104; r = 6:2%; t = 2 A = P ert = 12; 104e0:062(2) = $13; 701:92 75. P = $12,104; r = 6:2%; t = 4 81. A = $2000; r = 6%; t = 5; m = 1 ³ r ´¡tm P =A 1+ m μ ¶¡5(1) 0:06 = 2000 1 + 1 = 2000(1:06)¡5 = $1494:52 82. A = $10,000; r = 8%; m = 2; t = 6 ³ r ´¡tm P =A 1+ m μ ¶¡2(6) 0:08 = 10,000 1 + 2 A = P ert A = 12,104e0:062(4) = 12,104e0:248 = $15,510.79 = 10,000(1:04)¡12 = $6245:97 76. A = $1500; r = 0:06; t = 9 A = P ert = 1500e0:06(9) = 1500e0:54 = $2574:01 77. P = $12,000; r = 0:05; t = 8 A = 12,000e0:05(8) = 12,000e0:40 = $17,901.90 78. r = 7%; m = 4 ³ r ´m =1 rE = 1 + m μ ¶4 0:07 ¡1 = 1+ 4 = 0:0719 = 7:19% 79. r = 6%; m = 12 ³ r ´m ¡1 rE = 1 + m μ ¶12 0:06 ¡1 = 1+ 12 = 0:0617 = 6:17% 80. r = 5% compounded continuously rE = er ¡ 1 = e0:05 ¡ 1 = 0:0513 = 5:13% NONLINEAR FUNCTIONS 83. r = 7%; t = 8; m = 2; P = 10,000 ³ r ´tm A=P 1+ m μ ¶8(2) 0:07 = 10,000 1 + 2 = 10,000(1:035)16 = $17,339.86 84. P = $1; r = 0:08 A = P ert ; A = 3(1) 3 = 1e0:08t ln 3 = 0:08t ln 3 =t 0:08 13:7 = t It would take about 13.7 years. 85. P = $6000; A = $8000; t = 3 A = P ert 8000 = 6000e3r 4 = e3r 3 ln 4 ¡ ln 3 = 3r ln 4 ¡ ln 3 r= 3 r ¼ 0:0959 or about 9:59% Chapter 2 Review Exercises ³ r ´¡tm 86. P = A 1 + m μ ¶¡3(12) 0:06 P = 25,000 1 + 12 The maximum revenue is $56,250. (i) = 25,000(1:005)¡36 = $20,891.12 87. (a) n = 1000 ¡ (p ¡ 50)(10); p ¸ 50 = 1000 ¡ 10p + 500 = 1500 ¡ 10p (b) R = pn R = p(1500 ¡ 10p) (c) p ¸ 50 Since n cannot be negative, 1500 ¡ 10p ¸ 0 ¡10p ¸ ¡1500 p · 150: Therefore, 50 · p · 150: (d) Since n = 1500 ¡ 10p; (j) The revenue starts at $50,000 when the price is $50, rises to a maximum of $56,250 when the price is $75, and falls to 0 when the price is $150. 88. C(x) = 5x + 3 x+1 (a) 10p = 1500 ¡ n n p = 150 ¡ : 10 R = pn ³ n´ n R = 150 ¡ 10 (e) Since she can sell at most 1000 tickets, 0 · n · 1000: (f ) R = ¡10p2 + 1500p ¡1500 ¡b = = 75 2a 2(¡10) The price producing maximum revenue is $75. (g) R = ¡ 1 2 n + 150n 10 ¡150 ¡b = ¡ 1 ¢ = 750 2a 2 ¡ 10 The number of tickets producing maximum revenue is 750. (h) R(p) = ¡10p2 + 1500p R(75) = ¡10(75)2 + 1500(75) = ¡56,250 + 112,500 = 56,250 (b) C(x + 1) ¡ C(x) 5(x + 1) + 3 5x + 3 ¡ (x + 1) + 1 x+1 5x + 8 5x + 3 ¡ = x+2 x+1 (5x + 8)(x + 1) ¡ (5x + 3)(x + 2) = (x + 2)(x + 1) = 5x2 + 13x + 8 ¡ 5x2 ¡ 13x ¡ 6 (x + 2)(x + 1) 2 = (x + 2)(x + 1) = (c) A(x) = C(x) = x 5x+3 x+1 x 5x + 3 = x(x + 1) Chapter 2 NONLINEAR FUNCTIONS 90. (a) y = at (d) A(x + 1) ¡ A(x) 5x + 3 5(x + 1) + 3 ¡ (x + 1)[(x + 1) + 1] x(x + 1) 5x + 3 5x + 8 ¡ = (x + 1)(x + 2) x(x + 1) = = Let y0 = 29:6; the value of y at t = 0: Use the point (45; 195:3) to …nd a: x(5x + 8) ¡ (5x + 3)(x + 2) x(x + 1)(x + 2) 2 195:3 = 29:6a45 2 5x + 8x ¡ 5x ¡ 13x ¡ 6 x(x + 1)(x + 2) ¡5x ¡ 6 = x(x + 1)(x + 2) = 195:3 29:6 r 45 195:3 a= 29:6 a45 = 89. C(x) = x2 + 4x + 7 ¼ 1:0428 y = 29:6(1:0428)t (a) (b) y = 29:43(1:04606)t (c) (b) C(x + 1) ¡ C(x) = (x + 1)2 + 4(x + 1) + 7 ¡ (x2 + 4x + 7) = x2 + 2x + 1 + 4x + 4 + 7 ¡ x2 ¡ 4x ¡ 7 = 2x + 5 (c) A(x) = x2 + 4x + 7 C(x) = x x 7 =x+4+ x (d) A(x + 1) ¡ A(x) 7 = (x + 1) + 4 + x+1 μ ¶ 7 ¡ x+4+ x 7 7 ¡x¡4¡ =x+1+4+ x+1 x 7 7 ¡ =1+ x+1 x 7x ¡ 7(x + 1) =1+ x(x + 1) 7x ¡ 7x ¡ 7 =1+ x(x + 1) 7 =1¡ x(x + 1) Yes, the answers are close to each other. (d) y = 0:03687x2 + 2:2560x + 23:554 y = ¡0:002306x3 +0:19386x2 ¡0:39623x+28:677 (e) Chapter 2 Review Exercises 2 14 91. F (x) = ¡ x2 + x + 96 3 3 The maximum fever occurs at the vertex of the parabola. ¡ 14 7 ¡b = 34 = 2a 2 ¡3 μ ¶2 μ ¶ 2 7 14 7 y=¡ + + 96 3 2 3 2 μ ¶ 2 49 49 + 96 =¡ + 3 4 3 49 49 =¡ + + 96 6 3 49 98 576 625 =¡ + + = ¼ 104:2 6 6 6 6 x= The maximum fever occurs on the third day. It is about 104.2± F. 92. (a) 100% ¡ 87:5% = 12:5% 12:5% = 0:125 = 1 125 = 1000 8 The fraction let in is 1 over the SPF rating. (b) (c) UVB = 1 ¡ 1 SPF 1 = 87:5% 8 1 1 ¡ = 75:0% 4 (d) 1 ¡ The increase is 12.5%. 1 = 96:6% 30 1 1¡ = 93:3% 15 (e) 1 ¡ The increase is 3.3% or about 3.3%. (f ) The increase in percent protection decreases to zero. Chapter 2 NONLINEAR FUNCTIONS 93. (a) (b) Quadratic: y = 846:061x2 ¡ 10,579.2x + 46,970.6 Cubic: y = ¡194:777x3 + 3475.56x2 ¡ 19,558.4x + 51,879 Quartic: y = 8:89685x4 ¡ 354:921x3 + 4374:14x2 ¡ 21,159.8x + 52,263.4 (c) (d) Use the functions found in part (b), with x = 11: Quadratic: 32,973 Cubic: ¡1969 Quartic: 6635 94. (a) The …rst three years of infancy corresponds to 0 months to 36 months, so the domain is [0; 36]: (b) In both cases, the graph of the quadratic function in the exponent opens upward and the x coordinate of the vertex is greater than 36 (x ¼ 47 for the awake infants and x ¼ 40 for the sleeping infants). So the quadratic functions are both decreasing over this time. Therefore, both respiratory rates are decreasing. (c) (d) When x = 12; the waking respiratory rate is y ¼ 49:23 breaths per minute, and the sleeping respiratory rate is y ¼ 38:55. Therefore, for a 1-year-old infant in the 95th percentile, the waking respiratory rate is approximately 49:23 ¡ 38:55 ¼ 10:7 breaths per minute higher. Chapter 2 Review Exercises 98. I(x) ¸ 1 I(x) = 10e¡0:3x 95. 10e¡0:3x ¸ 1 e¡0:3x ¸ 0:1 ¡0:3x ¸ ln 0:1 x· This function has a maximum value at x ¼ 187:9: At x ¼ 187:9; y ¼ 345: The largest girth for which this formula gives a reasonable answer is 187.9 cm. The predicted mass of a polar bear with this girth is 345 kg. 96. y = 17,000; y0 = 15,000; t = 4 y = y0 ekt 17,000 = 15,000e4k (a) 17 = e4k 15 μ ¶ 17 ln = 4k 15 0:125 =k 4 0:0313 = k So, y = 15,000e0:0313t : (b) 45,000 = 15,000e0:0313t 3 = e0:0313t ln 3 = 0:0313t ln 3 =k 0:0313 35 = k It would take about 35 years. 97. p(t) = 1:79 ¢ 1011 (2026:87 ¡ t)0:99 (a) p(2005) ¼ 8:441 billion This is about 1.964 billion more than the estimate of 6.477 billion. (b) p(2020) ¼ 26:56 billion p(2025) ¼ 96:32 billion ln 0:1 ¼ 7:7 ¡0:3 The greatest depth is about 7.7 m: 99. Graph y = c(t) = e¡t ¡ e¡2t on a graphing calculator and locate the maximum point. A calculator shows that the x-coordinate of the maximum point is about 0.69, and the ycoordinate is exactly 0.25. Thus, the maximum concentration of 0.25 occurs at about 0.69 minutes. c ´ ¡at c ³ e 100. g(t) = + g0 ¡ a a (a) If g0 = 0:08; c = 0:1; and a = 1:3; the function becomes μ ¶ 0:1 ¡1:3t 0:1 + 0:08 ¡ : e g(t) = 1:3 1:3 Graph this function on a graphing calculator. Use a window with Xmin = 0, since this represents the time when the drug is …rst injected. A good choice for the viewing window is [0, 5] by [0.07, 0.11], Xscl = 0.5, Yscl = 0.01. From the graph, we see that the maximum value of g for t ¸ 0 occurs at t = 0; the time when the drug is …rst injected. The maximum amount of glucose in the bloodstream, given by G(0); is 0.08 gram. (b) From the graph, we see that the amount of glucose in the bloodstream decreases from the initial value of 0.08 gram, so it will never increase to 0.1 gram. We can also reach this conclusion by graphing y1 = G(t) and y2 = 0:1 on the same screen with the window given in (a) and observing that the graphs of y1 and y2 do not intersect. (c) From the graph, we see that as t increases, the graph of y1 = G(t) becomes almost horizontal, and G(t) approaches approximately 0.0769. Note that 0:1 c = ¼ 0:0769: a 1:3 The amount of glucose in the bloodstream after a long time approaches 0.0769 grams. Chapter 2 101. y = y0 e¡kt (a) (c) 100,000 = 128,000e¡k(5) 128,000 = 100,000e5k d v+w = 128 = e5k 100 μ ¶ 128 ln = 5k 100 = NONLINEAR FUNCTIONS 2d d + v¡w d v+w 2d (v + w)(v ¡ w) ¢ d + v¡w (v + w)(v ¡ w) 2d(v2 ¡ w2 ) d(v ¡ w) + d(v + w) 2d(v 2 ¡ w2 ) dv ¡ dw + dv + dw 2d(v2 ¡ w2 ) = 2dv w2 v 2 ¡ w2 =v¡ = v v = 0:05 ¼ k y = 100,000e¡0:05t (b) 70,000 = 100,000e¡0:05t 7 = e¡0:05t 10 7 = ¡0:05t ln 10 (d) vaver = v ¡ w2 v vaver will be greatest when w = 0: 7:1 ¼ t It will take about 7.1 years. £ ¡ A ¢¤ ln 1 + 8:33 K 102. t = (1:26 £ 10 ) ln 2 9 (a) A = 0; K > 0 t = (1:26 £ 109 ) 104. (a) x = 0:9 means the speed is 10% slower on the return trip. x = 1:1 means the speed is 10% faster on the return trip. (b) vaver = ln [1 + 8:33(0)] ln 2 ln [1 + 8:33(0:212)] ln 2 ln 2:76596 = (1:26 £ 109 ) ln 2 = 1,849,403,169 or about 1:85 £ 109 years t = (1:26 £ 109 ) (c) As r increases, t increases, but at a slower and slower rate. As r decreases, t decreases at a faster and faster rate. 103. (a) Since the speed in one direction is v + w and in the other direction is v ¡ w; the time in one d d and in the other direction is v¡w : direction is v+w d d So the total time is v+w + v¡w : (b) The average speed is the total distance divided by the total time. So vaver = d v+w 2d : d + v¡w 2d = d + xv d v 2d xv ¢ d xv + xv 2dxv 2dxv = dx + d d(x + 1) μ ¶ 2x 2xv = = v x+1 x+1 = = (1:26 £ 109 )(0) = 0 years (b) d v (c) The formula for vaver is a rational function with a horizontal asymptote at vaver = 2v: This means that as the return velocity becomes greater and greater, the average velocity approaches twice the velocity on the …rst part of the trip, and can never exceed twice that velocity. 105. (a) P = kD1 164:8 = k(30:1) k= 164:8 ¼ 5:48 30:1 For n = 1; P = 5:48D: P = kD1:5 164:8 = k(30:1)1:5 k= 164:8 ¼ 1:00 (30:1)1:5 For n = 1:5; P = 1:00D1:5 : Extended Application: Characteristics of the Monkeyface Prickleback P = kD2 164:8 = k(30:1)2 164:8 k= ¼ 0:182 (30:1)2 For n = 2; P = 0:182D2 : (b) Extended Application: Characteristics of the Monkeyface Prickleback 1. Lt Lt L4 L11 L17 = Lx (1 ¡ e¡kt ) = 71:5(1 ¡ e¡0:1t ) = 71:5(1 ¡ e¡0:4 ) ¼ 23:6 = 71:5(1 ¡ e¡1:1 ) ¼ 47:7 = 71:5(1 ¡ e¡1:7 ) ¼ 58:4 The estimates are low. 2. W = aLb W = 0:01289L2:9 P = 1:00D 1:5 appears to be the best …t. (c) P = 1:00(39:5)1:5 ¼ 248:3 years If L = 25; W = 0:01289(25)2:9 ¼ 146:0 If L = 40; W = 0:01289(40)2:9 ¼ 570:5 If L = 60; W = 0:01289(60)2:9 ¼ 1848:8 Compared to the curve, the answers are reasonable. (d) We obtain P = 1:00D1:5 : This is the same as the function found in part (b). 106. (a) (b) Using a graphing calculator, r = 0:63: (c) Yes, the data are now more linear than in part (a). (d) Using a graphing calculator, r = 0:91: r is closer to 1. (e) Using a graphing calculator, Y = 0:873 log x ¡ 0:0255: