5128_ch04ansTE_pp652-661

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Additional Answers 659 65. x 2 2, x 3 4, x 4 8, and x 5 16; ⏐x n ⏐ 2 n1 . [–10, 10] by [–3, 3] 66. (a) b 0 f(a), b 1 f (a), and b 2 f (a) . 2 (b) 1 x x 2 (c) As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical. (d) The quadratic approximation is 1 (x 1) (x 1) 2 . As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical. (e) The quadratic approximation is x x2 1 . 2 8 As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical. (f) For f :1 x ; for g:1 (x 1) 2 x; x for h:1 2 67. Finding a zero of sin x by Newton’s method would use the recursive formula sin( xn) x n1 x n x c os( xn) n tan x n , and that is exactly what the calculator would be doing. Any zero of sin x would be a multiple of p. 69. lim tan x lim sin x/ cos x x→0 x x→0 x lim x→0 1 sin x cos x x lim 1 x→0 cos x lim sin x x→0 x (1)(1) 1. 70. g(a) c, so if E(a) 0, then g(a) f(a) and c f(a). Then E(x) f(x) g(x) f(x) f(a) m(x a). E(x) Thus, f (x) f (a) m. x a x a lim f(x ) f(a) f (a), so if the limit of x→a x a E(x) is zero, then m f (a) and g(x) L(x). x a Section 4.6 Exercises 4.6 45. (a) The point being plotted would correspond to a point on the edge of the wheel as the wheel turns. (b) One possible answer: 16pt, where t is in seconds. (c) Assuming counterclockwise motion, the rates are as follows. p 4 ; d x 71.086 ft/sec dt d y 71.086 ft/sec dt 2 : d x 100.531 ft/sec dt d y 0 ft/sec dt p: d x 0 ft/sec dt d y 100.531 ft/sec dt 46. (a) One possible answer: x 30 cos , y 40 30 sin (b) Assuming counterclockwise motion, the rates are as follows. When t 5: d x 0 ft/sec dt d y 18.850 ft/sec dt when t 8: d x 17.927 ft/sec dt d y 5.825 ft/sec dt Quick Quiz (Sections 4.4–4.6) 4. (a) Since d y 1 1 dx 2 x1/2 , the slope at (25, 5) is . 1 0 1 The linearization is L(x) (x 25) 5, so L(26) 5.1. 1 0 f( 5) (b) x 2 5 5 52 26 5.1. f ( 5) 2(5) (c) The appropriate linearization is for the curve y 3 x at the point (27, 3). Since d y 1 1 dx 3 x2/3 , the slope at (27, 3) is . The linearization is 2 7 1 L(x) (x 27) 3, so L(26) 3 – 1/27 2.963. 2 7 Review Exercises 5. (a) Approximately (, 0.385] (b) Approximately [0.385, ) (c) None (d) (, ) (e) Local maximum at (0.385, 1.215) (f) None 6. (a) [1, ) (b) (, 1] (c) (, ) (d) None (e) Local minimum at (1, 0) (f ) None 7. (a) [0, 1) (b) (1, 0] (c) (1, 1) (d) None (e) Local minimum at (0, 1) (f) None 8. (a) (, 2 1/3 ] (, 0.794] (b) [2 1/3 ,1) [0.794, 1) and (1, ) (c) (, 2 1/3 ) (, 1.260) and (1, ) (d) (1.260, 1) (e) Local maximum at 21/3 , 2 3 21/3 (0.794, 0.529) (f) 21/3 , 1 3 21/3 (1.260, 0.420) 9. (a) None (b) [1, 1] (c) (1, 0) (d) (0, 1) (e) Local maximum at (1, p); local minimum at (1, 0) (f) 0, p 2
660 Additional Answers 10. (a) [3, 3] (e) Local maximum at (0.889, 1.011); (b) (,3] and [3, ) local minimum at (0, 0) (c) Approximately (2.584, 0.706) and (3.290, ) (f) (d) Approximately (, 2.584) and (0.706, 3.290) 2 9 , 0.667 (e) Local maximum at 16. (a) Approximately (, 0.215] 3, (b) Approximately [0.215, 2) and (2, ) 3 1 (1.732, 0.183); (c) Approximately (2, 3.710) 4 (d) (, 2) and approximately (3.710, ) local minimum at (e) Local maximum at (0.215, 2.417) 3, (f) (3.710, 3.420) (1.732, 0.683) 4 35. . y (f) (2.584, 0.573), (0.706, 0.338), and (3.290, 0.161) 2 11. (a) (0, 2] (b) [2, 0) (c) None (d) (2, 0) and (0, 2) (e) Local maxima at (2, ln 2) and (2, ln 2) x (f) None –3 3 12. (a) Approximately [0, 0.176], 0.994, p 2 , y = f(x) [2.148, 2.965], 3.834, 3 p 2 , and 5.591, 2p (b) Approximately [0.176, 0.994], p 2 , 2.148 , p [2.965, 3.834], and 3 , 5.591 2 (c) Approximately (0.542, 1.266), (1.876, 2.600), (3.425, 4.281), and (5.144, 6.000) (d) Approximately (0, 0.542), (1.266, 1.876), (2.600, 3.425), (4.281, 5.144), and (6.000, 2p) (e) Local maxima at (0.176, 1.266), p 2 ,0 p and (2.965, 1.266), 3 ,2 2 , and (2p, 1); local minima at (0, 1), (0.994, 0.513), (2.148, 0.513), (3.834, 1.806), and (5.591, 1.806) Note that the local extrema at x 3.834, x 3 p , and x 5.591 are also absolute extrema. 2 (f) (0.542, 0.437), (1.266, 0.267), (1.876, 0.267), (2.600, 0.437), (3.425, 0.329), (4.281, 0.120), (5.144, 0.120), and (6.000, 0.329) 2 13. (a) 0, (b) (, 0] and 2 3 , (c) (, 0) (d) (0, ) (e) Local maximum at 2 16 (1.155, 3.079) 3 3 3 1 , 33 (f ) None 14. (a) Approximately [0.578, 1.692] (b) Approximately (, 0.578] and [1.692, ) (c) Approximately (, 1.079) (d) Approximately (1.079, ) (e) Local maximum at (1.692, 20.517); local minimum at (0.578, 0.972) (f) (1.079, 13.601) 15. (a) 0, 8 9 (b) (, 0] and 8 , ) 9 (c) , 2 9 (d) 2 9 ,0 and (0, ) –3 36. (a) Absolute minimum is 2 at x 1; absolute maximum is 3 at x 3 (b) None (c) 1633001.59 41. (a) With some rounding, y 1 17.471e 0.06378t (b) The regression fit looks very good: [0, 80] by [0, 1600000] (c) 1,476,128 829,210 2,305,337 (d) About 1885; the graph has an inflection at this point. (e) The maximum population will be about 2,462,000. (f) There are many possible causes. Advances in transportation began drawing the population southward after 1920, and Tennessee was wellsituated geographically to become a crossroads of river, railroad, and automobile routes. By the year 2000 there had been numerous other demographic changes. It should also be pointed out that the census years in the data (1850 – 1910) include the years of the Civil War and Reconstruction, so the regression is based on unusual data to begin with. 51. (a) V(x) x(15 – 2x)(5 – x) (b) 0 x 5 [0, 5] by [10, 70]
Page 1 and 2: 652 Additional Answers 31. Maximum
Page 3 and 4: 654 Additional Answers k(b) f(b) g
Page 5 and 6: 656 Additional Answers 53. y 4 3 2
Page 7: 658 Additional Answers 63. (a) y(0)

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