Unformatted text preview: 444 Cumulative Review
(e) Local (and absolute) maximum of approximately 3.079 at x
2 3 6 40. (a) Increasing in [ 0.7, 2] (where f 0), decreasing in [ 2, 0.7] (where f 0), and has a local minimum at x 0.7. (b) y 2x 2 3x 3 and x 2 3 6 ; 0 and local (and absolute) minimum of 0 at x at x (f)
[ 3, 3] by [ 15, 10] 2 ( 1.042, 1.853) 1 and an absolute 2. 3,
3 , 2 (c) f (x) f (0) 41. f (x) f (x) 42. x2 x2 2 3 x 3 1: f (x) 3x 3x 3 2 x 3x C; choose C so that 2 2 3 3 2 x x 3x 1. 3 2 43. (a) f has an absolute maximum at x minimum at x 3. (b) f has a point of inflection at x 2: (c) The function f (x) cos x cos x C; choose C so that f (0) 1. 1 (x 1)2 2 7 x 2 2 1 2 x x 2 3 is one example of a function with the given properties. [ 2.35, 2.35] by [ 0.5, 3.5] f (x) is defined on [ 2, 2]. f (x) f (x) 2x 4 x
2 x3 4 x2 8x 4 3x 3 x2 [ 3.7, 5.7] by [ 3, 5] ; solve 44. y = 2 1 A(x) A (x) A (x) 4x 0 for x to find x 0, x 2 6 . 3 The graph of y f (x) is shown. 16 x2 , and the area of the rectangle for x 16 x2 1 x 16 x 2. 16 x2 2(8 x 2) 2 0 is x 16 x2 16 x2 , and so 0 when x = 2 2) 2 and y 2. The maximum possible area is A(2
[ 2.35, 2.35] by [ 10, 10] 8, with dimensions 4 45. f 2 by 2. 2 and f 2
4 (a) (b) 2, 2 6 2 6 , 0, 3 3 4 4 sec 2 x 2 4 4 tan or 4 2. The 2 6 2 6 ,0 , ,2 3 3 equation is y 0 for y 2 x Use NDER to plot f (x) and find that f (x) x 1.042. 1.414x 0.303 46. V s 3 dV 3s 2 ds Since ds 0.01s, the error of the volume calculation is approximately dV 3s 2(0.01s) 0.03s 3 0.03V, or 3%.
[ 2.35, 2.35] by [ 15, 5] (c) Approximately ( 1.042, 1.042) (d) Approximately ( 2, 1.042), (1.042, 2) ...
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 Spring '08
 GERMAN
 Critical Point

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