Unformatted text preview: Section 4.2
15. (a) f is continuous on [0, 1] and differentiable on (0, 1). (b) f (c) 2c 2 2c c
f (1) f (0) 1 0 2 ( 1) 1 135 20. (a) The secant line passes through (1, f (1)) (3, f (3))
2 3 0 1 1 2 (1, 0) and (3,
2 2 2), so its slope is
1 2 1 2 1 2 .
1 2 1
1 2 The equation is y or y x (x 1) 0.707x 0 0.707. . , or y 16. (a) f is continuous on [0, 1] and differentiable on (0, 1). (b) f (c)
2 1/3 c 3 f (1) f (0) 1 0 1 0 1 3 2 3 3 2 8 27 (b) We need to find c such that f (c)
1 2 c 1 1 2 c 1/3 2 c c 1 1 c f (c)
1 2 3 2 2 c c 17. (a) f is continuous on [ 1, 1] and differentiable on ( 1, 1). (b) f (c)
f (1) 1
2 f 3 2 1 2 1 1 2 f ( 1) ( 1)
2 The tangent line has slope
3 1 , . Its equation is y 2 2 1 1 and passes through
2 1 2 x 3 2 1 2 or 1 1 c
2 y x , or y 0.707x 0.354. 0. 2 2 4
2 2 2 2 1 2/3 x , f is not differentiable at x 3 1 1 c2 c2 c
2 21. (a) Since f (x) (b)
4
2 1 c 1 4
2 0.771
[ 1, 1] by [ 1, 1] 18. (a) f is continuous on [2, 4] and differentiable on (2, 4). (b) f (c)
1 c 1 f (4) 4 ln 3 2 2 ln 3 2 ln 3 f (2) 2 ln 1 (c) f (c)
1 2/3 c 3 1 2/3 c 3 f (1) f ( 1) 1 ( 1) 1 ( 1) 2 c 1 c 1 3 3
3/2 1 2.820 (0.5, 2.5) 2.5. c 2/3 19. (a) The secant line passes through (0.5, f (0.5)) and (2, f(2)) (2, 2.5), so its equation is y c 0.192 (b) The slope of the secant line is 0, so we need to find c such that f (c) 0. 1 c c
2 2 c f (c) 0 1 1 f (1) 2 The tangent line has slope 0 and passes through (1, 2), so its equation is y 2. ...
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 Fall '08
 GERMAN
 Derivative, Slope

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