Problem Set 11s

53 1 3 3 but 153 is not within the interval so c 153 f

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Unformatted text preview: 1) 1 10 − 2 7 7 3c 2 + 1 = = 8 ⇒ c2 = ⇒ c = ± = ±1.53 1 3 3 But − 1..53 is not within the interval so c = 1.53 f ' ( x) = 3x 2 + 1 ⇒ f ' (c) = 3c 2 + 1 = 1 x +1 c. f ( x ) = [0,2] Solution: 1 f ( x) = x +1 f ' ( x) = − f ' (c ) = − (c + 1)2 1 (x + 1)2 1 (c + 1) 2 = 1 / 3 −1 1 =− 2 3 3. Use the graph = 3 ⇒ c + 1 = ± 3 ⇒ c = −1 ± 3 Since c = −1 − 3 is not on the interval. c = −1 + 3 = 0.73 d. f (x) = Solution: x x+2 [1, 4] of the derivative of f to locate the critical points x0 at which f has either a loca...
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This document was uploaded on 02/27/2014 for the course M 408c at University of Texas.

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