HW11sol - M408C, Homework 11 Solution Instructor: Guoliang...

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M408C, Homework 11 — Solution Instructor: Guoliang Wu 1. Use Theorem 7 on P389 to find ( f - 1 ) 0 (2) where f ( x ) = x 3 + 3 sin x + 2 cos x. Solution: Recall ( f - 1 ) 0 ( f ( x )) = 1 f 0 ( x ) . We first find x that satisfies f ( x ) = 2. Thus x 3 + 3 sin x + 2 cos x = 2 x = 0 . Since f 0 ( x ) = 3 x 2 + 3 cos x - 2 sin x f 0 (0) = 3 6 = 0, we have ( f - 1 ) 0 (2) = 1 f 0 (0) = 1 3 . 2. Find the absolute minimum value of the function g ( x ) = e x x ,x > 0 . Solution: (1). g 0 ( x ) = xe x - e x x 2 = 0 e x ( x - 1) = 0 x = 1 is the only critical piont. (2). g 0 ( x ) = e x x 2 ( x - 1) > 0 if x > 1 and g 0 ( x ) < 0 if x < 1.
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This note was uploaded on 12/03/2010 for the course M 408c taught by Professor Mcadam during the Fall '06 term at University of Texas at Austin.

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