(a) Explanation Given: Every local maximum is a global maximum. Calculation: The given statement is false. Example: Consider the function f ( x ) = 2 x 4 − 12 x 3 + 16 x 2 on the interval [0,5] Differentiate f ( x ) = 2 x 4 − 12 x 3 + 16 x 2 with respect to x, get f ′ ( x ) = 8 x 3 − 36 x 2 + 32 x To find the critical points, equate f ′ ( x ) to zero and solve for x 8 x 3 − 36 x 2 + 32 x = 0 x = 0 , 1.22 , 3.3 Therefore, the critical points are x = 0 , 1.22 , 3.3 Find the second derivative of the function f ( x ) f ′ ′ ( x ) = 24 x 2 − 72 x + 32 Clearly, f ′ ( 0 ) = 0 and f ′ ′ ( 0 ) = 32 > 0 by second derivative test, conclude that the function f ( x ) has local minimum at x = 0 The local minimum is f ( 0 ) = 0 And f ′ ( 3.3 ) = 0 and f ′ ′ ( 3

14th Edition

ISBN: 9781464140051

Author: TAALMAN, Laura, Kohn, Peter

Publisher: W. H. Freeman

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Chapter 3.4, Problem 1C

(a)

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To find: whether the given statement is true or false

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Chapter 3.4, Problem 0C

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Chapter 3 Solutions

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whether the given statement is true or false

Solution Summary: The author explains how to find whether the given statement is true or false. Every local maximum is a global maximum.

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