Explanation Obtain the derivative f ′ ( x ) as follows. f ′ ( x ) = d d x ( − 2 x 3 + 3 x 2 + 72 x ) = − 2 ( 3 x 2 ) + 3 ( 2 x ) + 72 = − 6 x 2 + 6 x + 72 Thus, the value of f ′ ( x ) is − 6 x 2 + 6 x + 72 . Equate the derivative f ′ ( x ) to 0. − 6 x 2 + 6 x + 72 = 0 x 2 − x − 12 = 0 x 2 + 3 x − 4 x − 12 = 0 ( x + 3 ) ( x − 4 ) = 0 x = − 3 or x = 4 Thus, the zeros are − 3 and 4. Evaluate the function f ″ ( x ) by differentiating the function f ′ ( x ) as follows. f ″ ( x ) = d d x ( − 6 x 2 + 6 x + 72 ) = − 6 ( 2 x ) + 6 = − 12 x + 6 Obtain the value of f ″ ( x ) at zeros. Substitute the value x = − 3 in f ″ ( x ) = − 12 x + 6 , f ″ ( − 3 ) = − 12 ( − 3 ) + 6 = 36 + 6 = 42 Thus, the value of f ″ ( x ) at x = − 3 is f ″ ( 3 ) = 42

Finite Mathematics and Calculus with Applications (10th Edition)

10th Edition

ISBN: 9780321979407

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 13.3, Problem 5YT

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To find: The relative extrema of the function f(x)=−2x3+3x2+72x.

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Chapter 13.3, Problem 4YT

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Finite Mathematics and Calculus with Applications (10th Edition)

Ch. 13.1 - Find where the function is increasing and...Ch. 13.1 - Prob. 2YT Ch. 13.1 - Prob. 3YT Ch. 13.1 - Prob. 4YT Ch. 13.1 - Prob. 1WE Ch. 13.1 - Prob. 2WE Ch. 13.1 - Prob. 3WE Ch. 13.1 - Prob. 4WE Ch. 13.1 - Prob. 5WE Ch. 13.1 - Prob. 6WE

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The relative extrema of the function f ( x ) = − 2 x 3 + 3 x 2 + 72 x .

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