Explanation Given: The function is f ( x ) = x 7 3 + x 4 3 . Calculation: Evaluate f ′ ( x ) as follows. f ′ ( x ) = d d x ( x 7 3 + x 4 3 ) = ( 7 3 ) x 4 3 + ( 4 3 ) x 1 3 Thus, f ′ ( x ) = ( 7 3 ) x 4 3 + ( 4 3 ) x 1 3 . Equate f ′ ( x ) to 0 to find the critical number. ( 7 3 ) x 4 3 + ( 4 3 ) x 1 3 = 0 x 1 3 3 ( 7 x + 4 ) = 0 x 1 3 = 0 or 7 x + 4 = 0 x = 0 or x = − 4 7 The critical numbers are x = 0 and x = − 4 7 . The second derivative f ″ ( x ) can be obtained from f ′ ( x ) = ( 7 3 ) x 4 3 + ( 4 3 ) x 1 3 as follows. f ′ ′ ( x ) = d d x ( ( 7 3 ) x 4 3 + ( 4 3 ) x 1 3 ) = 7 3 ( 4 3 ) x 1 3 + 4 3 ( 1 3 ) x − 2 3 = 28 9 x 1 3 + 4 9 x − 2 3 Thus, the value of f ″ ( x ) is f ′ ′ ( x ) = 28 9 x 1 3 + 4 9 x − 2 3 . Substitute x = − 4 7 in f ′ ′ ( x ) = 28 9 x 1 3 + 4 9 x − 2 3 . f ′ ′ ( − 4 7 ) = 28 9 ( − 4 7 ) 1 3 + 4 9 ( − 4 7 ) − 2 3 = ( 3

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 65E

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To find: The critical numbers for the given function and corresponding relative maxima or minima using the suitable derivative test.

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Chapter 13.3, Problem 64E

Chapter 13.3, Problem 66E

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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 critical numbers for the given function and corresponding relative maxima or minima using the suitable derivative test.

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To find: The critical numbers for the given function and corresponding relative maxima or minima using the suitable derivative test.

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Chapter 13 Solutions