Explanation The function is f ( x ) = 5 x 2 3 + 2 x 5 3 and interval is [ − 2 , 1 ] . Find the critical number in the interval [ − 2 , 1 ] to compute the absolute minimum and absolute maximum. Differentiate the function f ( x ) = 5 x 2 3 + 2 x 5 3 as follows. f ′ ( x ) = d d x ( 5 x 2 3 + 2 x 5 3 ) = 5 ( 2 3 ) x 2 3 − 1 + 2 ( 5 3 ) x 5 3 − 1 = 10 3 x − 1 3 + 10 3 x 2 3 = 10 3 x 1 3 + 10 x 2 3 3 On simplify further, f ′ ( x ) = 10 3 x 1 3 + 10 x 2 3 3 = 10 + 10 x 3 x 1 3 = 10 ( x + 1 ) 3 x 3 Set f ′ ( x ) = 0 and solve for x . 10 ( x + 1 ) 3 x 3 = 0 x + 1 = 0 x = 1 The derivative of the function is undefined at x = 0 . Find the values of the function at the critical points and the end points of the given interval. Hence, the critical numbers are x = 0 and x = − 1 . Substitute x = − 2 in f ( x ) = 5 x 2 3 + 2 x 5 3 , f ( − 2 ) = 5 ( − 2 ) 2 3 + 2 ( − 2 ) 5 3 = 5 ( 4 ) 1 3 + 2 ( − 1 ) 5 3 ( 2 ) 5 3 = 5 ( 4 ) 1 3 + 2 ( − 1 ) 5 3 ( 2 ) 5 3 = 7

Calculus with Applications (11th Edition)

11th Edition

ISBN: 9780321979421

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

Publisher: PEARSON

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Chapter 6.1, Problem 23E

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To find: The absolute extrema of f(x)=5x23+2x53 on the domain [−2,1] if exist and verify the answer using the graphing calculator.

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Chapter 6.1, Problem 22E

Chapter 6.1, Problem 24E

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Chapter 6 Solutions

Calculus with Applications (11th Edition)

Ch. 6.1 - YOUR TURN 1 Find the absolute extrema of the...Ch. 6.1 - Prob. 2YT Ch. 6.1 - Prob. 1WE Ch. 6.1 - Prob. 2WE Ch. 6.1 - Prob. 1E Ch. 6.1 - Find the locations of any absolute extrema for the...Ch. 6.1 - Prob. 3E Ch. 6.1 - Prob. 4E Ch. 6.1 - Prob. 5E Ch. 6.1 - Prob. 6E

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The absolute extrema of f ( x ) = 5 x 2 3 + 2 x 5 3 on the domain [ − 2 , 1 ] if exist and verify the answer using the graphing calculator.

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To find: The absolute extrema of f(x)=5x23+2x53 on the domain [−2,1] if exist and verify the answer using the graphing calculator.

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Chapter 6 Solutions