Explanation The function is f ( x ) = x 3 + 2 x 2 − 15 x + 3 and interval is [ − 4 , 2 ] . Find the critical number in the interval [ − 4 , 2 ] to compute the absolute minimum and absolute maximum. Differentiate the function f ( x ) = x 3 + 2 x 2 − 15 x + 3 as follows. f ′ ( x ) = d d x ( x 3 + 2 x 2 − 15 x + 3 ) = ( 3 ) x 3 − 1 + 2 ( 2 ) x 2 − 1 − 15 ( 1 ) + ( 0 ) = 3 x 2 + 4 x − 15 That is, f ′ ( x ) = 3 x 2 + 4 x − 15 . Set f ′ ( x ) = 0 and solve for x . 3 x 2 + 4 x − 15 = 0 3 x ( x + 3 ) − 5 ( x + 3 ) = 0 ( x + 3 ) ( 3 x − 5 ) = 0 That is, x + 3 = 0 or 3 x − 5 = 0 . 3 x = 5 x = 5 3 Hence, the critical numbers are x = − 3 and x = 5 3 . Find the values of the function at the critical points and the end points of the given interval. Substitute x = − 4 in f ( x ) = x 3 + 2 x 2 − 15 x + 3 , f ( − 4 ) = ( − 4 ) 3 + 2 ( − 4 ) 2 − 15 ( − 4 ) + 3 = − 64 + 32 + 60 + 3 = 31 That is, f ( − 4 ) = 31

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, Problem 13RE

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To find: The absolute extrema of f(x)=x3+2x2−15x+3 on the domain [−4,2] if exist.

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Chapter 6, Problem 12RE

Chapter 6, Problem 14RE

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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 ) = x 3 + 2 x 2 − 15 x + 3 on the domain [ − 4 , 2 ] if exist.

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Chapter 6 Solutions