Explanation Given: y = f ( x ) = 3 x 4 + 8 x 3 − 6 x 2 − 24 x We know that the critical points are the points at which the derivatives of the function become zero or undefined. Here, the given function is y = f ( x ) = 3 x 4 + 8 x 3 − 6 x 2 − 24 x Differentiating the above function with respect to x, we get f ' ( x ) = d d x ( 3 x 4 + 8 x 3 − 6 x 2 − 24 x ) = 12 x 3 + 24 x 2 − 12 x − 24 To find out the critical points, we equate the above equation to zero. i.e. f ' ( x ) = 0 i.e. 12 x 3 + 24 x 2 − 12 x − 24 = 0 12 ( x + 2 ) ( x 2 − 1 ) = 0 12 ( x + 2 ) ( x + 1 ) ( x − 1 ) = 0 x = − 2 , x = − 1 , x = 1 Therefore, the critical points of the given function are x = − 2 , x = − 1 , x = 1 and the domain of the function is in the interval of -¥< x <¥

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 4.3, Problem 31E

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To find all the critical points of the function y=3x4+8x3−6x2−24x and the intervals on which the function is increasing or decreasing. Also determine the critical point is a local minimum or maximum.

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Chapter 4.3, Problem 30E

Chapter 4.3, Problem 32E

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Chapter 4 Solutions

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To find all the critical points of the function y = 3 x 4 + 8 x 3 − 6 x 2 − 24 x and the intervals on which the function is increasing or decreasing. Also determine the critical point is a local minimum or maximum.

Solution Summary: The author explains that the critical points are the points at which the derivatives of the function become zero or undefined.

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To find all the critical points of the function y=3x4+8x3−6x2−24x and the intervals on which the function is increasing or decreasing. Also determine the critical point is a local minimum or maximum.

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