Explanation Given: y = f ( x ) = x 3 − 12 x 2 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 ) = x 3 − 12 x 2 Differentiating the above function with respect to x, we get f ' ( x ) = d d x ( x 3 − 12 x 2 ) = ( 3 x 2 − 12 x ) To find out the critical points, we equate the above equation to zero. i.e. f ' ( x ) = 0 i.e. 3 x 2 − 12 x = 0 i.e. 3 x ( x − 8 ) = 0 x = 0 , x = 8 Therefore, the critical point of the given function is x = 0 , x = 8 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 29E

To determine

To find all the critical points of the function y=x3−12x2 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 28E

Chapter 4.3, Problem 30E

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To find all the critical points of the function y = x 3 − 12 x 2 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=x3−12x2 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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