Explanation Given: y = f ( x ) = 1 3 x 3 + 3 2 x 2 + 2 x + 4 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 f ( x ) = 1 3 x 3 + 3 2 x 2 + 2 x + 4 Differentiating the above function with respect to x, we get f ' ( x ) = d d x ( 1 3 x 3 + 3 2 x 2 + 2 x + 4 ) = x 2 + 3 x + 2 = ( x + 1 ) ( x + 2 ) To find out the critical points, we equate the above equation to zero. i.e. f ' ( x ) = 0 i.e. ( x + 1 ) ( x + 2 ) = 0 x = − 2 , x = − 1 Therefore, the critical points of the given function are x = − 2 , 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 33E

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To find all the critical points of the function y=13x3+32x2+2x+4 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 32E

Chapter 4.3, Problem 34E

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To find all the critical points of the function y = 1 3 x 3 + 3 2 x 2 + 2 x + 4 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=13x3+32x2+2x+4 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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