Explanation Given Information: The provided function is y = ( x 2 − 4 ) 2 and the derivative is d y d x = 4 x ( x + 2 ) ( x − 2 ) . Formula Used: To obtain the relative maxima and minima of a function, 1. Find the first derivative of the function. 2. Set the derivative equal to 0 to obtain the critical points. 3. Evaluate f ′ ( x ) at values of x to the left and one to the right of each critical point to develop a sign diagram. (a) If f ′ ( x ) > 0 to the left and f ′ ( x ) < 0 to the right of the critical value, the critical point is a relative maximum point. (b) If f ′ ( x ) < 0 to the left and f ′ ( x ) > 0 to the right of the critical value, the critical point is a relative minimum point. Calculation: Consider the provided function y = ( x 2 − 4 ) 2 , Now, equate the derivative to 0. d y d x = 0 4 x ( x + 2 ) ( x − 2 ) = 0 From the property of multiplication, 4 x = 0 x = 0 Or, x + 2 = 0 x = − 2 Or, x − 2 = 0 x = 2 Therefore, the critical values are 0 , − 2, and 2 . Substitute 0 for x in the function y = ( x 2 − 4 ) 2 . y = ( ( 0 ) 2 − 4 ) 2 = ( 0 − 4 ) 2 = 16 Therefore, ( 0 , 16 ) is a critical point. Substitute 2 for x in the function y = ( x 2 − 4 ) 2 . y = ( ( 2 ) 2 − 4 ) 2 = ( 4 − 4 ) 2 = 0 Therefore, ( 2 , 0 ) is a critical point. Substitute − 2 for x in the function y = ( x 2 − 4 ) 2 . y = ( ( − 2 ) 2 − 4 ) 2 = ( 4 − 4 ) 2 = 0 Therefore, ( − 2 , 0 ) is a critical point. Test the derivative at the left and right sides of all the critical points. Therefore the intervals to check the maxima, minima and point of inflection are ( − ∞ , − 2 ] , [ − 2 , 0 ] , [ 0 , 2 ] and [ 2 , ∞ ) . For interval ( − ∞ , − 2 ] : Substitute − 3 for x in the function y ′ = 4 x ( x + 2 ) ( x − 2 )

WebAssign Printed Access Card for Harshbarger/Reynolds' Mathematical Applications for the Management, Life, and Social Sciences, 12th Edition, Multi-Term

12th Edition

ISBN: 9781337630542

Author: Ronald J. Harshbarger, James J. Reynolds

Publisher: Cengage Learning

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Chapter 10.1, Problem 32E

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To calculate: The critical values, critical points, the relative maxima, relative minima, or horizontal points of inflection of function y=(x2−4)2 and its derivative dydx=4x(x+2)(x−2) and also sketch the graph.

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Chapter 10.1, Problem 31E

Chapter 10.1, Problem 33E

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Chapter 10 Solutions

WebAssign Printed Access Card for Harshbarger/Reynolds' Mathematical Applications for the Management, Life, and Social Sciences, 12th Edition, Multi-Term

Ch. 10.1 - 1. The x- values of critical points are found...Ch. 10.1 - Prob. 2CP Ch. 10.1 - Prob. 3CP Ch. 10.1 - Prob. 4CP Ch. 10.1 - In Problems 1 and 2, use the indicated points on...Ch. 10.1 - In Problems 1 and 2, use the indicated points on...Ch. 10.1 - 3. Use the graph of in Problem 1 to identify at...Ch. 10.1 - 4. Use the graph of in Problem 2 to identify at...Ch. 10.1 - In Problems 5 and 6, use the sign diagram for to...Ch. 10.1 - In Problems 5 and 6, use the sign diagram for to...

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The critical values, critical points, the relative maxima , relative minima , or horizontal points of inflection of function y = ( x 2 − 4 ) 2 and its derivative d y d x = 4 x ( x + 2 ) ( x − 2 ) and also sketch the graph.

Solution Summary: The author explains how to calculate the critical values, critical points, relative maxima, and relative minima of a function.

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To calculate: The critical values, critical points, the relative maxima, relative minima, or horizontal points of inflection of function y=(x2−4)2 and its derivative dydx=4x(x+2)(x−2) and also sketch the graph.

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Chapter 10 Solutions