Explanation Given Information: The provided function is y = ( x 2 − 2 x ) 2 and the derivative is d y d x = 4 x ( x − 1 ) ( x − 2 ) . Formula Used: The simple power rule of derivative, d d x ( x n ) = n x n − 1 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 − 2 x ) 2 , The critical values are the only values at which the graph can have turning points, the derivative cannot change sign anywhere except at the critical value. Hence, there will no change in the values of critical values as in the derivative graph. Now, equate the derivative to 0. d y d x = 0 4 x ( x − 1 ) ( x − 2 ) = 0 From the property of multiplication, 4 x = 0 x = 0 Or, x − 1 = 0 x = 1 Or, x − 2 = 0 x = 2 Therefore, the critical values are 0 , 1, and 2 . Substitute 0 for x in the function y = ( x 2 − 2 x ) 2 . y = ( ( 0 ) 2 − 2 ( 0 ) ) 2 = 0 − 0 = 0 Therefore, ( 0 , 0 ) is a critical point. Substitute 1 for x in the function y = ( x 2 − 2 x ) 2 . y = ( ( 1 ) 2 − 2 ( 1 ) ) 2 = ( 1 − 2 ) 2 = 1 Therefore, ( 1 , 1 ) is a critical point. Substitute 2 for x in the function y = ( x 2 − 2 x ) 2 . y = ( ( 2 ) 2 − 2 ( 2 ) ) 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 ( − ∞ , 0 ] , [ 0 , 1 ] , [ 1 , 2 ] and [ 2 , ∞ ) . For interval ( − ∞ , 0 ] : Substitute − 1 for x in the function y ′ = 4 x ( x − 1 ) ( 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 31E

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To calculate: The critical values, critical points, the relative maxima, relative minima, or horizontal points of inflection of function y=(x2−2x)2 and its derivative dydx=4x(x−1)(x−2) and also sketch the graph.

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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 − 2 x ) 2 and its derivative d y d x = 4 x ( x − 1 ) ( 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=(x2−2x)2 and its derivative dydx=4x(x−1)(x−2) and also sketch the graph.

Chapter 10 Solutions