Explanation Given Information: The provided function is y = 1 3 x 3 + x 2 − 24 x + 20 . 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. Explanation: Consider the provided function y = 1 3 x 3 + x 2 − 24 x + 20 , 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. Take out the first derivative of the equation by the power rule, y ′ = x 2 + 2 x − 24 = x 2 + 6 x − 4 x − 24 = ( x + 6 ) ( x − 4 ) Put the value of y ′ = 0 , y ′ = ( x + 6 ) ( x − 4 ) = 0 Evaluate the values of x from the equation: ( x + 6 ) ( x − 4 ) = 0 x = 4 , − 6 Hence, the value(s) of x are x = 4 , − 6 . Evaluate the values of the original functions with the critical values: Put x = 4 in the equation y = 1 3 x 3 + x 2 − 24 x + 20 , y = 1 3 ( 4 ) 3 + ( 4 ) 2 − 24 ( 4 ) + 20 = 64 3 + 16 − 96 + 20 = − 116 3 Hence, ( 4 , − 116 3 ) is a critical point. Put x = − 6 in the equation y = 1 3 x 3 + x 2 − 24 x + 20 , y = 1 3 ( − 6 ) 3 + ( − 6 ) 2 − 24 ( − 6 ) + 20 = − 216 3 + 36 + 144 + 20 = 128 Hence, ( − 6 , 128 ) is a critical point. Test the derivative at the left and right sides of all the critical points. Left of -6 is x = − 7 , On the right is x = 0

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 27E

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To calculate: The relative maxima, relative minima, or horizontal points of inflection by observing the graph y=13x3+x2−24x+20.

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Chapter 10.1, Problem 26E

Chapter 10.1, Problem 28E

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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 relative maxima , relative minima , or horizontal points of inflection by observing the graph y = 1 3 x 3 + x 2 − 24 x + 20 .

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To calculate: The relative maxima, relative minima, or horizontal points of inflection by observing the graph y=13x3+x2−24x+20.

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