Explanation Given Information: The provided function is y = x 3 ( x − 5 ) 2 27 and the derivative is d y d x = 5 x 2 ( x − 3 ) ( x − 5 ) 27 . 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 3 ( x − 5 ) 2 27 , and the derivative d y d x = 5 x 2 ( x − 3 ) ( x − 5 ) 27 , Now, equate the derivative to 0. d y d x = 0 5 x 2 ( x − 3 ) ( x − 5 ) 27 = 0 Evaluate the values of x from the equation: 5 x 2 ( x − 3 ) ( x − 5 ) 27 = 0 x = 0 , 5 , 3 Hence, the value(s) of x are x = 0 , 5 , 3 . Evaluate the values of the original functions with the critical values: Substitute 0 for x in the function y = x 3 ( x − 5 ) 2 27 . y = ( 0 ) 3 ( 0 − 5 ) 2 27 = 0 27 = 0 Hence, ( 0 , 0 ) is a critical point. Put x = 3 in the equation y = x 3 ( x − 5 ) 2 27 , y = ( 3 ) 3 ( 3 − 5 ) 2 27 = 27 ( 4 ) 27 = 4 Hence, ( 3 , 4 ) is a critical point. Put x = 5 in the equation y = x 3 ( x − 5 ) 2 27 , y = ( 5 ) 3 ( 5 − 5 ) 2 27 = 125 ( 0 ) 27 = 0 Hence, ( 5 , 0 ) is a critical point

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 33E

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To calculate: The critical values, critical points, the relative maxima, relative minima, or horizontal points of inflection of the function y=x3(x−5)227 and the derivative dydx=5x2(x−3)(x−5)27 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 the function y = x 3 ( x − 5 ) 2 27 and the derivative d y d x = 5 x 2 ( x − 3 ) ( x − 5 ) 27 and also sketch the graph.

Solution Summary: The author calculates the critical values, critical points, the relative maxima, relative minima and horizontal points of inflection of the function.

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To calculate: The critical values, critical points, the relative maxima, relative minima, or horizontal points of inflection of the function y=x3(x−5)227 and the derivative dydx=5x2(x−3)(x−5)27 and also sketch the graph.

Chapter 10 Solutions