Explanation Given Information: The provided equation is f ( x ) = 1 − 3 x + 3 x 2 − x 3 . Formula Used: 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. This, in an interval between two critical values, the sign of the derivative at any value in the interval will be the sign of the derivative at all values in the interval. As per the First Derivative Test, The first derivative of the function is evaluated. The first derivative is made equal to zero in order to get the critical points. The values of the critical values are kept inside the original function which gives the critical points. The intervals of the values of x are then evaluated for the relative maximum and minimum. Calculation: Consider the provided equation f ( x ) = 1 − 3 x + 3 x 2 − x 3 , 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, f ′ ( x ) = − 3 + 6 x − 3 x 2 Put the value of y ′ = 0 , f ′ ( x ) = − 3 + 6 x − 3 x 2 = 0 = − 3 ( 1 − 2 x + x 2 ) = 0 = 1 − 2 x + x 2 = 0 Hence, solve the quadratic equation : 1 − 2 x + x 2 = 0 1 − x − x + x 2 = 0 ( 1 − x ) − x ( 1 − x ) = 0 Hence, the values of x is x = 1

Mathematical Applications for the Management, Life, and Social Sciences

12th Edition

ISBN: 9781337625340

Author: Ronald J. Harshbarger, James J. Reynolds

Publisher: Cengage Learning

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Chapter 10, Problem 3RE

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To calculate: The critical values, relative maxima, relative minima, or horizontal points of inflection of the function f(x)=1−3x+3x2−x3 by using the derivative.

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Chapter 10, Problem 2RE

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Chapter 10 Solutions

Mathematical Applications for the Management, Life, and Social Sciences

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, relative maxima , relative minima , or horizontal points of inflection of the function f ( x ) = 1 − 3 x + 3 x 2 − x 3 by using the derivative.

Solution Summary: The author calculates the critical values, relative maxima and relative minima, and horizontal points of inflection of the function f(x) by using the derivative.

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To calculate: The critical values, relative maxima, relative minima, or horizontal points of inflection of the function f(x)=1−3x+3x2−x3 by using the derivative.

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Chapter 10 Solutions