Explanation Given Information: The provided function is f ( x ) = 3 x 5 − 20 x 3 , the first derivative is f ′ ( x ) = 15 x 2 ( x − 2 ) ( x + 2 ) and second derivative is f ″ ( x ) = 60 x ( x 2 − 2 ) . Formula used: To find relative maxima and minima of a function, Step 1. Set the first derivative of the function to zero, f ′ ( x ) = 0 , to find the critical values of the function. Step 2. Substitute the critical values into f ( x ) and calculate the critical points. Step 3. Evaluate f ″ ( x ) at each critical value for which f ′ ( x ) = 0 . If f ″ ( x 0 ) < 0 , a relative maximum occurs at x 0 . If f ″ ( x 0 ) > 0 , a relative minimum occurs at x 0 . If f ″ ( x 0 ) = 0 or f ″ ( x 0 ) is undefined, the second derivative test fails and then use the first derivative test. Calculation: Consider the provided function, f ( x ) = 3 x 5 − 20 x 3 Now, consider the first derivative f ′ ( x ) = 15 x 2 ( x − 2 ) ( x + 2 ) . Now, to obtain the critical values, set f ′ ( x ) = 0 as, 15 x 2 ( x − 2 ) ( x + 2 ) = 0 Thus, either ( x − 2 ) ( x + 2 ) = 0 or 15 x 2 = 0 . First consider ( x − 2 ) ( x + 2 ) = 0 . Thus, x + 2 = 0 . Subtract 2 from both sides as below, x + 2 − 2 = 0 − 2 x = − 2 Now consider x − 2 = 0 . Add 2 on both sides as below, x − 2 + 2 = 0 + 2 x = 2 And, 15 x 2 = 0 x 2 = 0 x = 0 Thus, the critical values of the function are at x = − 2 , x = 0 and x = 2 . Now, substitute 0 for x in the function f ( x ) = 3 x 5 − 20 x 3 , f ( x ) = 3 ( 0 ) 5 − 20 ( 0 ) 3 = 0 Substitute 2 for x in the function f ( x ) = 3 x 5 − 20 x 3 , f ( x ) = 3 ( 2 ) 5 − 20 ( 2 ) 3 = 96 − 160 = − 64 And, substitute − 2 for x in the function f ( x ) = 3 x 5 − 20 x 3 , f ( x ) = 3 ( − 2 ) 5 − 20 ( − 2 ) 3 = − 96 + 160 = 64 Thus, the critical points are ( 0 , 0 ) , ( 2 , − 64 ) and ( − 2 , 64 ) . Now, consider the second derivative f ″ ( x ) = 60 x ( x 2 − 2 ) . To obtain point of inflection, set f ″ ( x ) = 0 as, 60 x ( x 2 − 2 ) = 0 Thus, either x 2 − 2 = 0 or 60 x = 0 . First consider x 2 − 2 = 0 . Add 2 on both sides and take square root, x 2 − 2 + 2 = 0 + 2 x 2 = 2 x = ± 2 And, 60 x = 0 x = 0 Now, substitute 0 for x in the function f ( x ) = 3 x 5 − 20 x 3 , f ( x ) = 3 ( 0 ) 5 − 20 ( 0 ) 3 = 0 Now, substitute 2 for x in the function f ( x ) = 3 x 5 − 20 x 3 , f ( x ) = 3 x 5 − 20 x 3 = 3 ( 2 ) 5 − 20 ( 2 ) 3 = 3 ( 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 10.2, Problem 21E

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To calculate: The relative minimum, relative maximum and points of inflection for the provided function f(x)=3x5−20x3 and the first derivative f′(x)=15x2(x−2)(x+2) and second derivative f″(x)=60x(x2−2) also sketch its graph.

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Chapter 10.2, Problem 20E

Chapter 10.2, Problem 22E

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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 relative minimum, relative maximum and points of inflection for the provided function f ( x ) = 3 x 5 − 20 x 3 and the first derivative f ′ ( x ) = 15 x 2 ( x − 2 ) ( x + 2 ) and second derivative f ″ ( x ) = 60 x ( x 2 − 2 ) also sketch its graph.

Solution Summary: The author explains how to calculate the relative minimum, relative maximum, and points of inflection for the provided function.

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To calculate: The relative minimum, relative maximum and points of inflection for the provided function f(x)=3x5−20x3 and the first derivative f′(x)=15x2(x−2)(x+2) and second derivative f″(x)=60x(x2−2) also sketch its graph.

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Chapter 10 Solutions