Explanation Given: The function is, f ( x ) = x 2 e − x and the interval is, [-1, 3]. Calculation: Obtain the first derivative of the given function. f ' ( x ) = 2 x e − x − x 2 e − x Set f ′ ( x ) = 0 and obtain the critical numbers. 2 x e − x − x 2 e − x = 0 2 x e − x = x 2 e − x x = 0 , 2 The critical numbers are 0 and 2 which lies on the given interval [ − 1 , 3 ] . Apply the extreme values of the given interval and the critical numbers in f ( x ) . Substitute x = 0 in f ( x ) , f ( 0 ) = 0 2 e − 0 = 0 Substitute x = 2 in f ( x ) , f ( 2 ) = 2 2 e − 2 = 4 e 2 ≈ 0.54 Substitute x = − 1 in f ( x ) , f ( − 1 ) = ( − 1 ) 2 e − ( − 1 ) = e ≈ 2.72 Substitute x = 3 in f ( x ) , f ( 3 ) = 3 2 e − 3 = 9 e 3 ≈ 0

Single Variable Calculus: Early Transcendentals, Volume I

8th Edition

ISBN: 9781305270343

Author: James Stewart

Publisher: Cengage Learning

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Chapter 4, Problem 6RE

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To find: The local maximum and minimum and the absolute maximum and absolute minimum of the function f(x)=x2e−x on the interval [-1, 3].

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Chapter 4 Solutions

Single Variable Calculus: Early Transcendentals, Volume I

Ch. 4.1 - Explain the difference between an absolute minimum...Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Use the graph to state the absolute and local...Ch. 4.1 - Use the graph to state the absolute and local...Ch. 4.1 - Sketch the graph of a function f that is...Ch. 4.1 - Sketch the graph of a function f that is...Ch. 4.1 - Prob. 9E Ch. 4.1 - Prob. 10E

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The local maximum and minimum and the absolute maximum and absolute minimum of the function f ( x ) = x 2 e − x on the interval [-1, 3].

Solution Summary: The author explains how to find the local maximum, minimum, and absolute minimum of a given function.

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To find: The local maximum and minimum and the absolute maximum and absolute minimum of the function f(x)=x2e−x on the interval [-1, 3].

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Chapter 4 Solutions