Explanation Given: The function is f ( x ) = ( x − 1 ) e − x . Calculation: Consider the function. f ( x ) = ( x − 1 ) e − x (1) Differentiate Equation (1) with respect to x . f ′ ( x ) = e − x + ( 1 − x ) e − x = ( 2 − x ) e − x (2) Substitute 0 for f ′ ( x ) in Equation (2) 0 = ( 2 − x ) e − x x = 2 Substitute 2 for x in Equation (1). f ( 2 ) = ( 2 − 1 ) e − 2 = e − 2 Therefore, the critical points for the function are ( 2 , e − 2 ) . Thus, the function is increasing over the interval ( − ∞ , 2 ) and the function is decreasing over the interval ( 2 , ∞ ) . Differentiate Equation (2) with respect to x .

2nd Edition

ISBN: 9780321964038

Author: GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher: Pearson Addison Wesley,

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Chapter 5.4, Problem 27E

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To Graph:

The function f(x)=(x−1)e−x by considering the domain, critical points, symmetry, regions where the function is increasing or decreasing, inflection point, regions where the function is concave upward or concave downward, intercepts where possible and asymptotes where applicable.

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Chapter 5.4, Problem 26E

Chapter 5.4, Problem 28E

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Solution Summary: The author analyzes the function by considering the domain, critical points, symmetry, regions where it is increasing or decreasing, inflection point, intercepts where possible and asymptotes where applicable.

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To Graph: The function f(x)=(x−1)e−x by considering the domain, critical points, symmetry, regions where the function is increasing or decreasing, inflection point, regions where the function is concave upward or concave downward, intercepts where possible and asymptotes where applicable.

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Chapter 5 Solutions

To Graph:

The function f(x)=(x−1)e−x by considering the domain, critical points, symmetry, regions where the function is increasing or decreasing, inflection point, regions where the function is concave upward or concave downward, intercepts where possible and asymptotes where applicable.