(a) Explanation Given: The function is f ( x ) = x 2 ln x . Calculation: Obtain the derivative of f ( x ) . f ′ ( x ) = d d x ( x 2 ln x ) Apply the Product Rule, f ′ ( x ) = x 2 ( ln x ) ′ + ( ln x ) ( x 2 ) ′ = x 2 ( 1 x ) + ln x ( 2 x ) = x ( 1 + 2 ln x ) Set f ′ ( x ) = 0 and obtain the critical points. x ( 1 + 2 ln x ) = 0 x = 0 , 1 + 2 ln x = 0 x = 0 , ln x = − 1 2 x = 0 , x = e − 1 2 Hence, the critical points occurs at x = 0 , x = e − 1 2 ( x = 1 e ≈ 0.6065 ) and in the interval notation ( 0 , 1 e ) ∪ ( 1 e , ∞ ) . Determine the nature of f ′ ( x ) as follows (b) To determine To find: The local maximum and the minimum values of f ( x ) = x 2 ln x . (c) To determine To find: The intervals of concavity and the inflection points of f ( x ) = x 2 ln x .

Single Variable Calculus: Early Transcendentals

8th Edition

ISBN: 9781305270336

Author: James Stewart

Publisher: Cengage Learning

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Chapter 4.3, Problem 16E

(a)

To determine

To find: The intervals on which the function f(x)=x2lnx is increasing or decreasing.

(b)

To determine

To find: The local maximum and the minimum values of f(x)=x2lnx .

(c)

To determine

To find: The intervals of concavity and the inflection points of f(x)=x2lnx .

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Chapter 4.3, Problem 15E

Chapter 4.3, Problem 17E

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Chapter 4 Solutions

Single Variable Calculus: Early Transcendentals

Ch. 4.1 - Explain the difference between an absolute minimum...Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - For each of the numbers a, b, c, d, r, and s,...Ch. 4.1 - Prob. 5E Ch. 4.1 - Use the graph to state the absolute and local...Ch. 4.1 - Prob. 7E 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. 10E

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The intervals on which the function f ( x ) = x 2 ln x is increasing or decreasing.

Solution Summary: The author explains how to find the intervals on which the function f(x) is increasing or decreasing.

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To find: The intervals on which the function f(x)=x2lnx is increasing or decreasing.

To find: The local maximum and the minimum values of f(x)=x2lnx .

To find: The intervals of concavity and the inflection points of f(x)=x2lnx .

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Chapter 4 Solutions