   Chapter 4.5, Problem 72E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Analyzing a Graph In Exercises 69-72, analyze and sketch the graph of the function. Label any relative extrema, points of inflection, and asymptotes. y = ( ln x ) 2

To determine

To graph: The sketch of the function y=(lnx)2 and label the relative extrema, points of inflection and asymptotes.

Explanation

Given Information:

The provided function is y=(lnx)2.

Graph:

Consider the function, y=(lnx)2

Evaluate the first derivative of the function, by applying chain rule.

y=ddx(lnx)2=2lnx(1x)=2lnxx

Set the derivative equal to zero to compute the value of x.

2lnxx=02lnx=0lnx=0x=e0

Further solve as,

x=1

Hence, the critical number is x=1.

Now, evaluate the second derivative of the function by applying quotient rule.

y(x)=ddx[2lnxx]=xddx2lnx2lnxddxxx2=2x×1x2lnxx2=22lnxx2

Set the second derivative equal to zero, to determine the value of x.

22lnxx2=022lnx=02lnx=2lnx=1

Further solve as,

x=e1x=2.718

Substitute x=2.718 in the function.

y(2.718)=(ln2.718)2=(0.99989)21

Therefore, the point of inflection is (2

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