   Chapter 4.5, Problem 69E ### 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 = x ln   x

To determine

To graph: The sketch of the function y=xlnx and label the relative extrema, points of inflection and asymptotes.

Explanation

Given Information:

The provided function is y=xlnx.

Graph:

Consider the function y=xlnx.

Evaluate the first derivative of the function, by applying quotient rule and constant multiple rule.

y=ddx(xlnx)=lnxddxxxddxlnx(lnx)2=lnx×1x×1x(lnx)2=lnx1(lnx)2

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

lnx1(lnx)2=0lnx1=0lnx=1x=e1

Further solve as,

x2.718

Hence, the critical number is x=2.718.

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

y(x)=ddx[lnx1(lnx)2]=(lnx)2ddx(lnx1)(lnx1)ddx(lnx)2((lnx)2)2=(lnx)2×1x(lnx1)2lnx((lnx)2)2=(lnx)22(lnx)2+2lnxx((lnx)2)2

Further solve as,

y(x)=lnx(lnx+2)x((lnx)2)2=(lnx+2)x(lnx)3

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

(lnx+2)x(lnx)3=0lnx+2=0lnx=2x=e2

Further solve as,

x7.3890

Substitute x=7

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