   Chapter 4.5, Problem 71E ### 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 2 ln x 4

To determine

To graph: The sketch of the function y=x2lnx4 and label the relative extrema, points of inflection and asymptotes.

Explanation

Given Information:

The provided function is y=x2lnx4.

Graph:

Consider the function, y=x2lnx4

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

y=ddx(x2lnx4)=x2ddxlnx4+lnx4ddxx2=(14)x21x4+lnx42x=x+2xlnx4

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

x+2xlnx4=02xlnx4=xlnx4=x2xlnx4=12

Further solve as,

x4=e12x=4×0.6065x2.426

Hence, the critical number is x=2.426.

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

y(x)=ddx[x+2xlnx4]=ddxx+ddx[2xlnx4]=1+lnx4ddx(2x)+2xddxlnx4=1+2lnx4+2x×1x4×14

Further solve as,

y(x)=1+2lnx4+2=3+2lnx4

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

3+2lnx4=02lnx4=3lnx4=32x4=e32

Further solve as,

x=4e32x=4×0.2231x0.8924

Substitute x=0.8924 in f(x),

y(0.8924)=(0.8924)2ln0.89244=0.7963×(ln0.8924ln4)=0.7963×(0

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