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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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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 5 x x 2

To determine

To graph: The sketch of the function y=ln5xx2 and label the relative extrema, points of inflection and asymptotes.

Explanation

Given Information:

The provided function is y=ln5xx2.

Graph:

Consider the function, y=ln5xx2

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

y=ddx(ln5xx2)=x2ddxln5xln5xddxx2(x2)2=x2×55xln5x×2x(x)4=x2xln5x(x)4

Further solve as,

y=x(12ln5x)(x)4=12ln5x(x)3

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

12ln5x(x)3=012ln5x=02ln5x=1ln5x=12

Further solve as,

5x=e12x=15×1.6487x0.33

Hence, the critical number is x=0.33.

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

y(x)=ddx[12ln5x(x)3]=(x)3ddx(12ln5x)(12ln5x)ddx(x)3((x)3)2=(x)3×(2×55x)(12ln5x)×3x2x6=2x23x2+6x2ln5xx6

Further solve as,

y(x)=5x2+6x2ln5xx6=x2(5+6ln5x)x6=6ln5x5x4

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

6ln5x5x4=06ln5x5=06ln5x=5ln5x=56

Further solve as,

5x=e56x=15×2.3009x0

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