   Chapter 4.5, Problem 68E ### 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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# Finding Relative Extrema In Exercises 65-68, find all relative extrema of the function. Use a graphing utility to verify your result. See Example 6. y = ln ( x 3 + 3 x 2 + 2 )

To determine

To calculate: The relative extrema of the function y=ln(x3+3x2+2).

Explanation

Given Information:

The provided function is y=ln(x3+3x2+2).

Formula used:

Rules for derivatives:

Sum rule of derivative:

(f(x)+g(x))'=f'(x)+g'(x)

Derivative of function f(x)=ln x is,

ddxlnx=1x

Constant multiple rule is f'(cx)=cf(x) where c is constant.

Derivative of function y=xn using power rule is dydx=nxn1.

If f(x) and g(x) are two functions then quotient rule of derivative of functions is,

ddx[f(x)g(x)]=g(x)f(x)f(x)g(x)[g(x)]2

Calculation:

Consider the function y=ln(x3+3x2+2).

First, evaluate the first derivative of the function y=ln(x3+3x2+2).

dydx=ddxln(x3+3x2+2)

Apply the chain rule,

ddxln(x3+3x2+2)=ddxln(x3+3x2+2)×ddx(x3+3x2+2)=1(x3+3x2+2)[ddxx3+ddx3x2+ddx2]=1(x3+3x2+2)[3x2+6x+0]=3x2+6xx3+3x2+2

Hence, the first derivative of the function is dydx=3x2+6xx3+3x2+2.

Now, set the derivative equal to 0.

3x2+6xx3+3x2+2=03x2+6x=03x(x+2)=03x=0 or x+2=0

Further solve as,

x+2=0x=2

Hence, the critical numbers are x=0,x=2.

Evaluate the second derivative of y=ln(x3+3x2+2) as,

ddx(3x2+6xx3+3x2+2)=(x3+3x2+2)ddx(3x2+6x)(3x2+6x)ddx(x3+3x2+2)(x3+3x2+2)2=(x3+3x2+2)(6x+6)(3x2+6x)(3x2+6x)(x3+3x2+2)2=6x4+6x3+18x3+18x2+12x+12(9x4+18x3+18

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