   Chapter 4.5, Problem 68E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Use the guidelines of this section to sketch the curve. In guideline D find an equation of the slant asymptote. y = x 3 ( x + 1 ) 2

To determine

To Find: The equation of the slant asymptote and sketch the curve using the guide lines.

Explanation

Given:

The equation of the curve is,

f(x)=x3(x+1)2 (1).

Calculation:

A)

Compute the domain.

Considering the denominator in equation (1)

(x+1)2=0x2+1=0x=1

Therefore, domains are (,1) and (1,) .

B)

Compute y intercepts.

Substitute the value 0 for x in equation (1).

f(0)=03(0+1)2f(0)=0

Compute x intercepts.

Substitute the value 0 for f(x) in the numerator of equation (1).

f(x)=x3f(x)=0x=0

Therefore, x intercepts and y intercepts are 0 .

C)

Compute the symmetric:

Apply the negative and positive values for x in equation (1).

Substitute the 1 for x .

f(1)=13(1+1)2f(1)=

Substitute the 1 for x .

f(1)=13(1+1)2f(1)=14

f(x)f(x)

Therefore, the curve is no symmetry.

D)

Compute the vertical asymptote:

limx1x3+4x2=

limx1+x3+4x2=

Therefore, the vertical asymptote x is 1 .

Compute the horizontal asymptote:

In equation(1)  the value of numerator is greater than the denominator so the curve will not have the horizontal asymptote

Compute the slant asymptote using equation.

Rewrite the equation(1).

f(x)=x3(x+1)2f(x)=x3x2+2x+1

x2+2x+1x2x3(x3+2x2+x)_02x2x(2x24x2)_0+3x+2_

f(x)=(x2)+3x+2x2+2x+1f(x)=(x2)+3x+2(x+1)2

Therefore, the slant asymptote is y=(x2)

E)

Compute the interval of increase or decrease.

Apply UV method to differentiate the equation (2)

f'(x)=vu'uv'v2f'(x)=(x+1)23x2x32(x+1)((x+1)2)2f'(x)=x2(x+1)(3(x+1)2x)(x+1)4f'(x)=x3(x+3)(x+1)3

If

f'(x)>0

Substitute the interval <x<3 and 1<x< in the equation founded above,

f'(1)=13(1+3)(1+1)3f'(1)=48f'(1)=12

The f(x) increasing while substituting the limits (,3) and (1,) .

If

f'(x)<0

Substitute the interval 3<x<1 in the equation founded above,

f'(2)=23(2+3)(2+1)3f'(2)=81f'(2)=8

The f(x) decreasing while substituting the limit (3,1)

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