   Chapter 4.5, Problem 67E ### 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 + 4 x 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 as below.

f(x)=x3+4x2 (1).

Calculation:

A)

Compute the domain.

Considering the denominator in equation (1)

x2=0x=0

Therefore, domains are (,0) and (0,) .

B)

Compute y intercepts.

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

f(0)=03+402f(0)=

Compute x intercepts.

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

f(x)=x3+4f(x)=0x=43

Therefore, x intercepts 43 and no y intercepts.

C)

Compute the symmetric:

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

Substitute the 1 for x .

f(1)=(1)3+4(1)2f(1)=3

Substitute the 1 for x .

f(1)=(1)3+4(1)2f(1)=5

f(x)f(x)

Therefore, the curve is no symmetry.

D)

Compute the vertical asymptote:

limx0x3+4x2=

Therefore, the vertical asymptote x is 0 .

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.

f(x)=x3+4x2

x2xx3+4(x3)_0+4_

f(x)=x+4x2

Therefore, the slant asymptote is y=x

E)

Compute the interval of increase or decrease.

Apply UV method to differentiate the equation (2)

f'(x)=vu'uv'v2f'(x)=x2(3x2+0)(x3+4)2xx4f'(x)=3x4(x3+4)2xx4f'(x)=x38x3

If

f'(x)>0

Substitute the interval <x<0 and 2<x< in the equation founded above,

f'(3)=33833f'(3)=1927

The f(x) increasing while substituting the limits (,0) and (2,)

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