   Chapter 4.5, Problem 75E ### 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

# Discuss the asymptotic behavior of f(x) = (x4 + 1)/x in the same manner as in Exercise 74. Then use your results to help sketch the graph of f.

To determine

To show: Applying the expression limx[f(x)x3]=0 shows that the graph of f approaches the graph of y=x3 and y=f(x) is asymptotic to the parabola y=x3 and to sketch the curve.

Explanation

Given:

The equation of the curve is as below.

f(x)=(x4+1)x (1)

Calculation:

Apply the condition limx[f(x)x3]=0.

limx[f(x)x3]=limx(x4+1xx3)=limx(x4+1x4x)=limx1x=0

Therefore, the condition limx[f(x)x3]=0 is true.

Clearly the curve y=f(x) is asymptotic to y=x3.

Compute the domain.

Consider the denominator in equation (1)

x=0

Therefore, the domain is (,0)(0,).

Compute the y-intercepts.

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

f(0)=(04+1)0f(0)=

Compute x intercepts.

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

f(x)=0x4+1=0x4=1x=14

So the equation has no real solution.

Therefore, there are no x and y intercepts.

Compute the symmetric.

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

Substitute the value of 1 for x.

f(1)=(14+1)1f(1)=21f(1)=2

Substitute the value of 1 for x.

f(1)=(14+1)1f(1)=21f(1)=2

f(x)=f(x)f(x)

So the function is odd.

Therefore, the curve is symmetric about the origin.

Compute the vertical asymptote.

limx0+(x4+1)x=

limx0(x4+1)x=

Therefore, the vertical asymptote is x=0_.

Compute the horizontal asymptote.

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

Then, the curve y=f(x) is asymptotic to the parabola y=x3.

Compute the interval of increase or decrease.

Differentiate equation (1) as follows.

f'(x)=ddx((x4+1)x)f'(x)=x(4x3+0)(x4+1)x2f'(x)=3x41x2

Check whether the condition f'(x)>0 is true inside the limit <x<134.

Substitute the value 2 for x in the above equation.

f'(2)=3(2)41(2)2f'(2)=11.75

Hence, the condition f'(x)>0 is true.

Therefore, the function f(x) is increasing on (,134).

Check whether the condition f'(x)>0 is true in the limit 134<x<.

Substitute the value 2 for x in the above equation.

f'(2)=3(2)41(2)2f'(2)=11.75

Hence, the condition f'(x)>0 is true.

Therefore, the function f(x) is increasing on (134,).

Check whether the condition f'(x)<0 is true in the limit 134<x<0.

Substitute the value 0.5 for in the above equation.

f'(0.5)=3(0.5)41(0.5)2f'(0.5)=3.25<0

Hence, the condition f'(x)<0 is true.

Therefore, the function f(x) is decreasing inside the domain (134,0).

Check whether the condition f'(x)<0 is true in the limit 0<x<134

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