Chapter 4.5, Problem 18E

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

8th Edition
James Stewart
ISBN: 9781305270336

Chapter
Section

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

8th Edition
James Stewart
ISBN: 9781305270336
Textbook Problem

# Use the guidelines of this section to sketch the curve. y = x x 3 − 1

To determine

To Find: sketch the curve using the equation y=xx31 .

Explanation

Given:

The equation of the curve is as below.

f(x)=xx3âˆ’1 (1).

Calculation:

A)

Compute the domain.

Consider the denominator in equation (1).

x3âˆ’1=0x3=1x=13x=1

Therefore, domains are (âˆ’âˆž,1) and (1,âˆž) .

B)

Compute y intercepts.

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

f(0)=xx3âˆ’1f(0)=003âˆ’1f(0)=0

Compute the x intercept.

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

f(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 âˆ’1 for x .

f(âˆ’1)=âˆ’1âˆ’13âˆ’1f(âˆ’1)=12

Substitute the 1 for x .

f(1)=113âˆ’1f(1)=âˆž

Therefore, the curve has no symmetry.

D)

Compute the horizontal asymptote:

limxâ†’Â±âˆžxx3âˆ’1=0

Therefore, the horizontal asymptote y is 0 .

Compute the vertical asymptote:

limxâ†’1âˆ’xx3âˆ’1=âˆ’âˆž

limxâ†’1+xx3âˆ’1=âˆž

Therefore, the vertical asymptote x is 1 .

E)

Compute the interval of increase or decrease.

Apply UV method to differentiate the equation (2)

f'(x)=vu'âˆ’uv'v2f'(x)=(x3âˆ’1)(1)âˆ’x(3x2)(x3âˆ’1)2f'(x)=âˆ’2x3âˆ’1(x3âˆ’1)2

If the condition f'(x)>0 is true, substitute the interval x<âˆ’123 in the equation.

f'(âˆ’13)=âˆ’2(âˆ’13)3âˆ’1((âˆ’13)3âˆ’1)2f'(âˆ’13)=14

The f(x) increasing while substituting the limits (âˆ’âˆž,âˆ’123) .

Substitute the interval âˆ’123<x<1 in the equation founded above,

f'(0.5)=âˆ’2(0.5)3âˆ’1((0.5)3âˆ’1)2f'(0.5)=âˆ’8049

The function f(x) is decreasing while substituting the limits (âˆ’123,1) .

Then, the function is decreasing on (âˆ’123,1) and increasing on (âˆ’âˆž,âˆ’123) .

F)

Find the local maximum and minimum values.

For minima:

If the condition f(x)â‰¤0 is true, then the minima will occur.

f(1)=113âˆ’1f(1)=âˆž

If the condition f(x)â‰¥0 is true, then the maxima will occur.

f(âˆ’123)=âˆ’123âˆ’123âˆ’1f(âˆ’123)=23123

Therefore, the curve has no minima and the maxima is 23123

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