   Chapter 4.5, Problem 18E ### 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. 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)=xx31 (1).

Calculation:

A)

Compute the domain.

Consider the denominator in equation (1).

x31=0x3=1x=13x=1

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

B)

Compute y intercepts.

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

f(0)=xx31f(0)=0031f(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 1 for x .

f(1)=1131f(1)=12

Substitute the 1 for x .

f(1)=1131f(1)=

Therefore, the curve has no symmetry.

D)

Compute the horizontal asymptote:

limx±xx31=0

Therefore, the horizontal asymptote y is 0 .

Compute the vertical asymptote:

limx1xx31=

limx1+xx31=

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'uv'v2f'(x)=(x31)(1)x(3x2)(x31)2f'(x)=2x31(x31)2

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

f'(13)=2(13)31((13)31)2f'(13)=14

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

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

f'(0.5)=2(0.5)31((0.5)31)2f'(0.5)=8049

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

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

F)

Find the local maximum and minimum values.

For minima:

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

f(1)=1131f(1)=

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

f(123)=1231231f(123)=23123

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

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