   Chapter 4.5, Problem 50E ### 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 = ln(1 + x3)

To determine

To Find: The curve by plotting x and y coordinates using guideline.

Explanation

Given:

The given curve is as below.

f(x)=y=ln(1+x3) (1)

Calculation:

(a)

Calculate the domain.

The function f(x) is undefined for x<0 .

The domain is x(1,) .

(b)

Calculate the intercepts.

Calculate the y intercept.

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

f(x)=ln(1+x3)

f(0)=ln(1+03)=ln0=0

The y intercept occurs at y=0 .

The x intercept occurs at x=0 .

(c)

Calculate the symmetry.

Substitute x for x in the equation (1).

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

The function is neither even nor odd. Therefore, the curve has no y axis or rotational symmetry.

(d)

Calculate asymptotes.

Apply limit of x tends to (x) in the equation (1).

limxf(x)=ln(1+x3)=ln(1+3)=

Apply limit of x tends to (x) in the equation (1).

limxf(x)=ln(1+x3)=ln(1+()3)=

Horizontal asymptotes do not exist.

Apply limit of y tends to (y) in the equation (1).

limyf(x)=ln(1+x3)

=ln(1+x3)ln(0)=ln(1+x3)ln(0)=0=1+x3x3=1x=1

It has the vertical asymptote x=1 .

(e)

Calculate the intervals.

f(x)=ln(1+x3)

Differentiate the equation (1) with respect to x .

ddx(lnx)=1x+ddxx

Write the expression for f'(x) as below.

f'(x)=11+x3(3x2)

f'(x)=3x21+x3 (2)

Equate equation (2) to zero.

3x21+x3=0x=0

Consider the limit (0,) .

Substitute 1 for x in equation (2).

f'(1)=3(1)21+(1)3=32=1.5>0

The function f(x) is increasing for domain (0,) .

Consider the limit (,0) .

Substitute 2 for x in equation (2).

f'(2)=3(2)21+(2)3=127=1.71<0

The function f(x)<0 except when the condition x=1 is true

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