   Chapter 4.5, Problem 49E ### 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(sin x)

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=lnsinx (1)

Calculation:

(a)

Calculate the domain.

The function f(x) is undefined for sinx=0 .

Therefore, the domain is x(2nπ,π+2nπ) .

(b)

Calculate the intercepts.

Calculate the y intercept.

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

f(x)=lnsinx

f(0)=lnsin0=ln0=

The y intercept does not exist.

Calculate the value of x intercept.

Substitute 0 for y in the equation (1).

lnsinx=0

Here the term ln(1)=0 .

So sinx=1

Therefore, x intercept occurs when the condition x=π2+2nπ is true.

(c)

Calculate the symmetry.

Substitute x for x in the equation (1).

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

The function is neither even nor odd. Therefore, the curve is not symmetric about y axis and origin.

(d)

Calculate asymptotes.

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

limxnπf(x)=lnsin(nπ)=lnsin(nπ)=

It has the vertical asymptotes x=nπ . Horizontal asymptotes do not exist.

(e)

Calculate the intervals.

f(x)=lnsinx

Differentiate the equation (1) with respect to x .

f'(x)=dln(sinx)dsinx×dsinxdx=d(sinx)sinx

f'(x)=cotx (2)

So equate equation (2) to zero.

cotx=0x=cot1(0)=π2+nπ

Consider the limit (0,π)

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