   Chapter 4.5, Problem 54E ### 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 = tan − 1 ( x − 1 x + 1 )

To determine

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

Explanation

Given:

The given curve is y=tan1(x1x+1) (1)

Calculation:

(a)

Calculate the domain.

Consider the denominator x+1 which is not equal to zero,

x+10x1

Therefore, the domain is (,1) and (1,) .

(b)

Calculate the intercepts.

Substitute 0 for x in the equation (1).

y=f(x)=tan1(x1x+1)f(0)=tan1(010+1)=tan1(1)=π4

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

y=f(x)=tan1(x1x+1)0=tan1(x1x+1)tan0=(x1x+1)0=(x1x+1)x1=0x=1

Therefore, the intercept of x and y intercept are x=1 and y=π4 .

(c)

Calculate the symmetry.

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

Substitute -1 for x in the equation (1).

f(x)=tan1(x1x+1)=tan1(111+1)=

Substitute +1 for x in the equation (1).

f(x)=tan1(x1x+1)=tan1(111+1)=0

Therefore, the conditions f(x)f(x) and f(x)-f(x) are true and the symmetry condition is false. Hence, there is no symmetry about y axis and origin.

(d)

Calculate asymptotes.

Substitute ± for x using limit in the equation (1).

limx±f(x)=limxtan1(x1x+1)=tan1(1+1)=π4

Therefore, the horizontal asymptotes are y=π4 .

Substitute ±1 for x using limit in the equation (1).

limx1+f(x)=limx1tan1(x1x+1)=tan1(111+1)=π2limx1f(x)=limx1tan1(x1x+1)=tan1(111+1)=π2

Therefore, the point (1,π2) and (1,π2) are not defined

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