   Chapter 4.5, Problem 11E ### 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 2 2 − 3 x + x 2

To determine

To Find: Sketch the curve using the equation y=xx223x+x2 .

Explanation

Given:

The equation of the curve is as below.

y=xx223x+x2 (1).

Calculation:

Rewrite the equation (1) as below.

y=xx223x+x2y=x(1x)(1x)(2x)

y=x2x (2)

Find the hole in the curve.

Considering the numerator equation (1)

xx2=0x(1x)=0x=0(or)x=1

There is no curve in the point of x when it is 1 .

Substitute the value 1 for x in the equation (2).

y=121y=1

Therefore, the hole is at (1,1) .

A)

Compute the domain.

Considering the denominator equation (1).

23x+x2=0

Factorize the above equation.

(1x)(2x)=0x=1,2

Therefore, domains are (,1) , (1,2) and (2,) .

B)

Compute x and y intercepts.

Substitute the value 0 for x in the equation.

y=020y=0

Therefore, both x 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(x)=1(1)223(1)+(1)2f(x)=26f(x)=13

Substitute the value of 1 for x .

f(x)=1(1)223(1)+(1)2f(x)=26f(x)=02

The above value of f(x) is not same for positive and negative value of x .

Therefore, the curve is no symmetric.

D)

Compute the horizontal asymptote.

limx±x2x=1

Therefore, the horizontal asymptote y is 1 .

Compute the vertical asymptote.

limx2x2x=

limx2+x2x=

Therefore, the vertical asymptote x is 2 .

E)

Apply UV method to differentiate the equation (2).

f'(x)=vu'uv'v2f'(x)=(2x)1x(1)(2x)2f'(x)=2(2x)2>0

Apply the value of 1 for x in the above equation

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