   Chapter 4.5, Problem 26E ### 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 2 − x 2

To determine

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

Explanation

Given:

The given curve is y=x2x2 (1)

Calculation:

(a)

Calculate the domain.

Assume the value inside the radical is greater than or equal to zero.

2x2=0x2=2x=±2

Therefore the domain is (2,2)

(b)

Calculate the intercepts.

Calculate the value of x-intercept.

Take y=0 then,

y=x2x20=x2x2Equate zero for both the xx=02x2=02x2=02=x2x=±2

Therefore, the x -intercept is (0,±2)

(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)=x2x2f(1)=(1)2(1)2=3

Substitute +1 for x in the equation (1)

f(x)=x2x2f(1)=12(1)2=3

Hence, f(x)=f(x) so, function f is odd and the graph is symmetric about the origin.

(d)

Calculate asymptotes.

The function is defined over its entire domain that is the closed interval so there are no asymptotes.

(e)

Calculate the intervals.

Differentiate the equation (1).

f(x)=x2x2f'(x)=[x.12(2x2)12.(2x)]+[2x2.(1)]=x22x2+2x2=x2+(2x22x2)2x2=x2+2x22x2=2x2+22x2=2(1x2)2x2=2(1+x)(1x)2x2(a2b2=(a+b)(ab))

Therefore, f'(x) is negative for 2<x<1 and 1<x<2 , and f'(x) is positive for 1<x<1 . Thus, f'(x) is increasing on (1,1) and decreasing on (2,1) and (1,2)

(f)

Here, the minimum value appears when f'(x)<0 within the intervals and a maximum value appears when f'(x)>0 within the intervals.

Calculate the local minimum and maximum values.

Substitute -1 for x between the intervals (1,1) in the equation (1).

f(x)=x2x2=(1)2(1)2=1<0

Therefore, the local minimum at (1,1)

Substitute 1 for x between the intervals (1,2) in the equation (1)

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