   Chapter 4.5, Problem 15E ### 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 + 3

To determine

To find: sketch the curve using the equation y=x2x2+3 .

Explanation

Given:

The equation of the curve is,

y=x2x2+3 (1).

Calculation:

Rewrite the equation (1).

f(x)=x2x2+3f(x)=(x2+3)3(x2+3)f(x)=13x2+3

f(x)=13x2+3 (2)

A)

Compute the domain.

Considering the denominator in equation (1)

x2+3=0x=±

The denominator is never be zero.  So, D = R

Therefore, domains are (,) .

B)

Compute y intercepts.

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

y=0202+3y=0

Compute x intercepts.

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

x2=0x=0

Therefore, y intercepts and x intercepts are (0,0) .

C)

Compute the symmetric:

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

Substitute the 1 for x .

f(x)=x2x2+3f(1)=12(1)2+3f(1)=14

Substitute the 1 for x .

f(x)=x2x2+3f(1)=12(1)2+3f(1)=14

f(x)=f(x)

so, the function is even.

Therefore, the curve is symmetric about the y axis.

D)

Compute the horizontal asymptote:

limx±x2x2+3=1

Therefore, the horizontal asymptote y is 1 .

Compute the vertical asymptote:

The domain is all real number.  So, there are no vertical asymptotes

Therefore, there is no vertical asymptote.

E)

Compute the interval of increase or decrease.

Apply UV method to differentiate the equation (2)

f'(x)=vu'uv'v2f'(x)=2x(x2+3)(x2)(2x)(x2+3)2f'(x)=2x3+6x2x3(x2+3)2f'(x)=6x(x2+3)2

The equation has negative sign before, so, it is always negative.

Using reciprocal rule

The inequality got reversed.

0=6xx=0

For the interval substitute the negative value and positive value in the equation we founded above.

f'(1)=6×1(12+3)2f'(1)=38

f'(1)=6×1(12+3)2f'(1)=38

Then, the function is decreasing on (,0) and increasing on (0,)

F)

Find the local maximum and minimum values.

For maxima

If

f(x)0 then, the ,minima will occurs.

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

f'(0)=02(02+3)f'(0)=0

If

f(x)0 then, the maxima will occurs.

the value of f'(0) is zero

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