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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

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Section
BuyFindarrow_forward

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 4

To determine

To find: sketch the curve using the equation y=xx24 .

Explanation

Given:

The equation of the curve is,

y=xx24 (1).

Calculation:

Rewrite the equation (1).

f(x)=xx24f(x)=x(x2)2

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

A)

Compute the domain.

Considering the denominator equation (1)

x24=0x=±2

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

B)

Compute y intercepts.

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

y=0024y=0

Compute x intercepts.

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

x=0

The value is not real for x .

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

C)

Compute the curve is symmetric:

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

Substitute the value 1 for x .

f(x)=xx24f(1)=1(1)24f(1)=13

Substitute the value 1 for x .

f(x)=xx24f(1)=1(1)24f(1)=13

f(x)=f(x)

so, the function is odd.

Therefore, the curve is symmetric about the origin.

D)

Compute the horizontal asymptote:

limx±xx24=0

Therefore, the horizontal asymptote y is 0 .

Compute the vertical asymptote:

limx2xx24=

limx2+xx24=

limx2+xx24=

limx2xx24=

Therefore, the vertical asymptote x is ±2 .

E)

Compute the interval of increase or decrease.

Apply UV method to differentiate the equation (2) below

f'(x)=vu'uv'v2f'(x)=(x24)(1)x(2x)(x24)2f'(x)=(x24)2x2(x24)2f'(x)=x24(x24)2<0

The equation of  f'(x) have negative sign.  So for all the value of x the f'(x) will be smaller than zero.

Then, the function is decreasing on (,2) , (2,2) and (2,) .

F)

Find the local maximum and minimum values.

For maxima

If

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

If

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

In the curve there is no maxima and minima because the highest and lowest limit are and

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