   Chapter 4.5, Problem 31E ### 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 − 1 3

To determine

To Sketch: The curve by plotting x and y coordinates using guidelines.

Explanation

Given:

The Cartesian equation is as below.

y=x213 (1)

Calculation:

a)

Calculate the domain.

From the equation (1), the function is defined for all real values of x (there is no restriction on the value of x). Hence, the domain of f(x) is all real values of x .

For the polynomial the domain is (,) .

b)

Calculate the intercepts.

Calculate the value of x-intercept.

Substitute 0 for y in the equation (1).

x213=0x21=0x2=1x=±1

Hence, x -intercepts are (1,0) and (1,0) .

Calculate the y-intercept.

Substitute 0 for x in the equation (1).

f(0)=(0)213=13=1

Therefore, the y -intercept point is (0,1) .

c)

Calculate the symmetry.

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

Substitute 1 for x in the equation (1).

f(1)=(1)213=0

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

f(1)=(1)213=0

Here f(x)=f(x) , it is an even function. So it has y-axis symmetry.

d)

Calculate asymptotes.

Apply limit of x tends to (x) in the equation (1).

limxy=

Apply limit of x tends to (x) in the equation (1).

limxy=

Here, the value of limit gets infinity; this implies there is no horizontal asymptote or vertical asymptote.

e)

Calculate the intervals.

Differentiate the equation (1) with respect to x

Apply the chain rule.

df(u)dx=dfdududx

Substitute u3 for f and x21 for u in the above equation.

df(u)dx=ddu(u3)ddx(x21)=ddu(u13)ddx(x21)=13u(131)(2x)=13u23(2x)=2x3u23

Substitute x21 for u in the above equation.

f'(x)=2x3(x21)23 (2)

Substitute 0 for f'(x) in the equation (2).

2x3(x21)23=02x=0x=0

Take the interval as (,0) .

Substitute -2 for x in the equation (2).

f'(2)=2(2)3((2)21)23=23(41)23=23(3)23=0.32

Here f'(2)<0 , thus function f is decreasing in the interval (,0) .

Take the interval as (0,) .

Substitute 2 for x in the equation (2).

f'(2)=2(2)3((2)21)23=43(41)23=43(3)23=0.641

Here f'(2)>0 , thus the function f is increasing in the interval (0,) .

Thus the function f increasing on (0,) and decreasing on (,0) .

Note the function f'(x)= , when the value of x is -1 and 1.

f)

Calculate the local maximum and minimum value.

From the increasing and decreasing information in the part (e).

Calculate the local minimum value.

Substitute 0 for x in the equation (1).

f(0)=(0)213=1

Thus, the local minimum occurs at (0,1) .

There is no local maximum.

g)

Calculate the concavity.

Differentiate the equation (2) with respect to x using the quotient rule.

(uv)'=u'vv'uv2

Substitute 2x for u and 3(x21)23 for v in the above formula

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