Chapter 4.5, Problem 4E

### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336

Chapter
Section

### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336
Textbook Problem

# Use the guidelines of this section to sketch the curve.y = x4 − 8x2 + 8

To determine

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

Explanation

Given:

The Cartesian equation is as below.

y=x4âˆ’8x2+8 (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 (âˆ’âˆž,âˆž) .

b)

Calculate the intercepts.

Calculate the value of x-intercept.

Substitute 0 for y in the equation (1).

x4âˆ’8x2+8=0

Substitute u for x2 in the above equation.

u2âˆ’8u+8=0

Calculate the value of u using the quadratic formula.

u=âˆ’bÂ±b2âˆ’4ac2a

Substitute 1 for a , -8 for b and 8 for c in the above equation.

u=âˆ’(âˆ’8)Â±(âˆ’8)2âˆ’4(1)(8)2(1)=8Â±64âˆ’322=8Â±5.6572=8+5.6572,8âˆ’5.6572u=6.828,â€‰1.172

Substitute x2 for u in the above equation.

x2=6.828,â€‰1.172x=2.613,â€‰1.082

Hence, x -intercept are (2.613,0) and (1.082,0) .

Calculate the y-intercept.

Substitute 0 for x in the equation (1).

f(0)=(0)4âˆ’8(0)2+8=8

Therefore, the y -intercept is (0,8) .

c)

Calculate the symmetry.

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

Substitute âˆ’1 for x in the equation (1).

f(âˆ’1)=(âˆ’1)4âˆ’8(âˆ’1)2+8=1âˆ’8+8=1

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

f(1)=(1)4âˆ’8(1)2+8=1âˆ’8+8=1

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

d)

Calculate asymptotes.

Apply limit of x tends to âˆž (xâ†’âˆž) in the equation (1).

limxâ†’âˆžy=âˆž

Apply limit of x tends to âˆ’âˆž (xâ†’âˆ’âˆž) in the equation (1).

limxâ†’âˆ’âˆžy=âˆ’âˆž

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 .

f'(x)=4x3âˆ’16x (2)

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

4x3âˆ’16x=04x(x2âˆ’4)=0x2âˆ’4=0x2=4x=Â±2,â€‰0

Take the interval as (âˆ’âˆž,âˆ’2) .

Substitute -3 for x in the equation (2).

f'(âˆ’3)=4(âˆ’3)3âˆ’16(âˆ’3)=âˆ’108+48=âˆ’60

Here f'(âˆ’3)<0 , thus function f is decreasing in the interval (âˆ’âˆž,âˆ’2) .

Take the interval as (âˆ’2,0) .

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

f'(âˆ’1)=4(âˆ’1)3âˆ’16(âˆ’1)=âˆ’4+16=12

Here f'(âˆ’1)>0 , thus the function f is increasing in the interval (âˆ’2,0) .

Take the interval as (0,2) .

Substitute 1 for x in the equation (2).

f'(1)=4(1)3âˆ’16(1)=4âˆ’16=âˆ’12

Here f'(1)<0 , thus the function f is decreasing in the interval (0,2) .

Take the interval as (2,âˆž)

f'(3)=4(3)3âˆ’16(3)=108âˆ’48=60

Here f'(3)>0 , thus the function f is increasing in the interval (2,âˆž)

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