   Chapter 4.5, Problem 5E ### 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(x − 4)3

To determine

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

Explanation

Given:

The given curve is

f(x)=y=x(x4)3 (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).

x(x4)3=yx(x4)3=0

(x4)3=0

(x4)=0

x=4

Therefore, the x -intercept are (0,4) .

Calculate the value of y -intercept.

Substitute 0 for x in the equation (1).

x(x4)3=y0(04)3=y

y=0

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

(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)=x(x4)3f(1)=1(14)3f(1)=125

Substitute +1 for x in the equation (1)

f(x)=x(x4)3f(1)=1(14)3f(1)=27

Hence, f(x)f(x) so, there is no symmetry.

(d)

Calculate asymptotes.

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

limxx(x4)3=(4)3=

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

limxx(x4)3=(4)3=

Here, the value of limit gets infinity, so there is no horizontal and vertical asymptote.

(e)

Calculate the intervals.

Differentiate the equation (1).

f(x)=x(x4)3

f'(x)=3x(x4)2+(x4) (2)

Consider the limit (1,)

Substitute the 2 for the value of x in (2)

f'(x)=3(2)(24)2+(24)=22>0

Consider the limit (,1) .

Substitute the 0 for the value of x in (2)

f'(x)=3(0)(04)2+(04)=4<0

When the condition x>1 is true; f'(x)>0

When the condition x<1 is true; f'(x)<0

The increasing behavior is shown in the limit (1,)

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