   Chapter 4.5, Problem 22E ### 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 − 4 ) x 3

To determine

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

Explanation

Given:

The given curve is y=(x4)x3 (1)

Calculation:

Rewrite the equation (1).

y=f(x)=(x4)x3y=f(x)=(x4)x13=x434x13

The equation is y=f(x)=x434x13 (2)

(a)

Calculate the domain.

Here, the values inside the function must be greater or equal to zero. There is no denominator term in the given equation, so there is no fixed domain for the equation. Therefore, the domain is (,) .

(b)

Calculate the intercepts as below.

Calculate the value of x-intercept as below.

Take the value of y=0 .

y=(x4)x30=(x4)x3x3=0x=0x4=0x=4

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

Calculate the y-intercept.

y=(04)03y=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)=(x4)x3f(1)=(14)13=5

Substitute 1 for x in the equation (1).

f(x)=(x4)x3f(1)=(14)13=3

Hence, the condition f(x)f(x) is true and there is no symmetry.

(d)

Calculate asymptotes.

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

limx(x4)x3=(3)3=

Therefore, the value of limit for x gets to infinity and there is no horizontal asymptote. The function f starts at a point and goes up to infinity and there is no vertical asymptote.

(e)

Calculate the intervals for x, between which the value of f will increase or decrease.

Differentiate the equation (2).

f(x)=x434x13f'(x)=43x1343x23=43x23(x1)f'(x)=4(x1)3x23>0

The above equation with condition f'(x)>0 is true, and it implies that x>1 , so the function f increasing on (1,) and decreasing on (,1)

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