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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 3 ) x

To determine

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

Explanation

Given:

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

Calculation:

Rewrite the equation (1) as below.

y=f(x)=(x3)xy=f(x)=(x3)x12=x323x12

The equation is y=f(x)=x323x12 (2)

(a)

Calculate the domain.

Here, the values range inside the function must be greater or equal to zero. That is, there is no denominator in the given equation, so there is no domain. Therefore, the domain range of values will have the coordinates (0,) .

(b)

Calculate the intercepts.

Calculate the value of x-intercept.

Take the value of y=0 .

y=(x3)x0=(x3)xx=0x=0x3=0x=3

Therefore, the x -intercept is (0,3) .

Calculate the y-intercept.

y=(03)0y=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)=(x3)xf(1)=(13)1=4

Substitute 1 for x in the equation (1).

f(x)=(x3)xf(1)=(13)1=2

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

(d)

Calculate asymptotes.

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

limx(x3)x=(3)=

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

(e)

Calculate the intervals for the variable x .

Differentiate the equation (2).

f(x)=x323x12f'(x)=32x1232x12=32x12(x1)f'(x)=3(x1)2x>0

The above equation of f'(x)>0 which implies x>1 , so the function f increasing on (1,) and decreasing on (0,1)

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