   Chapter 4.5, Problem 30E ### 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 = x5/3 − 5x2/3

To determine

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

Explanation

Given:

The Cartesian equation

y=x5/35x2/3 (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).

x5/35x2/3=0x2/3(x5)=0x5=0x=5,0

Hence, x -intercept points are (5,0) and (0,0) .

Calculate the y-intercept.

Substitute 0 for x in the equation (1).

f(0)=(0)5/35(0)2/3=0

Therefore, the y -intercept point 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(1)=(1)5/35(1)2/3=1+5=4

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

f(1)=(1)5/35(1)2/3=15=4

Here, the graph is neither even nor odd. f(x) is even if f(x)=f(x) and odd if f(x)=f(x) .

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 .

f'(x)=53x(531)5(23x(231))=53x23103x13=53(x232x13)

f'(x)=53(x232x13)=53(x23x132x13)

f'(x)=5(x2)3x13 (2)

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

5(x2)3x13=05(x2)=0x2=0x=2

Take the interval as (2,) .

Substitute 3 for x in the equation (2).

f'(3)=5(32)3(3)13=53(1.442)=1.1556

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

Take the interval as (,0) .

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

f'(1)=5(12)3(1)13=153=5

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

Take the interval as (0,2) .

Substitute 1 for x in the equation (2).

f'(1)=5(12)3(1)13=53=1

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