   Chapter 4.5, Problem 7E ### 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 = 1 5 x 5 − 8 3 x 3 + 16 x

To determine

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

Explanation

Given:

The given curve is

f(x)=y=15x583x3+16x (1)

Calculation:

Calculate the domain.

From 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, which is (,).

Calculate the intercepts.

Calculate the y –intercept.

f(x)=15x583x3+16x

f(0)=15(0)583(0)3+16(0)=0

The y–intercept is x=0.

Calculate the value of x-intercept.

Substitute 0 for y in equation (1).

15x583x3+16x=yx(15x483x2+16)=0

x=0 or (15x483x2+16)=0

x=0 or (3x440x2+240)=0

Determine the factor of (3x440x2+240)=0

Substitute x2 by t in the above equation, (3t240t+240)=0.

This is of the form (at2bt+c)=0.

Comparing the coefficient of a, b and c.

a=3, b=40 and c=240

Determine the factors using the formula b±b24ac2a.

Substitute the value of a, b and c.in the above equation.

x=(40)±(40)24(240)(3)2(3)x=40±12806

so x=0 or x=40±12806

Since the square root contains a negative value, x cannot take the value 40±12806.

Therefore, the x-intercept is x=0.

Calculate the symmetry.

Substitute x for x in equation (1)

f(x)=15x583x3+16xf(x)=15(x)583(x)3+16(x)f(x)=15x5+83x316x

f(x)=(15x583x3+16x)

f(x)=f(x)

It is an odd function.

So, it has rotational symmetry about the origin

Calculate asymptotes.

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

limxf(x)=15x583x3+16x=(155833+16)=

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

limxf(x)=15x583x3+16x=(15()583()3+16())=

Hence, as the value of limit gets infinity, there is no horizontal and vertical asymptote.

Determine the intervals.

Differentiate the equation (1) with respect to x.

f(x)=15x583x3+16x

f'(x)=x48x2+16

f'(x)=(x24)2

f'(x)=(x+2)2(x2)2 (2)

Equate f'(x) to 0

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