   Chapter 10.5, Problem 11E ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# For each function in Problems 11-18, find any horizontal and vertical asymptotes, and use information from the first derivative to sketch the graph. f ( x ) = 2 x + 2 x − 3

To determine

To calculate: The horizontal and vertical for the provided function f(x)=2x+2x3 and sketch its graph.

Explanation

Given Information:

The provided function is f(x)=2x+2x3.

Formula used:

A vertical asymptote of a rational function h(x)=f(x)g(x) is x=a where g(a)=0 and f(a)0.

A horizontal asymptote of a rational function h(x)=f(x)g(x) is,

Step 1. A line y=0 if the degree of the numerator is less than the degree of the denominator.

Step 2. The line y=anbn ratio of the leading coefficients if the degree of the numerator is equal to the degree of the denominator.

Step 3. Does not exist if the degree of the numerator is greater than the degree of the denominator.

Calculation:

Consider the provided function,

f(x)=2x+2x3

Recall that a vertical asymptote of a rational function h(x)=f(x)g(x) is x=a where g(a)=0 and f(a)0.

Set the denominator x3 of the function equal to zero,

x3=0x=3

Substitute 3 for x in the numerator as below,

2x+2=2(3)+2=6+2=80

Thus, the vertical asymptote is the line x=3.

Degree of the numerator and denominator of the function f(x)=2x+2x3 is 1.

Thus, the horizontal asymptote is the line y=anbn ratio of the leading coefficients.

Since, the leading coefficient in the numerator is 2 and in the denominator is 1.

Thus, the horizontal asymptote is,

y=21=2

Now, calculate the first derivative of the function f(x)=2x+2x3.

f'(x)=2(x3)(2x+2)(x3)2=2x62x2(x3)2=8(x3)2

Now, to obtain the critical values, set f'(x)=0 as,

8(x3)2=08=0

Which is not possible.

Thus, there are no critical points and hence no relative maximum or minimum.

Calculate f(x) at x=0,

f(0)=2(0)+203=23

Thus, the point is (0,23)

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