   Chapter 10.5, Problem 18E ### 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 ) = 4 x 2 x 4 + 1

To determine

To calculate: The horizontal and vertical for the provided function f(x)=4x2x4+1 and sketch its graph.

Explanation

Given Information:

The provided function is f(x)=4x2x4+1.

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 the degree of the numerator is greater than the degree of the denominator.

To find relative maxima and minima of a function,

1. Set the first derivative of the function to zero f'(x)=0, to find the critical values of the function.

2. Substitute the critical values into f(x) and calculate the critical points.

3. Calculate the sign of the function to the left and right of the critical values.

If f(x)<0 and f(x)>0 on the left and right side respectively of a critical value, then the point is relative minimum.

If f(x)>0 and f(x)<0 on the left and right side respectively of a critical value, then the point is relative maximum.

Calculation:

Consider the provided function,

f(x)=4x2x4+1

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 x of the function equal to zero.

x4+1=0x4=1

Which is not possible.

Thus, there is no vertical asymptote.

Degree of the numerator is 2 and denominator is 4 of the function f(x)=4x2x4+1.

Since, the degree of the numerator is less than the denominator.

Thus, the horizontal asymptote is y=0.

Now, calculate the first derivative of the function f(x)=4x2x4+1.

f'(x)=8x×(x4+1)4x2×(4x3)(x4+1)2=8x8x5(x4+1)2=8x(1x4)(x4+1)2=8x(1x)(1+x)(1+x2)(x4+1)2

Now, to obtain the critical values, set y'=0 as,

8x(1x)(1+x)(1+x2)(x4+1)2=08x(1x)(1+x)(1+x2)=0

Thus, either 8x=0 or (1x)=0 or (1+x)=0.

First consider 8x=0.

Solve for x.

8x=0x=0

Now consider (1x)=0.

Solve for x.

(1x)=01=xx=1

Now consider (1+x)=0.

Solve for x.

(1+x)=0x=1

Thus, the critical points are x=0 x=1 and x=1.

Now, calculate the sign of the y' on either side of the critical points.

First consider x=0.

Calculate f'(x) at x=12.

f'(x)=8x(1x)(1+x)(1+x2)(x4+1)2=8(12)(1(12))(1+(12))(1+(12)2)((12)4+1)2=8(12)(1+12)(112)(1+14)(116+1)2

Thus, f'(x)<0 on left of x=0

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