Chapter 4.6, Problem 24E

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516

Chapter
Section

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516
Textbook Problem

# Analyzing the Graph of a Function In Exercises 5-34, analyze and sketch a graph of the function. Label any intercepts, relative extrema, points of inflection, and asymptotes. Use a graphing utility to verify your results. g ( x ) = x 9 − x 2

To determine

To graph: The function y=x9x2.

Explanation

Given: The function y=x9âˆ’x2.

Graph:

Consider the provided function.

The function has a domain of all real numbers less between âˆ’3 and 3. Thus the domain of the function would be [âˆ’3,3].

The range of the function would be (âˆ’4.5,4.5].

Now find the x and y intercepts by equating f(x) and x to zero respectively to obtain:

The x-intercepts would be (âˆ’3,0),(3,0),(0,0) and the y-intercept would be (0,0).

The function has no vertical or horizontal asymptotes

Now, differentiate the function with respect to x and equate it to zero to obtain the critical points.

9âˆ’2x29âˆ’x2=09âˆ’2x2=0x=92x=Â±322

This gives three test intervals (âˆ’3,âˆ’322),(âˆ’322,322),(322,3).

Let âˆ’2.5âˆˆ(âˆ’3,âˆ’322)

f'(âˆ’2.5)=9âˆ’2(âˆ’2.5)29âˆ’(âˆ’2.5)2<0

The function is decreasing in this interval.

Let 0âˆˆ(âˆ’322,322)

f'(0)=9âˆ’2(0)29âˆ’(0)2=3>0

The function is increasing in this interval.

Let 2.5âˆˆ(322,3)

f'(2.5)=9âˆ’2(2.5)29âˆ’(2.5)2<0

The function is decreasing in this interval.

Now obtain the value of the function at the critical points.

f(322)=3229âˆ’(322)2=92f(âˆ’322)=âˆ’3229âˆ’(âˆ’322)2=âˆ’92

Thus, the critical points of the function are (âˆ’322,âˆ’92),(322,92)

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