Chapter 4.6, Problem 18E

### 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. f ( x ) = x 3 x 2 − 9

To determine

To graph: The function y=x3x29.

Explanation

Given: The function y=x3x2âˆ’9.

Graph:

Consider the provided function.

The function being a rational function has a domain as all the real numbers except for âˆ’3 and 3.

The function will return all possible real values. This implies that the range of the function is (âˆ’âˆž,âˆž).

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

There y-intercept and x-intercept are both (0,0).

The function has two vertical asymptotes x=âˆ’3,x=3 as the denominator is not defined at those points.

The function can be rewritten as:

y=x3x2âˆ’9=x+9xx2âˆ’9

This implies that the line y=x is the slant asymptote of the provided function.

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

x2(x2âˆ’27)(x2âˆ’9)2=0x2(x2âˆ’27)=0x=0,âˆ’33,âˆ’33

This gives six test intervals (âˆ’âˆž,âˆ’33),(âˆ’33,âˆ’3),(âˆ’3,0),(0,3),(3,33),(33,âˆž).

Let âˆ’6âˆˆ(âˆ’âˆž,âˆ’33).

f'(âˆ’6)=(âˆ’6)2((âˆ’6)2âˆ’27)((âˆ’6)2âˆ’9)2>0

The function is increasing in this interval.

Let âˆ’4âˆˆ(âˆ’33,âˆ’3).

f'(âˆ’4)=(âˆ’4)2((âˆ’4)2âˆ’27)((âˆ’4)2âˆ’9)2<0

The function is decreasing in this interval.

Let âˆ’1âˆˆ(âˆ’3,0).

f'(âˆ’1)=(âˆ’1)2((âˆ’1)2âˆ’27)((âˆ’1)2âˆ’9)2<0

The function is decreasing in this interval.

Let 1âˆˆ(0,3).

f'(âˆ’1)=(1)2((1)2âˆ’27)((1)2âˆ’9)2<0

The function is decreasing in this interval.

Let 4âˆˆ(3,33).

f'(4)=(4)2((4)2âˆ’27)((4)2âˆ’9)2<0

The function is decreasing in this interval.

Let 6âˆˆ(33,âˆž).

f'(6)=(6)2((6)2âˆ’27)((6)2âˆ’9)2>0

The function is increasing in this interval.

Now obtain the value of the function at these critical points

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