Chapter 4.6, Problem 5E

### 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. y = 1 x − 2 − 3

To determine

To graph: The function y=1x23

Explanation

Given: The function y=1xâˆ’2âˆ’3.

Graph:

The function being a polynomial has a domain as all the real numbers except for 2.

The function canâ€™t take the value 3 and thus the range of the function would be (âˆ’âˆž,3)âˆª(3,âˆž).

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

The x-intercept is (73,0) and the y-intercept is (0,âˆ’72).

Now the function does have a vertical asymptote as x=2 because the denominator is not defined at 2.

Also,

limxâ†’âˆž(1xâˆ’2âˆ’3)=limxâ†’âˆž(1xâˆ’2)âˆ’3=limxâ†’âˆž1x1âˆ’2xâˆ’3=01âˆ’0âˆ’3=âˆ’3

This implies that the function has a horizontal asymptote as y=âˆ’3

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

âˆ’1(xâˆ’2)2=0

Clearly, this can never be zero and is not defined at 2 where the function is not defined itself. Hence, there are no critical points

This gives two test intervals (âˆ’âˆž,2),(2,âˆž).

Let 0âˆˆ(âˆ’âˆž,2).

f'(0)=âˆ’1(0âˆ’2)2=âˆ’14<0

The function is decreasing in this interval.

Let 3âˆˆ(2,âˆž)

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