Chapter 4.6, Problem 8E

### 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 = x − 4 x − 3

To determine

To graph: The function y=x4x3

Explanation

Given: The function y=xâˆ’4xâˆ’3.

Graph:

The function being a rational function has a domain as all the real numbers except 3.

The function can never take a value 1 as the numerator will never be equal to the denominator. This implies that the range of the function would be (âˆ’âˆž,1)âˆª(1,âˆž).

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

The x-intercept is (4,0) and the y-intercept is (0,43).

Now the function does have a vertical asymptote as x=3 because at that point the denominator is not defined.

Also,

limxâ†’âˆž(xâˆ’4xâˆ’3)=limxâ†’âˆž(1âˆ’4x1âˆ’3x)=1âˆ’limxâ†’âˆž4x1âˆ’limxâ†’âˆž1x=1âˆ’01âˆ’0=1

This implies that the function has a horizontal asymptote as y=1.

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

1(xâˆ’3)2=0

This can never be equal to zero and is not defined at 3 where the function itself is not defined. So there are no critical points.

This gives two test intervals (âˆ’âˆž,3),(3,âˆž).

Let âˆ’2âˆˆ(âˆ’âˆž,3).

f'(âˆ’2)=1(âˆ’2âˆ’3)2=125>0

The function is increasing in this interval.

Let 4âˆˆ(3,âˆž)

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