Chapter 4.6, Problem 43E

### 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 Transcendental Function In Exercises 43-54, 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 ) = e 3 x ( 2 − x )

To determine

To graph: The function f(x)=e3x(2x)

Explanation

Given: The function f(x)=e3x(2âˆ’x).

Graph: The function being an exponential function is defined for all real x and thus the domain of the function has will be (âˆ’âˆž,âˆž).

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

The x-intercepts is (2,0) and y-intercept is (0,2).

Also,

limxâ†’âˆ’âˆže3x(2âˆ’x)=limxâ†’âˆ’âˆže3xlimxâ†’âˆ’âˆž(2âˆ’x)=0(limxâ†’âˆž2âˆ’x)=0

The function has a horizontal asymptote as y=0 when x keeps on decreasing.

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

(5âˆ’3x)e3x=05âˆ’3x=03x=5x=53

This gives two test intervals (âˆ’âˆž,53),(53,âˆž).

Let 0âˆˆ(âˆ’âˆž,53).

f'(0)=(5âˆ’3(0))e3(0)=5>0

The function is increasing on this interval.

Let 2âˆˆ(53,âˆž).

f'(2)=(5âˆ’3(2))e3(2)=âˆ’e6<0

The function is decreasing on this interval.

This implies that the function has a relative maximum at 53.

Now find the value of the function at the critical point.

f(53)=e3(53)(2âˆ’53)=e53

The relative maximum is (53,e53)

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