Chapter 4.6, Problem 29E

### 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 = 3 x 4 + 4 x 3

To determine

To graph: The function y=3x4+4x3

Explanation

Given: The function y=3x4+4x3.

Graph:

The function, being a polynomial has a domain of all real numbers. Thus the domain of the function would be (âˆ’âˆž,âˆž).

The range of the function would be [âˆ’1,âˆž).

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

The x-intercepts would be (âˆ’43,0),(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.

f'(x)=12x2(x+1)12x2(x+1)=0x=0,âˆ’1

This gives three test intervals (âˆ’âˆž,âˆ’1),(âˆ’1,0),(0,âˆž).

Let âˆ’2âˆˆ(âˆ’âˆž,âˆ’1).

f'(âˆ’2)=12(âˆ’2)2(âˆ’2+1)=âˆ’48<0

The function is decreasing on this interval.

Let âˆ’0.5âˆˆ(âˆ’1,0).

f'(âˆ’0.5)=12(âˆ’0.5)2(âˆ’0.5+1)=1.5>0

The function is increasing on this interval.

Let 2âˆˆ(0,âˆž).

f'(2)=12(2)2(2+1)=144<0

The function is increasing on this interval.

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

f(0)=3(0)4+4(0)3=0f(âˆ’1)=3(âˆ’1)4+4(âˆ’1)3=âˆ’1

Thus, the critical points are (âˆ’1,âˆ’1),(0,0).

Now check the second derivative to obtain the inflection points.

f''(x)=12x(3x+2)12x(3x+2)=0x=0,âˆ’23

This gives three test intervals (âˆ’âˆž,âˆ’23),(âˆ’23,0),(0,âˆž)

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