Chapter 4.6, Problem 30E

### 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 = − 2 x 4 + 3 x 2

To determine

To graph: The function y=2x4+3x2

Explanation

Given: The function y=âˆ’2x4+3x2.

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 function is a polynomial and thus the range of the function would be (âˆ’âˆž,98].

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

The x-intercepts would be (âˆ’62,0),(0,0),(62,0) 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)=âˆ’2x(4x2âˆ’3)âˆ’2x(4x2âˆ’3)=0x=0,âˆ’32,32

This gives four test intervals (âˆ’âˆž,âˆ’32),(âˆ’32,0),(0,32)(32,âˆž).

Let âˆ’1âˆˆ(âˆ’âˆž,âˆ’32).

f'(âˆ’1)=âˆ’2(âˆ’1)(4(âˆ’1)2âˆ’3)=2>0

The function is increasing on this interval.

Let âˆ’0.5âˆˆ(âˆ’32,0).

f'(âˆ’0.5)=âˆ’2(âˆ’0.5)(4(âˆ’0.5)2âˆ’3)=âˆ’2<0

The function is decreasing on this interval.

Let 0.5âˆˆ(0,32).

f'(0.5)=âˆ’2(0.5)(4(0.5)2âˆ’3)=2>0

The function is increasing on this interval.

Let 1âˆˆ(32,âˆž).

f'(1)=âˆ’2(1)(4(1)2âˆ’3)=2<0

The function is increasing on this interval.

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

f(0)=âˆ’2(0)4+3(0)2=0f(32)=âˆ’2(32)4+3(32)2=98

And,

f(âˆ’32)=âˆ’2(âˆ’32)4+3(âˆ’32)2=98

Thus, the critical points are (âˆ’32,98),(32,98)

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