Chapter 4, Problem 86RE

### 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 81-92, 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 ) = 2 x 1 + x 2

To determine

To graph: The provided function f(x)=2x1+x2.

Explanation

Given:

The function f(x)=2x1+x2.

Graph:

Consider the provided function.

The function exists for all real values of x. Thus the domain of the function is (âˆ’âˆž,âˆž).

The numerator would always be less than the denominator The range of the function would be [âˆ’1,1].

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

The function has one x-intercept (0,0) and the y-intercept is (0,0).

The function has a horizontal asymptote y=0.

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

âˆ’2x2âˆ’2(x2+1)2=02x2=2x=âˆ’1,1

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

Let âˆ’2âˆˆ(âˆ’âˆž,âˆ’1).

f'(âˆ’2)=âˆ’2(âˆ’2)2âˆ’2((âˆ’2)2+1)2<0

The function is decreasing in this interval.

Let 0âˆˆ(âˆ’1,1).

f'(0)=âˆ’2(0)2âˆ’2((0)2+1)2>0

The function is increasing in this interval.

Let 2âˆˆ(1,âˆž).

f'(2)=âˆ’2(2)2âˆ’2((2)2+1)2<0

The function is decreasing in this interval.

Now see the second derivative to see if there are any inflection points.

f''(x)=4x(x2âˆ’3)(x2+1)34x(x2âˆ’3)(x2+1)3=0x=âˆ’3,0,3

This gives four test intervals (âˆ’âˆž,âˆ’3),(âˆ’3,0),(0,3),(3,âˆž).

Let âˆ’2âˆˆ(âˆ’âˆž,âˆ’3).

f''(âˆ’2)=4(âˆ’2)((âˆ’2)2âˆ’3)((âˆ’2)2+1)3<0

The function is concave downwards in this interval

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