- #1

Specter

- 120

- 8

## Homework Statement

For the following function ##f(x)=\frac {2x^2} {x^2-4}##, find the following:

a) The x and y intercepts

b) the horizontal and vertical asymptotes

c) the first and second derivatives

d) any local maximum or minimum points

e) the intervals of increasing and decreasing

f) any inflection points

g) the intervals of concavity

(solved some of these already but included the full questions just incase)

## Homework Equations

## The Attempt at a Solution

[/B]

I mostly need help with the second derivative.

a) The x and y intercepts are (0,0)

b) The horizontal and vertical asymptotes are

Vertical:

##x^2-4=0##

##(x+2)(x-2)##

x=2 and x=-2

Horizontal:

##\frac 2 {1}=2##

c) The first and second derivatives

I had a lot of trouble finding the second derivative, and still don't understand a few of the final steps. Here's my work.

##f(x)=\frac {2x^2} {x^2-4}##

##f'(x)=\frac {4x(x^2-4)-2x^2(2x)} {(x^2-4)^2 }##

##=\frac {4x^3-16x-4x^3} {(x^2-4)^2}##

##=\frac {-16} {(x^2-4)^2}##

##f''(x)=\frac {-16(x^2-4)^2-(-16x)(4x(x^2-4))} {(x^2-4)^4}## I solved up to here on my own, after this I am not sure where ##(x^2-4)## went, and why the denominator went from ##(x^2-4)^4## to ##(x^2-4)^3##. Also where did the square go on the ##(x^2-4)^2##?

##=\frac {-16(x^2-4)+64x^2} {(x^2-4)^3}##

##=\frac {-16x^2+64+64x^2} {(x^2-4)^3}##

##=\frac {48x^2+64} {(x^2-4)^3}##

d) any local maximum or minimum points.

Set the first derivative to 0 solve for x:

##f'(x)=\frac {-16x} {(x^2-4)^2}=0##

##=-16x=0##

##x=0##

Sub x value into original function

##y=\frac {2(0)^2} {(0)^2-4}##

##=\frac 0 {-4} =0##

(0,0) is a possible max or min point.