Finding x/y-intercepts, asymptotes, derivatives, and max min

[Specter]

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.

 

[RPinPA]

Simple matter of factoring.

You have $-16(x^2 - 4)^2 - (-16x)(4x)(x^2 - 4) = -16(x^2 - 4)^2 + 64x^2(x^2 - 4)$. Both terms have a $(x^2 - 4)$ factor, so you can factor it out.
$= (x^2 - 4) \left [ -16(x^2 - 4) + 64x^2 \right ]$
And that $(x^2 - 4)$ cancels out with one factor in the denominator, leaving you $(x^2 - 4)^3$ in the denominator and $\left [ -16(x^2 - 4) + 64x^2 \right ]$ in the numerator.

I think that answers the question you were asking.

If it's not clear, replace the expression $x^2 - 4$ with $y$.
$\frac {-16y^2 + 64x^2y} {y^4} = \frac{y (-16y + 64x^2)} {y^4} = \frac{-16y + 64x^2}{y^3}$

 

[Specter]

Simple matter of factoring.

You have $-16(x^2 - 4)^2 - (-16x)(4x)(x^2 - 4) = -16(x^2 - 4)^2 + 64x^2(x^2 - 4)$. Both terms have a $(x^2 - 4)$ factor, so you can factor it out.
$= (x^2 - 4) \left [ -16(x^2 - 4) + 64x^2 \right ]$
And that $(x^2 - 4)$ cancels out with one factor in the denominator, leaving you $(x^2 - 4)^3$ in the denominator and $\left [ -16(x^2 - 4) + 64x^2 \right ]$ in the numerator.

I think that answers the question you were asking.

If it's not clear, replace the expression $x^2 - 4$ with $y$.
$\frac {-16y^2 + 64x^2y} {y^4} = \frac{y (-16y + 64x^2)} {y^4} = \frac{-16y + 64x^2}{y^3}$

That kind of makes sense. What confuses me is the square on $(x^2-4)^2$. Now this is my first math course in a while, but wouldn't it just get factored from $(x^2-4)^2$ to$(x^2-4)$ or is that not how it works?

This is what I was thinking, but I haven't had to factor in a while so I'm probably wrong.

$f''(x)=\frac {-16(x^2-4)^2+64x^2(x^2-4)} {x^2-4)^4}$

$=\frac {(x^2-4) [-16(x^2-4)^2+64x^2} {(x^2-4)^4}$

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

 

[RPinPA]

Go the other way. What is $y (-16y^2 + 64x^2)$? By the distributive law it's $-16y^2*y + 64x^2*y = -16y^3 + 64x^2y$

Do that with your expression. Multiply the $x^2 - 4$ back by both of the terms inside the square brackets. What happens to the first term when you multiply it by $(x^2 - 4)$? What is $-16(x^2 - 4)^2$ times $(x^2 - 4)$?

I think I'm not quite understanding your question. You asked "wouldn't it get factored from $(x^2 - 4)^2$ to $(x^2 - 4)$? If you look at what I did, I had $(x^2 - 4)^2$ before factoring and $(x^2 - 4)$ inside the brackets after factoring. So it did in fact get factored from $(x^2 - 4)^2$ to $(x^2 - 4)$.

 

[Specter]

Go the other way. What is $y (-16y^2 + 64x^2)$? By the distributive law it's $-16y^2*y + 64x^2*y = -16y^3 + 64x^2y$

Do that with your expression. Multiply the $x^2 - 4$ back by both of the terms inside the square brackets. What happens to the first term when you multiply it by $(x^2 - 4)$? What is $-16(x^2 - 4)^2$ times $(x^2 - 4)$?

I think I'm not quite understanding your question. You asked "wouldn't it get factored from $(x^2 - 4)^2$ to $(x^2 - 4)$? If you look at what I did, I had $(x^2 - 4)^2$ before factoring and $(x^2 - 4)$ inside the brackets after factoring. So it did in fact get factored from $(x^2 - 4)^2$ to $(x^2 - 4)$.

Thank you, it just took some time for me to understand this.I get it now!