Concave up and down of f (solving inequality)

[cjh12398]

1. Homework Statement
f(x)=(2x^3+2x^2-5x-2) / 2(x^(2)-1)

f''(x)=(-12x^5-24x^3+36x)/(4x^8-16x^6+24x^4-16x^2+4)

Find the intervals where f is concave up.

2. The attempt at a solution (I am having trouble interpreting the results at the end or if I've made a mistake somewhere):

Attempt at finding intervals of concave up:
f''(x)>0 where f(x) is concave up.
Therefore,
(-12x^5-24x^3+36x)/(4x^8-16x^6+24x^4-16x^2+4)>0

-12x^5-24x^3+36x > 0 (multiplied denominator up)

x(-12x^4-24x^2+36) > 0 (factored out x)

So, I've stated that there are 2 cases where this is true, when the left side is + , + or
-, -.

So, for case 1 (+, +):
x>0 and -12x^4-24x^2+36>0
let x^2=u
-12u^2-24u+36>0
-12(u^2+2u-3)>0
u^2+2u-3<0
(u+3)(u-1)<0. At this point, I have again stated that there are 2 cases for where this is true, when the left side is +, + or -, - (am I supposed to do this?)

So, for case 1 of this factored expression in case 1 of the entire inequality:
u+3>0 and u-1>0
u>-3 and u>1
x^2>-3 (ignore due to non-real solutions) and x^2>1
x>+-sqrt(1)
Therefore, x>1 and x<-1.

And for case 2 of this factored expression in case 1 of the entire inequality:
u+3<0 and u-1<0
u<-3 and u<1
x^2<-3 (ignore again) and x^2<1
x<+-sqrt(1)
Therefore, x<1 and x>-1

So overall from case 1 of the entire inequality (+,+), I have 2 sets of solutions.
1) x>0, x>1 and x<-1
2) x>0, x<1 and x>-1

I don't want to type out the entire of case 2, however the general jist of what I have done is similar to the case 1 working out and from case 2 of the entire inequality (-, -) I have also have 2 sets of solutions:
1) x<0, x<-1 and x>1
2) x<0, x>-1 and x<1

At this point, I'm having trouble determining the overall intervals of where concave up occurs. I have checked the actual solution, and it is concave up when x<-1 and 0<x<1.

Any help would be greatly appreciated.

 

[gopher_p]

Do you remember using sign charts in pre-calc to solve rational inequalities? That's the way to go here.