1. The problem statement, all variables and given/known data 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.