Given is the density $$f(x)=\begin{cases} \frac{3}{4}\left(2x-x^2\right)&\mbox{if }x \in (0,2) \\ 0&\mbox{else}\end{cases}$$
Find a random variable for the density.
There are probably several methods for doing this but I only know of inverse transform and I like to try to use it here.
Firstly, we need the distribution function of the density. It is (just assume this is correct):
$$F(x)=\begin{cases} \frac{3}{4}x^2-\frac{1}{4}x^3&\mbox{if }0<x\leq 2 \\ 1 &\mbox{if }x>2\\ 0 &\mbox{if }x<0\end{cases}$$
Now we need to inverse this distribution function: $$y=\frac{3}{4}x^2-\frac{1}{4}x^3$$
I used a software for this one because it got very complicated when I tried to do it by hand :(
$$x= \sqrt[3]{2\sqrt{y^2-y}-2y+1}+\frac{1}{\sqrt[3]{2\sqrt{y^2-y}-2y+1}}$$
While the other party of the distribution function didn't need any change except for the cases, so for the inverse we have
$$F^{-1}(y)=\begin{cases} \sqrt[3]{2\sqrt{y^2-y}-2y+1}+\frac{1}{\sqrt[3]{2\sqrt{y^2-y}-2y+1}}&\mbox{if }\\ 1 &\mbox{if }\\ 0 &\mbox{if }\end{cases}$$
I didn't expect it will end up that complicated, I don't even know what cases I need to use and how the inverse of $0$ will be treated as it doesn't seem to be defined : /
Maybe there is an easier way of solving the problem or did I miss an important step / did a mistake somewhere? :s