I have to solve the following problem:
Let $X$ be a continuous random variable with PDF $$f(x)= \begin{cases} x+1, &-1\leq x<0\\ 1-x, &0\leq x\leq1\\ 0, &\text{otherwise} \end{cases}.$$ Let $Y=X(X-1)$. Determine the PDF of $Y$.
I have no idea how I'm supposed to solve this. Any help is appreciated.
Edit: I found the CDF to be $$F_X(x) = \begin{cases} 0, & x\leq -1\\ \frac{1}{2}(1+x)^{2}, &-1\leq x\leq 0\\ \frac{1}{2}\left(1+2x-x^{2}\right), &0\leq x\leq 1\\ 1,&x>1 \end{cases}.$$
For $-1<x<0 \ y$ varies between $[0,2]$, so by solving $x^2 -x - y= 0$ you need to use the negative root to find the CDF of $Y$: $$ F_Y(y) = P(Y \leq y) = P \bigg(X>\frac{1-\sqrt{1+4y}}{2}\bigg) = \int_{\frac{1-\sqrt{1+4y}}{2}}^{0}(1+x)dx $$ For $x \in [0,1] y \in [-\frac{1}{4}, 0]$, it's first a decreasing, and then an increasing function, you need to use the second (positive) root. For $x \in [0, \frac{1}{2}]$,
$$ F_Y(y) = P(X>h^{-1}(y)) = \int_{\frac{1+\sqrt{1+4y}}{2}}^{\frac{1}{2}}f(x)dx $$ For the third interval, $x \in [\frac{1}{2}, 1]$, you need $P(X\leq h^{-1}(y))$
$$ F_Y(y) = P\bigg(X < \frac{1+\sqrt{1+4y}}{2}\bigg) = P(X \leq h^{-1}(y)) = \int_{\frac{1}{2}}^{h^{-1}(y)}f(x)dx $$ In the two last cases $f(x)=1-x$