Hello everyone!

I got this transformation problem here:

$\displaystyle \text{let} \ f(x)=\frac{2}{5}|x|, \text{if} \ -1<x<2, \text{be the pdf of a continuous random variable} \ X$. $\displaystyle \text{find the pdf of } Y=X^2$.

The solution starts by mentioning that $\displaystyle g(x)=x^2$ is a two-to-one transformation. I'm a bit lost here, both $\displaystyle f(x) $ and $\displaystyle g(x)$ are two to one? What do we do??... The answer says:

$\displaystyle h(y)= \frac{2}{5}, 0<y<1$ and $\displaystyle h(y)=\frac{1}{5}, 1<y<4$.

Any help is appreciated