Assume $X$ to be a random variable whose probability density function is given by \begin{align*} f_{X}(x) = \begin{cases} \displaystyle\frac{3x^{2}}{2} & \text{if}\,\,\,-1\leq x \leq 1\\\\ 0 & \text{otherwise} \end{cases} \end{align*}
Find the probability density function of
a) $Y = 3X$
b) $W = -3 - X$
c) $V = X^{2}$
MY SOLUTION
a) Since $Y = g(X) = 3X$ is bijective and the support of $Y$ is by $S_{Y} = [-3,3]$, \begin{align*} f_{Y}(y) = f_{X}(g^{-1}(y))\left|\frac{\mathrm{d}}{\mathrm{d}y}g^{-1}(y)\right| = \frac{1}{3}f_{X}\left(\frac{y}{3}\right) = \frac{y^{2}}{18} \end{align*}
b) Once $W = h(X) = -3-x$ is bijective and the support of $W$ is $S_{W} = [-4,-2]$, \begin{align*} f_{W}(w) = f_{X}(h^{-1}(w))\left|\frac{\mathrm{d}}{\mathrm{d}w}h^{-1}(w)\right| = |-1|f_{X}(-3-w) = f_{X}(-3-w) = \frac{3(3+w)^{2}}{2} \end{align*}
c) In this case, the function $V = X^{2}$ is not bijective on the interval $[-1,1]$, where the support of $V$ is $S_{V} = [0,1]$. Based on this, I've tried the following approach \begin{align*} F_{V}(v) & = F_{X}(X^{2} \leq v) = F_{X}(|X|\leq\sqrt{v}) = F_{X}(-\sqrt{v}\leq X \leq \sqrt{v})\\\\ & = F_{X}(\sqrt{v}) - F_{X}(-\sqrt{v}) \Rightarrow f_{V}(v) = \frac{f_{X}(\sqrt{v}) + f_{X}(-\sqrt{v})}{2\sqrt{v}} = \frac{3\sqrt{v}}{2} \end{align*}
Besides checking the previous results, I would like to know if there is another possible way to tackle the last part through change of variables, just as it has been done at (a) and (b). Thanks in advance!
Everything is ok.
One useful sanity check in this kind of problem is to integrate the resulting density (it should be one).
Regarding the last part: the formula for transformation of variables can be generalized for non-bijective transformations (see eg here ) but your approach is perfectly fine and I think it's the most direct/safe way.