Question: Let $X ∼ Unif[−3, 3]$, and $Y = (X + 1)^2$. Find $f_{Y}(1)$ and $f_{Y}(9)$, where $f_{Y}(·)$ is the probability density function of Y.
My Approach: The pdf of $X$ is the following: $\frac{1}{6}$ if $ -3 \le x \le 3 $ and 0 otherwise
$$F_{Y}(Y) = P(Y \le y) = P((X+1)^2 \le y) = P(-\sqrt{y}-1 \le X \le \sqrt{y}-1)$$ $$ = F_{x}(\sqrt{y}-1) - F_{x}(-\sqrt{y}-1)$$ $$ f_{Y}(Y) = f_{x}(\sqrt{y}-1)\frac{d}{dy}(\sqrt{y}-1) - f_{x}(-\sqrt{y}-1)\frac{d}{dy}(-\sqrt{y}-1)$$
The pdf $ f_{Y}(Y) $ is then:
$$ f_{Y}(Y) = f_{x}(\sqrt{y}-1)\frac{1}{2\sqrt{y}} - f_{x}(-\sqrt{y}-1)\frac{-1}{2\sqrt{y}}$$
I substitute Y=1 and Y=9 to get
$$ f_{Y}(1) = f_{x}(\sqrt{1}-1)\frac{1}{2\sqrt{1}} - f_{x}(-\sqrt{1}-1)\frac{-1}{2\sqrt{1}}$$
$$ f_{Y}(1) = (\frac{1}{6})\frac{1}{2} + (\frac{1}{6})\frac{1}{2}$$ $$ f_{Y}(1) = \frac{2}{12} $$
Which is just $\frac{1}{6}$. I repeated this same process with $f_{Y}(9)$ and got $\frac{1}{36}$, and both of those answers are incorrect. If anyone can help me understand where I went wrong I would greatly appreciate it. Thank you!
$Y=(X+1)^2$ for $X\sim\mathcal{U}[-3..3]$ means the transformation will fold the support of $[-3..-1)\cup[-1..3]$ onto $[0..16]$. Specifically it will map $[-3..-1)$ onto $(0..4]$ and $[-1..3]$ onto $[0..16]$.
Thus I obtain that:
$$\begin{align}f_{\small Y}(y)&=\left\lvert\dfrac{1}{2\surd y}\right\rvert\,(f_{\small X}(-\sqrt y-1)\,\mathbf 1_{y\in(0..4]}+f_{\small X}(\sqrt y-1)\,\mathbf 1_{y\in(0..16]})\\[1ex]&=\dfrac{\mathbf 1_{y\in [0..4]}}{6\surd y}+\dfrac{\mathbf 1_{y\in(4..16]}}{12\surd y}\end{align}$$
So $f_{\small Y}(1) = \tfrac 1{6}$ and $f_{\small Y}(9)=\tfrac 1{36}$ as you had.