# pdf of transformed variable

Given pdf $$f_X(x)=\frac{x+2}{18}$$ where $$-2 < x < 4$$, I wanted to find another r.v. $$Y = \frac{12}{|X|}$$. I think the support of $$Y$$ would be $$3 < y < \infty$$ but I wasn't super sure. I found the cdf of $$X$$, which was $$F_X(x) = \int_{-2}^x\frac{x+2}{18}dx=\frac{x^2}{36}+\frac{x}{9}+\frac{1}{9}$$, $$-2 < x < 4$$. I tried using the cdf of $$X$$ to find the cdf of $$Y$$ and consequently find the pdf of $$Y$$ but have been struggling to do so. Could anyone help me with the derivation?

\begin{align*} F_Y(y)&=P(Y \leq y) \\ &=P(\frac{12}{|X|}\leq y) \\ &= P(|X| \geq \frac{12}{y}) \\ &= 1 - P(|X| \leq \frac{12}{y})\\ &=1- P(-\frac{12}{y} \leq X \leq \frac{12}{y}) \\ &= 1 - F_X(\frac{12}{y}) + F_X(-\frac{12}{y})\\ &= 1 - \frac{8}{3y} \end{align*} Then I just differentiate to get the pdf: \begin{align*} f_Y(y)&=\frac{d}{dy}F_Y(y) \\ &= \frac{8}{3y^2} \end{align*}

But this pdf doesn't integrate to $$1$$ so I'm not sure what's wrong.

Note that $$12/\lvert X\rvert$$ folds both the domains $$(-2;0)$$ and $$(0;2)$$ onto $$(6;\infty)$$, and the domain $$[2;4)$$ onto $$(3;6]$$.
More clearly, only when $$Y>6$$ it is mapped to by a positive and negative value for $$X$$. When $$3 then $$Y$$ is mapped to by only a positive value of $$X$$ (on $$[2;4)$$).
$$\begin{split}\mathsf P(Y\leq y) &=\mathsf P(\lvert X\rvert\leq 12/y)\mathbf 1_{6
$$\begin{split}f_Y(y)&=\begin{vmatrix}\dfrac{\partial (-12/y)}{\partial y}\end{vmatrix}f_X(-12/y)\mathbf 1_{y\in(6;\infty)}+\begin{vmatrix}\dfrac{\partial (12/y)}{\partial y}\end{vmatrix}f_X(12/y)\mathbf 1_{y\in(3;\infty)}\\ &=\dfrac{(-24+4y)}{3y^3}\mathbf 1_{6