An object has random velocity $V$ and kinetic energy $K = \frac12mV^2$, where $m$ is the mass of the object. Suppose that the velocity has the Laplacian distribution with probability density function $f_V(x) = \frac12e^{-\lvert x\rvert}$, $x \in R$. Show that the kinetic energy has a Weibull distribution: its probability density function is of the form $$f_K(x) = \begin{cases}\alpha \beta x^{\beta-1}\exp(-\alpha x^\beta) & \text{if }x \ge 0\\ 0 & \text{if } x < 0\text{,}\end{cases}$$
I'm working on this problem, but can't quite seem to get the right answer out. If anyone can spot where I'm going wrong in my working, I'd appreciate some pointers!
My work:
If we let $g$ be the function transforming $V$ as above, it is clear that $g$ is not monotone, but we can consider the cases where $x<0$ and $x\geq0$ separately.
Note that the cdf of this Laplacian distribution is $$F_{V}(x)=\begin{cases}\frac{1}{2}\exp(x) & x<0\\1-\frac{1}{2}\exp(-x) & x\geq0\end{cases}$$ and also note that $$g^{-1}(x)=\sqrt{\frac{2x}{m}}$$ So for the first case, where $x<0$, we have a monotone decreasing transform, so \begin{align*} F_{K}(x)&=1-F_{V}(g^{-1}(x))\\ &=1-\frac{1}{2}\exp\left(\sqrt{\frac{2x}{m}}\right) \end{align*} But this makes no sense as $x<0$ and we are assuming positive mass so that takes care of $f_{K}$ having $0$ probability measure when $x<0$.
For the case where $x\geq0$ we have monotone increasing transform so \begin{align*} F_{K}(x)&=F_{V}(g^{-1}(x))\\ &=1-\frac{1}{2}\exp\left(-\sqrt{\frac{2x}{m}}\right) \end{align*} If I now differentiate to get the pdf of $F_{K}$, I get \begin{align*} f_{K}(x)&=-\frac{1}{2}\exp\left(-\sqrt{\frac{2x}{m}}\right)\cdot\left(-\frac{1}{2}\right)\sqrt{\frac{2}{m}}\frac{1}{\sqrt{x}}\\ &=\frac{\sqrt{2}}{4\sqrt{mx}}\exp\left(-\sqrt{\frac{2x}{m}}\right) \end{align*} And if I let $\beta=\frac{1}{2}$ and $\alpha=\sqrt{\frac{2}{m}}$ I have $$f_{K}(x)=\frac{1}{2}\alpha\beta x^{\beta-1}\exp(-\alpha x^{\beta})\text{ for }x\geq0$$
My question is, how in the living daylight do I remove that $\frac{1}{2}$!?! In fact, the Weibull distribution should be of the form $F_{K}(x)=1-\exp(-\alpha x^{\beta})$, so alarm bells were already going off before I performed the differentiation above. I cannot pinpoint the errors I've made. Any hints would be appreciated.
Thanks for your time and patience.