Find the probability density function of the random variable $u=u(y)=\sqrt{y+1} \;$, y with known probability density function
$$f(y) = \begin{cases} \frac{1}{2\beta}e^{-\frac{y}{\beta}} &\quad\text{if} \; y \gt 0 \\ \frac{1}{2} &\quad\text{if} \; -1 \lt y \lt 0 \\ \end{cases}$$
What I did is, I used the change of variable theorem; $ g(u)=\rvert \frac{dy}{du}\lvert f(y(u))$
$$u=\sqrt{1+y} \implies y=u^{2}-1 \\ \rvert \frac{dy}{du}\lvert = 2u \\ \implies g(u)= \begin{cases} \frac{2u}{2\beta}e^{\frac{-u^{2}+1}{\beta}} &\quad\text{if} \; u \gt 1 \\ \frac{2u}{2} &\quad\text{if} \; 0 \lt u \lt 1 \\ \end{cases} \\ \implies g(u)= \begin{cases} \frac{u}{\beta}e^{\frac{-u^{2}+1}{\beta}} &\quad\text{if} \; u \gt 1 \\ u &\quad\text{if} \; 0 \lt u \lt 1 \\ \end{cases} $$
Finally, I don't know if the procedure is correct or I'm missing something since u is defined as a square root. Also don't know if the limits are correct either. Thanks in advance.