Question
Let $X$ have cumulative distribution function
$$F_X(x) = \begin{cases} 0\qquad \mathrm{if}\ x < 0,\\ \frac 1 8 x^3\quad \mathrm{if}\ 0 \leq x \leq 2,\\ 1\qquad \mathrm{if}\ x > 2.\\ \end{cases}$$
Find the cumulative distribution function of $Z = X^2$.
My working
I attempted to use to the CDF method to compute the CDF of $Z$ as
$$\begin{aligned} F_Z(z) & = \mathbb{P}(Z \leq z)\\ & = \mathbb{P}(X^2 \leq z)\\ & = \mathbb{P}(-\sqrt z \leq X \leq \sqrt z)\\ & = F_X(\sqrt z) - F_X(-\sqrt z)\\ & = \frac 1 8 (\sqrt z)^3 - \frac 1 8 (-\sqrt z)^3\\ & = \frac 1 4 (\sqrt z)^3\ \forall\ z \in [0, 4] \end{aligned}$$
However, I know this is incorrect as $F_Z(4)$ gives me $2$, but I am not sure where I have gone wrong.
Any intuitive explanations will be greatly appreciated :)
Edit
As mentioned in the comments, I was careless in that I did not realise that $F_X(-\sqrt z) = 0$!
The transformation function $Z=X^2$ is monotone in X - support thus
$$F_Y(y)=F_X(\sqrt{z})=\frac{1}{8}(\sqrt{z})^3\cdot\mathbb{1}_{(0;4)}(z)$$