Finding a tricky composition of two piecewise functions I have a question about finding the formula for a composition of two piecewise functions. The functions are defined as follows:
$$f(x) =
\begin{cases}
2x+1,  & \text{if $x \le 0$} \\
x^2, & \text{if $x > 0$}
\end{cases}$$
$$g(x) =
\begin{cases}
-x,  & \text{if $x < 2$} \\
5, & \text{if $x \ge 2$}
\end{cases}$$
My main question lies in how to approach finding the formula for the composition $g(f(x))$. I have seen a couple of other examples of questions like this online, but the domains of each piecewise function were the same, so the compositions weren't difficult to determine. 
In this case, I have assumed that, in finding $g(f(x))$, one must consider only the domain of $f(x)$. Thus, I think it would make sense to test for individual cases: for example, I would try to find $g(f(x))$ when $x <= 0$. $g(f(x))$ when $x <= 0$ would thus be $-2x-1$, right? However, I feel like I'm missing something critical, because I'm just assuming that the condition $x < 2$ for $g(x)$ can just be treated as $x <= 0$ in this case. Sorry for my rambling, and many thanks for anyone who can help lead me to the solution.

$$g(x) = \begin{cases} -x, & x < 2 \\ 5, & x \ge 2 \end{cases} $$
Therefore
$$g(f(x)) = \begin{cases} -f(x), & f(x) < 2 \\ 5, & f(x) \ge 2 \end{cases} $$
So now we need to know when $f(x) < 2$ and when $f(x) \ge 2$.
$$f(x) = \begin{cases} 2x + 1, & x \le 0 \\ x^2, & x > 0 \end{cases} $$
Let's look at one piece of $f$ at a time.  On the first piece, $2x + 1 \ge 2$ means $x \ge 1/2$.  But this is impossible because $x \le 0$ on the first piece.  Also on the first piece, $2x + 1 < 2$ means $x < 1/2$.  Well, on the first piece we have $x \le 0 < 1/2$, therefore $f(x) < 2$ on the entire first piece, i.e., $f(x) < 2$ if $x \le 0$.
On the second piece, $x^2 \ge 2$ means $x \ge \sqrt{2}$ or $x \le -\sqrt{2}$.  On the second piece we always have $x > 0$, therefore we have $x^2 \ge 2$ when $x \ge \sqrt{2}$.  Also on the second piece, $x^2 < 2$ means $-\sqrt{2} < x < \sqrt{2}$.  And since on the second piece we always have $x > 0$, then we must have $0 < x < \sqrt{2}$ in order to have $x^2 < 2$.
Putting it all together so far, we have the following:
$$ f(x) \ge 2 \text{ if and only if } x \ge \sqrt{2}$$
$$ f(x) < 2 \text{ if and only if } x \le 0 \text{ or } 0 < x < \sqrt{2} $$
Notice that this last one can be simplified but we need to keep them separate.  Why is this?  We'll see as we continue.  Recall:
$$g(f(x)) = \begin{cases} -f(x), & f(x) < 2 \\ 5, & f(x) \ge 2 \end{cases} $$
Therefore:
$$g(f(x)) = \begin{cases} -f(x), & x \le 0 \text{ or } 0 < x < \sqrt{2} \\ 5, & x \ge \sqrt{2} \end{cases} $$
Separating the conditions gives us:
$$g(f(x)) = \begin{cases} -f(x), & x \le 0\\ -f(x), & 0 < x < \sqrt{2} \\ 5, & x \ge \sqrt{2} \end{cases} $$
And we need to do this because $f(x)$ itself is different for $x \le 0$ and $0 < x < \sqrt{2}$.  Finally, we end up with:
$$ h(x) := g(f(x)) = \begin{cases} -(2x+1), & x \le 0 \\ -x^2, & 0 < x < \sqrt{2} \\ 5, & x \ge \sqrt{2} \end{cases} $$
