Let \[f(x) = \left\{
\begin{array}{cl}
-x - 3 & \text{if } x \le 1, \\
\frac{x}{2} + 1 & \text{if } x > 1.
\end{array}
\right.\]Find the sum of all values of $x$ such that $f(x) = 0$.
I need to find the values of $x$ that make $f(x) = 0$. Since $f(x)$ is defined piecewise, I have to consider each case separately. First, let me look at the case where $x \le 1$. In this case, $f(x) = -x - 3$, so I need to solve $-x - 3 = 0$. Adding $x$ to both sides, I get $-3 = x$. This is a valid solution, since $-3 \le 1$. Next, let me look at the case where $x > 1$. In this case, $f(x) = \frac{x}{2} + 1$, so I need to solve $\frac{x}{2} + 1 = 0$. Subtracting $1$ from both sides, I get $\frac{x}{2} = -1$. Multiplying both sides by $2$, I get $x = -2$. However, this is not a valid solution, since $-2 \not > 1$. Therefore, the only value of $x$ that satisfies $f(x) = 0$ is $x = -3$. To find the sum of all such values, I just add up $-3$. So the answer is $-3$. 