### Instruction: 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 $f(x) = 0$.### Response: 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 $-3$.