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 $f(x) = 0$.
We solve the equation $f(x) = 0$ on the domains $x \le 1$ and $x > 1.$

If $x \le 1,$ then $f(x) = -x - 3,$ so we want to solve $-x - 3 = 0.$ The solution is $x = -3,$ which satisfies $x \le 1.$

If $x > 1,$ then $f(x) = \frac{x}{2} + 1,$ so we want to solve $\frac{x}{2} + 1 = 0.$ The solution is $x = -2,$ but this value does not satisfy $x > 1.$

Therefore, the only solution is $x = \boxed{-3}.$
-3
Level 3
Algebra
Problem: 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 $f(x) = 0$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the values of \( x \) such that \( f(x) = 0 \), we need to analyze both cases defined in the piecewise function. 

First, consider the case when \( x \le 1 \). In this case, the function is defined as \( f(x) = -x - 3 \). We set this equal to zero:

\[
-x - 3 = 0
\]

Solving for \( x \):

\[
-x = 3 \\
x = -3
\]

Since \( -3 \) is less than \( 1 \), it is a valid solution for this case.

Next, consider the case when \( x > 1 \). Here, the function is defined as \( f(x) = \frac{x}{2} + 1 \). We set this equal to zero:

\[
\frac{x}{2} + 1 = 0
\]

Solving for \( x \):

\[
\frac{x}{2} = -1 \\
x = -2
\]

However, \( -2 \) does not satisfy the condition \( x > 1 \), so it is not a valid solution.

Now, we have only one valid solution, which is \( x = -3 \). The problem asks for the sum of all values of \( x \) such that \( f(x) = 0 \). Since there is only one solution, the sum is simply:

\[
-3
\]

Thus, the final answer is \( -3 \).<|END_OF_LATENT|> The final answer is $-3$.