### Instruction: Let \[f(x) = \left\{
\begin{array}{cl}
2x + 1 & \text{if } x \le 3, \\
8 - 4x & \text{if } x > 3.
\end{array}
\right.\]Find the sum of all values of $x$ such that $f(x) = 0.$.### Response: This problem involves a piecewise function, which means that the function has different rules for different intervals of $x$.
To find the values of $x$ that make $f(x) = 0$, I need to consider both cases: when $x \le 3$ and when $x > 3$.
I can use algebra to solve for $x$ in each case.
For the first case, $x \le 3$, I have $2x + 1 = 0$.
Subtracting $1$ from both sides gives me $2x = -1$.
Dividing both sides by $2$ gives me $x = -\frac{1}{2}$.
This is one possible value of $x$, but I need to check if it satisfies the condition $x \le 3$.
It does, since $-\frac{1}{2}$ is less than $3$.
So $-\frac{1}{2}$ is one of the values of $x$ I'm looking for.
For the second case, $x > 3$, I have $8 - 4x = 0$.
Adding $4x$ to both sides gives me $8 = 4x$.
Dividing both sides by $4$ gives me $x = 2$.
This is another possible value of $x$, but I need to check if it satisfies the condition $x > 3$.
It does not, since $2$ is not greater than $3$.
So $2$ is not one of the values of $x$ I'm looking for.
Therefore, the only value of $x$ that makes $f(x) = 0$ is $-\frac{1}{2}$.
The problem asks for the sum of all such values, so I just need to add up this one value.
The sum is $-\frac{1}{2}$.