### Instruction: Let $f(x) = x^2-3x$. For what values of $x$ is $f(f(x)) = f(x)$?  Enter all the solutions, separated by commas.### Response: I need to find the values of $x$ that make the expression $f(f(x))$ equal to the expression $f(x)$.
To do that, I can start by substituting $f(x)$ for $x$ in the definition of $f(x)$, and simplify.
That gives me $f(f(x)) = (f(x))^2 - 3f(x)$.
Now I can write an equation by equating $f(f(x))$ and $f(x)$: $(f(x))^2 - 3f(x) = f(x)$.
To solve for $f(x)$, I can rearrange this equation and factor: $(f(x))^2 - 4f(x) = 0$ implies $f(x)(f(x) - 4) = 0$.
So $f(x) = 0$ or $f(x) = 4$ are the possible values of $f(x)$ that make the equation true.
But I want to find the values of $x$, not $f(x)$.
To do that, I need to plug in these values of $f(x)$ into the original definition of $f(x)$ and solve for $x$.
So for $f(x) = 0$, I have $x^2 - 3x = 0$.
This implies $x(x-3) = 0$, so $x = 0$ or $x = 3$ are possible solutions.
For $f(x) = 4$, I have $x^2 - 3x = 4$.
This implies $x^2 - 3x - 4 = 0$, so $(x-4)(x+1) = 0$, and $x = 4$ or $x = -1$ are possible solutions.
Therefore, the values of $x$ that make $f(f(x)) = f(x)$ are $x = 0, 3, 4, -1$.