### 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 satisfy the equation $f(f(x)) = f(x)$.
To do that, I will first simplify the expression $f(f(x))$ by plugging in the definition of $f(x)$.
So, $f(f(x)) = f(x^2-3x) = (x^2-3x)^2 - 3(x^2-3x)$, using the same rule as before.
Now I have a quadratic equation in $x^2-3x$, which I can solve by setting it equal to $f(x)$ and rearranging.
So, $(x^2-3x)^2 - 3(x^2-3x) = x^2-3x$ implies $(x^2-3x)^2 - 4(x^2-3x) = 0$.
I can factor out $x^2-3x$ from both terms, and get $(x^2-3x)(x^2-3x-4) = 0$.
Now, by the zero product property, either $x^2-3x = 0$ or $x^2-3x-4 = 0$.
The first equation gives me $x = 0$ or $x = 3$ as solutions, and the second one gives me $x = -1$ or $x = 4$ as solutions, by using the quadratic formula or completing the square.
So, I have four possible values of $x$ that make $f(f(x)) = f(x)$ true.
I can check them by plugging them back into the original equation and verifying that they work.