### 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 $f(f(x))$ and $f(x)$ equal, so I will start by writing an equation: $f(f(x)) = f(x)$.
To simplify this equation, I need to substitute $f(x)$ with its definition, $x^2-3x$, in both sides.
This gives me: $(x^2-3x)^2-3(x^2-3x) = x^2-3x$.
Now I have a quadratic equation in terms of $x^2-3x$, so I will let $y = x^2-3x$ and rewrite the equation as: $y^2-3y = y$.
This is easier to solve, so I will move everything to one side and factor: $y^2-4y = 0$, which means $y(y-4) = 0$.
From this, I can see that $y = 0$ or $y = 4$ are the possible solutions for $y$, but I need to find the corresponding values for $x$.
So I will plug in each value of $y$ into the original definition of $y$ and solve for $x$.
If $y = 0$, then $x^2-3x = 0$, which means $x(x-3) = 0$.
This implies that $x = 0$ or $x = 3$ are the solutions for $x$ when $y = 0$.
If $y = 4$, then $x^2-3x = 4$, which means $x^2-3x-4 = 0$.
This is a quadratic equation that I can solve by factoring: $(x-4)(x+1) = 0$.
This implies that $x = 4$ or $x = -1$ are the solutions for $x$ when $y = 4$.
Therefore, the four values of $x$ that make $f(f(x)) = f(x)$ are $0, 3, 4, -1$.