### 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: So we want to solve $f(f(x)) = f(x)$.
That's right. Let's first find $f(f(x))$
Well $f(x) = x^2-3x$, so $f(f(x)) = (f(x))^2 - 3(f(x))$.
And $f(x) = x^2-3x$, so $(f(x))^2-3(f(x)) = (x^2-3x)^2-3(x^2-3x)$.
But we want to solve $f(f(x)) = f(x)$, so we want to solve $(x^2-3x)^2-3(x^2-3x) = x^2-3x$
Let's call $x^2-3x$ as y. Then we want to solve $y^2-3y=y$
We could also write this as $y^2-4y=0$.
That factors as $y(y-4)=0$.
So either y is 0 or y is 4.
But y is $x^2-3x$, so we need to solve $x^2-3x=0$ and $x^2-3x=4$.
The first equation is true when $x=0$ or $x=3$.
The second equation can be rewritten as $x^2-3x-4=0$ and it is true when $x=4$ or $x=-1$