### 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: To solve this problem, I need to find the values of $x$ that make $f(f(x))$ and $f(x)$ equal.
That means I need to set up an equation: $f(f(x)) = f(x)$.
To simplify this equation, I need to substitute $f(x)$ with its expression: $x^2-3x$.
So, $f(f(x))$ becomes $f(x^2-3x)$.
That means I need to plug in $x^2-3x$ wherever I see $x$ in the expression for $f(x)$.
That gives me $(x^2-3x)^2 - 3(x^2-3x)$.
Now I have an equation that looks like this: $(x^2-3x)^2 - 3(x^2-3x) = x^2-3x$.
This is a quadratic equation in terms of $x^2-3x$.
I can make it easier to work with by letting $y = x^2-3x$.
Then, the equation becomes $y^2-3y = y$.
To solve for $y$, I need to bring everything to one side and factor: $y^2-4y = 0$.
This means that $y(y-4) = 0$.
So, either $y = 0$ or $y = 4$.
But I don't want to find $y$, I want to find $x$.
So, I need to substitute back $y = x^2-3x$ and solve for $x$.
If $y = 0$, then $x^2-3x = 0$.
This means that $x(x-3) = 0$.
So, either $x = 0$ or $x = 3$.
These are two possible solutions for $x$.
If $y = 4$, then $x^2-3x = 4$.
This means that $x^2-3x-4 = 0$.
This is a quadratic equation that I can solve by factoring: $(x-4)(x+1) = 0$.
So, either $x = 4$ or $x = -1$.
These are two more possible solutions for $x$.
Therefore, there are four values of $x$ that make $f(f(x)) = f(x)$.
They are $x = 0, 3, 4, -1$.