### 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 start, I'll try to write $f(f(x))$ in terms of $x$.
I can do this by replacing every $x$ in $f(x)$ with $f(x)$.
So I get:

$$f(f(x)) = (f(x))^2 - 3f(x)$$
Now I have to solve for $x$ when this expression is equal to $f(x)$.
I can do this by subtracting $f(x)$ from both sides and simplifying.
So I get:

$$f(f(x)) - f(x) = (f(x))^2 - 4f(x) = 0$$
Now I have a quadratic equation in $f(x)$.
To solve it, I'll try to factor it.
I see that both terms have a common factor of $f(x)$, so I can pull that out:

$$(f(x))^2 - 4f(x) = f(x)(f(x) - 4) = 0$$
Now I can use the zero product property, which says that if a product of two factors is zero, then one of the factors must be zero.
So I have two possibilities:

$$f(x) = 0 \text{ or } f(x) - 4 = 0$$
I'll solve for $x$ in each case.
For the first one, I have to find the zeros of $f(x)$.
I can do this by setting $f(x) = 0$ and factoring:

$$f(x) = x^2 - 3x = x(x-3) = 0$$
So the solutions are $x = 0$ or $x = 3$.
These are two values of $x$ that make $f(f(x)) = f(x)$.
For the second case, I have to find the values of $x$ that make $f(x) = 4$.
I can do this by setting $f(x) = 4$ and solving for $x$:

$$f(x) = 4$$
$$x^2 - 3x = 4$$
$$x^2 - 3x - 4 = 0$$
This is another quadratic equation, which I can try to factor.
I look for two numbers that multiply to -4 and add to -3.
I see that -4 and 1 work, so I can write:

$$x^2 - 3x - 4 = (x-4)(x+1) = 0$$
So the solutions are $x = 4$ or $x = -1$.
These are two more values of $x$ that make $f(f(x)) = f(x)$.
I have found four values of $x$ that satisfy the given equation: 0, 3, 4, and -1.
I'll enter them in ascending order, separated by commas.