Finding the preimage I want to find the preimage of $]-2,4]$ for the function $f(x)=x^2-x$
This is what I have done so far:
We have $0=x^2-x-y$ and therefore the inverse is:
$$f^{-1}(y)=\frac{-1\pm \sqrt{1+4y}}{2}$$
And how do I find out the boundaries of the preimage? If I put the boundaries $-2$ and $4$ into the function $f^{-1}$, I will probably not get them. Also I'm not sure which function I have to take when, the one with plus sign or the one with a minus?

$$f(x) = x^2-x=x^2-x+\frac14-\frac14 = (x-1)^2 - \frac14 \geq -\frac14 $$
you are asked to find $x$ such that $-2\leq f(x) \leq 4$, but $f \geq -\frac14$ so the first inequality doesn't restrict us at any way.
$$ f(x) \leq 4 \Rightarrow (x-1)^2 -\frac14 \leq 4 \Rightarrow (x-1)^2 \leq \frac{17}4\Rightarrow$$
$$ -\frac{\sqrt{17}}{4} \leq x-1 \leq \frac{\sqrt{17}}{4} $$
$$ 1-\frac{\sqrt{17}}{4} \leq x \leq 1+\frac{\sqrt{17}}{4} $$
