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?

The first thing I would do is graph y= x^2- x.  That is a parabola crossing the x-axis at x= 0 and x= 1 and vertex (1/2, -1/4). That does not go down to -2 so this is the same as the pre-image of [-1/4, 4].  But looking at the graph, it is easy to see that the condition is that f(x) be less than 4.  Solving $x^2- x= 4$ we get $x= \frac{1\pm\sqrt{17}}{2}$.  The pre-image is $]\frac{1- \sqrt{17}}{2}, \frac{1+ \sqrt{17}}{2}[$
