Function with an $x$ not in simple form I've stumbled upon a practice example in an old textbook which I find confusing. Maybe it's because I haven't reached part of an explanation yet (went through pages, haven't found anything of help). Also, task doesn't have a solution offered. Some do, some don't - it's a weird textbook of ye old age (and country).
Here's what it states (sorry for possible rough translation): 
For a given function:
$$f(x)=\sqrt{x^2-6x+9}-\sqrt{x^2+6x+9}$$
a) How much is $f(x)$, for $-2 < x < 2$?
b) Calculate $f(\sqrt{2}-\sqrt{3})$
I admit I haven't seen this before, so I would really appreciate if someone would guide me through this, step by step preferable since I have no one to ask. I struggle with the meaning here of how much is $-2 < x < 2$. I understand this can't be an $x$ value itself. I guess it's supposed to be a range of a given function? So what is it then? A range of values I should put into a function or something?
What I did is I have "simplified" the function first (don't know if there's a term for it?). 
here's what I did so far: 
1: $f(x)=\sqrt{x^2-6x+9}-\sqrt{x^2+6x+9}$
2: $f(x)=\sqrt{{(x-3)}^2}-\sqrt{{(x+3)}^2}$ <- factored quadratics basically
3: $f(x)=\lvert(x-3)\rvert-\lvert(x+3)\rvert$ <- cancelled out sqrts with ^2 exponents
but how would I proceed now with given tasks? 
PS
Sorry I couldn't think of a better title.

It's badly worded, but I believe the first question is asking you to find $f((-2,2))$ (in case you aren't familiar with the notation, that's the range of $f$ where its domain is taken to be the interval from $-2$ to $2$). For the second, substituting it in should be fine. 
