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.
 A: 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. 
A: This is an exercise in carefully working through the meaning of the square root symbol and keeping track of sign changes.  You've done an excellent job of removing the square roots to obtain the expression
$$f(x)=|x-3|-|x+3|$$
The only thing you have to do now keep track of sign changes in the absolute value expressions.  For example, $|x-3|=x-3$ when $x\ge3$ and $|x-3|=3-x$ when $x\le3$.  Likewise, $|x+3|=x+3$ when $x\ge-3$ and $|x+3|=-(x+3)$ when $x\le-3$.  Since part a) is only asking about the interval $-2\lt x\lt2$, we have
$$f(x)=3-x-(x+3)\quad\text{for }x\in(-2,2)$$
and this simplifies to
$$f(x)=-2x$$
Part b) should now be very easy, once you notice that $x=\sqrt2-\sqrt3$ falls in the interval $-2\lt x\lt2$.
In point of fact, the sign changes at $x=3$ and $-3$ mean that
$$f(x)=
\begin{cases}
6\quad&\text{for }x\le-3\\
-2x\quad&\text{for }-3\lt x\lt 3\\
-6\quad&\text{for }3\le x
\end{cases}
$$
