Finding domain of $\sqrt{4-2\sqrt{x^2 - 1}}$ Find the (maximum) domain of: $\sqrt{4-2\sqrt{x^2 - 1}}$
Well, I guess it is $2 - \sqrt2 \sqrt{x^2 - 1}$ = $2- \sqrt{2(x^2 - 1)}$ 
from here, $x^2 - 1 \ge 0$ and then $ x \ge 1$ and $ x \ge -1$
What do you guys think?

You are right on the first restriction, we need $|x|\ge 1$. We also need $4-2\sqrt{x^2-1}\ge 0$. This inequality is equivalent to $2\sqrt{x^2-1}\le 4$, or equivalently $x^2-1\le 4$, that is, $|x|\le \sqrt{5}$. 
Thus a way of describing the (maximum) domain is that it is $\{x: 1\le |x|\le \sqrt{5}\}$.
