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?

$$x^2-1 \geq 0 \Rightarrow x^2 \geq  1 \Rightarrow x \geq 1 \text{ or } x \leq -1$$
Also:
$$4-2 \sqrt{x^2-1} \geq 0 \Rightarrow 4 \geq 2 \sqrt{x^2-1} \Rightarrow 2 \geq \sqrt{x^2-1} \Rightarrow 4 \geq x^2-1 \Rightarrow 5 \geq x^2 \\ \Rightarrow \sqrt{-5} \leq x \leq \sqrt{5}$$
So:
$$1 \leq x \leq \sqrt{5}  \ \text{  or } -\sqrt{5} \leq x \leq -1$$
Therefore the domain is:
$$\{ x \in \mathbb{R}: -\sqrt{5} \leq x \leq -1 \text{ or } 1 \leq x \leq \sqrt{5} \}$$
