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?
 A: 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 $\{x: 1\le |x|\le \sqrt{5}\}$.
A: $$x^2-1 \geq 0 \Rightarrow x^2 \geq  1 \Rightarrow x \geq 1 \text{ or } x \leq -1$$
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:
$$\{ x \in \mathbb{R}: -\sqrt{5} \leq x \leq -1 \text{ or } 1 \leq x \leq \sqrt{5} \}$$
A: Your first step is wrong: $\sqrt{A-B}$ does not simplify to $\sqrt{A}-\sqrt{B}$ .
You have two places that can give trouble in calculation: the two square roots. Both expressions inside the square roots must be nonnegative. So your first inequality is $x^2-1 \ge 0$. Can you proceed from there?
A: What can limit the domain of the expression is when you attempt to take the derivative of a strictly negative number.
The inner radical requires $$x^2-1\ge0,$$
i.e.
$$x\le-1\lor x\ge1.$$
And the outer radical requires
$$4-2\sqrt{x^2-1}\ge0,$$
or
$$2\sqrt{x^2-1}\le4,$$
 by squaring,
$$x^2-1\le4,$$
i.e.
$$-\sqrt5\le x\le\sqrt5.$$
In conclusion, by "anding" the conditions,
$$-\sqrt5\le x\le-1\lor1\le x\le\sqrt5.$$
