Domain of definition I need to find the domain of definition of the following functions:
$$
f(x)=\arccos\left(1/(x^2+1)\right), \quad g(x)=\arcsin\left(2x\sqrt{1-x^2}\right).
$$
For $f$, we have to find $x$ such that $-1\leq 1/(x^2+1)\leq 1$. I do this:
$$\left|\frac{1}{x^2+1}\right|\leq 1 \Longleftrightarrow x^2+1\geq 1\Longleftrightarrow |x|\geq0.$$
Does this mean that $D_f=\mathbb{R}$?
For $g$:
$$\left|2x\sqrt{1-x^2}\right|\leq 1 \Longleftrightarrow 4x^2(1-x^2)\leq 1 \Longleftrightarrow -4x^4+4x^2-1\leq 0.$$
I find two solutions $-1/\sqrt2$ and $1/\sqrt{2}$, and the condition $1-x^2\geq 0$ that is $x\in [-1,1]$. So for me the domain is $(-\infty,-1/\sqrt{2}]\cup [1/\sqrt{2},+\infty) \cap [-1,1]$, hence $[1/\sqrt{2},1]$, but I am not sure.
Can you help me please ?

Yes, the domain of $f$ is all of $\mathbb{R}$. For $g$ you need to solve the system
$$
\begin{cases}
1-x^2 \geq 0 \\
-1 \leq 2x\sqrt{1-x^2} \leq 1.
\end{cases}
$$
The first one gives $-1 \leq x \leq 1$. You can split the second getting two inequalities. Take $2x\sqrt{1-x^2}\leq 1$ first. Note that for $x\leq 0$ this is always true, so you can focus on $0<x\leq 1$. Square of both sides gives $4x^2(1-x^2) \leq 1$, namely $4x^4-4x^2+1 = (2x^2-1)^2\geq 0$, which is always true. Now note that the function $f(x) = 2x\sqrt{1-x^2}$ is odd, so solutions to the other inequality are the same. Therefore the final solution is just $|x| \leq 1$.
