I should find the domain of the function$$ f(x) = \arcsin\frac{x}{\sqrt{1+x^2}} . $$
How would one go about that?
I can "see" that this is defined on the interval:
$$ (-\infty, \infty) , $$
so
$$ D_f = (-\infty, \infty) $$
because the quotient is always between $$ [-1, 1] . $$
So, how would I prove this?
Well you can define $f$ for all $x\in \mathbb R$, where $\sqrt{1+x^2} \neq 0$ (because dviding by $0$ is not defined, and all other operations are allowed for all $x \in \mathbb R$).
Following that up, we get that $f$ is defined at all $x \in \mathbb R$, where $x^2 \neq -1$. So for this part, all $x \in \mathbb R$ are allowed.
As Chris pointed out, we have to check for $\arcsin (z)$, where $z = \frac{x}{\sqrt{x^2+1}}$. We know that $\arcsin(z)$ is only defined for $z \in [-1,1]$. Let's show $|z|\leq 1$, if $z = \frac{x}{\sqrt{x^2+1}}$. As $|x| = \sqrt{x^2} <\sqrt{x^2+1}$, and $\sqrt{x^2+1} > 0 \forall x\in \mathbb R$, we have $|z| = |\frac{x}{\sqrt{x^2+1}}|<1 \forall x \in \mathbb R$.
Because of this, $D_f = \mathbb R$.
