# Define $D_f$ for the function…

I should find for which x:es this function is defined:

$$f(x) = \arcsin\frac{x}{\sqrt{1+x^2}}$$

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 mathematically prove this?

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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, $D_f = \mathbb R$.
$\arcsin(x)$ is not defined for all $x$. – Chris Eagle Oct 11 '12 at 16:00