Find the domain of the function $f(x) = \frac{x}{\sqrt{x^2+9}}$ For some reason I can't find domain to this function, but it is pretty clear that it's (ALL X)
$$f(x) = \frac{x}{\sqrt{x^2+9}}$$
I can say that: $$\space x^2+9 > 0\space$$
$$x^2 > -9$$
$$x = \pm \sqrt{-9}$$
I don't get it..   
 A: $$f(x) = \frac{x}{\sqrt{x^2+9}}$$
Only $\sqrt{x^2+9}=0$ could be a problem for your domain. But as $x^2+9>0$ (sum of two squares), this cannot happen. Hence, the domain of $f$ is $R$. 
If you also need to find the map of your function:
First of all we evaluate the derivative of $f(x)$
$$f'(x)=\frac{1\cdot\sqrt{x^2+9}-x\frac{2x}{2\sqrt{x^2+9}}}{(x^2+9)}$$
$$=\frac{\sqrt{x^2+9}-\frac{x^2}{\sqrt{x^2+9}}}{(x^2+9)}=\frac{\frac{(x^2+9)-x^2}{\sqrt{x^2+9}}}{(x^2+9)}=\frac{9}{\sqrt{x^2+9}(x^2+9)}$$
As $9>0$, $(x^2+9)>0$ and $\sqrt{x^2+9}>0$, we can conclude that the function is strictly monotonic for all $x \in R$.
Now we need to evaluate the limits of $f(x)$ for $x \to \pm \infty$
$$\lim_{x\to \infty}\frac{x}{\sqrt{x^2+9}}=\lim_{x\to \infty}\frac{x}{|x|\sqrt{1+\frac{9}{x^2}}}=\lim_{x\to \infty}\frac{x}{x\sqrt{1+\frac{9}{x^2}}}=1$$
$$\lim_{x\to -\infty}\frac{x}{\sqrt{x^2+9}}=\lim_{x\to \infty}\frac{x}{|x|\sqrt{1+\frac{9}{x^2}}}=\lim_{x\to \infty}\frac{x}{-x\sqrt{1+\frac{9}{x^2}}}=-1$$
Hence, we conclude the function maps from $R \to (-1,1)$.
A: I suppose that you want $f(x)$ as a real valued function, and you are right. The domain is $\mathbb{R}$ since $x^2+9> 0 \forall x \in \mathbb{R}$, so $\sqrt{x^2+9}$ is a real number and not null for all real $x$ .
