Let's say we have two function $f(x) = \sqrt{x-3}$ and $g(x) = \sqrt{16-x^2}$, when finding the domain of $\frac{f}{g}$ do you find the domain of $\frac{\sqrt{3-x}}{\sqrt{16-x^2}}$ so that $x$ is an element of $(-4,3]$ or would you find the domain of $\sqrt{\frac{3-x}{16-x^2}}$ where $x$ is an element of $(-4,3] \cup (4, +\infty)$?
Thanks, any help is appreciated.
Best Answer
It must be $f\geq 0$ and $g > 0$, that is $x-3\geq 0$ and $16-x^2>0$. The solutions are $x\geq 3$ and $-4<x<4$, so the domain is $[3,4)$.