Let us say that I have a function $f(x)$ that we know is a concave. And let us also say that we have another function $g(x)$ that is a convex.
If I make a new function, $h(x) = \frac{f(x)}{g(x)}$, will this new function be convex or concave?
I was wondering is there is an easy way to determine this, instead of deriving it from scratch, (like maybe there is a property or rules relating to operations on con-vex/cave functions or something). Thanks.
On
In many cases, the ratio of a concave and a convex function is quasiconcave. Quasiconcavity and quasiconvexity are discussed in, e.g., Boyd & Vandenberghe; consult them for details.
A function $f$ is quasiconvex if its superlevel sets $\{x\,|\,f(x)\leq\alpha\}$ are convex for all fixed $\alpha$; a function $g$ is quasiconcave if its sublevel sets $\{x\,|\,g(x)\geq\alpha\}$ are convex for all fixed $\alpha$. This is not the same as convexity/concavity; for instance, a function $f$ is convex if its epigraph $\{(x,y)\,|\,f(x)\leq y\}$ is a convex set. Clearly, convexity implies quasiconvexity, but not the other way around; and similarly for concavity and quasiconcavity.
Because they are not convex, they cannot be used in typical convex optimization software; but because they represent convex sets, they can be converted to compliant form. Returning to your case of a concave function $f$ divided by a convex function $g$. If the convex function $g$ is positive on its domain, then
$$f(x)/g(x) \geq \alpha \quad\Longleftrightarrow\quad f(x) \geq \alpha g(x)$$
When $\alpha\geq 0$, this is a convex inequality that can be compatible with convex programming software. For $\alpha<0$, it is not convex.
If you wish to maximize a quasiconcave function (or minimize a quasiconvex function), you can solve a sequence of convex feasibility problems. Again, I would refer you to Boyd & Vandenberghe for details.
There is no general rule. Let $f(x)=1$; then $f$ is both concave and convex. Let $g(x)=e^{x^2}$, which is convex. However $$ \frac{f(x)}{g(x)}=e^{-x^2} $$ is neither concave nor convex.