Q:

Sufficient condition for $f(x)$ and $g(x)$ for it to exists $x$ s.t. $f(x) - g(x) = c$

I wonder if it is possible to find some conditions on the functions $f(x)$ and $g(x)$ such that I can be certain that there will exists an x such that $f(x) - g(x) = c$, where $c\in\mathbb{R}$.
I can think in many cases that would satisfy this, but I'm not sure about general conditions. Thank you for your answers :D

A:

First, setting $h(x)=f(x)-g(x)-c$, we simplify the task: find $x$ such that $h(x)=0$. Now, it is impossible to say anything unless we put some restrictions on $h$, and there are many possibilities; I'll list a couple.
If we are provided that $h$ is continuous (in particular, if $f$ and $g$ are), then a theorem by Weierstrass can be used: if $h(a)<0$ and $h(b)>0$ (or vise versa) then there exists $x \in (a,b)$ such that $h(x)=0$.
If we know that $h$ is differentiable, if $h(a)<0$ and $\forall x > a: \,h'(x)>A>0$ (where $A$ is just some positive constant), then, as an application of mean value theorem, there exists $x \in \left(a, a+\frac{-h(a)}{A}\right)$ such that $h(x)=0$.
These criteria can be easily transformed into theorems stating that $\exists x:\,f(x)-g(x)=c$ simply by suиstituting the definition of $h$.

