Increasing convex function eventually greater than increasing concave function

Question

Suppose that $f, g$ are non-negative and infinitely differentiable from $\mathbb{R}_+$ to $\mathbb{R}_+.$

Suppose that:

• $f,g$ are increasing.
• $f$ is strictly concave and $g$ is strictly convex.
• $f(0) = g(0)$.
• There exists $\epsilon > 0$ such that $f(\epsilon) > g(\epsilon)$

I want to show that there exists $M > \epsilon$ such that $x \ge M$ implies that $f(x) \le g(x)$, perhaps after adding some extra conditions (see answer below). Intuitively, it makes sense: $f$ will increase at a lower rate than $g$ more and more as $x$ increases. I've tried formalizing this argument with a Taylor expansion, but I didn't succeed.

Answer 1

This is not true, for example, if the functions are both straight lines.

Even if the functions are strictly convex and strictly concave, I would not expect this to be true, for example, if the slopes converge to certain values for both functions: the concave function can still have the slope greater than the convex function for all time $>0$. I have not worked this out though, completely, and I might be wrong.