If $f(0) > g(0)$, and $\exists N > 0$ such that if $k\geq N$ then $f(k)>g(k)$ and $f,g$ are monotonic increasing, is $f(x)>g(x)$ $\forall x\geq 0$?

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Suppose we have two monotonically increasing functions $f,g$ with $f(0)>g(0)$ and the existence of a finite $N>0$ such that if $k\geq N$ then $f(k) > g(k)$

I now want to show that $f(x) > g(x)$ for all values of $x\geq 0$. What extra criteria would I need? I know that with this alone its not enough to prove it because, which I was able to visualize just by drawing a simple picture.

The obvious answer is that if $f'(x) \geq g'(x)$ for all $x\geq 0$ then we are good, but I was looking for a more lax condition.

My intuition tells me that if I can show that the derivatives themselves are monotonic then that is sufficient because we will not have any locations in which $f$ slows down and dips below $g$ and then increases at a faster rate to become greater again, however how can I write this out in a tight way?