Let $C\geq 2$ and $L>0$. Does there exist $g \in C^1([0,L])$ such that \begin{equation*} g(x)>0, \qquad g'(x)>0, \qquad g'(x) > (g(L) - g(x))C \end{equation*} holds for any $x \in [0,L]$? How large can the length $L$ of the interval be chosen?

**First example:** if
\begin{equation*}
g(x) = ax + b
\end{equation*}
with $a, b>0$, then $$g'(x) > (g(L) - g(x)) C$$ is equivalent to $L < \frac{1}{C}$.

**Second example:** let $f \colon [0,L] \rightarrow [\varepsilon, \frac\pi2]$ be defined by
\begin{equation*}
f(x) = \frac{\pi x}{2L} + \varepsilon \Big(1 - \frac{x}{L}\Big).
\end{equation*}
Then, if we choose
\begin{equation*}
g(x) = -\cot(f(x)) + \cot(\varepsilon) + \delta
\end{equation*}
with $\delta>0$ and $\varepsilon \in (0, \frac\pi2)$, we have $$g'(x) = \Big(\frac{\pi}{2} - \varepsilon\Big) \frac{1}{L \sin^2(f(x))},$$ while $$g(L) - g(x) = \cot(f(x)).$$ Hence $g'(x) > (g(L) - g(x)) C$ is equivalent to $L < \frac{\pi - 2\varepsilon}{C}$.

But is there a function $g$ allowing $L$ to be larger?

This question is also posted on Math Stack Exchange: https://math.stackexchange.com/questions/3280984/find-function-with-inequality-for-derivative