Trying to find a concave function defined on the positive reals, satisfying some inequalities, I came up with the following relation

$f\left(x\right) = 1 - \left(1 - f\left(x+1\right)\right)^{\frac{x}{x+1}}$

where $x \geq 0$. The only progress I could make was figuring that $f(0) = 0$, and if I postulate some value for $f(1)$ I can in principle calculate $f(n)$, where $n$ is a positive integer. Any hints for further progress are appreciated.

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    $\begingroup$ The general solution is $f(x)=1-c(x)^x$ where $c:\mathbb{R}_{\geq 0}\to\mathbb{R}_{\geq 0}$ is any function satisfying $c(x)=c(x+1)$ for all $x$. $\endgroup$ – Redundant Aunt May 7 at 13:27
  • $\begingroup$ @Redundant Aunt Thanks for the answer. I now realise that a better formulation would have been in terms of function $g$ where $g(x) = 1-f(x)$. $\endgroup$ – dayandnightatom May 9 at 7:01
  • $\begingroup$ @RedundantAunt post as an answer? $\endgroup$ – user574848 May 25 at 0:22

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