Is there a function that satisfies $f(\log \frac{p}{1-p}) = p(1-p)$?

Question

Let $p\in (0,1)$ I am wondering there exists a function $f:\mathbb{R}\to[0,{1\over 2}]$ that satisfies $$ f\left(\log \frac{p}{1-p}\right) = p(1-p) $$

I've tried messing around with exponentials but nothing has worked so far.

Answer 1

Write $x= \log \frac{p}{1-p}$ then $p = {e^x\over e^x+1}$ so $$f(x) = {e^x\over e^x+1}\left(1- {e^x\over e^x+1}\right) = {e^x\over (e^x+1)^2}$$

and clearly $f(x)\in [0,{1\over 2}]$.