We recall that $ V_0^x $ is monotone increasing, thus f is monotone decreasing, and $ V_0^x = f(0)-f(x) $.
Then we have
$ f(0)-f(x) = (f(0)-f(x))^{\frac{1}{2}} \Rightarrow (f(0)-f(x))^2 = (f(0)-f(x)) $
Applying the quadratic formula yields $ f(x) = f(0) $ or $ f(x) = f(0)-1 $, with two compatible functions:
$ f_1(x) = \frac{1}{3} $,
$ f_2(x) = \frac{1}{3}, \ 0<x\leq 1, f_2(0) = \frac{4}{3} $
-pw
