Solving the functional equation $f(1-x)+f(x)=1$ Let $f(1-x)+f(x)=1$ for a function $f:\mathbb{R}\to\mathbb{R}$.
What would the be the value of a possible $f(x)$? (I just need an example).
 A: Call $\tau(x)=x-\frac12$. Notice that $f$ satisfies your equation if and only if $g:=\tau\circ f$ satisfies $$g(1-x)+g(x)=0$$
Notice that $\tau^{-1}(x)=x+\frac12$ and notice that a function $g$ satisfies this equation if and only if $h=g\circ\tau^{-1}$ satisfies $$h\left(\frac12-x\right)+h\left(x-\frac12\right)=0$$
In other words, if and only if $h$ is an odd function.
So, $f$ solves your equation if and only if $\tau\circ f\circ \tau^{-1}$ is odd; id est, if and only if there is some odd function $H$ such that $f(x)=\tau^{-1}\circ H\circ \tau(x)=\frac12+ H\left(x-\frac12\right)$.
For instance:


*

*$f(x)=\frac12+\sin\left(x-\frac12\right)$

*$f(x)=\frac12+\frac18(2x-1)^3$

*$f(x)=\frac12+e^{x-\frac12}-e^{\frac12-x}$

*$f(x)=\frac12+\cos(\pi x)\quad$ °__°' $\quad\ddot\smile$
A: If you put $x=\frac 12$ you find that $f(\frac 12)=\frac 12$.
For $x\lt \frac 12$ you can freely choose any value for $f(x)$ using your favourite scheme for choosing.
Then for $x\gt \frac 12$ you have $f(x)=1-f(1-x)$, and you have defined $f(1-x)$ already because $1-x\lt \frac 12$.
You can adjust your choice to obtain continuity or differentiability at $x=\frac 12$
The functions $f(x)=x$ and $f(x)=1-x$ are monotonic increasing/decreasing
One of the other answers identifies how $f$ can be built out of any odd function, and gives a convenient construction.
