Find all solutions to the functional equation $f(1-x) = f(x) + 1 - 2x.$ Source: M&IQ
I first tried to use the fact that $1-x$ is cyclic, and that failed. I then tried to apply the fact that when $f(1-x) = f(x),$ we must have $f(x) = ax^2 - ax + b,$ where $a$ and $b$ are arbitrary numbers. How should I move on from here, or is there a better method to solve this problem?
Let $g(x)=f(x)-x$. Then the given functional equation translates to \begin{align} &g(1-x)=g(x)&&\forall x\in\Bbb R\\ \implies &g\left(\frac12+\left(\frac12-x\right)\right)=g\left(\frac12-\left(\frac12-x\right)\right)&&\forall x\in\Bbb R\\ \implies &g\left(\frac12+y\right)=g\left(\frac12-y\right)&&\forall y\in\Bbb R \end{align} Take any function symmetric about $\frac12$ $\left(\left(x-\frac12\right)^{2n}, \text{ say}\right)$ and that will correspond to a solution of the given functional equation.
Discussion of techniques to solve functional equations are given e.g. by Evan Chen in "Introduction to Functional Equations" and "Monsters". Mostly aimed at Math Olympiad competitors, they are quite accessible.
