Solutions to the functional equation $f(1-x) = f(x) + 1 - 2x$

Question

Find all solutions to the functional equation $f(1-x) = f(x) + 1 - 2x.$ Source: M&IQ

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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?

Answer 1

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.

Answer 2

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.