The functional equation $f(-x+b)=f(x)$

Question

I can solve the (periodic) functional equation $f(x+b)=f(x)$ completely ($x\in \mathbb{R}$ and $b\neq 0$). Indeed, its general solution is $f=\phi o (\; )_b$, where $(\; )_b$ is the $b$-decimal (fractional) part function defined by $$(\; )_b(x) =(x)_b:=x-b\lfloor \frac{x}{b}\rfloor,$$ (see http://nntdm.net/papers/nntdm-19/NNTDM-19-4-04-15.pdf) and $\phi$ every real function defined on $b[0,1)$. In general, it is solved on arbitrary groups (see http://www.ijmex.com/index.php/ijmex/article/viewFile/194/115).

Now, can somebody solve the functional equation $f(-x+b)=f(x)$? (is there such a general solution for it?)

Answer 1

In fact this functional equation belongs to the form of http://eqworld.ipmnet.ru/en/solutions/fe/fe1113.pdf.

The general solution is $f(x)=\Theta(x,b-x)$ , where $\Theta(u,v)$ is any symmetric function.