If $f(2a-b) = f(a) \cdot f(b)$ for all a and b, and the function is never equal to 0, find the value of f(5).
As such, what I've already tried is simply eliminating all possibilities of what type of functions I could be working with.
However, when relations such as piecewise functions come into play, I can't even eliminate the possibility of functions that are to an odd degree. Unless it of course has no removable discontinuity. However, these are relatively new terms for myself and I may simply be overcomplicating the question.
So if anyone has any thoughts or insight as to how I should proceed, it would be greatly appreciated.
If $a=b$ (and since $f(a)\ne 0$ for all $a$) you get $$f(a) = f(a)^2\implies f(a)=1;\;\ \forall a $$
So $ f(5) = 1$