Cyclic function in functional equation

Question

$f(x) + f \left( \frac{x - 1}{x} \right) = \frac{5x^2 - x - 5}{x}$

I proved $f(\frac{x-1}{x})$ was cyclic in cycles of 3, becoming $f(\frac{-1}{x-1})$, then becoming $f(x)$ but can't see how to apply this and get solutions.

Answer 1

As you said this becomes cyclic.

Replacing $x \to \frac{x-1}{x}$ in the functional equation gives $$f(\frac{x - 1}{x}) + f \left( \frac{- 1}{x-1} \right) = \frac{5(\frac{x - 1}{x})^2 - (\frac{x - 1}{x}) - 5}{\frac{x - 1}{x}}$$

Replacing $x \to \frac{-1}{x-1}$ in the functional equation gives $$f( \frac{-1}{x-1}) + f \left( x \right) = \frac{5( \frac{-1}{x-1})^2 - ( \frac{-1}{x-1}) - 5}{ \frac{-1}{x-1}}$$

Therefore, if I didn't do any mistake calculating the RHS: $$f(x) + f \left( \frac{x - 1}{x} \right) = \frac{5x^2 - x - 5}{x} \\ f(\frac{x - 1}{x}) + f \left( \frac{- 1}{x-1} \right) = \frac{-x^2-9x+1}{x(x-1)}\\ f( \frac{-1}{x-1}) + f \left( x \right) = \frac{5x^2-11x+1}{ x-1}$$

Now add the three relations together to find $f(x) + f \left( \frac{x - 1}{x} \right) + f \left( \frac{- 1}{x-1} \right)$. Once you find this sum, subtract the second relation.