Question

Find all functions `f : R` \ `{-1} -> R` such that `[f(x)f(z)]/[y+1] = [f(y)f(z)]/[x+1]` whenever `x,y` and `z` do not equal `-1`?

Answer 1

`f(x) = C_0/(x+1)`

Explanation:

Supposing `f(z) ne 0` we have

`f(x)=(y+1)/(x+1)f(y)` now making `y = x-1`

`f(x)=x/(x+1)f(x-1)` Solving this difference equation we get

`f(x) = C_0/(x+1)`