Determine all possible real valued functions f, with x being a real number belonging to (0,1), such that:
f(x) + f(1/(1-x)) = 2(1 - 2x)/(x(1 -x))
Note: The interval related symbols are in consonance with this
article.
Let g(x) = 2(1 - 2x)/(x(1 -x))
Use three transformations:
a) x = y
then f(y) + f(1/(1-y)) = g(y)
b) x = (1/(1-y))
then f(1/(1-y)) + f((y-1)/y) = g(1/(1-y))
c) x= ((y-1)/y)
then f((y-1)/y) + f(y) = g((y-1)/y)
Add a and c, and subtract b, to get
2f(y) = g(y) + g((y-1)/y) - g(1/(1-y))
Thus, there is one and only one function that works:
f(x) = (g(x) + g((x-1)/x) - g(1/(1-x)))/2
where g(x) = 2(1 - 2x)/(x(1 -x))
I am too lazy to simplify the above, but I look forward to the answer.
Actually, I am not too lazy to remove the 2.
f(x) = h(x) + h((x-1)/x) - h(1/(1-x))
where h(x) = (1 - 2x)/(x(1 -x))
An approach (spoiler)
