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.
(In reply to
An approach (spoiler) by Steve Herman)
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))
Lets start simplifying this by using fraction decomposition on g(x): 2(1-2x)/(x(1-x)) = 2/x - 2/(1-x)
Then g((x-1)/x) = 2/((x-1)/x) - 2/(1-((x-1)/x)) = 2 + 2/(x-1) - 2x
And g(1/(1-x)) = 2/(1/(1-x)) - 2/(1-(1/(1-x))) = -2x + 2/x
So then f(x) = ((2/x - 2/(1-x)) + (2 + 2/(x-1) - 2x) - (-2x + 2/x))/2 = 1 + 2/(x-1)
So lets check f(x) = 1 + 2/(x-1) then:
f(x) + f(1/(1-x))
= 1 + 2/(x-1) + 1 + 2/(1/(1-x)-1)
= 2 + 2/(x-1) - 2(x-1)/x
= (2x(x-1) + 2x - 2(x-1)^2) / (x(x-1))
= (4x-2) / (x(x-1))
= 2(1-2x)/(x(x-1))
re: An approach (spoiler)
