Find h(x), for x≠0, x≠1, given that: h(x)+h(1/(1−x))=1−x−1/(1−x)

In order to do this question, we need to find some way of eliminating the h(1/(1−x)), so that we are left with h(x), thus solving it. One way to do this, is to set up simultaneous equations. We may need to set up more than two, but that will become clear later. To set up simultaneous equations we need an equally true equation with a h(x) and a h(1/(1-x)) term in it (think of these as variables, like x or y), but it can't just be the same equation. The only way we can generate a new equation with a h(x) term in it, is to substitute some function of x, into h(1/(1-x)) so that it becomes h(x). So, lets call this substitute u and say that 1/(1-u) = x. Solving for u gives: u=(x-1)/x. Now substitute this into the original equation for all x: h((x-1)/x)+h(x)=1-(x-1)/x-x.This is good, but we now have a new term, h((x-1)/x), which we don't want. It's not all bad though, because all this means is that we are going to have to set up a third simultaneous equation (3 equations for 3 variables/unknowns) with this term in it. The hope is, that this third equation will not introduce a fourth unknown. If it did, then we may have to repeat again (if it gets that messy though, then we may have gone wrong somewhere). So now the question is, is there some substitution into 1/(1-x) that gives (x-1)/x. Let's try it: 1/(1-u) = (x-1)/x. Solving for u gives: u = 1/(1-x). This is very good, because when we substitue this into the original equation, we get: h(1/(1-x))+h((x-1)/x)=1-1/(1-x)-(x-1)/x. This equation has no new terms in it, so we have 3 valid equations, with 3 unknowns in them.Here they are:h(x)+h(1/(1−x))=1−x−1/(1−x)h((x-1)/x)+h(x)=1-(x-1)/x-xh(1/(1-x))+h((x-1)/x)=1-1/(1-x)-(x-1)/xSolving these, treating each h( ) term like a variable, x,y,etc., gives the solution:h(x) = (1-2x)/2