You should come up with the following substitution such that:

`x = (x-1)/x =gt f((x-1)/x) + f(((x-1)/x - 1)/x) = 1 + (x-1)/x`

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

`f((x-1)/x) + f(1/(1 - x)) = (2x-1)/x`

You need to substitute `(x-1)/x` for x in equation `f((x-1)/x) + f(1/(1 - x)) = (2x-1)/x` such that:

`f(1/(1-x)) + f(x) = 2 - 1/((x-1)/x)`

`f(1/(1-x)) + f(x) = (2x - 2 - x)/(x-1)`

`f(1/(1-x)) + f(x) = (x-2)/(x-1)`

You need to subtract `f((x-1)/x) + f(1/(1 - x)) =(2x-1)/x` from `f(x) + f((x-1)/x) = 1 + x` such that:

`f(x) + f((x-1)/x) -f((x-1)/x)- f(1/(1 - x)) = 1 + x - (2x-1)/x `

`f(x) - f(1/(1 - x)) = 1 + x - (2x-1)/x `

You need to add `f(x) - f(1/(1 - x)) = 1 + x - (2x-1)/x` to `f(1/(1-x)) + f(x) = (x-2)/(x-1)` such that:

`f(1/(1-x)) + f(x) + f(x) - f(1/(1 - x))= (x-2)/(x-1) +1 + x - (2x-1)/x`

`2f(x) = (x-2)/(x-1) + 1 + x - (2x-1)/x`

`f(x) = (x(x-2) + x(x+1)(x-1) - (2x-1)(x-1))/(2x(x-1))`

`f(x) = (x^2 - 2x + x^3 - x - 2x^2 + 3x - 1)/(2x(x-1))`

`f(x) = (x^3 - x^2 - 1)/(2x(x-1))`

**Hence, evaluating f(x) under the given conditions yields `f(x) = (x^3 - x^2 - 1)/(2x(x-1)).` **