Find a solution for $f\left(\frac{1}{x}\right)+f(x+1)=x$ Title says all. If $f$ is an analytic function on the real line, and $f\left(\dfrac{1}{x}\right)+f(x+1)=x$, what, if any, is a possible solution for $f(x)$?
Additionally, what are any solutions for $f\left(\dfrac{1}{x}\right)-f(x+1)=x$?

A few hints that might help...


*

*$1/x = x+1$ when $x = \frac{\pm\sqrt{5}-1}2$

*Differentiating gives: $-\frac{f'(1/x)}{x^2}+f'(1+x)=1$

*Differentiating again gives: $f''(1+x)+\frac{f''(1/x)}{x^4}+\frac{2f'(1/x)}{x^3}=0$ - this can then be continued.

*An "analytic function" has a Taylor series at any point that is convergent within a non-zero region around the point. So what would the Taylor series look like at the points given in hint 1?


ADDED:
A consideration of limits may also be useful. Indeed, with a substitution of $x=1/y-1$, you have $$f\left(\frac{y}{1-y}\right)+f\left(\frac1y\right)=\frac1y-1$$
We can then cancel out the $\frac1y$ term by first replacing $y$ with $x$, and limits from here may be useful.
