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: Not an answer but maybe something to consider for your second functional equation,
Let $\phi$ denote the golden ratio so that we have $\frac{1}{\phi}+1=\phi$
Then by the second functional equation if we set $x=\frac{1}{\phi}$ we have:
$$f(\phi)-f(\frac{1}{\phi}+1)=\frac{1}{\phi}$$
$$f(\phi)-f(\phi)=\frac{1}{\phi}$$
$$0=\frac{1}{\phi}$$
Which obviously isn't true so $f(x)$ isn't properly defined at $x=\phi$

In addition either $f(x)$ isn't analytic at $x=0$ or we must have that:
$$f(x)\sim -x$$
Because under the substitution $x\rightarrow x-1$ we have:
$$f(\frac{1}{x-1})-f(x)=x-1$$
$$-f(x)=x-1-f(\frac{1}{x-1})$$
$$f(x)=-x+1+f(\frac{1}{x-1})$$
$$f(x)=-x+O(1)$$
Where $\lim_{x\to\infty}1+f(\frac{1}{x-1})=1+f(0)=O(1)$ because by assumption $f$ is analytic at $0$ and therefore continuous at $0$, so we are able to interchange the limits.
A: 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.
A: $$f(x)+f\left(\frac{x+1}{x}\right)=\frac1x\\
f\left(\frac{x+1}{x}\right)+f\left(\frac{2x+1}{x+1}\right)-\frac1\phi=\frac{x}{x+1}-\frac1\phi\\
f\left(\frac{2x+1}{x+1}\right)+f\left(\frac{3x+2}{2x+1}\right)-\frac1\phi=\frac{x+1}{2x+1}-\frac1\phi$$
If $f$ is continuous at $\phi$, then 
$$f(x)=(\frac1x-\frac1{2\phi})-(\frac{x}{x+1}-\frac1{\phi})+(\frac{x+1}{2x+1}-\frac1{\phi})-...\\
=(\frac1x-\frac1{2\phi})-(1-\frac1{\phi}-\frac1{x+1})+(\frac12+\frac1{2(2x+1)}-\frac1{\phi})-...\\
=C+\frac1x+\frac1{x+1}+\frac1{2(2x+1)}+\frac1{3(3x+2)}+\frac1{5(5x+3)}+...
$$
for $C-\frac1{2\phi}-\frac1{1\times2}-\frac1{3\times5}-\frac1{8\times13}-...$
A: Just looking at $f(1/x)+f(x+1)=x$. My assumption is that you mean a function defined and analytic at every point on the real line. If so, then there is no such function. 
You would have
$$
\begin{align}
\lim_{x\to\infty}f(x)&=\lim_{x\to0^+}f(1/x)\\
&=\lim_{x\to0^+}\left(x-f(x+1)\right)\\
&=-f(1)
\end{align}
$$
And then:
$$
\begin{align}
f(0)&=\lim_{x\to0^+}f(x)\\
&=\lim_{x\to\infty}f(1/x)\\
&=\lim_{x\to\infty}\left(x-f(x+1)\right)&\Big(\lim_{x\to\infty}f(x+1)&=\lim_{x\to\infty}f(x)=-f(1)\Big)\\
&=\lim_{x\to\infty}(x)+f(1)
\end{align}
$$
So $f(0)$ does not exist. Now, if you wanted a function that is analytic on the positive reals, then that would be something else.
