How to solve $f(x)+f\left(\frac{1}{x}\right)=e^{x+\frac{1}{x}}$

Question

Given that $f:\mathbb{R}_0 \rightarrow \mathbb{R}_0$ find such $f$ that $$f(x)+f\left(\frac{1}{x}\right)=e^{x+\frac{1}{x}}$$

Note that I came up with this question, and personally am not sure that there exists a closed form solution, as my efforts seemed ineffective.

The only things that I believe I can say is that $f(1)=\frac{e^2}{2}, f(-1)=\frac{1}{2e^2}$.

If there exists no closed form solution, I would appreciate some more information on $f$, such as if it is differentiable.

Answer 1

There will be a very large number of solutions, but an obvious one is $$f(x)=\tfrac{1}{2}e^{x+\frac{1}{x}}$$ for $x \not = 0$

More generally take any $g(x)$ and define $f(x)=g(x)$ when $|x| \gt 1$ and $f(x)=e^{x+\frac{1}{x}} - g\left(\tfrac{1}{x}\right)$ when $0 \lt |x| \lt 1$, plus the observations you have made. $f(0)$ is arbitrary