So basically, a function $f$ with $f(\frac{1}{x}) = - f(x)$. Additionally, it should also be strictly increasing.
I know that the logarithm has this property, but I'm looking for a function with different boundary conditions. Namely: f(0) = -1 (and $f(x -> \infty) = 1$).
I know one solution to this: $f(x) = \frac{x-1}{x+1}$, but I am wondering:
Is there a general method to find such functions?
Is my given solution unique?
For $x>0$, substitute $x = e^t$ or $t = \log x$. Then, we have
$$
g(t):=f(e^t) = -f(e^{-t})=-g(-t),\quad \forall t\in\mathbb{R},
$$and $g(\infty) = 1$, $g(-\infty)=-1$ (assuming continuity of $f$ at $0$.) Hence $f\big|_{(0,\infty)}$ corresponds to an odd function $g$ satisfying the above boundary condition at $\pm\infty$. Similarly, $f\big|_{(-\infty,0)}$ corresponds to some odd $h$ in the same manner.
This shows there are as many solutions as there are odd functions $g,h$ for which $\lim\limits_{x\to\infty} g(x)=\lim\limits_{x\to\infty}h(x)=1$. One example other than $f(x) = \frac{x-1}{x+1}$ is $$ f(x) = \frac{2}{\pi}\arctan(\log |x|),\quad x\neq 0 $$ and $f(0)=-1$. Also notice that the OP's solution corresponds to $g(t) = \frac{e^t-1}{e^t+1}=\tanh \frac{t}{2}$.
