Solve $x \tanh(x) = constant$

Question

Does the following equation admit a real solution:

$x\cdot \tanh(x) = C$

with $C$ a constant.

While I was not able to find a specific answer with symbolic calculations, this solutions seems to exist graphically ...

Answer 1

Note that $f(x)=x\tanh x = (-x)\tanh (-x)=f(-x)$ so the function is even. Also $0\leq f(x)$, and if $0<x$: $$ (x\tanh x)' = \tanh x + x \operatorname{sech}^2x >0 . $$ It's not hard to see that since $\lim_{x\to\infty}\tanh x = 1$ the function is unbounded. From all this we can conclude that the equation has two solutions for every $C\in\mathbb{R}^+\cup \{0\}$.

For $1<<C$ (actually for $C$ around $2$ this approximation starts working), the solution will get very close to $x=C$ (I'd like to see if I can add some math to this assertion later).

$$ %\frac{1-e^{-2x}}{1+e^{-2x}} = \frac{1}{1+e^{-2x}} - \frac{1}{1+e^{2x}} = \sum_{k=0}^\infty %e^{-2xk} - \frac{1}{1+e^{2x}} $$