Consider the following function $\ f:\mathbb{R} \to \mathbb{R}$ given by:
$$f(x) = - \tanh\left(3 x - \tanh^{-1}( u_{0} )\right)$$
In the above, $u_{0}$ is a constant where $u_{0} \in (-\infty, -1) \cup (1, \infty)$.
When I plot this function, it is clear that there is one singularity. I'm trying to find out for which value of $x$ does this singularity occur.
I've been scratching my head about this for too long, so I thought I'd ask here.
The singularities of $\tanh{z}=\displaystyle\frac{e^{2z}-1}{e^{2z}+1}$ occur when $e^{2z}+1=0$, then you need to solve
$$\displaystyle e^{2(3z-\tanh^{-1}{u_0})}+1=0,$$
when $z\in \Bbb{C}.$