The question is whether the concatenation of two sigmoid function must again be a sigmoid function.
Definition. A bounded and two-times differentiable function $f:\Bbb R\to\Bbb R$ is called sigmoid function if $f'\not=0$ and there is exactly one $x\in\Bbb R$ with $f''(x)=0$.
Here is an example:
Given two sigmoid-functions $f$ and $g$, the function $h:=f\circ g$ is obviously bounded and two times differentiable. Also $h'\not=0$ is not hard to see. I even know that there is an inflection point (i.e. a value $x$ with $h''(x)=0$). But I failed to show that there is only one!
I admit that my initial motivation for this question comes from this post. I thought this would be one way to attack it in an elegant way. Still hard, though. Also no idea if there might be counterexamples.
Define $g(x)=10\tanh(x/10)+\tanh(10x)$ and $f(x)=g(x-5)$.
$g(x)$:
$g'(x):$
$g''(x):$
$f(g(x)):$
Note that despite how the graphs look, all derivatives are obviously bounded.