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Consider the bounded, positive and monotonic functions $f(x) = \frac{x^k}{a^k+x^k}$ and $g(x) = \frac{b^h}{b^h+x^h}$ with $a,b>0,$ and $k,h > 1$ and defined for $x\in \mathbb{R}^+$. Using simulations, it looks to me that the second derivative $(f\circ g\circ f \circ g)”$ has at most one zero.

I tried to prove it analytically using the chain rule and the fact that $f'(x) = \frac{k}{x}f(x)\big(1-f(x)\big)$ and similarly for $g'(x)$ but with no success. I can see that the number of zeros (taking their multiplicity into account) must be odd because the second derivative is positive at $x=0$ and negative at $x\rightarrow \infty$. It’s probably irrelevant, but I can also show that the second derivative can be expressed as the sum of four functions $h_{1}, h_{2},h_{3}, h_{4}$ with $h_{1},h_{2}$ having two zeros and $h_{3},h_{4}$ having one zeros and $h_{1},h_{2},h_{3}$ having a common zero.

Any help towards a proof or a counter-example would be much appreciated.

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