How to find "width" of this function of exponentials?

Question

I have a function $$ f(x) = \frac{p + q e^{s x}}{e^{s x} + 1} $$ where $p > q$ and $s > 0$. It's derivative is $$ f'(x) = - \frac{s \left(p - q\right) e^{s x}}{e^{2 s x} + 2 e^{s x} + 1} $$ The plot of the function and its derivative are as shown below By varying $s$, I can control how sharp the transition from $y=p$ to $y=q$ is. But I want to know over what length does the switch takes place. Thus, I am trying to find an expression for the "width" of the negative bump in the $f'(x)$ i.e. the region where $f'(x)$ is non-zero. I can see that it is probably asymptotically decaying to zero as we go away from $x=0$. I can numerically find out the width but does anyone see how I can write down an expression for the width?

Answer 1

Of course, $f'$ is nonzero everywhere (assumng $s(p-q)\neq 0$). But you probably really just want to find where $|f'| > \epsilon$ for some small $\epsilon$. You can even simplify by assuming $p>q$, in which case you solve $f' < \epsilon$, or $p<q$, which corresponds to $f'>\epsilon$. Note the sign of $f'$ is constant.

Note too that you can write $f'$ as $$f'(x)= \frac{cu}{(u+1)^2}$$ where $u=e^{sx}$ and $c = -s(p-q)$. Can you then solve the quadratic inequality $$\frac{cu}{(u+1)^2}>\epsilon$$ (and don't forget to rewrite the solution in terms of $x$)?