# Find set of $a$ satisfying $\lvert w \sigma^{'}(wa+b) \rvert \geq 1$

Let's consider the product $\lvert w \sigma^{'}(z) \rvert=\lvert w \sigma^{'}(wa+b) \rvert$ where $\sigma(z)={1 \over 1+ e^{-z}}$ is the sigmoid function. We know $\sigma^{'}(wa+b) \leq {1 \over 4}$ and so we can only have $\lvert w \sigma^{'}(wa+b) \rvert \geq 1$ when $\lvert w \rvert \geq 4$. Given this, we want to show that the interval of $a$ satisfying $\lvert w \sigma^{'}(wa+b) \rvert \geq 1$ is no greater in width than $${2 \over \lvert w \rvert} \ln{\left( {{\lvert w \rvert\left(1+\sqrt{1-{4 \over \lvert w \rvert}}\right)} \over 2}-1\right)}$$ To show this, I use $\sigma^{'}(z)={e^{-z} \over (1+e^{-z})^2}>0$ to get $${e^{-z} \over (1+e^{-z})^2} \geq {1 \over \lvert w \rvert}$$ $$e^{-2z}+(2 -\lvert w \rvert)e^{-z}+1 \leq 0$$ $\Delta = \lvert w \rvert (\lvert w \rvert - 4) \geq 0$, so the solutions to the inequality are between $$e^{-z} = {{\lvert w \rvert\left(1\pm\sqrt{1-{4 \over \lvert w \rvert}}\right)} \over 2}-1$$ and we have $a$ between $$-{1 \over w}\ln{\left( {{\lvert w \rvert\left(1\pm\sqrt{1-{4 \over \lvert w \rvert}}\right)} \over 2}-1\right)}-{b \over w}$$ I don't quite see how I can relate this to the expected result and would appreciate any hints.

Let $$x_{1,2}={{\lvert w \rvert\left(1\pm\sqrt{1-{4 \over \lvert w \rvert}}\right)} \over 2}-1$$ be the roots of the equation $$x^2+(2 -\lvert w \rvert)x+1=0.$$ By Vieta’s formula $x_2=1/x_1$. So
$$\frac {-1}w(\ln x_1+b-\ln x_2-b)=\frac {-2}w\ln x_1.$$