Let $f : \mathbb{R} \to (0, \infty)$ be a
D4364: Real function such that
| (i) |
\begin{equation}
f(x)
= \frac{1}{1 + e^{- x}}
\end{equation}
|
We have
\begin{equation}
\begin{split}
1 - f(x)
= 1 - \frac{1}{1 + e^{- x}}
& = \frac{1 + e^{- x}}{1 + e^{- x}} - \frac{1}{1 + e^{- x}} \\
& = \frac{1 + e^{- x} - 1}{1 + e^{- x}} \\
& = \frac{e^{- x}}{1 + e^{- x}} \\
& = \frac{e^x}{e^x} \frac{e^{- x}}{1 + e^{- x}} \\
& = \frac{1}{e^x + 1} \\
& = f(-x)
\end{split}
\end{equation}
$\square$
