What is the inverse of f(x)=(e^{-x})/(1+e^{-x)} ?

The inverse of f(x)=(e^{-x})/(1+e^{-x)} is -ln(-x/(x-1))

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inverse of f(x)=(e^{-x})/(1+e^{-x)}

inverse

f (x) = \frac{e ^{- x}}{1 + e ^{- x}}

- ln (- \frac{x}{x - 1})

Solution steps

y = \frac{e ^{- x}}{1 + e ^{- x}}

Replace

x

with

y

x = \frac{e ^{- y}}{1 + e ^{- y}}

Solve for

y, x = \frac{e ^{- y}}{1 + e ^{- y}}

- ln (- \frac{x}{x - 1})

f (x) = tan (\frac{3.14 x}{2}) + 3

f (x) = f (x) = 2 x^{5} + 7

\frac{x - 3}{x + 5}

15 n^{2} \cdot 10 n^{3}

f (x) = 9 \cdot 6^{2 x}

What is the inverse of f(x)=(e^{-x})/(1+e^{-x)} ?
The inverse of f(x)=(e^{-x})/(1+e^{-x)} is -ln(-x/(x-1))