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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

y = \frac{1}{x ^{2}}

f (x) = - x^{2} + 6 x - 9

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

f (x) = x^{3} + 6 x^{2}

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