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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

f (x) = arctan (- \frac{3}{3})

\frac{9 + 2 x}{10 - x}

- \frac{1 - 2 z ^{- 1}}{z ^{- 1} - 2}

f (x) = \frac{3}{6 + 4 ^{x + 1}}

f (x) = 2 \frac{1}{3} x + 1

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