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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

coth (x)

5 + 2 e^{- t}

f (x) = 3 \cdot x

f (2) = 3^{x + 2} + 1

ln (2 x) - 3

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