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})

f (x) = - \frac{4}{5} x + 12

f (x) = \frac{5 - 3 ^{x}}{2}

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

f (x) = \frac{2 x + 9}{7 x}

f (x) = (x + 3)^{2} - 4

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))