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{x ^{5}}{10})^{\frac{1}{7}}

f (n) = n + 1

f (x) = 12 x - 5

3 x^{2} + 4

tan (2)

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))