What is the inverse of y=(e^x+e^{-x})/2 ?

The inverse of y=(e^x+e^{-x})/2 is ln(x+sqrt(x^2-1)),ln(x-sqrt(x^2-1))

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inverse of y=(e^x+e^{-x})/2

inverse

y = \frac{e ^{x} + e ^{- x}}{2}

ln (x + x^{2} - 1), ln (x - x^{2} - 1)

Solution steps

y = \frac{e ^{x} + e ^{- x}}{2}

Replace

x

with

y

x = \frac{e ^{y} + e ^{- y}}{2}

Solve for

y, x = \frac{e ^{y} + e ^{- y}}{2}

ln (x + x^{2} - 1), ln (x - x^{2} - 1)

4 cos (3 x - \frac{1}{2} π)

f (x) = \frac{x ^{2} - 8 x + 15}{x ^{2} - 1}

f (x) = \frac{4 x - 5}{3}

\frac{- 3 x}{2 x + 5}

f (x) = 3^{x + 1} - 4

What is the inverse of y=(e^x+e^{-x})/2 ?
The inverse of y=(e^x+e^{-x})/2 is ln(x+sqrt(x^2-1)),ln(x-sqrt(x^2-1))