How do you find the inverse of $f (x) = e^{x} - e^{- x}$ $f(x) = e^x - e^-x$ ?

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Answer 1 · 2016-01-30T01:03:16

Let $y = f (x)$ $y=f(x)$ and rearrange into a quadratic in $e^{x}$ $e^x$ to find:

$f^{- 1} (y) = ln (\frac{y + \sqrt{y^{2} + 4}}{2})$ $f^(-1)(y) = ln((y+sqrt(y^2+4))/2)$

Let $y = e^{x} - e^{- x}$ $y = e^x-e^-x$

Then $y (e^{x}) = {(e^{x})}^{2} - 1$ $y(e^x) = (e^x)^2-1$

So ${(e^{x})}^{2} - y (e^{x}) - 1 = 0$ $(e^x)^2-y(e^x)-1 = 0$

$e^{x} = \frac{y \pm \sqrt{y^{2} + 4}}{2}$ $e^x = (y+-sqrt(y^2+4))/2$

One of these roots is negative, requiring $x$ $x$ to be non-Real.

So only the positive root is useful for $e^{x}$ $e^x$ as a Real valued function of Real values of $x$ $x$ :

$x = ln (\frac{y + \sqrt{y^{2} + 4}}{2})$ $x = ln((y+sqrt(y^2+4))/2)$

That is:

$f^{- 1} (y) = ln (\frac{y + \sqrt{y^{2} + 4}}{2})$ $f^(-1)(y) = ln((y+sqrt(y^2+4))/2)$

How do you find the inverse of f(x) = e^x - e^-xf(x)=ex−e−xf(x) = e^x - e^-x?