Find the inverse of f(x) in 4 minutes!

[Lorena_Santoro]

 

[skeeter]

why a time limit?

$y=\dfrac{e^x-e^{-x}}{e^x+e^{-x}} = \dfrac{e^{2x}-1}{e^{2x}+1}$

$x = \dfrac{e^{2y}-1}{e^{2y}+1}$

$xe^{2y}+x = e^{2y}-1$

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

$e^{2y}=\dfrac{x+1}{1-x} \implies y = \dfrac{1}{2}\ln\left(\dfrac{x+1}{1-x}\right)$

 

[Lorena_Santoro]

It's a math-quiz like channel.

 

[I like Serena]

We might recognize the definition of $\sinh x=\frac 12(e^x-e^{-x})$ and its associates $\cosh x$ and $\tanh x$.
We have $\tanh x=\frac{\sinh x}{\cosh x}=\frac{e^x-e^{-x}}{e^x+e^{-x}}$, which is the same as the given $f(x)$.
Therefore $f^{-1}(x)=\tanh^{-1} x=\artanh x$, which happens to be the same as $\frac 12\ln\left(\frac{1+x}{1-x}\right)$ as skeeter showed. See Inverse hyperbolic tangent on wiki.

 

[HOI]

What should I do with the other three minutes?

 

[Lorena_Santoro]

What should I do with the other three minutes?

Enjoy being so smart! ;-)

 

[HOI]

And a nice cup of Earl Grey tea!

 

[Lorena_Santoro]

I'd totally agree!