Find the inverse of f, where $f(x)= \frac {e^x - e^{-x}} {e^x + e^{-x}} $

Question

I have tried letting $y= f(x)^{-1}$

So, f(y) = x

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

Then, multiplying the top and bottom by $e^y$, ${\frac {e^y(e^y - e^{-y})}{e^y(e^y + e^{-y})}= x}$

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

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

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

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

Therefore, $e^{2y} = \frac {x+1}{1-x}$

$2y=ln (\frac {x+1}{1-x})$

$y= \frac{1}{2} ( \frac {x+1}{1-x})$

$f^{-1} (x) =\frac {1}{2} ln (\frac {1+x}{1-x})$

The answer for the inverse from the book is $f^{-1} (x) =\frac {1}{2} ln (\frac {1+x}{x-1})$

So, I can’t seem to get the positive sign for x in the denominator.

There’s another very similar question in Maths Stack Exchange, but the positive and negative signs are different : What is the inverse of $f(x)=\frac{e^x+e^{-x}}{e^x-e^{-x}}$?

Please help me solve this. Thank you

Answer 1

The book is wrong. If you put y=f(x) then $$ 1-y= 2e^{-x}/(e^x+e^{-x}) $$ , which is positive, and $$ 1+y = 2e^x/(e^x+e^{-x}) $$ ,which is also positive so the book wouldd be trying totake the ln of a negative number. Your methodology is completely correct and your answer is right.