Hi I am trying to prove that the inverse of $f(t) = \frac{1-e^t}{1+e^t}$ is $F^{-1}(t) = \ln\left(\frac{1-t}{1+t} \right )$
But I don't quite know where to start? Do I just sub $\frac{1-e^t}{1+e^t}$ for all t in $F^{-1}(t) = \ln \left(\frac{1-t}{1+t} \right)$?
From the original equation, we want to solve for $t$ in terms of $y=f(t)$.
$\displaystyle y = \frac{1-e^t}{1+e^t} $
$\displaystyle (1+e^t)y = 1-e^t$
$\displaystyle y + ye^t = 1 - e^t$ (*)
$(1+y)e^t = 1-y$ $\quad$ (grouping $e^t$ terms together, attempting to solve for $e^t$)
$\displaystyle e^t = \frac{1-y}{1+y}$
I can divide by $1+y$ because $y\neq-1$. You can plug $y=-1$ into the original equation and has no solution.
Now you can take the natural logarithm on both sides, switch y and t, and obtain the desired result.
I hope this helps.