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

[Lorena_Santoro]

https://youtu.be/eMbBK0IcSNI

[Guest]

You have [tex]f(x)= y= \frac{e^x- e^{-x}}{e^x+ e^{-x}}[/tex]. Then [tex]f^{-1}(y)= x= \frac{e^y- e^{-x}}{e^y+ e^{-y}}[/tex]. To write it as [tex]y= f^{-x}[/tex], solve for y. Let [tex]z= e^y[/tex] so [tex]x= \frac{z- \frac{1}{z}}{z+ \frac{1}{z}}[/tex].

Multiply numerator and denominator on the right by z: [tex]x= \frac{z^2- 1}{z^2+ 1}[/tex].

Multiply both sides by [tex]z^2+ 1[/tex]: [tex]x(z^2+ 1)= xz^2+ x= z^2- 1[/tex]. Subtract [tex]z^2+ x[/tex] from both sides:
[tex](x- 1)z^2= -x- 1[/tex] so [tex](1- x)z^2= x+ 1[/tex].

Divide both sides by [tex]1- x[/tex]:
[tex]z^2= \frac{1}{1- x}[/tex].
(This requires x not equal to 1.)

Take the square root of both sides:
[tex]z= \frac{\pm 1}{\sqrt{1- x}}[/tex]
(This is a real number only for [tex]x< 1[/tex].)

Replace z with [tex]e^y[/tex]:
[tex]e^y= \frac{\pm 1}{\sqrt{1- x}}[/tex]
Finally, take the logarithm of both sides:
(This requires that we take the positive root)
[tex]f^{-1}(x)= y= ln\left(\frac{1}{\sqrt{1- x}}\right)[/tex].

(And f has an inverse only for x< 1.)

Now, what should I do with the other three minutes?