Inverse of s(x)=[1+f(x)]/[1-f(x)]

[caitlinbennet]

f is a one to one function with inverse f^-1, and we are asked to find the inverse of s(x)=[1+f(x)]/[1-f(x)]

My attempts leave me with s^-1(x)= [f^-1(1-x)]/[f^-1(-x-1)] and I don't think this is correct. I can't find any examples of problems like this online.

 

[tiny-tim]

welcome to pf!

hi caitlinbennet! welcome to pf!

(try using the X² button just above the Reply box )

s(x)=[1+f(x)]/[1-f(x)]

so f(x) = … ?

and what does that tell you about s^-1 ?

 

[caitlinbennet]

hi caitlinbennet! welcome to pf!

(try using the X² button just above the Reply box )


so f(x) = … ?

and what does that tell you about s^-1 ?

Is s^-1 a composition of f^-1.. I'm definitely missing something silly here...

 

[tiny-tim]

f(x) = … ?

 

[caitlinbennet]

f(x) = … ?

(1+s(x))/1-s(x)?

 

[caitlinbennet]

really have no idea where to start..

 

[caitlinbennet]

I found the answer on wolfram alpha, I can invert normal functions but this one has me totally stumped.

 

[tiny-tim]

hi caitlinbennet!

(just got up :zzz:)

(1+s(x))/1-s(x)?

ok, so suppose s(y) = x

then y = s^-1(x), and y = f^-1(f(y)) = … ?

 

[caitlinbennet]

f(x)?

 

[tiny-tim]

f^-1(f(y)) = … ?

i mean, in terms of s(y) ?