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

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Discussion Overview

The discussion revolves around finding the inverse of the function s(x) = [1 + f(x)] / [1 - f(x)], where f is a one-to-one function with an inverse f^-1. Participants explore the steps and reasoning involved in deriving s^-1(x) and express uncertainty about their approaches.

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant attempts to derive the inverse and arrives at s^-1(x) = [f^-1(1 - x)] / [f^-1(-x - 1)], expressing doubt about its correctness.
  • Several participants inquire about the form of f(x) and its implications for finding s^-1, suggesting that understanding f(x) is crucial.
  • Another participant mentions having difficulty starting the problem and feels stumped by the function.
  • One participant references finding an answer on Wolfram Alpha, indicating that they can invert normal functions but struggle with this specific case.
  • There are repeated prompts for clarification on the relationship between s^-1 and f^-1, indicating a search for a clearer understanding of the composition involved.

Areas of Agreement / Disagreement

Participants express uncertainty and confusion about the problem, with no consensus on the correct approach or solution. Multiple viewpoints and methods are presented without resolution.

Contextual Notes

Participants have not established the specific form of f(x), which may affect their ability to derive s^-1(x). There are also unresolved questions about the relationships between the functions involved.

caitlinbennet
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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.
 
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welcome to pf!

hi caitlinbennet! welcome to pf! :smile:

(try using the X2 button just above the Reply box :wink:)
caitlinbennet said:
s(x)=[1+f(x)]/[1-f(x)]

so f(x) = … ?

and what does that tell you about s-1 ? :wink:
 


tiny-tim said:
hi caitlinbennet! welcome to pf! :smile:

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


so f(x) = … ?

and what does that tell you about s-1 ? :wink:


Is s-1 a composition of f-1.. I'm definitely missing something silly here...
 
f(x) = … ?
 
tiny-tim said:
f(x) = … ?
(1+s(x))/1-s(x)?
 
really have no idea where to start..
 
I found the answer on wolfram alpha, I can invert normal functions but this one has me totally stumped.
 
hi caitlinbennet! :smile:

(just got up :zzz:)
caitlinbennet said:
(1+s(x))/1-s(x)?

ok, so suppose s(y) = x

then y = s-1(x), and y = f-1(f(y)) = … ? :wink:
 
f(x)?
 
  • #10
tiny-tim said:
f-1(f(y)) = … ? :wink:

i mean, in terms of s(y) ?
 

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