Grade 12 Math (inverse of functions)

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SUMMARY

The discussion focuses on finding the inverse function F-1(x) for the equation f(x) = [((x-1)/(x+1)) + ((x-1)/(x+1))]^(1/2). Participants clarify that the innermost expression simplifies to 2 * ((x-1)/(x+1)). The process of isolating y after switching x's with y's is outlined, leading to the derived inverse function x = -((y^2 + 1)/(y^2 - 1)). This solution is confirmed as a common problem in Grade 12 mathematics.

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f(x)=[((x-1)/(x+1))+((x-1)/(x+1))]1/2

What is F-1(x)

No matter what I try I am unable to isolate for y after switching x's with y's. Any ideas?
 
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Ethan_Tab said:
f(x)=[((x-1)/(x+1))+((x-1)/(x+1))]1/2
The innermost expression is ##\frac{x - 1}{x + 1} + \frac{x - 1}{x + 1}##, which can be simplified. Is that what you meant to write?
Ethan_Tab said:
What is F-1(x)

No matter what I try I am unable to isolate for y after switching x's with y's. Any ideas?

BTW, this looks like a homework problem, so I'm moving this thread.
 
This is a pretty common problem. I'll work on a similar one as an example.
##y = \sqrt{ \frac{x-1}{x+1} } ##
##y^2 = \frac{x-1}{x+1},\, y\geq 0##
## y^2 = \frac{x+1}{x+1}-\frac{2}{x+1}##
## y^2-1 = -\frac{2}{x+1}##
##x+1 = -\frac{2}{y^2-1}##
##x= -\frac{2}{y^2-1}-1=-\frac{2}{y^2-1}-\frac{y^2-1}{y^2-1}=-\frac{y^2+1}{y^2-1}##
 

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