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Define \(f(x)=\frac{1+x}{1-x}\) and \(g(x)=\frac{-2}{x+1}\). Find the value of \(g(f(g(f(\dotsb g(f(12)) \dotsb ))))\) where the function f is applied 8 times, and the function g is applied 8 times, alternating between the two.

 Dec 5, 2019
 #1
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1st  iteration

 

f(12)  =  13/-11 =   -13/11

 

g (f(12))  =  -2  / [ -13/11 + 1 ]   =   -  2 / [ -2/11]   =  11

 

2nd iteration

 

f ( g ( f (12) ))   =   f (11)   =   12 / - 10  =   - 6/ 5

 

g ( f ( g ( f (12) )) )  =  g (-6/5)   = -2 / [ -6/5 + 1]   =  -2 / [ -1/5]  =  10

 

Continuing  like this

 

3rd iteration

 

f(10)  = -11/9

g (-11/9)  =  -2 / [ -11/9 + 1 ] =  -2 /[ -2/9]  =   9

 

So...it appears that each iteration  lessens the result   by  1

 

So

 

11,10, 9, 8 ,7, 6, 5, 4

 

So....after 8 iterations....we get    4

 

 

 

cool cool cool

 Dec 5, 2019
edited by CPhill  Dec 5, 2019

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