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Define \(f(x)=\frac{1+x}{1-x}\) and \(g(x)=\frac{-2}{x+1}\). Find the value of \(g(f(g(f(\ldots g(f(12)) \ldots ))))\) where there are 16 compositions of the functions g, and f, alternating between the two.

ant101  Aug 28, 2018
 #1
avatar+19994 
+1

Define
\(f(x)=\dfrac{1+x}{1-x}\)
and
\(g(x)=\dfrac{-2}{x+1}\).
Find the value of
\(\large{g(f(g(f(\ldots g(f(12)) \ldots ))))}\)
where there are 16 compositions of the functions g, and f,
alternating between the two.

 

\(\begin{array}{|rcll|} \hline f(x) &=& \dfrac{x+1}{1-x} \\\\ g(f(x)) = \dfrac{-2}{\dfrac{x+1}{1-x}+1} &&=& x-1 \\\\ f(g(f(x))) = \dfrac{x-1+1}{1-(x-1)} &=& \dfrac{x+0}{2-x} \\\\ g(f(g(f(x)))) = \dfrac{-2}{\dfrac{x}{2-x}+1}& &=&x-2 \\\\ f(g(f(g(f(x))))) = \dfrac{x-2+1}{1-(x-2)} &=& \dfrac{x-1}{3-x} \\\\ g(f(g(f(g(f(x))))))=\dfrac{-2}{\dfrac{x-1}{3-x}+1} & &=& x-3 \\\\ f(g(f(g(f(g(f(x))))))) = \dfrac{x-3+1}{1-(x-3)} &=& \dfrac{x-2}{4-x} \\\\ g(f(g(f(g(f(g(f(x))))))))=\dfrac{-2}{\dfrac{x-2}{4-x}+1} & &=& x-4 \\\\ \ldots \\ \hline \end{array}\)

 

\(\text{When there are $\mathbf{16}$ compositions of the functions $g$, and $f$, alternating between the two.} \)

\(\begin{array}{|rcll|} \hline g(f(g(f(\ldots g(f(12)) \ldots )))) &=& x-16 \quad & | \quad x = 12 \\ &=& 12-16 \\ &=& -4 \\ \hline \end{array}\)

 

laugh

heureka  Aug 28, 2018
 #2
avatar+553 
+1

Nice solution! But, the answer is different help!

ant101  Aug 28, 2018

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