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# Define..!!!!

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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.

Aug 28, 2018

#1
+21848
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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.

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

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

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

Aug 28, 2018
#2
+853
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Nice solution! But, the answer is different help!

ant101  Aug 28, 2018