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.

mathtoo Aug 29, 2018

#1**0 **

The obvious approach is to define \(h(x)=g(f(x))\) and then apply \(h(x)\) to the number 12 just 8 times.

First you need to work out \(h(x)\) algebraically by replacing the x in \(g(x)\) by the formula for \(f(x)\). Nobody will complain if you do a bit of Mathematics before you do the calculation! The result turns out to be simpler than you might expect, because the setter tried to make things easy for you! Now apply \(h(x)\) to 12 just 8 times. You don't need a calculator.

To say any more would be to give the game away (or to score points).

Guest Aug 29, 2018

#2**+4 **

**Define \(f(x)=\dfrac{1+x}{1-x}\) and \(g(x)=\dfrac{-2}{x+1}\). 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.**

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

If **\(h(x) = g(f(x))\) **8 times, then:

\(\begin{array}{|rcll|} \hline g(f(g(f(\ldots g(f(12)) \ldots )))) &=& x-8\quad & | \quad x = 12 \\ &=& 12-8 \\ &=& 4 \\ \hline \end{array} \)

heureka Aug 30, 2018