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2. Diketahui f(11)=11 dan
$f(x+3)=\frac{f(x)-1}{f(x)+1}$
tentukan nilai dari f(2012) ?

Penyelesaian:

$f(14)=\frac{f(11)-1}{f(11)+1}=\frac{11-1}{11+1}=\frac{10}{12}=\frac{5}{6}$

   $f(17)=\frac{f(14)-1}{f(14)+1}=\frac{\frac{5}{6}-1}{\frac{5}{6}+1}=\frac{-1/6}{11/6}=-\frac{1}{11}$
   $f(20)=\frac{f(17)-1}{f(17)+1}=\frac{-\frac{1}{11}-1}{-\frac{1}{11}+1}=\frac{-12/11}{10/11}=-\frac{12}{10}=-\frac{6}{5}$
   $f(23)=\frac{f(20)-1}{f(20)+1}=\frac{-\frac{6}{5}-1}{-\frac{6}{5}+1}=\frac{-11/5}{-1/5}=11$
Artinya
$f(2011)=f(4.501)= f(4)=f(20)=-\frac{6}{5}$

Daftar Hasil Penyelesaian

Minggu, 22 April 2012

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