First slide

Functions (XII)

Question

Let $f (x) = \frac{x - 3}{x + 1}, x \neq - 1$ . Then $f^{2010} (2014)$ (where $f^{n} (x) = f o f . . .$ of $(x)$ ( $n$ times)) is

Moderate

A

2010
B

4020
C

4028
D

2014

Solution

$\begin{aligned} f^{2} (x) = f o f (x) = f (f (x)) = f (\frac{x - 3}{x + 1}) \\ = \frac{\frac{x - 3}{x + 1} - 3}{\frac{x - 3}{x + 1} + 1} = - \frac{x + 3}{x - 1} \\ f^{3} (x) = f (- \frac{x + 3}{x - 1}) = \frac{\frac{- x - 3}{x - 1} - 3}{\frac{- x - 3}{x - 1} + 1} = x . \end{aligned}$
Hence $f^{3 k} (x) = x$
$f^{2010} (2014) = f^{3.670} (2014) = 2014$ .

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