Question
Let $f:(1,\infty) \rightarrow \mathbf{R}$ be function defined as $f(x) = \frac{x - 1}{x + 1}$. Let $f^{i + 1}(x) = f\left( f^{i}(x) \right)$,
$i = 1,2,\ldots,25$, where $f^{1}(x) = (x)$. If $g(x) + f^{26}(x) = 0$, $x \in (1,\infty)$, then the area of the region bounded by the curves $y = g(x),2y = 2x - 3,y = 0$ and $x = 4$ is:
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Solution
Sol. [Key Concept: Iterating $f(x) = \frac{x - 1}{x + 1}$ gives period $4$, so $f^{26}(x) = - \frac{1}{x} = g(x)$; the area bounded by $y = g(x),\ 2y = 2x - 3,\ y = 0,\ x = 4$ is $\frac{1}{8} + ln2$.]
$${f^{2}(x) = f(f(x)) = \frac{\frac{x - 1}{x + 1} - 1}{\frac{x - 1}{x + 1} + 1} = \frac{- 1}{x} }{f^{3}(x) = f\left( f^{2}(x) \right) = f\left( \frac{- 1}{x} \right) = \frac{1 + x}{1 - x} }{f^{4}(x) = f\left( f^{3}(x) \right) = f\left( f\left( \frac{- 1}{x} \right) \right) = x }{f^{5}(x) = f\left( f^{4}(x) \right) = f(x) }{f^{6}(x) = f\left( f^{5}(x) \right) = f(f(x)) = \frac{- 1}{x} }$$•
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$${f^{26}(x) = \frac{- 1}{x} }{g(x) = \frac{1}{x} }$$
$${x - 3/2 = 1/x \Rightarrow 2x^{2} - 3x - 2 = 0 }{\Rightarrow x = - 1/2,2 }$$Req. area $= \int_{2}^{4}\mspace{2mu}\frac{1}{x}dx + \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$
$$= \frac{1}{8} + (\mathcal{l}nx)_{2}^{4} = \frac{1}{8} + \mathcal{l}n2 $$
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