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Question Number 136740 by bramlexs22 last updated on 25/Mar/21

Given f ({2 f}^{- 1} (x)) = \frac{x}{2 - x} .

what is f (x) ?

Answered by Olaf last updated on 25/Mar/21

f (2 f^{- 1} (x)) = \frac{x}{2 - x} (1)

Let y = 2 f^{- 1} (x) \Rightarrow x = f (\frac{y}{2})

(1) : f (y) = \frac{f (\frac{y}{2})}{2 - f (\frac{y}{2})}

\frac{1}{f (y)} = \frac{2 - f (\frac{y}{2})}{f (\frac{y}{2})} = \frac{2}{f (\frac{y}{2})} - 1 (2)

Let g (y) = \frac{1}{f (y)}

(2) : g (y) = 2 g (\frac{y}{2}) - 1 (3)

\Rightarrow g^{'} (y) = 2 \times \frac{1}{2} g^{'} (\frac{y}{2}) = g^{'} (\frac{y}{2})

g^{'} (y) = g^{'} (\frac{y}{2}) = \dots = g^{'} (\frac{y}{2^{n}})

g^{'} (y) = \lim_{n \to \infty} g^{'} (\frac{y}{2^{n}}) = g^{'} (0) = cst a

\Rightarrow g (y) = a y + b

(3) : a y + b = 2 (a \frac{y}{2} + b) - 1

b = 1, g (y) = a y + 1

f (y) = \frac{1}{g (y)} = \frac{1}{a y + 1}

Let x = f (y) \Rightarrow y = f^{- 1} (x)

x = \frac{1}{a y + 1}

y = \frac{1}{a} (\frac{1}{x} - 1) = f^{- 1} (x)

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