f-x-2-f-1-x-1-x-64x-x-D-f-x-

Question Number 140470 by EDWIN88 last updated on 08/May/21

{(f (x))}^{2} . f (\frac{1 - x}{1 + x}) = 64 x, \forall x \in D

\Rightarrow f (x) = ?

Answered by benjo_mathlover last updated on 08/May/21

(1) {(f (x))}^{2} . f (\frac{1 - x}{1 + x}) = 64 x

(2) replace x \to \frac{1 - x}{1 + x}

\Rightarrow {(f (\frac{1 - x}{1 + x}))}^{2} . f (x) = \frac{64 (1 - x)}{1 + x}

eq (1) \Rightarrow squaring

∵ {(f (x))}^{4} . {(f (\frac{1 - x}{1 + x}))}^{2} = {(64 x)}^{2}

\Rightarrow \frac{{(f (x))}^{4} . {(f (\frac{1 - x}{1 + x}))}^{2}}{f (x) . {(f (\frac{1 - x}{1 + x}))}^{2}} = \frac{64^{2} x^{2} (1 + x)}{64 (1 - x)}

\Rightarrow {(f (x))}^{3} = \frac{{64 x}^{2} (1 + x)}{1 - x}

∴ \Rightarrow f (x) = 4 \sqrt[3]{\frac{x^{3} + x^{2}}{1 - x}}

Answered by mathmax by abdo last updated on 09/May/21

\frac{1 - x}{1 + x} = t \Rightarrow 1 - x = t + tx \Rightarrow (1 + t) x = 1 - t \Rightarrow x = \frac{1 - t}{1 + t} \Rightarrow

f^{2} (\frac{1 - t}{1 + t}) . f (t) = 64 . \frac{1 - t}{1 + t} and f^{2} (t) . f (\frac{1 - t}{1 + t}) = 64 t \Rightarrow

f (\frac{1 - t}{1 + t}) = \frac{64 t}{f^{2} (t)} \Rightarrow {(\frac{64 t}{f^{2} (t)})}^{2} . f (t) = 64 \frac{1 - t}{1 + t} \Rightarrow

\frac{{64 t}^{2}}{f^{3} (t)} = \frac{1 - t}{1 + t} \Rightarrow \frac{f^{3} (t)}{{64 t}^{2}} = \frac{1 + t}{1 - t} \Rightarrow f^{3} (t) = 64 \frac{t^{2} + t^{3}}{1 - t} \Rightarrow

f (t) =^{3} \sqrt{64 . \frac{t^{2} + t^{3}}{1 - t}}