Given-f-x-2x-1-2x-1-and-f-1-x-2-1-f-1-x-2-1-1-x-2-then-x-

Question Number 129213 by bramlexs22 last updated on 13/Jan/21

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

then x = ?

Answered by liberty last updated on 13/Jan/21

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

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

(∵) (∴) \Rightarrow \frac{f (\frac{1}{x^{2}}) - 1}{f (\frac{1}{x^{2}}) + 1} = - \frac{1}{2} x^{2} = \frac{1}{x^{2}}

x^{4} + 2 = 0 \Rightarrow (x^{2} - i \sqrt{2}) (x^{2} + i \sqrt{2}) = 0

x = \pm \sqrt{i \sqrt{2}}; x = \pm \sqrt{- i \sqrt{2}}

Answered by mr W last updated on 14/Jan/21

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

l e t t = \frac{1}{x^{2}}

\frac{f (t) - 1}{f (t) + 1} = \frac{1 - \frac{2}{2 t + 1} - 1}{1 - \frac{2}{2 t + 1} + 1} = \frac{- 1}{2 t} = t

\Rightarrow t^{2} = - \frac{1}{2}

\Rightarrow \frac{1}{x^{4}} = - \frac{1}{2}

\Rightarrow x^{4} = - 2 = 2 e^{π i}

\Rightarrow x = \sqrt[4]{2} e^{(\frac{k π}{2} + \frac{π}{4}) i} (k = 0, 1, 2, 3)

= \frac{1}{\sqrt[4]{2}} (1 + i), \frac{1}{\sqrt[4]{2}} (- 1 + i), \frac{1}{\sqrt[4]{2}} (- 1 - i), \frac{1}{\sqrt[4]{2}} (1 - i)

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