f-x-f-1-x-f-x-

Question Number 197585 by Erico last updated on 23/Sep/23

f (x) = f (\frac{1}{x})

\Rightarrow f (x) = ?

Answered by EdwarT last updated on 24/Sep/23

- \frac{1}{x^{2}}

Commented by mr W last updated on 24/Sep/23

h a v e y o u c h e c k e d y o u r s o l u t i o n ?

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

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

b u t f^{'} (x) = \frac{2}{x^{3}} \neq f (\frac{1}{x})

Answered by Erico last updated on 24/Sep/23

May be like this

f (x) = 2 \sqrt{x} \cos (\frac{\sqrt{3}}{2} lnx - \frac{π}{3})

Answered by Frix last updated on 24/Sep/23

For x > 0

f (x) = 2 \sqrt{x} \sin \frac{2 π + 3 \sqrt{3} \ln x}{6}

Answered by witcher3 last updated on 24/Sep/23

f^{'} (x) = f (\frac{1}{x})

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

\Leftrightarrow x^{2} f^{″} (x) + f (x) = 0

f (x) = x^{r}

\Rightarrow (r (r - 1) + 1) x^{r} = 0

\Rightarrow r^{2} - r + 1 = 0

r = \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}

f (x) = x^{\frac{1}{2}} ({ax}^{\frac{i}{2 \sqrt{3}}} + {bx}^{- \frac{i}{2 \sqrt{3}}})

= \sqrt{x} (asin (\frac{\ln (x)}{2} \sqrt{3}) + bcos (\frac{\ln (x)}{2} \sqrt{3})

f^{'} (x) = f (\frac{1}{x})

\Rightarrow

\frac{1}{\sqrt{x}} ((\frac{a}{2} - \frac{b}{2} \sqrt{3}) \sin (\frac{\ln (x)}{2}] \sqrt{3}) + (\frac{b}{2} + \frac{a}{2} \sqrt{3}) \cos (\frac{\ln (x)}{2} \sqrt{3}))

= \frac{1}{\sqrt{x}} (asin (\frac{\ln (\frac{1}{x})}{2} \sqrt{3}) + bcos (\frac{\ln (\frac{1}{x})}{2} \sqrt{3})), \forall x > 0

\Rightarrow \frac{a}{2} - \frac{b}{2} \sqrt{3} = - a

\frac{b}{2} + \frac{a}{2} \sqrt{3} = b

b = a \sqrt{3}

f (x) = a \sqrt{x} (\sin (\frac{\ln (x) \sqrt{3}}{2}) + \sqrt{3} \cos (\frac{\ln (x) \sqrt{3}}{2}))

= 2 a \sqrt{x} (\sin (\frac{\ln (x) \sqrt{3}}{2} + \frac{π}{3})), a \in R

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