Question Number 207402 by MATHEMATICSAM last updated on 13/May/24
$${f}\left({x}\right)\:+\:\mathrm{2}{f}\left(\frac{\mathrm{1}}{{x}}\right)\:=\:\mathrm{3}{x}. \\ $$$$\:{f}\:'\left({x}\right)\:=\:? \\ $$
Answered by Frix last updated on 13/May/24
![x=t: f(t)+2f((1/t))=3t x=(1/t): f((1/t))+2f(t)=(3/t) ⇒ f(t)=(2/t)−t [∧ f((1/t))=2t−(1/t)] ⇒ f′(x)=−(2/x^2 )−1](https://www.tinkutara.com/question/Q207403.png)
$${x}={t}:\:{f}\left({t}\right)+\mathrm{2}{f}\left(\frac{\mathrm{1}}{{t}}\right)=\mathrm{3}{t} \\ $$$${x}=\frac{\mathrm{1}}{{t}}:\:{f}\left(\frac{\mathrm{1}}{{t}}\right)+\mathrm{2}{f}\left({t}\right)=\frac{\mathrm{3}}{{t}} \\ $$$$\Rightarrow\:{f}\left({t}\right)=\frac{\mathrm{2}}{{t}}−{t}\:\left[\wedge\:{f}\left(\frac{\mathrm{1}}{{t}}\right)=\mathrm{2}{t}−\frac{\mathrm{1}}{{t}}\right] \\ $$$$\Rightarrow\:{f}'\left({x}\right)=−\frac{\mathrm{2}}{{x}^{\mathrm{2}} }−\mathrm{1} \\ $$