Question Number 108193 by bemath last updated on 15/Aug/20
$$\:\:\frac{\heartsuit\mathcal{B}{e}\mathcal{M}{ath}\heartsuit}{\blacklozenge} \\ $$$$\:{Given}\:\mathrm{2}{f}\left({x}\right)+\mathrm{3}{f}\left(\frac{\mathrm{1}}{{x}}\right)=\mathrm{2}{x}+\mathrm{3} \\ $$$${find}\:{f}\left(\mathrm{3}\right)\:?\: \\ $$
Commented by bemath last updated on 15/Aug/20
$${thank}\:{you}\:{both} \\ $$
Answered by john santu last updated on 15/Aug/20
$$\:\:\:\frac{\checkmark\mathcal{JS}\checkmark}{\spadesuit} \\ $$$${put}\:{x}=\mathrm{3}\:\Rightarrow\mathrm{2}{f}\left(\mathrm{3}\right)+\mathrm{3}{f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{9} \\ $$$${put}\:{x}=\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow\mathrm{2}{f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{3}{f}\left(\mathrm{3}\right)=\frac{\mathrm{2}}{\mathrm{3}}+\mathrm{3} \\ $$$$\begin{cases}{\mathrm{4}{f}\left(\mathrm{3}\right)+\mathrm{6}{f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{18}}\\{\mathrm{9}{f}\left(\mathrm{3}\right)+\mathrm{6}{f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{11}}\end{cases} \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{2}\right)\Rightarrow−\mathrm{5}{f}\left(\mathrm{3}\right)=\mathrm{7} \\ $$$$\Rightarrow{f}\left(\mathrm{3}\right)=−\frac{\mathrm{7}}{\mathrm{5}} \\ $$
Answered by mr W last updated on 15/Aug/20
$$\mathrm{2}{f}\left({x}\right)+\mathrm{3}{f}\left(\frac{\mathrm{1}}{{x}}\right)=\mathrm{2}{x}+\mathrm{3}\:\:\:…\left({i}\right) \\ $$$${replace}\:{x}\:{with}\:\frac{\mathrm{1}}{{x}} \\ $$$$\mathrm{2}{f}\left(\frac{\mathrm{1}}{{x}}\right)+\mathrm{3}{f}\left({x}\right)=\frac{\mathrm{2}}{{x}}+\mathrm{3}\:\:\:…\left({ii}\right) \\ $$$$\mathrm{3}\left({ii}\right)−\mathrm{2}\left({i}\right): \\ $$$$\mathrm{5}{f}\left({x}\right)=\mathrm{3}+\frac{\mathrm{6}}{{x}}−\mathrm{4}{x} \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\mathrm{3}}{\mathrm{5}}+\frac{\mathrm{6}}{\mathrm{5}{x}}−\frac{\mathrm{4}{x}}{\mathrm{5}} \\ $$$$\Rightarrow{f}\left(\mathrm{3}\right)=−\frac{\mathrm{7}}{\mathrm{5}} \\ $$
Answered by Dwaipayan Shikari last updated on 15/Aug/20
$$\mathrm{2}{f}\left(\mathrm{3}\right)+\mathrm{3}{f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{9}\Rightarrow\mathrm{4}{f}\left(\mathrm{3}\right)+\mathrm{6}{f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{18} \\ $$$$\mathrm{3}{f}\left(\mathrm{3}\right)+\mathrm{2}{f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{11}}{\mathrm{3}}\Rightarrow\mathrm{9}{f}\left(\mathrm{3}\right)+\mathrm{6}{f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{11} \\ $$$$−\mathrm{5}{f}\left(\mathrm{3}\right)=\mathrm{7} \\ $$$${f}\left(\mathrm{3}\right)=−\frac{\mathrm{7}}{\mathrm{5}} \\ $$$$ \\ $$