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Question Number 108193 by bemath last updated on 15/Aug/20

((♥BeMath♥)/⧫) Given 2f(x)+3f((1/x))=2x+3 find f(3) ?

$$\:\:\frac{\heartsuit\mathcal{B}{e}\mathcal{M}{ath}\heartsuit}{\blacklozenge} \\ $$$$\:{Given}\:\mathrm{2}{f}\left({x}\right)+\mathrm{3}{f}\left(\frac{\mathrm{1}}{{x}}\right)=\mathrm{2}{x}+\mathrm{3} \\ $$$${find}\:{f}\left(\mathrm{3}\right)\:?\: \\ $$

Commented by bemath last updated on 15/Aug/20

thank you both

$${thank}\:{you}\:{both} \\ $$

Answered by john santu last updated on 15/Aug/20

((✓JS✓)/♠) put x=3 ⇒2f(3)+3f((1/3))=9 put x=(1/3)⇒2f((1/3))+3f(3)=(2/3)+3 { ((4f(3)+6f((1/3))=18)),((9f(3)+6f((1/3))=11)) :} (1)−(2)⇒−5f(3)=7 ⇒f(3)=−(7/5)

$$\:\:\:\frac{\checkmark\mathcal{JS}\checkmark}{\spadesuit} \\ $$$${put}\:{x}=\mathrm{3}\:\Rightarrow\mathrm{2}{f}\left(\mathrm{3}\right)+\mathrm{3}{f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{9} \\ $$$${put}\:{x}=\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow\mathrm{2}{f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{3}{f}\left(\mathrm{3}\right)=\frac{\mathrm{2}}{\mathrm{3}}+\mathrm{3} \\ $$$$\begin{cases}{\mathrm{4}{f}\left(\mathrm{3}\right)+\mathrm{6}{f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{18}}\\{\mathrm{9}{f}\left(\mathrm{3}\right)+\mathrm{6}{f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{11}}\end{cases} \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{2}\right)\Rightarrow−\mathrm{5}{f}\left(\mathrm{3}\right)=\mathrm{7} \\ $$$$\Rightarrow{f}\left(\mathrm{3}\right)=−\frac{\mathrm{7}}{\mathrm{5}} \\ $$

Answered by mr W last updated on 15/Aug/20

2f(x)+3f((1/x))=2x+3 ...(i) replace x with (1/x) 2f((1/x))+3f(x)=(2/x)+3 ...(ii) 3(ii)−2(i): 5f(x)=3+(6/x)−4x ⇒f(x)=(3/5)+(6/(5x))−((4x)/5) ⇒f(3)=−(7/5)

$$\mathrm{2}{f}\left({x}\right)+\mathrm{3}{f}\left(\frac{\mathrm{1}}{{x}}\right)=\mathrm{2}{x}+\mathrm{3}\:\:\:...\left({i}\right) \\ $$$${replace}\:{x}\:{with}\:\frac{\mathrm{1}}{{x}} \\ $$$$\mathrm{2}{f}\left(\frac{\mathrm{1}}{{x}}\right)+\mathrm{3}{f}\left({x}\right)=\frac{\mathrm{2}}{{x}}+\mathrm{3}\:\:\:...\left({ii}\right) \\ $$$$\mathrm{3}\left({ii}\right)−\mathrm{2}\left({i}\right): \\ $$$$\mathrm{5}{f}\left({x}\right)=\mathrm{3}+\frac{\mathrm{6}}{{x}}−\mathrm{4}{x} \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\mathrm{3}}{\mathrm{5}}+\frac{\mathrm{6}}{\mathrm{5}{x}}−\frac{\mathrm{4}{x}}{\mathrm{5}} \\ $$$$\Rightarrow{f}\left(\mathrm{3}\right)=−\frac{\mathrm{7}}{\mathrm{5}} \\ $$

Answered by Dwaipayan Shikari last updated on 15/Aug/20

2f(3)+3f((1/3))=9⇒4f(3)+6f((1/3))=18 3f(3)+2f((1/3))=((11)/3)⇒9f(3)+6f((1/3))=11 −5f(3)=7 f(3)=−(7/5)

$$\mathrm{2}{f}\left(\mathrm{3}\right)+\mathrm{3}{f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{9}\Rightarrow\mathrm{4}{f}\left(\mathrm{3}\right)+\mathrm{6}{f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{18} \\ $$$$\mathrm{3}{f}\left(\mathrm{3}\right)+\mathrm{2}{f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{11}}{\mathrm{3}}\Rightarrow\mathrm{9}{f}\left(\mathrm{3}\right)+\mathrm{6}{f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{11} \\ $$$$−\mathrm{5}{f}\left(\mathrm{3}\right)=\mathrm{7} \\ $$$${f}\left(\mathrm{3}\right)=−\frac{\mathrm{7}}{\mathrm{5}} \\ $$$$ \\ $$

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