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Question Number 91578 by jagoll last updated on 01/May/20

  f((1/x))+2f(x)= ((4x^3 +6x)/(3x^2 ))  f(x)=?

$$ \\ $$$${f}\left(\frac{\mathrm{1}}{{x}}\right)+\mathrm{2}{f}\left({x}\right)=\:\frac{\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}}{\mathrm{3}{x}^{\mathrm{2}} } \\ $$$${f}\left({x}\right)=? \\ $$

Commented by john santu last updated on 01/May/20

replace x by (1/x)  ⇒f(x)+2f((1/x))= ((((4/x^3 ))+((6/x)))/(((3/x^2 ))))  f(x)+2f((1/x)) = ((4+6x^2 )/(3x))...(ii)  (i)×2⇒ 4f(x)+2f((1/x))=((8x^3 +12x)/(3x^2 ))  (i)−(ii)  3f(x) = ((8x^3 +12x)/(3x^2 ))−((6x^3 +4x)/(3x^2 ))  3f(x) = ((2x^3 +8x)/(3x^2 ))=((2x^2 +8)/(3x))  ∴f(x)=((2x^2 +8)/(9x))

$${replace}\:{x}\:{by}\:\frac{\mathrm{1}}{{x}} \\ $$$$\Rightarrow{f}\left({x}\right)+\mathrm{2}{f}\left(\frac{\mathrm{1}}{{x}}\right)=\:\frac{\left(\frac{\mathrm{4}}{{x}^{\mathrm{3}} }\right)+\left(\frac{\mathrm{6}}{{x}}\right)}{\left(\frac{\mathrm{3}}{{x}^{\mathrm{2}} }\right)} \\ $$$${f}\left({x}\right)+\mathrm{2}{f}\left(\frac{\mathrm{1}}{{x}}\right)\:=\:\frac{\mathrm{4}+\mathrm{6}{x}^{\mathrm{2}} }{\mathrm{3}{x}}...\left({ii}\right) \\ $$$$\left({i}\right)×\mathrm{2}\Rightarrow\:\mathrm{4}{f}\left({x}\right)+\mathrm{2}{f}\left(\frac{\mathrm{1}}{{x}}\right)=\frac{\mathrm{8}{x}^{\mathrm{3}} +\mathrm{12}{x}}{\mathrm{3}{x}^{\mathrm{2}} } \\ $$$$\left({i}\right)−\left({ii}\right) \\ $$$$\mathrm{3}{f}\left({x}\right)\:=\:\frac{\mathrm{8}{x}^{\mathrm{3}} +\mathrm{12}{x}}{\mathrm{3}{x}^{\mathrm{2}} }−\frac{\mathrm{6}{x}^{\mathrm{3}} +\mathrm{4}{x}}{\mathrm{3}{x}^{\mathrm{2}} } \\ $$$$\mathrm{3}{f}\left({x}\right)\:=\:\frac{\mathrm{2}{x}^{\mathrm{3}} +\mathrm{8}{x}}{\mathrm{3}{x}^{\mathrm{2}} }=\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{8}}{\mathrm{3}{x}} \\ $$$$\therefore{f}\left({x}\right)=\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{8}}{\mathrm{9}{x}} \\ $$

Commented by jagoll last updated on 01/May/20

thank you

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

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