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Question Number 155809 by zainaltanjung last updated on 05/Oct/21
  Given that f○g = (x^2 /(2x^2  − x + 4))  and    g(x) = (x/(x − 2)), find f(x) ?
$$\:\:\mathrm{Given}\:\mathrm{that}\:{f}\circ{g}\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} \:−\:{x}\:+\:\mathrm{4}}\:\:\mathrm{and} \\ $$$$\:\:{g}\left({x}\right)\:=\:\frac{{x}}{{x}\:−\:\mathrm{2}},\:\mathrm{find}\:{f}\left({x}\right)\:? \\ $$$$ \\ $$
Answered by PRITHWISH SEN 2 last updated on 05/Oct/21
put t= (x/(x−2))  x=(2/(t−1))+2=((2t)/(t−1))           f(t)= (((((2t)/(t−1)))^2 )/(2(((2t)/(t−1)))^2 −((2t)/(t−1))+4))       = ((4t^2 )/(10t^2 −6t+4))  ∴ f(x)=((4x^2 )/(10x^2 −6x+4))
$$\mathrm{put}\:\mathrm{t}=\:\frac{\mathrm{x}}{\mathrm{x}−\mathrm{2}} \\ $$$$\mathrm{x}=\frac{\mathrm{2}}{\mathrm{t}−\mathrm{1}}+\mathrm{2}=\frac{\mathrm{2t}}{\mathrm{t}−\mathrm{1}}\:\:\:\:\:\:\:\:\: \\ $$$$\mathrm{f}\left(\mathrm{t}\right)=\:\frac{\left(\frac{\mathrm{2t}}{\mathrm{t}−\mathrm{1}}\right)^{\mathrm{2}} }{\mathrm{2}\left(\frac{\mathrm{2t}}{\mathrm{t}−\mathrm{1}}\right)^{\mathrm{2}} −\frac{\mathrm{2t}}{\mathrm{t}−\mathrm{1}}+\mathrm{4}} \\ $$$$\:\:\:\:\:=\:\frac{\mathrm{4t}^{\mathrm{2}} }{\mathrm{10t}^{\mathrm{2}} −\mathrm{6t}+\mathrm{4}} \\ $$$$\therefore\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{4x}^{\mathrm{2}} }{\mathrm{10x}^{\mathrm{2}} −\mathrm{6x}+\mathrm{4}} \\ $$

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