Question Number 152799 by nadovic last updated on 01/Sep/21
$$\:\:\mathrm{Given}\:\mathrm{that}\:{f}\circ{g}\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} \:−\:{x}\:+\:\mathrm{4}}\:\:\mathrm{and} \\ $$$$\:\:{g}\left({x}\right)\:=\:\frac{{x}}{{x}\:−\:\mathrm{2}},\:\mathrm{find}\:{f}\left({x}\right). \\ $$
Commented by otchereabdullai@gmail.com last updated on 08/Sep/21
$$\mathrm{nice}! \\ $$
Answered by Olaf_Thorendsen last updated on 01/Sep/21
$${g}\left({x}\right)\:=\:\frac{{x}}{{x}−\mathrm{2}}\:=\:{y} \\ $$$$\Rightarrow\:{x}\:=\:\:\frac{\mathrm{2}{y}}{{y}−\mathrm{1}} \\ $$$$ \\ $$$${f}\mathrm{o}{g}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} −{x}+\mathrm{4}} \\ $$$${f}\mathrm{o}{g}\left({x}\right)\:=\:\frac{\frac{\mathrm{4}{y}^{\mathrm{2}} }{\left({y}−\mathrm{1}\right)^{\mathrm{2}} }}{\frac{\mathrm{8}{y}^{\mathrm{2}} }{\left({y}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{2}{y}}{{y}−\mathrm{1}}+\mathrm{4}} \\ $$$${f}\mathrm{o}{g}\left({x}\right)\:=\:\frac{\mathrm{4}{y}^{\mathrm{2}} }{\mathrm{8}{y}^{\mathrm{2}} −\mathrm{2}{y}\left({y}−\mathrm{1}\right)+\mathrm{4}\left({y}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${f}\mathrm{o}{g}\left({x}\right)\:=\:\frac{\mathrm{4}{y}^{\mathrm{2}} }{\mathrm{8}{y}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} +\mathrm{2}{y}+\mathrm{4}{y}^{\mathrm{2}} −\mathrm{8}{y}+\mathrm{4}} \\ $$$${f}\mathrm{o}{g}\left({x}\right)\:=\:\frac{\mathrm{2}{y}^{\mathrm{2}} }{\mathrm{5}{y}^{\mathrm{2}} −\mathrm{3}{y}+\mathrm{2}} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}} \\ $$
Commented by SANOGO last updated on 01/Sep/21
$${vous}\:{etes}\:{geniale} \\ $$
Commented by puissant last updated on 01/Sep/21
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Commented by peter frank last updated on 03/Sep/21
$$\mathrm{good} \\ $$