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Question Number 55621 by Tawa1 last updated on 28/Feb/19
$$\mathrm{The}\:\mathrm{function}\:\:\:\mathrm{pogof}\left(\mathrm{x}\right)\:\:=\:\:\mathrm{x}^{\mathrm{4}} \:+\:\mathrm{2x}^{\mathrm{3}} \:+\:\mathrm{2x}^{\mathrm{2}} \:\:\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{the}\:\:\mathrm{half}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{function}\:\mathrm{of}\:\:\mathrm{p}.\:\:\mathrm{Find}\:\:\mathrm{g}\left(\mathrm{x}\right). \\ $$
Answered by kaivan.ahmadi last updated on 28/Feb/19
$$\mathrm{2}{p}\left({gof}\left({x}\right)\right)=\mathrm{2}{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right) \\ $$$${divide}\:{p}\left({x}\right)\Rightarrow{p}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} \\ $$$$\Rightarrow{g}\left({f}\left({x}\right)\right)={x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}=\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1} \\ $$$${then}\:\:{g}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{1}\:{and}\:{f}\left({x}\right)={x}+\mathrm{1}\:{is} \\ $$$${one}\:{of}\:{the}\:{answers}. \\ $$$${generally}\:\:{g}\left({x}\right)=\left({x}−{k}\right)^{\mathrm{2}} +\mathrm{1}\: \\ $$$${and}\:{f}\left({x}\right)={x}+{k}+\mathrm{1}\:{is}\:{answer} \\ $$$$ \\ $$$${this}\:{question}\:{has}\:{no}\:{unique}\:{answer} \\ $$$$ \\ $$$${t}={f}\left({x}\right)\Rightarrow{f}^{−\mathrm{1}} \left({t}\right)={x} \\ $$$${g}\left({t}\right)=\left({f}^{−\mathrm{1}} \left({t}\right)+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1} \\ $$