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Question Number 195391 by mathlove last updated on 01/Aug/23 | ||
$${f}^{\mathrm{2}} \left({x}\right)+\mathrm{2}{f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{15} \\ $$$${f}\left({x}\right)=? \\ $$ | ||
Answered by MM42 last updated on 01/Aug/23 | ||
$$\left({f}+\mathrm{1}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{16}=\left({x}−\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{f}=\mid{x}−\mathrm{4}\mid−\mathrm{1}\:\checkmark \\ $$ | ||
Answered by kapoorshah last updated on 01/Aug/23 | ||
$${let}\:{f}\left({x}\right)\:=\:{ax}\:+\:{b} \\ $$$$\left({ax}\:+\:{b}\right)^{\mathrm{2}} \:+\:\mathrm{2}\left({ax}\:+\:{b}\right)\:=\:{x}^{\mathrm{2}} \:−\:\mathrm{8}{x}\:+\mathrm{15} \\ $$$${a}^{\mathrm{2}} {x}^{\mathrm{2}} \:+\:\mathrm{2}{abx}\:+\:{b}^{\mathrm{2}} \:+\:\mathrm{2}{ax}\:+\:\mathrm{2}{b}\:=\:{x}^{\mathrm{2}} \:−\:\mathrm{8}{x}\:+\mathrm{15} \\ $$$${a}^{\mathrm{2}} {x}^{\mathrm{2}} \:+\:\left(\mathrm{2}{ab}\:+\:\mathrm{2}{a}\right){x}\:+\:{b}^{\mathrm{2}} \:+\:\mathrm{2}{b}\:=\:{x}^{\mathrm{2}} \:−\:\mathrm{8}{x}\:+\mathrm{15} \\ $$$$ \\ $$$${a}^{\mathrm{2}} \:=\:\mathrm{1} \\ $$$${a}\:=\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}\:=\:−\mathrm{1} \\ $$$$\mathrm{2}{b}\:+\:\mathrm{2}\:=\:−\mathrm{8}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}{b}\:−\:\:\mathrm{2}\:=\:−\mathrm{8} \\ $$$${b}\:=\:−\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}\:=\:\mathrm{3} \\ $$$$\therefore\:{f}\left({x}\right)\:=\:{x}\:−\:\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:{f}\left({x}\right)\:=\:−{x}\:+\:\mathrm{3} \\ $$$$ \\ $$ | ||