Question Number 205825 by tri26112004 last updated on 31/Mar/24
![[f′(x)]^2 +4f(x)=8x^2 −32x+28 ⇒f(x)=¿](https://www.tinkutara.com/question/Q205825.png)
$$\left[{f}'\left({x}\right)\right]^{\mathrm{2}} +\mathrm{4}{f}\left({x}\right)=\mathrm{8}{x}^{\mathrm{2}} −\mathrm{32}{x}+\mathrm{28} \\ $$$$\Rightarrow{f}\left({x}\right)=¿ \\ $$
Commented by mr W last updated on 31/Mar/24
$${f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3} \\ $$$${or} \\ $$$${f}\left({x}\right)=−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{8}{x}−\mathrm{9} \\ $$$${or}\:… \\ $$
Answered by MM42 last updated on 31/Mar/24
$${f}={ax}^{\mathrm{2}} +{bx}+{c}\Rightarrow{f}'=\mathrm{2}{ax}+{b} \\ $$$$\Rightarrow\mathrm{4}{a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{4}{abx}+{b}^{\mathrm{2}} +\mathrm{4}{ax}^{\mathrm{2}} +\mathrm{4}{bx}+\mathrm{4}{c}=\mathrm{8}{x}^{\mathrm{2}} −\mathrm{32}{x}+\mathrm{28} \\ $$$$\Rightarrow\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{a}=\mathrm{8}\Rightarrow{a}=\mathrm{1}\:{or}\:{a}=−\mathrm{2} \\ $$$$\mathrm{4}{ab}+\mathrm{4}{b}=−\mathrm{32}\:\:\&\:\:{b}^{\mathrm{2}} +\mathrm{4}{c}=\mathrm{28} \\ $$$${if}\:{a}=\mathrm{1}\Rightarrow{b}=−\mathrm{4}\:\&\:{c}=\mathrm{3} \\ $$$$\Rightarrow{f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3}\:\:\checkmark \\ $$$${if}\:{a}=−\mathrm{2}\Rightarrow{b}=\mathrm{8}\:\&\:{c}=−\mathrm{9} \\ $$$$\Rightarrow{f}\left({x}\right)=−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{8}{x}−\mathrm{9}\:\:\checkmark \\ $$$$ \\ $$