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

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