Question Number 146062 by mnjuly1970 last updated on 10/Jul/21
$$ \\ $$$$\:\:\:\:{find}\:\:{values}\:\:{a}\:,\:{b}\:,\:{c}\:\:{such}\:{that}: \\ $$$$\:\:\:\:−\mathrm{1}\leqslant\:{ax}\:^{\mathrm{2}} +{bx}\:+{c}\:\leqslant\:\mathrm{1} \\ $$$$\:\:\:\:\:\:{and}\:\:\frac{\mathrm{6}{b}^{\:\mathrm{2}} +\:\mathrm{8}\:{a}^{\:\mathrm{2}} }{\mathrm{3}}\:{is}\:{Max}... \\ $$
Commented by mr W last updated on 10/Jul/21
$${it}'{s}\:{not}\:{possible}\:{with}\:{a}\neq\mathrm{0}\:{and}\:{b}\neq\mathrm{0} \\ $$$${that}\:\:−\mathrm{1}\leqslant\:{ax}\:^{\mathrm{2}} +{bx}\:+{c}\:\leqslant\:\mathrm{1}. \\ $$$${that}\:{means},\:{such}\:{that} \\ $$$$−\mathrm{1}\leqslant\:{ax}\:^{\mathrm{2}} +{bx}\:+{c}\:\leqslant\:\mathrm{1} \\ $$$${we}\:{must}\:{have}\:{a}={b}=\mathrm{0}. \\ $$$$−\mathrm{1}\leqslant{c}\leqslant\mathrm{1}. \\ $$$${since}\:{max}=\frac{\mathrm{6}{b}^{\mathrm{2}} +\mathrm{8}{a}^{\mathrm{2}} }{\mathrm{3}}\:=\mathrm{0}\:\Rightarrow{c}=\mathrm{0} \\ $$