Question Number 150119 by mathdanisur last updated on 09/Aug/21
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{value}\:\mathrm{of}\:\mathrm{a}\:\mathrm{given} \\ $$$$\mathrm{expression}: \\ $$$$\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{6x}\:+\:\mathrm{8}\right)^{\mathrm{2}} \:+\:\mathrm{5} \\ $$
Answered by Ar Brandon last updated on 09/Aug/21
$${f}\left({x}\right)=\left({x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{8}\right)^{\mathrm{2}} +\mathrm{5} \\ $$$${f}\:'\left({x}\right)=\mathrm{2}\left(\mathrm{2}{x}+\mathrm{6}\right)\left({x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{8}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=−\mathrm{3},\:{x}=−\mathrm{2},\:{x}=−\mathrm{4} \\ $$$${f}\:''\left({x}\right)=\mathrm{2}\left(\mathrm{2}{x}+\mathrm{6}\right)\left(\mathrm{2}{x}+\mathrm{6}\right)+\mathrm{4}\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{8}\right) \\ $$$${f}\:''\left(−\mathrm{4}\right)>\mathrm{0}\:\Rightarrow{f}\left(−\mathrm{4}\right)\:\mathrm{is}\:\mathrm{minimum}\:\mathrm{value} \\ $$$${f}\left(−\mathrm{4}\right)=\left(\mathrm{16}−\mathrm{24}+\mathrm{8}\right)^{\mathrm{2}} +\mathrm{5}=\mathrm{5} \\ $$
Commented by mathdanisur last updated on 09/Aug/21
$$\mathrm{Thank}\:\mathrm{You}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$
Answered by ajfour last updated on 09/Aug/21
$${y}=\left\{\left({x}+\mathrm{3}\right)^{\mathrm{2}} −\mathrm{1}\right\}^{\mathrm{2}} +\mathrm{5} \\ $$$$\:\:{y}_{{min}} =\mathrm{5} \\ $$$${when}\:\:{x}=−\mathrm{3}\pm\mathrm{1}\:=−\mathrm{2},\:−\mathrm{4} \\ $$
Commented by mathdanisur last updated on 09/Aug/21
$$\mathrm{Thankyou}\:\mathrm{Ser} \\ $$