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Question Number 150119 by mathdanisur last updated on 09/Aug/21

Find the smallest value of a given  expression:  (x^2  + 6x + 8)^2  + 5

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{value}\:\mathrm{of}\:\mathrm{a}\:\mathrm{given} \\ $$$$\mathrm{expression}: \\ $$$$\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{6x}\:+\:\mathrm{8}\right)^{\mathrm{2}} \:+\:\mathrm{5} \\ $$

Answered by Ar Brandon last updated on 09/Aug/21

f(x)=(x^2 +6x+8)^2 +5  f ′(x)=2(2x+6)(x^2 +6x+8)=0  ⇒x=−3, x=−2, x=−4  f ′′(x)=2(2x+6)(2x+6)+4(4x^2 +6x+8)  f ′′(−4)>0 ⇒f(−4) is minimum value  f(−4)=(16−24+8)^2 +5=5

$${f}\left({x}\right)=\left({x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{8}\right)^{\mathrm{2}} +\mathrm{5} \\ $$$${f}\:'\left({x}\right)=\mathrm{2}\left(\mathrm{2}{x}+\mathrm{6}\right)\left({x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{8}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=−\mathrm{3},\:{x}=−\mathrm{2},\:{x}=−\mathrm{4} \\ $$$${f}\:''\left({x}\right)=\mathrm{2}\left(\mathrm{2}{x}+\mathrm{6}\right)\left(\mathrm{2}{x}+\mathrm{6}\right)+\mathrm{4}\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{8}\right) \\ $$$${f}\:''\left(−\mathrm{4}\right)>\mathrm{0}\:\Rightarrow{f}\left(−\mathrm{4}\right)\:\mathrm{is}\:\mathrm{minimum}\:\mathrm{value} \\ $$$${f}\left(−\mathrm{4}\right)=\left(\mathrm{16}−\mathrm{24}+\mathrm{8}\right)^{\mathrm{2}} +\mathrm{5}=\mathrm{5} \\ $$

Commented by mathdanisur last updated on 09/Aug/21

Thank You Ser

$$\mathrm{Thank}\:\mathrm{You}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$

Answered by ajfour last updated on 09/Aug/21

y={(x+3)^2 −1}^2 +5    y_(min) =5  when  x=−3±1 =−2, −4

$${y}=\left\{\left({x}+\mathrm{3}\right)^{\mathrm{2}} −\mathrm{1}\right\}^{\mathrm{2}} +\mathrm{5} \\ $$$$\:\:{y}_{{min}} =\mathrm{5} \\ $$$${when}\:\:{x}=−\mathrm{3}\pm\mathrm{1}\:=−\mathrm{2},\:−\mathrm{4} \\ $$

Commented by mathdanisur last updated on 09/Aug/21

Thankyou Ser

$$\mathrm{Thankyou}\:\mathrm{Ser} \\ $$

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