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Question Number 114181 by bemath last updated on 17/Sep/20
Given a function f(x) = x^2 +(1/x^2 )+4x+(4/x) ; where x>0. find the minimum value of f(x)
$${Given}\:{a}\:{function}\: \\ $$$${f}\left({x}\right)\:=\:{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{4}{x}+\frac{\mathrm{4}}{{x}}\:;\:{where}\:{x}>\mathrm{0}. \\ $$$${find}\:{the}\:{minimum}\:{value}\:{of}\:{f}\left({x}\right) \\ $$
Answered by bobhans last updated on 17/Sep/20
recall p+(1/p) ≥ 2 ; p > 0 now f(x)= (x^2 +(1/x^2 ))+4(x+(1/x)) ⇒f(x)= (x+(1/x))^2 −2+4(x+(1/x)) f(x)∣_(minimum) = 2^2 −2+4.2 = 10.
$${recall}\:{p}+\frac{\mathrm{1}}{{p}}\:\geqslant\:\mathrm{2}\:;\:{p}\:>\:\mathrm{0} \\ $$$${now}\:{f}\left({x}\right)=\:\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)+\mathrm{4}\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$$\Rightarrow{f}\left({x}\right)=\:\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}+\mathrm{4}\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$${f}\left({x}\right)\mid_{{minimum}} \:=\:\mathrm{2}^{\mathrm{2}} −\mathrm{2}+\mathrm{4}.\mathrm{2}\:=\:\mathrm{10}. \\ $$
Answered by mathmax by abdo last updated on 17/Sep/20
x^(2 ) +(1/x^2 ) ≥2 and x+(1/x) ≥2 ⇒4(x+(1/x))≥8 ⇒x^2 +(1/x^2 ) +4(x+(1/x))≥10 ⇒ min_R^+ f(x) =10
$$\mathrm{x}^{\mathrm{2}\:} \:+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:\geqslant\mathrm{2}\:\:\mathrm{and}\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\:\geqslant\mathrm{2}\:\Rightarrow\mathrm{4}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)\geqslant\mathrm{8}\:\Rightarrow\mathrm{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:+\mathrm{4}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)\geqslant\mathrm{10}\:\Rightarrow \\ $$$$\mathrm{min}_{\mathrm{R}^{+} } \mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{10} \\ $$
Answered by 1549442205PVT last updated on 18/Sep/20
f(x)=x^2 +(1/x^2 )+4(x+(1/x)) =(x−(1/x))^2 +4((√x)−(1/( (√x))))^2 +10≥10 The equality ocurrs if and only if x=1.Hence,f(x)_(min) =10 when x=2
$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }+\mathrm{4}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$$$=\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} +\mathrm{4}\left(\sqrt{\mathrm{x}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}\right)^{\mathrm{2}} +\mathrm{10}\geqslant\mathrm{10} \\ $$$$\mathrm{The}\:\mathrm{equality}\:\mathrm{ocurrs}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if} \\ $$$$\mathrm{x}=\mathrm{1}.\mathrm{Hence},\mathrm{f}\left(\mathrm{x}\right)_{\mathrm{min}} =\mathrm{10}\:\mathrm{when}\:\mathrm{x}=\mathrm{2} \\ $$

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