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Q-179570-Posted-by-Infinityaction-30-10-2022-find-the-minimum-of-f-x-f-x-x-2-4-x-2-8x-12-x-25-x-2-4-x-2-16x-16-x-80-f-x-x-4-2-2-x-3-2-

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Question Number 179717 by a.lgnaoui last updated on 01/Nov/22
Q 179570 (Posted by Infinityaction 30.10.2022) find the minimum of f(x) f(x)=(√(x^2 +(4/x^2 )−8x−((12)/x)+25)) +(√(x^2 +(4/x^2 )−16x−((16)/x)+80)) −−−−−−−−−−−−−−−−−− f(x)=(√((x−4)^2 +((2/x)−3)^2 )) +(√((x−8)^2 +((2/x)−4)^2 )) Df=R−{0} f(x)≥0 Min(f(x))=x / f(x)=0 (√((x−4)^2 +((2/x)−3)^2 )) +(√((x−8)^2 +((2/x)−4)^2 )) =0 (x−4)^2 +((2/x)−3)^2 =(x−8)^2 +((2/x)−4)^2 (1) x−8=(x−4)−4 and ((2/x)−4)=((2/x)−3)−1 x^2 −((55)/8)x+(1/2)=0 x=0,07351334 and x=6,80148666} x=0,073513334 f(x)=49,044679(rejete) x=6,80148666 f(x)=7,7899055 Donc 7,7899055 est minimum de f(x)
$$\mathrm{Q}\:\mathrm{179570}\:\left({Posted}\:{by}\:\mathrm{Infinityaction}\:\mathrm{30}.\mathrm{10}.\mathrm{2022}\right) \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{4}}{\mathrm{x}^{\mathrm{2}} }−\mathrm{8x}−\frac{\mathrm{12}}{\mathrm{x}}+\mathrm{25}}\:+\sqrt{\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{4}}{\mathrm{x}^{\mathrm{2}} }−\mathrm{16x}−\frac{\mathrm{16}}{{x}}+\mathrm{80}}\: \\ $$$$−−−−−−−−−−−−−−−−−− \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\left(\mathrm{x}−\mathrm{4}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}}{\mathrm{x}}−\mathrm{3}\right)^{\mathrm{2}} }\:\:+\sqrt{\left(\mathrm{x}−\mathrm{8}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}}{\mathrm{x}}−\mathrm{4}\right)^{\mathrm{2}} }\: \\ $$$$\mathrm{D}{f}=\mathbb{R}−\left\{\mathrm{0}\right\}\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)\geqslant\mathrm{0}\: \\ $$$$\mathrm{Min}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\mathrm{x}\:/\:{f}\left(\mathrm{x}\right)=\mathrm{0} \\ $$$$\sqrt{\left(\mathrm{x}−\mathrm{4}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}}{\mathrm{x}}−\mathrm{3}\right)^{\mathrm{2}} \:}\:+\sqrt{\left(\mathrm{x}−\mathrm{8}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}}{\mathrm{x}}−\mathrm{4}\right)^{\mathrm{2}} \:}\:=\mathrm{0} \\ $$$$\left(\mathrm{x}−\mathrm{4}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}}{\mathrm{x}}−\mathrm{3}\right)^{\mathrm{2}} =\left(\mathrm{x}−\mathrm{8}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}}{\mathrm{x}}−\mathrm{4}\right)^{\mathrm{2}} \:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{x}−\mathrm{8}=\left(\mathrm{x}−\mathrm{4}\right)−\mathrm{4}\:\:\:\:\mathrm{and}\:\:\left(\frac{\mathrm{2}}{\mathrm{x}}−\mathrm{4}\right)=\left(\frac{\mathrm{2}}{\mathrm{x}}−\mathrm{3}\right)−\mathrm{1} \\ $$$$\left.\:\:\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{55}}{\mathrm{8}}\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0}\:\:\:{x}=\mathrm{0},\mathrm{07351334}\:{and}\:{x}=\mathrm{6},\mathrm{80148666}\right\} \\ $$$$\mathrm{x}=\mathrm{0},\mathrm{073513334} \\ $$$${f}\left(\mathrm{x}\right)=\mathrm{49},\mathrm{044679}\left({rejete}\right) \\ $$$$\mathrm{x}=\mathrm{6},\mathrm{80148666}\:\:\:\:\:{f}\left(\mathrm{x}\right)=\mathrm{7},\mathrm{7899055} \\ $$$${Donc}\:\:\mathrm{7},\mathrm{7899055}\:\:\:{est}\:{minimum}\:{de}\:{f}\left({x}\right) \\ $$$$ \\ $$
Commented by a.lgnaoui last updated on 01/Nov/22
Q179570 (look to actual anser)
$${Q}\mathrm{179570}\:\:\:\left({look}\:{to}\:\:{actual}\:{anser}\right) \\ $$
Commented by mr W last updated on 01/Nov/22
wrong! what′s the reason that (√(a(x)))+(√(b(x))) =0 means mininum of (√(a(x)))+(√(b(x)))? it′s paradox that you set (√(a(x)))+(√(b(x)))=0 and at same time you get (√(a(x)))+(√(b(x))) =minimum≠0. besides you set (√(a(x)))+(√(b(x)))=0 and at same time a(x)=b(x). your logic is weird.
$${wrong}! \\ $$$${what}'{s}\:{the}\:{reason}\:{that}\:\sqrt{{a}\left({x}\right)}+\sqrt{{b}\left({x}\right)}\:=\mathrm{0} \\ $$$${means}\:{mininum}\:{of}\:\sqrt{{a}\left({x}\right)}+\sqrt{{b}\left({x}\right)}? \\ $$$${it}'{s}\:{paradox}\:{that}\:{you}\:{set}\: \\ $$$$\sqrt{{a}\left({x}\right)}+\sqrt{{b}\left({x}\right)}=\mathrm{0}\:{and}\:{at}\:{same}\:{time}\:{you} \\ $$$${get}\:\sqrt{{a}\left({x}\right)}+\sqrt{{b}\left({x}\right)}\:={minimum}\neq\mathrm{0}. \\ $$$${besides}\:{you}\:{set}\:\sqrt{{a}\left({x}\right)}+\sqrt{{b}\left({x}\right)}=\mathrm{0}\:{and} \\ $$$${at}\:{same}\:{time}\:{a}\left({x}\right)={b}\left({x}\right).\:{your}\:{logic}\: \\ $$$${is}\:{weird}. \\ $$
Commented by a.lgnaoui last updated on 01/Nov/22
We can find de value by derivation of f(x) i have trying it but its long I try aigain I hope that the resultat were similaire.
$$\mathrm{We}\:\mathrm{can}\:\mathrm{find}\:\mathrm{de}\:\mathrm{value}\:\mathrm{by}\:\mathrm{derivation}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\mathrm{i}\:\mathrm{have}\:\mathrm{trying}\:\mathrm{it}\:\mathrm{but}\:\mathrm{its}\:\mathrm{long} \\ $$$$\mathrm{I}\:\mathrm{try}\:\mathrm{aigain}\:\mathrm{I}\:\mathrm{hope}\:\mathrm{that}\:\mathrm{the}\: \\ $$$$\mathrm{resultat}\:\mathrm{were}\:\mathrm{similaire}. \\ $$
Commented by mr W last updated on 01/Nov/22
i think it′s impossible to find the minimum exactly in this question.
$${i}\:{think}\:{it}'{s}\:{impossible}\:{to}\:{find}\:{the} \\ $$$${minimum}\:{exactly}\:{in}\:{this}\:{question}. \\ $$
Commented by a.lgnaoui last updated on 01/Nov/22
(√(a(x)^2 +b(x)^2 )) +(√(c(x)^2 +d(x)^2 )) ≥0 pour x∈Df so the minimum is the value of x wich a(x)^2 +b(x)^2 =0 and c(x)^2 +d(x)^2 =0 (√(a(x)^2 +b(x)^2 )) =−(√(c(x)^2 +d(x)^2 )) a(x)^2 +b(x)^2 =c(x)^2 +d(x)^2 lim(f(x)_(x→+∞) =lim(f(x)_(x→−∞) )=∞ im(f(x))_(x→±0) =+∞
$$\sqrt{{a}\left({x}\right)^{\mathrm{2}} +{b}\left({x}\right)^{\mathrm{2}} }\:+\sqrt{{c}\left({x}\right)^{\mathrm{2}} +{d}\left({x}\right)^{\mathrm{2}} }\:\geqslant\mathrm{0}\:\:{pour}\:{x}\in{Df} \\ $$$${so}\:{the}\:{minimum}\:{is}\:{the}\:{value}\:{of}\:{x} \\ $$$${wich}\:{a}\left({x}\right)^{\mathrm{2}} +{b}\left({x}\right)^{\mathrm{2}} =\mathrm{0}\:\:{and}\:\:{c}\left({x}\right)^{\mathrm{2}} +{d}\left({x}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\sqrt{{a}\left({x}\right)^{\mathrm{2}} +{b}\left({x}\right)^{\mathrm{2}} }\:=−\sqrt{{c}\left({x}\right)^{\mathrm{2}} +{d}\left({x}\right)^{\mathrm{2}} \:} \\ $$$${a}\left({x}\right)^{\mathrm{2}} +{b}\left({x}\right)^{\mathrm{2}} ={c}\left({x}\right)^{\mathrm{2}} +{d}\left({x}\right)^{\mathrm{2}} \\ $$$$\mathrm{lim}\left(\mathrm{f}\left(\mathrm{x}\right)_{\mathrm{x}\rightarrow+\infty} =\mathrm{lim}\left(\mathrm{f}\left(\mathrm{x}\right)_{\mathrm{x}\rightarrow−\infty} \right)=\infty\right. \\ $$$$\mathrm{im}\left(\mathrm{f}\left(\mathrm{x}\right)\right)_{\mathrm{x}\rightarrow\pm\mathrm{0}} =+\infty \\ $$
Commented by mr W last updated on 01/Nov/22
it′s clearly wrong. just look at example: f(x)=(√(x^2 +4))+(√(x^2 +8))
$${it}'{s}\:{clearly}\:{wrong}. \\ $$$${just}\:{look}\:{at}\:{example}: \\ $$$${f}\left({x}\right)=\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}+\sqrt{{x}^{\mathrm{2}} +\mathrm{8}} \\ $$
Answered by manxsol last updated on 01/Nov/22
Commented by manxsol last updated on 01/Nov/22
Commented by mr W last updated on 01/Nov/22
Commented by manxsol last updated on 02/Nov/22
ok,AP≠PB.use f′(x) o .......
$${ok},{AP}\neq{PB}.{use}\:{f}'\left({x}\right)\:\:{o}\:\:\:……. \\ $$
Commented by manxsol last updated on 02/Nov/22
thanks
$${thanks} \\ $$
Commented by mr W last updated on 02/Nov/22
f′(x)=0 can not be exactly solved, therefore the minimum of f(x) can not be calculated exactly.
$${f}'\left({x}\right)=\mathrm{0}\:{can}\:{not}\:{be}\:{exactly}\:{solved}, \\ $$$${therefore}\:{the}\:{minimum}\:{of}\:{f}\left({x}\right)\:{can}\: \\ $$$${not}\:{be}\:{calculated}\:{exactly}. \\ $$

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