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Question Number 115332 by bobhans last updated on 25/Sep/20
Minimum value of function f(x)= ((16x^2 cos^2 x+4)/(x cos x)) where −π<x<0
$${Minimum}\:{value}\:{of}\:{function}\: \\ $$$${f}\left({x}\right)=\:\frac{\mathrm{16}{x}^{\mathrm{2}} \:\mathrm{cos}\:^{\mathrm{2}} {x}+\mathrm{4}}{{x}\:\mathrm{cos}\:{x}}\:{where}\:−\pi<{x}<\mathrm{0} \\ $$
Commented by bemath last updated on 25/Sep/20
⇔ f(x)=16x cos x + 4x^(−1) sec x f ′(x)=16cos x−16xsin x+(−4x^(−2) sec x+4x^(−1) sec x tan x )=0 4cos x−4xsin x = x^(−1) sec x (x^(−1) −tan x) 4cos x−4xsin x = (1/(xcos x))((1/x)−((sin x)/(cos x))) 4cos x−4xsin x = ((cos x−xsin x)/(xcos x)) (cos x−xsin x)(4−(1/(x cos x)))=0 (cos x−xsin x)(((4x cos x−1)/(x cos x)))=0 { ((cos x = x sin x ⇒tan x = (1/x))),((4x cos x = 1⇒ cos x = (1/(4x)))) :} (1) for tan x = (1/x) ⇒tan^2 x = (1/x^2 ) sec^2 x = 1+(1/x^2 ) = ((x^2 +1)/x^2 ) cos^2 x = (x^2 /(x^2 +1)) ⇒ cos x = ± (√(x^2 /(x^2 +1))) f(x)= ((16x^2 ((x^2 /(x^2 +1)))+4)/(± x(√(x^2 /(x^2 +1))) ))=((16x^4 +4x^2 +4)/(±x^2 (√(x^2 +1)))) (2) for cos x = (1/(4x)) f(x) = ((16x^2 ((1/(16x^2 ))) +4)/(x((1/(4x))))) = (5/(((1/4)))) =20
$$\Leftrightarrow\:{f}\left({x}\right)=\mathrm{16}{x}\:\mathrm{cos}\:{x}\:+\:\mathrm{4}{x}^{−\mathrm{1}} \:\mathrm{sec}\:{x} \\ $$$${f}\:'\left({x}\right)=\mathrm{16cos}\:{x}−\mathrm{16}{x}\mathrm{sin}\:{x}+\left(−\mathrm{4}{x}^{−\mathrm{2}} \:\mathrm{sec}\:{x}+\mathrm{4}{x}^{−\mathrm{1}} \:\mathrm{sec}\:{x}\:\mathrm{tan}\:{x}\:\right)=\mathrm{0} \\ $$$$\mathrm{4cos}\:{x}−\mathrm{4}{x}\mathrm{sin}\:{x}\:=\:{x}^{−\mathrm{1}} \mathrm{sec}\:{x}\:\left({x}^{−\mathrm{1}} −\mathrm{tan}\:{x}\right) \\ $$$$\mathrm{4cos}\:{x}−\mathrm{4}{x}\mathrm{sin}\:{x}\:=\:\frac{\mathrm{1}}{{x}\mathrm{cos}\:{x}}\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}\right) \\ $$$$\mathrm{4cos}\:{x}−\mathrm{4}{x}\mathrm{sin}\:{x}\:=\:\frac{\mathrm{cos}\:{x}−{x}\mathrm{sin}\:{x}}{{x}\mathrm{cos}\:{x}} \\ $$$$\left(\mathrm{cos}\:{x}−{x}\mathrm{sin}\:{x}\right)\left(\mathrm{4}−\frac{\mathrm{1}}{{x}\:\mathrm{cos}\:{x}}\right)=\mathrm{0} \\ $$$$\left(\mathrm{cos}\:{x}−{x}\mathrm{sin}\:{x}\right)\left(\frac{\mathrm{4}{x}\:\mathrm{cos}\:{x}−\mathrm{1}}{{x}\:\mathrm{cos}\:{x}}\right)=\mathrm{0} \\ $$$$\begin{cases}{\mathrm{cos}\:{x}\:=\:{x}\:\mathrm{sin}\:{x}\:\Rightarrow\mathrm{tan}\:{x}\:=\:\frac{\mathrm{1}}{{x}}}\\{\mathrm{4}{x}\:\mathrm{cos}\:{x}\:=\:\mathrm{1}\Rightarrow\:\mathrm{cos}\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{4}{x}}}\end{cases} \\ $$$$\left(\mathrm{1}\right)\:{for}\:\mathrm{tan}\:{x}\:=\:\frac{\mathrm{1}}{{x}}\:\Rightarrow\mathrm{tan}\:^{\mathrm{2}} {x}\:=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\mathrm{sec}\:^{\mathrm{2}} {x}\:=\:\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:=\:\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\mathrm{cos}\:^{\mathrm{2}} {x}\:=\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}\:\Rightarrow\:\mathrm{cos}\:{x}\:=\:\pm\:\sqrt{\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${f}\left({x}\right)=\:\frac{\mathrm{16}{x}^{\mathrm{2}} \left(\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}\right)+\mathrm{4}}{\pm\:{x}\sqrt{\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}}\:\:}=\frac{\mathrm{16}{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}}{\pm{x}^{\mathrm{2}} \:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:{for}\:\mathrm{cos}\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{4}{x}} \\ $$$$\:\:\:{f}\left({x}\right)\:=\:\frac{\mathrm{16}{x}^{\mathrm{2}} \:\left(\frac{\mathrm{1}}{\mathrm{16}{x}^{\mathrm{2}} }\right)\:+\mathrm{4}}{{x}\left(\frac{\mathrm{1}}{\mathrm{4}{x}}\right)}\:=\:\frac{\mathrm{5}}{\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}\:=\mathrm{20} \\ $$$$ \\ $$
Commented by soumyasaha last updated on 25/Sep/20
f(x) = 16xcosx + (4/(xcosx)) Now, −π <x <0 ⇒ cosx < 0 and x ≠(π/2) ⇒ xcosx > 0 We know, A.M. ≥ G.M. ⇒ ((16cosx + (4/(xcosx)))/2) ≥ (√(16cosx.(4/(xcosx)))) ⇒ ((f(x))/2) ≥ (√(64)) ⇒ f(x) ≥ 16 ∴ minimum value of f(x) is 16.
$$\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{16xcosx}\:+\:\frac{\mathrm{4}}{\mathrm{xcosx}}\: \\ $$$$\:\:\mathrm{Now},\:−\pi\:<\mathrm{x}\:<\mathrm{0}\:\Rightarrow\:\mathrm{cosx}\:<\:\mathrm{0}\:\mathrm{and}\:\mathrm{x}\:\neq\frac{\pi}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{xcosx}\:>\:\mathrm{0} \\ $$$$\:\mathrm{We}\:\mathrm{know},\:\:\mathrm{A}.\mathrm{M}.\:\geqslant\:\mathrm{G}.\mathrm{M}. \\ $$$$\:\:\:\Rightarrow\:\frac{\mathrm{16cosx}\:+\:\frac{\mathrm{4}}{\mathrm{xcosx}}}{\mathrm{2}}\:\geqslant\:\sqrt{\mathrm{16cosx}.\frac{\mathrm{4}}{\mathrm{xcosx}}} \\ $$$$\:\:\:\Rightarrow\:\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{2}}\:\geqslant\:\sqrt{\mathrm{64}} \\ $$$$\:\:\:\Rightarrow\:\mathrm{f}\left(\mathrm{x}\right)\:\geqslant\:\mathrm{16} \\ $$$$\:\:\therefore\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{16}. \\ $$$$ \\ $$
Commented by bemath last updated on 25/Sep/20
Answered by soumyasaha last updated on 25/Sep/20
Commented by bemath last updated on 25/Sep/20
when x = ? sir for f(x) minimum i don′t have exact value
$${when}\:{x}\:=\:?\:{sir}\:{for}\:{f}\left({x}\right)\:{minimum} \\ $$$${i}\:{don}'{t}\:{have}\:{exact}\:{value} \\ $$
Answered by MJS_new last updated on 25/Sep/20
cos −(π/2) =0 ⇒ absolute minima/maxima don′t exist you can only find local minima/maxima
$$\mathrm{cos}\:−\frac{\pi}{\mathrm{2}}\:=\mathrm{0}\:\Rightarrow\:\mathrm{absolute}\:\mathrm{minima}/\mathrm{maxima} \\ $$$$\mathrm{don}'\mathrm{t}\:\mathrm{exist} \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{only}\:\mathrm{find}\:\mathrm{local}\:\mathrm{minima}/\mathrm{maxima} \\ $$
Answered by TANMAY PANACEA last updated on 25/Sep/20
((16t^2 +4)/t) t=xcosx 4(4t+(1/t)) 4[(2(√t) −(1/( (√t))))^2 +2×2(√t) ×(1/( (√t)))] 4[(2(√t) −(1/( (√t))))^2 +4] 16+4(2(√t) −(1/( (√t))))^2 →min value =16
$$\frac{\mathrm{16}{t}^{\mathrm{2}} +\mathrm{4}}{{t}}\:\:\:\:\:\:{t}={xcosx} \\ $$$$\mathrm{4}\left(\mathrm{4}{t}+\frac{\mathrm{1}}{{t}}\right) \\ $$$$\mathrm{4}\left[\left(\mathrm{2}\sqrt{{t}}\:−\frac{\mathrm{1}}{\:\sqrt{{t}}}\right)^{\mathrm{2}} +\mathrm{2}×\mathrm{2}\sqrt{{t}}\:×\frac{\mathrm{1}}{\:\sqrt{{t}}}\right] \\ $$$$\mathrm{4}\left[\left(\mathrm{2}\sqrt{{t}}\:−\frac{\mathrm{1}}{\:\sqrt{{t}}}\right)^{\mathrm{2}} +\mathrm{4}\right] \\ $$$$\mathrm{16}+\mathrm{4}\left(\mathrm{2}\sqrt{{t}}\:−\frac{\mathrm{1}}{\:\sqrt{{t}}}\right)^{\mathrm{2}} \rightarrow{min}\:{value}\:=\mathrm{16} \\ $$
Answered by MJS_new last updated on 25/Sep/20
local maximum at (((−π)),((16π+(4/π))) ) local minimum at (((≈−1.84520)),((16)) ) local maximum at (((≈−1.09801)),((−16)) ) local minimum at (((≈−.860334)),((≈−16.1064)) ) local maximum at (((≈−.610031)),((−16)) ) absolute min/max doesn′t exist because −∞<f(x)<∞ proof: lim_(x→(−(π/2))^− ) f(x) =+∞ lim_(x→(−(π/2))^+ ) f(x)=−∞ lim_(x→0^− ) f(x) =−∞
$$\mathrm{local}\:\mathrm{maximum}\:\mathrm{at}\:\begin{pmatrix}{−\pi}\\{\mathrm{16}\pi+\frac{\mathrm{4}}{\pi}}\end{pmatrix} \\ $$$$\mathrm{local}\:\mathrm{minimum}\:\mathrm{at}\:\begin{pmatrix}{\approx−\mathrm{1}.\mathrm{84520}}\\{\mathrm{16}}\end{pmatrix} \\ $$$$\mathrm{local}\:\mathrm{maximum}\:\mathrm{at}\:\begin{pmatrix}{\approx−\mathrm{1}.\mathrm{09801}}\\{−\mathrm{16}}\end{pmatrix} \\ $$$$\mathrm{local}\:\mathrm{minimum}\:\mathrm{at}\:\begin{pmatrix}{\approx−.\mathrm{860334}}\\{\approx−\mathrm{16}.\mathrm{1064}}\end{pmatrix} \\ $$$$\mathrm{local}\:\mathrm{maximum}\:\mathrm{at}\:\begin{pmatrix}{\approx−.\mathrm{610031}}\\{−\mathrm{16}}\end{pmatrix} \\ $$$$\mathrm{absolute}\:\mathrm{min}/\mathrm{max}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{exist}\:\mathrm{because} \\ $$$$−\infty<{f}\left({x}\right)<\infty \\ $$$$\mathrm{proof}: \\ $$$$\underset{{x}\rightarrow\left(−\frac{\pi}{\mathrm{2}}\right)^{−} } {\mathrm{lim}}\:{f}\left({x}\right)\:=+\infty \\ $$$$\underset{{x}\rightarrow\left(−\frac{\pi}{\mathrm{2}}\right)^{+} } {\mathrm{lim}}\:{f}\left({x}\right)=−\infty \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:{f}\left({x}\right)\:=−\infty \\ $$

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