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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 bybemath 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 bysoumyasaha 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 bybemath last updated on 25/Sep/20

Answered by soumyasaha last updated on 25/Sep/20

Commented bybemath 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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