what-minimum-value-of-f-x-9tan-2-x-4cot-2-x-

Question Number 78207 by jagoll last updated on 15/Jan/20

$${what}\:{minimum}\: \\ $$$${value}\:{of}\:{f}\left({x}\right)=\:\mathrm{9tan}\:^{\mathrm{2}} \left({x}\right)+\mathrm{4cot}\:^{\mathrm{2}} \left({x}\right) \\ $$

Commented by jagoll last updated on 15/Jan/20

$${i}'{m}\:{got}\:\mathrm{24}.\:{it}\:{right}\:{answer}? \\ $$

Commented by john santu last updated on 15/Jan/20

$${sir}\:{Mjs}\:{it}\:{answer}\:{is}\:{correct}? \\ $$

Commented by MJS last updated on 15/Jan/20

$$\mathrm{I}\:\mathrm{get}\:\mathrm{12} \\ $$

Commented by jagoll last updated on 15/Jan/20

$${how}\:{sir}? \\ $$

Commented by MJS last updated on 15/Jan/20

let tan x =t f(t)=9t^2 +(4/t^2 ) f′(t)=18t−(8/t^3 )=(2/t^3 )(3t^2 −2)(3t^2 +2)=0 ⇒ t=±((√6)/3) f′′(t)=(6/t^4 )(3t^4 +4) >0∀t∈R ⇒ minima at ±((√6)/3) tan x =±((√6)/3) ⇒ x=nπ±arctan ((√6)/3) f(nπ±arctan ((√6)/3))=12

$$\mathrm{let}\:\mathrm{tan}\:{x}\:={t} \\ $$$${f}\left({t}\right)=\mathrm{9}{t}^{\mathrm{2}} +\frac{\mathrm{4}}{{t}^{\mathrm{2}} } \\ $$$${f}'\left({t}\right)=\mathrm{18}{t}−\frac{\mathrm{8}}{{t}^{\mathrm{3}} }=\frac{\mathrm{2}}{{t}^{\mathrm{3}} }\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{2}\right)\left(\mathrm{3}{t}^{\mathrm{2}} +\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{t}=\pm\frac{\sqrt{\mathrm{6}}}{\mathrm{3}} \\ $$$${f}''\left({t}\right)=\frac{\mathrm{6}}{{t}^{\mathrm{4}} }\left(\mathrm{3}{t}^{\mathrm{4}} +\mathrm{4}\right)\:>\mathrm{0}\forall{t}\in\mathbb{R}\:\Rightarrow\:\mathrm{minima}\:\mathrm{at}\:\pm\frac{\sqrt{\mathrm{6}}}{\mathrm{3}} \\ $$$$\mathrm{tan}\:{x}\:=\pm\frac{\sqrt{\mathrm{6}}}{\mathrm{3}}\:\Rightarrow\:{x}={n}\pi\pm\mathrm{arctan}\:\frac{\sqrt{\mathrm{6}}}{\mathrm{3}} \\ $$$${f}\left({n}\pi\pm\mathrm{arctan}\:\frac{\sqrt{\mathrm{6}}}{\mathrm{3}}\right)=\mathrm{12} \\ $$

Commented by jagoll last updated on 15/Jan/20

$${thanks}\:{sir}.\:{my}\:{answer}\:{is}\:{wrong} \\ $$

Answered by behi83417@gmail.com last updated on 15/Jan/20

9tg^2 x.4cotg^2 x=36=cons. ⇒f(x) is minimum when:9tg^2 x=4cotg^2 x ⇒tg^4 x=(4/9)=((36)/(81))⇒tgx=±((√6)/3) ⇒f_(min) =9×(6/9)+4×(9/6)=6+6=12 .■

$$\mathrm{9tg}^{\mathrm{2}} \mathrm{x}.\mathrm{4cotg}^{\mathrm{2}} \mathrm{x}=\mathrm{36}=\mathrm{cons}. \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{minimum}\:\mathrm{when}:\mathrm{9tg}^{\mathrm{2}} \mathrm{x}=\mathrm{4cotg}^{\mathrm{2}} \mathrm{x} \\ $$$$\Rightarrow\mathrm{tg}^{\mathrm{4}} \mathrm{x}=\frac{\mathrm{4}}{\mathrm{9}}=\frac{\mathrm{36}}{\mathrm{81}}\Rightarrow\mathrm{tgx}=\pm\frac{\sqrt{\mathrm{6}}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{f}_{\mathrm{min}} =\mathrm{9}×\frac{\mathrm{6}}{\mathrm{9}}+\mathrm{4}×\frac{\mathrm{9}}{\mathrm{6}}=\mathrm{6}+\mathrm{6}=\mathrm{12}\:.\blacksquare \\ $$

Commented by jagoll last updated on 15/Jan/20

$${thanks}\:{sir} \\ $$

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