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Question Number 162721 by mnjuly1970 last updated on 31/Dec/21

calculate f (x )= (( 1)/(4(1+cos ((x/2))) )) +(1/(9(1−cos ((x/2))))) ( x ≠ 2k π , k ∈ Z) f_( min) = ? Adapted From Instagram

$$ \\ $$$$\:\:\:\:\:\:\:{calculate}\: \\ $$$$\:\:\:\:\:\:{f}\:\left({x}\:\right)=\:\frac{\:\mathrm{1}}{\mathrm{4}\left(\mathrm{1}+{cos}\:\left(\frac{{x}}{\mathrm{2}}\right)\right)\:}\:+\frac{\mathrm{1}}{\mathrm{9}\left(\mathrm{1}−{cos}\:\left(\frac{{x}}{\mathrm{2}}\right)\right)}\:\:\left(\:{x}\:\neq\:\mathrm{2}{k}\:\pi\:,\:{k}\:\in\:\mathbb{Z}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{f}_{\:{min}} =\:? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathscr{A}{dapted}\:\mathscr{F}{rom}\:\mathscr{I}{nstagram}\: \\ $$$$ \\ $$

Answered by Ar Brandon last updated on 31/Dec/21

f(x)=(1/(4(1+cos(x/2))))+(1/(9(1−cos(x/2)))) =(1/(8cos^2 (x/4)))+(1/(18sin^2 (x/4)))=((8+10sin^2 (x/4))/(36sin^2 (x/2))) g(x)=36sin^2 (x/2)⇒g′(x)=18sinx=0⇒x=π f(x) min ⇒ g(x) max ⇒x=π f_(min) =((8+10sin^2 (π/4))/(36sin^2 (π/2)))=((8+5)/(36))=((13)/(36))

$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{1}+\mathrm{cos}\frac{{x}}{\mathrm{2}}\right)}+\frac{\mathrm{1}}{\mathrm{9}\left(\mathrm{1}−\mathrm{cos}\frac{{x}}{\mathrm{2}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{8cos}^{\mathrm{2}} \frac{{x}}{\mathrm{4}}}+\frac{\mathrm{1}}{\mathrm{18sin}^{\mathrm{2}} \frac{{x}}{\mathrm{4}}}=\frac{\mathrm{8}+\mathrm{10sin}^{\mathrm{2}} \frac{{x}}{\mathrm{4}}}{\mathrm{36sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}} \\ $$$$\mathrm{g}\left({x}\right)=\mathrm{36sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\Rightarrow\mathrm{g}'\left({x}\right)=\mathrm{18sin}{x}=\mathrm{0}\Rightarrow{x}=\pi \\ $$$${f}\left({x}\right)\:\mathrm{min}\:\Rightarrow\:\mathrm{g}\left({x}\right)\:\mathrm{ma}{x}\:\Rightarrow{x}=\pi \\ $$$${f}_{{min}} =\frac{\mathrm{8}+\mathrm{10sin}^{\mathrm{2}} \frac{\pi}{\mathrm{4}}}{\mathrm{36sin}^{\mathrm{2}} \frac{\pi}{\mathrm{2}}}=\frac{\mathrm{8}+\mathrm{5}}{\mathrm{36}}=\frac{\mathrm{13}}{\mathrm{36}} \\ $$

Answered by mr W last updated on 31/Dec/21

f(x)=(1/(4×2 cos^2 (x/4)))+(1/(9×2 sin^2 (x/4))) f(x)=(1/8)((1/(1−sin^2 (x/4)))+(4/(9 sin^2 (x/4)))) f(t)=(1/8)((1/(1−t))+(4/(9t))) with t=sin^2 (x/4) ∈(0,1) f′(t)=0 ⇒(1/((1−t)^2 ))−(4/(9t^2 ))=0 5t^2 +8t−4=0 (5t−2)(t+2)=0 ⇒t=(2/5), −2 (rejected ) f_(min) =(1/8)((1/(1−(2/5)))+(4/(9×(2/5))))=((25)/(72)) ✓

$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}×\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\frac{{x}}{\mathrm{4}}}+\frac{\mathrm{1}}{\mathrm{9}×\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\frac{{x}}{\mathrm{4}}} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\frac{{x}}{\mathrm{4}}}+\frac{\mathrm{4}}{\mathrm{9}\:\mathrm{sin}^{\mathrm{2}} \:\frac{{x}}{\mathrm{4}}}\right) \\ $$$${f}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{\mathrm{1}}{\mathrm{1}−{t}}+\frac{\mathrm{4}}{\mathrm{9}{t}}\right)\:{with}\:{t}=\mathrm{sin}^{\mathrm{2}} \:\frac{{x}}{\mathrm{4}}\:\in\left(\mathrm{0},\mathrm{1}\right) \\ $$$${f}'\left({t}\right)=\mathrm{0}\:\Rightarrow\frac{\mathrm{1}}{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }−\frac{\mathrm{4}}{\mathrm{9}{t}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\mathrm{5}{t}^{\mathrm{2}} +\mathrm{8}{t}−\mathrm{4}=\mathrm{0} \\ $$$$\left(\mathrm{5}{t}−\mathrm{2}\right)\left({t}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{\mathrm{2}}{\mathrm{5}},\:−\mathrm{2}\:\left({rejected}\:\right) \\ $$$${f}_{{min}} =\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{5}}}+\frac{\mathrm{4}}{\mathrm{9}×\frac{\mathrm{2}}{\mathrm{5}}}\right)=\frac{\mathrm{25}}{\mathrm{72}}\:\checkmark \\ $$

Commented by mnjuly1970 last updated on 31/Dec/21

bravo sir W nice solution as always...grateful...

$$\:\:\:\:{bravo}\:{sir}\:\:{W}\:\:{nice}\:{solution} \\ $$$$\:\:\:{as}\:\:{always}...{grateful}... \\ $$

Answered by mnjuly1970 last updated on 31/Dec/21

f(x)= (1/8) +(1/8) tan^( 2) ((x/4))+(1/(18)) +(1/(18)) cot^( 2) ((x/4)) ≥((13)/(72)) + 2 (√((1/8) .(1/(18)))) =((13)/(72)) +(1/6) f_( min) = ((25)/(72))

$$\:\:\:\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{8}}\:+\frac{\mathrm{1}}{\mathrm{8}}\:{tan}^{\:\mathrm{2}} \left(\frac{{x}}{\mathrm{4}}\right)+\frac{\mathrm{1}}{\mathrm{18}}\:+\frac{\mathrm{1}}{\mathrm{18}}\:{cot}^{\:\mathrm{2}} \left(\frac{{x}}{\mathrm{4}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\geqslant\frac{\mathrm{13}}{\mathrm{72}}\:+\:\mathrm{2}\:\sqrt{\frac{\mathrm{1}}{\mathrm{8}}\:.\frac{\mathrm{1}}{\mathrm{18}}}\:=\frac{\mathrm{13}}{\mathrm{72}}\:+\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:{f}_{\:{min}} \:=\:\frac{\mathrm{25}}{\mathrm{72}} \\ $$

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