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Question Number 132474 by bemath last updated on 14/Feb/21

maximum value of f(x)=(√(4sin (x)+1)) −cos (x) is

$$\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{4sin}\:\left(\mathrm{x}\right)+\mathrm{1}}\:−\mathrm{cos}\:\left(\mathrm{x}\right) \\ $$$$\mathrm{is}\: \\ $$

Commented by EDWIN88 last updated on 14/Feb/21

Commented by bemath last updated on 14/Feb/21

approximation

$$\mathrm{approximation} \\ $$

Answered by mr W last updated on 14/Feb/21

((df(x))/dx)=((2 cos x)/( (√(4 sin x+1))))+sin x=0 ((4 cos^2 x)/( 4 sin x+1))=sin^2 x ((4(1−sin^2 x))/( 4 sin x+1))=sin^2 x 4sin^3 x+5 sin^2 x−4=0 (1/(sin^3 x))−(5/(4 sin x))−1=0 Δ=(−(1/2))^2 +(−(5/(12)))^3 =((307)/(1728))>0 ⇒(1/(sin x))=(1/2)(((((√(921))/9)+4))^(1/3) −((((√(921))/9)−4))^(1/3) ) ⇒sin x=(2/( ((((√(921))/9)+4))^(1/3) −((((√(921))/9)−4))^(1/3) )) cos x=±(√(1−sin^2 x)) f(x)_(max) =(√(4 sin x+1))+(√(1−sin^2 x))≈2.663 815 973

$$\frac{{df}\left({x}\right)}{{dx}}=\frac{\mathrm{2}\:\mathrm{cos}\:{x}}{\:\sqrt{\mathrm{4}\:\mathrm{sin}\:{x}+\mathrm{1}}}+\mathrm{sin}\:{x}=\mathrm{0} \\ $$$$\frac{\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:{x}}{\:\mathrm{4}\:\mathrm{sin}\:{x}+\mathrm{1}}=\mathrm{sin}^{\mathrm{2}} \:{x} \\ $$$$\frac{\mathrm{4}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{x}\right)}{\:\mathrm{4}\:\mathrm{sin}\:{x}+\mathrm{1}}=\mathrm{sin}^{\mathrm{2}} \:{x} \\ $$$$\mathrm{4sin}^{\mathrm{3}} \:{x}+\mathrm{5}\:\mathrm{sin}^{\mathrm{2}} \:{x}−\mathrm{4}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{3}} \:{x}}−\frac{\mathrm{5}}{\mathrm{4}\:\mathrm{sin}\:{x}}−\mathrm{1}=\mathrm{0} \\ $$$$\Delta=\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(−\frac{\mathrm{5}}{\mathrm{12}}\right)^{\mathrm{3}} =\frac{\mathrm{307}}{\mathrm{1728}}>\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{sin}\:{x}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{921}}}{\mathrm{9}}+\mathrm{4}}−\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{921}}}{\mathrm{9}}−\mathrm{4}}\right) \\ $$$$\Rightarrow\mathrm{sin}\:{x}=\frac{\mathrm{2}}{\:\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{921}}}{\mathrm{9}}+\mathrm{4}}−\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{921}}}{\mathrm{9}}−\mathrm{4}}} \\ $$$$\mathrm{cos}\:{x}=\pm\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{x}} \\ $$$${f}\left({x}\right)_{{max}} =\sqrt{\mathrm{4}\:\mathrm{sin}\:{x}+\mathrm{1}}+\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{x}}\approx\mathrm{2}.\mathrm{663}\:\mathrm{815}\:\mathrm{973} \\ $$

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