maximum-value-of-f-x-4sin-x-1-cos-x-is-

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

maximum value of

f (x) = \sqrt{4 \sin (x) + 1} - \cos (x)

is

Commented by EDWIN88 last updated on 14/Feb/21

Commented by bemath last updated on 14/Feb/21

approximation

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

\frac{d f (x)}{d x} = \frac{2 \cos x}{\sqrt{4 \sin x + 1}} + \sin x = 0

\frac{4 \cos^{2} x}{4 \sin x + 1} = \sin^{2} x

\frac{4 (1 - \sin^{2} x)}{4 \sin x + 1} = \sin^{2} x

{4 \sin}^{3} x + 5 \sin^{2} x - 4 = 0

\frac{1}{\sin^{3} x} - \frac{5}{4 \sin x} - 1 = 0

Δ = {(- \frac{1}{2})}^{2} + {(- \frac{5}{12})}^{3} = \frac{307}{1728} > 0

\Rightarrow \frac{1}{\sin x} = \frac{1}{2} (\sqrt[3]{\frac{\sqrt{921}}{9} + 4} - \sqrt[3]{\frac{\sqrt{921}}{9} - 4})

\Rightarrow \sin x = \frac{2}{\sqrt[3]{\frac{\sqrt{921}}{9} + 4} - \sqrt[3]{\frac{\sqrt{921}}{9} - 4}}

\cos x = \pm \sqrt{1 - \sin^{2} x}

f {(x)}_{m a x} = \sqrt{4 \sin x + 1} + \sqrt{1 - \sin^{2} x} \approx 2.663 815 973

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