f(x) = 1+2 sin x+3 cos2x, 0

Question

f(x) = 1+2 sin x+3 cos2x, 0<=x<=(2pi)/3 is ________________ .

KollegeApply · Accepted Answer

\text { Minimum at } x = $\pi$/2 \text { Given }: f\left( x \right) = 1 + 2 \sin x + 3 \cos^2 x \Rightarrow f'\left( x \right) = 2 \cos x - 6 \cos x \sin x \Rightarrow f'\left( x \right) = 2 \cos x\left( 1 - 3 \sin x \right) \text { For a local maxima or a local minima, we must have } f'\left( x \right) = 0 \Rightarrow 2 \cos x\left( 1 - 3 \sin x \right) = 0 \Rightarrow 2 \cos x = 0 or \left( 1 - 3 \sin x \right) = 0 \Rightarrow \cos x = 0 \ or \sin x = $\frac{1}{3}$ \Rightarrow x = $\pi$/2 or x = \sin^{- 1} \left( $\frac{1}{3}$ \right) \text { Now,} f''\left( x \right) = - 2 \sin x - 6 \cos 2x \Rightarrow f''\left( $\pi$/2 \right) = - 2 \sin $\pi$/2 - 6 \cos \left( 2 $\times$ $\pi$/2 \right) = - 2 + 6 = 4 > 0 \text { So, x } = $\pi$/2 \text { is a local minima }. \text { Also }, f''\left( \sin^{- 1} \left( $\frac{1}{3}$ \right) \right) = - 2 \sin \left( \sin^{- 1} \left( $\frac{1}{3}$ \right) \right) - 6 \cos \left( \sin^{- 1} \left( $\frac{1}{3}$ \right) \right) = - $\frac{2}{3}$ - 6 $\times$ 2$\sqrt$(2)/3 = - \left( $\frac{2}{3}$ + 4$\sqrt$(2) \right) < 0 \text { So,} x = \sin^{- 1} \left( $\frac{1}{3}$ \right)\text { is a local maxima }.

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