Find the point where f(x) = 2sin x + cos 2x h | vedclass

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(C) Given $f(x) = 2\sin x + \cos 2x$ for $0 \leq x \leq 2\pi$.First,find the derivative: $f'(x) = 2\cos x - 2\sin 2x$.Set $f'(x) = 0$ to find critical

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$x = \pi / 6$

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(C) Given $f(x) = 2\sin x + \cos 2x$ for $0 \leq x \leq 2\pi$.First,find the derivative: $f'(x) = 2\cos x - 2\sin 2x$.Set $f'(x) = 0$ to find critical points:$2\cos x - 4\sin x \cos x = 0$$2\cos x(1 - 2\sin x) = 0$This gives $\cos x = 0$ or $\sin x = 1/2$.In the interval $[0, 2\pi]$,$\cos x = 0$ at $x = \pi/2, 3\pi/2$ and $\sin x = 1/2$ at $x = \pi/6, 5\pi/6$.Now,find the second derivative: $f''(x) = -2\sin x - 4\cos 2x$.Check the values:At $x = \pi/6$: $f''(\pi/6) = -2(1/2) - 4\cos(\pi/3) = -1 - 4(1/2) = -3 At $x = \pi/2$: $f''(\pi/2) = -2(1) - 4\cos(\pi) = -2 + 4 = 2 > 0$ (Local Minimum).At $x = 5\pi/6$: $f''(5\pi/6) = -2(1/2) - 4\cos(5\pi/3) = -1 - 4(1/2) = -3 At $x = 3\pi/2$: $f''(3\pi/2) = -2(-1) - 4\cos(3\pi) = 2 + 4 = 6 > 0$ (Local Minimum).Comparing values at critical points: $f(\pi/6) = 2(1/2) + 1/2 = 1.5$,$f(5\pi/6) = 1.5$,$f(\pi/2) = 2(1) - 1 = 1$,$f(3\pi/2) = 2(-1) - 1 = -3$.The maximum value is $1.5$ at $x = \pi/6$ and $x = 5\pi/6$.

Find the point where $f(x) = 2\sin x + \cos 2x$ has a maximum value in the interval $0 \leq x \leq 2\pi$.

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