The function f(x) = sin x(1 + cos x) at x = f | vedclass

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(A) Let $f(x) = \sin x + \sin x \cos x = \sin x + \frac{1}{2} \sin 2x$. Find the first derivative: $f'(x) = \cos x + \cos 2x$. Find the second derivat

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(A) Let $f(x) = \sin x + \sin x \cos x = \sin x + \frac{1}{2} \sin 2x$. Find the first derivative: $f'(x) = \cos x + \cos 2x$. Find the second derivative: $f''(x) = -\sin x - 2 \sin 2x$. At $x = \frac{\pi}{3}$: $f''\left(\frac{\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right) - 2 \sin\left(\frac{2\pi}{3}\right)$. Since $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$ and $\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}$: $f''\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} - 2\left(\frac{\sqrt{3}}{2}\right) = -\frac{\sqrt{3}}{2} - \sqrt{3} = -\frac{3\sqrt{3}}{2}$. Since $f''\left(\frac{\pi}{3}\right) < 0$,the function has a local maximum at $x = \frac{\pi}{3}$.

The function $f(x) = \sin x(1 + \cos x)$ at $x = \frac{\pi}{3}$ is:

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