Let f(x)=4 cos ^3 x+3 sqrt{3} cos ^2 x-10. Th | vedclass

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(B) Given $f(x)=4 \cos ^3 x+3 \sqrt{3} \cos ^2 x-10$ for $x \in(0, 2 \pi)$.First,find the derivative $f^{\prime}(x)$:$f^{\prime}(x) = 4(3 \cos ^2 x)(-

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$2$

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(B) Given $f(x)=4 \cos ^3 x+3 \sqrt{3} \cos ^2 x-10$ for $x \in(0, 2 \pi)$.First,find the derivative $f^{\prime}(x)$:$f^{\prime}(x) = 4(3 \cos ^2 x)(-\sin x) + 3 \sqrt{3}(2 \cos x)(-\sin x)$$f^{\prime}(x) = -12 \cos ^2 x \sin x - 6 \sqrt{3} \cos x \sin x$$f^{\prime}(x) = -6 \sin x \cos x (2 \cos x + \sqrt{3})$$f^{\prime}(x) = -3 \sin(2x) (2 \cos x + \sqrt{3})$To find critical points,set $f^{\prime}(x) = 0$:$-3 \sin(2x) = 0 \Rightarrow 2x = 0, \pi, 2\pi, 3\pi, 4\pi \Rightarrow x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$$2 \cos x + \sqrt{3} = 0 \Rightarrow \cos x = -\frac{\sqrt{3}}{2} \Rightarrow x = \frac{5\pi}{6}, \frac{7\pi}{6}$Critical points in $(0, 2\pi)$ are $\frac{\pi}{2}, \frac{5\pi}{6}, \pi, \frac{7\pi}{6}, \frac{3\pi}{2}$.Analyzing the sign of $f^{\prime}(x)$:For $x \in (0, \frac{\pi}{2})$,$f^{\prime}(x) For $x \in (\frac{\pi}{2}, \frac{5\pi}{6})$,$f^{\prime}(x) > 0$.For $x \in (\frac{5\pi}{6}, \pi)$,$f^{\prime}(x) For $x \in (\pi, \frac{7\pi}{6})$,$f^{\prime}(x) > 0$.For $x \in (\frac{7\pi}{6}, \frac{3\pi}{2})$,$f^{\prime}(x) For $x \in (\frac{3\pi}{2}, 2\pi)$,$f^{\prime}(x) > 0$.Local maxima occur where $f^{\prime}(x)$ changes from positive to negative. This happens at $x = \frac{5\pi}{6}$ and $x = \frac{3\pi}{2}$.Thus,there are $2$ points of local maxima.

Let $f(x)=4 \cos ^3 x+3 \sqrt{3} \cos ^2 x-10$. The number of points of local maxima of $f$ in the interval $(0, 2 \pi)$ is:

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