If f(x)=sin x+2 cos ^{2} x for frac{pi}{4} le | vedclass

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(C) Given, $f(x)=\sin x+2 \cos ^{2} x$ for $x \in \left[\frac{\pi}{4}, \frac{3 \pi}{4}\right]$.$f'(x) = \cos x - 4 \cos x \sin x = \cos x(1 - 2 \sin 2

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minimum at $x=\frac{\pi}{2}$

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(C) Given, $f(x)=\sin x+2 \cos ^{2} x$ for $x \in \left[\frac{\pi}{4}, \frac{3 \pi}{4}\right]$.$f'(x) = \cos x - 4 \cos x \sin x = \cos x(1 - 2 \sin 2x)$ is incorrect. Let us differentiate correctly:$f'(x) = \cos x + 2(2 \cos x)(-\sin x) = \cos x - 2 \sin 2x$.Setting $f'(x) = 0$, we have $\cos x(1 - 4 \sin x) = 0$.This gives $\cos x = 0$ or $\sin x = \frac{1}{4}$.For $x \in \left[\frac{\pi}{4}, \frac{3 \pi}{4}\right]$, $\cos x = 0 \Rightarrow x = \frac{\pi}{2}$.Also, $\sin x = \frac{1}{4}$ has no solution in this interval since $\sin x \geq \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \approx 0.707 > 0.25$.Checking the critical point and endpoints:$f\left(\frac{\pi}{4}\right) = \sin \frac{\pi}{4} + 2 \cos^2 \frac{\pi}{4} = \frac{1}{\sqrt{2}} + 2(\frac{1}{2}) = \frac{1}{\sqrt{2}} + 1 \approx 1.707$.$f\left(\frac{\pi}{2}\right) = \sin \frac{\pi}{2} + 2 \cos^2 \frac{\pi}{2} = 1 + 0 = 1$.$f\left(\frac{3 \pi}{4}\right) = \sin \frac{3 \pi}{4} + 2 \cos^2 \frac{3 \pi}{4} = \frac{1}{\sqrt{2}} + 2(\frac{1}{2}) = \frac{1}{\sqrt{2}} + 1 \approx 1.707$.Thus, $f(x)$ attains its minimum at $x = \frac{\pi}{2}$.

If $f(x)=\sin x+2 \cos ^{2} x$ for $\frac{\pi}{4} \leq x \leq \frac{3 \pi}{4}$, then $f$ attains its

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