Question
At what points in the interval [0, 2$\pi$], does the function sin 2x attain its maximum value?

Answer

Given that f(x) = sin 2x, $x \in[0,2 \pi]$
$f^\prime$(x) = 2 cos 2x
Now, f'(x) = 0
$\Rightarrow$ cos 2x = 0
$\Rightarrow$ 2x = 0
$\Rightarrow x=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$
Now, we evaluate the value of f at critical point $x=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$ and at end points of the interval [0, 2$\pi$]
$f\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{2}=1$
$f\left(\frac{3 \pi}{4}\right)=\sin \frac{3 \pi}{2}=-1$
$f\left(\frac{5 \pi}{4}\right)=\sin \frac{5 \pi}{2}=1$
$f\left(\frac{7 \pi}{4}\right)=\sin \frac{7 \pi}{2}=-1$
f(0) = sin 0, and f(2$\pi$) = sin 4$\pi$ = 0
Therefore, the absolute maximum value of f on $[0,2 \pi]$ is 1 occuring at $x=\frac{\pi}{4}$ and $x=\frac{5 \pi}{4}$.

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