At what points in the interval [0, 2pi],does | vedclass

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(A) Let $f(x) = \sin 2x$.Then,$f'(x) = 2 \cos 2x$.For critical points,set $f'(x) = 0$,which implies $2 \cos 2x = 0$,so $\cos 2x = 0$.In the interval $

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$x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$

Vedclass · Answer

(A) Let $f(x) = \sin 2x$.Then,$f'(x) = 2 \cos 2x$.For critical points,set $f'(x) = 0$,which implies $2 \cos 2x = 0$,so $\cos 2x = 0$.In the interval $[0, 2\pi]$,$2x$ lies in $[0, 4\pi]$. Thus,$2x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$.This gives $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.Evaluating $f(x)$ at these points:$f(\frac{\pi}{4}) = \sin(\frac{\pi}{2}) = 1$$f(\frac{3\pi}{4}) = \sin(\frac{3\pi}{2}) = -1$$f(\frac{5\pi}{4}) = \sin(\frac{5\pi}{2}) = 1$$f(\frac{7\pi}{4}) = \sin(\frac{7\pi}{2}) = -1$Also,at the endpoints $f(0) = 0$ and $f(2\pi) = 0$.The maximum value is $1$,which occurs at $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$.

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

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