For the function f(x) = sin 3x,where x in [0, | vedclass

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(D) To determine the intervals of increase and decrease,we find the derivative of $f(x) = \sin 3x$.$f'(x) = \frac{d}{dx}(\sin 3x) = 3 \cos 3x$.For the

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increasing on $[0, \frac{\pi}{6})$ and decreasing on $(\frac{\pi}{6}, \frac{\pi}{2}]$

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(D) To determine the intervals of increase and decrease,we find the derivative of $f(x) = \sin 3x$.$f'(x) = \frac{d}{dx}(\sin 3x) = 3 \cos 3x$.For the function to be increasing,$f'(x) > 0$,so $3 \cos 3x > 0$,which implies $\cos 3x > 0$.Since $x \in [0, \frac{\pi}{2}]$,$3x \in [0, \frac{3\pi}{2}]$.$\cos 3x > 0$ when $3x \in [0, \frac{\pi}{2})$,which means $x \in [0, \frac{\pi}{6})$.For the function to be decreasing,$f'(x) $\cos 3x Thus,the function is increasing on $[0, \frac{\pi}{6})$ and decreasing on $(\frac{\pi}{6}, \frac{\pi}{2}]$.Therefore,the correct option is $D$.

For the function $f(x) = \sin 3x$,where $x \in [0, \frac{\pi}{2}]$,which of the following is true?

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