The function $f(x)=\sin 3x$ on $\left[0, frac{\pi}{2} ight]$ is incr…

Question

The function $f(x)=\sin 3x$ on $\left[0,	frac{\pi}{2}ight]$ is increasing on the interval

Accepted Answer

B. $\left[0,	frac{\pi}{6}ight)$

1. $f'(x)=3\cos 3x$.
2. $f'(x)=0$ when $\cos 3x=0$, i.e. $3x=	frac{\pi}{2}$, so $x=	frac{\pi}{6}$.
3. On $\left[0,	frac{\pi}{6}ight)$: $3x\in[0,	frac{\pi}{2})$, $\cos 3x>0$, so $f'(x)>0$ (increasing).
4. On $\left(	frac{\pi}{6},	frac{\pi}{2}ight)$: $\cos 3x<0$, decreasing, ruling out A.
_Source: syllabus Mathematics Ch 6 "Application of Derivatives", p.9_

Answer

$\left(	frac{\pi}{6},	frac{\pi}{2}ight)$

Answer

$\left[0,	frac{\pi}{2}ight]$

Answer

$\left(	frac{\pi}{3},	frac{\pi}{2}ight)$

The function $f(x)=\sin 3x$ on $\left[0,\tfrac{\pi}{2}\right]$ is increasing on the interval

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