Within (0, π/2), on which interval is f(x) = sin 3x strictly decreasing?

Increasing & Decreasing Functions · Question 59 of 74

Question 59

MediumOriginal question

Within $\left(0, \frac{\pi}{2}\right)$ , on which interval is $f(x) = \sin 3x$ strictly decreasing?

(0, \frac{\pi}{6})

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(0, \frac{\pi}{3})

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(\frac{\pi}{6}, \frac{\pi}{2})

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(\frac{\pi}{12}, \frac{\pi}{3})

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Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

$f'(x) = 3\cos 3x$ , which is negative when $\frac{\pi}{2} < 3x < \frac{3\pi}{2}$ .
That is $\frac{\pi}{6} < x < \frac{\pi}{2}$ , all inside $\left(0, \frac{\pi}{2}\right)$ .

Worth noting

$\left(0, \frac{\pi}{6}\right)$ is where $\sin 3x$ rises to its peak at $3x = \frac{\pi}{2}$ .

Ten questions from Increasing & Decreasing Functions, at your own pace.

Ten questions from Increasing & Decreasing Functions, at your own pace.