Question
Find intervals in which the function given by f (x) = sin 3x, x $\in\left[0, \frac{\pi}{2}\right]$ is (a) increasing (b) decreasing.
✓
Answer
$f\left( x \right) = \sin 3x$
$f'\left( x \right) = 3\cos 3x$
$f'\left( x \right) = 0$
$\cos 3x = 0$
$3x = \frac{\pi }{2}$
$x = \frac{\pi }{6}$
$f'\left( x \right) = 3\cos 3x$
$f'\left( x \right) = 0$
$\cos 3x = 0$
$3x = \frac{\pi }{2}$
$x = \frac{\pi }{6}$
| S | A |
| T | C |
| int. | Sign of f’(x) | Result |
|---|---|---|
| $\left[ {0,\frac{\pi }{6}} \right)$ | +tive | increase |
| $\left( {\frac{\pi }{6},\frac{\pi }{2}} \right]$ | -tive | Decrease |
Hence, f(x) is increasing on$\left( {0,\frac{\pi }{6}} \right)$and decreasing on $\left( {\frac{\pi }{6},\frac{\pi }{2}} \right)$
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