The problem is asking me to find the Critical Points of :

$$f(x) = \sin(3x)$$ on the closed interval: $$[\frac{-\pi}{4},\, \frac {\pi}{3}]$$

I know that $f'= 3\cos(3x)$. the problem I seem to be having here is remembering back to pre-calculus. I'm missing some kind of rule that lets me set $f'= 3\cos(3x)$ equal to zero.

I also know that the answer turns out to be $\pm\frac{\pi}{6}$. I just can't seem to remember how to get there. If someone could please explain that to me it would be much appreciated! Thanks!

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    $\begingroup$ Do you know how solve $\cos(\alpha)=0$? $\endgroup$ – Git Gud Feb 26 '15 at 22:05
  • $\begingroup$ Unless I'm mistaken, $Cos(x) = 0$ @ $\frac{\pi}{2}$ and $\frac{3\pi}{2}$. Right? $\endgroup$ – Michael Martin Feb 26 '15 at 22:07
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    $\begingroup$ you are new to mse, so you may not know. it is a courtesy to accept/upvote an answer if you are happy with it. $\endgroup$ – abel Feb 26 '15 at 23:05
  • $\begingroup$ I tried! it said I need more reputation to up vote! @abel EDIT: Oh I see I can click the "check mark." Thanks! $\endgroup$ – Michael Martin Feb 27 '15 at 2:55

Here is how: You set $3\cos(3x)=0$. All you have to do now is to solve for $x$. Solving:

$3\cos(3x)=0$ means $\cos(3x)=0$, taking the inverse cosine on both sides of the last equation we get:

$3x=\frac{\pi}{2}$ and $3x=-\frac{\pi}{2}$. (the choice of $\pi /2$ and $-\pi/2$ here is due to the given interval)

Clearly, $x=\pi /6$ and $-\pi/6$.


$$\cos 3x=0\iff 3x\equiv \frac\pi 2 \mod \pi\iff x\equiv \frac\pi6 \mod \frac\pi3. $$ In the interval $\Bigl[-\dfrac\pi 4,\dfrac\pi 3\Bigr]$, you find only $\,\pm\dfrac\pi6$.

  • $\begingroup$ I knew it was something silly I forgot like that! Thank you! $\endgroup$ – Michael Martin Feb 26 '15 at 22:11
  • $\begingroup$ You're welcome! $\endgroup$ – Bernard Feb 26 '15 at 22:16

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