$$y=\sqrt{3}\cos(3x)+\sin(3x); 0\le{x}\le{\frac{2\pi}{3}}$$
I know that the local extrema can be determined by using the first derivative test. I took the derivative of $y$ and got $$y'= -3\sqrt{3}\sin(3x)+3\cos(3x)$$ I then solved the derivative for when it's value is $0$ and got $x=\dfrac{\pi}{2}$ . I then used this critical point and subdivided the interval. I found that there were 3 points of local extrema after doing all my work, local maximum $x=0, \dfrac{2\pi}{3}$ and local minimum $x=\dfrac{\pi}{2}$. However according to the online homework, there were 4 different points of extrema. Local maximum $x=\dfrac{\pi}{18},\dfrac{2\pi}{3}$ and local minimum $x=0,\dfrac{7\pi}{18}$. I am really confused as to how there are 4 points of local extrema, did I leave out an answer somewhere? I am also confused as to how they got $x=\dfrac{\pi}{18},\dfrac{7\pi}{18}$ as points of local extrema. Could someone explain this to me?
Best Answer
As I noted in my comment: At $x = \pi/2, \quad y' \neq 0;$ ... $y'(\pi/2) = 3\sqrt{3}$
You need to solve for $$y'= -3\sqrt{3}\sin(3x)+3\cos(3x) = 0 \quad \iff \quad 3\sqrt{3}\sin(3x) = 3\cos(3x)$$ $$\iff \quad \sqrt 3 \sin(3x) = \cos(3x),\quad x\in \left[0, \frac{2\pi}{3}\right]$$
Note that $$\sqrt 3 \sin(3x) = \cos(3x) \iff \sqrt 3 \dfrac{\sin(3x)}{\cos(3x)} = \sqrt 3 \tan(3x) = 1\iff \tan(3x) = \frac{1}{\sqrt 3}$$
Solving for $x$ will give you 2 potential critical points on your interval; then recall that you need to also check endpoints of an interval as potential extrema.