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$$y=\sqrt{3}\cos(3x)+\sin(3x); 0\le{x}\le{\frac{2\pi}{3}}$$

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?

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    $\begingroup$ At $x = \pi/2, y' \neq 0$... $y'(\pi/2) = 3\sqrt{3}$ $\endgroup$ – Namaste Apr 29 '13 at 3:44
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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.

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  • $\begingroup$ How do you get 3 potential critical points when solving for $x$? I only get one point which is $\frac{\pi}{2}$. $\endgroup$ – Kot Apr 29 '13 at 4:24
  • $\begingroup$ Did you read the first sentence, or my comment? $y'(x)\neq 0$ when $x = \pi/2$. I say three potential critical points, but one will outside your interval...Follow what I've done to solve $y' = 0$. Then: Solve for $x$ given $3x = \tan^{-1}(1/\sqrt 3)$...then find which value is outside of $x \in [0, 2\pi/3]$. You will be left with two solutions to x: one will be a maximum, and one will be a minimum. The other max is the endpoint of the interval on which $x$ is defined: $2 \pi/3$, and the other minimum at the endpoint $x = 0$ $\endgroup$ – Namaste Apr 29 '13 at 4:28
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    $\begingroup$ I see what I did wrong, I multiplied by $3$ instead of dividing by $3$. I now have $x=\frac{\pi}{18}$. I do not understand where the two other points would come from. Does the inverse tangent function give multiple answers? $\endgroup$ – Kot Apr 29 '13 at 4:31
  • $\begingroup$ Yes, the other critical point in your interval is at $\pi/18 + \pi/3 = 7\pi/18$. The calculator typically only gives one value...but you can test to confirm that $x = 7\pi/18$ also makes $y' = 0$ $\endgroup$ – Namaste Apr 29 '13 at 4:33
  • $\begingroup$ I am a little confused as to why you added $\pi/3$ to the first answer. If it was tangent don't you add $\pi$ to get the other value? $\endgroup$ – Kot Apr 29 '13 at 4:37
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$ y=\sqrt{3}\cos(3x)+\sin(3x)=2(\dfrac{\sqrt{3}}{2}\cos(3x)+\dfrac{1}{2}\sin(3x)=2(\sin\dfrac{\pi}{3}\cos(3x)+cos\dfrac{\pi}{3}\sin(3x))=2\sin(3x+\dfrac{\pi}{3}) $

$\dfrac{\pi}{3} \leq 3x+\dfrac{\pi}{3} \leq 2\pi+\dfrac{\pi}{3}$, so there is 2 peaks when $3x+\dfrac{\pi}{3}=\dfrac{\pi}{2}$ or $\dfrac{3 \pi}{2} $, another 2 are the end points ie:$x=0$ or $x=\dfrac{3 \pi}{2}$

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  • $\begingroup$ An excellent proof showing calculus is unnecessary. $\endgroup$ – Stefan Smith Apr 29 '13 at 15:29
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$y=\sqrt 3\cos(3x)+\sin(3x)=2\sin(3x+\frac\pi3)$

$y'=2\cdot3\cos(3x+\frac\pi3)$

For the extreme values of $y, y'=0\implies \cos(3x+\frac\pi3)=0$

$\implies 3x+\frac\pi3=(2n+1)\frac\pi2$ whether $n$ is any integer

$\implies 3x =(6n+1)\frac\pi6 $

As $0\le x\le\frac{2\pi}3, 0\le 3x\le 2\pi$

$\implies 0\le (6n+1)\frac\pi6\le 2\pi \implies0\le 6n+1\le 12\implies n=0,1$

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