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October 19th, 2017, 09:07 PM   #1
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Joined: Apr 2017
From: New York

Posts: 155
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Critical numbers

Hi guys I was working on this question but couldn't conclude.Any idea/help is greatly appreciated.

I found a weird number as my critical number and could not plug in into my trig function since there is no such trig value that I know

and tried end point and found zero.
so what is my absolute min and max.

THANKS
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 October 19th, 2017, 09:22 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,552 Thanks: 1402 $f(x) = 12 \cos(x) + 6 \sin(2x)$ $f^\prime(x) = -12 \sin(x) + 12 \cos(2x)$ $f^\prime = 0 \Rightarrow$ $12 \sin(x) = 12 \cos(2x)$ $\sin(x) = \cos(2x)$ $\sin(x) = \cos^(x) - \sin^2(x) = 1-2\sin^2(x)$ $2\sin^2(x) + \sin(x) - 1 = 0$ $u=\sin(x)$ $2u^2 +u - 1 = 0$ $(2u - 1 )(u +1) = 0$ $u = \dfrac 1 2,~-1$ $\sin(x) = \dfrac 1 2 \Rightarrow x = \dfrac {\pi}{6},~\dfrac{5\pi}{6}$ $\sin(x) = -1 \Rightarrow x = \dfrac{3\pi}{2}$
 October 19th, 2017, 10:08 PM #3 Senior Member   Joined: Apr 2017 From: New York Posts: 155 Thanks: 6 Thank you very much Romsek. I liked your steps better. Could you check my steps and tell me where I made a mistake? thanks.
 October 19th, 2017, 10:35 PM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 2,552 Thanks: 1402 $\cos(2x) = 1 - 2\sin^2(x)$ you have it as $\cos(2x) = 1 - \sin^2(x)$

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