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October 19th, 2017, 09:07 PM   #1
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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.

Please see attachment
THANKS
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October 19th, 2017, 09:22 PM   #2
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$f(x) = 12 \cos(x) + 6 \sin(2x)$

$f^\prime(x) = -12 \sin(x) + 12 \cos(2x)$

$f^\prime = 0 \Rightarrow $

$12 \sin(x) = 12 \cos(2x)$

$\sin(x) = \cos(2x)$

$\sin(x) = \cos^(x) - \sin^2(x) = 1-2\sin^2(x)$

$2\sin^2(x) + \sin(x) - 1 = 0$

$u=\sin(x)$

$2u^2 +u - 1 = 0$

$(2u - 1 )(u +1) = 0$

$u = \dfrac 1 2,~-1$

$\sin(x) = \dfrac 1 2 \Rightarrow x = \dfrac {\pi}{6},~\dfrac{5\pi}{6}$

$\sin(x) = -1 \Rightarrow x = \dfrac{3\pi}{2}$
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October 19th, 2017, 10:08 PM   #3
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Thank you very much Romsek. I liked your steps better.

Could you check my steps and tell me where I made a mistake?
thanks.
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October 19th, 2017, 10:35 PM   #4
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$\cos(2x) = 1 - 2\sin^2(x)$

you have it as

$\cos(2x) = 1 - \sin^2(x)$
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