October 19th, 2017, 09:07 PM  #1 
Senior Member Joined: Apr 2017 From: New York Posts: 155 Thanks: 6  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 
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 $ $12 \sin(x) = 12 \cos(2x)$ $\sin(x) = \cos(2x)$ $\sin(x) = \cos^(x)  \sin^2(x) = 12\sin^2(x)$ $2\sin^2(x) + \sin(x)  1 = 0$ $u=\sin(x)$ $2u^2 +u  1 = 0$ $(2u  1 )(u +1) = 0$ $u = \dfrac 1 2,~1$ $\sin(x) = \dfrac 1 2 \Rightarrow x = \dfrac {\pi}{6},~\dfrac{5\pi}{6}$ $\sin(x) = 1 \Rightarrow x = \dfrac{3\pi}{2}$ 
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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