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 Calculus Calculus Math Forum

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.

THANKS Attached Images Screen Shot 2017-10-20 at 1.02.32 AM.jpg (23.4 KB, 6 views) 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 $\cos(2x) = 1 - \sin^2(x)$ Tags critical, numbers Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jkh1919 Calculus 10 July 24th, 2012 08:46 PM lovetolearn Calculus 1 April 5th, 2012 07:44 PM tsl182forever8 Calculus 7 March 3rd, 2012 04:41 PM Calc12 Calculus 9 December 14th, 2010 08:18 PM lovetolearn Algebra 0 December 31st, 1969 04:00 PM

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