My Math Forum horizontal tangent line

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 October 5th, 2017, 07:21 PM #1 Member   Joined: Apr 2017 From: New York Posts: 77 Thanks: 6 horizontal tangent line Hi guys, I was wondering how I can solve this question attached. Let me show what I did here but couldn't conclude. Thanks all. f(x) = 2cos(x) + cos²(x) f'(x) = 2sin(x) - 2sin(x)cos(x) => f '(x) = -2sin(x)(1 + cos(x)) and then here sin(x) = 0 OR cos(x) = -1 sin(x) is 0 at pi, 3pi, 5pi, etc. cos(x) is -1 at 0, 2pi, 4pi, ... that is it and then how could I continue? Last edited by skipjack; October 6th, 2017 at 02:51 AM.
 October 5th, 2017, 08:00 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,851 Thanks: 1075 Math Focus: Elementary mathematics and beyond Some errors. The second line of your work is missing a minus sign but you've got it right in your third line. Also, $$\sin(x) = 0 \implies x = k\pi, \, k \in \mathbb{Z}$$ and $$\cos(x) = -1 \implies x = (2k - 1)\pi, \, k \in \mathbb{Z}$$ Can you think of what to do next? (Hint: there are two extremes so you need to find two values to substitute into f(x).) Last edited by greg1313; October 5th, 2017 at 08:11 PM.
 October 6th, 2017, 07:49 AM #3 Member   Joined: Apr 2017 From: New York Posts: 77 Thanks: 6 Thanks Greg, how about the y values where tangent is horizontal? I want to compare with my work. thanks you were very helpful
 October 6th, 2017, 08:22 AM #4 Member   Joined: Apr 2017 From: New York Posts: 77 Thanks: 6 Am I going to plug in x value into original function or into derivative.

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