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October 5th, 2017, 07:21 PM   #1
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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.
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October 5th, 2017, 08:00 PM   #2
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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.
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October 6th, 2017, 07:49 AM   #3
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Thanks Greg,
how about the y values where tangent is horizontal?

I want to compare with my work.

thanks you were very helpful
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October 6th, 2017, 08:22 AM   #4
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Am I going to plug in x value into original function or into derivative.
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