What is the derivative of f(θ)= 2cosθ+ cos^2 θ?
I got -2sinθ-2sinθcosθ but i think it's wrong because when i get critical numbers by setting the equation to 0, i get π. My problem asks to intervals of increase and decrease, but my answers are just wrong. So i'm asking if i got the right derivative. And if you want you could find the intervals of increase of decrease. Also the problem is restricted to the interval 0 ≤ θ ≤ 2π. Thanks
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f(Θ) = 2cosΘ + cos²Θ
f '(Θ) = -2sinΘ + 2cosΘ(-sinΘ)
= -2sinΘ - 2cosΘsinΘ
= -2sinΘ(1 + cosΘ)
Set equal to 0:
-2sinΘ(1 + cosΘ) = 0
-2sinΘ = 0 and 1+cosΘ=0
sinΘ = 0 and cosΘ = -1
Θ = 0, π, 2π and Θ = 2π/3 and 4π/3
(0, 2π/3) (2π/3, π) (π, 4π/3) (4π/3, 2π)
Use test points:
π/2.........5π/6........7π/6.......3π/2
Find the value sign of f '(test point)
neg.........neg..........neg........pos
dec..........dec.........dec.........inc