Find the critical numbers of the function on the interval 0 ≤ θ < 2π. f(θ) = 2cos(θ) + (sin(θ))^2??
To find critical numbers you want to take the derivative and set it equal to 0. f(θ) = 2cos(θ) + (sin(θ))^2 f '(θ) = -2sin(θ) + 2sin(θ)cos(θ)
by critical number you mean the angle?
yea it is just cos and sin is what gets me. they confuse me.
critical numbers are where f'(x) is either 0 or undefined, so first we need the derivative using the chain rule: f'(x)=-2sinx+2sinx(cosx)=0 cosx=1 x=2(pi)n where n=0,1,2,...
0 = -2sin(θ) + 2sin(θ)cos(θ) 0 = 2sin(θ)(-1 + cos(θ)) So you either want cos(θ) = 1 or sin(θ)=0
cos(θ) = 1 at θ = π/2 or sin(θ)=0 at θ = 0, π The critical values should be 0, π/2, and π
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