Final answer:
To find the critical points of the function 2sec(θ)+tan(θ), set the derivative equal to 0 or undefined. The critical points are the values of θ that satisfy those conditions. The critical points are θ = π/2 and θ = π.
Step-by-step explanation:
To find the critical points of the function 2sec(θ)+tan(θ), we need to find the values of θ that make the derivative of the function equal to 0 or undefined. The derivative of 2sec(θ)+tan(θ) can be found using the chain rule and trigonometric identities. The critical points are the values of θ where the derivative is 0 or undefined.
Using the chain rule, the derivative of sec(θ) is sec(θ)tan(θ) and the derivative of tan(θ) is sec2(θ). The derivative of 2sec(θ)+tan(θ) is therefore 2sec(θ)tan(θ)+sec2(θ).
Setting the derivative equal to 0 and solving for θ, we have 2sec(θ)tan(θ)+sec2(θ) = 0. This equation can be rearranged to sec(θ)(2tan(θ)+sec(θ)) = 0. The critical points occur when sec(θ) = 0 or when 2tan(θ)+sec(θ) = 0.
For option a) θ = π/4, we have sec(π/4) = 1/√2 and tan(π/4) = 1, so 2tan(π/4)+sec(π/4) = 2+1/√2, which is not equal to 0. Therefore, θ = π/4 is not a critical point.
For option b) θ = 3π/4, we have sec(3π/4) = -1/√2 and tan(3π/4) = -1, so 2tan(3π/4)+sec(3π/4) = -2-1/√2, which is not equal to 0. Therefore, θ = 3π/4 is not a critical point.
For option c) θ = π/2, we have sec(π/2) = ∞ and tan(π/2) = ∞, so 2tan(π/2)+sec(π/2) = ∞+∞, which is undefined. Therefore, θ = π/2 is a critical point.
For option d) θ = π, we have sec(π) = -1 and tan(π) = 0, so 2tan(π)+sec(π) = 0. Therefore, θ = π is a critical point.