Find the critical numbers of the function on the interval 0 ≤ θ < 2π.
f(θ) = 2cos(θ) + (sin(θ))^2
Critical point:
A critical point at the interior of the domain of a function is the point where the derivative is zero or undefined.
f(θ) = 2cos(θ) + (sin(θ))^2
does not have any undefined points for θ∈ℝ.
So we only need to find the values of θ which make f'(θ)=0.
Differentiate f(θ) and equate to zero. Solve for all roots for the equation and these are the critical points.
The following graph might help you check you answer:
http://img256.imageshack.us/img256/11/1291497419.png
solve on the interval of 0,2π, cos θ / 3 =-1
To find the critical numbers of the function f(θ) = 2cos(θ) + (sin(θ))^2 on the interval 0 ≤ θ < 2π, we need to find the values of θ where the derivative of the function is equal to zero or undefined.
First, let's find the derivative of f(θ) with respect to θ:
f'(θ) = -2sin(θ) + 2sin(θ)cos(θ)
To find the critical numbers, we need to solve the equation f'(θ) = 0:
-2sin(θ) + 2sin(θ)cos(θ) = 0
Factoring out 2sin(θ), we get:
2sin(θ)(cos(θ) - 1) = 0
This equation is satisfied when sin(θ) = 0 or cos(θ) - 1 = 0.
For sin(θ) = 0, the solutions on the interval 0 ≤ θ < 2π are θ = 0 and θ = π.
For cos(θ) - 1 = 0, the solution is cos(θ) = 1, which is satisfied when θ = 0.
Therefore, the critical numbers of the function on the interval 0 ≤ θ < 2π are θ = 0 and θ = π.
To find the critical numbers of the function f(θ) = 2cos(θ) + (sin(θ))^2 on the interval 0 ≤ θ < 2π, we need to find the values of θ where the derivative of the function is equal to zero or does not exist.
Step 1: Find the derivative of f(θ). Let's call it f'(θ).
f'(θ) = -2sin(θ) + 2sin(θ)cos(θ)
Step 2: Set f'(θ) = 0 and solve for θ.
-2sin(θ) + 2sin(θ)cos(θ) = 0
Factor out 2sin(θ):
2sin(θ)(cos(θ) - 1) = 0
Now, set each factor equal to zero:
2sin(θ) = 0 or cos(θ) - 1 = 0
For 2sin(θ) = 0, we have sin(θ) = 0. This occurs when θ = 0 or θ = π.
For cos(θ) - 1 = 0, we have cos(θ) = 1. This occurs when θ = 0.
So, the critical numbers of the function on the interval 0 ≤ θ < 2π are θ = 0 and θ = π.