Find the critical numbers of the function. (Enter your answers as a comma-separated list. Use n to denote any arbitrary integer values. If an answer does not exist, enter DNE.)
f(θ) = 4 cos θ + 2 sin^2 θ
f = 4cosθ + 2sin^2θ
= 4cosθ + 2 - 2cos^2θ
f' = -4sinθ + 4cosθsinθ
critical numbers are when f'=0, so
4sinθ(cosθ-1) = 0
so, sinθ=0 or cosθ=1
...
To find the critical numbers of the function f(θ) = 4 cos θ + 2 sin^2 θ, we need to find the values of θ for which the derivative of f(θ) equals zero or is undefined.
First, let's find the derivative of f(θ) with respect to θ:
f'(θ) = -4 sin θ + 4 sin θ cos θ
To find the critical numbers, we set f'(θ) equal to zero and solve for θ:
-4 sin θ + 4 sin θ cos θ = 0
Factoring out sin θ:
sin θ (-4 + 4 cos θ) = 0
Setting each factor equal to zero gives us two equations:
sin θ = 0 or -4 + 4 cos θ = 0
For sin θ = 0, the solutions are θ = n * π, where n is an integer.
For -4 + 4 cos θ = 0, we can solve for cos θ:
4 cos θ = 4
cos θ = 1
The only solution to cos θ = 1 is θ = 2nπ, where n is an integer.
Combining the solutions, the critical numbers of the function f(θ) = 4 cos θ + 2 sin^2 θ are θ = n * π and θ = 2nπ.
To find the critical numbers of the function f(θ) = 4 cos θ + 2 sin^2 θ, we need to find the values of θ where the derivative of f is equal to zero or does not exist.
Step 1: Find the derivative of f(θ) with respect to θ.
f'(θ) = -4 sin θ + 4 sin θ cos θ
Step 2: Set the derivative equal to zero and solve for θ.
-4 sin θ + 4 sin θ cos θ = 0
Factor out sin θ:
-4 sin θ (1 - cos θ) = 0
This equation is satisfied when either sin θ = 0 or (1 - cos θ) = 0.
For sin θ = 0, we have θ = nπ, where n is an integer.
For (1 - cos θ) = 0, we solve for θ as follows:
1 - cos θ = 0
cos θ = 1
This equation is satisfied when θ = 2nπ, where n is an integer.
Combining both solutions, the critical numbers of f(θ) are θ = nπ and θ = 2nπ, where n is an integer.
Therefore, the critical numbers of the function f(θ) = 4 cos θ + 2 sin^2 θ are θ = nπ and θ = 2nπ, where n is an integer.