Find the critical numbers of the function on the interval 0 ≤ θ < 2π. Small & large value
f(θ) = 2cos(θ) + (sin(θ))^2
let's take f '(Ø)
f '(Ø) = -2sinØ + 2sinØcosØ
= 0 for max/mins
sinØ - sinØcosØ = 0
sinØ(1 + cosØ) = 0
sinØ = 0 or cosØ = -1
Ø = 0,π,2π or Ø = 3π/2
f(0) = 2+0 = 2
f(π) = -2+0 = -2
f(2π) = 2 + 0 = 2
f(3π/2) = 0 + 1 = 0
max value of the function is 2 when Ø = 0 or 2π
min value of the function is -2 when Ø = 3π/2
2sinθcosθ-1=0
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 f(θ) is equal to zero or does not exist.
Step 1: Find the derivative of f(θ) with respect to θ.
f'(θ) = -2sin(θ) + 2sin(θ)cos(θ)
Step 2: Set f'(θ) equal to zero and solve for θ.
-2sin(θ) + 2sin(θ)cos(θ) = 0
2sin(θ)(cos(θ) - 1) = 0
Step 3: Solve for the values of θ that make f'(θ) equal to zero.
sin(θ) = 0 or cos(θ) - 1 = 0
For sin(θ) = 0, the solutions are θ = 0, π, 2π.
For cos(θ) - 1 = 0, the solutions are cos(θ) = 1, which occurs when θ = 0, 2π.
Step 4: Determine if the derivative is undefined at any point in the interval.
The derivative f'(θ) is defined for all values of θ in the interval 0 ≤ θ < 2π. Therefore, there are no values of θ where f'(θ) does not exist.
Step 5: Gather all the critical numbers.
The critical numbers of the function f(θ) = 2cos(θ) + (sin(θ))^2 on the interval 0 ≤ θ < 2π are θ = 0, π, 2π.
Now, let's find the small and large values of the function f(θ) at these critical numbers.
For θ = 0:
f(0) = 2cos(0) + (sin(0))^2 = 2(1) + 0 = 2.
For θ = π:
f(π) = 2cos(π) + (sin(π))^2 = 2(-1) + 0 = -2.
For θ = 2π:
f(2π) = 2cos(2π) + (sin(2π))^2 = 2(1) + 0 = 2.
Therefore, the small value of f(θ) on the interval 0 ≤ θ < 2π is -2, and the large value is 2.
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 at which the derivative of the function is equal to zero or does not exist.
Step 1: Find the derivative of f(θ).
f'(θ) = -2sin(θ) + 2sin(θ)cos(θ)
Step 2: Set the derivative equal to zero and solve for θ.
-2sin(θ) + 2sin(θ)cos(θ) = 0
Step 3: Factor out sin(θ).
sin(θ)(-2 + 2cos(θ)) = 0
Step 4: Set each factor equal to zero and solve for θ.
a) sin(θ) = 0
This occurs when θ is 0 or π (or any integer multiple of π), since sin(0) = 0 and sin(π) = 0.
b) -2 + 2cos(θ) = 0
Adding 2 to both sides, we get:
2cos(θ) = 2
Dividing both sides by 2, we have:
cos(θ) = 1
This occurs when θ is 0 (or any integer multiple of 2π), since cos(0) = 1.
Therefore, the critical numbers of the function f(θ) = 2cos(θ) + (sin(θ))^2 on the interval 0 ≤ θ < 2π are:
- θ = 0, π, and any integer multiple of 2π.
To find the small and large values of the function on this interval, we can evaluate the function at the critical numbers as well as the endpoints of the interval.
f(0) = 2cos(0) + (sin(0))^2 = 2(1) + (0)^2 = 2
f(π) = 2cos(π) + (sin(π))^2 = 2(-1) + (0)^2 = -2
f(2π) = 2cos(2π) + (sin(2π))^2 = 2(1) + (0)^2 = 2
Therefore, the small value of the function is -2 and the large value is 2 on the interval 0 ≤ θ < 2π.