Answer:
To find the maximum and minimum values of the function g(θ) = 2θ - 6sin(θ) on the interval 0, π, we need to find the critical points and then evaluate the function at these points as well as at the endpoints of the interval.
First, we find the derivative of g(θ):
g'(θ) = 2 - 6cos(θ)
Setting g'(θ) = 0 to find the critical points:
0 = 2 - 6cos(θ)
6cos(θ) = 2
cos(θ) = 2/6
cos(θ) = 1/3
We can solve for θ in the interval 0, π using inverse cosine:
θ = arccos(1/3) ≈ 1.23 radians
To check if this is a maximum, minimum, or neither, we can use the second derivative test.
g''(θ) = 6sin(θ)
Since g''(θ) > 0 for 0 < θ < π, we confirm that θ = arccos(1/3) is a local minimum.
Now, we need to evaluate g(θ) at the critical point and the endpoints of the interval:
g(0) = 0
g(π) = 2π
Therefore, the minimum value occurs at θ = arccos(1/3) and the maximum value occurs at θ = π.
The minimum value is:
g(arccos(1/3)) ≈ 1.85
The maximum value is:
g(π) = 2π ≈ 6.28
So,
Minimum value ≈ 1.85
Maximum value ≈ 6.28
