To solve this question, we need to find where the derivative of the function equals to zero (critical points), and then evaluate the function at these points and at the end points of the interval.
We first find the derivative of g(θ) = 7θ - 8sin(θ).
Using the rules of derivative, we know the derivative of θ is 1 and the derivative of sin(θ) is cos(θ). Hence, the derivative g'(θ) = 7 - 8 cos(θ).
Setting this equal to zero to find the critical points, we get 7 - 8 cos(θ) = 0. From this, we can solve for θ.
We now have a list of critical points. Recall our original interval was [0, 2π]. Because these are the boundaries of our allowed values for θ, we must also include 0 and 2π in our list of numbers to check.
To obtain the global maximum and minimum values, we can plug each of these points into the original function g(θ) and evaluate.
g(0) = 7(0) - 8[sin(0)] = 0
g(2π) = 7(2π) - 8[sin(2π)] = 14π
Plus the value of the function at the critical points. As a result, a list of function values has been obtained.
The global maximum and minimum of our function on the interval [0,2π] will simply be the largest and smallest number in this list we produced.
The global minimum value of the function is approximately -0.335 and the global maximum value is approximately 44.318 to the nearest thousandth.