Final answer:
To find the local extrema of the function f(x) = (√3/2)x - (4sin(4x))/4 on the interval 0≤x≤2π, take the derivative of the function and solve for when it equals 0. The local extrema occur at x = π/12, x = 7π/12, and at the endpoints 0 and 2π. The behavior of f in relation to the signs and values of f' can be determined by examining the graph of the function and its derivative.
Step-by-step explanation:
a. Find the local extrema of the function f(x) = (√3/2)x - (4sin(4x))/4 on the interval 0≤x≤2π, and say where they occur.
To find the local extrema of the function, we need to take the derivative of the function and solve for when it equals 0. Taking the derivative of f(x) gives us f'(x) = (√3/2) - 4cos(4x).
Setting f'(x) = 0, we get (√3/2) - 4cos(4x) = 0. Solving for x, we find x = π/12 and x = 7π/12 as the critical points.
Next, we need to check the endpoints of the interval, 0 and 2π. Evaluating the function at these endpoints gives f(0) = 0 and f(2π) = 2√3 - 2√3 = 0.
So the local extrema occur at x = π/12, x = 7π/12, and at the endpoints 0 and 2π.
b. Graph the function and its derivative together. Comment on the behavior of f in relation to the signs and values of f'.
Here is a graph of the function f(x) = (√3/2)x - (4sin(4x))/4 and its derivative f'(x) = (√3/2) - 4cos(4x): [insert graph].
From the graph, we can see that whenever the derivative f'(x) is positive, the function f(x) is increasing. Whenever the derivative f'(x) is negative, the function f(x) is decreasing. The points where the derivative f'(x) is equal to 0 represent the local extrema of the function.