Find the critical numbers of f, find the open intervals on which the function is increasing or decreasing and use the first derivative test to identify all relative extremes

f(x) = 3 + cos^2 x 0<x<2pie

To find the critical numbers of the function f(x) = 3 + cos^2(x), we need to find the values of x where the derivative of the function is equal to 0 or undefined.

Step 1: Find the derivative of f(x):
f'(x) = -2sin(x)cos(x)

Step 2: Set f'(x) equal to 0 and solve for x:
-2sin(x)cos(x) = 0

To find the critical numbers, we need to solve for x in the equation -2sin(x)cos(x) = 0. Since sin(x) and cos(x) can both be equal to 0 or |sin(x)| and |cos(x)| can equal 1, we have the following possibilities:

1. sin(x) = 0 and cos(x) ≠ 0: This occurs at x = 0 and x = π.

2. sin(x) ≠ 0 and cos(x) = 0: This occurs at x = π/2.

So, the critical numbers of f(x) are x = 0, x = π/2, and x = π.

Next, we need to determine the intervals on which the function is increasing or decreasing. To do this, we can use the first derivative test:

Step 3: Choose test points in each interval and evaluate the sign of the derivative at those points.

We divide the interval (0, 2π) into three subintervals: (0, π/2), (π/2, π), and (π, 2π).

For the interval (0, π/2), we can choose the test point x = π/4.
Evaluating the derivative at x = π/4:
f'(π/4) = -2sin(π/4)cos(π/4) = -(√2/2)(√2/2) = -1/2
Since f'(x) is negative in this interval at the test point x = π/4, f(x) is decreasing on (0, π/2).

For the interval (π/2, π), we can choose the test point x = 3π/4.
Evaluating the derivative at x = 3π/4:
f'(3π/4) = -2sin(3π/4)cos(3π/4) = -(-√2/2)(-√2/2) = -1/2
Since f'(x) is negative in this interval at the test point x = 3π/4, f(x) is decreasing on (π/2, π).

For the interval (π, 2π), we can choose the test point x = 5π/4.
Evaluating the derivative at x = 5π/4:
f'(5π/4) = -2sin(5π/4)cos(5π/4) = -(-√2/2)(√2/2) = 1/2
Since f'(x) is positive in this interval at the test point x = 5π/4, f(x) is increasing on (π, 2π).

Now we can use the first derivative test to identify the relative extrema.

Since f'(x) changes sign from negative to positive at x = π, it indicates a relative minimum at x = π.

Therefore, the critical numbers are x = 0, x = π/2, and x = π, and the relative minimum is at x = π.

To summarize:
- The critical numbers of f(x) = 3 + cos^2(x) are x = 0, x = π/2, and x = π.
- The function is decreasing on the interval (0, π/2) and (π/2, π).
- The function is increasing on the interval (π, 2π).
- The relative minimum is at x = π.