Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 5

You need to evaluate the extreme values of the function on the interval [-pi,pi], hence, you need to evaluate the zeroes of the function f'(x).
You need to find the derivative of the function:
f'(x) = 1-2sin x
You need to solve for x the equation f'(x) = 0, such that:
1 - 2sin x = 0 => 2sin x = 1 => sin x = 1/2 => x = pi/6 and x = 5pi/6
You need to evaluate the values of the function at critical points x = pi/6 and x = 5pi/6.
f(pi/6) = pi/6+ 2cos(pi/6) = pi/6 + sqrt3
Since 5pi/6 = pi - pi/6 and 2cos(5pi/6) = -2cos(pi/6)
f(5pi/6) = pi - pi/6 - 2cos(pi/6) =pi - pi/6 -sqrt 3
You need to evaluate the function at the end points:
f(pi) = pi + 2cos pi = pi - 2
f(-pi) = - pi + 2cos(-pi) = -pi - 2
Hence, evaluating the absolute maximum of the function, on interval [-pi,pi], yields pi/6+sqrt3 and it occurs at x = pi/6 and evaluating the minimum of the function yields -pi-2 and it occurs at x = -pi.