find the absolute maximum and minimum for f(x) = x+ sin(x) on the closed interval [0, 2pi]

Answer 1

f' = 1 + cos(x)
f'=0 when cos(x) = -1, or x=pi

however,
f''(x) = -sin(x) which is also 0 at x=pi.

So, (pi,pi) is an inflection point, not a max or min.

f(2pi) = 2pi

so that is the absolute max on [0,2pi]

Note that f'(x) >= 0 for all x, so there is no max or min, which require changing direction.