Answer:
y = x + sin(x)
y' = 1 + cos(x)
Setting y' to zero, we have:
y' = 0
1 + cos(x) = 0
cos(x) = -1
x = pi, on the interval [0, 2pi]
y'' = -sin(x)
When x = pi, y'' = -sin(pi) = 0
Thus, we have an extremum at x = pi, but it is neither a local maxima nor a local minima.
Notice as well that y' = 1 + cos(x) >= 1 for all real values of x.
Thus, y is an increasing function.
This implies that on the interval [0, 2pi], the absolute minima is at x = 0, where y = 0 + sin(0) = 0; and the absolute maxima is at x = 2pi, where y = 2pi + sin(2pi) = 2pi.
