Find all extrema in the interval 0 2pi if y=x+sinx

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.