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

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.

Im not that good with division but i think 5 is -6but dont quote me on thatim not the best in division but thats my answer