Find extrema on interval [0, 2Pi]?

Question

I need to state all extrema of y=x+sinx in the interval [0, 2Pi]

I think the first derivative f'(x)= 1+cosx and that the function f(x) has a relative maximum at (Pi, Pi)

The function f'(x) has maxima at (0, 2) (2Pi, 2) and minima at (Pi, 0) I think.

Answers I can choose from:
a. (-1,-1+(3Pi/2), (0, 0)
b. (2Pi, 2Pi), (0, 0)
c. (2Pi, 2Pi), (Pi, Pi)
d. (Pi, Pi), (0, 0)
e. none of these

It's not a I think because that point lies outside the limit. Do I need to look at the graph of f'(x) or f(x)??

Answer 1

f'(x) = 0 at x = pi
Notice that f''(x) = - sin(x) is 0 at x = pi, so (pi, pi) is a point of inflection. The function goes from being concave down to concave up. You can confirm this by looking at f'(x). f'(x) > 0 excect at x = pi, so it is increasing over the interval [0, 2pi] except at x = pi where it flattens out temporarily.

That says the min is going to be at (0, 0) and the max is going to be at (2pi, 2pi).

Answer 2

f(x)=1/2x^2-2x on interval [0,5)