This function is differentiable on `(0, 2pi).` It is increasing where its derivative is positive and is decreasing when the derivative is negative.

`f'(x) = 1 + 2cos(x),` its roots on `(0, 2pi)` are `(2pi)/3` and `(4pi)/3.`

f'(x) is negative between these roots and is positive outside. Therefore f(x) is...

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This function is differentiable on `(0, 2pi).` It is increasing where its derivative is positive and is decreasing when the derivative is negative.

`f'(x) = 1 + 2cos(x),` its roots on `(0, 2pi)` are `(2pi)/3` and `(4pi)/3.`

f'(x) is negative between these roots and is positive outside. Therefore f(x) is increasing on `(0, (2pi)/3)` and on `((4pi)/3, 2pi)` and f(x) is decreasing on `((2pi)/3, (4pi)/3).` And this means `(2pi)/3` is a relative maximum and `(4pi)/3` is a relative minimum.