`f(x) = x + 2sin(x)` Consider the function on the interval (0, 2pi). Find the open intervals on which the function is increasing or decreasing, apply first derivative test to identify all relative extrema.
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.