`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...

`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.

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Chapter 3, 3.3 - Problem 44 - Calculus of a Single Variable (10th Edition, Ron Larson).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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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.

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