`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.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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...

Unlock
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Start your 48-Hour Free Trial

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.

Approved by eNotes Editorial Team