`y = x - 2cos(x), 0<x<2pi` Identify the open intervals on which the function is increasing or decreasing.

Textbook Question

Chapter 3, 3.3 - Problem 15 - Calculus of a Single Variable (10th Edition, Ron Larson).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to determine the intervals on which the function increases or decreases, hence, you need to determine where the first derivative is positive or negative.

You need to evaluate the first derivative, hence you need to differentiate with respect to x, such that:

`(dy)/(dx) = 1 + 2sin x`

You need to solve for x the equation `(dy)/(dx) = 0` , such that:

` 1 + 2sin x = 0 => 2sin x = -1 => sin x = -1/2`

You need to remember that the sine function is negative over the interval (pi,2pi).

`sin x = -1/2 => x = pi + pi/6 => x = 7pi/6`

`sin x = -1/2 => x = 2pi - pi/6 => x = 11pi/6`

Hence, the function is decreasing over` (7pi/6, 11pi/6)` and it is increasing over `(0,2pi)-(7pi/6, 11pi/6).`

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