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

Textbook Question

Chapter 3, 3.3 - Problem 14 - 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 is increasing or decreasing, hence, you need to find out intervals on which the first derivative is positive or negative.

You need to determine the first derivative, using the chain rule, such that:

`f'(x) = (cos(x/2))' => f'(x) = -(sin(x/2))*(x/2)'`

`f'(x) = -(sin(x/2))/2`

Now, you need to solve for x the equation f'(x) = 0:

`-(sin(x/2))/2 = 0 => (sin(x/2)) = 0`

You need to remember that the sine function is equal to 0, in `(0,2pi), ` at `x/2 = pi.`

`x = 2pi`

Hence, the derivative is negative for `x/2 in (pi,2pi)` and it is positive for` x/2 in (0,pi).`

Hence, the function increases for `x in (0,2pi)` and it decreases for (2pi,4pi), but, since the behavior of the function is analyzed over (0,2pi), then the function increases over `(0,2pi).`

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