Question

How do you find the intervals of increasing and decreasing using the first derivative given `y=x/2+cosx`?

Answer 1

The function is increasing `iff x in (pi/6+2pi k, 5 pi/6 + 2 pi k)`
The function is constant `iff x = pi/6 + 2 pi k` or `x = 5pi/6 + 2pi k`
The function is decreasing `iff x in RR - [pi/6+2pi k, 5 pi/6 + 2 pi k]`

Explanation:

If `y=x/2+cos(x)` then `y' = 1/2 - sin(x)`

if `y'>0` then the function at that point is increasing.
If `y'<0` then the function at that point is decreasing.
If `y'=0` then the function at that point is constant.

The function `sin(x)` has a periodic behavior.

Lets construct a values table for the function `sin(x)`
`x=0=>sin(x)=0`
`x=pi/6=>sin(x)=1/2`
`x=pi/4=>sin(x)=sqrt(3)/2`
`x=pi/2=>sin(x)=1`

We also know that `sin(-x)=-sin(x)` and `sin(pi-theta)=sin(theta)`

To the derivative has a positive value, we must have `sin(x)>1/2`.

To have this we must have `pi-pi/6 > x > pi/6`
That means exactly the same as `5 pi / 6 > x > pi/6`

Because `sin(x)` is periodical, we could have

`5pi/6 + 2 pi k > x > pi/6 + 2 pi k, k in ZZ`

The points where the derivative has the exactly value of zero is when `x=pi/6 + 2 pi k" or " x = 5pi/6 + 2 pi k`

So, now we have that:

For any `k in ZZ` it's true that:
The function is increasing `iff x in (pi/6+2pi k, 5 pi/6 + 2 pi k)`
The function is constant `iff x = pi/6 + 2 pi k` or `x = 5pi/6 + 2pi k`

So, we must have the function decreasing in all the other possible values

The function is decreasing `iff x in RR - [pi/6+2pi k, 5 pi/6 + 2 pi k]`

We can see this in the graph of the function:
graph{y=x/2 + cos(x) [-8.21, 10.14, -3.56, 5.61]}

And here is the derivative:
graph{y=1/2 - sin(x) [-8.21, 10.14, -3.56, 5.61]}