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

1 Answer
Feb 9, 2018

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]}