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

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Feb 9, 2018

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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]}

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