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

Question

Emilia Aguilar · Accepted Answer

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]# 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# Thus, we now possess 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# Therefore, the function must be decreasing for all other possible values. The function is decreasing #iff x in RR - [pi/6+2pi k, 5 pi/6 + 2 pi k]# The function's graph, graph{y=x/2 + cos(x) [-8.21, 10.14, -3.56, 5.61]}, illustrates this. Graph{y=1/2 - sin(x) [-8.21, 10.14, -3.56, 5.61]} shows the derivative.

Ezra Adams · Answer

undefined To find the intervals of increasing and decreasing for the function $ y = \frac{x}{2} + \cos(x) $, follow these steps:

1. Find the first derivative of the function.
2. Set the first derivative equal to zero and solve for $ x $. These are the critical points.
3. Test the intervals between the critical points by using test points in the first derivative.
4. If the first derivative is positive in an interval, the function is increasing in that interval. If the first derivative is negative, the function is decreasing.

Let's go through the steps:

1. First derivative of $ y = \frac{x}{2} + \cos(x) $:
   $ y' = \frac{1}{2} - \sin(x) $

2. Set $ y' = 0 $ and solve for $ x $:
   $ \frac{1}{2} - \sin(x) = 0 $
   $ \sin(x) = \frac{1}{2} $
   Solving for $ x $, we get $ x = \frac{\pi}{6} + 2\pi n $ and $ x = \frac{5\pi}{6} + 2\pi n $, where $ n $ is an integer.

3. Test intervals:
   Choose test points between the critical points and plug them into $ y' = \frac{1}{2} - \sin(x) $.
   For example, if we choose $ x = 0 $, $ \sin(0) = 0 $, so $ y' = \frac{1}{2} $, which is positive. Therefore, the function is increasing on the interval $ (-\infty, \frac{\pi}{6}) $.

4. Analyze the intervals:
   - The function is increasing on $ (-\infty, \frac{\pi}{6}) $ and $ (\frac{5\pi}{6}, \infty) $.
   - The function is decreasing on $ (\frac{\pi}{6}, \frac{5\pi}{6}) $.

These intervals represent where the function $ y = \frac{x}{2} + \cos(x) $ is increasing and decreasing, respectively.