How do you find the critical points of #y=x-2sinx# on the interval #[0, pi/2]#?

Question

William Abdalla · Accepted Answer

There is one critical point of #y=x-2sinx# on the interval #[0,pi/2]# at #(pi/3, pi/3 - sqrt3) = (1.047, -.685)#

The critical points of a function occur where the derivative of the function is zero on the specified interval.

Find the derivative of #y#:
#y=x-2sinx#

#dy/dx = 1 - 2cosx#

Set the derivative, #dy/dx# equal to zero:
#0=1-2cosx#

#2cosx = 1#

#cosx = 1/2#

Within the interval from #[0,pi/2]#, picture the unit circle, and where is #"cosine of x"# equal to #1/2#

#x=pi/3#

#x approx 1.047#

Therefore, there is only one critical point of #y# on the provided interval, and this is where #x=pi/3#.

The #y# value at this point is:

#y=x-2sinx#

#y=frac{pi}{3} - 2sin(pi/3)#

#y= pi/3 - 2(sqrt3 / 2)#

#y = pi/3 - sqrt3#

#y approx -.685#

We can check our answer graphically. The slope changes from negative to positive at approximately 1.05.
graph{x-2sinx [-10, 10, -5, 5]}

Elijah Abram · Answer

undefined To find the critical points of $ y = x - 2 \sin(x) $ on the interval $ [0, \frac{\pi}{2}] $, we need to first find the derivative of the function and then set it equal to zero to solve for critical points.

The derivative of the function is $ y' = 1 - 2 \cos(x) $.

Setting the derivative equal to zero and solving for $ x $, we get:

$ 1 - 2 \cos(x) = 0 $

$ 2 \cos(x) = 1 $

$ \cos(x) = \frac{1}{2} $

This equation is satisfied when $ x = \frac{\pi}{3} $ and $ x = \frac{5\pi}{3} $ within the interval $ [0, \frac{\pi}{2}] $.

So, the critical points of $ y = x - 2 \sin(x) $ on the interval $ [0, \frac{\pi}{2}] $ are $ x = \frac{\pi}{3} $ and $ x = \frac{5\pi}{3} $.