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

1 Answer
Mar 2, 2018

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

Explanation:

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
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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]}