Question

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

Answer 1

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`

`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]}