# Use the closed-interval method to find the absolute maximum and minimum values of the function f(x)=x-2sinx on the interval [-pi/4, pi/2]I understand that I need to identify the end points and find...

Use the closed-interval method to find the absolute maximum and minimum values of the function f(x)=x-2sinx on the interval [-pi/4, pi/2]

I understand that I need to identify the end points and find my critical numbers but I am having a hard time going about this. I found the derivate to be f'(x)=0-2cos x, is this correct?

Could you please show work and how to go about it, instead of just the answer please! Thank you in advance!!

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To use the closed interval method, we need to find the values of the function at the endpoints and also where the derivative of the function equals zero. The largest y-value of these is the absolute maximum in the interval and the lowest y-value is the absolute minimum in the interval.

The values of the endpoints are:

`f(-pi/4)=-pi/4-2sin(-pi/4)=-pi/4+2/sqrt2 approx 0.63`

`f(pi/2)=pi/2-2sin(pi/2)=pi/2-2 approx -0.43`

To get the derivative, we use the power rule and the derivative of `cosx` to get:

`f(x)=x-2sinx`

`f'(x)=1-2cos x` which is zero when

`2cosx=1` divide by 2

`cosx=1/2`

which happens at `x=pi/3` in the interval given.

We need to evaluate the function at this point to get:

`f(pi/3)=pi/3-2sin(pi/3)=pi/3-2sqrt3 / 2 =pi/3-sqrt3 approx -0.68`

**This means the absolute maximum is at `(-pi/4,-pi/4+2/sqrt2)` and the absolute minimu is at `(pi/3,pi/3-sqrt3)` .**