# 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!!

*print*Print*list*Cite

### 1 Answer

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)` .**