Question

How do you use the closed-interval method to find the absolute maximum and minimum values of the function `f(x)=x-2sinx` on the interval `[-π/4, π/2]`?

Answer 1

The answer is:

`A(pi/3,pi/3-sqrt3)` absolute minimum;

`B(-pi/4,-pi/4+2sqrt2)` absolute maximum.

First of all, let's see if there are some local maximum o minimum in that interval.

`y'=1-2cosx`

`y'>=0hArr1-2cosx>=0hArrcosx<=1/2hArrpi/3<=x<=pi/2`

So the function grows in `pi/3<=x<=pi/2`

and decreases in `-pi/4<=x<=pi/3`, as you can see from the graph:

graph{x-2sinx [-4.93, 4.94, -2.465, 2.466]}

So the point `A(pi/3,pi/3-sqrt3)` is a local minimum and there not exists a local maximum.

Now we have to calculate the ordinate of both the extremes of the interval, because the absolute maximum and minimum could be in that points.

The points are:

`B(-pi/4,-pi/4+2sqrt2)` and `C(pi/2,pi/2-2)`.

So the absolute maximum is `B` and the absolute minimum is `A`.

Answer 2

First of all, let's recall how the method works: if you have a continuous function (which is `f(x)=x-2\sin(x)`), and a closed and bounded interval (which is `[-\pi/4,\pi/2]`), then the function will have global maximum and minimum. This two points can either be one of the ends of the interval, or a critical, internal point. A point `x` is said to be critical for the function `f` if you have `f'(x)=0`.

So, first of all, let's derive the function: since the derivative of a sum is the sum of the derivatives, you have that

`Df= Dx - D(2\sin(x))`

Now, let's recall that we can factor out constants:

`Df= Dx-2D\sin(x)`

These are both elementary derivatives, and we have `Dx=1`, and `D\sin(x)=\cos(x)`. So,

`Df=f'=1-2\cos(x)`

This function has zeroes if and only if `\cos(x)=1/2`. The only point of the domain in which this happens is `\pi/3`.

The only thing left is thus to compare the values of the function in `-\pi/4`, `\pi/3` and `\pi/2`. Out of these three values, the smaller will be the global minimum, and the largest will be the global maximum.

Some easy computations show that:

`f(-\pi/4)=\sqrt(2)-\pi/4=0.628...`
`f(\pi/3)=\pi/3-\sqrt(3)=-0.684...`
`f(\pi/2)=\pi/2-2=-0.429...`

And so the minimum value of `f` is `f(\pi/3)`, while the maximum value is `f(-\pi/4)`