# 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]?

Mar 16, 2015

$A \left(\frac{\pi}{3} , \frac{\pi}{3} - \sqrt{3}\right)$ absolute minimum;

$B \left(- \frac{\pi}{4} , - \frac{\pi}{4} + 2 \sqrt{2}\right)$ absolute maximum.

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

$y ' = 1 - 2 \cos x$

$y ' \ge 0 \Leftrightarrow 1 - 2 \cos x \ge 0 \Leftrightarrow \cos x \le \frac{1}{2} \Leftrightarrow \frac{\pi}{3} \le x \le \frac{\pi}{2}$

So the function grows in $\frac{\pi}{3} \le x \le \frac{\pi}{2}$

and decreases in $- \frac{\pi}{4} \le x \le \frac{\pi}{3}$, as you can see from the graph:

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

So the point $A \left(\frac{\pi}{3} , \frac{\pi}{3} - \sqrt{3}\right)$ 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:

$B \left(- \frac{\pi}{4} , - \frac{\pi}{4} + 2 \sqrt{2}\right)$ and $C \left(\frac{\pi}{2} , \frac{\pi}{2} - 2\right)$.

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

Mar 16, 2015

First of all, let's recall how the method works: if you have a continuous function (which is $f \left(x\right) = x - 2 \setminus \sin \left(x\right)$), and a closed and bounded interval (which is $\left[- \setminus \frac{\pi}{4} , \setminus \frac{\pi}{2}\right]$), 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 $f ' \left(x\right) = 0$.

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

$D f = D x - D \left(2 \setminus \sin \left(x\right)\right)$

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

$D f = D x - 2 D \setminus \sin \left(x\right)$

These are both elementary derivatives, and we have $D x = 1$, and $D \setminus \sin \left(x\right) = \setminus \cos \left(x\right)$. So,

$D f = f ' = 1 - 2 \setminus \cos \left(x\right)$

This function has zeroes if and only if $\setminus \cos \left(x\right) = \frac{1}{2}$. The only point of the domain in which this happens is $\setminus \frac{\pi}{3}$.

The only thing left is thus to compare the values of the function in $- \setminus \frac{\pi}{4}$, $\setminus \frac{\pi}{3}$ and $\setminus \frac{\pi}{2}$. 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:

$f \left(- \setminus \frac{\pi}{4}\right) = \setminus \sqrt{2} - \setminus \frac{\pi}{4} = 0.628 \ldots$
$f \left(\setminus \frac{\pi}{3}\right) = \setminus \frac{\pi}{3} - \setminus \sqrt{3} = - 0.684 \ldots$
$f \left(\setminus \frac{\pi}{2}\right) = \setminus \frac{\pi}{2} - 2 = - 0.429 \ldots$

And so the minimum value of $f$ is $f \left(\setminus \frac{\pi}{3}\right)$, while the maximum value is $f \left(- \setminus \frac{\pi}{4}\right)$