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]#?

2 Answers
Mar 16, 2015

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#.

Mar 16, 2015

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)#