Finding Global Maximum and Minimum on a Given Interval

[tunabeast]

[SOLVED] Maximum and minimum problem

Homework Statement

Determine the global maximum and global minimum of the following func-
tions on the given interval, or indicate why these do not exist.

f(x) = $$x^{2}e^{x}$$ on [2, 1].

Homework Equations

The Attempt at a Solution

I do not understand the way the lecturer solves this kind of question. I am used to finding values of f'(x) and then substituting these into f''(x) to find the local maxima/minima, he is using some kind of number scale which is hard to illustrate. As for the actual solution i think there is a local maximum at x=0, and a minimum at x=-2.

 

[sokratesla]

As for the actual solution i think there is a local maximum at x=0, and a minimum at x=-2.

# You found the answer by yourself. The question is whether there are any extremum points in the given interval (from 1 to 2) and the points you found are not in the given interval.

 

[tunabeast]

Just reading over the question i realize i made a mistake, the interval is in fact [-2, 1]. Do i need to substitute all of the numbers in the interval to the equation, ie. -2,-1,0,1?

 

[sokratesla]

Just reading over the question i realize i made a mistake, the interval is in fact [-2, 1]. Do i need to substitute all of the numbers in the interval to the equation, ie. -2,-1,0,1?

# OK, then. As you said the derivative of the function is 0 at x=0, and x=-2 and they are in the interval [-2,1]. By further calculating the second derivative or just plotting, or sketching the function, you can see whether these points are minima or maxima (or saddle points etc). And f(0) and f(-2) are the numbers you want.
# You can not substitute all of the numbers in the interval to the function if the numbers are real. There are infinite numbers :-) Differentiation is devised for this purpose.

 

[tunabeast]

Thanks for your help :)

 

[HallsofIvy]

A continuous function always has "global" minimum and maximum on a closed and bounded interval! Such maximum and minimum must a one of these kinds of places:
1. In the interior of the interval where the derivative is 0.
2. In the interior of the interval where the derivative does not exist.
3. At one of the endoints.

Here you have found that
1. The derivative is 0 at -2 and 0.
2. The derivative always exists.
3. The end points are -2 and 1.

The maximum and minimum must occur at x= -2, or x= 0, or x= 1. I suspect that is the "kind of number scale" your instructor is using.

I wondered why you had included x= 1 in that. Surely you are not just checking all integer values of x in the interval? That's wrong for two reasons. In the first place there is no need to check

I was wondering why you had included x= 1 in your list. Surely you are not checking at all integer values of x in the interval? That's wrong for two reasons. First you don't need to check x= 1: it doesn't meet any of the requirements above. Far more importantly, there is no requirement that a maximum or minimum must be at an integer value of x (except that it makes the function simpler for students to work with)! You might be missing the correct value completely. If I were your teacher and saw you doing that, I would give you a problem to work in which the max and min were NOT at integer values of x.