# Maximum and minimum values

1. Apr 23, 2014

### Mr. Fest

Hello,

This might seem like a very basic problem for the most of you but I am a little bit confused as to the problem of maximum and minimum values.

For, example I do understand why y = x^2 has no absolute maximum as y-->$\infty$ as x-->$\infty$...

However, in the closed interval, say, [0, 2], does the function y = x^2 have a local maximum = 4 and a local minimum = 2 at it's endpoints? In this case it so happens that y'(0) = 0 and we have a local minimum there but would it have been the same for the closed interval [1, 2] (with min = 1 and max = 4)?

If someone asks you to motivate why a certain function has a maximum or minimum in a certain interval what would you say? In the latter of above examples, would it suffice to say, y is continuous and y>0 in our interval --> there be a number such that y(m) ≤ y(x) ≤ y(M) for all x in our closed interval. These max and min occurs either in critical points, where y'(x) = 0, in singular points, where y'(x) is not defined or in endpoints. And thereafter do calculations to show that y'(x) = 0 for x = 0, not in our interval and since y'(x) is defined for all x this means that our max and min is found in the endpoints.

Of course the above example is a very simplified one, but I want to try and understand the idea.

What about for an open interval?

Let's take this function as an example:

f(x) = $\frac{x}{(x-1)(x-4)}$, motivate why the function must have a largest value in the interval 1<x<4 and find this value.

f(x) is continuous and |f(x)| > 0 (f(x) < 0) in the interval (1, 4). As we approach 1+ and 4- f(x) --> -$\infty$. Therefore, f(x) will have a maximum value in the interval (1, 4) in either a point where f'(x) = 0 or where f'(x) is not defined. Since f'(x) = $\frac{-x^2 + 4}{((x-1)(x-4))^2}$ and is continuous in the interval (1, 4) we have no singular points where f'(x) is not defined. This means, that the maximum value will be found in a point where f'(x) = 0.

For f'(x) to be zero it is required that its numerator, -x^2 + 4 = 0 --> x = $\pm$2 and since only 2 is in the interval we have f(2) = -1 which is the maximum value of f(x) in the open interval (1,4).

I'm trying to learn the idea behind the mathematics and not just to learn how to solve the problems. I'm in my first year of BSc in mathematics so try to not explain in a too advanced way.

I appreciate any help and corrections.

Sincerely,
Mr. Fest

PS. What I mostly want to understand is the answer to the following question: "How do I answer the question: Why must a given function f(x) have a smallest or largest value or both in a given closed or open interval?"

Last edited: Apr 23, 2014
2. Apr 23, 2014

### SteamKing

Staff Emeritus
For y = x^2 on [0,2], the local minimum occurs at x = 0 and is y = 0.
The local maximum occurs at x = 2 and y = 4.

Not sure what you mean by 'local minimum = 2'.

3. Apr 23, 2014

### Mr. Fest

Typo, what I meant was "local minimum = 0" as in the value of the local minimum is 0.

Maybe you can also enlighten me regarding the rest of the post as well? Would be really appreciated.

What I mostly want to understand is the answer to the following question: "How do I answer the question: Why must a given function f(x) have a smallest or largest value or both in a given closed or open interval?"

Last edited: Apr 23, 2014
4. Apr 23, 2014

### DrewD

Are you asking about the extreme value theorem? A max/min is only guaranteed for a continuous function on a closed interval.

On an open interval there is no guarantee, so you need to be more careful. There isn't a simple theorem in this case. You need to find local max/min (using critical points as you have) and then show that the function is always decreasing/increasing away from the local extrema. I don't know of any simple criteria that will tell you if a local extremum is absolute on an open interval.

Your proof seems to work, but it is needlessly complicated. All you need to show is that $f$ is increasing on $(1,2)$ and decreasing on $(2,4)$. That is how I would proceed because it is easier IMO when there are multiple local extrema.

Why are you interested in $|f(x)|>0$?

5. Apr 23, 2014

### Mr. Fest

So in other words I should make use of f''(x)? When f'(x) = 0 and when f''(x) < 0 then we have a local maximum?

Yeah I'm not sure why I wanted $|f(x)|>0$, basically I wanted to say that f(x) < 0 for that interval.

6. Apr 23, 2014

### AlephZero

The way you discussed the examples in your posts only applies to functions that are continuous and/or differentiable. In general, anything can happen.

For example consider the function f(x) = x when x is rational and f(x) = 0 when x is irrational. f(x) does not have a smallest or largest value in either the closed interval [-1,1] or the open interval (-1,1). But it does have both a smallest and largest value in the closed interval [-π,π] (note, π is an irrational number).

Studying math at university level, it's useful to make your own collection of "pathological" functions like f(x), to check conjectures against. Other examples are functions like $\sin(1/x)$, $(1/x)\sin(1/x)$, etc, which don't necessarily play nice when x is close to 0.

7. Apr 23, 2014

### Mr. Fest

Since I am in uni I need to be able to answer such questions at an upcoming exam. Since it's still early on we're not playing around with irrational numbers yet.

Looking at old exams in this course all have had 1 or 2 questions of the type:
Motivate why function f(x) must take a largest or smallest value in the open interval (a, b) and find this value.

Or

Motivate why function f(x) must take an absolute maximum value in the closed interval [a, b] and find this value.

Or

Motivate why function f(x) must take a smallest and largest value in the closed interval [a, b] and find this value.