Theory

This section is broken down into the following topics:

1. Absolute maxima and minima
2. Local maxima and minima
3. Critical numbers
4. The Closed Interval Method

I. Absolute maxima and minima

The absolute maximum of a function is the highest value that it reaches over a closed interval. Similarly, the absolute minimum of a function is the lowest value that it reaches over a closed interval.

Stated in more mathematical terms, the absolute maximum is some number f(c) such that f(c) >= f(x) for all x in the closed interval. The absolute minimum is similar, except f(c) <= f(x) for all x.

The Extreme Value Theorem requires that such absolute maxima and minima exist. It can be stated as follows:

One easy way to find these absolute maxima and minima is by looking at a graph of the function in question. If you have such a graph available, just look for highest and lowest value of y - these are the absolute maximum and minimum values, respectively.

Let's look at a simple example. Suppose you were asked the following question:

We consider the graph of y = f(x) = x²:

The absolute maximum is the highest y-value that the function reaches on the interval [3,3]. Looking at the graph, this value is y = 9, which occurs at x = ±3.

Notice that this absolute maximum is only valid for the given interval. If we were to extend the graph to the left or right, the function would take on values far greater than
y = 9. (Don't believe us? Try it for yourself!)

The absolute minimum is the lowest value that the function reaches on the interval [3,3]. Looking at the graph, this value is y = 0, which occurs only when x = 0.

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II. Local maxima and minima

The local maximum of a function is a value that is greater than all values that are near it; the local minimum is a value that is less than all values near it.

Notice that these local maxima and minima are not necessarily the absolute maxima and minima - but they could be!

Let's consider the following example:

We begin, as with Example 1, with a graph of the function over the specified domain:

One way to visualize the concept of local maxima and minima is to consider this problem in terms of helium balloons and marbles.

Suppose we were to release a handful of balloons from below the graph, and suppose we dropped a handful of marbles onto the graph from above. What would happen? The balloons would naturally rise as high as possible, and the marbles would be pulled as low as possible by gravity.

The end result would look something like this:

The balloons rise to find points that are higher than all other nearby points on the curve. In this case, if we examine only the points very close to x = 0, we see that the value of the function at x = 0 is higher than the value of the function on either side. Thus, the balloon has found a local maximum of the function. The local maximum appears to be
0 when x = 0.

Next, we consider the marbles. These have fallen to points that are lower than all nearby points on the curve. If we constrain our view to the area near x = ±1.7, we note that the value of the function at these points is lower than the value of the function on either side. Thus, the marbles have found the local minima of the function. These local minima appear to be approximately -7.5 and -10.5, respectively.

Wondering how we can find the exact values of these maxima and minima? That's where the calculus comes in - read on!

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III. Critical numbers

You may have noticed in the graphs above that the slope of the tangent line at the local maxima and minima is zero (i.e., the graph appears "flat" at these points). This is, in fact, true, and has been formalized as Fermat's Theorem:

Notice that Fermat's Theorem does not state that there will always be a maximum or minimum where the derivative is 0, nor does it handle the cases where the derivative is undefined at c.

Based on Fermat's Theorem, we know to start looking for maxima and minima at points where f'(c) = 0 or f'(c) does not exist. Because of the importance of these two types of points, we give them a special name: critical numbers. A critical number is a number in the domain of f such that f'(c) = 0 or does not exist. (In graphical terms, these points are often referred to as critical points.)

Before we move on to using these critical numbers to find maxima and minima, let's take a look at a simple example to illustrate the concept:

We start by finding f'(x) = 2x - 4 (using the power rule). We then set f'(c) = 0,
so 2c - 4 = 0 and thus c = 4/2 = 2. Since f'(c) is defined everywhere (it is a continuous linear function), the only critical is 2.

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IV. The Closed Interval Method

We can now take what we know about critical points and use them to find the exact absolute maxima and minima of functions.

Notice that the absolute maxima and minima are always local maxima and minima, too, in which case they occur at critical numbers, or they occur at the endpoints of the interval. Based on this observation, we can formulate the Closed Interval Method, a three-step process that is guaranteed to find the absolute maximum and minimum of a continuous function:

This sounds a lot trickier than it actually is. To convince you of that fact, let's take a look at a typical example (not at all unlike what you'll find in Homework Assignment 8):

We'll go through the Closed Interval Method step-by-step. Step one is to find the values of f at the critical numbers; before we can do that, though, we must first identify the critical numbers. As above, we do so by differentiating and checking where the derivative is zero or undefined.

Using the power rule, f'(x) = 6x - 12. f'(x) = 0 when 6x - 12 = 0, which occurs if and only if x = 2. We thus have 2 as a critical number of f(x). Since f'(x) is linear, it is continuous and defined for all real numbers, so there are no points to consider where f'(x) is undefined.

We now find the value of f(x) at the critical number 2. Substituting 2 for x, we find that f(2) = 3(2)² - 12(2) + 5 = -7.

Step two is to find the values of f at the endpoints of the interval. In this case, we evaluate f(0) and f(3).

f(0) = 3(0)² - 12(0) + 5 = 5.
f(3) = 3(3)² - 12(3) + 5 = -4.

In step three, we compare the three values obtained above: -7, 5, and -4. The largest value is 5, so 5 is the absolute maximum of f(x) over the given interval. The smallest value is -7, so -7 is the absolute minimum of f(x) over the given interval.

It's that simple! :)

source: internet