The Extreme Value Theorem

The Extreme Value Theorem, sometimes abbreviated EVT, says that a continuous function has a largest and smallest value on a closed interval. This is used to show thing like:

There is a way to set the price of an item so as to maximize profits.
Among all ellipses enclosing a fixed area there is one with a smallest perimeter. (The circle, in fact.)
What goes up must come down.

Like the IVT, the EVT tells us that something exists but does not say where. It is of some comfort to know that we can maximize profits, but the shrewd merchant wants to know how to set the price. Calculus gives us methods to solve problems of this nature and these methods are the subject of Stage 8.

For now, we should understand what the Extreme Value Theorem says and the role of BOTH hypotheses: that the function be continuous and that the interval be closed.

Examples.

The following three examples illustrate

A standard application of the Extreme Value Theorem.
How the Extreme Value Theorem can fail if the interval is not closed, and
How the Extreme Value Theorem can fail if the function is not continuous.

Example 1. The function sin(x) - cos(x) has a largest value on the interval [0, 2].

Justification: sin(x) - cos(x) is a continuous function and [0,2] is a closed interval.
Where does this maximum occur and what is the maximum value? These questions are not answered by the EVT. We can get approximate answers by examining the graph.

Graph of sin(x) - cos(x) on [0,2]

Example 2. The function f(x) = x does NOT attain a minimum value on the interval (0,1].

Explanation: The reason that f, which is a continuous function, can get away with not attaining a minimum value is that the interval (0,1] is not closed. If zero were in the interval, then f(0)=0 would be the minimum value. To formally show that there can be no minimum value we argue by contradiction.
Suppose were the minimum value. Since f(x) is the identity function, would then be an element of the interval (0,1], so > 0. But f(/2) = /2 is also a value of f(x) on (0,1]. Since /2 < , this contradicts the assumption that was the minimum value of f(x) on (0,1].

Example 3. Let h(x) be the piecewise function

	{	1/x	x 0
h(x)=
		0	x = 0

Then h(x) is defined everywhere but is not continuous at x=0. Now consider h(x) on the interval [-1,1]. We have that

lim	h(x) = and	lim	h(x) = -
x --> 0⁺		x --> 0^-

It follows that h(x) assumes both arbitrarily large and arbitrarily negative values on the interval [-1,1], so it has neither a maximum nor a minumum on this closed interval. This shows the role which continuity plays in the EVT.

Equipment Check: Decide whether each of the following is a valid argument. Be careful! Just because the premise and conclusion are correct does not mean that the argument itself is correct. (For example, the argument "There is only one Tuesday in each week because balsa wood floats" is NOT a valid argument.)

Statement 1:The Extreme Value Theorem guarantees that the function sqrt( \|cos(x³)\| ) has a minimum value on the interval [-2,7].
Valid Argument Invalid Argument

Statement 2:The Extreme Value Theorem guarantees that the function 1 - t² has a maximum value on the interval (0,1).
Valid Argument Invalid Argument

Statement 3:The Extreme Value Theorem guarantees that the function x³ - 1 has a zero on the interval [0,2].
Valid Argument Invalid Argument

Statement 4:If a function h(u) does not achieve a maximum on the interval [4,5], then h(u) is not continuous.
Valid Argument Invalid Argument

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