Extreme Value Theorem

A company wants its highest possible profit. An engineer wants the lowest possible cost. Before trying to find the best value, it helps to know that one exists. The extreme value theorem guarantees it: a continuous function on a closed interval always reaches a highest and a lowest value. Then a short method using critical points finds them. This is the core of every optimization problem later in the chapter.

Absolute vs Local Extrema

On an interval, the absolute maximum of is the largest value takes anywhere on that interval, and the absolute minimum is the smallest. Together they’re the absolute extrema (also called global extrema).

A local maximum only has to beat the points near it. The difference matters. In the graph, is a local maximum with , a little peak. But the function climbs much higher by the right end of the interval, reaching . The absolute maximum is at an endpoint, where there’s no peak at all.

The Extreme Value Theorem

If is continuous on a closed interval , then has an absolute maximum value and an absolute minimum value on .

The theorem promises that the extremes exist. It doesn’t say where they are or what they are; the closed interval method below finds that. Like the intermediate value theorem, it’s an existence theorem, and like that theorem it depends on continuity.

Why Both Conditions Matter

• The interval must be closed. on the open interval gets closer and closer to 2 but never reaches it, because isn’t included. There’s no largest value: whatever value you pick, there’s a bigger one. The same goes for the minimum.
• The function must be continuous. With a jump, the top of a piece can be an open dot, a value the function approaches but never takes. Then there’s no maximum.
• The interval must be bounded. On , has no maximum at all.

Failing a condition doesn’t mean there can’t be extremes, only that they aren’t guaranteed. on the open interval still has maximum 1 and minimum .

Where Can the Extremes Be?

On , an absolute maximum or minimum happens either

• at an endpoint, or , or
• at an interior point, where it is also a local extreme, and so, by Fermat’s theorem from the critical points lesson, at a critical point.

That short list of candidates is the whole idea behind the method.

How to Find the Absolute Maximum and Minimum

The closed interval method, for continuous on :

1. Find the critical points of that lie inside . Ignore any outside the interval.
2. Evaluate at each of those critical points.
3. Evaluate at the endpoints, and .
4. Compare. The largest value is the absolute maximum; the smallest is the absolute minimum.

A table of candidates keeps the work organised, as in the examples.

Example 1: A Cubic With a Tie

Find the absolute maximum and minimum of on .

Step 1 — critical points:

Both are inside .

Steps 2 and 3 — evaluate the candidates:

Step 4 — compare. The absolute maximum is 17, at . The absolute minimum is , reached at both and . The critical point is a local maximum but neither absolute extreme.

Example 2: A Critical Point Outside the Interval

Find the absolute extrema of on .

gives , which is not in , so it’s not a candidate. Only the endpoints are left: and .

The absolute minimum is at , and the absolute maximum is at . Including would give the wrong minimum: the function never takes that value on .

Example 3: A Cusp

Find the absolute extrema of on .

is never 0 and is undefined at , which is inside the interval. Candidates:

Absolute maximum 4 at ; absolute minimum 0 at , the cusp. If you only solved , you’d miss the minimum entirely.

Example 4: A Trig Function

Find the absolute extrema of on .

when , so in this interval.

Absolute maximum at ; absolute minimum at .

Example 5: Exponential and Log Functions

Find the absolute extrema of on .

at . Then , and . The maximum is at and the minimum is 0 at .

Find the absolute extrema of on .

at . Then , and . The maximum is at the right endpoint and the minimum is at .

Why the Theorem Is True

A full proof needs a property of the real numbers called completeness (there are no “gaps” in the number line), which is covered in analysis courses rather than Calculus 1. The idea has two steps. First, the values of are bounded: if they weren’t, you could find points in with growing without limit, those points would pile up at some in the interval (this is where “closed and bounded” is used), and continuity at would force to stay close to , a contradiction. Second, the values then have a least upper bound . Pick points where gets closer and closer to ; they pile up at some in , and continuity gives , so the maximum is actually reached. The same argument gives the minimum.

Common Mistakes

• Forgetting the endpoints. The absolute extremes are often there, as in Examples 1, 2 and 5.
• Using critical points outside the interval. In Example 2, isn’t in , so isn’t a candidate.
• Missing critical points where is undefined, like the cusp in Example 3.
• Answering with the -value instead of the value. “The absolute maximum is 17, at .” The maximum is the -value; is where it happens.
• Applying the method on an open interval or to a discontinuous function, where the theorem’s guarantee doesn’t hold.

Practice Problems

Try each one before opening the answer. The practice set has 12 more with instant checking.

Problem 1. Find the absolute maximum and minimum of on .

Problem 2. Find the absolute extrema of on .

Problem 3. Find the absolute extrema of on .

Problem 4. Find the absolute extrema of on .

Problem 5. Find the absolute extrema of on .

Problem 6. Does the extreme value theorem guarantee that has an absolute maximum on ?

The extreme value theorem finds the extremes on a closed interval by checking a short list of candidates. Next, Rolle’s theorem and the mean value theorem use it to connect a function’s average rate of change to its derivative.