Testing All Candidates for Absolute Extrema

Why Compare Every Candidate?

A derivative can help locate possible extrema, but it does not by itself tell you the absolute maximum or minimum on an interval. An interior critical number may correspond to a local extremum, a stationary point that is not an extremum, or an extremum that is not absolute. Endpoints also matter: an absolute extremum can occur there even though no derivative test applies.

For a continuous function on a closed interval, the reliable method is to identify all interior critical numbers, evaluate the function at those numbers and at both endpoints, and compare the resulting values. The conclusion depends on checking the complete candidate set, not just finding a point where the derivative is zero.

Definitions and the Closed-Interval Test

An absolute maximum value of \(f\) on \([a,b]\) is a value \(f(c)\) such that \(f(c)\geq f(x)\) for every \(x\in[a,b]\). An absolute minimum value is a value \(f(c)\) such that \(f(c)\leq f(x)\) for every \(x\in[a,b]\). A point where one of these values occurs is an absolute maximum point or absolute minimum point, respectively. There can be more than one such point.

For this test, an interior critical number is a number \(c\in(a,b)\) where \(f'(c)=0\) or \(f'(c)\) does not exist. The function must be defined at \(c\) for \(c\) to be a candidate. Some conventions call endpoints critical numbers too; the test lists endpoints separately so they cannot be overlooked.

The continuity condition matters. The Extreme Value Theorem guarantees that the extrema exist under this condition; it does not say where they occur. The candidate conclusion also uses the fact that a differentiable function at an interior local maximum or minimum has derivative zero. If the derivative does not exist at an interior extremum, that point is included by the critical-number test instead.

Why the Candidate Test Works

Here is the justification behind the method. Suppose \(f\) is continuous on \([a,b]\). By the Extreme Value Theorem, there is at least one point \(c_{\max}\in[a,b]\) where \(f\) reaches its absolute maximum, and at least one point \(c_{\min}\in[a,b]\) where it reaches its absolute minimum.

Consider either extremum point \(c\). If \(c=a\) or \(c=b\), it is an endpoint candidate. Otherwise, \(c\) is in the interior. An absolute maximum or minimum is also a local maximum or minimum: nearby points in the interval cannot have a more extreme value. If \(f\) is differentiable at \(c\), Fermat’s theorem says \(f'(c)=0\). If it is not differentiable there, \(c\) is still a critical number by definition. Thus each absolute extremum occurs among the listed candidates. Since every candidate is itself in \([a,b]\), comparing their function values identifies the greatest and least values on the interval.

Worked Examples

Why a Candidate Is Not Automatically an Absolute Extremum

Being on the candidate list means that a point must be checked; it does not mean the point must be an absolute maximum or minimum. The function \((x^2-1)^2\) illustrates this: \(x=0\) is critical, but its value \(1\) is neither the greatest nor the least candidate value. The derivative identifies possible locations; the value comparison makes the global decision.

The method also allows ties. If two or more candidates give the same greatest value, all corresponding points are absolute maximum points. The same is true for the least value. Report every point at which the extreme value occurs, not just the first one found.

A Common Mistake: Omitting the Endpoints

Consider \(f(x)=x^2\) on \([-2,1]\). A common incorrect approach is to find only the stationary point and conclude that its value determines the absolute extrema. Since \(f'(x)=2x\), the only interior critical number is \(x=0\), with \(f(0)=0\). If we compare no other values, we might incorrectly report \(0\) as both the maximum and minimum.

The correction is to include the endpoints:

Now the greatest value is \(4\), at \(x=-2\), and the least is \(0\), at \(x=0\). The error was not in differentiating; it was in leaving out valid candidates.

Using the Test Carefully

In many AP problems, the candidate list is finite, so a table of candidates and function values makes the comparison clear. If there are infinitely many critical numbers, do not claim that a finite table has tested them all. The theorem still locates an absolute extremum among the endpoints and critical numbers, but additional analysis may be needed to compare an infinite set of candidate values.

For a finite candidate list, a clear written justification can be organized as claim, evidence, and conclusion: claim that the greatest or least value occurs at a specified point; provide the full list and computed values as evidence; then state the conclusion after comparing them. This makes it possible to verify that no endpoint or interior critical number was missed.