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Absolute Extrema and the Candidates Test

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Local versus absolute, and why the interval matters

A local extremum is a value larger or smaller than all nearby values; an absolute extremum is the largest or smallest value on the entire interval under consideration. The two questions have different methods, and confusing them costs points. The Extreme Value Theorem guarantees that a function continuous on a closed, bounded interval [a, b] attains both an absolute maximum and an absolute minimum on that interval. Both hypotheses are load-bearing: on the open interval (0, 1) the function f(x) = x has no absolute maximum, and on [0, 1] a function with a jump discontinuity may have none either. When a question specifies a closed interval, it is inviting the Candidates Test.

The Candidates Test in three steps

On a closed interval, absolute extrema can occur only at critical points or at the endpoints, which turns the problem into finite arithmetic. Step 1: find all critical points in the interval — where f′ = 0 or f′ is undefined. Step 2: evaluate f (not f′) at every critical point and at both endpoints. Step 3: the largest of those output values is the absolute maximum and the smallest is the absolute minimum. Notice what the method does not require: no first or second derivative test, no sign chart, no concavity. Comparing the actual function values settles it, which is why this is the most reliable method available and why the exam expects a table of candidates as the justification.

Open and unbounded intervals need a different argument

If the interval is open, half-open, or infinite, the Candidates Test does not directly apply because there are no endpoints to evaluate — and an extremum may simply not exist. Two legitimate substitutes. First, a first-derivative sign analysis: if f′ > 0 on the entire interval to the left of c and f′ < 0 on the entire interval to the right, then f(c) is the absolute maximum on that interval, and the words "on the entire interval" are what make the conclusion absolute rather than local. Second, an argument from end behavior: examine the limits as x approaches the open ends and show the function does not exceed the candidate value. If a question gives an interval like (0, ∞), expect one of these two and say which you are using.

Candidates Test
On [a, b] with f continuous: compare f(a), f(b), and f(c) at every critical c in (a, b)
Evaluate f, not f′. The largest value is the absolute maximum; the smallest is the absolute minimum.
Worked example

Find the absolute maximum and minimum of f(x) = x³ − 3x² + 1 on the interval [−1, 3].

  1. 1.Differentiate: f′(x) = 3x² − 6x = 3x(x − 2), so critical points are x = 0 and x = 2. Both lie in [−1, 3].
  2. 2.Evaluate f at all four candidates. f(−1) = (−1)³ − 3(−1)² + 1 = −1 − 3 + 1 = −3.
  3. 3.f(0) = 0 − 0 + 1 = 1.
  4. 4.f(2) = 8 − 3(4) + 1 = 8 − 12 + 1 = −3.
  5. 5.f(3) = 27 − 3(9) + 1 = 27 − 27 + 1 = 1.
  6. 6.Compare the four values −3, 1, −3, 1. The largest is 1 and the smallest is −3.
Answer: The absolute maximum is 1, attained at both x = 0 and x = 3. The absolute minimum is −3, attained at both x = −1 and x = 2. Ties are permitted — the question asks for the value, and more than one location can achieve it.
On the exam

The most common way to lose this question is to forget the endpoints. A critical point is a candidate, not an answer, and on many problems the absolute extremum sits at an endpoint where f′ is not zero at all. Write all candidates in a table before comparing.

Checkpoint

The Extreme Value Theorem guarantees an absolute maximum and minimum for f provided that

Checkpoint

Using the Candidates Test on [a, b], you should evaluate

Answer the 2 checkpoints as you read.

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