Calculus AB — absolute extrema and the second derivative test

Finding the largest or smallest value of a function is one of the most common tasks in AP Calculus AB, and one of the most commonly lost points, because the answer depends on where you look and on the reason you give. Topics 5.2, 5.5, 5.7 and the optimization topics of Unit 5 build a small toolkit: the Extreme Value Theorem says when an absolute maximum and minimum must exist, the candidates test finds them on a closed interval, and the second derivative test classifies a critical point when it works. This material trains all three and the arguments that go with them.

The twelve quiz questions open with the theorem itself: which of four functions is guaranteed both extrema on its interval, where the three that fail break continuity, leave an endpoint open or grow without bound. The candidates test then comes in several forms. You find the absolute minimum of x³ − 3x² on [−1, 4], which is reached twice, once at an endpoint; the absolute minimum of x + 2 cos x on [0, π], which needs both solutions of sin x = 1/2; the absolute maximum of x e⁻ˣ on [0, 3]; and the absolute maximum on [0, 4] of a function known only through its derivative and one value, which again is reached twice. A table of values at the critical points and endpoints shows that sometimes nothing needs computing at all, and x³ − 3x on [0, 2] shows an absolute maximum sitting at an endpoint.

The second derivative test appears as a quick classification, as the case where it says nothing because f″(c) = 0, and inside the argument that settles x + 16/x for x > 0: on an interval that is not closed, one critical point that is a relative minimum is the absolute minimum. Two applied problems close the set, fencing a rectangle against a wall and building an open box with a square base, each with a single critical point.

The wrong options come from the habits that cost points: forgetting an endpoint, reporting the x-value where the value of f was asked for, missing a second solution in the interval, trusting the second derivative test when it is inconclusive, and assuming no minimum can exist because the interval is open.

The flashcards give the theorem and its hypotheses, the steps of the candidates test, the second derivative test and its inconclusive case, the definition of a critical point, the single-critical-point argument and the justification wording graders look for.

This is independent practice written by Zestly, based on the published AP Calculus AB course framework; it is not produced or endorsed by the College Board.

  • State the hypotheses of the Extreme Value Theorem and recognize when they fail
  • Find absolute extrema on a closed interval with the candidates test
  • Find extrema of a function given its derivative and one value, or given a table of candidate values
  • Apply the second derivative test and recognize when it is inconclusive
  • Use the single-critical-point argument on an interval that is not closed
  • Solve optimization problems and justify the extremum

Practice material written by Zestly, based on the College Board AP Calculus AB course framework (Unit 5: Analytical Applications of Differentiation, topics 5.2, 5.5, 5.7, 5.10–5.11), 2026–27 course and exam description.

Sample question

For which function does the Extreme Value Theorem guarantee both an absolute maximum and an absolute minimum on the stated interval?

See the answer

$f(x) = \sqrt{x}$ on $[0, 4]$

The theorem needs a function continuous on a closed interval $[a, b]$. $\sqrt{x}$ is continuous on $[0, 4]$, so it has an absolute maximum ($2$ at $x = 4$) and minimum ($0$ at $x = 0$). $\frac{1}{x}$ is not continuous at $x = 0$; $(0, 2)$ is not closed, and indeed $x^2$ has no minimum there because $0$ is never reached; $\left[0, \frac{\pi}{2}\right)$ is not closed and $\tan x$ grows without bound on it.

← Calculus AB

↑ AP