Critical Points and the Extreme Value Theorem
Where can a function's biggest and smallest values hide? Only in two kinds of places: critical points and endpoints. That short list turns "find the extreme values" into a bookkeeping exercise: evaluate f at a handful of candidates and compare.
By the end you'll be able to find every critical point of a function (where \(f' = 0\) or \(f'\) is undefined), state when the Extreme Value Theorem guarantees absolute extrema, and run the closed-interval method to locate them.
Predict: on your current interval, does the absolute maximum of \(f(x) = x^3 - 3x\) sit at a critical point or at an endpoint? Drag the two endpoint handles (or use the sliders) and watch which candidate wins.
The curve has two critical points, where \(f'(x) = 3x^2 - 3 = 0\) — that is, \(x = -1\) and \(x = 1\). Your interval \([a, b]\) is the shaded band. Every candidate — each critical point inside the band, plus the two endpoints — gets a dot; the green dot holds the absolute max and the orange dot the absolute min. Hover or focus any dot for its exact value.
On [0.0, 2.0]: absolute max f = 2.00 at x = 2.0 (endpoint) · absolute min f = −2.00 at x = 1.0 (critical point)
Critical points occur where \(f' = 0\) or \(f'\) is undefined; the Extreme Value Theorem guarantees that a function continuous on a closed interval attains an absolute maximum and minimum, found by comparing f's values at the critical points and the endpoints.
A critical point of \(f\) is an interior point \(c\) where \(f'(c) = 0\) or \(f'(c)\) does not exist. By Fermat's Theorem, any local extremum at an interior point must land on a critical point — so critical points are the ONLY interior candidates. The Extreme Value Theorem (EVT) adds the guarantee: if \(f\) is continuous on a closed interval \([a, b]\), it attains both an absolute maximum and an absolute minimum there. Both hypotheses are essential — recall from your work on continuity that removing an endpoint or allowing a jump lets the extreme value slip away unattained.
This is the closed-interval method, and it's how you actually find extreme values of a continuous \(f\) on \([a, b]\): (1) find all critical points in \((a, b)\) — differentiate and solve \(f' = 0\), and note anywhere \(f'\) is undefined; (2) evaluate \(f\) at each critical point; (3) evaluate \(f\) at both endpoints; (4) the largest value wins the absolute max, the smallest the absolute min. No graphing, no guesswork — a finite list of candidates and a comparison. That's exactly what the demo above does every time you drag: it re-collects the candidates and crowns the winners. Notice how often an endpoint wins even though it's not a critical point.
Find the absolute extrema of \(f(x) = x^3 - 3x\) on \([0, 2]\).
Critical points: \(f'(x) = 3x^2 - 3 = 0 \Rightarrow x = \pm 1\); only \(x = 1\)
lies in \((0, 2)\). \(f'\) exists everywhere, so that's the full list.
Candidates: \(f(0) = 0\), \(f(1) = 1 - 3 = -2\), \(f(2) = 8 - 6 = 2\).
Compare: absolute max \(= 2\) at \(x = 2\) (an endpoint!); absolute min \(=
-2\) at \(x = 1\) (the critical point). Set \(a = 0\), \(b = 2\) in the demo above to
see it.
Find the absolute extrema of \(g(x) = x^3 - 12x\) on \([0, 3]\).
Critical points: \(g'(x) = 3x^2 - 12 = 0 \Rightarrow x = \pm 2\); only \(x =
2\) lies in \((0, 3)\).
Candidates: \(g(0) = 0\), \(g(2) = \) ____, \(g(3) = 27 - 36 = -9\).
Compare: the absolute max is ____ at \(x = \) ____, and the absolute min is
____ at \(x = \) ____.
Reveal the answer
\(g(2) = 8 - 24 = -16\). Comparing \(g(0) = 0\), \(g(2) = -16\), and \(g(3) = -9\): the absolute max is \(0\) at \(x = 0\) (the LEFT endpoint this time), and the absolute min is \(-16\) at \(x = 2\) (the critical point). If you'd skipped the endpoints, you'd have missed the maximum entirely.
More info — three pitfalls that lose points
Ignoring endpoints. Endpoints routinely hold the absolute extremum even
though they're never critical points — in the worked example above the max lived at
\(x = 2\), where \(f'(2) = 9 \neq 0\).
Forgetting where \(f'\) is undefined. A corner like \(f(x) = |x|\) at \(x =
0\) is a critical point with no horizontal tangent — this is the same
differentiability failure you met when studying corners and cusps, and skipping it
can skip the minimum.
Confusing local with absolute. A critical point is only a candidate: it may
be a local max, a local min, or neither (a flat spot). Deciding which — using the
sign of \(f'\) on either side, as you did when reading shape from the derivative's
sign — comes next in this chapter; the closed-interval comparison sidesteps the
question entirely by just evaluating everyone. The OpenStax section in Dive deeper
below walks through more examples of each pitfall.
Check your understanding
Use the closed-interval method to find the absolute extrema of \(f(x) = x^2 - 4x\) on \([0, 5]\).
Why does the Extreme Value Theorem require a CLOSED interval \([a, b]\) rather than an open one?
To run the closed-interval method on \(f(x) = x^3 - 6x^2 + 9x + 2\), you first need its critical points. Differentiate term by term: what is \(f'(x)\), and where does \(f'(x) = 0\)?
\(f(x) = |x - 2|\) on \([0, 3]\) has \(f'(x)\) never equal to \(0\). Where can its absolute minimum occur?
Recap
- A critical point is an interior point \(c\) where \(f'(c) = 0\) OR \(f'(c)\) does not exist — both kinds count.
- Extreme Value Theorem: \(f\) continuous on a CLOSED interval \([a, b]\) attains an absolute max and an absolute min there. Drop continuity or closedness and the guarantee fails.
- Closed-interval method: evaluate \(f\) at every critical point in \((a, b)\) and at both endpoints; the largest value is the absolute max, the smallest the absolute min.
- Endpoints are candidates even though they're not critical points — they often win.
Dive deeper
- OpenStax Calculus Volume 1 — 4.3 Maxima and Minima State the Extreme Value Theorem and locate critical points
- Paul's Online Notes — Minimum and Maximum Values Distinguish absolute from relative extrema on an interval
Sources
- Critical Points and the Extreme Value Theorem