Derivatives turn a curve from a picture into a machine you can interrogate. Where a graph peaks, where it bends, where it hides a surprise — every one of those features is read off f′ and f″. This chapter is the toolkit for questioning a function hard: finding every peak and valley, proving theorems that guarantee hidden points, and judging the shape of a curve before you ever plot it.
1 · Global vs local extrema
A local maximum at x = c means f(c) ≥ f(x) for all x near c — a peak of the neighbourhood. A global (absolute) maximum means f(c) ≥ f(x) for every x in the domain — the tallest point overall. Every global extremum in the interior of the domain is also local, but a local extremum need not be global, and a global extremum may sit on the boundary, where f′ has no reason to vanish.
The Extreme Value Theorem guarantees that a continuous function on a closed, bounded interval [a, b] attains both a global maximum and a global minimum. Both hypotheses matter: x² on the open interval (0, 1) never attains its supremum; 1/x on (0, 1] grows without bound.
2 · Finding the global max and min
The candidate method is mechanical and reliable on any closed interval:
- Find where f′(x) = 0 or f′(x) fails to exist — the critical points — and keep only those inside [a, b].
- Keep the endpoints a and b as candidates as well.
- Evaluate f at every candidate. The largest value is the global maximum; the smallest is the global minimum.
On an open or infinite interval there may be no global extremum at all; running the candidate check and inspecting behaviour at infinity is how you discover that.
Global extrema of f(x) = x³ − 3x on [0, 2]
- Differentiate: f′(x) = 3x² − 3 = 3(x − 1)(x + 1). Critical points: x = 1 and x = −1. Only x = 1 lies inside [0, 2].
- List candidates: x = 0, x = 1, x = 2.
- Evaluate: f(0) = 0, f(1) = 1 − 3 = −2, f(2) = 8 − 6 = 2.
- Conclude: global minimum −2 at x = 1; global maximum 2 at x = 2. Note the global max occurs at an endpoint, not at a critical point — skipping endpoints is the classic error here.
3 · Rolle's theorem and the Mean Value Theorem
Rolle's theorem: if f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there is at least one c in (a, b) with f′(c) = 0. Geometrically: a curve that starts and ends level must flatten out somewhere in between.
The Mean Value Theorem (MVT) drops the equal-height requirement. Under the same continuity and differentiability hypotheses, there exists c in (a, b) with
f′(c) = (f(b) − f(a)) / (b − a)
Rolle is the special case f(a) = f(b). A useful consequence: the average rate of change over [a, b] is always achieved exactly somewhere inside. This is also the engine behind error estimates for linear approximation — remember it for the next chapter.
MVT demands differentiability on the open interval and continuity on the closed one. f(x) = |x| on [−1, 1] satisfies f(1) = f(−1) = 0, but no f′ vanishes — MVT is not violated, because f fails to be differentiable at 0. Likewise a single jump discontinuity anywhere in [a, b] voids the whole conclusion, even if f is smooth elsewhere.
4 · The second derivative and concavity
f″(x) > 0 on an interval means f′ is increasing and the graph is concave up — it cups, like a bowl collecting water. f″(x) < 0 means concave down — it caps, like an umbrella. A point where f″ changes sign is a point of inflection: the curve switches between cupping and capping, and the tangent line crosses from one side of the curve to the other. An inflection point requires an actual sign change — a zero of f″ alone is not enough, as x⁴ at x = 0 shows.
5 · Classifying critical points: the two tests
The first derivative test classifies a critical point by the sign change of f′: positive → negative means a local maximum; negative → positive means a local minimum; no sign change means a stationary inflection, like x³ at 0.
The second derivative test uses f″ directly at c: f′(c) = 0 with f″(c) > 0 gives a local minimum; f′(c) = 0 with f″(c) < 0 gives a local maximum; f″(c) = 0 is inconclusive, and you fall back on the first derivative test.
Classify the critical points of f(x) = 2x³ − 9x² + 12x + 1
- Find critical points: f′(x) = 6x² − 18x + 12 = 6(x − 1)(x − 2), so x = 1 and x = 2.
- Apply f″: f″(x) = 12x − 18. Then f″(1) = −6 < 0, so x = 1 is a local maximum with f(1) = 6; and f″(2) = 6 > 0, so x = 2 is a local minimum with f(2) = 5.
- Inflection check: f″(x) = 0 gives x = 3/2. Since f″ is a non-constant linear function it changes sign there, so (3/2, f(3/2)) = (3/2, 11/2) is a genuine inflection point — every cubic has exactly one.
Candidates: f′(x) = 0, f′ undefined, endpoints a, b
Rolle: f(a) = f(b) ⇒ ∃ c ∈ (a, b) with f′(c) = 0
MVT: f′(c) = (f(b) − f(a)) / (b − a)
f″ > 0 concave up; f″ < 0 concave down
Inflection point: f″ changes sign
f′(c) = 0, f″(c) < 0 ⇒ local max; f″(c) > 0 ⇒ local min
Common mistakes
- Forgetting endpoints when hunting global extrema on [a, b] — the largest value often lives there.
- Assuming every f′(x) = 0 is a max or min; it can be a stationary inflection, as x³ at 0.
- Calling a zero of f″ an inflection point without verifying the sign change.
- Applying Rolle's theorem or MVT without checking continuity on [a, b] and differentiability on (a, b).
- Reading a local maximum as the global maximum on a closed interval without comparing endpoint values.
- Treating f″(c) = 0 as a dead end instead of switching to the first derivative test.