Calculus I › Applications of derivatives › full formula sheet

f′(c) = 0:  f″(c) > 0 ⇒ local min,  f″(c) < 0 ⇒ local maxSay it: at a critical point c: a positive second derivative means a local minimum, a negative one a local maximum.

The second derivative test

Read the bend at a flat point — smile means valley, frown means peak.

Requires f′(c) = 0 (a genuinely flat point). If f″(c) = 0, the test goes silent — fall back to the first derivative test.

Before this lesson: First derivative test · Concavity

Where it comes from

The problem: classify the critical points of x³ − 3x without building a full sign chart. The first derivative test works but costs a chart per point — the second derivative test classifies each critical point with a single evaluation. The intuition is pure picture: at a flat point, the bend decides everything.

x² at 0
=
f′(0) = 0, f″(0) = 2 > 0
Flat point on a smile ∪ — the valley bottom. Local min.
−x² at 0
=
f′(0) = 0, f″(0) = −2 < 0
Flat point on a frown ∩ — the peak. Local max.
x4 at 0
=
f′(0) = 0, f″(0) = 0
Silence. Flat, but the bend is momentarily zero too — the picture doesn’t resolve. (It’s a min, but the test can’t see it.)

So the test is a shortcut with a blind spot: blazing fast when f″(c) ≠ 0, mute when f″(c) = 0. The naive “it always works” dies on x4.

Before reading on: at a flat point (f′(c) = 0) the curve smiles. Is that a max or a min — and which sign of f″ paints the smile?

Derivation

Assume f′(c) = 0 and f″(c) > 0. Write f″(c) via the limit definition — and watch it force a −→+ sign change in f′.

f″(c)
=
limh→0 [f′(c+h) − f′(c)] / h = limh→0 f′(c+h) / h
Step 1 — limit form. f″ is the derivative of f′. Since f′(c) = 0, the numerator simplifies to f′(c+h).
limit > 0
⇒
f′(c+h)/h > 0  for small h ≠ 0
Step 2 — the ratio stays positive. If a limit is positive, the expression is positive near (but not at) the limit point.
⇒
f′ < 0 left of c, > 0 right of c
Step 3 — signs forced. f′(c+h)/h > 0 means f′(c+h) has the same sign as h: negative for h < 0 (left), positive for h > 0 (right).
f′: −→+
⇒
local min at c
Step 4 — first derivative test. A −→+ change is a valley. The second derivative test is secretly the first test, fast-forwarded. ∎

The mirror and the silence: f″(c) < 0 forces f′(c+h) to take the opposite sign from h — + left, − right, a +→− change, a local max. And f″(c) = 0? The limit is 0, Step 2 collapses, no sign is forced — x4 (min), −x4 (max), x³ (neither) all sit at 0. Silence isn’t “neither” — it’s “ask the first derivative test.”

Before reading on: f(x) = x⁴ at x = 0 gives f″(0) = 0: the test is silent. Predict the verdict anyway — and name the test you would use instead.

How to use it

The procedure, every time:

  1. Find critical points with f′ = 0. (This test can’t start at undefined-derivative points like cusps — it needs a flat f′(c) = 0.)
  2. Compute f″ once.
  3. Evaluate at each critical point: f″(c) > 0 ⇒ local min; f″(c) < 0 ⇒ local max; f″(c) = 0 ⇒ silent — fall back to the first derivative test.
  4. Report f(c) for each classified point.

Speed vs. coverage

One f″ evaluation per point beats a whole sign chart — that’s the pitch. The price: silence at f″(c) = 0 and inapplicability at cusps. Fast when it talks; the first test is the backup when it doesn’t.

The direction trap

f″ > 0 ⇒ min (smile ∪ holds water; water pools at the bottom). Burn this in — “positive means max” is the #1 mix-up.

Common mistake: reading f″(c) = 0 as “neither max nor min.” Silence is not a verdict — x4 is a min and the test says nothing. Always fall back to the first derivative test.

Worked examples

Four classifications, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: f(x) = x³ − 3x

  1. Critical points. f′(x) = 3x² − 3 = 0 at x = ±1.
  2. Second derivative. f″(x) = 6x.
  3. Evaluate. f″(−1) = −6 < 0 → local max, f(−1) = 2. f″(1) = 6 > 0 → local min, f(1) = −2.
  4. Compare. Two evaluations did what a full sign chart did on the first-derivative-test page — same answers, less writing.
Common mistake: “f″(1) = 6 > 0, so local max.” Positive = smile = valley = min. Chant: “cup holds water; water pools at the minimum.”
Your turn: f(x) = x³ − 12x — classify via the second derivative test.

Answer: Local max at x = −2 (f = 16); local min at x = 2 (f = −16).

f′ = 3x² − 12 = 0 at x = ±2; f″(x) = 6x. f″(−2) = −12 < 0 (max), f″(2) = 12 > 0 (min) ✓.

Example 2 — three points: f(x) = x4 − 2x²

  1. Critical points. f′(x) = 4x³ − 4x = 4x(x−1)(x+1) = 0 at x = −1, 0, 1.
  2. Second derivative. f″(x) = 12x² − 4.
  3. Evaluate. f″(±1) = 12 − 4 = 8 > 0 → local min at both, f(±1) = 1 − 2 = −1. f″(0) = −4 < 0 → local max, f(0) = 0.
Common mistake: evaluating f″ at only one of ±1 (“they’re symmetric”) and forgetting the other. Symmetry suggests, it doesn’t classify — evaluate every point.
Your turn: f(x) = x⁴ − 8x² — classify each critical point.

Answer: Local min at x = ±2 (f = −16); local max at x = 0 (f = 0).

f′ = 4x³ − 16x = 4x(x² − 4) = 0 at x = 0, ±2; f″(x) = 12x² − 16. f″(±2) = 32 > 0 (min, f = 16 − 32 = −16); f″(0) = −16 < 0 (max) ✓.

Example 3 — the silence: f(x) = x4

  1. Critical points. f′(x) = 4x³ = 0 at x = 0.
  2. Second derivative. f″(x) = 12x²; f″(0) = 0 — silent.
  3. Fall back. First derivative test: f′(x) = 4x³ is − left of 0, + right — −→+ → local min, f(0) = 0.
  4. The lesson. Silence meant “I can’t tell,” not “neither.” (−x4 is silent-and-max; x³ is silent-and-neither — all three live at f″(0) = 0.)
Common mistake: “f″(0) = 0, so 0 is neither.” The test abstained — you don’t get to record its abstention as a verdict. Run the first derivative test.
Your turn: f(x) = −x⁴ — what does the second derivative test say at x = 0?

Answer: Silent: f″(0) = 0. Fall back to the first test: local max, f(0) = 0.

f′(x) = −4x³ = 0 at x = 0; f″(x) = −12x², f″(0) = 0. First test: f′ is + left of 0 and − right — +→−, a max.

Example 4 — mixed verdicts: f(x) = 3x5 − 5x³

  1. Critical points. f′(x) = 15x4 − 15x² = 15x²(x−1)(x+1) = 0 at x = −1, 0, 1.
  2. Second derivative. f″(x) = 60x³ − 30x = 30x(2x²−1).
  3. Evaluate. f″(−1) = −30 < 0 → local max, f(−1) = −3 + 5 = 2. f″(1) = 30 > 0 → local min, f(1) = 3 − 5 = −2. f″(0) = 0 — silent.
  4. Fall back at 0. f′ = 15x²(x−1)(x+1): the x² never flips, and (x−1)(x+1) is negative on both sides of 0 — no sign change → neither (terrace).
Common mistake: forcing a verdict at x = 0 from the second derivative test (“f″(0) = 0 means flat means min”). Silence plus the x² factor means: check the first test, find no change, report neither.
Your turn: f(x) = x5 − 5x³ — classify each critical point.

Answer: Local max at x = −√3 (f = 6√3); local min at x = √3 (f = −6√3); neither at x = 0.

f′ = 5x⁴ − 15x² = 5x²(x² − 3) = 0 at x = 0, ±√3; f″(x) = 20x³ − 30x. f″(√3) = 30√3 > 0 (min, f = 9√3 − 15√3 = −6√3); f″(−√3) = −30√3 < 0 (max, f = 6√3); f″(0) = 0 silent — first test: 5x²(x²−3) keeps its sign across 0, so neither ✓.

Memorization tips

  • Cup holds water: f″ > 0 is a smile ∪; water pools at the bottom — the min. Frown ∩ spills from the peak — the max.
  • Flat first, bend second: the test only opens with f′(c) = 0. No flat point, no test (cusps need the first test).
  • Silence ≠ neither: f″(c) = 0 means “ask the first derivative test” — x4, −x4, x³ prove all three outcomes live there.
  • One evaluation per point: that’s the whole pitch — f″(c)’s sign replaces an entire sign chart.
  • Secret identity: the test is the first derivative test fast-forwarded — positive f″(c) forces a −→+ change. Knowing why beats memorizing what.
  • Evaluate every point: symmetry suggests but never classifies — plug in each critical point.

Final challenge

Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and the second derivative test is yours.

← Back to the Calculus I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

What is the second derivative test?

At a critical point with f′(c) = 0: f″(c) > 0 ⇒ local min, f″(c) < 0 ⇒ local max, f″(c) = 0 ⇒ the test is silent (fall back to the first derivative test).

Why does f″(c) > 0 give a local min?

A flat point (f′(c) = 0) on a smile (f″ > 0) is the bottom of the valley. Rigorously: f″(c) > 0 forces f′ negative just left of c and positive just right — a −→+ change, i.e. a local min by the first derivative test.

When does the second derivative test fail?

When f″(c) = 0 — it goes silent, saying nothing. x4 (min), −x4 (max), and x³ (neither) all have f′(0) = f″(0) = 0. It also can’t start when f′(c) is undefined (cusps).

First vs. second derivative test — which is better?

The second is faster (one f″ evaluation per critical point vs. a whole sign chart) but narrower: it needs f′(c) = 0 and f″(c) ≠ 0. The first always works. Use the second for speed; fall back to the first on silence.

Does f″(c) > 0 mean max or min?

Min — smile at a flat point is a valley bottom. Mnemonic: “∪ holds water, and water pools at the minimum.” f″ < 0 (frown at flat) is the peak.

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