Calculus I › Applications of derivatives › full formula sheet
Inflection point
Where the curve changes its mind — from smile to frown, or frown to smile.
An inflection point is a point (c, f(c)) on the graph where concavity flips. Two requirements: f defined at c, and f″ changing sign across c.
Before this lesson: Concavity
Where it comes from
The problem: where does y = x³ switch from bending downward to bending upward? The naive rule — “f″ = 0 means inflection” — is executed by f(x) = x4 at 0:
The right intuition: drive through an S-curve. Your steering wheel turns left, passes through center, turns right — the instant it crosses center, the bend flips. That crossing instant is the inflection point. For x³, f″(x) = 6x crosses from − to + at 0: frown ∩ becomes smile ∪, and (0, 0) is the inflection point.
Before reading on: f″(c) = 0 at a candidate. Is that enough to declare an inflection point? Which counterexample should you test first?
Derivation
This one is a definition plus sign logic — short, but every word matters.
Why “or undefined”? f(x) = ∛√x has f″ undefined at 0, yet concavity flips from up (x < 0) to down (x > 0) — (0, 0) is a genuine inflection point with a vertical tangent. Undefined isn’t disqualified; it’s a candidate.
Before reading on: f(x) = ∛√x at x = 0: f″ is undefined there. Can a point where the second derivative doesn’t exist still be an inflection point?
How to use it
The procedure, every time:
- Compute f″.
- Find candidates: where f″ = 0 or f″ is undefined — but only where f itself is defined. (x = 0 is not a candidate for 1/x: there’s no point there.)
- Sign-chart f″ around each candidate.
- Keep only the sign flips. No flip, no inflection — x4 at 0 says hello.
- Report the point (c, f(c)) — an inflection point is a point, not an x-value.
Inflection ≠ extremum ≠ critical point
Three independent ideas. f(x) = x³ + x has an inflection at 0 (f″ = 6x flips) but no critical points at all (f′ = 3x² + 1 > 0). Don’t hunt inflections among critical points — hunt them in the f″ sign chart.
When to reach for it
Completing a concavity analysis and curve sketches — the inflection points are where the drawn curve visibly changes its bend.
Worked examples
Four bends that flip (or don’t). In each one, read every step — the why of each move is the lesson.
Example 1 — the basic move: f(x) = x³
- Differentiate twice. f″(x) = 6x.
- Candidates. f″ = 0 at x = 0; f defined there. ✓
- Signs. 6x < 0 for x < 0, > 0 for x > 0 — a genuine −→+ flip.
- Conclude. Inflection point at (0, 0) — frown becomes smile.
Your turn: f(x) = −x³ — any inflection point?
Answer: (0, 0).
f″(x) = −6x flips +→− at x = 0, and f(0) = 0 is defined ✓.
Example 2 — the acquittal: f(x) = x4
- Differentiate twice. f″(x) = 12x².
- Candidates. f″ = 0 at x = 0.
- Signs. 12x² ≥ 0 on both sides — + and +, no flip.
- Conclude. No inflection point at 0 (or anywhere). The bowl never changes its mind.
Your turn: f(x) = x6 — any inflection point?
Answer: None.
f″(x) = 30x⁴ ≥ 0 with no sign change — the bowl never changes its mind.
Example 3 — undefined but genuine: f(x) = ∛√x
- Differentiate twice. f′(x) = (1/3)x−2/3, f″(x) = −(2/9)x−5/3 = −2/(9x5/3).
- Candidates. f″ undefined at x = 0 — but f(0) = 0 is defined. ✓ (Why keep it? Undefined-derivative points are candidates, not disqualifications.)
- Signs. For x < 0: x5/3 < 0, so −2/(negative) > 0 — concave up. For x > 0: −2/(positive) < 0 — concave down. A real flip.
- Conclude. Inflection point at (0, 0), with a vertical tangent (f′(0) undefined too).
Your turn: f(x) = x1/5 — any inflection point?
Answer: (0, 0).
f″(x) = −(4/25)x−9/5: for x < 0 the denominator x9/5 < 0, so f″ > 0; for x > 0, f″ < 0. A genuine flip, and f(0) = 0 is defined ✓.
Example 4 — trigonometry: f(x) = sin x on [0, 2π]
- Differentiate twice. f″(x) = −sin x.
- Candidates. −sin x = 0 at x = 0, π, 2π.
- Signs. On (0, π): sin > 0 → f″ < 0. On (π, 2π): sin < 0 → f″ > 0. Flip at π; at 0 and 2π there’s no “both sides” inside the interval.
- Conclude. Inflection point at (π, 0) — the arch hands off to the bowl.
Your turn: f(x) = cos x on [0, 2π] — inflection points?
Answer: (π/2, 0) and (3π/2, 0).
f″(x) = −cos x flips at π/2 (+→−) and 3π/2 (−→+); f is defined at both ✓.
Memorization tips
- The steering-wheel image: left, center, right — the crossing instant is the inflection point.
- Suspects vs. verdict: f″ = 0 (or undefined) opens the trial; the sign flip convicts. x4 is the acquittal you must remember.
- Must be on the graph: no f(c), no point, no inflection — 1/x at 0 is the warning.
- Report (c, f(c)): an inflection point is a point — compute the y-value.
- Not an extremum: bend flips, height doesn’t peak — x³ + x inflects at 0 with zero critical points.
- Same chart as concavity: you already built the f″ sign chart — inflection points are just where it flips.
Final challenge
Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and inflection points are yours.
← Back to the Calculus I formula sheet
How to learn a formula here
- 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.
- Work the examples with the answers covered, then uncover one step at a time and compare.
- Finish with the final challenge — five mixed questions including the classic traps.
- Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.
Frequently asked questions
What is an inflection point?
A point where the curve changes concavity — smile to frown or frown to smile. f must be defined (and continuous) there, and f″ must change sign across it.
Does f″(c) = 0 mean c is an inflection point?
Not by itself. f(x) = x4 has f″(0) = 0 but is concave up on both sides — no sign flip, no inflection. Only a sign change of f″ delivers the verdict.
Can an inflection point have an undefined second derivative?
Yes. f(x) = ∛√x has f″ undefined at 0, yet concavity flips from up to down there — (0, 0) is an inflection point with a vertical tangent.
Is 0 an inflection point of 1/x?
No — f(0) doesn’t exist, so there is no point (0, f(0)) on the graph. Concavity changes across x = 0, but an inflection point must be an actual point of the curve.
Is an inflection point a max or min?
No — it’s where the bend flips, not where the height peaks. x³ + x has an inflection at 0 but no critical points at all; the concepts are independent.
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