IB Maths AI HL
Further Differentiation
Paper 1 & 2
~8 min read
Stationary Points
A stationary point is anywhere the curve is momentarily flat — f′(x) = 0. Most are turning points (a local max or min), but some are points of inflection. Now that you have the second derivative, classifying them is fast: the sign of f′′(x) usually tells you the nature in one step.
📘 What you need to know
- Stationary point: where f′(x) = 0 — the tangent is horizontal.
- Turning point: a stationary point where the curve changes direction (local max or min).
- Not all stationary points turn: a horizontal point of inflection is stationary but isn’t a turning point.
- Second derivative test: f′′ > 0 → min; f′′ < 0 → max; f′′ = 0 → inconclusive.
- First derivative test: check the sign of f′ just either side — only needed when f′′ = 0.
- “Classify” = “find the nature” — same instruction, different wording.
Stationary vs turning points
Every turning point is stationary, but not every stationary point turns. The difference is whether the curve actually changes direction there.
turning point
curve turns
Changes increasing↔decreasing — a local max or min.
horizontal inflection
flat but no turn
Stationary (f′ = 0) but the curve carries on the same direction.
Second derivative test
f′′(x) > 0 → min · f′′(x) < 0 → max · f′′(x) = 0 → inconclusive
✗ not in booklet — remember the directions
🧠 “Smile is a min, frown is a max”
Positive f′′ bends the curve into a smile ⌣ (concave up → minimum); negative f′′ bends it into a frown ⌢ (concave down → maximum).
Local vs global: a local min/max is the lowest/highest point nearby. Many curves run off to ±∞, so a local extreme often isn’t the function’s overall (“global”) extreme.
Finding and classifying them
🧭 Recipe — coordinates & nature
- Find f′(x) and solve f′(x) = 0 for the x-coordinate(s).
- Find f′′(x) and evaluate it at each stationary point.
- Read the sign: + → local min, − → local max, 0 → use the first derivative test.
- First derivative test (if needed): check f′ just below and above; +→− is a max, −→+ is a min, no sign change is a point of inflection.
- Find y by substituting each x into f(x).
| Point type | f′ before | at point | f′ after | f′′ |
|---|
| Local minimum | − | 0 | + | > 0 |
| Local maximum | + | 0 | − | < 0 |
| Inflection (stationary) | same sign | 0 | same sign | = 0 |
🤔 Why is f′′ = 0 not enough on its own?
A zero second derivative could be an inflection — but it could also be a max or min where the bend just happens to flatten momentarily. The second derivative test simply can’t tell which, so you fall back to the first derivative test: watching whether the gradient actually changes sign as you pass through. Only a genuine sign change confirms a turning point.
Worked examples
All parts use f(x) = 2x3 − 3x2 − 36x + 25.
WE 1Find f′(x) and the x-coordinates of the stationary points
Differentiate and solve f′(x) = 0.
f′(x) = 6x² − 6x − 36 = 6(x² − x − 6)
6(x − 3)(x + 2) = 0
x = 3 and x = −2
WE 2Find the y-coordinates
Substitute each x into f(x).
f(3) = 2(27) − 3(9) − 36(3) + 25 = 54 − 27 − 108 + 25 = −56
f(−2) = 2(−8) − 3(4) − 36(−2) + 25 = −16 − 12 + 72 + 25 = 69
(3, −56) and (−2, 69)
WE 3Find f′′(x) and classify x = 3
Differentiate again and check the sign at x = 3.
f″(x) = 12x − 6 = 6(2x − 1)
f″(3) = 6(6 − 1) = 30
30 > 0 → concave up
(3, −56) local minimum
Same second derivative, evaluated at x = −2.
f″(−2) = 6(−4 − 1) = −30
−30 < 0 → concave down
(−2, 69) local maximum
Both stationary points are turning points here.
WE 5When the test fails: classify the stationary point of y = x4
Here f′′ = 0 at the stationary point, so use the first derivative test.
f′(x) = 4x³ = 0 → x = 0; f″(x) = 12x², f″(0) = 0 (inconclusive)
f′(−1) = −4 (< 0), f′(1) = 4 (> 0)
− → + ⟹ minimum
(0, 0) local minimum
💡 Top tips
- Second derivative first — it’s usually the quickest classifier.
- Fall back to the first derivative test only when f′′ = 0.
- Always give coordinates — substitute back into f(x) for the y-value.
- “Classify” = “nature” — don’t be thrown by the wording.
- GDC sketch is valid unless the question says “show that” or asks for an algebraic method.
- Even for algebra, sketch on the GDC first to know what you’re aiming for.
⚠ Common mistakes
- Assuming f′′ = 0 means inflection — it might be a flat-bending max or min; test the first derivative.
- Reversing the test — positive f′′ is a min (smile), negative is a max (frown).
- Forgetting the y-coordinate — a stationary point is a full coordinate.
- Substituting into f′ for y — that gives 0 (the gradient), not the point; use f(x).
- Missing a stationary point — solve f′(x) = 0 fully (factorise carefully).
Next up — Concavity & Points of Inflection. You’ve used f′′(x) to classify turning points; the final topic of the unit zooms in on what f′′ reveals about the shape of the whole curve — where it’s concave up or concave down, and the points of inflection where that concavity flips (which, crucially, need not be stationary at all).
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