IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 ~8 min read

Stationary Points

A stationary point is where f′(x) = 0 — the curve has a horizontal tangent there. Solve f′(x) = 0 to locate them, then use the second derivative f″(x) to classify each: positive → local min, negative → local max, zero → inconclusive (use the first-derivative sign test). Optimisation problems are stationary-point problems in disguise.

📘 What you need to know

Stationary point: any point where f′(x) = 0. The curve’s tangent is horizontal there.
Three types: local minimum, local maximum, or (horizontal) point of inflection.
Find them: solve f′(x) = 0 for the x-coordinate(s); substitute into f to get the y-coordinate.
Second derivative test: f″(x) > 0 → local min; f″(x) < 0 → local max; f″(x) = 0 → inconclusive.
First derivative sign test (when f″ = 0): check the sign of f′ just before and just after the stationary point.
Optimisation: write the quantity to optimise as a function of one variable, find stationary point of that function, classify with f″.
Local vs global: local min/max may not be the smallest/largest value over the whole domain — check endpoints separately.
GDC verification: graphing mode usually pinpoints stationary points and tells you the nature directly.

The three types of stationary point

All three have a horizontal tangent (f′ = 0). The second derivative tells them apart: f″ < 0 → max (concave down); f″ > 0 → min (concave up); f″ = 0 → may be a horizontal point of inflection.

Classification — second derivative test

Second derivative test f″(x) > 0 ⇒ LOCAL MIN | f″(x) < 0 ⇒ LOCAL MAX | f″(x) = 0 ⇒ INCONCLUSIVE

When f″(x) = 0 at a stationary point, the test fails — the point could still be a min, max, or inflection. Switch to the first-derivative sign test: evaluate f′ just left and just right of the stationary point.

First derivative sign test: if f′ changes from − to + → local min. From + to − → local max. No sign change → horizontal point of inflection.

🧭 Recipe — find & classify stationary points

Differentiate to get f′(x); set f′(x) = 0 and solve for x.
Find y-coordinates by substituting each x-value into f(x).
Differentiate again to get f″(x); evaluate at each stationary point.
Classify: f″ > 0 → min, f″ < 0 → max, f″ = 0 → use first-derivative sign test.
State the answer as coordinates with nature; verify with GDC graph if not “show that”.

Worked examples

WE 1

Quadratic — single stationary point

Find and classify the stationary point of f(x) = x² − 8x + 11.

Find f′(x) and solve f′(x) = 0 f′(x) = 2x − 8 2x − 8 = 0 → x = 4 y-coordinate f(4) = 16 − 32 + 11 = −5 point: (4, −5) Classify with f″(x) f″(x) = 2 > 0 (constant) → local minimum (4, −5) is a local minimum parabola with positive leading coefficient always has a single local min — the vertex

WE 2

Cubic — two stationary points

Find the coordinates and nature of the stationary points of f(x) = 2x³ − 9x² + 12x + 1.

Differentiate and factor f′(x) = 0 f′(x) = 6x² − 18x + 12 = 6(x² − 3x + 2) = 6(x − 1)(x − 2) → x = 1 or x = 2 Find y-coordinates f(1) = 2 − 9 + 12 + 1 = 6 → (1, 6) f(2) = 16 − 36 + 24 + 1 = 5 → (2, 5) Classify with f″(x) = 12x − 18 f″(1) = 12 − 18 = −6 < 0 → local max f″(2) = 24 − 18 = 6 > 0 → local min (1, 6) is a local max; (2, 5) is a local min cubic with positive leading coeff → max comes first (smaller x), then min

WE 3

Quartic — when f″ = 0, use first-derivative sign test

Find and fully classify the stationary points of f(x) = x⁴ − 4x³ + 6.

Differentiate and factor f′(x) = 4x³ − 12x² = 4x²(x − 3) → x = 0 (double root) or x = 3 y-coordinates f(0) = 6 → (0, 6) f(3) = 81 − 108 + 6 = −21 → (3, −21) Try second derivative: f″(x) = 12x² − 24x f″(0) = 0 → INCONCLUSIVE for x = 0 f″(3) = 108 − 72 = 36 > 0 → local min at (3, −21) First-derivative sign test for x = 0 f′(−0.1) = 4(0.01)(−3.1) ≈ −0.12 < 0 (decreasing left of 0) f′(0.1) = 4(0.01)(−2.9) ≈ −0.12 < 0 (still decreasing right of 0) no sign change → horizontal point of inflection (0, 6) is a horizontal point of inflection; (3, −21) is a local min when f′ has a double root (x² factor), it touches zero without crossing — that’s why no sign change

WE 4

Function with exponential — needs product rule on f′

The function f(x) = x e^−x. Find and classify the stationary point.

Differentiate using product rule u = x, v = e^(-x); u′ = 1, v′ = -e^(-x) f′(x) = e^(-x) + x·(-e^(-x)) = e^(-x)(1 – x) Solve f′(x) = 0 e^(-x) ≠ 0 (always positive) → 1 – x = 0 → x = 1 f(1) = 1·e^(-1) = 1/e → (1, 1/e) Classify with f″(x) — differentiate f′ again f″(x) = -e^(-x)(1 – x) + e^(-x)(-1) = e^(-x)(x – 2) f″(1) = e^(-1)·(-1) = -1/e < 0 → local max (1, 1/e) is a local maximum the global maximum of x·e^(-x) on (0, ∞) — appears in many probability and physics models

WE 5

Reverse: find parameters so a stationary point exists at a given location

The curve y = x³ + ax² + bx has a stationary point at (1, −2). Find a and b and classify the stationary point.

Apply the two conditions y(1) = -2: 1 + a + b = -2 → a + b = -3 …(1) y′(x) = 3x² + 2ax + b y′(1) = 0: 3 + 2a + b = 0 …(2) Subtract (1) from (2) (3 + 2a + b) − (a + b) = 0 − (−3) 3 + a = 3 → a = 0 From (1): b = -3 − 0 = -3 Classify: y = x³ − 3x; y″ = 6x y″(1) = 6 > 0 → local minimum a = 0, b = −3; (1, −2) is a local minimum verify: y′ = 3x² − 3 = 3(x − 1)(x + 1) — stationary at x = ±1 ✓

WE 6

Optimisation — minimum surface area of an open box

An open box (no lid) has a square base of side x cm and volume 256 cm³. Find the value of x that minimises the surface area, and find this minimum surface area.

Set up: let height = h. Volume constraint gives h in terms of x V = x²h = 256 → h = 256/x² Surface area = base + 4 sides (no lid) A = x² + 4xh = x² + 4x · 256/x² = x² + 1024/x Differentiate and find stationary point dA/dx = 2x − 1024/x² 2x = 1024/x² → 2x³ = 1024 → x³ = 512 → x = 8 Classify with d²A/dx² d²A/dx² = 2 + 2048/x³ at x = 8: 2 + 2048/512 = 2 + 4 = 6 > 0 → local min ✓ Minimum surface area A(8) = 64 + 1024/8 = 64 + 128 = 192 x = 8 cm gives minimum A = 192 cm² (height h = 4 cm) classic optimisation: write A as a function of one variable, minimise, classify with f″

💡 Top tips

Always factor f′(x) before solving — it’s often the fastest way to spot the roots and check for double roots (which signal points of inflection).
e^kx is never zero — when f′ has an exp factor, ignore it when solving f′ = 0; only the OTHER factor matters.
For optimisation: identify the constraint (volume, perimeter, etc.) and the objective (area, cost, etc.); use the constraint to reduce to one variable.
Always classify stationary points unless the question only asks for “coordinates”. “Find the stationary points” usually means coordinates AND nature.
GDC backup — graph the function and use the max/min finder to verify your algebraic work, even on “show that” questions.

⚠ Common mistakes

Forgetting y-coordinates — “find the stationary points” usually asks for both x AND y. Substitute back into f, NOT into f′.
Stopping at “f″ = 0” — this is INCONCLUSIVE, not “inflection”. You MUST do the first-derivative sign test to confirm.
Using f instead of f′ for stationary point — set the DERIVATIVE to zero, not the function.
Ignoring constraints in optimisation — for a real-world problem, x = 8 means an 8 cm side; check the answer makes physical sense.
Misreading f″ sign — f″ > 0 means concave UP and so a local MIN (smile shape). f″ < 0 means concave down → local MAX.

Up next: Concavity & Points of Inflection. The second derivative tells you more than just “min vs max” — its sign across an interval determines whether the curve is concave up (smiling) or concave down (frowning), and points where concavity changes are points of inflection. Crucially, points of inflection don’t need to be stationary; many curves change concavity while still rising or falling.

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