Key Idea: This topic uses the derivative to find the peaks, troughs and bends of a curve and to solve real max/min problems. It's almost all Paper 1 by-hand: differentiate, set the derivative to zero, solve, then justify the nature.
⛰️ Stationary points: find, classify, locate
A stationary point is where the gradient is zero. Find it by solving f′(x) = 0 (factor the derivative first), classify it with the table below, then get the y-coordinate by substituting x into the original f(x) — not into f′(x), which just gives 0.
| Second derivative | Concavity | Nature |
|---|---|---|
| f″(x) > 0 | Concave up ∪ | Minimum (holds water) |
| f″(x) < 0 | Concave down ∩ | Maximum |
| f″(x) = 0 | Test the sign of f′ either side instead | Inconclusive — check by hand |
🔧 Optimisation & inflexion
Optimisation recipe — real-world max/min
- Model the quantity to optimise (area, cost, volume…).
- Use the constraint to get it in one variable.
- Differentiate and solve f′(x) = 0.
- Classify the solution (f″, or sign of f′).
- Answer the question asked — usually the max/min value, with units.
- Reciprocal model T = ax + b/x: write b/x as bx⁻¹, then keep the positive root.
A point of inflexion is where the curve changes concavity. Solve f″(x) = 0 for candidate x-values, then confirm f″ changes sign across each. f″ = 0 alone is not enough — e.g. y = x⁴ has f″(0) = 0 but no inflexion.
✏️ IB-style worked examples
IB-style question — find and classify stationary points
Find and classify the stationary points of f(x) = x³ − 3x² − 9x + 2, giving their coordinates.
Step by step:
Differentiate and solve f′(x) = 0 (factor first).
Classify with the second derivative.
Substitute into the ORIGINAL f for each y-coordinate.
Maximum (−1, 7); minimum (3, −25).
IB-style question — maximise an area (optimisation)
A rectangular enclosure uses a wall as one side and 60 m of fencing for the other three sides. With width x, find the width that gives the maximum area, and that maximum area.
Step by step:
Model with the constraint (two widths + one length = 60).
Differentiate, solve A′(x) = 0, and justify the maximum.
Answer what's asked — the maximum area, not just x.
Width 15 m gives the maximum area of 450 m².
IB-style question — point of inflexion (confirm the sign change)
Find the point of inflexion of f(x) = x³ − 6x² + 4x + 1.
Step by step:
Differentiate twice and solve f″(x) = 0.
Confirm f″ changes sign across x = 2.
Get y from the original function.
Point of inflexion at (2, −7).
Important: Don't stop at f′(x) = 0. You must justify whether it's a max or min (f″, or the sign of f′), and then give what's actually asked — the maximum volume or the dimensions, with units, not just the value of x. For inflexion, always confirm f″ changes sign.
Tap each card to reveal the answer.
How do you find stationary points? Solve f′(x) = 0 for the x-values.
f″(x) > 0 at a stationary point — max or min? Minimum (concave up ∪, holds water). f″ < 0 is a maximum.
Where does the y-coordinate of a stationary point come from? Substitute x into the original f(x) — substituting into f′ just gives 0.
First step of any optimisation problem? Use the constraint to write the quantity in one variable, then differentiate.
T = x + 25/x, x > 0 — value of x that minimises T? T′ = 1 − 25x⁻² = 0 ⇒ x² = 25 ⇒ x = 5 (keep the positive root).
f″(0) = 0 for y = x⁴ — is x = 0 an inflexion? No — f″ > 0 on both sides, so no sign change, so no inflexion.
Exam tips
- Stationary points: solve f′(x) = 0 (factor the derivative first).
- Classify with f″ (>0 min, <0 max); if f″ = 0, test the sign of f′ either side.
- y-coordinate always comes from the ORIGINAL function f, never f′.
- Optimisation: get to ONE variable via the constraint, then answer what's asked (with units).
- Inflexion needs f″(x) = 0 AND a sign change of f″ — check both sides every time.