IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 ~7 min read

Concavity & Points of Inflection

The sign of f″(x) determines whether the curve “smiles” or “frowns”. f″(x) > 0 means concave up (smile shape); f″(x) < 0 means concave down (frown shape). A point of inflection is where the concavity changes — both f″ = 0 AND a sign change of f″ are needed. Crucially, points of inflection don’t need to be stationary points.

📘 What you need to know

Concave up: f″(x) > 0 throughout an interval — curve bends upward (☺ shape).
Concave down: f″(x) < 0 throughout an interval — curve bends downward (☹ shape).
Point of inflection (POI): a point where concavity changes; requires BOTH f″(x) = 0 AND f″ changes sign through it.
f″(x) = 0 alone is NOT enough — could be a local min, max, or actual POI. Must verify the sign change.
POI need NOT be stationary: f′(x) at a POI can be any value (positive, negative, or zero).
Horizontal POI (also called stationary POI): both f′(x) = 0 AND f″(x) = 0 with f″ sign change. The tangent is horizontal there.
Test at a single point: just evaluate f″(x₁) — sign tells you concavity at that point.
Find an interval of concavity: solve f″(x) > 0 (concave up) or f″(x) < 0 (concave down) — gives a range of x-values.

What concavity looks like

The tangent lines sit on the inside of the bend. If they lie below the curve → concave up. If they lie above → concave down.

Memory trick: concave UP looks like a cUp ☕ (or a smile); concave Down looks like a froWN. The second derivative sign matches: positive = up, negative = down.

Points of inflection

At a point of inflection the curve transitions from one concavity to the other. The tangent crosses through the curve at this point.

Conditions for a point of inflection at x = a f″(a) = 0 AND f″ changes sign through x = a

Both conditions matter: if f″(a) = 0 but f″ has the same sign on both sides (e.g. f″(x) = x² touches zero at 0 but is non-negative everywhere), then x = a is NOT a point of inflection — concavity didn’t change.

Stationary vs non-stationary inflection

Stationary (horizontal) POI

f′(a) = 0 & f″(a) = 0

tangent is horizontal at the POI
example: y = x³ at x = 0

Non-stationary POI

f′(a) ≠ 0 & f″(a) = 0

tangent is sloped at the POI
example: y = x³ + x at x = 0

🧭 Recipe — find & justify points of inflection

Differentiate twice to get f″(x); solve f″(x) = 0 for the candidate x-values.
Test concavity on each side of each candidate by evaluating f″ slightly left and slightly right.
Confirm sign change: if concavity flips through x = a, then x = a is a point of inflection.
Find the y-coordinate by substituting into the original f(x).
State the answer fully: “(a, f(a)) is a POI because f″(a) = 0 AND concavity changes from … to …”.

Worked examples

WE 1

Concavity at specific points

The function f(x) = 2x³ − x² + 5. Determine whether the curve is concave up or concave down at: (a) x = 1; (b) x = −1.

Find f″(x) f′(x) = 6x² − 2x f″(x) = 12x − 2 (a) Evaluate at x = 1 f″(1) = 12 − 2 = 10 > 0 → concave up (b) Evaluate at x = −1 f″(−1) = −12 − 2 = −14 < 0 → concave down (a) concave up at x = 1; (b) concave down at x = −1 at a single point, just check the SIGN of f″ — magnitude doesn’t matter

WE 2

Find intervals of concavity

For f(x) = x³ + 2x² − 7, find the values of x for which the graph is (a) concave up; (b) concave down.

Find f″(x) f′(x) = 3x² + 4x f″(x) = 6x + 4 (a) Concave up: solve f″(x) > 0 6x + 4 > 0 → 6x > −4 → x > −2/3 (b) Concave down: solve f″(x) < 0 6x + 4 < 0 → x < −2/3 (a) concave up for x > −2/3; (b) concave down for x < −2/3 at x = −2/3 the concavity changes — this is the POI of any cubic with positive leading coeff

WE 3

Find the POI of a cubic — fully justified

Find the coordinates of the point of inflection on the graph of y = x³ − 6x² + 9x + 2. Fully justify your answer.

Differentiate twice y′ = 3x² − 12x + 9 y″ = 6x − 12 Solve y″(x) = 0 6x − 12 = 0 → x = 2 Test concavity each side of x = 2 y″(1.9) = 11.4 − 12 = −0.6 < 0 (concave down) y″(2.1) = 12.6 − 12 = 0.6 > 0 (concave up) → concavity changes through x = 2 ✓ y-coordinate y(2) = 8 − 24 + 18 + 2 = 4 (2, 4) is a point of inflection (since y″(2) = 0 AND concavity changes) a cubic with non-zero leading coefficient ALWAYS has exactly one POI

WE 4

Non-stationary POI — POI ≠ stationary point

For f(x) = x³ + x: (a) Show that (0, 0) is a point of inflection. (b) Show that (0, 0) is NOT a stationary point.

Compute f′ and f″ f′(x) = 3x² + 1 f″(x) = 6x (a) f″(0) = 0; check concavity changes f″(−1) = −6 < 0 (concave down) f″(1) = 6 > 0 (concave up) → concavity changes through x = 0; f(0) = 0 → (0, 0) is a POI ✓ (b) Test if it’s a stationary point f′(0) = 3·0 + 1 = 1 ≠ 0 → tangent has slope 1 at (0, 0), NOT horizontal (a) (0, 0) is a POI; (b) NOT a stationary point — slope is 1, tangent is y = x key takeaway: POI is about CONCAVITY change, NOT about gradient being zero

WE 5

Quartic — multiple points of inflection

Find all points of inflection on the graph of f(x) = x⁴ − 6x².

Compute f″(x) f′(x) = 4x³ − 12x f″(x) = 12x² − 12 = 12(x² − 1) = 12(x − 1)(x + 1) Solve f″(x) = 0 → x = −1 or x = 1 Test concavity in each region x < −1: f″(−2) = 36 > 0 (concave up) −1 < x < 1: f″(0) = −12 < 0 (concave down) x > 1: f″(2) = 36 > 0 (concave up) → concavity changes at BOTH x = −1 and x = 1 y-coordinates f(−1) = 1 − 6 = −5 f(1) = 1 − 6 = −5 Points of inflection: (−1, −5) and (1, −5) a quartic can have up to 2 POIs (since f″ is quadratic, with up to 2 roots)

WE 6

Application — kinematics: when does the acceleration change sign?

A car’s displacement (in metres) from a starting point at time t seconds is given by s(t) = −t³ + 6t² + 2t, for 0 ≤ t ≤ 5.

(a) Find the velocity v(t) and acceleration a(t). (b) Find the time at which the acceleration is zero, and the displacement at that time. (c) Interpret what happens physically at this moment.

(a) Differentiate v(t) = s′(t) = −3t² + 12t + 2 a(t) = s″(t) = −6t + 12 (b) Solve a(t) = 0 −6t + 12 = 0 → t = 2 seconds s(2) = −8 + 24 + 4 = 20 metres Check concavity change a(1) = 6 > 0 (s concave up; car speeding up) a(3) = −6 < 0 (s concave down; car slowing down) → POI on s(t) at t = 2 (c) v(2) = −12 + 24 + 2 = 14 m/s — maximum velocity (a) v(t) = −3t² + 12t + 2, a(t) = −6t + 12; (b) t = 2 s, s = 20 m; (c) the car reaches its maximum speed (14 m/s) and then starts decelerating in kinematics: s″(t) is acceleration. POI on s ↔ acceleration crosses zero ↔ maximum velocity

💡 Top tips

“Smile or frown” — concave up smiles (☺), concave down frowns (☹). Match the shape to the f″ sign.
For “show that it’s a POI”: ALWAYS state both conditions — f″ = 0 AND f″ changes sign. Just one isn’t enough.
For real-world contexts: f″ is often a rate of a rate (acceleration = rate of change of velocity = rate of change of rate of change of position). POI marks a transition.
Sign-test using nearby values: pick numbers like ±0.1 from the candidate; compute f″ to see the sign change.
GDC trick: graph f″(x) and look for places it crosses (not touches) the x-axis — those are the POI x-coordinates.

⚠ Common mistakes

“f″ = 0 means inflection” — NOT true on its own. Must verify the sign change.
Confusing POI with stationary points — POI is about CONCAVITY change, stationary point is about gradient being zero. A POI may or may not be stationary.
Sign confusion: f″ > 0 means concave UP (smile, like a CUP holding water). The intuition “positive = up” matches.
Forgetting the y-coordinate — “find the point of inflection” wants both x and y, not just x.
Forgetting “fully justify” means showing BOTH the concavity-change argument and computing f″ on each side. Just stating “f″(2) = 0” loses marks.

Up next: Derivatives & Graphs. Putting it all together — given the graph of f, sketch the graphs of f′ and f″, and vice versa. Roots of f′ correspond to stationary points of f; roots of f″ correspond to points of inflection of f. The interplay between f, f′, and f″ in a single problem is a Paper 1 favourite.

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