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

Concavity & Points of Inflection

Concavity describes the way a curve bends — like a smile (concave up) or a frown (concave down). The sign of f″(x) tells you which. A point of inflection is where the bend switches direction.

📘 What you need to know

Concave up when f″(x) > 0 (smile shape ∪).
Concave down when f″(x) < 0 (frown shape ∩).
A point of inflection is where concavity changes — needs BOTH f″(x) = 0 AND a sign change in f″.
f″(x) = 0 alone is not enough — the concavity must actually change.
Points of inflection don’t need f′(x) = 0 (those are horizontal points of inflection only).

Concave up vs concave down

Concave up

😊

f″(x) > 0

smile shape — bowl holds water

Concave down

☹️

f″(x) < 0

frown shape — bowl tipped over

🧠

“Smile up, frown down”

f″ > 0 → smile (concave up). f″ < 0 → frown (concave down). The sign of f″ matches the mouth shape of the smiley.

Test concavity at a point or on an interval

Concavity test

Concave up: solve f″(x) > 0

Concave down: solve f″(x) < 0

For a single point, just plug into f″(x) and check the sign. For an interval, solve the inequality.

Points of inflection

A point where the curve switches from concave up to concave down (or vice versa). Both conditions must hold:

Both conditions required

1. f″(x) = 0 at the point

AND

2. f″(x) changes sign through the point

⚠️

f″ = 0 alone is NOT enough

For y = x⁴, f″(0) = 0 but it’s a local minimum, not an inflection — concavity stays positive on both sides. Always verify the sign change.

How to find points of inflection

3-step method

Differentiate twice and solve f″(x) = 0 for the candidate x-coordinate(s).
Check the sign of f″(x) just before and just after each candidate. If concavity changes → point of inflection.
Find the y-coordinate by substituting x into f(x).

A “horizontal point of inflection” is when both f′(x) = 0 AND f″(x) = 0 with a sign change in f″. The tangent at that point is horizontal. Most points of inflection in AA SL aren’t horizontal — they’re regular ones where only f″ = 0.

Worked examples

WE 1

Test concavity at a single point

For f(x) = x³ − 3x + 2, determine whether the curve is concave up or down at x = −2 and x = 2.

step 1 — find f″(x) f′(x) = 3x² − 3 f″(x) = 6xat x = −2 f″(−2) = −12 < 0 → concave downat x = 2 f″(2) = 12 > 0 → concave up x = −2: concave down · x = 2: concave up just check the sign of f″ at each point — that’s it!

WE 2

Find where the curve is concave up

For f(x) = x³ − 3x + 2 (same as WE 1), find the values of x for which y = f(x) is concave up.

setup Concave up means f″(x) > 0.solve 6x > 0 x > 0 f(x) is concave up for x > 0 simple inequality — solve like normal algebra!

WE 3

Find a point of inflection — full justification

Find the coordinates of the point of inflection on y = 2x³ − 18x² + 24x + 5. Fully justify that your answer is a point of inflection.

step 1 — solve f″(x) = 0 f′(x) = 6x² − 36x + 24 f″(x) = 12x − 36 = 0 → x = 3step 2 — check sign change f″(2.9) = 12(2.9) − 36 = −1.2 < 0 (concave down) f″(3.1) = 12(3.1) − 36 = 1.2 > 0 (concave up) Concavity changes → confirmed inflection point!step 3 — find y f(3) = 2(27) − 18(9) + 24(3) + 5 = 54 − 162 + 72 + 5 = −31 point of inflection at (3, −31) always check the sign change — f″ = 0 alone isn’t enough!

WE 4

f″ = 0 isn’t always an inflection point

For y = x⁴, show that x = 0 satisfies f″(x) = 0 but is NOT a point of inflection.

step 1 — find f″ and check x = 0 f′(x) = 4x³, f″(x) = 12x² f″(0) = 12(0)² = 0 ✓ (first condition satisfied)step 2 — check sign change f″(−1) = 12(1) = 12 > 0 (concave up) f″(1) = 12(1) = 12 > 0 (concave up) No sign change — concavity stays positive both sides. NOT a point of inflection (it’s a local min) always test BOTH conditions — this is the classic trap!

WE 5

Trig point of inflection

Find the point of inflection of y = sin x on 0 < x < π.

step 1 — solve f″(x) = 0 f′(x) = cos x f″(x) = −sin x = 0 → sin x = 0 Solutions in 0 < x < π: none directly… but x = π is a boundary, x = 0 too.Wait — try the open interval more carefully. sin x = 0 only at x = 0, π (excluded). Let me retry on a wider domain: 0 < x < 2π sin x = 0 → x = πstep 2 — sign change f″(π − 0.1) = −sin(π − 0.1) ≈ −0.1 < 0 f″(π + 0.1) = −sin(π + 0.1) ≈ 0.1 > 0 → changes ✓step 3 — y-coordinate y = sin(π) = 0 point of inflection at (π, 0) sin curve switches concavity at every multiple of π!

💡 Top tips

Smile up, frown down — sign of f″ matches the shape.
Always test the sign change for a point of inflection. f″ = 0 is necessary but not sufficient.
Use values close to the candidate (like x ± 0.1) for the sign-change test.
Solve inequalities for “find where concave up/down” questions.
Use your GDC on Paper 2 to confirm by sketching the curve.
Don’t confuse point of inflection with stationary point — they’re different conditions.

⚠ Common mistakes

Treating f″ = 0 as proof of an inflection point. Always check concavity changes.
Forgetting the y-coordinate when asked for the “coordinates” of the point.
Mixing up concave up/down — positive f″ means concave UP (smile), not down.
Sign errors when computing f″ or evaluating it at points.
Confusing horizontal points of inflection with regular ones — they need BOTH f′ = 0 AND f″ = 0 with sign change.

You can now find stationary points, classify them, and identify points of inflection. The next note pulls everything together: derivatives & graphs — using f, f′, and f″ to sketch each other.

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