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

Stationary Points

A stationary point is where the curve is momentarily flat — gradient = 0. To find them, set f′(x) = 0 and solve. To classify them as a max, min, or point of inflection, use the second derivative. Together, they pin down everything you need to know about the shape of a curve.

📘 What you need to know

Stationary point: f′(x) = 0 (curve is flat there).
Three types: local maximum, local minimum, or stationary point of inflection.
Find them: solve f′(x) = 0 for x-coordinates. Sub back into f(x) for y-coordinates.
Classify them with the second derivative test: f″(x) > 0 → minimum · f″(x) < 0 → maximum · f″(x) = 0 → use first derivative test instead.

The three types

Local maximum

f″(x) < 0

curve goes up then down

Local minimum

f″(x) > 0

curve goes down then up

Inflection

f″(x) = 0

flat moment, same direction either side

A “turning point” means a max or min — the curve actually turns direction. A “stationary point of inflection” is flat for a moment but doesn’t turn — it keeps going the same way.

How to find stationary points

3-step method

Differentiate to get f′(x).
Solve f′(x) = 0 for the x-coordinates.
Substitute each x-value into f(x) for the y-coordinates.

Classify with the second derivative

At a stationary point, if…	Then it is a…
f″(x) > 0	Local minimum (curve concave up)
f″(x) < 0	Local maximum (curve concave down)
f″(x) = 0	Inconclusive — use the first derivative test instead

🧠

“Positive = pit, negative = peak”

f″ > 0 → bowl shape (pit) → minimum at the bottom. f″ < 0 → upside-down bowl (peak) → maximum at the top. The sign of f″ matches the shape.

Backup: the first derivative test

If f″(x) = 0 at a stationary point, you can’t classify it with the second derivative. Instead, check the sign of f′(x) just before and just after the point.

Sign change in f′(x)	Type
− then + (down then up)	Local minimum
+ then − (up then down)	Local maximum
same sign both sides	Stationary point of inflection

📍

Use your GDC if allowed

On Paper 2, your GDC will solve f′(x) = 0 and find max/min coordinates directly in graphing mode. Use it to check answers — but for “show that” questions, you must do it algebraically.

Worked examples

WE 1

Find a stationary point

Find the coordinates of the stationary point on the curve y = x² − 6x + 5.

step 1 — differentiate dy/dx = 2x − 6step 2 — set = 0 2x − 6 = 0 → x = 3step 3 — find y y = (3)² − 6(3) + 5 = 9 − 18 + 5 = −4 stationary point at (3, −4) d²y/dx² = 2 > 0 → it’s a minimum (positive quadratic, makes sense!)

WE 2

Full classification of a cubic

Find the coordinates and nature of the stationary points on y = 2x³ − 3x² − 36x + 25.

step 1 — differentiate & solve f′(x) = 6x² − 6x − 36 = 6(x² − x − 6) = 6(x − 3)(x + 2) f′(x) = 0 → x = 3 or x = −2step 2 — y-coordinates f(3) = 2(27) − 3(9) − 108 + 25 = −56 f(−2) = 2(−8) − 3(4) + 72 + 25 = 69step 3 — classify (second derivative) f″(x) = 12x − 6 f″(3) = 30 > 0 → minimum f″(−2) = −30 < 0 → maximum (3, −56) local min · (−2, 69) local max always factor f′(x) to find the roots quickly!

WE 3

Stationary point of a trig function

Find and classify the stationary point of y = sin x + cos x for 0 ≤ x ≤ π.

step 1 — differentiate & solve dy/dx = cos x − sin x = 0 → tan x = 1 x = π/4step 2 — y-coordinate y = sin(π/4) + cos(π/4) = √2/2 + √2/2 = √2step 3 — second derivative d²y/dx² = −sin x − cos x at x = π/4: −√2/2 − √2/2 = −√2 < 0 → maximum local max at (π/4, √2) remember to keep your GDC in radians!

WE 4

Stationary point of f(x) = x e^(−x)

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

step 1 — differentiate (product rule) u = x, v = e^−x; u′ = 1, v′ = −e^−x f′(x) = x(−e^−x) + e^−x(1) = e^−x(1 − x)step 2 — solve f′(x) = 0 e^−x ≠ 0, so 1 − x = 0 → x = 1 f(1) = 1 · e⁻¹ = 1/estep 3 — classify f″(x) = e^−x(x − 2) f″(1) = e⁻¹(−1) = −1/e < 0 → maximum local max at (1, 1/e) e^−x is always positive — it never makes the derivative zero on its own!

WE 5

When the second derivative test fails

Find and classify the stationary point of y = x⁴.

step 1 — find it dy/dx = 4x³ = 0 → x = 0 y = 0step 2 — second derivative test d²y/dx² = 12x² at x = 0: 12(0)² = 0 → INCONCLUSIVEstep 3 — first derivative test f′(−1) = 4(−1)³ = −4 < 0 f′(1) = 4(1)³ = 4 > 0 Sign goes − to + → minimum local min at (0, 0) when f″ = 0, fall back to the first derivative test!

💡 Top tips

Factor f′(x) before solving = 0. It usually gives multiple roots cleanly.
Always find the y-coordinate by subbing into f(x) — not f′(x) or f″(x).
“Positive = pit, negative = peak” for the second derivative test.
If f″ = 0, switch to the first derivative test by checking signs either side.
Use your GDC to verify max/min coordinates on Paper 2.
Watch the language: “find” needs coordinates; “classify” or “nature” needs the type too.

⚠ Common mistakes

Stopping at the x-coordinate. A “stationary point” is (x, y) — sub back to find y.
Subbing into f′(x) or f″(x) for the y-coordinate instead of f(x).
Forgetting to classify when the question asks for the nature.
Using f″(x) = 0 as proof of inflection — it’s only a possibility, you must check sign change.
Sign errors when factorising f′(x) or computing f″ at a point.

Stationary points are about flat points. Next: concavity & points of inflection — looking at how the whole curve bends, not just specific spots.

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