IB Maths AA SLTopic 5 — CalculusPaper 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).
Solvef′(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 — differentiatedy/dx = 2x − 6step 2 — set = 02x − 6 = 0 → x = 3step 3 — find yy = (3)² − 6(3) + 5 = 9 − 18 + 5 = −4stationary 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.
Find and classify the stationary point of y = sin x + cos x for 0 ≤ x ≤ π.
step 1 — differentiate & solvedy/dx = cos x − sin x = 0 → tan x = 1x = π/4step 2 — y-coordinatey = sin(π/4) + cos(π/4) = √2/2 + √2/2 = √2step 3 — second derivatived²y/dx² = −sin x − cos xat x = π/4: −√2/2 − √2/2 = −√2 < 0 → maximumlocal 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−xf′(x) = x(−e−x) + e−x(1) = e−x(1 − x)step 2 — solve f′(x) = 0e−x ≠ 0, so 1 − x = 0 →x = 1f(1) = 1 · e−1 = 1/estep 3 — classifyf″(x) = e−x(x − 2)f″(1) = e−1(−1) = −1/e < 0 → maximumlocal 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 itdy/dx = 4x³ = 0 → x = 0y = 0step 2 — second derivative testd²y/dx² = 12x²at x = 0: 12(0)² = 0 → INCONCLUSIVEstep 3 — first derivative testf′(−1) = 4(−1)³ = −4 < 0f′(1) = 4(1)³ = 4 > 0Sign goes − to + → minimumlocal 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.
Need help with Stationary Points?
Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.
