IB Maths AA SL Topic 5 — Calculus Paper 1 & 2 🎯 Skill ~4 min practice

AA SL Stationary Points skills

A stationary point is where the gradient is zero — a max, a min, or a flat point of inflection. Set f‘(x) = 0 to find them, then use the second derivative to classify. Three steps, every time.

The Method

f‘(x) = 0 stationary point ⇔ gradient is zero · classify with f”

Find f‘(x) and set it equal to zero. Solve for x — these are your stationary x-values.
Find the y-values by plugging each x back into f(x) (not the derivative!).
Classify each point using the second derivative test: compute f'' at each x, then read off max/min/inflection from the sign.

Type 1

Local Maximum

f”(x) < 0

curve concave down

Type 2

Local Minimum

f”(x) > 0

curve concave up

Type 3

Point of Inflection

f”(x) = 0

need further checks

The second derivative test — classify each stationary point

Compute f''(x) by differentiating f‘(x) again.
Plug in each stationary x-value into f”.
If f''(x) < 0 → local maximum (concave down).
If f''(x) > 0 → local minimum (concave up).
If f''(x) = 0 → inconclusive. Check the sign of f‘ on either side instead.

Worked examples

WE 1 EASY

Find the stationary point of y = x² − 6x + 5 and classify it.

step 1 — find f’ and set to 0 dy/dx = 2x − 6 = 0 x = 3step 2 — find y y = 9 − 18 + 5 = −4step 3 — classify with f” d²y/dx² = 2 > 0 → minimumlocal min at (3, −4) positive parabola = always opens up = always a minimum!

WE 2 MEDIUM

Find and classify the stationary points of y = x³ − 3x² − 9x + 5.

step 1 — derivative = 0 dy/dx = 3x² − 6x − 9 = 0 divide by 3: x² − 2x − 3 = 0 (x − 3)(x + 1) = 0 x = 3 or x = −1step 2 — find y values at x = 3: y = 27 − 27 − 27 + 5 = −22 at x = −1: y = −1 − 3 + 9 + 5 = 10step 3 — classify with f” d²y/dx² = 6x − 6 at x = 3: 12 > 0 → min at x = −1: −12 < 0 → maxmax at (−1, 10) · min at (3, −22) cubics typically have 2 stationary points — one max, one min!

WE 3 HARD

Find the stationary points of y = x⁴ − 4x³ and classify each.

step 1 — derivative = 0 dy/dx = 4x³ − 12x² = 0 4x²(x − 3) = 0 x = 0 or x = 3step 2 — find y values at x = 0: y = 0 at x = 3: y = 81 − 108 = −27step 3 — classify d²y/dx² = 12x² − 24x at x = 0: 0 → inconclusive! check f’ sign around x = 0: at x = −0.5: 4(−0.5)³ − 12(0.25) = −0.5 − 3 = −3.5 (neg) at x = 0.5: 0.5 − 3 = −2.5 (neg) same sign → point of inflection at (0, 0) at x = 3: 108 − 72 = 36 > 0 → mininflection at (0, 0) · min at (3, −27) when f”=0, fall back on the f’ sign-change test — the only safe way!

Practice questions

Try each one yourself first, then click the question to reveal the worked answer. Always set f‘(x) = 0 first, then classify with f”.

Q1 EASY Find and classify the stationary point of y = x² + 4x − 1. Show answer ▼Hide answer ▲

dy/dx = 2x + 4 = 0 → x = −2 y = 4 − 8 − 1 = −5 d²y/dx² = 2 > 0 → minimum min at (−2, −5)

Q2 EASY Find and classify the stationary point of y = 5 − x². Show answer ▼Hide answer ▲

dy/dx = −2x = 0 → x = 0 y = 5 d²y/dx² = −2 < 0 → maximum max at (0, 5)

Q3 MEDIUM Find and classify the stationary points of y = x³ − 12x + 4. Show answer ▼Hide answer ▲

dy/dx = 3x² − 12 = 0 x² = 4 → x = ±2 at x = 2: y = 8 − 24 + 4 = −12 at x = −2: y = −8 + 24 + 4 = 20 d²y/dx² = 6x x=2: 12 > 0 → min x=−2: −12 < 0 → max max at (−2, 20) · min at (2, −12)

Q4 MEDIUM Find the stationary point of y = x + 4/x for x > 0, and classify it. Show answer ▼Hide answer ▲

rewrite y = x + 4x⁻¹ dy/dx = 1 − 4/x² = 0 x² = 4 → x = 2 (since x > 0) y = 2 + 4/2 = 4 d²y/dx² = 8/x³ at x=2: 8/8 = 1 > 0 → min min at (2, 4) always rewrite fractions as powers — easier to differentiate twice!

Q5 HARD Find the stationary points of y = 2x³ − 9x² + 12x − 3 and identify the local max value. Show answer ▼Hide answer ▲

dy/dx = 6x² − 18x + 12 = 0 divide by 6: x² − 3x + 2 = 0 (x − 1)(x − 2) = 0 → x = 1, 2 at x = 1: y = 2 − 9 + 12 − 3 = 2 at x = 2: y = 16 − 36 + 24 − 3 = 1 d²y/dx² = 12x − 18 x=1: −6 < 0 → max x=2: 6 > 0 → min local max value = 2 (at x = 1) “max value” means the y-coordinate, not the point itself!

⚠ Common mistakes

Plugging x back into the derivative instead of the original function. y₁ = f(x₁), not f‘(x₁) — that’s always 0!
Mixing up max and min signs. f” > 0 → MIN (smile shape). f” < 0 → MAX (frown shape).
Stopping when f” = 0. The test is inconclusive — fall back on the sign of f‘ on either side instead.
Reporting only x or only y. A stationary point is the full coordinate (x, y) — give both.
Calling “max value” the point. Max value = the y-coordinate. Max point = the (x, y). Read the question.
Forgetting to check the domain. If x is restricted (e.g. x > 0), discard solutions that fall outside.

📖

Want the theory?

Read the full Stationary Points notes for the link to concavity, the proof of the second derivative test, and how stationary points appear in optimisation problems.

Need help with Stationary Points?

Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.

Book Free Session →