Mathematics Advanced • Year 11 • Module 3 • Lesson 8

Stationary Points

Build procedural fluency in finding and classifying stationary points using the second-derivative test or a sign table.

Build · Skill Drill

1. Quick recall

Answer each in the space provided. 1 mark each

Q1.1 Define a stationary point in one line, using algebraic language.

A stationary point is a point on a curve where ______________________.

Q1.2 Complete the second-derivative test. Assume f′(a) = 0.

If f″(a) ______ 0, then (a, f(a)) is a local maximum.

If f″(a) ______ 0, then (a, f(a)) is a local minimum.

If f″(a) ______ 0, the test is inconclusive; use a sign table for f′(x).

Q1.3 A stationary point is a point (x, y). The y-coordinate is found by substituting the stationary x-value back into ____________ (write "f(x)" or "f′(x)").

Stuck? Revisit lesson § Key Terms and § Concept (formula box).

2. Worked example — classify the stationary points of f(x) = x³ − 3x²

Follow the find → coordinate → classify pattern.

Problem. Find and classify all stationary points of f(x) = x³ − 3x².

Step 1 — Differentiate and set f′(x) = 0.

f′(x) = 3x² − 6x = 3x(x − 2) = 0 ⇒ x = 0 or x = 2

Reason: factorising before solving is faster and cleaner than the quadratic formula.

Step 2 — Find each y-coordinate.

f(0) = 0; f(2) = 8 − 12 = −4

Reason: stationary points are (x, y), not just x-values.

Step 3 — Classify using the second derivative test.

f″(x) = 6x − 6

f″(0) = −6 < 0 ⇒ (0, 0) is a local MAXIMUM

f″(2) = 6 > 0 ⇒ (2, −4) is a local MINIMUM

Reason: concave down at a stationary point means it bends like a cap (max); concave up means it bends like a cup (min).

Conclusion. Local max at (0, 0); local min at (2, −4).

3. Faded example — fill in the missing steps

Find and classify the stationary points of f(x) = x³ − 3x + 2. Fill in each blank. 4 marks

Step 1 — Differentiate:

f′(x) = ______________________

Step 2 — Solve f′(x) = 0:

______________________ = 0 ⇒ x = ______ or x = ______

Step 3 — Find each y-coordinate:

f( ____ ) = ________ ; f( ____ ) = ________

Step 4 — Second derivative: f″(x) = ____________

Step 5 — Classify using f″:

f″( ____ ) = ______, which is ______ 0, so ( ____ , ____ ) is a local __________.

Stuck? Revisit lesson § Worked Example 2 (second derivative test).

4. Graduated practice — find and classify

For each, find all stationary points (with coordinates) and classify each.

Foundation — single stationary point (4 questions)

Q	Function	Stationary point(s) (x, y)	Nature
4.1 1	f(x) = x² − 4x + 3
4.2 1	f(x) = −x² + 6x
4.3 1	f(x) = x² + 2x + 5
4.4 1	f(x) = 5 − (x − 1)²

Standard — cubics and beyond (6 questions)

4.5 Find and classify the stationary points of f(x) = x³ − 6x² + 9x + 1. 3 marks

4.6 Find and classify the stationary points of f(x) = x³ − 3x. 3 marks

4.7 Find and classify the stationary points of f(x) = x⁴ − 2x². 3 marks

4.8 Show that f(x) = x³ has a stationary point at x = 0 and classify it (you will need a sign table for f′ because f″(0) = 0). 2 marks

4.9 For f(x) = x³ − 3x² + 4, find both stationary points and classify each. 3 marks

4.10 Find and classify the stationary points of f(x) = x⁴ − 4x³. (Watch for a horizontal inflection.) 3 marks

Extension — work backwards (2 questions)

4.11 The curve y = x³ + ax² + bx has a stationary point at (1, −4). Find a and b, and use the second derivative test to determine the nature of the stationary point at (1, −4). 3 marks

4.12 Sketch the curve y = x³ − 3x showing all stationary points clearly. Indicate where the curve is concave up and concave down. 3 marks

Stuck on 4.8 or 4.10? When f″(x) = 0, examine the sign of f′(x) just before and just after the stationary x-value.

5. Self-check the easy 3

Tick the first three once you have checked.

For 4.1 I wrote the stationary point as (2, −1) with the y-coordinate (not just x = 2).

For 4.2 I noticed the coefficient of x² is negative, so the parabola opens down and the stationary point is a maximum.

For 4.3 my discriminant of f′(x) = 2x + 2 has only one solution, x = −1, so there is exactly one stationary point.

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What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Definition of stationary point

A point on a curve where f′(x) = 0 (the gradient / first derivative is zero, i.e. the tangent is horizontal).

Q1.2 — Second derivative test

f″(a) < 0 ⇒ local maximum. f″(a) > 0 ⇒ local minimum. f″(a) = 0 ⇒ inconclusive.

Q1.3 — Finding the y-coordinate

Substitute back into f(x) (the original function), not into f′(x).

Q3 — Faded example f(x) = x³ − 3x + 2

Step 1: f′(x) = 3x² − 3. Step 2: 3x² − 3 = 0 ⇒ x² = 1 ⇒ x = 1 or x = −1. Step 3: f(1) = 1 − 3 + 2 = 0; f(−1) = −1 + 3 + 2 = 4. Step 4: f″(x) = 6x. Step 5: f″(1) = 6, > 0 ⇒ (1, 0) is a local minimum. f″(−1) = −6, < 0 ⇒ (−1, 4) is a local maximum.

Q4.1 — f(x) = x² − 4x + 3

f′(x) = 2x − 4 = 0 ⇒ x = 2; f(2) = 4 − 8 + 3 = −1. f″(x) = 2 > 0 ⇒ local min at (2, −1).

Q4.2 — f(x) = −x² + 6x

f′(x) = −2x + 6 = 0 ⇒ x = 3; f(3) = −9 + 18 = 9. f″(x) = −2 < 0 ⇒ local max at (3, 9).

Q4.3 — f(x) = x² + 2x + 5

f′(x) = 2x + 2 = 0 ⇒ x = −1; f(−1) = 1 − 2 + 5 = 4. f″(x) = 2 > 0 ⇒ local min at (−1, 4).

Q4.4 — f(x) = 5 − (x − 1)²

f′(x) = −2(x − 1) = 0 ⇒ x = 1; f(1) = 5. f″(x) = −2 < 0 ⇒ local max at (1, 5).

Q4.5 — f(x) = x³ − 6x² + 9x + 1

f′(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3) = 0 ⇒ x = 1 or x = 3. f(1) = 1 − 6 + 9 + 1 = 5; f(3) = 27 − 54 + 27 + 1 = 1. f″(x) = 6x − 12; f″(1) = −6 < 0 ⇒ local max at (1, 5). f″(3) = 6 > 0 ⇒ local min at (3, 1).

Q4.6 — f(x) = x³ − 3x

f′(x) = 3x² − 3 = 3(x² − 1) = 0 ⇒ x = ±1. f(−1) = −1 + 3 = 2; f(1) = 1 − 3 = −2. f″(x) = 6x; f″(−1) = −6 < 0 ⇒ local max at (−1, 2). f″(1) = 6 > 0 ⇒ local min at (1, −2).

Q4.7 — f(x) = x⁴ − 2x²

f′(x) = 4x³ − 4x = 4x(x² − 1) = 4x(x − 1)(x + 1) = 0 ⇒ x = 0, ±1. f(0) = 0; f(±1) = 1 − 2 = −1. f″(x) = 12x² − 4; f″(0) = −4 < 0 ⇒ local max at (0, 0). f″(±1) = 12 − 4 = 8 > 0 ⇒ local min at (1, −1) and (−1, −1).

Q4.8 — f(x) = x³ at x = 0

f′(x) = 3x²; f′(0) = 0, so there is a stationary point at (0, 0). f″(x) = 6x; f″(0) = 0 (test inconclusive). Sign table for f′: at x = −0.1, f′ = 0.03 > 0; at x = +0.1, f′ = 0.03 > 0. f′ does not change sign, so (0, 0) is a horizontal point of inflection (a stationary inflection).

Q4.9 — f(x) = x³ − 3x² + 4

f′(x) = 3x² − 6x = 3x(x − 2) = 0 ⇒ x = 0 or x = 2. f(0) = 4; f(2) = 8 − 12 + 4 = 0. f″(x) = 6x − 6; f″(0) = −6 < 0 ⇒ local max at (0, 4). f″(2) = 6 > 0 ⇒ local min at (2, 0).

Q4.10 — f(x) = x⁴ − 4x³

f′(x) = 4x³ − 12x² = 4x²(x − 3) = 0 ⇒ x = 0 (double root) or x = 3. f(0) = 0; f(3) = 81 − 108 = −27. f″(x) = 12x² − 24x; f″(0) = 0 (test further). Sign of f′: at x = −1, f′ = −4 − (−12) = −16 (−); at x = 1, f′ = 4 − 12 = −8 (−). f′ does not change sign at x = 0, so (0, 0) is a horizontal inflection. f″(3) = 108 − 72 = 36 > 0 ⇒ local min at (3, −27).

Q4.11 — y = x³ + ax² + bx with stationary point (1, −4)

Two conditions: y(1) = 1 + a + b = −4 ⇒ a + b = −5. y′(x) = 3x² + 2ax + b; y′(1) = 3 + 2a + b = 0 ⇒ 2a + b = −3. Subtracting: a = 2, hence b = −7. y″(x) = 6x + 2a = 6x + 4; y″(1) = 10 > 0 ⇒ (1, −4) is a local minimum.

Q4.12 — Sketch of y = x³ − 3x

From Q4.6: local max (−1, 2), local min (1, −2). y″ = 6x: concave down for x < 0 (left half) and concave up for x > 0 (right half). Point of inflection at (0, 0). Sketch is a standard cubic rising through the max, falling through the inflection, falling further to the min, then rising again. x-intercepts: x³ = 3x ⇒ x(x² − 3) = 0 ⇒ x = 0, ±√3.