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NotesMath AI HLTopic 5.6Stationary Points and Their Nature
Back to Math AI HL Topics
5.6.12 min read

Stationary Points and Their Nature

IB Mathematics: Applications and Interpretation • Unit 5

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Contents

  • What is a stationary point?
  • Finding stationary points
  • Classifying with the second derivative
  • Classifying with a sign diagram
Definition: A stationary point is any x-value where f′(x) = 0.

At a stationary point, the tangent to the curve is horizontal. The function is momentarily neither increasing nor decreasing.
TypeWhat it looks likeSign of f′ either side
Local maximumCurve peaks and turns down+ then −
Local minimumCurve dips and turns up− then +
Point of inflectionCurve flattens but keeps direction+ then + (or − then −)

The word 'local' means it is the highest or lowest point in a nearby region — there may be higher or lower points elsewhere on the curve.

Stationary ≠ turning point: All turning points are stationary points.

But not all stationary points are turning points — a point of inflection is stationary but does not change direction.

[Diagram: math-stationary-points] - Available in full study mode

Method: find all stationary points

  • Differentiate: find f′(x).
  • Set f′(x) = 0 and solve for x.
  • Find the y-coordinate: substitute each x into f(x).
  • State the coordinates.

Example

Step by step

  1. f(x) = 2x³ − 9x² + 12x − 1
  2. Step 1 — f′(x) = 6x² − 18x + 12 = 6(x² − 3x + 2) = 6(x − 1)(x − 2)
  3. Step 2 — Set = 0: x = 1 or x = 2
  4. Step 3 — f(1) = 2 − 9 + 12 − 1 = 4. f(2) = 16 − 36 + 24 − 1 = 3.
  5. Result — Stationary points at (1, 4) and (2, 3).
Always find y-coordinates: Many students forget to substitute x back into f(x).

The question asks for coordinates — both x and y are required.

IB-style question — stationary points of a cubic [7 marks]

A function is defined by f(x) = 2x³ − 3x² − 36x + 5.

(a) Find the coordinates of the stationary points of f.

(b) Classify each stationary point.

(c) State the interval on which f is decreasing.

Step by step

  1. (a) Differentiate, then set the derivative equal to zero.
  2. Take out the common factor and factorise the quadratic.
  3. Substitute each x into f(x) for the y-coordinates.
  4. (b) Graph f on the GDC, or use a sign diagram of f′. The left point is a peak, the right point is a valley.
  5. (c) f is decreasing where f′(x) < 0, which is between the two stationary x-values.

Final answer

(a) (−2, 49) and (3, −76). (b) (−2, 49) is a local maximum; (3, −76) is a local minimum. (c) f is decreasing for −2 < x < 3.

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Classifying with the GDC

Two AI SL ways to classify: Once you have found a stationary point, decide whether it is a local maximum or minimum using EITHER the GDC graph OR a first-derivative sign diagram (next section).

In AI SL you do not use the second derivative.
The GDC method: Graph y = f(x) on the GDC.

A local maximum is a peak (the curve turns from rising to falling); a local minimum is a valley (falling to rising).

Use the GDC's maximum and minimum tools to read the coordinates directly.

Classifying with the GDC

Classify the stationary points of f(x) = 2x³ − 9x² + 12x − 1 (found at x = 1 and x = 2).

Step by step

  1. Graph y = 2x³ − 9x² + 12x − 1 on the GDC.
  2. The GDC maximum tool gives a peak at (1, 4).
  3. The GDC minimum tool gives a valley at (2, 3).

Final answer

Local maximum at (1, 4); local minimum at (2, 3).

GDC allowed on both papers: For a quick classification, graph the function and read which stationary point is the peak and which is the valley.

For an analytic justification, use the first-derivative sign diagram (next section).

The sign diagram of f′(x) is the most reliable classification method — it always works, even when the second derivative test fails.

Classification rules from sign diagram: At a stationary point x = a: • f′ changes + → − : local maximum • f′ changes − → + : local minimum • f′ stays + → + (or − → −) : point of inflection

Same example — sign diagram approach

Step by step

  1. Stationary points at x = 1 and x = 2
  2. Test x = 0 (region x < 1) — f′(0) = 12 > 0 → +
  3. Test x = 1.5 (region 1 < x < 2) — f′(1.5) = 6(2.25) − 27 + 12 = −1.5 < 0 → −
  4. Test x = 3 (region x > 2) — f′(3) = 54 − 54 + 12 = 12 > 0 → +
  5. Sign diagram — x: 1 2 f′: + | − | + max min
  6. Result — Local maximum at x = 1. Local minimum at x = 2.
IB exam: both methods accepted: You can use either the second derivative or a sign diagram.

Many students prefer the sign diagram because it shows the full picture.

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f(x) = x² + 4x − 12. Find the minimum value of f. [2 marks]

Related Math AI HL Topics

Continue learning with these related topics from the same unit:

5.1.1Introduction to Limits
5.10.1The second derivative & concavity
5.11.1Integration techniques
5.12.1Area under and between curves
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