SL 5.6 — Stationary points, local maxima & minima

Stationary points occur where the instantaneous rate of change (the derivative) is zero: f'(x) = 0.
These points can be local maxima, local minima, or points of inflection with horizontal tangent.

Term / Concept Definition / Short explanation
Stationary point Point x = c where f'(c) = 0 (horizontal tangent). Must check classification using tests.
Local maximum A point where f(c) is greater than nearby values. Occurs when derivative changes + → − (first derivative) or f”(c) < 0 (second derivative test).
Local minimum A point where f(c) is less than nearby values. Occurs when derivative changes − → + (first derivative) or f”(c) > 0.
Point of inflection (horizontal tangent) Stationary point where concavity changes sign (f” changes sign). The tangent is horizontal but the point is neither max nor min.
First derivative test Examine sign of f'(x) either side of the stationary point to classify it.
Second derivative test If f”(c) > 0 → local min; f”(c) < 0 → local max; if f”(c)=0 test inconclusive (use first derivative test).

📌 1. Finding stationary points

  1. Compute f'(x).
  2. Solve f'(x) = 0 for candidate x-values (stationary points).
  3. Classify each candidate using either the first derivative test (check sign change of f’ around the point) or the second derivative test (compute f”(c)).
  4. Include endpoints of domain when identifying absolute extrema on a closed interval.

🔍 TOK Perspective

Stationary points are mathematical predictions about local behaviour — consider limitations: are assumptions (smoothness, differentiability) justified when modelling physical systems? How does measurement noise affect detection of stationary points?

📌 2. First derivative test & second derivative test

First derivative test (most reliable when you can evaluate sign): check sign of f'(x) just left and right of x = c:

  • If f’ changes from + to − → local maximum at c.
  • If f’ changes from − to + → local minimum at c.
  • If f’ does not change sign → not a max/min (possible horizontal inflection).

Curve sketching

🌍 Real-world connections (when each case occurs)

Local maximum (application):

Example — profit optimization: if P(x) is profit for producing x units, a local maximum stationary point gives the production level where marginal profit = 0 and profit is locally highest. Use when marginal cost/revenue curves meet.

Local minimum (application):

Example — engineering stress: if S(x) is stress vs design parameter x, a local minimum of S may indicate the safest design locally. Designers look for minima of cost, stress or error functions.

Horizontal inflection (application):

Example — kinematics: displacement s(t) may have a horizontal tangent (velocity 0) at turnaround point; if curvature changes sign it indicates motion changing acceleration pattern (e.g., from speeding up to slowing down) — not necessarily a max/min of displacement in a larger domain.

🧠 Examiner Tip

  • Always state: (1) derivative f'(x), (2) solutions f'(x)=0, (3) classification (first/second derivative test) and (4) coordinates (x, f(x)) — this full structure earns method marks.
  • If f”(c)=0 do not panic — explicitly perform sign check on f'(x) around c (show small table of signs or use sample points).

Second derivative test (faster when f” is easy to evaluate):

  • Compute f”(c). If f”(c) > 0 → local minimum (concave up).
  • If f”(c) < 0 → local maximum (concave down).
  • If f”(c) = 0 → constant change; use first derivative test or higher derivatives.

Second Derivative Test: Meaning, Types & Formula with Examples

Worked example 1 (polynomial)

Let f(x) = x3 − 3x2 − 9x + 5.
1) f'(x) = 3x2 − 6x − 9 = 3(x2 − 2x − 3) = 3(x−3)(x+1).
2) Solve f'(x)=0 ⇒ x = 3, x = −1 (stationary candidates).
3) f”(x) = 6x − 6. Evaluate: f”(3)=12 >0 → local minimum at x=3. f”(−1)=−12 <0 → local maximum at x=−1.
4) Compute values if needed: f(3)=33−3·32−9·3+5 = 27−27−27+5 = −22 (local min value = −22). f(−1)=−1−3−(−9)+5 = 10 (local max value = 10).

Worked example 2 (horizontal inflection)

f(x) = x3. f'(x) = 3x2. Stationary candidate: x=0 because f'(0)=0. f”(x) = 6x so f”(0)=0 (test inconclusive). Check sign of f’ either side: for x<0 f’ >0? (3x2 always ≥0), actually f’≥0 both sides (nonnegative) and function changes from negative to positive values; x=0 is a point of inflection with horizontal tangent (not a max/min).

📱 GDC Tips — Stationary Points

  • TI-Nspire CX II (Graphical / CAS):Enter the function f(x) in the Graphs application.
    Open Menu → Analyse Graph → Calculus → d/dx to display the derivative curve.
    Stationary points occur where this derivative graph crosses the x-axis.
  • To find exact stationary x-values, use
    Menu → Analyse Graph → Zero on the derivative graph,
    or use the CAS command solve(f′(x)=0, x).
    Record all solutions within the given domain.
  • To classify the stationary point, evaluate the second derivative:
    Menu → Analyse Graph → Calculus → d²/dx²,
    then substitute the stationary x-value.
    If the result is positive → minimum, negative → maximum.
  • If the second derivative is zero, plot f′(x) and check sign changes visually
    (or test values on either side of the point) to determine whether the point is
    a maximum, minimum, or horizontal point of inflection.
  • Casio fx-CG50 / CG100:Enter the function in the GRAPH menu.
    Use G-Solv → d/dx to compute the gradient at points,
    or use G-Solv → Root on the derivative to find where f′(x)=0.
  • After locating stationary x-values, use G-Solv → Min or G-Solv → Max
    to confirm classification, but always support your answer
    with derivative reasoning in written solutions.
  • For IB exams, explicitly state when GDC is used
    (e.g. “Stationary points found using GDC”)
    and still show the analytical derivative to earn full method marks.

📐 IA Spotlight

  • IA idea: fit a polynomial or smoothing function to experimental data (e.g., speed vs time) and find stationary points to discuss turning points or peaks — always discuss error and model choice.
  • Show both algebraic work and technology output (plots, tables), and discuss whether stationary points are robust to small changes in data.

Stationary Points — Practice & Examination Questions

📌 Multiple Choice Questions (MCQs)

MCQ 1.
A stationary point occurs at x = c when:

  • A. f(c) = 0
  • B. f′(c) ≠ 0
  • C. f′(c) = 0
  • D. f″(c) = 0
Show Answer

Correct answer: C

A stationary point occurs where the gradient of the curve is zero, meaning the tangent is horizontal.
This condition is mathematically expressed as f′(c) = 0.

MCQ 2.
If f′(x) changes from positive to negative at x = a, the stationary point at x = a is:

  • A. A local minimum
  • B. A local maximum
  • C. A point of inflection
  • D. Undefined
Show Answer

Correct answer: B

A change in the derivative from positive to negative indicates the function increases before x = a
and decreases after x = a, which defines a local maximum.

MCQ 3.
If f′(c) = 0 and f″(c) = 0, then:

  • A. The point must be a minimum
  • B. The point must be a maximum
  • C. The test is inconclusive
  • D. The function has no stationary point
Show Answer

Correct answer: C

When the second derivative equals zero, the second derivative test cannot classify the stationary point.
Further analysis using the first derivative test or higher derivatives is required.

📌 Short Answer Questions

Question 1.
Explain why f′(x) = 0 does not always indicate a maximum or minimum.

Show Answer

The condition f′(x) = 0 indicates a horizontal tangent, but this alone does not determine the nature
of the stationary point. The curve may change concavity without changing direction, resulting in
a horizontal point of inflection rather than a maximum or minimum.

To classify the point, additional tests such as the first derivative sign test or second derivative
test must be applied.

Question 2.
State one advantage of the first derivative test over the second derivative test.

Show Answer

The first derivative test remains valid even when the second derivative equals zero.
It directly examines changes in the sign of the gradient, making it reliable for identifying
horizontal points of inflection where the second derivative test fails.

📌 Long Answer / Explainer Questions

Question 1.
Let f(x) = x3 − 6x2 + 9x + 1.

  1. Find the stationary points.
  2. Classify each stationary point.
  3. State the coordinates of each point.
Show Full Solution

Step 1: Find the derivative

f′(x) = 3x2 − 12x + 9

Step 2: Solve f′(x) = 0

3x2 − 12x + 9 = 0
x2 − 4x + 3 = 0
(x − 1)(x − 3) = 0

Stationary points at x = 1 and x = 3.

Step 3: Second derivative test

f″(x) = 6x − 12

f″(1) = −6 < 0 → local maximum
f″(3) = 6 > 0 → local minimum

Step 4: Coordinates

f(1) = 1 − 6 + 9 + 1 = 5
f(3) = 27 − 54 + 27 + 1 = 1

Local maximum at (1, 5)
Local minimum at (3, 1)

Question 2.
The displacement of a particle is given by s(t) = t4 − 4t2.

  1. Find when the particle is stationary.
  2. Determine the nature of each stationary point.
  3. Explain the physical meaning of each result.
Show Full Solution

Step 1: Velocity function

v(t) = s′(t) = 4t3 − 8t

Step 2: Solve v(t) = 0

4t(t2 − 2) = 0
t = 0, ±√2

Step 3: Acceleration

a(t) = v′(t) = 12t2 − 8

a(0) = −8 → local maximum
a(±√2) = 16 > 0 → local minima

Step 4: Interpretation

At t = 0, the particle momentarily stops and changes direction, reaching a maximum displacement.
At t = ±√2, the particle reverses direction again, corresponding to minimum displacement values.