Practice Maths

Stationary Points and the First Derivative Test

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Key Terms

Stationary point
: a point (a, f(a)) where f′(a) = 0. The tangent is horizontal.
First Derivative Test
: examine the sign of f′(x) just to the left and right of x = a:
   — f′ changes −→+: local minimum
   — f′ changes +→−: local maximum
   — f′ same sign on both sides: stationary point of inflection (horizontal inflection)
Increasing
: f′(x) > 0 on an interval; Decreasing: f′(x) < 0 on an interval.
📚 First Derivative Test Summary
Sign of f′ leftSign of f′ rightClassification
−+Local minimum
+−Local maximum
++Inflection (horizontal)
−−Inflection (horizontal)
x y y=x³−3x −2 −1 0 1 2 max(−1,2) min(1,−2) Sign of f′(x): + − + f′=0 f′=0 increasing decreasing increasing ↑ local max ↓ local min
Hot Tip A stationary point where f′ = 0 on both sides is a stationary point of inflection (horizontal inflection) — not a turning point. Classic example: y = x³ at x = 0.

Worked Example 1 — Finding and Classifying Stationary Points

Question: Find and classify the stationary points of y = x³ − 3x.

Step 1 — Differentiate: f′(x) = 3x² − 3 = 3(x − 1)(x + 1)

Step 2 — Solve f′ = 0: x = 1 or x = −1

Step 3 — y-values: f(−1) = −1 + 3 = 2;   f(1) = 1 − 3 = −2

Step 4 — Sign test:

  • Test x = −2: f′(−2) = 3(4) − 3 = 9 > 0 → + (increasing before x = −1)
  • Test x = 0: f′(0) = −3 < 0 → − (decreasing between −1 and 1)
  • Test x = 2: f′(2) = 9 > 0 → + (increasing after x = 1)

Conclusion: x = −1 is a local maximum (f(−1) = 2); x = 1 is a local minimum (f(1) = −2).

Worked Example 2 — Stationary Point of Inflection

Question: Find and classify the stationary point of y = x³.

f′(x) = 3x² = 0 at x = 0. f(0) = 0. Point: (0, 0).

Sign test: f′(−1) = 3 > 0; f′(1) = 3 > 0. Sign does not change.

Conclusion: (0, 0) is a stationary point of inflection.

Introduction: What is a Stationary Point?

At a stationary point, the gradient of the tangent is zero — the curve is momentarily “flat”. This can mean the function has reached a local high point (local maximum), a local low point (local minimum), or is briefly horizontal before continuing in the same direction (stationary inflection).

The First Derivative Test — Step by Step

  1. Find f′(x) by differentiating.
  2. Solve f′(x) = 0 to find the x-values of stationary points.
  3. Find y-values by substituting into f(x).
  4. Build a sign chart: pick test values just left and right of each stationary point. Evaluate f′ at each test value.
  5. Classify using the sign change rule.

Increasing and Decreasing Intervals

Once you have the sign chart from the first derivative test, you can immediately read off the increasing and decreasing intervals:

  • Where f′(x) > 0: f is increasing.
  • Where f′(x) < 0: f is decreasing.
Worked Example — Four-step Classification

Find and classify all stationary points of f(x) = 2x³ − 3x² − 12x + 1.

Step 1: f′(x) = 6x² − 6x − 12 = 6(x² − x − 2) = 6(x − 2)(x + 1)

Step 2: f′ = 0 at x = 2 and x = −1.

Step 3: f(−1) = −2 − 3 + 12 + 1 = 8. f(2) = 16 − 12 − 24 + 1 = −19.

Step 4 sign chart: Test x = −2: f′(−2) = 6(−4)(−1) = 24 > 0 (+). Test x = 0: f′(0) = 6(0−2)(0+1) = −12 < 0 (−). Test x = 3: f′(3) = 6(1)(4) = 24 > 0 (+).

Conclusion: x = −1 is a local maximum (f = 8); x = 2 is a local minimum (f = −19). Increasing on (−∞, −1) ∪ (2, ∞); decreasing on (−1, 2).

Worked Example — f(x) = x&sup4; − 4x³

Find and classify all stationary points.

Step 1: f′(x) = 4x³ − 12x² = 4x²(x − 3)

Step 2: f′ = 0 at x = 0 and x = 3.

Step 3: f(0) = 0. f(3) = 81 − 108 = −27.

Step 4: Test x = −1: f′(−1) = 4(1)(−4) = −16 < 0 (−). Test x = 1: f′(1) = 4(1)(−2) = −8 < 0 (−). Test x = 4: f′(4) = 4(16)(1) = 64 > 0 (+).

Conclusion: x = 0: sign is − on both sides → stationary point of inflection. x = 3: sign changes −→+ → local minimum (f = −27).

💡 Key Reminder: f′(a) = 0 is necessary but not sufficient for a turning point. Always check the sign of f′ on both sides. If the sign doesn’t change, it’s an inflection, not a turning point.

Summary

To find and classify stationary points: differentiate, solve f′ = 0, find y-values, build a sign chart with test values on each side, and classify by sign change. The first derivative test is the definitive classification method at Year 11 level.

Mastery Practice

See Answers ➔
  1. Fluency

    Find all stationary points of each function (give both x and y coordinates).

    1. (a) y = x² − 6x + 5
    2. (b) y = x³ − 12x
    3. (c) y = 2x³ − 3x²
    4. (d) y = x4 − 8x²
  2. Fluency

    For each stationary point you found in Question 1, classify it as a local maximum, local minimum, or stationary point of inflection using the first derivative test.

  3. Fluency

    Using the sign charts from Question 2, state the intervals on which each function is increasing and decreasing.

  4. Understanding

    A function has f′(x) = x(x − 2)(x + 3). Without finding f(x) itself, determine the x-values of all stationary points and classify each one.

  5. Understanding

    Find the local maximum and local minimum values of f(x) = 2x³ − 9x² + 12x − 4.

  6. Understanding

    Show that y = x³ + x has no stationary points and is therefore always increasing for all real x.

  7. Understanding

    For f(x) = x4 − 4x³:

    1. (a) Find f′(x) and all stationary points.
    2. (b) Classify each stationary point using the first derivative test.
    3. (c) State the intervals where f is increasing and decreasing.
  8. Problem Solving

    Given f(x) = ax³ + bx² + 3x − 2, the function has a stationary point at x = 1 with f(1) = 0. Find the values of a and b.

  9. Problem Solving

    For f(x) = x³ − 3kx + 2, find the value(s) of k for which f has no stationary points.

    Hint: Consider when f′(x) = 0 has no real solutions.
  10. Problem Solving

    The area of a region bounded by a river and a straight road is modelled by A(x) = 3x² − x³⁄3 for 0 ≤ x ≤ 9, where x is measured in kilometres.

    1. (a) Find A′(x) and solve A′(x) = 0.
    2. (b) Verify this is a maximum using the first derivative test.
    3. (c) Find the maximum area.