Loav · VCE Mathematical Methods · Units 3 & 4
D04 · Differential calculus · Current curriculum
Use derivative signs to distinguish a turn from a horizontal pause, compare endpoints, and locate points where the gradient itself is locally greatest or least.
1. A stationary point is a candidate, not a classification
A stationary point occurs at an input a where f′(a) = 0. Its coordinates are (a, f(a)). Solving the derivative equation finds the inputs; substituting into the original function finds the heights. A zero derivative tells you that the tangent is horizontal. It does not, by itself, tell you whether the point is a maximum, a minimum or a stationary point of inflection.
To classify a stationary point, examine the sign of f′ on the two sides. A change from positive to negative means the function rises then falls, giving a local maximum. A change from negative to positive means it falls then rises, giving a local minimum. If the derivative keeps the same sign, the graph does not turn there. For a smooth curve, a horizontal passage through such a point can be a stationary point of inflection.
A local maximum is at least as high as nearby values; an absolute maximum is at least as high as every value in the stated domain. Those are different comparisons. A local minimum can be higher than an endpoint elsewhere in the interval. Always read whether a question asks for local features, the greatest value on an interval, or a complete sketch.
2. Build a sign table from reliable information
Find all zeros of the derivative and any inputs where it is undefined. Put them in increasing order and choose a test input in each intervening interval. Substitute into a factored derivative where possible. Factors whose signs are already known, such as eˣ or a squared non-zero denominator, can simplify the reasoning. The table is valid only on intervals where the derivative formula and the original function are defined.
Do not assume the sign changes at every zero. An even power such as (x − 1)² is non-negative on both sides of its zero. A repeated factor can therefore produce a stationary point without a turn. Nor should a discontinuity be labelled stationary merely because a simplified derivative suggests zero there: the original function and derivative must both exist at the input.
Worked example 1: classify two stationary points
Let f(x) = x³ − 3x² − 9x + 2. Differentiate and factor: f′(x) = 3x² − 6x − 9 = 3(x − 3)(x + 1). The stationary inputs are x = −1 and x = 3.
For x < −1, both linear factors are negative, so f′ is positive. For −1 < x < 3, one is negative and the other positive, so f′ is negative. For x > 3, both are positive. The signs are therefore +, −, + across the three intervals.
At x = −1, positive changes to negative, so there is a local maximum. Its height is f(−1) = 7. At x = 3, negative changes to positive, so there is a local minimum with height f(3) = −25. Report the points (−1, 7) and (3, −25), with their classifications. Since the cubic is unbounded above and below on ℝ, neither is an absolute extremum on the whole real line.
3. Strict increase can include an isolated horizontal tangent
The definition of strictly increasing is that f(x₁) < f(x₂) whenever x₁ < x₂ in the interval. It compares function values. A positive derivative throughout an interval is a sufficient way to establish this property, but it is not necessary at every single input. The function f(x) = x³ is strictly increasing on ℝ even though f′(0) = 0.
For x³, the derivative 3x² is positive on either side of zero and zero only at zero. The graph keeps rising through the origin; it does not flatten across a whole interval. Splitting the increasing interval at zero is unnecessary if the question asks for the largest interval of strict increase. In contrast, a genuinely constant segment prevents strict increase over any interval containing two distinct points within that segment.
The same distinction applies to strict decrease. A derivative that is negative except at an isolated stationary input can still describe a strictly decreasing function. In an explanation, state the behaviour across the zero rather than claiming that the derivative must be strictly negative everywhere. The sign table helps identify whether a zero interrupts the direction of travel.
4. Inflection points as local extremes of gradient
For the smooth functions considered here, identify a point of inflection by examining where the tangent gradient changes from locally increasing to locally decreasing, or vice versa. In other words, the derivative function has a local maximum or minimum. This viewpoint directly connects inflection to a locally greatest rate of increase or decrease. It also works when the original curve has no horizontal tangent at the point.
If f′(a) = 0 at such a point, the inflection point is stationary. For x³, the derivative 3x² has a minimum of zero at x = 0, and the original point (0, 0) is a stationary point of inflection. If f′(a) ≠ 0, the inflection point is non-stationary. Its tangent has a non-zero gradient even though that gradient is locally smallest or largest.
To locate an extremum of gradient, you can inspect the graph of r(x) = f′(x), use algebra such as completing the square, or differentiate r and test its sign changes where appropriate. Merely solving r′(x) = 0 produces candidates, just as solving f′(x) = 0 does for the original function. Check that the gradient really has a local extremum. A separate concavity classification is not needed for this approach.
Worked example 2: a non-stationary inflection point
Consider h(x) = x³ + 3x² + 5x + 2. Its gradient function is r(x) = h′(x) = 3x² + 6x + 5 = 3(x + 1)² + 2.
The completed-square form shows that the gradient has its minimum value 2 at x = −1. It decreases towards that input and increases afterwards. Thus the original curve has a point of inflection at x = −1. Its height is h(−1) = −1, giving the point (−1, −1).
Because h′(−1) = 2 rather than zero, the inflection point is non-stationary. In fact h′(x) ≥ 2 for every real x, so h is strictly increasing everywhere. The graph never turns, yet it has a point where its upward gradient is locally smallest. This example separates an extremum of the function from an extremum of its rate of change.
5. Absolute extrema on a closed interval
For a continuous function on a closed, bounded interval [a, b], an absolute maximum and minimum are attained. To find them, collect interior stationary inputs, interior points where the derivative does not exist, and both endpoints. Evaluate the original function at every valid candidate and compare the values. Classification of each stationary point is helpful, but a complete endpoint comparison is still necessary.
On an open interval, a bound might be approached without being attained. For example, f(x) = x on 0 < x < 1 has no maximum or minimum, even though its values lie between 0 and 1. An excluded endpoint cannot supply an attained extremum. In an unbounded domain, inspect the long-term behaviour before claiming that a local feature is a global optimum.
Worked example 3: compare an interior maximum with endpoints
Find the absolute maximum and minimum of f(x) = xe−x on [0, 4]. The derivative is f′(x) = e−x(1 − x). Since the exponential is positive, the derivative is positive for x < 1 and negative for x > 1. The only interior stationary input is x = 1.
Evaluate all candidates: f(0) = 0, f(1) = 1/e and f(4) = 4/e⁴. The sign change establishes that f increases to x = 1 then decreases. Its maximum is therefore 1/e, attained at x = 1. Since f is positive for x > 0 in this interval, its minimum is 0, attained at x = 0.
Numerically, 1/e ≈ 0.367879 and 4/e⁴ ≈ 0.073263. These values support the exact comparison. Reporting only the stationary point would miss the endpoint minimum and would not fully answer the interval question.
6. Assemble a graph from compatible features
A useful sketch combines the domain, intercepts, stationary points, inflection points and any relevant asymptotic behaviour. Find intercepts from the original function, stationary inputs from its derivative, and gradient extrema from the derivative’s behaviour. Label each point with coordinates from the correct function. An inflection input found using r belongs to the original curve at height f(a), not at height r(a).
Draw the curve so that its rise and fall agree with the derivative signs. Check the order and relative heights of labelled points before adding smooth connecting strokes. If exact intercepts are unavailable and technology is allowed, mark numerical approximations to a stated accuracy. The purpose of the sketch is to communicate verified features, not to reproduce every pixel of a calculator display.
7. Common errors and a compact checking routine
A derivative zero is not automatically a maximum. A local minimum is not automatically the smallest value on the given interval. A point of inflection need not be stationary. A function can remain strictly increasing through an isolated zero derivative. These four distinctions prevent a large proportion of curve-analysis mistakes.
Before finishing, ask: did I include every allowed endpoint, keep every domain exclusion, classify by surrounding behaviour, and substitute into the original function for coordinates? If the task concerns maximum rate, inspect the derivative rather than maximising the original function. If it concerns maximum amount, do the opposite. State both the relevant input and the quantity achieved there, with contextual units when given.
Practice: seven original questions
- Classify the stationary point of f(x) = x⁴ using the sign of f′.
- Explain why g(x) = x³ is strictly increasing on ℝ despite having a stationary point. Classify that point.
- Locate and classify the point of inflection of h(x) = x³ + 3x using its gradient function.
- Find the absolute maximum and minimum of p(x) = x² − 6x + 2 on [0, 5].
- A differentiable function has f′(x) = (x + 2)(x − 1)². Classify its stationary inputs and state a largest interval on which f is strictly increasing.
- Find the absolute maximum and minimum values of sin x on [0, 2π], stating where they occur.
- For q(x) = x³ − 6x² + 13x + 1, find the input where its gradient is least, the least gradient, and the corresponding inflection point.
Worked answers and reasoning
- f′(x) = 4x³ is negative for x < 0 and positive for x > 0. The sign changes from − to + at zero, so (0, 0) is a local minimum. Since x⁴ ≥ 0 for all real x, it is also the absolute minimum on ℝ. A zero derivative alone would not establish this classification.
- g′(x) = 3x² is positive except at zero. The function continues increasing through the origin, so its largest interval of strict increase is ℝ. The derivative has a local minimum of zero there, and the original graph does not turn: (0, 0) is a stationary point of inflection.
- h′(x) = 3x² + 3 has its minimum value 3 at x = 0. Its values decrease towards zero and increase afterwards. Therefore h has a non-stationary point of inflection at (0, 0), because h′(0) = 3 ≠ 0. There is no local maximum or minimum of h there.
- p′(x) = 2x − 6, so the interior stationary input is 3. Compare p(0) = 2, p(3) = −7 and p(5) = −3. The absolute maximum is 2 at x = 0; the absolute minimum is −7 at x = 3. The endpoint x = 5 must be checked even though it is neither answer.
- The stationary inputs are −2 and 1. The squared factor is positive except at 1, so the sign is negative for x < −2 and positive for x > −2 apart from zero at 1. Thus −2 gives a local minimum, while 1 is a stationary inflection input. The function is strictly increasing throughout [−2, ∞), or on (−2, ∞) if open intervals are requested; the isolated zero at 1 does not split that behaviour.
- The derivative is cos x, zero at π/2 and 3π/2 in the interval. The candidate function values are 0 at both endpoints, 1 at π/2 and −1 at 3π/2. Hence the absolute maximum is 1 at π/2 and the absolute minimum is −1 at 3π/2. All arguments are in radians.
- q′(x) = 3x² − 12x + 13 = 3(x − 2)² + 1. The least gradient is 1, attained at x = 2. Since q(2) = 8 − 24 + 26 + 1 = 11, the inflection point is (2, 11). It is non-stationary because its gradient is 1 rather than zero.
Curriculum and further study
Aligned to the current VCAA VCE Mathematics Study Design, Mathematical Methods Units 3 and 4, Calculus area of study and relevant outcome knowledge and skills. The VCAA Mathematical Methods FAQ, updated February 2026, clarifies inflection points and strict increasing or decreasing behaviour. Curriculum checked 15 October 2026. Unit 3 is the suggested teaching sequence; the official design specifies combined Units 3 and 4 coverage.
Further study: Evans, Lipson and Greenwood, Cambridge Senior Mathematics VCE: Mathematical Methods Units 3 & 4, second edition (2023), sections 10C–10D and 10F. Independent Loav teaching material, not a VCAA or Cambridge publication. Explanations, worked examples, practice questions, answers and diagrams are original.
