Determining the Nature of Stationary Points Revision Notes | GCSE AQA Higher Further Maths

1. Why Classifying Stationary Points Matters

You already know how to find where $\frac{dy}{dx} = 0$ — that locates a stationary point. But finding it is only half the job. The exam will ask you to determine its nature: is it a peak, a trough, or something else entirely?

There are two methods for doing this, and the specification requires you to know both. The first uses the behaviour of the gradient on either side of the point — called the increasing and decreasing functions method. The second uses the second derivative, $\frac{d^2y}{dx^2}$. Both lead to the same conclusion; the skill is knowing when each is most efficient.

2. Method 1 — Increasing and Decreasing Functions

Think about what happens to the gradient as you walk along a curve towards a peak. Before the peak, the curve is rising — the gradient is positive. At the peak, the gradient is zero. After the peak, the curve falls — the gradient is negative. That sign change tells you everything.

To use this method, pick a value of $x$ slightly to the left of the stationary point and a value slightly to the right. Substitute each into $\frac{dy}{dx}$ and record the sign. The pattern of signs reveals the nature of the point.

A point of inflection is where the gradient is momentarily zero but the curve does not actually turn — it flattens and then continues in the same direction. This is a common source of lost marks when students assume every stationary point must be a maximum or minimum.

Worked example: Classifying a Stationary Point Using Gradient Signs

A curve has equation $y = x^3 - 3x + 2$. Find the stationary points and determine their nature using increasing and decreasing functions.

1. $\frac{dy}{dx} = 3x^2 - 3$ — Differentiate each term using the power rule. [M1]
2. $3x^2 - 3 = 0$ $3x^2 = 3$ $x^2 = 1$ $x = 1$ or $x = -1$ — Stationary points occur where the gradient is zero. [M1]
3. At $x = 1$: $y = (1)^3 - 3(1) + 2 = 1 - 3 + 2 = 0$ At $x = -1$: $y = (-1)^3 - 3(-1) + 2 = -1 + 3 + 2 = 4$ — Substitute each $x$-value back into the original equation. [A1]
4. At $x = 0$: $\frac{dy}{dx} = 3(0)^2 - 3 = -3 < 0$ At $x = 2$: $\frac{dy}{dx} = 3(2)^2 - 3 = 12 - 3 = 9 > 0$ Sign change: $(-) \to (+)$ — Negative then positive means the curve decreases then increases — local minimum. [A1]
5. At $x = -2$: $\frac{dy}{dx} = 3(-2)^2 - 3 = 12 - 3 = 9 > 0$ At $x = 0$: $\frac{dy}{dx} = 3(0)^2 - 3 = -3 < 0$ Sign change: $(+) \to (-)$ — Positive then negative means the curve increases then decreases — local maximum. [A1]

Answer: Local maximum at $(-1, 4)$; local minimum at $(1, 0)$.

3. Method 2 — The Second Derivative Test

The second derivative, $\frac{d^2y}{dx^2}$, measures how the gradient itself is changing. At a stationary point, if the gradient is decreasing (going from positive to negative), the curve is bending downwards — that is a maximum. If the gradient is increasing, the curve bends upwards — that is a minimum.

A useful memory anchor: think of a hill (maximum) as a curve that is concave down — it bends like an upside-down bowl. The second derivative is negative there. A valley (minimum) is concave up — it bends like a right-way-up bowl. The second derivative is positive there.

Worked example: Classifying Stationary Points Using the Second Derivative

Use the second derivative to classify the stationary points of $y = x^3 - 3x + 2$.

1. $\frac{dy}{dx} = 3x^2 - 3$ — Already found in the previous example. Stationary points at $x = 1$ and $x = -1$.
2. $\frac{d^2y}{dx^2} = 6x$ — Differentiate $\frac{dy}{dx}$ with respect to $x$. [M1]
3. $\frac{d^2y}{dx^2}\bigg|_{x=1} = 6(1) = 6 > 0$ — Positive second derivative at $x = 1$ confirms a local minimum. [A1]
4. $\frac{d^2y}{dx^2}\bigg|_{x=-1} = 6(-1) = -6 < 0$ — Negative second derivative at $x = -1$ confirms a local maximum. [A1]

Answer: Local maximum at $(-1, 4)$; local minimum at $(1, 0)$. Both methods agree.

4. Visualising the Two Methods Together

Seeing both methods on the same curve makes the connection concrete. The graph below shows $y = x^3 - 3x + 2$ with its maximum at $(-1, 4)$ and minimum at $(1, 0)$. Notice how the curve rises, peaks, falls, troughs, and then rises again — exactly what the sign patterns and second derivative values predicted.

5. Choosing Between the Two Methods

Both methods are valid, but each has a situation where it is the better choice. The second derivative test is faster when $\frac{d^2y}{dx^2}$ is straightforward to compute and gives a non-zero result. The gradient sign method is the reliable fallback when the second derivative equals zero at the stationary point — because in that case the second derivative test gives no information at all.

For example, the curve $y = x^4$ has a stationary point at $x = 0$. Here $\frac{d^2y}{dx^2} = 12x^2$, which equals zero at $x = 0$. The second derivative test is inconclusive. Checking gradient signs — negative to the left, positive to the right — confirms it is a minimum. The gradient sign method always works; the second derivative test sometimes does not.

6. Guided Practice

Now apply both methods to a fresh curve. Work through the step-by-step below, covering each step and attempting it before revealing the working.

Worked example: Full Classification of Stationary Points

Find and classify the stationary points of $y = 2x^3 - 9x^2 + 12x - 4$, using both methods.

1. $\frac{dy}{dx} = 6x^2 - 18x + 12$ — Apply the power rule to each term. [M1]
2. $6x^2 - 18x + 12 = 0$ $x^2 - 3x + 2 = 0$ $(x - 1)(x - 2) = 0$ $x = 1$ or $x = 2$ — Divide through by 6 first to simplify, then factorise. [M1, A1]
3. At $x = 1$: $y = 2(1)^3 - 9(1)^2 + 12(1) - 4 = 2 - 9 + 12 - 4 = 1$ At $x = 2$: $y = 2(2)^3 - 9(2)^2 + 12(2) - 4 = 16 - 36 + 24 - 4 = 0$ — Substitute each $x$-value into the original equation. [A1]
4. $\frac{d^2y}{dx^2} = 12x - 18$ At $x = 1$: $12(1) - 18 = -6 < 0 \Rightarrow$ maximum At $x = 2$: $12(2) - 18 = 24 - 18 = 6 > 0 \Rightarrow$ minimum — Differentiate $\frac{dy}{dx}$ and substitute each stationary $x$-value. [M1, A1]
5. At $x = 0$: $\frac{dy}{dx} = 6(0)^2 - 18(0) + 12 = 12 > 0$ At $x = 1.5$: $\frac{dy}{dx} = 6(1.5)^2 - 18(1.5) + 12 = 13.5 - 27 + 12 = -1.5 < 0$ Sign: $(+) \to (-)$ confirms maximum. — Both methods agree — this is a local maximum at $(1, 1)$. [A1]

Answer: Local maximum at $(1, 1)$; local minimum at $(2, 0)$.

7. Exam Technique and Common Mistakes

Examiners award marks at specific stages: differentiating correctly, setting $\frac{dy}{dx} = 0$, finding the $y$-coordinate, and then providing clear evidence of classification. A bare statement — 'it is a maximum' — without supporting working will not earn the classification mark. You must show either the second derivative value or the gradient sign table.

When the question says 'determine the nature', that is your signal to go beyond just finding the stationary point. Write out your evidence explicitly: state the value of $\frac{d^2y}{dx^2}$ and its sign, or write a clear sign table with values substituted. Either approach earns full marks if the working is shown.