Finding Maxima and Minima Using Differentiation Revision Notes | GCSE AQA Higher Further Maths

1. Why Maxima and Minima Matter

Imagine you are designing a box to hold the maximum possible volume, or finding the point where a profit function peaks. These are real problems where you need to locate the highest or lowest point on a curve — and differentiation gives you the exact tool to do it.

At any maximum or minimum point on a smooth curve, the tangent is horizontal. A horizontal tangent has gradient zero. So the key idea is this: set $\frac{dy}{dx} = 0$ and solve. The $x$-values you find are the locations of stationary points.

Finding where $\frac{dy}{dx} = 0$ tells you the $x$-coordinate of each stationary point. Substituting that $x$-value back into the original equation $y = f(x)$ gives you the $y$-coordinate. Together, these form the coordinates of the stationary point.

2. The Method: Finding Stationary Points

The process always follows the same sequence. Differentiate the function, set the derivative equal to zero, solve for $x$, then find the corresponding $y$-values. Here is that method applied to a concrete example.

Worked example: Finding the Stationary Points of a Curve

Find the coordinates of the stationary points of $y = x^3 - 3x^2 - 9x + 5$.

1. $\frac{dy}{dx} = 3x^2 - 6x - 9$ — Differentiate each term using the power rule: bring the power down and reduce it by 1. [M1]
2. $3x^2 - 6x - 9 = 0$ — Stationary points occur where the gradient is zero.
3. $x^2 - 2x - 3 = 0$ $(x - 3)(x + 1) = 0$ — Divide through by 3 first to make factorising easier. [M1]
4. $x = 3$ or $x = -1$ — Each bracket gives one stationary point location.
5. When $x = 3$: $y = (3)^3 - 3(3)^2 - 9(3) + 5 = 27 - 27 - 27 + 5 = -22$ When $x = -1$: $y = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10$ — Substitute each x-value into $y = x^3 - 3x^2 - 9x + 5$. Verify: $3 \times 3 \times 3 = 27$; $3 \times 9 = 27$; $9 \times 3 = 27$. For $x = -1$: $(-1)^3 = -1$; $3 \times (-1)^2 = 3 \times 1 = 3$; $9 \times (-1) = -9$. [A1]

Answer: Stationary points at $(3, -22)$ and $(-1, 10)$.

3. Classifying Stationary Points Using the Second Derivative

Knowing where a stationary point is located is only half the job. You also need to know whether it is a maximum or a minimum. The second derivative test gives you a reliable method to classify each point.

Differentiate $\frac{dy}{dx}$ again to get the second derivative $\frac{d^2y}{dx^2}$. Substitute the $x$-coordinate of each stationary point into $\frac{d^2y}{dx^2}$ and check the sign of the result.

A useful memory anchor: a minimum looks like a valley — the curve dips down then rises back up, which corresponds to a positive (upward) curvature. A maximum looks like a hill — the curve rises then falls, corresponding to negative (downward) curvature.

4. Full Worked Example: Find and Classify

Now that you have both tools — finding stationary points and classifying them — here is a complete worked example that ties the full method together. This mirrors the style of question you will see in an exam.

Worked example: Full Method: Find and Classify Stationary Points

Find the coordinates of the stationary points of $y = 2x^3 - 3x^2 - 12x + 4$ and determine whether each is a maximum or minimum.

1. $\frac{dy}{dx} = 6x^2 - 6x - 12$ — Apply the power rule to each term. [M1]
2. $6x^2 - 6x - 12 = 0$ — Stationary points occur where the gradient equals zero.
3. $x^2 - x - 2 = 0$ $(x - 2)(x + 1) = 0$ — Divide through by 6. Then factorise: two numbers that multiply to $-2$ and add to $-1$ are $-2$ and $+1$. [M1]
4. $x = 2$ or $x = -1$ — Two stationary points to investigate.
5. When $x = 2$: $y = 2(2)^3 - 3(2)^2 - 12(2) + 4 = 2(8) - 3(4) - 24 + 4 = 16 - 12 - 24 + 4 = -16$ When $x = -1$: $y = 2(-1)^3 - 3(-1)^2 - 12(-1) + 4 = 2(-1) - 3(1) + 12 + 4 = -2 - 3 + 12 + 4 = 11$ — Verify: $2^3 = 2 \times 2 \times 2 = 8$; $2 \times 8 = 16$; $3 \times 4 = 12$; $12 \times 2 = 24$. For $x = -1$: $(-1)^3 = -1$; $(-1)^2 = 1$. [A1]
6. $\frac{d^2y}{dx^2} = 12x - 6$ — Differentiate $\frac{dy}{dx} = 6x^2 - 6x - 12$ again. [M1]
7. At $x = 2$: $\frac{d^2y}{dx^2} = 12(2) - 6 = 24 - 6 = 18 > 0 \Rightarrow$ minimum At $x = -1$: $\frac{d^2y}{dx^2} = 12(-1) - 6 = -12 - 6 = -18 < 0 \Rightarrow$ maximum — Positive second derivative means minimum; negative means maximum. [A1]

Answer: Maximum at $(-1, 11)$; minimum at $(2, -16)$.

The graph confirms what the algebra found: the curve rises to a local maximum at $(-1, 11)$, then falls to a local minimum at $(2, -16)$, before rising again. Notice how the curve is flat (horizontal tangent) at both of those points — that is exactly what $\frac{dy}{dx} = 0$ captures.

5. Practice: Apply the Full Method

The best way to consolidate this method is to work through problems yourself before checking. Try each question below, then reveal the solution.

6. Exam Technique and Common Mistakes

Examiners award marks at specific stages of this method, so knowing where the marks sit helps you avoid losing them unnecessarily. The method marks typically go to: differentiating correctly, setting $\frac{dy}{dx} = 0$ and solving, and finding the second derivative. The accuracy marks go to the correct coordinates and correct classification.

One of the most common errors is finding the $x$-coordinates of stationary points but forgetting to substitute back to find the $y$-coordinates. The question asks for coordinates, so both values are required for full marks. Always write your answer as a coordinate pair.

Every question on maxima and minima at GCSE follows this same five-step sequence. Practise it until the order is automatic, and you will be able to handle any variation the examiner presents — whether the function is a cubic, a quartic, or set in a real-world context.