# Stationary points and turning points

> Edexcel International GCSE Further Pure Mathematics · 4PM1
> Source: https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s9-stationary-points-and-turning-points/

This guide covers all content required for Edexcel IGCSE Further Pure Math (4PM1) section 9D on stationary and turning points, including finding points and classifying their nature using two validated exam techniques.

**Prerequisites:** [Differentiate functions using power, chain, product, quotient rules for $x^n$, $\sin(kx)$, $\cos(kx)$, $e^{kx}$](https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s9-rules-of-differentiation/); [Solve linear and quadratic equations](https://www.owlsprep.com/study/edexcel-igcse-algebra-solving-quadratic-equations/)

## Learning objectives

- Solve dy/dx = 0 to find coordinates of stationary points for all 4PM1-approved differentiable functions
- Classify stationary points using both the second derivative test and first derivative sign change test
- Present solutions clearly to earn full marks on structured exam questions for this topic

## Finding Stationary Points by Solving $\frac{dy}{dx} = 0$

A stationary point on a curve $y = f(x)$ is any point where the gradient of the curve equals zero, meaning the tangent to the curve is perfectly horizontal at that position.

**Stationary Point** — A point $(x, y)$ on the curve $y = f(x)$ where $\frac{dy}{dx} = 0$, so the gradient of the curve is zero.

**Worked example:** Find the coordinates of the stationary points on the curve $y = x^3 - 6x^2 + 9x + 2$.

1. 1. Differentiate $y$ with respect to $x$ to find the first derivative:

   $$\frac{dy}{dx} = 3x^2 - 12x + 9$$
2. 2. Set $\frac{dy}{dx} = 0$ and solve for $x$:

   $$3x^2 - 12x + 9 = 0 \implies x^2 - 4x + 3 = 0 \implies (x-1)(x-3) = 0 \implies x = 1 \text{ or } x = 3$$
3. 3. Substitute each $x$ value back into the original function to find corresponding $y$ values: For $x=1$, $y = 1 - 6 + 9 + 2 = 6$. For $x=3$, $y = 27 - 54 + 27 + 2 = 2$.
4. 4. State the stationary points: $(1, 6)$ and $(3, 2)$.

> **Exam tip:** Always substitute your $x$ values back into the original function, not the derivative, to get the correct $y$ coordinate for full marks.

*Calculator:* allowed

## Classifying Points with the Second Derivative Test

Once you have found stationary points, you need to classify their type: local maximum, local minimum, or point of inflexion. The second derivative test is the fastest method when $\frac{d^2y}{dx^2}$ is non-zero at the point.

**Second Derivative Test** — A test to classify stationary points using the sign of the second derivative: if $\frac{d^2y}{dx^2} < 0$, the point is a local maximum; if $\frac{d^2y}{dx^2} > 0$, it is a local minimum; if $\frac{d^2y}{dx^2} = 0$, the test is inconclusive.

**Worked example:** Classify the stationary points $(1, 6)$ and $(3, 2)$ from the previous example using the second derivative test.

1. 1. Differentiate the first derivative $\frac{dy}{dx} = 3x^2 - 12x + 9$ to get the second derivative:

   $$\frac{d^2y}{dx^2} = 6x - 12$$
2. 2. Evaluate $\frac{d^2y}{dx^2}$ at $x=1$: $6(1) - 12 = -6$. Since $-6 < 0$, $(1, 6)$ is a local maximum.
3. 3. Evaluate $\frac{d^2y}{dx^2}$ at $x=3$: $6(3) - 12 = 6$. Since $6 > 0$, $(3, 2)$ is a local minimum.
4. 4. State your conclusion clearly, citing the second derivative test as your method.

> **warning**
>
> If the second derivative equals zero at a stationary point, you cannot use this test. You must use the first derivative sign change test instead, as the point could be any of the three types.

*Calculator:* allowed

## First Derivative Sign Change Test for Stationary Points

This test is used when the second derivative test is inconclusive, or if calculating the second derivative is time-consuming for complex functions. It works by checking the sign of $\frac{dy}{dx}$ just left and right of the stationary point to observe how the gradient changes.

**First Derivative Sign Change Test** — A test that checks the sign of $\frac{dy}{dx}$ at points marginally smaller and larger than the stationary point $x$-coordinate: gradient change from + to - = local maximum; - to + = local minimum; no sign change = point of inflexion.

**Worked example:** A function has a stationary point at $x=0$ where $\frac{d^2y}{dx^2} = 0$. The first derivative is $\frac{dy}{dx} = x^3$. Classify the stationary point using the first derivative test.

1. 1. Choose a test point just left of $x=0$, e.g. $x = -0.1$. Calculate $\frac{dy}{dx}$: $(-0.1)^3 = -0.001$, which is negative.
2. 2. Choose a test point just right of $x=0$, e.g. $x = 0.1$. Calculate $\frac{dy}{dx}$: $(0.1)^3 = 0.001$, which is positive.
3. 3. Observe the sign change: gradient changes from negative to positive as you pass $x=0$.
4. 4. Conclusion: The stationary point at $x=0$ is a local minimum.

> **Exam tip:** When choosing test points for this method, pick values very close to the stationary point $x$-coordinate to ensure there are no other stationary points between your test point and the target point.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Setting $\frac{d^2y}{dx^2} = 0$ to find stationary points
  - Why it fails: Stationary points are defined by a first derivative of zero, not second derivative
  - Correct: Always solve $\frac{dy}{dx} = 0$ to find $x$-coordinates of stationary points first
- **Wrong:** Guessing the point type when $\frac{d^2y}{dx^2} = 0$
  - Why it fails: The second derivative test is inconclusive in this case, so guessing will lose marks
  - Correct: Switch to the first derivative sign change test to classify the point correctly
- **Wrong:** Substituting $x$ values into $\frac{dy}{dx}$ to get $y$-coordinates of stationary points
  - Why it fails: $\frac{dy}{dx}$ gives the gradient, not the $y$-value of the original curve
  - Correct: Always substitute $x$ values into the original function $y = f(x)$ to calculate corresponding $y$ coordinates
- **Wrong:** Using test points too far from the stationary point for the first derivative test
  - Why it fails: There may be other stationary points between your test point and the target point, leading to incorrect sign observations
  - Correct: Select test points very close to the stationary $x$ value, e.g. $\pm 0.1$ for a point at $x=2$, to avoid errors

## Cheatsheet

| Task | Method | Key Rule |
| --- | --- | --- |
| Find stationary points | Solve $\frac{dy}{dx} = 0$ | Substitute $x$ into original $f(x)$ for $y$ value |
| Classify point (fast method) | Second derivative test | $\frac{d^2y}{dx^2} < 0 = $ max; $>0 = $ min; $0 = $ inconclusive |
| Classify point (backup method) | First derivative sign test | + to - = max; - to + = min; no change = inflexion |
| Exam presentation | State test used | Write a clear conclusion for each stationary point |

## What's next

Now that you can find and classify stationary points, the next topic in the Edexcel IGCSE Further Pure Math calculus unit is applied optimisation, where you will use these skills to solve real-world problems requiring maximum or minimum values. You will also need to use your stationary point knowledge when studying gradients of tangents and normals to curves. Mastering this topic is critical for scoring high marks on calculus questions, which typically make up 15-20% of the total marks on each 4PM1 paper. Make sure to practice past paper questions to build speed and accuracy in both finding and classifying points under exam conditions.

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