--- title: "Stationary Points and the Second Derivative" description: "Solve f'(x) = 0 to locate maximum, minimum and inflection points, then use the sign of f''(x) to decide which is which." canonical: https://lightmysky.com/learn/mathematics/stationary-points-and-the-second-derivative-mt_VbGJEFFgfs source: https://lightmysky.com/learn/mathematics/stationary-points-and-the-second-derivative-mt_VbGJEFFgfs.md retrieved: 2026-09-02 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Stationary Points and the Second Derivative Solve f'(x) = 0 to locate maximum, minimum and inflection points, then use the sign of f''(x) to decide which is which. Subject: Mathematics · Area: Calculus & Analysis · Ages 17 to 18 Page: https://lightmysky.com/learn/mathematics/stationary-points-and-the-second-derivative-mt_VbGJEFFgfs ## Ready when they can - Find the stationary points of y = x³ - 3x and classify each one - Use f''(x) to distinguish a maximum from a minimum - Explain what f''(x) = 0 leaves undecided and what to check instead ## Lesson: Stationary Points and the Second Derivative Last stop you split a curve into rising stretches and falling ones, and left the boundaries between them alone. Those boundaries are where f'(x) = 0, and the curve is flat there for an instant. They are called stationary points, and on the Ridgeway plan they are the places that matter most: the top of the ramp and the bottom of the bowl. Finding them is easy, since it is one equation. Saying which kind each one is takes one more idea, and that is what this stop is for. The method has three steps. Solve f'(x) = 0 to get the x values. Put each x back into the original curve, never into the derivative, to get its y value, so each point comes out as a coordinate pair. Then classify it. Differentiating a second time gives f''(x), the second derivative, which is the rate at which the gradient itself is changing. At a stationary point, f''(x) > 0 means a minimum and f''(x) < 0 means a maximum. *(drawing: Three steps, three different rules. The middle one is the one people skip.)* **Example.** Take y = x³ - 12x² + 36x. Then f'(x) = 3x² - 24x + 36, which factorises to 3(x - 2)(x - 6), so the stationary points are at x = 2 and x = 6. Heights come from the curve: at x = 2, 8 - 48 + 72 = 32, and at x = 6, 216 - 432 + 216 = 0. Now f''(x) = 6x - 24. At x = 2 that is -12, which is negative, so (2, 32) is a maximum. At x = 6 it is 12, positive, so (6, 0) is a minimum. The test is not a rule to memorise, it is a reading. f''(x) reports whether the gradient is rising or falling. Walk through a minimum and the gradient goes from negative, through zero, to positive: it is increasing, so f''(x) is positive. Walk through a maximum and the gradient goes from positive, through zero, to negative: it is decreasing, so f''(x) is negative. The sign of the second derivative is just the sign of the first derivative's own trend. *(drawing: The second derivative watches the gradient, not the curve.)* One case the test cannot settle. If f''(x) comes out as 0 at a stationary point, it reports nothing, and both answers stay open. For y = x⁴ the point at x = 0 has f'(0) = 0 and f''(0) = 0, yet it is a minimum. For y = x³ both are zero at x = 0 too, and it is neither: the curve flattens and carries straight on: a point of inflection. When f''(x) = 0, go back to the sign of f'(x) just either side. If the sign changes, the point turns; if it stays the same, it is a point of inflection. **Recap.** A stationary point is where f'(x) = 0, so solve that equation for x, then put each x into the original curve to get its y. Classify with the second derivative: f''(x) > 0 gives a minimum and f''(x) < 0 gives a maximum, because f''(x) reports whether the gradient is rising or falling. If f''(x) = 0 the test decides nothing, and you check the sign of f'(x) on either side instead: a change of sign means the point turns, and no change means a point of inflection. ## Practice 24 questions on this page, each with its working shown. ## Needs first - [Increasing and Decreasing Functions](https://lightmysky.com/learn/mathematics/increasing-and-decreasing-functions-mt_KOrXxQvBrZ) - [Sketching a Curve from Its Factorised Form](https://lightmysky.com/learn/mathematics/sketching-a-curve-from-its-factorised-form-mt_x4tZLdH8fh) ## Opens up - [Optimisation with Calculus](https://lightmysky.com/learn/mathematics/optimisation-with-calculus-mt_9I0ql39Lzw)