Flat tangents mark the turning points.

A derivative isn't only a gradient — it's a detector. Where it is zero, the curve stops climbing or falling. Those moments are its peaks and troughs, and finding them is how you find the best of anything.

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Wherever the tangent goes flat, the lower graph crosses zero. For $y = x^3 - 3x$ that happens twice: at $x = -1$ and $x = 1$.

The second derivative says which way the curve bends. At $x = -1$ it's $-6 < 0$: bending down, so $(-1,\,2)$ is a maximum. At $x = 1$ it's $6 > 0$: bending up, so $(1,\,-2)$ is a minimum.

A 12 cm square of card. Cut a square of side $x$ from each corner and fold up the sides: the base is $(12 - 2x)^2$, the height $x$,

Small $x$: a flat tray. Big $x$: a tall, thin tube. The best box is in between.

Cut 2 cm squares and the box holds 128 cm³ — more than any other cut. Check it really is a maximum: $\tfrac{d^2V}{dx^2} = -96 + 24x = -48 < 0$ at $x = 2$.

Set the derivative to zero. Then check.

Tangents and normals, classifying stationary points, more optimisation, and twelve graded exam questions with full worked solutions are waiting.

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