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Pure · Differentiation

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Stationary points (maxima and minima)

Where a curve is momentarily flat — found by setting dy/dx = 0 — and how the second derivative tells a maximum from a minimum, plus reading off where a curve is increasing or decreasing.

Pure · Differentiation

Stationary points (maxima and minima)

Where a curve is momentarily flat — found by setting dy/dx = 0 — and how the second derivative tells a maximum from a minimum, plus reading off where a curve is increasing or decreasing.

Why it works

A stationary point is a spot where the curve is, for an instant, perfectly flat — its gradient is zero. Since the gradient is dydx\frac{dy}{dx}, finding stationary points is a single step: set dydx=0\frac{dy}{dx} = 0 and solve. (Not y=0y = 0 — that finds where the curve crosses the xx-axis, a completely different thing.)

That usually gives more than one answer — a cubic's gradient is a quadratic, so expect up to two stationary points — and each xx needs its yy put back in (using the original equation) to give full coordinates.

Then: peak or trough? Look at how the gradient is changing as you pass through the point — and the rate of change of the gradient is the second derivative, d2ydx2\frac{d^2y}{dx^2} (just differentiate again).
  • At a maximum, the gradient runs positive → 00 → negative: it is
decreasing, so d2ydx2<0\frac{d^2y}{dx^2} < 0.
  • At a minimum, the gradient runs negative → 00 → positive: it is
increasing, so d2ydx2>0\frac{d^2y}{dx^2} > 0.

A memory hook: a minimum holds water (concave up, ++); a maximum spills it (concave down, −-). If the second derivative comes out as exactly 00 the test is inconclusive — fall back to checking the gradient's sign just before and just after the point.

The same derivative reads off where a curve is increasing or decreasing without finding the turning points at all: dydx>0\frac{dy}{dx} > 0 means rising, dydx<0\frac{dy}{dx} < 0 means falling. Solve that inequality and you have the intervals.-2-112-55max (-1, 2)min (1, -2)xyAbove, y=x3−3xy = x^3 - 3x has a maximum at (−1,2)(-1, 2) and a minimum at (1,−2)(1, -2) — at both, the tangent would be horizontal (dydx=0\frac{dy}{dx} = 0).