A-Level Maths 9709 / Topic 1.7 / AS & A Level

Where the curve turns flat.

At the top of a hill or the bottom of a valley the tangent is horizontal. Find where \(\dfrac{dy}{dx} = 0\), decide whether each point is a maximum or a minimum, and use them to sketch curves and solve optimisation problems.

Full-screen labFunctions and differentiation lab → Topic 1.7All differentiation lessons →

The key idea

Stationary points are where \(\dfrac{dy}{dx} = 0\). If \(\dfrac{d^{2}y}{dx^{2}} < 0\) there, the point is a maximum; if \(\dfrac{d^{2}y}{dx^{2}} > 0\), a minimum. If \(\dfrac{d^{2}y}{dx^{2}} = 0\), check the sign of \(\dfrac{dy}{dx}\) on either side instead.

Fig. 1: y = x³ − 3x² − 9x + 5 has a maximum at (−1, 10) and a minimum at (3, −22); the tangent is horizontal at both

Section 01

Where the gradient is zero.

In the lab, slide the point along the curve and watch the tangent. Stop wherever it lies flat: the gradient readout shows 0. Note whether the curve goes up then down there, or down then up.

Worked example Locate and classify

Find the stationary points of \(y = x^{3} - 3x^{2} - 9x + 5\) and determine their nature.

\(\dfrac{dy}{dx} = 3x^{2} - 6x - 9 = 3(x - 3)(x + 1)\).
\(\dfrac{dy}{dx} = 0\) gives \(x = -1\) or \(x = 3\).
Coordinates from the curve: \(x = -1\): \(y = -1 - 3 + 9 + 5 = 10\); \(\ x = 3\): \(y = 27 - 27 - 27 + 5 = -22\).
\(\dfrac{d^{2}y}{dx^{2}} = 6x - 6\): at \(x = -1\) it is \(-12 < 0\), a maximum; at \(x = 3\) it is \(12 > 0\), a minimum.

Answer: Maximum \((-1, 10)\); minimum \((3, -22)\) (Fig. 1).

Examiner trap

A question that says 'find the stationary points' wants both coordinates. Leaving the answer as \(x = -1\) and \(x = 3\) loses the accuracy marks.

Section 02

Maximum or minimum?

Two tests are allowed. The second derivative is quicker; the sign of the gradient always works. If a question names a method, use it.

Test	Maximum	Minimum
\(\dfrac{d^{2}y}{dx^{2}}\) at the point	negative	positive
sign of \(\dfrac{dy}{dx}\) just left, then just right	\(+\), then \(-\)	\(-\), then \(+\)

Worked example Negative powers

Find the stationary points of \(y = x + \dfrac{4}{x}\) and determine their nature.

\(y = x + 4x^{-1}\), so \(\dfrac{dy}{dx} = 1 - 4x^{-2} = 1 - \dfrac{4}{x^{2}}\).
\(1 - \dfrac{4}{x^{2}} = 0\) gives \(x^{2} = 4\), so \(x = 2\) or \(x = -2\).
\(x = 2\): \(y = 2 + 2 = 4\); \(\ x = -2\): \(y = -2 - 2 = -4\).
\(\dfrac{d^{2}y}{dx^{2}} = 8x^{-3} = \dfrac{8}{x^{3}}\): at \(x = 2\) it is \(1 > 0\); at \(x = -2\) it is \(-1 < 0\).

Answer: Minimum \((2, 4)\); maximum \((-2, -4)\). The maximum is lower than the minimum: they are on different branches of the curve.

Worked example When the second derivative is zero

Find the nature of the stationary point of \(y = (x - 1)^{4} + 2\).

\(\dfrac{dy}{dx} = 4(x - 1)^{3} = 0\) only at \(x = 1\), where \(y = 2\).
\(\dfrac{d^{2}y}{dx^{2}} = 12(x - 1)^{2} = 0\) at \(x = 1\): this test gives no answer.
Check the gradient either side: at \(x = 0.9\), \(4(-0.1)^{3} = -0.004 < 0\); at \(x = 1.1\), \(4(0.1)^{3} = 0.004 > 0\).
Gradient goes from negative to positive.

Answer: \((1, 2)\) is a minimum.

Your turn

Find the stationary points of \(y = 2x^{3} - 6x + 1\) and determine their nature.

Examiner trap

\(\dfrac{d^{2}y}{dx^{2}} = 0\) does NOT tell you the nature. Go back to the sign of \(\dfrac{dy}{dx}\) either side of the point, choosing \(x\)-values close to it.

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

Which condition defines a stationary point?

Quick check+10 XP

At a stationary point, \(\dfrac{d^{2}y}{dx^{2}} = -5\). What kind of point is it?

Quick check+10 XP

Find the stationary point of \(y = x^{2} - 6x + 1\).

Quick check+10 XP

At a stationary point \(x = a\), \(f''(a) = 0\). What should you do next?

Section 03

Sketching and optimising.

Stationary points, together with where the curve meets the axes, are enough to sketch most curves. In a problem in context, build a formula for the quantity to be maximised or minimised in ONE variable, then find its stationary point.

Worked example The largest box

An open box is made from a square sheet of card of side \(12\) cm by cutting a square of side \(x\) cm from each corner and folding up the sides. Find the maximum volume.

The base is \((12 - 2x)\) cm square and the height is \(x\) cm, so \(V = x(12 - 2x)^{2} = 4x^{3} - 48x^{2} + 144x\).
\(\dfrac{dV}{dx} = 12x^{2} - 96x + 144 = 12(x - 2)(x - 6)\).
\(x = 6\) leaves no base (\(V = 0\)), so \(x = 2\) for \(0 < x < 6\).
\(\dfrac{d^{2}V}{dx^{2}} = 24x - 96 = -48 < 0\) at \(x = 2\): a maximum.
\(V = 2 \times 8^{2} = 128\).

Answer: Maximum volume \(128\ \text{cm}^{3}\) (when \(x = 2\)).

Examiner trap

Answer the question asked: 'find the maximum volume' wants \(V = 128\), not \(x = 2\). And always show that the point IS a maximum, with the second derivative or a sign check.

Stage 2 · Exam

Exam-style questions

Unlocks once the four checks above are done. Worth more XP, written to AS & A Level Paper 1 to 6 standard.

Finish the four checks above to unlock the exam questions

Exam style+20 XP

Find the stationary points of the curve \(y = x + \dfrac{4}{x}\) and their nature.

Exam style+20 XP

Squares of side \(x\) cm are cut from the corners of a \(12\) cm by \(12\) cm sheet, which is folded into an open box. Find the maximum volume of the box, in \(\text{cm}^{3}\).

Exam style+20 XP

The curve \(y = x^{3} + ax^{2} + bx\) has a stationary point at \((1, -5)\). Find \(a\) and \(b\).

Exam style+20 XP

Find the coordinates of the maximum point on \(y = 2x^{3} - 6x + 1\).

Skill unlocked

Stationary points, classified.

This skill is now recorded on your profile. Keep the chain going.

-Rank -Level -Score -Topics

Go deeper · practice

Six original Cambridge-style questions

Locating and classifying stationary points with both tests, finding unknown coefficients, sketching a cubic and minimising the surface area of a can.