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AS & A Level · Practice questions · Stationary points and their nature

Flat spots, found and named.

Six original Cambridge-style questions on stationary points: finding them, deciding their nature, using them to sketch, and solving a maximum and minimum problem.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
The method

Gradient zero, then test the nature.

01
Recall
[3 marks]

Find the coordinates of the stationary point on \(y = 2x^{2} - 8x + 3\) and determine its nature.

  • \(\dfrac{dy}{dx} = 4x - 8 = 0\) gives \(x = 2\); \(y = 8 - 16 + 3 = -5\). ✓
  • \(\dfrac{d^{2}y}{dx^{2}} = 4 > 0\), so \((2, -5)\) is a minimum. ✓
02
Calculation
[5 marks]

Find the stationary points on the curve \(y = x^{3} - 12x + 4\) and determine their nature.

  • \(\dfrac{dy}{dx} = 3x^{2} - 12 = 0\) gives \(x = \pm 2\). ✓
  • \(x = 2\): \(y = 8 - 24 + 4 = -12\). \(\ x = -2\): \(y = -8 + 24 + 4 = 20\). ✓
  • \(\dfrac{d^{2}y}{dx^{2}} = 6x\): \(12 > 0\) at \(x = 2\), \(-12 < 0\) at \(x = -2\). ✓
  • Minimum \((2, -12)\); maximum \((-2, 20)\). ✓
03
Calculation
[4 marks]

A curve has equation \(y = 4\sqrt{x} - x\) for \(x > 0\). Find the coordinates of the stationary point and determine its nature.

  • \(y = 4x^{\frac{1}{2}} - x\), so \(\dfrac{dy}{dx} = 2x^{-\frac{1}{2}} - 1\). ✓
  • \(\dfrac{2}{\sqrt{x}} = 1\) gives \(\sqrt{x} = 2\), \(x = 4\), \(y = 8 - 4 = 4\). ✓
  • \(\dfrac{d^{2}y}{dx^{2}} = -x^{-\frac{3}{2}} = -\tfrac{1}{8} < 0\) at \(x = 4\). ✓
  • \((4, 4)\) is a maximum. ✓
04
Analysis
[5 marks]

The curve \(y = x^{3} + px^{2} + q\) has a stationary point at \((2, -1)\).

(a) Find the values of the constants \(p\) and \(q\).

(b) Find the other stationary point and determine the nature of both.

  • (a) \(\dfrac{dy}{dx} = 3x^{2} + 2px\); at \(x = 2\): \(12 + 4p = 0\), so \(p = -3\). ✓
  • On the curve: \(8 - 12 + q = -1\), so \(q = 3\). ✓
  • (b) \(3x^{2} - 6x = 3x(x - 2) = 0\) gives \(x = 0\), \(y = 3\). ✓
  • \(\dfrac{d^{2}y}{dx^{2}} = 6x - 6\): \(-6 < 0\) at \(x = 0\), so \((0, 3)\) is a maximum; \(6 > 0\) at \(x = 2\), so \((2, -1)\) is a minimum. ✓
05
Analysis
[5 marks]

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

(b) Sketch the curve, showing the coordinates of the stationary points and of the points where it meets the axes.

  • (a) \(y = x^{3} - 3x^{2}\), \(\dfrac{dy}{dx} = 3x^{2} - 6x = 3x(x - 2)\), zero at \(x = 0\) and \(x = 2\). ✓
  • \(y(0) = 0\), \(y(2) = 8 - 12 = -4\). \(\dfrac{d^{2}y}{dx^{2}} = 6x - 6\): \(-6\) at \(x = 0\) (maximum), \(6\) at \(x = 2\) (minimum). ✓
  • (b) Meets the axes at \((0, 0)\) and \((3, 0)\); touches the \(x\)-axis at the maximum \((0, 0)\). ✓
  • Positive cubic: falls in from the bottom left, up to the maximum at \((0, 0)\), down to the minimum at \((2, -4)\), then up through \((3, 0)\). ✓
06
Analysis
[7 marks]

A closed cylindrical can has radius \(r\) cm and height \(h\) cm. Its volume is \(128\pi\ \text{cm}^{3}\).

(a) Show that the total surface area is \(A = 2\pi r^{2} + \dfrac{256\pi}{r}\).

(b) Find the value of \(r\) for which \(A\) is a minimum, and the minimum surface area.

  • (a) \(\pi r^{2}h = 128\pi\), so \(h = \dfrac{128}{r^{2}}\). ✓
  • \(A = 2\pi r^{2} + 2\pi rh = 2\pi r^{2} + 2\pi r \times \dfrac{128}{r^{2}} = 2\pi r^{2} + \dfrac{256\pi}{r}\). ✓
  • (b) \(\dfrac{dA}{dr} = 4\pi r - \dfrac{256\pi}{r^{2}} = 0\) gives \(r^{3} = 64\), so \(r = 4\). ✓
  • \(\dfrac{d^{2}A}{dr^{2}} = 4\pi + \dfrac{512\pi}{r^{3}} > 0\), so this is a minimum. ✓
  • \(A = 32\pi + 64\pi = 96\pi \approx 302\ \text{cm}^{2}\) (3 s.f.). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.