The shape of the lesson
By the end of the lesson, learners can
- Where the gradient is zero: In the lab, slide the point along the curve and watch the tangent.
- Maximum or minimum?: Two tests are allowed.
- Sketching and optimising: Stationary points, together with where the curve meets the axes, are enough to sketch most curves.
- Find: solve \(\dfrac{dy}{dx} = 0\); get \(y\) from the curve.
- Second derivative: negative means maximum, positive means minimum.
- If it is zero: check the sign of \(\dfrac{dy}{dx}\) just either side.
- Sketch: stationary points, axis crossings, and the behaviour for large \(|x|\).
- Optimise: one-variable formula, differentiate, solve, justify max or min, answer the question asked.
One picture every idea returns to
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.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Where the curve turns flat.. At the top of a hill or the bottom of a valley the tangent is horizontal. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Where the gradient is zero | In the lab, slide the point along the curve and watch the tangent. Project tab 6 of the lab (Functions and differentiation lab, stationary points) and let a student drive it while the class predicts. Model the worked example Locate and classify on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Maximum or minimum? | Two tests are allowed. Model the worked example Negative powers on the board, one line at a time. Your turn: Find the stationary points of \(y = 2x^{3} - 6x + 1\) and determine their nature. | Pairs |
| 21 to 30 min | Teach and check: Sketching and optimising | Stationary points, together with where the curve meets the axes, are enough to sketch most curves. Model the worked example The largest box on the board, one line at a time. | Think-Pair-Share |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 29 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Sketching and optimising, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Think-Pair-Share
Pose the question, give thirty seconds of silent thinking, then a minute to compare with a partner before a few pairs share with the class. The silent minute is the part that cannot be cut.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| A question that says 'find the stationary points' wants both coordinates. | Leaving the answer as \(x = -1\) and \(x = 3\) loses the accuracy marks. |
| \(\dfrac{d^{2}y}{dx^{2}} = 0\) 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. |
| 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. |
Stationary points and their nature: worksheet
6 questions, 29 marks. Written in the style of the papers; no past paper question is reproduced.
Find the coordinates of the stationary point on \(y = 2x^{2} - 8x + 3\) and determine its nature.
Find the stationary points on the curve \(y = x^{3} - 12x + 4\) and determine their nature.
A curve has equation \(y = 4\sqrt{x} - x\) for \(x > 0\). Find the coordinates of the stationary point and determine its nature.
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) 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 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.
Answer key and marking notes
- \(\dfrac{dy}{dx} = 4x - 8 = 0\) gives \(x = 2\); \(y = 8 - 16 + 3 = -5\).
- \(\dfrac{d^{2}y}{dx^{2}} = 4 > 0\), so \((2, -5)\) is a minimum.
- \(\dfrac{dy}{dx} = 3x^{2} - 12 = 0\) 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 \((-2, 20)\).
- \(y = 4x^{\frac{1}{2}} - x\), 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.
- (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.
- (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 \((3, 0)\).
- (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.).
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
For a curve \(y = f(x)\), \(f'(2) = 0\) and \(f''(2) = 4\). Which statement is correct?
- AThe point where \(x = 2\) is a minimum point
- BThe point where \(x = 2\) is a maximum point
- CThe point where \(x = 2\) is not a stationary point
- DThe curve has gradient \(4\) at \(x = 2\)
Find the coordinates of the stationary point of the curve \(y = 2x^{2} + 8x - 3\).
- A\((-2, -11)\)
- B\((2, 21)\)
- C\((-2, 0)\)
- D\((-4, -3)\)
Find the coordinates of the minimum point on the curve \(y = x^{3} - 12x + 1\).
- A\((2, -15)\)
- B\((-2, 17)\)
- C\((0, 1)\)
- D\((4, 17)\)
Find the stationary point of the curve \(y = 5 + 4x - x^{2}\) and determine its nature.
- A\((2, 9)\), a maximum
- B\((2, 9)\), a minimum
- C\((-2, -7)\), a maximum
- D\((4, 5)\), a maximum
Find the stationary point of the curve \(y = 2x + \dfrac{8}{x^{2}}\) and determine its nature.
- A\((2, 6)\), a minimum
- B\((2, 6)\), a maximum
- C\((-2, -2)\), a minimum
- D\((2, 8)\), a minimum
The curve \(y = ax^{3} + bx^{2}\), where \(a\) and \(b\) are constants, has a stationary point at \((1, 4)\). Find \(a\) and \(b\).
- A\(a = -8,\ b = 12\)
- B\(a = 8,\ b = -12\)
- C\(a = 4,\ b = -4\)
- D\(a = -2,\ b = 6\)
A farmer has \(60\) m of fencing to make a rectangular pen against a long straight wall, so only three sides need fencing. The two sides at right angles to the wall are each \(x\) m long. Find the maximum possible area of the pen.
- A\(450\ \text{m}^{2}\)
- B\(900\ \text{m}^{2}\)
- C\(225\ \text{m}^{2}\)
- D\(15\ \text{m}^{2}\)
Find the coordinates of the maximum point on the curve \(y = (x^{2} - 4)^{2}\).
- A\((0, 16)\)
- B\((2, 0)\)
- C\((0, -4)\)
- D\((4, 144)\)
Quiz key
\(f'(2) = 0\) makes it a stationary point, and \(f''(2) > 0\) means the gradient is increasing through zero (negative, then positive), so it is a minimum. A maximum needs \(f''(2) < 0\). The gradient at \(x = 2\) is \(f'(2) = 0\); the 4 is the second derivative, not the gradient.
\(\dfrac{dy}{dx} = 4x + 8 = 0\) gives \(x = -2\); then \(y = 8 - 16 - 3 = -11\). \((2, 21)\) comes from a sign slip solving \(4x + 8 = 0\); \((-2, 0)\) substitutes \(x\) back into \(\dfrac{dy}{dx}\), which is always 0 there; \((-4, -3)\) differentiates \(2x^{2}\) as \(2x\).
\(\dfrac{dy}{dx} = 3x^{2} - 12 = 0\) gives \(x^{2} = 4\), \(x = \pm 2\). \(\dfrac{d^{2}y}{dx^{2}} = 6x\) is \(12 > 0\) at \(x = 2\) (minimum) and \(-12 < 0\) at \(x = -2\) (maximum). Minimum: \(y = 8 - 24 + 1 = -15\). \((-2, 17)\) is the maximum; \((0, 1)\) is where \(\dfrac{d^{2}y}{dx^{2}} = 0\), not a stationary point; \((4, 17)\) forgets to square-root \(x^{2} = 4\).
\(\dfrac{dy}{dx} = 4 - 2x = 0\) gives \(x = 2\), \(y = 5 + 8 - 4 = 9\). \(\dfrac{d^{2}y}{dx^{2}} = -2 < 0\), so it is a maximum (as expected for a curve with a negative \(x^{2}\) term). "Minimum" misreads the sign of the second derivative; \((-2, -7)\) is a sign slip solving \(4 - 2x = 0\); \((4, 5)\) differentiates \(x^{2}\) as \(x\).
\(\dfrac{dy}{dx} = 2 - 16x^{-3} = 0\) gives \(x^{3} = 8\), \(x = 2\), and \(y = 4 + \dfrac{8}{4} = 6\). \(\dfrac{d^{2}y}{dx^{2}} = 48x^{-4} = 3 > 0\) at \(x = 2\): a minimum. \((-2, -2)\) comes from \(2 + 16x^{-3} = 0\), losing the minus sign from the power; \((2, 8)\) evaluates \(\dfrac{8}{x}\) instead of \(\dfrac{8}{x^{2}}\); "maximum" misreads the positive second derivative.
On the curve: \(a + b = 4\). Stationary: \(\dfrac{dy}{dx} = 3ax^{2} + 2bx\), so \(3a + 2b = 0\). Then \(3a + 2(4 - a) = 0\), \(a = -8\), \(b = 12\). (Check: \(\dfrac{d^{2}y}{dx^{2}} = 6a + 2b = -24\), a maximum.) \(a = 4, b = -4\) swaps the two conditions (gradient 4, \(y = 0\)); \(a = -2, b = 6\) differentiates \(bx^{2}\) as \(bx\); \(a = 8, b = -12\) has both signs wrong.
\(A = x(60 - 2x) = 60x - 2x^{2}\). \(\dfrac{dA}{dx} = 60 - 4x = 0\) gives \(x = 15\), and \(\dfrac{d^{2}A}{dx^{2}} = -4 < 0\) confirms a maximum. \(A = 15 \times 30 = 450\ \text{m}^{2}\). \(900\) uses \(60 - x\) for the long side, forgetting there are two sides of length \(x\); \(225\) fences all four sides (a \(15\) by \(15\) square); \(15\) is the value of \(x\), not the area.
\(\dfrac{dy}{dx} = 2(x^{2} - 4) \times 2x = 4x(x^{2} - 4)\), zero at \(x = 0, \pm 2\). \(\dfrac{d^{2}y}{dx^{2}} = 12x^{2} - 16\) is \(-16 < 0\) at \(x = 0\) (maximum) and \(32 > 0\) at \(x = \pm 2\) (minima). At \(x = 0\): \(y = (-4)^{2} = 16\). \((2, 0)\) is a minimum; \((0, -4)\) forgets to square; \((4, 144)\) solves \(x^{2} = 4\) as \(x = 4\).
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.