The shape of the lesson
By the end of the lesson, learners can
- Where the gradient is zero: In the lab's stationary points tab, slide the point along the curve and watch the tangent.
- Maximum or minimum?: The syllabus accepts any correct method.
- The biggest and the smallest: Many problems ask for the largest area, the greatest volume or the least cost.
- Differentiate: expand first if needed, then find \(\dfrac{dy}{dx}\).
- Solve \(\dfrac{dy}{dx} = 0\); a cubic gives a quadratic, so look for two answers.
- Coordinates: substitute each \(x\) into the original equation.
- Classify: \(\dfrac{d^2y}{dx^2} > 0\) minimum, \(< 0\) maximum; or check the gradient either side.
- In context: build a one-variable formula, reject impossible values, and answer the actual question.
One picture every idea returns to
Stationary points are where \(\dfrac{dy}{dx} = 0\). Solve for \(x\), then find \(y\) from the original equation. If \(\dfrac{d^2y}{dx^2} > 0\) it is a minimum; if \(\dfrac{d^2y}{dx^2} < 0\) it is a maximum.
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.. Extended. 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's stationary points tab, 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 A quadratic on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Maximum or minimum? | The syllabus accepts any correct method. Model the worked example Using the second derivative on the board, one line at a time. Your turn: Find the stationary point of \(y = 5 + 4x - x^2\) and show that it is a maximum. | Pairs |
| 21 to 30 min | Teach and check: The biggest and the smallest | Many problems ask for the largest area, the greatest volume or the least cost. Model the worked example Largest area on the board, one line at a time. | Anticipation / Reaction Guide |
| 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, 19 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 The biggest and the smallest, 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.
Anticipation / Reaction Guide
Before the lesson, students mark five statements true or false. After the teaching they mark them again and justify any change of mind in one sentence each. The statements come from the examiner traps below.
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 |
|---|---|
| \(x^2 = 1\) has TWO solutions, \(x = 1\) and \(x = -1\). | Missing the negative root loses a whole stationary point. |
| Decide using the second derivative or the gradient, not the size of \(y\). | Here the maximum \((1, 5)\) is only slightly higher than the minimum \((2, 4)\), and on other curves a local maximum can even be LOWER than a minimum. |
| Answer the question that was asked. | Finding \(x = 10\) is not the end: the question wants the AREA, \(200\ \text{m}^2\). Also reject values that make no physical sense, such as \(x = 6\) for the box. |
Stationary points: worksheet
6 questions, 19 marks. Written in the style of the papers; no past paper question is reproduced.
Find the coordinates of the stationary point of \(y = x^2 + 6x - 1\).
Find the coordinates of the two stationary points of \(y = x^3 - 6x^2 + 9x\).
Use the second derivative to determine the nature of each stationary point of \(y = x^3 - 6x^2 + 9x\).
Find the stationary points of \(y = 12x - x^3\) and state which is the maximum.
The curve \(y = x^4 - 8x^2\) has a stationary point at the origin.
By finding the gradient at \(x = -1\) and \(x = 1\), decide whether the origin is a maximum or a minimum.
An open box is made from a square sheet of card 18 cm by 18 cm by cutting a square of side \(x\) cm from each corner and folding up the sides.
(a) Show that the volume is \(V = 4x^3 - 72x^2 + 324x\).
(b) Find the value of \(x\) that gives the greatest volume, and the greatest volume.
(c) Show that your answer is a maximum.
Answer key and marking notes
- \(\dfrac{dy}{dx} = 2x + 6 = 0\), so \(x = -3\).
- \(y = 9 - 18 - 1 = -10\): the point is \((-3, -10)\).
- \(\dfrac{dy}{dx} = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3)\).
- \(\dfrac{dy}{dx} = 0\) when \(x = 1\) or \(x = 3\).
- \(x = 1\): \(y = 1 - 6 + 9 = 4\). \(x = 3\): \(y = 27 - 54 + 27 = 0\).
- The points are \((1, 4)\) and \((3, 0)\).
- \(\dfrac{d^2y}{dx^2} = 6x - 12\).
- At \(x = 1\): \(-6 < 0\), so \((1, 4)\) is a maximum. At \(x = 3\): \(6 > 0\), so \((3, 0)\) is a minimum.
- \(\dfrac{dy}{dx} = 12 - 3x^2 = 0\), so \(x^2 = 4\) and \(x = \pm 2\).
- \(x = 2\): \(y = 24 - 8 = 16\). \(x = -2\): \(y = -24 + 8 = -16\).
- \(\dfrac{d^2y}{dx^2} = -6x\) is \(-12\) at \(x = 2\), so \((2, 16)\) is the maximum (and \((-2, -16)\) the minimum).
- \(\dfrac{dy}{dx} = 4x^3 - 16x\).
- At \(x = -1\): \(-4 + 16 = 12\), positive. At \(x = 1\): \(4 - 16 = -12\), negative.
- The gradient goes \(+, 0, -\), so the origin is a maximum.
- (a) Base \((18 - 2x)\) by \((18 - 2x)\), height \(x\): \(V = x(324 - 72x + 4x^2) = 4x^3 - 72x^2 + 324x\).
- (b) \(\dfrac{dV}{dx} = 12x^2 - 144x + 324 = 12(x^2 - 12x + 27) = 12(x - 3)(x - 9)\).
- \(x = 9\) uses the whole sheet (no base), so \(x = 3\).
- \(V = 3 \times 12^2 = 3 \times 144 = 432\ \text{cm}^3\).
- (c) \(\dfrac{d^2V}{dx^2} = 24x - 144 = 72 - 144 = -72 < 0\) at \(x = 3\), so it is a maximum.
6 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.
Find the stationary point of \(y = 5 + 6x - x^2\) and state whether it is a maximum or a minimum.
- A\((3, 14)\), maximum
- B\((3, 14)\), minimum
- C\((-3, -22)\), maximum
- D\((3, 0)\), maximum
Find the coordinates of the stationary points of \(y = x^3 - 6x^2 + 9x + 1\).
- A\((1, 5)\) and \((3, 1)\)
- B\((1, 5)\) only
- C\((-1, -15)\) and \((-3, -107)\)
- D\((1, 0)\) and \((3, 0)\)
\(y = x^3 - 6x^2 + 9x + 1\) has a stationary point at \((3, 1)\). The gradient is \(-0.57\) at \(x = 2.9\) and \(0.63\) at \(x = 3.1\). What does this show?
- AThe gradient goes from negative to positive, so \((3, 1)\) is a minimum
- BThe gradient goes from negative to positive, so \((3, 1)\) is a maximum
- C\((3, 1)\) is not a stationary point, because the gradients are not 0
- DThe gradient goes from positive to negative, so \((3, 1)\) is a maximum
A cuboid has a square base of side \(x\) cm and a height of \((12 - 2x)\) cm, so its volume is \(V = 12x^2 - 2x^3\). Find the greatest possible volume.
- A\(64\ \text{cm}^3\)
- B\(4\ \text{cm}^3\)
- C\(32\ \text{cm}^3\)
- D\(0\ \text{cm}^3\)
The curve \(y = x^3 + kx\) has a stationary point where \(x = 2\). Find the value of \(k\).
- A\(k = -12\)
- B\(k = 12\)
- C\(k = -4\)
- D\(k = -6\)
The graph of \(y = x^3 - 3x^2\) is shown. Which statement is correct?
- A\(\dfrac{dy}{dx} = 0\) at \(x = 0\) and \(x = 2\), and \(\dfrac{d^2y}{dx^2} < 0\) at \(x = 0\)
- B\(\dfrac{dy}{dx} = 0\) at \(x = 0\) and \(x = 3\)
- C\(\dfrac{dy}{dx} = 0\) at \(x = 0\) and \(x = 2\), and \(\dfrac{d^2y}{dx^2} > 0\) at \(x = 0\)
- D\(\dfrac{dy}{dx} = 0\) only at \(x = 2\), because the curve is lowest there
Quiz key
\(\dfrac{dy}{dx} = 6 - 2x = 0\) gives \(x = 3\), and \(y = 5 + 18 - 9 = 14\). \(\dfrac{d^2y}{dx^2} = -2 < 0\) (a ∩-shaped curve), so \((3, 14)\) is a maximum. Calling it a minimum mixes up the rule. \((-3, -22)\) comes from solving \(6 - 2x = 0\) as \(x = -3\). \((3, 0)\) gives the value of \(\dfrac{dy}{dx}\) at the point instead of \(y\).
\(\dfrac{dy}{dx} = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3) = 0\), so \(x = 1\) or \(x = 3\). Then \(y = 1 - 6 + 9 + 1 = 5\) and \(y = 27 - 54 + 27 + 1 = 1\). A quadratic gives two values of \(x\), so stopping at \((1, 5)\) loses a point. \((-1, -15)\) and \((-3, -107)\) take the roots of \((x - 1)(x - 3) = 0\) as \(-1\) and \(-3\). \((1, 0)\) and \((3, 0)\) give the gradient, not the \(y\)-coordinate.
Negative gradient (downhill) then positive (uphill) makes a valley, so \((3, 1)\) is a minimum. This agrees with \(\dfrac{d^2y}{dx^2} = 6x - 12 = 6 > 0\). Negative then positive does NOT mean a maximum: a maximum goes uphill then downhill. The gradients either side are not 0; only the gradient AT \(x = 3\) is. Reading the gradients in the wrong order (positive then negative) would describe a maximum, but \(x = 2.9\) comes before \(x = 3.1\).
\(\dfrac{dV}{dx} = 24x - 6x^2 = 6x(4 - x) = 0\), so \(x = 0\) or \(x = 4\). \(x = 0\) gives no box, so \(x = 4\): \(V = 12(16) - 2(64) = 192 - 128 = 64\ \text{cm}^3\). \(\dfrac{d^2V}{dx^2} = 24 - 12x = -24 < 0\) confirms a maximum. 4 is the value of \(x\), not the volume. 32 comes from solving \(6x^2 = 24x\) as \(x^2 = 4\), \(x = 2\). 0 uses the root \(x = 0\), which is the minimum.
\(\dfrac{dy}{dx} = 3x^2 + k\). At \(x = 2\): \(12 + k = 0\), so \(k = -12\). \(k = 12\) loses the sign when solving. \(k = -4\) differentiates \(x^3\) as \(x^2\), giving \(4 + k = 0\). \(k = -6\) uses \(3x\) instead of \(3x^2\), giving \(6 + k = 0\).
\(\dfrac{dy}{dx} = 3x^2 - 6x = 3x(x - 2) = 0\) at \(x = 0\) and \(x = 2\). \(\dfrac{d^2y}{dx^2} = 6x - 6\) is \(-6 < 0\) at \(x = 0\), so \((0, 0)\) is a maximum, as the graph shows; at \(x = 2\) it is \(6 > 0\), the minimum \((2, -4)\). \(x = 3\) is where the curve crosses the \(x\)-axis (a root), not a turning point. \(\dfrac{d^2y}{dx^2} > 0\) at \(x = 0\) would make it a minimum, which contradicts the peak in the graph. A maximum has zero gradient just as a minimum does.
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.