The shape of the lesson
By the end of the lesson, learners can
- Sketch first: how many roots?: In the lab, choose a curve and move the line across it.
- The change of sign test: Rearrange so that one side is zero.
- Accuracy, and when the test fails: To confirm a root to a given accuracy, test the bounds of the rounding interval.
- Sketch both sides of the equation to count roots and see roughly where they are.
- Rearrange to \(f(x) = 0\) and check \(f\) is continuous on the interval.
- Evaluate \(f(a)\) and \(f(b)\), in radians for trig, and quote the values.
- Conclude: "sign change, and \(f\) continuous, so a root lies between \(a\) and \(b\)".
- Accuracy: test the rounding bounds, e.g. 1.505 and 1.515 for 1.51.
One picture every idea returns to
Write the equation as \(f(x) = 0\). If \(f\) is continuous on \([a, b]\) and \(f(a)\) and \(f(b)\) have opposite signs, then there is at least one root between \(a\) and \(b\). To show a root is \(\alpha\) correct to 2 d.p., show the sign changes between \(\alpha - 0.005\) and \(\alpha + 0.005\).
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, Find where the sign flips.. Papers 2 and 3. 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: Sketch first: how many roots? | In the lab, choose a curve and move the line across it. Project tab 3 of the lab (Graphs of functions lab, solving graphically) and let a student drive it while the class predicts. Model the worked example Graphs and a sign change on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: The change of sign test | Rearrange so that one side is zero. Model the worked example Between consecutive integers on the board, one line at a time. Your turn: Show that the equation \(x^3 + x = 5\) has a root between 1.5 and 1.6. | Pairs |
| 21 to 30 min | Teach and check: Accuracy, and when the test fails | To confirm a root to a given accuracy, test the bounds of the rounding interval. Model the worked example Correct to 2 decimal places on the board, one line at a time. | Cooperative Concept Mapping |
| 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, 21 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 Accuracy, and when the test fails, 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.
Cooperative Concept Mapping
Teams receive the key terms on cards and arrange them on a large sheet, drawing and labelling the links. Compare maps between teams before showing the model map.
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 |
|---|---|
| In a sketch-and-count question the marks are for two correct graph shapes in the right places AND a stated conclusion ("one point of intersection, so one root"). | A sketch with no statement does not earn the final mark. |
| Compare with zero, not with each other: you must rearrange to \(f(x) = 0\) first. | Checking that \(\ln 2 3 - 3\) is fine, but only if you say clearly that the sign of \(\ln x - (3 - x)\) has changed. |
| No sign change does NOT mean no root. \(f(x) = (x-1)^2\) has a root at \(x = 1\) but is positive on both sides; and two roots between \(a\) and \(b\) cancel out the sign change. | A sketch protects you from both traps. |
Locating roots by a change of sign: worksheet
6 questions, 21 marks. Written in the style of the papers; no past paper question is reproduced.
Show that the equation \(x^3 - 3x - 1 = 0\) has a root between 1 and 2.
By sketching the graphs of \(y = e^{-x}\) and \(y = 3 - x^2\) on the same diagram, state the number of real roots of the equation \(e^{-x} = 3 - x^2\). Show that one root lies between 1 and 2.
Show that the equation \(x = 2\sin x\) has a root between 1.8 and 2.
A student says: "\(g(x) = \dfrac{1}{x-1}\) has \(g(0) = -1\) and \(g(2) = 1\), so the equation \(g(x) = 0\) has a root between 0 and 2." Explain why the student is wrong.
The equation \(e^{x} = 3x\) has a root \(\beta\) between 0 and 1. Show that \(\beta = 0.62\) correct to 2 decimal places.
A chord \(AB\) of a circle with centre \(O\) and radius \(r\) subtends an angle of \(\theta\) radians at \(O\). The area of the minor segment cut off by \(AB\) is one quarter of the area of the circle.
(a) Show that \(\theta = \sin\theta + \dfrac{\pi}{2}\).
(b) Show that this equation has a root between 2.3 and 2.4.
(c) Show that \(\theta = 2.31\) correct to 2 decimal places.
Answer key and marking notes
- \(f(x) = x^3 - 3x - 1\) is continuous.
- \(f(1) = 1 - 3 - 1 = -3\) and \(f(2) = 8 - 6 - 1 = 1\).
- Sign change, so there is a root between 1 and 2.
- \(y = 3 - x^2\) is an inverted parabola with maximum \((0, 3)\); \(y = e^{-x}\) is decreasing through \((0, 1)\).
- At \(x = 0\) the parabola is above the curve; far to the left \(e^{-x}\) is huge, and to the right the parabola falls below the \(x\)-axis. So they cross twice: two real roots.
- \(f(x) = e^{-x} - 3 + x^2\): \(f(1) = 0.368 - 2 = -1.63\) and \(f(2) = 0.135 + 1 = 1.14\).
- Sign change and \(f\) continuous, so a root lies between 1 and 2.
- \(f(x) = x - 2\sin x\), continuous; radians.
- \(f(1.8) = 1.8 - 1.9477 = -0.148\) and \(f(2) = 2 - 1.8186 = 0.181\).
- Sign change, so a root lies between 1.8 and 2.
- \(\dfrac{1}{x-1}\) can never be 0, because its numerator is never 0.
- \(g\) is not continuous on \([0, 2]\): it has a vertical asymptote at \(x = 1\).
- The sign change comes from the jump across the asymptote, so the test does not apply.
- \(f(x) = e^{x} - 3x\); the bounds are 0.615 and 0.625.
- \(f(0.615) = 1.84966 - 1.845 = 0.00466\) and \(f(0.625) = 1.86825 - 1.875 = -0.00675\).
- Sign change, so \(0.615 < \beta < 0.625\), and \(\beta = 0.62\) to 2 d.p.
- (a) Segment area \(= \dfrac{1}{2}r^2\theta - \dfrac{1}{2}r^2\sin\theta = \dfrac{1}{4}\pi r^2\).
- Divide by \(\dfrac{1}{2}r^2\): \(\theta - \sin\theta = \dfrac{\pi}{2}\), so \(\theta = \sin\theta + \dfrac{\pi}{2}\).
- (b) \(f(\theta) = \theta - \sin\theta - \dfrac{\pi}{2}\): \(f(2.3) = -0.0165\) and \(f(2.4) = 0.154\). Sign change, so a root lies between 2.3 and 2.4.
- (c) \(f(2.305) = -0.00816\) and \(f(2.315) = 0.00858\). Sign change, so \(2.305 < \theta < 2.315\) and \(\theta = 2.31\) to 2 d.p.
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.
The equation \(x^3 + x - 3 = 0\) has exactly one real root. In which interval does it lie?
- A\(1.2 < x < 1.3\)
- B\(1.1 < x < 1.2\)
- C\(1.3 < x < 1.4\)
- D\(1.0 < x < 1.1\)
You want to show that a root \(\alpha\) of \(f(x) = 0\) is 2.35 correct to 2 decimal places. Which two values of \(x\) should you substitute into \(f(x)\)?
- A\(2.345\) and \(2.355\)
- B\(2.34\) and \(2.36\)
- C\(2.3\) and \(2.4\)
- D\(2.35\) and \(2.355\)
The diagram shows the graphs of \(y = e^x\) and \(y = 4 - x^2\). How many real roots does the equation \(e^x = 4 - x^2\) have, and between which consecutive integers does the positive root lie?
- A2 roots; the positive root is between 1 and 2
- B2 roots; the positive root is between 0 and 1
- C1 root; it is between 1 and 2
- D3 roots; the positive root is between 1 and 2
\(\tan 1 = 1.557\) and \(\tan 2 = -2.185\) (radians). Does the equation \(\tan x = 0\) have a root between 1 and 2?
- ANo: \(\tan x\) is not continuous on this interval, because of the asymptote at \(x = \dfrac{\pi}{2}\)
- BYes: the sign changes, so there must be a root
- CYes: the root is at \(x = \dfrac{\pi}{2}\)
- DIt cannot be decided without evaluating \(\tan 1.5\)
The equation \(\ln x = 4 - 2x\) has one real root. Between which pair of values does it lie?
- A\(1.7\) and \(1.8\)
- B\(1.6\) and \(1.7\)
- C\(1.8\) and \(1.9\)
- D\(1.9\) and \(2.0\)
Let \(f(x) = x^3 - 2x - 5\). Which calculation proves that the root of \(f(x) = 0\) is 2.09 correct to 2 decimal places?
- A\(f(2.085) = -0.106\) and \(f(2.095) = 0.00501\)
- B\(f(2.08) = -0.161\) and \(f(2.1) = 0.061\)
- C\(f(2.09) = -0.0507\), which is close to 0
- D\(f(2.085) = -0.106\) and \(f(2.09) = -0.0507\)
The equation \(2\sin x = x\) has one positive root, where \(x\) is in radians. Between which pair of values does it lie?
- A\(1.8\) and \(1.9\)
- B\(1.9\) and \(2.0\)
- C\(1.7\) and \(1.8\)
- DThere is no positive root.
Let \(f(x) = x^2 - 3x + 2.1\). Then \(f(1) = 0.1\) and \(f(2) = 0.1\). Which statement is correct?
- AThere are two roots of \(f(x) = 0\) between 1 and 2.
- BThere is no root of \(f(x) = 0\) between 1 and 2.
- CThere is exactly one root of \(f(x) = 0\) between 1 and 2.
- DThere is a root of \(f(x) = 0\) at \(x = 1.5\).
Quiz key
\(f(1.2) = 1.728 + 1.2 - 3 = -0.072\) and \(f(1.3) = 2.197 + 1.3 - 3 = 0.497\). \(f\) is continuous and changes sign, so the root lies between 1.2 and 1.3. The other intervals show no change: \(f(1.0) = -1\), \(f(1.1) = -0.569\) are both negative, and \(f(1.4) = 1.144\) has the same sign as \(f(1.3)\). The tempting \(1.1 < x < 1.2\) stops as soon as \(f\) is "nearly zero", but \(f(1.2)\) is still negative.
Numbers that round to 2.35 lie in \(2.345 \le x < 2.355\). A sign change between \(f(2.345)\) and \(f(2.355)\) traps \(\alpha\) in that interval, so \(\alpha = 2.35\) to 2 d.p. The tempting 2.34 and 2.36 only show \(\alpha\) is between 2.34 and 2.36, which could round to 2.34, 2.35 or 2.36. 2.3 and 2.4 is even wider, and 2.35 to 2.355 covers only half of the rounding interval.
The curves cross twice, once for \(x < 0\) and once for \(x > 0\), so there are 2 roots. With \(f(x) = e^x - 4 + x^2\): \(f(1) = e - 3 = -0.282\) and \(f(2) = e^2 = 7.39\), a change of sign, so the positive root is between 1 and 2 (it is about 1.06). The tempting "between 0 and 1" trusts the picture near \(x = 1\), but \(f(0) = -3\) and \(f(1) < 0\) show no change of sign there. "1 root" misses the intersection near \(x = -2\), where \(e^x\) is small but still positive.
\(\tan x\) has a vertical asymptote at \(x = \dfrac{\pi}{2} \approx 1.571\), which lies between 1 and 2. The sign changes by jumping from \(+\infty\) to \(-\infty\), not by passing through 0. The roots of \(\tan x = 0\) are \(0, \pi, 2\pi, \ldots\), none of them between 1 and 2. The tempting "yes" applies the test without checking continuity. \(\tan\dfrac{\pi}{2}\) is undefined, not zero, and evaluating \(\tan 1.5 = 14.1\) would not settle anything.
\(f(1.7) = 0.531 + 3.4 - 4 = -0.069\) and \(f(1.8) = 0.588 + 3.6 - 4 = 0.188\). \(f\) is continuous for \(x > 0\) and changes sign, so the root lies between 1.7 and 1.8 (it is about 1.73). The tempting 1.6 and 1.7 gives \(f(1.6) = -0.330\) and \(f(1.7) = -0.069\), both negative. Between 1.8 and 1.9 and between 1.9 and 2.0 both values are positive.
Numbers that round to 2.09 lie between 2.085 and 2.095. \(f(2.085) < 0 < f(2.095)\), and \(f\) is continuous, so the root lies in that interval and equals 2.09 to 2 d.p. The tempting pair 2.08 and 2.1 only shows the root is between 2.08 and 2.10, so it could be 2.08, 2.09 or 2.10 to 2 d.p. A single value close to 0 proves nothing, and the last pair has no change of sign.
In radians: \(f(1.8) = 2(0.9738) - 1.8 = 0.148\) and \(f(1.9) = 2(0.9463) - 1.9 = -0.0074\). The sign changes, so the root is between 1.8 and 1.9 (about 1.896). The tempting 1.9 and 2.0 assumes that because \(f(1.9)\) is so small the root must be just above 1.9, but \(f(2.0) = -0.181\) has the same sign. 1.7 and 1.8 gives two positive values, and "no positive root" comes from working in degrees, where \(2\sin x\) is tiny and \(f\) is negative for every value tried.
\(f(1.5) = 2.25 - 4.5 + 2.1 = -0.15\), so the sign changes between 1 and 1.5 and again between 1.5 and 2: two roots. (By the formula they are \(\dfrac{3 \pm \sqrt{0.6}}{2} = 1.11\) and \(1.89\).) The tempting "no root" assumes that equal signs at the ends mean no root, but an even number of roots also gives no overall change of sign. "Exactly one" is impossible without a sign change at the ends, and \(f(1.5) = -0.15 \ne 0\).
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.