Looks Simple — Can You Solve It?
Take $x^{3}-x-2=0$ and try to solve it exactly. Factorising turns up no pair that works, and there is no cubic formula you would reach for in an exam. So a fair question: does this equation even have a solution — and if it does, how can we pin it down?
Three lines of reasoning below use these ideas. Predict each one first, then click Correct step or Wrong step.
Those two moves answer the opening question. The sign change traps the root in $(1,2)$; feeding each answer back through the rearrangement gives $x_1=1.5,\ x_2=1.51829,\ x_3=1.52094,\ x_4=1.52132,\ldots$, which homes in on the root $1.521$ — accurate enough to use.
Try each in your head before clicking an answer — this is a test, not a reveal. See how many you get first try. The clock starts on your first answer.
R1. A root of $f(x)=0$ is a value of $x$ where:
R2. The function $f$ is continuous with $f(1)=-2$ and $f(2)=5$. What follows?
R3. Make $x$ the subject of $x^{3}=x+2$:
If $f$ is continuous on $[a,b]$ and $f(a)$ and $f(b)$ have opposite signs, then $f(x)=0$ has at least one root in $(a,b)$:
Evaluate at consecutive integers and look for the change of sign. For $f(x)=x^{3}-x-2$:
| x | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| f(x) | -2 | -2 | 4 | 22 |
The sign changes between $1$ and $2$, so a root lies in the unit interval $(1,2)$. A sign change only traps the root — it does not yet tell us its value.
Continuity is essential: a function that jumps over the axis can change sign without ever crossing it.
From $x^{3}-x-2=0$ we get $x^{3}=x+2$, and taking cube roots gives:
Now feed each answer back into the same formula, $x_{n+1}=F(x_n)=\sqrt[3]{x_n+2}$, starting from $x_1=1.5$:
| n | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| xₙ | 1.5 | 1.51829 | 1.52094 | 1.52132 | 1.52137 |
| xₙ₊₁ | 1.51829 | 1.52094 | 1.52132 | 1.52137 | 1.52138 |
The iterates settle on $x=1.521\ (3\ \text{d.p.})$.
A value that satisfies $x=F(x)$ is a fixed point of $F$ — and it is exactly a root of the original equation, because the two equations are equivalent.
A different rearrangement of $x^{3}-x-2=0$ is $x=x^{3}-2$, i.e. $x_{n+1}=x_n^{3}-2$. From $x_1=1.5$:
The values move apart, so this iteration diverges and never reaches the root. In general the iteration converges when the iterates settle down; near the root this happens when $|F'(x)|\lt 1$.
Pairs pick a starting value x₁, then press Iterate to feed each answer back through x_{n+1} = F(x_n). Use the buttons to switch between two rearrangements of x³ − x − 2 = 0: the live plot of y = F(x) with y = x, and the table, show the iterates stepping toward the fixed point for one rearrangement and away from it for the other. Take turns: one partner sets the start and presses Iterate, the other predicts the next value first.
Each card shows a sign-change claim, a rearrangement or an iteration result. Decide whether it is correct, then explain the slip out loud to a partner.
(i) Show that $x^{3}-x-2=0$ has a root between 1 and 2. (ii) The equation is rearranged into the form $x=F(x)$ with $F(x)=\sqrt[3]{x+2}$. Use the resulting iterative formula to find the root correct to 3 decimal places.
Trap the root with a sign change first. Then feed each answer back into $F(x)=\sqrt[3]{x+2}$ — never back into the original start.
$f(1)=1-1-2=-2\lt 0$ and $f(2)=8-2-2=4\gt 0$, so by the sign-change criterion a root lies in $(1,2)$.
From $x^{3}=x+2$, $x=\sqrt[3]{x+2}$, so $x_{n+1}=\sqrt[3]{x_n+2}$.
$x_1=1.5$
$x_2=1.51829$
$x_3=1.52094$
$x_4=1.52132$
$x_5=1.52137$
which has settled at
$x=1.521\ (3\ \text{d.p.})$
(i) Show that $\mathrm{e}^{x}+x-6=0$ has a root in the interval $[1,2]$. (ii) Show that the equation may be written $x=\ln(6-x)$. (iii) Use an iterative formula based on this equation to find the root correct to 3 decimal places.
Rearrange to isolate $\mathrm{e}^{x}$, then take natural logs of both sides to free the $x$.
$f(1)=\mathrm{e}+1-6\approx -2.282\lt 0$ and $f(2)=\mathrm{e}^{2}+2-6\approx 3.389\gt 0$, so a root lies in $(1,2)$.
$\mathrm{e}^{x}=6-x\ \Rightarrow\ x=\ln(6-x)$, giving $x_{n+1}=\ln(6-x_n)$.
$x_1=1.5$
$x_2=1.50408$
$x_3=1.50317$
$x_4=1.50337$
$x_5=1.50333$
so
$x=1.503\ (3\ \text{d.p.})$
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. A rearrangement of $x^{3}-8x-13=0$ into the form $x=F(x)$ is:
Q2. The first iterate of $x_{n+1}=\sqrt[3]{x_n+2}$ from $x_1=1.5$ is:
Q3. Iterating $x_{n+1}=x_n^{3}-2$ from $x_1=1.5$:
Q4. The equation $x=\cos x$ ($x$ in radians) is solved by iterating $x_{n+1}=\cos x_n$. Starting from $x_1=0.7$, the root correct to 3 d.p. is:
Work each on paper, then reveal the solution to check your working.
(i) Show that $\mathrm{e}^{-x}-x+2=0$ has a root in the interval $[2,3]$. (ii) The equation is rearranged as $x=\mathrm{e}^{-x}+2$. Use the resulting iterative formula to find the root correct to 3 decimal places.
Reveal Solution
$f(2)=\mathrm{e}^{-2}\approx 0.135\gt 0$ and $f(3)=\mathrm{e}^{-3}-1\approx -0.950\lt 0$, so a root lies in $(2,3)$.
Iterating $x_{n+1}=\mathrm{e}^{-x_n}+2$ from $x_1=2.5$:
$x_1=2.5$
$x_2=2.08208$
$x_3=2.12467$
$x_4=2.11947$
$x_5=2.12009$
$x_6=2.12002$
$x=2.120\ (3\ \text{d.p.})$
The equation $x^{3}-8x-13=0$ has one real root. (i) Find the two consecutive integers between which this root lies. (ii) Use the iterative formula $x_{n+1}=\sqrt[3]{8x_n+13}$ to find this root correct to 2 decimal places, giving each iteration to 4 decimal places.
Reveal Solution
$f(3)=27-24-13=-10\lt 0$ and $f(4)=64-32-13=19\gt 0$, so the root lies between $3$ and $4$.
Iterating from $x_1=3.5$:
$x_1=3.5$
$x_2=3.4482$
$x_3=3.4366$
$x_4=3.4339$
$x_5=3.4333$
$x_6=3.4332$
$x=3.43\ (2\ \text{d.p.})$
The curve $y=x\mathrm{e}^{2x}$ has a minimum point $M$. (i) Find the exact coordinates of $M$. (ii) Show that the curve meets the line $y=20$ at the point whose $x$-coordinate is a root of $x=\dfrac{1}{2}\ln\!\left(\dfrac{20}{x}\right)$. (iii) Use the iterative formula $x_{n+1}=\dfrac{1}{2}\ln\!\left(\dfrac{20}{x_n}\right)$ from $x_1=1.3$ to find the root correct to 2 decimal places.
Reveal Solution
$\dfrac{dy}{dx}=\mathrm{e}^{2x}(1+2x)=0$ at $x=-\dfrac{1}{2}$, giving $M\left(-\dfrac{1}{2},\ -\dfrac{1}{2\mathrm{e}}\right)$.
Setting $x\mathrm{e}^{2x}=20$ gives $\mathrm{e}^{2x}=\dfrac{20}{x}$, and taking logs: $x=\dfrac{1}{2}\ln\!\left(\dfrac{20}{x}\right)$.
Iterating from $x_1=1.3$:
$x_1=1.3$
$x_2=1.3667$
$x_3=1.3417$
$x_4=1.3509$
$x_5=1.3475$
$x_6=1.3487$
$x=1.35\ (2\ \text{d.p.})$
Exit Ticket
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