C3 June 2008 Q7
7.\[\mathrm{f}(x) = 3x^3 - 2x - 6\]
(a) Show that \(\mathrm{f}(x) = 0\) has a root, \(\alpha\), between \(x = 1.4\) and \(x = 1.45\) (2)
(b) Show that the equation \(\mathrm{f}(x) = 0\) can be written as\[x = \sqrt{\left(\frac{2}{x} + \frac{2}{3}\right)}, \qquad x \neq 0.\] (3)
(c) Starting with \(x_0 = 1.43\), use the iteration\[x_{n+1} = \sqrt{\left(\frac{2}{x_n} + \frac{2}{3}\right)}\]to calculate the values of \(x_1\), \(x_2\) and \(x_3\), giving your answers to 4 decimal places. (3)
(d) By choosing a suitable interval, show that \(\alpha = 1.435\) is correct to 3 decimal places. (3)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(1.4) = -0.568\ \ldots < 0\) \(\mathrm{f}(1.45) = 0.245\ \ldots > 0\) | M1 |
| Change of sign (and continuity) \(\Rightarrow \alpha \in (1.4, 1.45)\) | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(3x^3 = 2x + 6\) \(x^3 = \dfrac{2x}{3} + 2\) \(x^2 = \dfrac{2}{3} + \dfrac{2}{x}\) | M1 A1 |
| \(x = \sqrt{\left(\dfrac{2}{x} + \dfrac{2}{3}\right)}\ \ \ast\) cso | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(x_1 = 1.4371\) | B1 |
| \(x_2 = 1.4347\) | B1 |
| \(x_3 = 1.4355\) | B1 |
| (3) |
| Scheme | Marks |
|---|---|
| Choosing the interval \((1.4345, 1.4355)\) or appropriate tighter interval. | M1 |
| \(\mathrm{f}(1.4345) = -0.01\ \ldots\) \(\mathrm{f}(1.4355) = 0.003\ \ldots\) | M1 |
| Change of sign (and continuity) \(\Rightarrow \alpha \in (1.4345, 1.4355)\) \(\Rightarrow \alpha = 1.435\), correct to 3 decimal places \(\ast\) cso | A1 |
| (3) | |
| (11 marks) |