C3 January 2009 Q7
7.\[\mathrm{f}(x) = 3x\mathrm{e}^x - 1\]The curve with equation \(y = \mathrm{f}(x)\) has a turning point \(P\).
(a) Find the exact coordinates of \(P\). (5)
The equation \(\mathrm{f}(x) = 0\) has a root between \(x = 0.25\) and \(x = 0.3\)
(b) Use the iterative formula\[x_{n+1} = \frac{1}{3}\mathrm{e}^{-x_n}\]with \(x_0 = 0.25\) to find, to 4 decimal places, the values of \(x_1\), \(x_2\) and \(x_3\). (3)
(c) By choosing a suitable interval, show that a root of \(\mathrm{f}(x) = 0\) is \(x = 0.2576\) correct to 4 decimal places. (3)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}^{\prime}(x) = 3\mathrm{e}^x + 3x\mathrm{e}^x\) | M1 A1 |
| \(3\mathrm{e}^x + 3x\mathrm{e}^x = 3\mathrm{e}^x(1 + x) = 0\) \(x = -1\) | M1 A1 |
| \(\mathrm{f}(-1) = -3\mathrm{e}^{-1} - 1\) | B1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(x_1 = 0.2596\) | B1 |
| \(x_2 = 0.2571\) | B1 |
| \(x_3 = 0.2578\) | B1 |
| (3) |
| Scheme | Marks |
|---|---|
| Choosing \((0.257\,55, 0.257\,65)\) or an appropriate tighter interval. | M1 |
| \(\mathrm{f}(0.257\,55) = -0.000\,379\ \ldots\) \(\mathrm{f}(0.257\,65) = 0.000\,109\ \ldots\) | A1 |
| Change of sign (and continuity) \(\Rightarrow\) root \(\in (0.257\,55, 0.257\,65)\ \ast\) cso (\(\Rightarrow x = 0.2576\), is correct to 4 decimal places) | A1 |
| (3) | |
| (11 marks) |