14 The function f is defined by
$$f ( x ) = 3 ^ { x } \sqrt { x } - 1 \quad \text { where } x \geq 0$$
14
- \(\quad \mathrm { f } ( x ) = 0\) has a single solution at the point \(x = \alpha\)
By considering a suitable change of sign, show that \(\alpha\) lies between 0 and 1
14 - Show that
$$\mathrm { f } ^ { \prime } ( x ) = \frac { 3 ^ { x } ( 1 + x \ln 9 ) } { 2 \sqrt { x } }$$
14- (ii) Use the Newton-Raphson method with \(x _ { 1 } = 1\) to find \(x _ { 3 }\), an approximation for \(\alpha\).
| | Give your answer to five decimal places. | | [2 marks] \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) | 14- (iii) Explain why the Newton-Raphson method fails to find \(\alpha\) with \(x _ { 1 } = 0\)
| | [2 marks] \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) |
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