9.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{5abaa077-1da4-4023-b442-194f6972095b-26_659_783_287_641}
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\caption{Figure 3}
\end{figure}
Figure 3 shows a sketch of part of the curve \(C\) with equation
$$y = \sqrt { 3 + 4 \mathrm { e } ^ { x ^ { 2 } } } \quad x \geqslant 0$$
- Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), giving your answer in simplest form.
The point \(P\) with \(x\) coordinate \(\alpha\) lies on \(C\).
Given that the tangent to \(C\) at \(P\) passes through the origin, as shown in Figure 3, - show that \(x = \alpha\) is a solution of the equation
$$4 x ^ { 2 } e ^ { x ^ { 2 } } - 4 e ^ { x ^ { 2 } } - 3 = 0$$
- Hence show that \(\alpha\) lies between 1 and 2
- Show that the equation in part (b) can be written in the form
$$x = \frac { 1 } { 2 } \sqrt { 4 + 3 \mathrm { e } ^ { - x ^ { 2 } } }$$
The iteration formula
$$x _ { n + 1 } = \frac { 1 } { 2 } \sqrt { 4 + 3 \mathrm { e } ^ { - x _ { n } ^ { 2 } } }$$
with \(x _ { 1 } = 1\) is used to find an approximation for \(\alpha\).
- Use the iteration formula to find, to 4 decimal places, the value of
- \(X _ { 3 }\)
- \(\alpha\)