6.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{2ce10759-9ce6-47a1-b55d-d22082f88f55-12_518_670_248_740}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{figure}
Figure 1 shows a sketch of the curves with equations \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\) where
$$\begin{array} { l l }
\mathrm { f } ( x ) = \mathrm { e } ^ { 4 x ^ { 2 } - 1 } & x > 0
\mathrm {~g} ( x ) = 8 \ln x & x > 0
\end{array}$$
- Find
- \(\mathrm { f } ^ { \prime } ( x )\)
- \(\mathrm { g } ^ { \prime } ( x )\)
Given that \(\mathrm { f } ^ { \prime } ( x ) = \mathrm { g } ^ { \prime } ( x )\) at \(x = \alpha\)
- show that \(\alpha\) satisfies the equation
$$4 x ^ { 2 } + 2 \ln x - 1 = 0$$
The iterative formula
$$x _ { n + 1 } = \sqrt { \frac { 1 - 2 \ln x _ { n } } { 4 } }$$
is used with \(x _ { 1 } = 0.6\) to find an approximate value for \(\alpha\)
- Calculate, giving each answer to 4 decimal places,
- the value of \(x _ { 2 }\)
- the value of \(\alpha\)