- The curve with equation \(y = \mathrm { f } ( x )\) where
$$f ( x ) = x ^ { 2 } + \ln \left( 2 x ^ { 2 } - 4 x + 5 \right)$$
has a single turning point at \(x = \alpha\)
- Show that \(\alpha\) is a solution of the equation
$$2 x ^ { 3 } - 4 x ^ { 2 } + 7 x - 2 = 0$$
The iterative formula
$$x _ { n + 1 } = \frac { 1 } { 7 } \left( 2 + 4 x _ { n } ^ { 2 } - 2 x _ { n } ^ { 3 } \right)$$
is used to find an approximate value for \(\alpha\).
Starting with \(x _ { 1 } = 0.3\) - calculate, giving each answer to 4 decimal places,
- the value of \(x _ { 2 }\)
- the value of \(x _ { 4 }\)
Using a suitable interval and a suitable function that should be stated,
- show that \(\alpha\) is 0.341 to 3 decimal places.