6.
Curve \(C\) has equation
$$y = \left( x ^ { 2 } - 5 x + 8 \right) \mathrm { e } ^ { x ^ { 2 } } \quad x \in \mathbb { R }$$
- Show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \left( 2 x ^ { 3 } - 10 x ^ { 2 } + 18 x - 5 \right) \mathrm { e } ^ { x ^ { 2 } }$$
Given that
- \(C\) has only one stationary point
- the stationary point has \(x\) coordinate \(\alpha\)
- \(\frac { \mathrm { d } y } { \mathrm {~d} x } \approx - 0.5\) at \(x = 0.3\)
- \(\frac { \mathrm { d } y } { \mathrm {~d} x } \approx 0.9\) at \(x = 0.4\)
- explain why \(0.3 < \alpha < 0.4\)
- Show that \(\alpha\) is a solution of the equation
$$x = \frac { 5 \left( 2 x ^ { 2 } + 1 \right) } { 2 \left( x ^ { 2 } + 9 \right) }$$ - Using the iteration formula
$$x _ { n + 1 } = \frac { 5 \left( 2 x _ { n } ^ { 2 } + 1 \right) } { 2 \left( x _ { n } ^ { 2 } + 9 \right) } \quad \text { with } x _ { 1 } = 0.3$$
find
- the value of \(x _ { 3 }\) to 4 decimal places,
- the value of \(\alpha\) to 4 decimal places.