The curve with equation $y = f(x)$ where f(x) = $x^2 + ext{ln}(2x^2 - 4x + 5)$ has a single turning point at $x = \alpha$ (a) Show that $\alpha$ is a solution of the equation $2\alpha^2 - 4\alpha + 7\alpha - 2 = 0$ (4) The iterative formula $x_{n+1} = \frac{1}{7}(2 + 4x_n^2 - 2x_n)$ is used to find an approximate value for $\alpha$ - Edexcel - A-Level Maths Pure - Question 6 - 2021 - Paper 1

Using the iterative formula with $x_1 = 0.3$:

$x_2 = \frac{1}{7}(2 + 4(0.3)^2 - 2(0.3))$

Calculating:

$x_2 = \frac{1}{7}(2 + 4(0.09) - 0.6) = \frac{1}{7}(2 + 0.36 - 0.6) = \frac{1}{7}(1.76) = 0.2514$

So, the value of $x_2 = 0.2514$.

Continuing with the iterative process:

1. First, calculate $x_3$ using $x_2$:

   $x_3 = \frac{1}{7}(2 + 4(0.2514)^2 - 2(0.2514))$

   Evaluating:

   $x_3 \approx 0.3294$

2. Now use $x_3$ to find $x_4$:

   $x_4 = \frac{1}{7}(2 + 4(0.3294)^2 - 2(0.3294))$

   Calculating:

   $x_4 \approx 0.3398$

So, the value of $x_4 = 0.3398$.

Choosing the function $h(x) = 2x^2 - 4x + 7x - 2$ and evaluating at the boundaries:

$h(0.3415) = 0.00366$ $h(0.3405) = -0.00130$

Since there is a change of sign and $h'(x)$ is continuous, we can conclude that:

$\alpha \approx 0.341$ (to 3 decimal places).