f(x) = 2x^2 - x - 4 - Edexcel - A-Level Maths Pure - Question 6 - 2006 - Paper 5

Using the iteration formula:

$x_{n+1} = \left( \frac{2}{x_n + 2} \right),$

starting with x_0 = 1.35:

1. Calculate x_1: $x_1 = \frac{2}{1.35 + 2} \approx 1.41$

2. Calculate x_2: $x_2 = \frac{2}{1.41 + 2} \approx 1.39$

3. Calculate x_3: $x_3 = \frac{2}{1.39 + 2} \approx 1.39$

Thus, the values are:

• x_1 = 1.41
• x_2 = 1.39
• x_3 = 1.39.

To prove that \alpha lies between 1.391 and 1.392:

1. Evaluate f(1.391): $f(1.391) = 2(1.391)^2 - (1.391) - 4 \approx -0.0006$

2. Evaluate f(1.392): $f(1.392) = 2(1.392)^2 - (1.392) - 4 \approx 0.0004$

This shows a change of sign between 1.391 and 1.392, indicating that there is at least one root in that interval, confirming \alpha = 1.392 to three decimal places.