Let f(x) = x³ + 2x² - 3x - 11 - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 2

1. Start with the initial value:

   $x_1 = 0$

2. Apply the iterative formula:

   $x_{n+1} = \frac{3x_n + 11}{x_n + 2}$

   For x₂:

   $x_2 = \frac{3(0) + 11}{0 + 2} = \frac{11}{2} = 5.5$

   For x₃:

   $x_3 = \frac{3(5.5) + 11}{5.5 + 2} = \frac{16.5 + 11}{7.5} = \frac{27.5}{7.5} \approx 3.6667$

   For x₄:

   $x_4 = \frac{3(3.6667) + 11}{3.6667 + 2} \approx \frac{11.0001}{5.6667} \approx 1.9432$

   So, we have:

   $x_2 \approx 5.5, x_3 \approx 3.667, x_4 \approx 1.943$ (rounded to 3 decimal places)

To show that a = 2.057 correct to three decimal places:

1. Evaluate f(2.056) and f(2.057):

   $f(2.056) \approx -0.0413781637$

   $f(2.057) \approx 0.0014041940$

2. Noticing a sign change between f(2.056) and f(2.057) indicates that there is a root between these two values.

3. Hence, the value of a is approximately 2.057, which is between the values 2.056 and 2.058, confirming it is correct to 3 decimal places.