g(x) = e^(x) + x - 6 (a) Show that the equation g(x) = 0 can be written as x = ln(6 - x) + 1, x < 6 (2) The root of g(x) = 0 is α - Edexcel - A-Level Maths Pure - Question 23 - 2013 - Paper 1

Using the iterative formula provided: $x_{n+1} = ln(6 - x_n) + 1$

Calculation of x_1:

Starting with x_0 = 2:

$x_{1} = ln(6 - 2) + 1 = ln(4) + 1 ≈ 2.3863$

Calculation of x_2:

Using x_1 to find x_2:

$x_{2} = ln(6 - 2.3863) + 1 = ln(3.6137) + 1 ≈ 2.2847$

Calculation of x_3:

Using x_2 to find x_3:

$x_{3} = ln(6 - 2.2847) + 1 = ln(3.7153) + 1 ≈ 2.3125$

Thus the values are:

• x_1 ≈ 2.3863
• x_2 ≈ 2.2847
• x_3 ≈ 2.3125

To find α correctly to three decimal places, we start by finding an interval that contains the root α.

Choosing the interval:

Based on previous calculations:

• g(2.3065) = e^(2.3065) + 2.3065 - 6 ≈ -0.0002
• g(2.3075) = e^(2.3075) + 2.3075 - 6 ≈ 0.0044

Since g(2.3065) is slightly less than 0 and g(2.3075) is slightly more than 0, it shows that the root lies in the interval [2.3065, 2.3075].

Finalizing the estimate for α:

Using bisection or any iterative method on this interval would bring us to a value α = 2.307 (when rounded to three decimal places). Thus, we have shown that α = 2.307.