Figure 2 shows a sketch of part of the curve with equation y = 2 \, ext{cos} \, igg(\frac{1}{2} x^2 \bigg) + x^3 - 3x - 2 The curve crosses the x-axis at the point Q and has a minimum turning point at R - Edexcel - A-Level Maths Pure - Question 8 - 2014 - Paper 5

To prove that the x coordinate of R satisfies the equation given, we start with the expression:

$x = \sqrt{1 + \frac{2}{3} \sin \bigg( \frac{1}{2} x^2 \bigg)}$

By substituting the derivative of y into our analysis:

1. Differentiate y with respect to x: $\frac{dy}{dx} = -2 \sin \bigg( \frac{1}{2} x^2 \bigg) + 3x^2 - 3$
2. Set (\frac{dy}{dx} = 0) to find critical points where $-2 \sin \bigg( \frac{1}{2} x^2 \bigg) + 3x^2 - 3 = 0$ and find intersections accordingly. This ensures that R satisfies the equation and boundaries are established.

To find the values of x_1 and x_2 using the iterative formula, we begin with:

$x_{n+1} = \sqrt{1 + \frac{2}{3} \sin \bigg( \frac{1}{2} x_n^2 \bigg)}$

Using the initial value x_0 = 1.3:

1. For n = 0: $x_1 = \sqrt{1 + \frac{2}{3} \sin \bigg( \frac{1}{2} (1.3)^2 \bigg)}$ Calculate: $x_1 \approx 1.284\ (to \, 3 \, decimal \, places)$
2. For n = 1: $x_2 = \sqrt{1 + \frac{2}{3} \sin \bigg( \frac{1}{2} (1.284)^2 \bigg)}$ Calculate: $x_2 \approx 1.276 \ (to \, 3 \, decimal \, places)$

Thus, the values are:

• x_1 = 1.284
• x_2 = 1.276.