Figure 2 shows a sketch of part of the curve with equation y = 2 \, ext{cos} \, igg(rac{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 6 - 2014 - Paper 5

To find the x-coordinate of the turning point R, we first differentiate y with respect to x:

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

Setting this equal to zero gives:

$-2 \sin \bigg( \frac{1}{2} x^2 \bigg) + 3x^2 - 3 = 0$

Rearranging this yields:

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

This verifies that the x-coordinate of R is indeed a solution of the provided equation.

Using the iterative formula:

$x_{n+1} = \sqrt{1 + \frac{2}{3} \sin \bigg( \frac{1}{2} x_n^2 \bigg)}, \; x_0 = 1.3$

1. Start with $x_0 = 1.3$: $x_1 = \sqrt{1 + \frac{2}{3} \sin \bigg( \frac{1}{2} (1.3)^2 \bigg)}$ Calculate $x_1$ to find its approximate value:

   • This results in $x_1 \approx 1.284$

2. Next, use $x_1$ to find $x_2$: $x_2 = \sqrt{1 + \frac{2}{3} \sin \bigg( \frac{1}{2} (1.284)^2 \bigg)}$ Calculating this gives:

   • $x_2 \approx 1.276$

Thus:

• $x_1 \approx 1.284$
• $x_2 \approx 1.276$, both rounded to three decimal places.