Figure 2 shows a sketch of part of the curve with equation $y = 2 \, ext{cos}\left( \frac{1}{2} x^2 \right) + x^3 - 3x - 2$ The curve crosses the x-axis at the point Q and has a minimum turning point at R - Edexcel - A-Level Maths Pure - Question 7 - 2014 - Paper 5

To find the x-coordinate of R, we start with the expression:

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

Now we differentiate the original equation:

1. Taking the derivative:

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

   Then set this equal to zero to find critical points (potential minimum or maximum):

   $-2 \sin\left( \frac{1}{2} R^2 \right) + 3R^2 - 3 = 0$

2. Assuming R lies in a numerical range, substituting yields:

   $x = R \text{ is a solution}$

as derived from the original problem's requirements.

Starting with the iterative formula:

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

1. Begin with $x_0 = 1.3$:

   Calculate $x_1$:

   $x_1 = \sqrt{1 + \frac{2}{3} \sin\left( \frac{1}{2} (1.3)^2 \right)}$

   Evaluating gives:

   $x_1 \approx 1.284$

2. Now iterate to find $x_2$ using $x_1$:

   $x_2 = \sqrt{1 + \frac{2}{3} \sin\left( \frac{1}{2} (1.284)^2 \right)}$

   This results in:

   $x_2 \approx 1.276$

Final results to 3 decimal places:

• $x_1 \approx 1.284$
• $x_2 \approx 1.276$.