Figure 2 shows a sketch of part of the curve with equation y = 2 \, \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 demonstrate that the x coordinate of R satisfies the given equation:

1. Differentiate the function to find $\frac{dy}{dx}$:

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

2. Set the derivative equal to zero to find critical points:

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

3. Rearrange this to find the solution for R:

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

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

Using the iterative formula:

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

Starting with $x_0 = 1.3$:

1. Calculate $x_1$:

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

   Performing the calculation, we find:

   $x_1 \approx 1.284$

2. Now calculate $x_2$:

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

   From this, we find:

   $x_2 \approx 1.276$

Thus, the values are:

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