a) Show that the equation f(x)=0 has a solution in the interval 0.8 < x < 0.9 - Edexcel - A-Level Maths Pure - Question 7 - 2012 - Paper 6

To find the x-coordinate of the minimum point P, we need to first find the critical points by setting the derivative f'(x) to zero:

1. Differentiate f(x):

   $f'(x) = 2x - 3 - \frac{1}{2} \cos \left( \frac{x}{2} \right)$

2. Set the derivative equal to zero:

   $0 = 2x - 3 - \frac{1}{2} \cos \left( \frac{x}{2} \right)$

   Rearranging gives:

   $2x = 3 + \frac{1}{2} \cos \left( \frac{x}{2} \right)$

   Or, expressed differently:

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

Thus, it has been shown that the x-coordinate of P is indeed the solution of the equation.

1. For x_0 = 2:

   $x_1 = \frac{3 + \sin(2)}{2} \approx \frac{3 + 0.9093}{2} \approx 1.9546 \approx 1.955$

2. For x_1:

   $x_2 = \frac{3 + \sin(1.955)}{2} \approx \frac{3 + 0.9114}{2} \approx 1.9557 \approx 1.956$

3. For x_2:

   $x_3 = \frac{3 + \sin(1.956)}{2} \approx \frac{3 + 0.9116}{2} \approx 1.9560 \approx 1.956$

Thus, the values are:

• $x_1 \approx 1.955$
• $x_2 \approx 1.956$
• $x_3 \approx 1.956$.

To show that the x-coordinate of P is 1.9078 to 4 decimal places:

1. We first choose an interval around this value, such as (1.9075, 1.9080).

2. Evaluate f(1.9075) and f(1.9080):

   • $f(1.9075) \approx 0$
   • $f(1.9080) < 0$

3. Since there is a change in sign, by the Intermediate Value Theorem, we can conclude there is a root in the interval (1.9075, 1.9080).

4. Refine the interval further, if needed, to confirm that the x-coordinate of P is indeed 1.9078 correct to 4 decimal places.