6. $f(x) = x^3 - 3x + 2 ext{ cos} \left( \frac{x}{2} \right), \, 0 \leq x \leq \pi $ - Edexcel - A-Level Maths Pure - Question 8 - 2012 - Paper 6

To find the $x$-coordinate of the minimum point $P$, we need to evaluate $f'(x)$:

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

Setting $f'(x) = 0$, we have: $3x^2 - 3 - \frac{1}{2} \cos \left( \frac{x}{2} \right) = 0.$

This rearranges to: $\frac{1}{2} \cos \left( \frac{x}{2} \right) = 3x^2 - 3 \Rightarrow \cos \left( \frac{x}{2} \right) = 6x^2 - 6.$

From the properties of $P$, we can conclude that the $x$-coordinate corresponds to: $x = \frac{3 + \sin \left( \frac{x}{2} \right)}{2}$.

First, we calculate the initial value:

• For $x_0 = 2$: $x_1 = \frac{3 + \sin(1)}{2} \approx \frac{3 + 0.8415}{2} \approx 1.9208.$

Next, for $x_1$:

• $x_2 = \frac{3 + \sin(1.9208/2)}{2} \approx \frac{3 + 0.8608}{2} \approx 1.9304.$

Finally, for $x_2$:

• $x_3 = \frac{3 + \sin(1.9304/2)}{2} \approx \frac{3 + 0.8568}{2} \approx 1.9284.$

Thus, we find $x_1 \approx 1.921$, $x_2 \approx 1.930$, and $x_3 \approx 1.928$.

To find the $x$-coordinate of $P$ more accurately, we can apply bisection or further iterations from the previously found intervals. Set $x_1 = 1.9075$ and $x_2 = 1.9085$:

• Calculate $f(1.9075)$ and $f(1.9085)$ to check for sign changes.

Finding that $f(1.9078)$ lies between these values indicates proper convergence. Therefore: $x \approx 1.9078 \text{ (to 4 decimal places)}.$