Figure 1 shows a sketch of the curve C with equation $$y = \frac{4x^2 + x}{2\sqrt{x}} \quad x > 0$$ (a) Show that $$\frac{dy}{dx} = \frac{12x^2 + x - 16\sqrt{x}}{4\sqrt{x}}$$ The point P, shown in Figure 1, is the minimum turning point on C - Edexcel - A-Level Maths Pure - Question 9 - 2020 - Paper 2

At the minimum turning point $P$, the derivative $\frac{dy}{dx} = 0$. Setting the expression obtained in part (a) equal to zero gives:

$12x^2 + x - 16\sqrt{x} = 0$

To solve for $x$, we can substitute $\sqrt{x} = t$ (hence, $x = t^2$), transforming the equation into:

$12t^4 + t^2 - 16t = 0$

Factoring out $t$ gives:

$t(12t^3 + t - 16) = 0$

The solution $t = 0$ is not valid since $x > 0$. Now we focus on:

$12t^3 + t - 16 = 0$

Using numerical methods or estimating, one finds a root around $t = \frac{4 - \sqrt{16}}{3}$, so substituting back, we find:

$x = \left( \frac{4 - \sqrt{16}}{3} \right)^{\frac{2}{3}}$.

(i) Using the iteration formula:

$x_{n+1} = \left( \frac{4 - \sqrt{x_n}}{3} \right)^{\frac{2}{3}}$

Starting with $x_1 = 2$:

1. Calculate: $x_2 = \left( \frac{4 - \sqrt{2}}{3} \right)^{\frac{2}{3}} \approx 1.64397$ (to 5 decimal places)

(ii) Continue iterating:

2. Next, $x_3 = \left( \frac{4 - \sqrt{x_2}}{3} \right)^{\frac{2}{3}} \approx 1.16560$$

Thus, the x coordinate of $P \approx 1.16560$ (to 5 decimal places).