Figure 1 shows a sketch of part of the curve with equation $y = f(x)$, where $f(x) = (8 - x) ext{ln } x, ext{ for } x > 0$ The curve cuts the x-axis at the points A and B and has a maximum turning point at Q, as shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 7 - 2011 - Paper 4

To find the derivative $f'(x)$, we apply the product rule. Let:

v = ext{ln } x$$ Then, $$f'(x) = u \frac{dv}{dx} + v \frac{du}{dx}$$ Calculating the derivatives: - $rac{dv}{dx} = \frac{1}{x}$ - $rac{du}{dx} = -1$ Thus, $$f'(x) = (8 - x) \left( \frac{1}{x} \right) - ext{ln } x\cdot 1$$ Which simplifies to: $$f'(x) = \frac{8 - x}{x} - ext{ln } x$$

Evaluating $f(3.5)$ and $f(3.6)$:

• $f(3.5) = (8 - 3.5) ext{ln } 3.5 \approx 0.3295...$
• $f(3.6) = (8 - 3.6) ext{ln } 3.6 \approx -0.0578...$

Since $f(3.5) > 0$ and $f(3.6) < 0$, by the Intermediate Value Theorem, the x-coordinate of Q lies between 3.5 and 3.6.

At Q, we have $f'(x) = 0$, leading to:

\Rightarrow \text{ln } x = \frac{8 - x}{x}$$ By rearranging, we find: $$x \cdot \text{ln } x + x = 8 \ \Rightarrow x = \frac{8}{1 + \text{ln } x}$$

Using the iteration formula:

$x_{n+1} = \frac{8}{1 + \text{ln } x_n}$

1. For $x_0 = 3.55$: $x_1 = \frac{8}{1 + \text{ln } 3.55} \approx 3.528974374...$
2. For $x_1 = 3.528974374$: $x_2 = \frac{8}{1 + \text{ln } 3.528974374} \approx 3.532486401...$
3. For $x_2 = 3.532486401$: $x_3 = \frac{8}{1 + \text{ln } 3.532486401} \approx 3.538...$

Hence, the values are:

• $x_1 \approx 3.529$
• $x_2 \approx 3.532$
• $x_3 \approx 3.538$, rounded to three decimal places.