Figure 1 shows a sketch of part of the curve with equation $y = f(x)$, where $$f(x)=(8-x) ext{ln}x, ext{ } x > 0$$ 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 6 - 2011 - Paper 4

To find the derivative of $f(x)$, we apply the product rule:

Let:

• $u = (8 - x)$, therefore $u' = -1$
• $v = ext{ln} x$, therefore $v' = \frac{1}{x}$

Using the product rule:

$f'(x) = u'v + uv' = (-1)( ext{ln} x) + (8 - x) \cdot \frac{1}{x}$

Simplifying, we get:

$f'(x) = - ext{ln} x + \frac{8 - x}{x}$

So, $f'(x) = - ext{ln} x + \frac{8}{x} - 1$

We will evaluate $f'(3.5)$ and $f'(3.6)$ to show a sign change, indicating a root in the interval (3.5, 3.6).

First,

1. Calculate $f'(3.5)$:

   • Substitute $x = 3.5$ into $f'(x)$: $f'(3.5) = - ext{ln}(3.5) + \frac{8}{3.5} - 1 \approx 0.0295$

2. Then calculate $f'(3.6)$:

   • Substitute $x = 3.6$ into $f'(x)$: $f'(3.6) = - ext{ln}(3.6) + \frac{8}{3.6} - 1 \approx -0.0587$

The change in sign between $f'(3.5)$ and $f'(3.6)$ confirms a turning point in the interval (3.5, 3.6). Thus the x-coordinate of Q lies between 3.5 and 3.6.

At the turning point Q, $f'(x) = 0$.

From earlier, we have:

$-\text{ln} x + \frac{8 - x}{x} = 0$

Rearranging gives:

$\text{ln} x = \frac{8 - x}{x}$

Multiplying through by $x$:

$x \cdot \text{ln} x = 8 - x$

Rearranging this yields:

$x = \frac{8}{1 + \text{ln} x}$

The equation is satisfied, so the x-coordinate of Q is indeed the solution.

Using the iterative formula:

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

1. For $x_1$: $x_1 = \frac{8}{\text{ln}(3.55) + 1} \approx 3.529$

2. For $x_2$: $x_2 = \frac{8}{\text{ln}(3.529) + 1} \approx 3.538$

3. For $x_3$: $x_3 = \frac{8}{\text{ln}(3.538) + 1} \approx 3.536$

Final answers:

• $x_1 = 3.529$
• $x_2 = 3.538$
• $x_3 = 3.536$ (all rounded to 3 decimal places).