Figure 1 shows a sketch of part of the curve with equation $y = f(x)$, where $f(x) = (8 - x) ext{ln} x$, $x > 0$ 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 $f'(x)$, we apply the product rule: Let $u = (8 - x)$ and $v = ext{ln} x$.

Then, $f(x) = u imes v,$ so applying the product rule:

$f'(x) = u'v + uv'$

Calculating:

• $u' = -1$, v' = \frac{1}{x}$,

Thus, we have:

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

To show this, we evaluate $f(3.5)$ and $f(3.6)$:

1. Calculate $f(3.5)$: $f(3.5) = (8 - 3.5) ext{ln}(3.5) \approx 4.5 \times 1.25276 \approx 5.13142$

2. Calculate $f(3.6)$: $f(3.6) = (8 - 3.6) ext{ln}(3.6) \approx 4.4 \times 1.28093 \approx 5.63588$

Since $f(3.5) > 0$ and $f(3.6) < 0$, by the Intermediate Value Theorem, there exists a root between 3.5 and 3.6.

At the point Q, we have $f'(x) = 0$:

This implies:

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

Rearranging this gives:

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

Therefore, the equivalent expression is:

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

Using the iterative formula $x_{n+1} = \frac{8}{\text{ln} x_n + 1}$:

1. For $x_0 = 3.55$, $x_1 = \frac{8}{\text{ln}(3.55) + 1} \approx 3.529\ldots$

2. For $x_1 = 3.529$, $x_2 = \frac{8}{\text{ln}(3.529) + 1} \approx 3.538\ldots$

3. For $x_2 = 3.538$, $x_3 = \frac{8}{\text{ln}(3.538) + 1} \approx 3.548\ldots$

Thus, to 3 decimal places:

• $x_1 \approx 3.529,$
• $x_2 \approx 3.538,$
• $x_3 \approx 3.548.$