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 5 - 2011 - Paper 4

To find the derivative of $f(x)$, we will use the product rule for differentiation. Let:

$u = (8 - x), \, v = \text{ln} x$

Then,

$f(x) = u v$

Applying the product rule:

$f^{\prime}(x) = u v^{\prime} + v u^{\prime}$

Where:

$v^{\prime} = \frac{1}{x} \, \text{and} \, u^{\prime} = -1$

Thus, we have:

$f^{\prime}(x) = (8 - x) \cdot \frac{1}{x} + \text{ln} x \cdot (-1) = \frac{8 - x}{x} - \text{ln} x$

We evaluate $f(3.5)$ and $f(3.6)$:

1. Calculate $f(3.5)$: $f(3.5) = (8 - 3.5) \cdot \text{ln}(3.5) = 4.5 \cdot 1.25276... = 5.65042...$

2. Calculate $f(3.6)$: $f(3.6) = (8 - 3.6) \cdot \text{ln}(3.6) = 4.4 \cdot 1.28096... = 5.62923...$

Since $f(3.5) > 0$ and $f(3.6) < 0$, and noting that $f(x)$ is continuous, by the Intermediate Value Theorem, there exists a root in the interval $(3.5, 3.6)$.

At Q, $f^{\prime}(x) = 0$:

We take:

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

Rearranging, we find:

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

Multiplying by $x$ results in:

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

Thus, rearranging gives the equation:

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

Using the iterative formula:

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

Starting with $x_0 = 3.55$:

1. Calculate $x_1$: $x_1 = \frac{8}{\text{ln}(3.55) + 1} = \frac{8}{1.27065 + 1} = 3.529...$

2. Calculate $x_2$: $x_2 = \frac{8}{\text{ln}(3.529...) + 1} = \frac{8}{1.27021 + 1} = 3.528...$

3. Calculate $x_3$: $x_3 = \frac{8}{\text{ln}(3.528...) + 1} = \frac{8}{1.27041 + 1} = 3.528...$

Therefore, to three decimal places:

• $x_1 \approx 3.529$
• $x_2 \approx 3.528$
• $x_3 \approx 3.528$