4. (a) Differentiate to find $f'(x)$ - Edexcel - A-Level Maths Pure - Question 6 - 2005 - Paper 5

To show that $\alpha = \frac{1}{6} e^{-\alpha}$, we rearrange the expression obtained from setting $f'(x) = 0$:

1. From the previous derivation, we have:
   $6\alpha e^{\alpha} = 1$
2. This rearranges to give:
   $\alpha = \frac{1}{6} e^{-\alpha}$
   , confirming the required relation.

Given the iterative formula:
$x_{n+1} = \frac{1}{6} e^{-x_n}, \quad x_0 = 1$
we perform the iterations:

1. For $x_1$:
   $x_1 = \frac{1}{6} e^{-1} \approx 0.0613$
2. For $x_2$:
   $x_2 = \frac{1}{6} e^{-0.0613} \approx 0.5683$
3. For $x_3$:
   $x_3 = \frac{1}{6} e^{-0.5683} \approx 0.1425$
4. For $x_4$:
   $x_4 = \frac{1}{6} e^{-0.1425} \approx 0.1444$
   Thus, we find the values for $x_1, x_2, x_3$, and $x_4$ as 0.0613, 0.5683, 0.1425, and approximately 0.1444 respectively.

To demonstrate that $\alpha = 0.1443$ is correct to four decimal places:

1. Evaluate $f'(x)$ at points around $x = 0.1443$:
   • If $x = 0.14425$, compute:
     $f'(0.14425) \approx 0.0007$
   • If $x = 0.14435$, compute:
     $f'(0.14435) \approx -0.0021$
2. Observe that $f'(0.14425) > 0$ and $f'(0.14435) < 0$.
3. This indicates a sign change around $x = 0.1443$, hence confirming the root.
   In conclusion, with change of sign confirmed, $\alpha = 0.1443$ is accurate to four decimal places.