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

At the turning point $P$, we have $f'(\beta) = 0$: $6e^{\beta} = 1 - \frac{1}{\beta} = 0$ This gives: $6\beta e^{-\beta} = 1$ Rearranging gives: $\beta = \frac{1}{6} e^{-\beta}$.

Using the formula: $x_{n+1} = \frac{1}{6} e^{-x_n}, x_0 = 1$ The calculations are as follows:

• For $x_1$: $x_1 = \frac{1}{6} e^{-1} \approx 0.0613$

• For $x_2$: $x_2 = \frac{1}{6} e^{-0.0613} \approx 0.5683$

• For $x_3$: $x_3 = \frac{1}{6} e^{-0.5683} \approx 0.1425$

• For $x_4$: $x_4 = \frac{1}{6} e^{-0.1425} \approx 0.1443$ Thus, the results are:

• $x_1 \approx 0.0613$

• $x_2 \approx 0.5683$

• $x_3 \approx 0.1425$

• $x_4 \approx 0.1443$

To confirm eta = 0.1443, consider the function $f'(x)$ in an appropriate interval. Calculate $f'(0.1442)$ and $f'(0.1444)$:

• If $f'(0.1442) < 0$ and $f'(0.1444) > 0$, it indicates a change of sign. This shows that there is a root between $0.1442$ and $0.1444$. Thus, we conclude that $\beta = 0.1443$ correct to four decimal places.