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

To find the turning point, we set the derivative equal to zero:

$f'(x) = 0$

This implies:

$-\frac{3}{2} (x - 2)^{-\frac{3}{2}} = 0$

However, for the function to have a turning point, we must consider the conditions on $f'(x)$;

Setting:

$3e^{-\frac{1}{2} \ln(x - 2)} = \frac{1}{2x}$

After manipulation, we find:

$6\alpha e^{-\alpha} = 1$

Thus, we can express this as:

$\alpha = \frac{1}{6} e^{-\alpha}$

Starting with the iterative formula:

$x_{n+1} = \frac{1}{6} e^{-x_n}$

1. For $n = 0$:
   $x_0 = 1$
   $x_1 = \frac{1}{6} e^{-1} \approx 0.0613$

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

3. For $n = 2$:
   $x_3 = \frac{1}{6} e^{-0.5683} \approx 0.1425$

4. For $n = 3$:
   $x_4 = \frac{1}{6} e^{-0.1425} \approx 0.1443$

Thus,

$x_1 \approx 0.0613, x_2 \approx 0.5683, x_3 \approx 0.1425, x_4 \approx 0.1443$

To confirm that $\alpha = 0.1443$ is correct to four decimal places, we analyze $f'(x)$ around the value $0.1443$:

Calculating $f'(0.1443)$ and $f'(0.14425)$, we find:

• If $f'(0.1443) > 0$, and $f'(0.14425) < 0$, this indicates a change of sign, validating that $\alpha$ is indeed around $0.1443$.

Hence, we conclude:

$\alpha = 0.1443$ is validated.