The function f is defined by $$f: x \mapsto \ln(4 - 2x), \quad x < 2 \quad \text{and} \quad x \in \mathbb{R}.$$ (a) Show that the inverse function of f is defined by $$f^{-1}: x \mapsto 2 - \frac{1}{2} e^x$$ and write down the domain of f^{-1} - Edexcel - A-Level Maths Pure - Question 6 - 2007 - Paper 6

The range of f^{-1} is determined by the domain of the original function f. Since f is defined for all x where x < 2, the output of f will be:

$f(x) \in (-\infty, \ln(4)),$ where \ln(4) occurs at x approaching 2 from the left. Therefore, the range of f^{-1} is:

$f^{-1}(x) \in (2 - \frac{1}{2} e^{\ln(4)}, 2) = (0, 2).$

To sketch the graph of the inverse function:

1. y-intercept: When x = 0:

   $y = f^{-1}(0) = 2 - \frac{1}{2} e^0 = 1.$
   Coordinate: (0, 1).

2. x-intercept: When y = 0:

   $0 = 2 - \frac{1}{2} e^x \Rightarrow \frac{1}{2} e^x = 2 \Rightarrow e^x = 4 \Rightarrow x = \ln(4).$
   Coordinate: (\ln(4), 0).

The graph will have the shape of the function decreasing and asymptotic toward y = 2 as x approaches minus infinity.

Using the iterative formula:

$x_{n+1} = -\frac{1}{2} e^{x_n}$

1. Start with x_0 = -0.3:

   $x_1 = -\frac{1}{2} e^{-0.3} \approx -0.35092688 \text{ (to 4 decimal places: -0.3509)}$

2. Next, calculate x_2:

   $x_2 = -\frac{1}{2} e^{x_1} \approx -0.35201671 \text{ (to 4 decimal places: -0.3520)}$

For finding k, after multiple iterations using the earlier results:

Continuing with the iterative formula until the desired accuracy, we find:

$k \approx -0.352$ to 3 decimal places.