The function f is defined by f: x ↦ ln(4 − 2x), x < 2 and x ∈ ℝ - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 6

The range of f^(-1) can be determined from our earlier findings. As x approaches -∞, f^(-1)(x) approaches 2. Thus, we conclude that:

$ext{Range of } f^{-1} = (2, ext{∞})$

The graph of y = f^(-1)(x) is a transformed exponential curve. The intersection with the y-axis occurs when x = 0:

f^{-1}(0) = 2 - rac{1}{2}e^0 = 2 - rac{1}{2} = 1.5

So, the point is (0, 1.5).

The intersection with the x-axis occurs when y = 0:

0 = 2 - rac{1}{2}e^x

Solving gives:

Using the iterative formula:

1. Start with x_0 = -0.3: $$$$

ewline ≈ -0.3515$$

2. For x_1 = -0.3515: $$$$

ewline ≈ -0.3522$$

Both values should be provided to 4 decimal places:

Thus, x_1 ≈ -0.3515 and x_2 ≈ -0.3522.

To find k, we determine the limit of x_n as n approaches infinity. Notice that this requires additional iterations beyond x_2:

Assuming we continue the iteration until convergence, we see that:

$k ext{ converges to } -0.352$

Thus, the value is: $k = -0.352$