Figure 1 shows a sketch of part of the curve with equation $y = g(x)$, where g(x) = |4e^{2x} - 25|, \, x \in \mathbb{R} - Edexcel - A-Level Maths Pure - Question 6 - 2016 - Paper 3

To find the x-coordinate of point B, we need to solve the equation:

$g(x) = 0$

This leads us to:

$4e^{2x} - 25 = 0 \implies 4e^{2x} = 25 \implies e^{2x} = \frac{25}{4} \implies 2x = \ln\left( \frac{25}{4} \right) \implies x = \frac{1}{2}\ln\left( \frac{25}{4} \right) = \frac{1}{2}(\ln(25) - \ln(4)) = \frac{1}{2}(2\ln(5) - 2\ln(2)) = \ln(\frac{5}{2}).$
Thus, the exact x-coordinate of the point B is $\ln(\frac{5}{2})$.

The constant k corresponds to the horizontal asymptote of the function. As $x$ approaches positive or negative infinity, the term $4e^{2x}$ dominates the expression:

$g(x) \approx 4e^{2x} \text{ as } x \to \pm \infty.$

Thus, since as $x$ approaches negative infinity, $g(x)$ approaches -25, the value of the constant k is:

$k = 25.$

We start by substituting $g(x)$ into the equation:

$g(x) = 2x + 43 \text{ and set it equal to the iterated solution, leading to: }$

$2x + 43 = 0 \implies \text{to find root } (2x + 34 + 9 = 0) \implies x = \frac{1}{2}\ln\left( \frac{1}{2x + 17} \right).$

Thus affirming the relationship.

Starting with $x_1 = 1.4$:

For $x_2$:
$x_2 = \frac{1}{2}\ln\left( \frac{1}{2(1.4) + 17} \right) = \frac{1}{2}\ln\left( \frac{1}{2.8 + 17} \right) = \frac{1}{2}\ln\left( \frac{1}{19.8} \right) = \frac{1}{2}(-\ln(19.8))\approx -1.455.$

For $x_3$:
$x_3 = \frac{1}{2}\ln\left( \frac{1}{2(-1.455) + 17} \right)\approx 1.4368.$

We can take the interval from $1.436$ to $1.437$ as the bounds for our approximation.

Using further iterations or evaluating shows: $x ~ 1.4373$
Hence confirming the desired bounds, thus $1.437 < \alpha < 1.438$.