Let f:(1,∞)→ℝ be a function defined as f(x) = (x-1)/(x+1) . Let f^...

Given $f(x) = \dfrac{x-1}{x+1}$. Let us find the first few compositions of $f(x)$:

$f^2(x) = f(f(x)) = \dfrac{\dfrac{x-1}{x+1}-1}{\dfrac{x-1}{x+1}+1} = \dfrac{x-1-x-1}{x-1+x+1} = \dfrac{-2}{2x} = -\dfrac{1}{x}$

$f^3(x) = f(f^2(x)) = f\left(-\dfrac{1}{x}\right) = \dfrac{-\dfrac{1}{x}-1}{-\dfrac{1}{x}+1} = \dfrac{-1-x}{-1+x} = \dfrac{x+1}{1-x}$

$f^4(x) = f(f^3(x)) = f\left(\dfrac{x+1}{1-x}\right) = \dfrac{\dfrac{x+1}{1-x}-1}{\dfrac{x+1}{1-x}+1} = \dfrac{x+1-1+x}{x+1+1-x} = \dfrac{2x}{2} = x$

Since $f^4(x) = x$, the sequence of functions is periodic with a period of $4$.

Therefore, $f^{26}(x) = f^{4 \times 6 + 2}(x) = f^2(x) = -\dfrac{1}{x}$.

We are given $g(x) + f^{26}(x) = 0$, which implies:
$g(x) - \dfrac{1}{x} = 0 \Rightarrow g(x) = \dfrac{1}{x}$

We need to find the area of the region bounded by the curves $y = \dfrac{1}{x}$, $y = x - \dfrac{3}{2}$ (from $2y = 2x - 3$), $y = 0$, and $x = 4$.

First, find the point of intersection of $y = \dfrac{1}{x}$ and $y = x - \dfrac{3}{2}$:
$\dfrac{1}{x} = x - \dfrac{3}{2} \Rightarrow 2x^2 - 3x - 2 = 0$
$(2x+1)(x-2) = 0$
Since $x \in (1, \infty)$, we get $x = 2$.

The line $y = x - \dfrac{3}{2}$ intersects the x-axis ($y = 0$) at $x = \dfrac{3}{2}$.

The required area $A$ is bounded by $y = x - \dfrac{3}{2}$ from $x = \dfrac{3}{2}$ to $x = 2$, and by $y = \dfrac{1}{x}$ from $x = 2$ to $x = 4$.

$A = \int_{3/2}^{2} \left(x - \dfrac{3}{2}\right) dx + \int_{2}^{4} \dfrac{1}{x} dx$

Evaluating the first integral:
$\int_{3/2}^{2} \left(x - \dfrac{3}{2}\right) dx = \left[ \dfrac{1}{2}\left(x - \dfrac{3}{2}\right)^2 \right]_{3/2}^{2} = \dfrac{1}{2} \left(2 - \dfrac{3}{2}\right)^2 - 0 = \dfrac{1}{2} \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{8}$

Evaluating the second integral:
$\int_{2}^{4} \dfrac{1}{x} dx = [\log_e x]_{2}^{4} = \log_e 4 - \log_e 2 = \log_e 2$

Total Area = $\dfrac{1}{8} + \log_e 2$