The function $\mathrm{f}$ is defined by $\mathrm{f} : x \mapsto \frac{2x}{x^2 - 1}, \quad x \in \mathbb{R}, \ x > 1$.

1. Identify the asymptotes of the graph for the given domain.

For $x > 1$:
As $x \to 1^+$, the denominator $(x^2 - 1) \to 0^+$ and the numerator $2x \to 2$, so $\mathrm{f}(x) \to \infty$.
Thus, the vertical asymptote is $x = 1$.
As $x \to \infty$, $\mathrm{f}(x) = \frac{\frac{2}{x}}{1 - \frac{1}{x^2}} \to 0$.
Thus, the horizontal asymptote is $y = 0$.

2. Sketch the graph of $\mathrm{f}$ restricted to the domain.

Using a graphing calculator or by analyzing the behavior, we sketch the curve for $x > 1$.
The graph is a single smooth curve in the first quadrant, strictly decreasing from the vertical asymptote $x = 1$ towards the horizontal asymptote $y = 0$ (the $x$-axis).
(The sketch must strictly start to the right of $x = 1$, with axes and asymptotes $x = 1$, $y = 0$ labeled.)

3. Show that the function has an inverse by referring to its graph.

From the sketched graph, any horizontal line $y = k$ intersects the curve at most once.
By the Horizontal Line Test, $\mathrm{f}$ is a one-to-one function.
Therefore, $\mathrm{f}$ has an inverse.

1. Set up the equation $y = \mathrm{f}(x)$ and rearrange it to form a quadratic equation in $x$.

Let $y = \frac{2x}{x^2 - 1}$.
$y(x^2 - 1) = 2x$
$yx^2 - y = 2x$
$yx^2 - 2x - y = 0$

2. Solve for $x$ in terms of $y$ using the quadratic formula.

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(y)(-y)}}{2y}$
$x = \frac{2 \pm \sqrt{4 + 4y^2}}{2y}$
$x = \frac{2 \pm 2\sqrt{1 + y^2}}{2y}$
$x = \frac{1 \pm \sqrt{1 + y^2}}{y}$

3. Determine the correct sign for the inverse function using the domain of $\mathrm{f}$.

Since the domain of $\mathrm{f}$ is $x > 1$, we must have $x > 0$.
For $y > 0$, $\sqrt{1 + y^2} > 1$, so $1 - \sqrt{1 + y^2} < 0$, which would result in a negative $x$.
Thus, we reject the negative root.
$x = \frac{1 + \sqrt{1 + y^2}}{y}$
$\mathrm{f}^{-1}(x) = \frac{1 + \sqrt{1 + x^2}}{x}$

4. Determine the range of $\mathrm{f}$ to state the domain of $\mathrm{f}^{-1}$.

The domain of $\mathrm{f}^{-1}$ is the range of $\mathrm{f}$.
For $x > 1$, $x^2 - 1 > 0$ and $2x > 2$, so $\mathrm{f}(x) > 0$.
As $x \to 1^+$, $\mathrm{f}(x) \to \infty$.
As $x \to \infty$, $\mathrm{f}(x) = \frac{2/x}{1 - 1/x^2} \to 0$.
Therefore, the range of $\mathrm{f}$ is $(0, \infty)$.
The domain of $\mathrm{f}^{-1}$ is $x > 0$.

1. State the equation of the line of reflection between a function and its inverse.

The graph of $\mathrm{f}$ must be reflected in the line $y = x$ to obtain the graph of $\mathrm{f}^{-1}$.

2. Set up the equation to find the solution, using the 'hence' method.

Since the graphs intersect on the line of reflection, the solution to $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$ is the same as the solution to $\mathrm{f}(x) = x$.
$\frac{2x}{x^2 - 1} = x$

3. Solve the equation for $x$ and reject invalid roots using the domain of $\mathrm{f}$.

$2x = x(x^2 - 1)$
$x^3 - 3x = 0$
$x(x^2 - 3) = 0$
$x = 0 \quad \text{or} \quad x = \sqrt{3} \quad \text{or} \quad x = -\sqrt{3}$
Since the domain of $\mathrm{f}$ is $x > 1$, we reject $x = 0$ and $x = -\sqrt{3}$.
$x = \sqrt{3}$