函数 $\text{f}$ 的定义为 $\text{f} : x \mapsto \frac{x^2 + 4}{x}$， $x \in \mathbb{R}$。

1. Demonstrate that the function is not one-to-one by providing a counterexample.

Evaluate $\text{f}(x)$ for two different values of $x$, for instance $x=1$ and $x=4$:
$\text{f}(1) = \frac{1^2 + 4}{1} = 5$
$\text{f}(4) = \frac{4^2 + 4}{4} = 5$
Since $\text{f}(1) = \text{f}(4)$, there are distinct values of $x$ that map to the same value of $y$. Therefore, $\text{f}$ is not a one-to-one function (it is a many-to-one function), and hence does not have an inverse.

1. Set $y = \mathrm{f}(x)$ and rearrange into a quadratic equation in $x$.

$y = \frac{x^2 + 4}{x}$
$xy = x^2 + 4$
$x^2 - yx + 4 = 0$

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

$x = \frac{-(-y) \pm \sqrt{(-y)^2 - 4(1)(4)}}{2(1)}$
$x = \frac{y \pm \sqrt{y^2 - 16}}{2}$

3. Determine the correct sign for the square root based on the restricted domain.

Since the domain is restricted to $0 < x \le 2$, we can verify which sign to take by substituting a value within the domain. Let's test $x = 1$, which gives $y = \frac{1^2 + 4}{1} = 5$.
Using the positive root: $x = \frac{5 + \sqrt{25 - 16}}{2} = 4$ (Reject, since $4 \notin (0, 2]$).
Using the negative root: $x = \frac{5 - \sqrt{25 - 16}}{2} = 1$ (Accept).
Therefore, we must take the negative root:
$x = \frac{y - \sqrt{y^2 - 16}}{2}$

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

To find the range of $\mathrm{f}$ for $0 < x \le 2$:
$\mathrm{f}(x) = x + \frac{4}{x}$
$\mathrm{f}'(x) = 1 - \frac{4}{x^2} = \frac{x^2 - 4}{x^2}$
For $0 < x < 2$, $x^2 - 4 < 0$, so $\mathrm{f}'(x) < 0$. Thus, $\mathrm{f}$ is strictly decreasing.
At $x = 2$, $\mathrm{f}(2) = 4$.
As $x \to 0^+$, $\mathrm{f}(x) \to \infty$.
Thus, the range of $\mathrm{f}$ is $y \ge 4$.
(Alternatively, since $x$ is real, the discriminant $y^2 - 16 \ge 0 \implies y \le -4 \text{ or } y \ge 4$. Since $x>0$, $y>0$, so $y \ge 4$.)

5. State the final expression for the inverse function and its domain.

$\mathrm{f}^{-1}(x) = \frac{x - \sqrt{x^2 - 16}}{2}$
Domain: $x \ge 4$