函数 $\text{f}$ 的定义为
$\text{f} : x \to 2x - \frac{1}{2x}, \quad 0 < x < 2.$
已知 $\text{f}^{-1}$ 存在。

1. Set $y = \text{f}(x)$ and rearrange the expression into a quadratic equation in terms of $x$.

Let $y = 2x - \frac{1}{2x}$.
Multiplying throughout by $2x$:
$2xy = 4x^2 - 1$
$4x^2 - 2yx - 1 = 0$

2. Solve for $x$ using the quadratic formula and deduce the correct sign for the root.

Using the quadratic formula:
$x = \frac{-(-2y) \pm \sqrt{(-2y)^2 - 4(4)(-1)}}{2(4)}$
$x = \frac{2y \pm \sqrt{4y^2 + 16}}{8}$
$x = \frac{2y \pm 2\sqrt{y^2 + 4}}{8}$
$x = \frac{y \pm \sqrt{y^2 + 4}}{4}$

Since the domain of $\text{f}$ is $0 < x < 2$, we require $x > 0$.
Notice that $\sqrt{y^2 + 4} > \sqrt{y^2} = |y| \ge y$, which means $y - \sqrt{y^2 + 4} < 0$.
Thus, we must reject the negative root to ensure $x$ is positive:
$x = \frac{y + \sqrt{y^2 + 4}}{4}$

3. Find the domain of $\text{f}^{-1}$ by determining the range of $\text{f}$, and write the function in the required mapping form.

The domain of $\text{f}^{-1}$ is exactly the range of $\text{f}$.
$\text{f}(x) = 2x - \frac{1}{2x}$
Since $\text{f}'(x) = 2 + \frac{1}{2x^2} > 0$ for $x > 0$, the function $\text{f}$ is strictly increasing.
As $x \to 0^+$, $\text{f}(x) \to -\infty$.
As $x \to 2^-$, $\text{f}(x) \to 2(2) - \frac{1}{2(2)} = 4 - \frac{1}{4} = \frac{15}{4}$.

The range of $\text{f}$ is therefore $\left(-\infty, \frac{15}{4}\right)$.
Consequently, the domain of $\text{f}^{-1}$ is $x < \frac{15}{4}$.

Defining $\text{f}^{-1}$ in a similar mapping form:
$\text{f}^{-1} : x \to \frac{x + \sqrt{x^2 + 4}}{4}, \quad x < \frac{15}{4}$

1. Identify the key features of the graph of $y = \text{f}(x)$.

For $y = 2x - \frac{1}{2x}$ where $0 < x < 2$:

• As $x \to 0^+$, $y \to -\infty$, so the $y$-axis ($x = 0$) is a vertical asymptote.
• To find the $x$-intercept, set $y = 0$:
  $2x - \frac{1}{2x} = 0 \implies x^2 = \frac{1}{4} \implies x = \frac{1}{2} \quad \text{(since } x > 0\text{)}$
  So the $x$-intercept is $(\frac{1}{2}, 0)$.
• As $x \to 2^-$, $y \to 2(2) - \frac{1}{4} = \frac{15}{4}$. The graph ends at an open circle $(2, \frac{15}{4})$.

2. Identify the corresponding key features of $y = \text{f}^{-1}(x)$ by reflection in $y = x$.

Reflecting the features of $\text{f}$ across the line $y = x$ gives the properties of $\text{f}^{-1}$:

• The horizontal asymptote is the $x$-axis ($y = 0$).
• The $y$-intercept is $(0, \frac{1}{2})$.
• The graph ends at an open circle $(\frac{15}{4}, 2)$.

3. Find the point of intersection of the two graphs to ensure an accurate sketch.

The graphs of a function and its inverse intersect on the line $y = x$. Set $\text{f}(x) = x$:
$2x - \frac{1}{2x} = x \implies x = \frac{1}{2x} \implies x^2 = \frac{1}{2}$
Since $x > 0$, $x = \frac{1}{\sqrt{2}}$.
The intersection point is $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$.

4. Describe the final sketch of both graphs.

On the same axes, draw the line $y = x$ as a reference.

• Sketch $y = \text{f}(x)$ starting downwards asymptotically to $x=0$, passing through $(\frac{1}{2}, 0)$ and $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$, ending at an open circle $(2, \frac{15}{4})$.
• Sketch $y = \text{f}^{-1}(x)$ starting leftwards asymptotically to $y=0$, passing through $(0, \frac{1}{2})$ and $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$, ending at an open circle $(\frac{15}{4}, 2)$.

1. Find the limits of integration by determining the intersections of $y = \text{f}(x)$ with $y = x$ and the $x$-axis.

$\text{f}(x) = x \implies 2x - \frac{1}{2x} = x$
$x = \frac{1}{2x} \implies 2x^2 = 1 \implies x = \frac{1}{\sqrt{2}} \text{ (since } x > 0)$
$\text{f}(x) = 0 \implies 2x - \frac{1}{2x} = 0$
$2x = \frac{1}{2x} \implies 4x^2 = 1 \implies x = \frac{1}{2} \text{ (since } x > 0)$

2. Relate the area of region $R$ to the graph of $y = \text{f}(x)$ using symmetry.

By symmetry about the line $y = x$, the region $R$ bounded by $y = \text{f}^{-1}(x)$, $y = x$ and the $y$-axis ($x = 0$) has the exact same area as the reflected region bounded by $y = \text{f}(x)$, $y = x$ and the $x$-axis ($y = 0$).
Thus, the area is the difference between the area of the triangle under $y = x$ (from $x=0$ to $x=\frac{1}{\sqrt{2}}$) and the area under $y = \text{f}(x)$ (from $x=\frac{1}{2}$ to $x=\frac{1}{\sqrt{2}}$):
$\text{Area} = \int_{0}^{\frac{1}{\sqrt{2}}} x \, \mathrm{d}x - \int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \left( 2x - \frac{1}{2x} \right) \, \mathrm{d}x$

3. Evaluate the definite integrals to find the exact area.

$\int_{0}^{\frac{1}{\sqrt{2}}} x \, \mathrm{d}x = \left[ \frac{1}{2}x^2 \right]_{0}^{\frac{1}{\sqrt{2}}} = \frac{1}{2}\left(\frac{1}{2}\right) = \frac{1}{4}$
$\int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \left( 2x - \frac{1}{2x} \right) \, \mathrm{d}x = \left[ x^2 - \frac{1}{2}\ln x \right]_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}}$
$= \left( \frac{1}{2} - \frac{1}{2}\ln(2^{-1/2}) \right) - \left( \frac{1}{4} - \frac{1}{2}\ln(2^{-1}) \right)$
$= \left( \frac{1}{2} + \frac{1}{4}\ln 2 \right) - \left( \frac{1}{4} + \frac{1}{2}\ln 2 \right) = \frac{1}{4} - \frac{1}{4}\ln 2$
$\text{Area of } R = \frac{1}{4} - \left( \frac{1}{4} - \frac{1}{4}\ln 2 \right) = \frac{1}{4}\ln 2$

1. Find the range of the inner function $\text{f}$.

For $\text{f}(x) = 2x - \frac{1}{2x}$ with domain $(0, 2)$, the function is strictly increasing since $\text{f}'(x) = 2 + \frac{1}{2x^2} > 0$ for all $x \in (0, 2)$.
As $x \to 0^+$, $\text{f}(x) \to -\infty$.
As $x \to 2^-$, $\text{f}(x) \to 2(2) - \frac{1}{2(2)} = \frac{15}{4}$.
Therefore, the range of $\text{f}$ is $R_\text{f} = \left(-\infty, \frac{15}{4}\right)$.

2. Show that the composite function $\text{gf}$ exists.

From the definition of $\text{g}$, its domain is $D_\text{g} = (-\infty, 3] \cup (3, 5] = (-\infty, 5]$.
For the composite function $\text{gf}$ to exist, we must satisfy the condition $R_\text{f} \subseteq D_\text{g}$.
Since $R_\text{f} = \left(-\infty, \frac{15}{4}\right)$ and $\frac{15}{4} = 3.75 \le 5$, we have $\left(-\infty, \frac{15}{4}\right) \subseteq (-\infty, 5]$.
Thus, $R_\text{f} \subseteq D_\text{g}$, showing that $\text{gf}$ exists.

3. Evaluate the range of $\text{g}(x)$ for the subset of $R_\text{f}$ where $x \le 3$.

To find the range of $\text{gf}$, we map $R_\text{f}$ through $\text{g}$. We split $R_\text{f}$ at the boundary $x=3$.
For $x \in (-\infty, 3]$, $\text{g}(x) = \frac{3}{2} + \frac{3}{3x-11}$.
This rational function is strictly decreasing for $x < \frac{11}{3}$.
As $x \to -\infty$, $\frac{3}{3x-11} \to 0^-$, so $\text{g}(x) \to \left(\frac{3}{2}\right)^-$.
At $x = 3$, $\text{g}(3) = \frac{3}{2} + \frac{3}{3(3)-11} = \frac{3}{2} - \frac{3}{2} = 0$.
Thus, the range of $\text{g}$ for this interval is $\left[0, \frac{3}{2}\right)$.

4. Evaluate the range of $\text{g}(x)$ for the remaining subset of $R_\text{f}$ where $3 < x < \frac{15}{4}$.

For $x \in \left(3, \frac{15}{4}\right)$, $\text{g}(x) = \left| x - \frac{1}{3}x^2 \right|$.
The quadratic curve $y = x - \frac{1}{3}x^2 = -\frac{1}{3}x(x-3)$ is an inverted parabola with roots at $x=0$ and $x=3$, and a maximum turning point at $x=\frac{3}{2}$.
For $x \in \left(3, \frac{15}{4}\right)$, $y < 0$ and is strictly decreasing, so its absolute value $\text{g}(x) = \frac{1}{3}x^2 - x$ is strictly increasing.
As $x \to 3^+$, $\text{g}(x) \to 0$.
As $x \to \left(\frac{15}{4}\right)^-$, $\text{g}(x) \to \frac{1}{3}\left(\frac{15}{4}\right)^2 - \frac{15}{4} = \frac{75}{16} - \frac{60}{16} = \frac{15}{16}$.
Thus, the range of $\text{g}$ for this interval is $\left(0, \frac{15}{16}\right)$.

5. Determine the overall range of the composite function $\text{gf}$.

The overall range of $\text{gf}$ is the union of the ranges obtained from the two sub-intervals:
$R_{\text{gf}} = \left[0, \frac{3}{2}\right) \cup \left(0, \frac{15}{16}\right) = \left[0, \frac{3}{2}\right)$