函数 $f$ 的定义为 $f: x \mapsto 2x - \dfrac{1}{2x},\quad 0 < x < 2$。 已知 $f^{-1}$ 存在。

1. Set up a quadratic equation in terms of $x$.

Let $y = 2x - \frac{1}{2x}$.
Multiplying throughout by $2x$, we obtain:
$2xy = 4x^2 - 1$
Rearranging to form a quadratic equation in $x$:
$4x^2 - 2yx - 1 = 0$

2. Solve for $x$ using the quadratic formula and reject the invalid root.

Using the quadratic formula:
$x = \frac{-(-2y) \pm \sqrt{(-2y)^2 - 4(4)(-1)}}{2(4)}$
$x = \frac{2y \pm \sqrt{4y^2 + 16}}{8}$
$x = \frac{2y \pm 2\sqrt{y^2 + 4}}{8} = \frac{y \pm \sqrt{y^2 + 4}}{4}$
Since $x > 0$ for the given domain and $\sqrt{y^2 + 4} > \sqrt{y^2} = |y| \ge y$, we have $y - \sqrt{y^2 + 4} < 0$.
Thus, the negative root must be rejected:
$x = \frac{y + \sqrt{y^2 + 4}}{4}$

3. Determine the domain of the inverse function and state the final answer in similar form.

The domain of $f^{-1}$ corresponds to the range of $f$.
Since $f'(x) = 2 + \frac{1}{2x^2} > 0$, $f(x)$ is strictly increasing.
As $x \to 0^+$, $f(x) \to -\infty$.
As $x \to 2^-$, $f(x) \to 2(2) - \frac{1}{4(2)} = \frac{15}{4}$.
The range of $f$ is $(-\infty, \frac{15}{4})$.
Replacing $y$ with $x$ to express the inverse function in similar form:
$f^{-1}: x \mapsto \frac{x + \sqrt{x^2 + 4}}{4}, \quad x < \frac{15}{4}$

1. Identify the key features of the graph of $y=f(x)$.

For $f(x) = 2x - \frac{1}{2x}$ where $0 < x < 2$:

• As $x \to 0^+$, $f(x) \to -\infty$. Hence, $x=0$ (the $y$-axis) is a vertical asymptote for $y<0$.
• $x$-intercept: let $f(x) = 0 \implies 2x = \frac{1}{2x} \implies 4x^2 = 1 \implies x = 0.5$ (since $x>0$).
• Endpoint: at $x=2$, $f(2) = 2(2) - \frac{1}{4} = 3.75$. The graph ends at $(2, 3.75)$ with an empty circle (since $x<2$).

2. Identify the key features of $y=f^{-1}(x)$ using the property of inverse graphs.

The graph of $y=f^{-1}(x)$ is a reflection of $y=f(x)$ across the line $y=x$. Therefore:

• The vertical asymptote $x=0$ becomes a horizontal asymptote $y=0$ for $x<0$.
• The $x$-intercept $(0.5, 0)$ becomes the $y$-intercept $(0, 0.5)$.
• The endpoint $(2, 3.75)$ becomes $(3.75, 2)$, marked with an empty circle.

3. Find the point of intersection between the two curves and the line $y=x$.

Set $f(x) = x$ to find intersection with the reflection line:
$2x - \frac{1}{2x} = x \implies x = \frac{1}{2x} \implies 2x^2 = 1 \implies x = \frac{1}{\sqrt{2}}$.
The point of intersection is $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$.

4. Sketch both graphs on the same coordinate axes.

Draw the $x$ and $y$ axes, and the dashed line $y=x$.
Sketch $y=f(x)$ passing through $(0.5,0)$ and intersecting $y=x$ at $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$, stopping at the open circle $(2, 3.75)$.
Sketch $y=f^{-1}(x)$ passing through $(0,0.5)$ and intersecting $y=x$ at the same point, stopping at the open circle $(3.75, 2)$.
Label both curves, the asymptotes, the intercepts, and endpoints.

1. Relate the area of $R$ to the graph of $f(x)$ using symmetry.

By symmetry about the line $y = x$, the region $R$ bounded by $y = f^{-1}(x)$, $y = x$ and the $y$-axis ($x = 0$) has the same area as the reflected region bounded by $y = f(x)$, $y = x$ and the $x$-axis ($y = 0$).

2. Find the limits of integration by finding the intersection of $y = f(x)$ with $y = x$ and the $x$-axis.

Intersection of $y = f(x)$ and $y = x$:
$2x - \frac{1}{2x} = x$
$x = \frac{1}{2x} \implies x^2 = \frac{1}{2}$
Since $0 < x < 2$, $x = \frac{1}{\sqrt{2}}$.

Intersection of $y = f(x)$ and the $x$-axis ($y = 0$):
$2x - \frac{1}{2x} = 0 \implies x^2 = \frac{1}{4}$
Since $0 < x < 2$, $x = \frac{1}{2}$.

3. Set up the definite integrals for the required area.

The reflected area is the area under $y=x$ from $x=0$ to $x=\frac{1}{\sqrt{2}}$, minus the area under $y=f(x)$ from its $x$-intercept at $x=\frac{1}{2}$ to $x=\frac{1}{\sqrt{2}}$.
$\text{Area} = \int_{0}^{\frac{1}{\sqrt{2}}} x \, dx - \int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \left( 2x - \frac{1}{2x} \right) \, dx$

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

$\int_{0}^{\frac{1}{\sqrt{2}}} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{\frac{1}{\sqrt{2}}} = \frac{1}{4}$

$\int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \left( 2x - \frac{1}{2x} \right) \, dx = \left[ x^2 - \frac{1}{2}\ln x \right]_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}}$
$= \left( \frac{1}{2} - \frac{1}{2}\ln 2^{-1/2} \right) - \left( \frac{1}{4} - \frac{1}{2}\ln 2^{-1} \right)$
$= \left( \frac{1}{2} + \frac{1}{4}\ln 2 \right) - \left( \frac{1}{4} + \frac{1}{2}\ln 2 \right)$
$= \frac{1}{4} - \frac{1}{4}\ln 2$

$\text{Exact Area} = \frac{1}{4} - \left( \frac{1}{4} - \frac{1}{4}\ln 2 \right) = \frac{1}{4}\ln 2$

1. Find the range of the inner function $f$.

Since $f'(x) = 2 + \frac{1}{2x^2} > 0$ for $0 < x < 2$, $f$ is a strictly increasing function.
As $x \to 0^+$, $f(x) \to -\infty$.
As $x \to 2^-$, $f(x) \to 2(2) - \frac{1}{2(2)} = \frac{15}{4}$.
Thus, the range of $f$ is $R_f = \left(-\infty, \frac{15}{4}\right)$.

2. Show that the composite function $g \circ f$ exists.

The domain of $g$ is $D_g = (-\infty, 3] \cup (3, 5] = (-\infty, 5]$.
For the composite function $g \circ f$ to exist, we require the range of $f$ to be a subset of the domain of $g$ ($R_f \subseteq D_g$).
Since $R_f = \left(-\infty, \frac{15}{4}\right)$ and $\frac{15}{4} = 3.75 \le 5$, we have $R_f \subseteq D_g$.
Therefore, the composite function $g \circ f$ exists.

3. Find the range of $g \circ f$ for the first piece of $g$'s domain.

The range of $g \circ f$ is given by the image of $R_f$ under $g$, i.e., $g\left(\left(-\infty, \frac{15}{4}\right)\right)$. Because $g$ is a piecewise function split at $x = 3$, we split $R_f$ into $(-\infty, 3]$ and $\left(3, \frac{15}{4}\right)$.
For $x \in (-\infty, 3]$, $g(x) = \frac{3}{2} + \frac{3}{3x-11}$.
Since $g'(x) = -\frac{9}{(3x-11)^2} < 0$, $g$ is strictly decreasing on this interval.
At the upper limit $x = 3$, $g(3) = \frac{3}{2} + \frac{3}{9-11} = \frac{3}{2} - \frac{3}{2} = 0$.
As $x \to -\infty$, $\frac{3}{3x-11} \to 0$, so $g(x) \to \frac{3}{2}$.
Hence, the image of $(-\infty, 3]$ under $g$ is $\left[0, \frac{3}{2}\right)$.

4. Find the range of $g \circ f$ for the second piece of $g$'s domain and state the overall range.

For $x \in \left(3, \frac{15}{4}\right)$, $g(x) = \frac{x - 1 - x^2}{3} = \frac{-\left(x - \frac{1}{2}\right)^2 - \frac{3}{4}}{3}$.
This represents a downward-facing parabola with its vertex at $x = \frac{1}{2}$. Thus, $g(x)$ is strictly decreasing for $x > 3$.
As $x \to 3^+$, $g(x) \to \frac{3 - 1 - (3)^2}{3} = -\frac{7}{3}$.
As $x \to \left(\frac{15}{4}\right)^-$, $g(x) \to \frac{\frac{15}{4} - 1 - \left(\frac{15}{4}\right)^2}{3} = \frac{\frac{60-16-225}{16}}{3} = -\frac{181}{48}$.
Hence, the image of $\left(3, \frac{15}{4}\right)$ under $g$ is $\left(-\frac{181}{48}, -\frac{7}{3}\right)$.
Combining both images, the overall range of $g \circ f$ is $\left(-\frac{181}{48}, -\frac{7}{3}\right) \cup \left[0, \frac{3}{2}\right)$.