The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by $$\begin{aligned} \mathrm{f} &: x \mapsto x - \frac{1}{x}, \quad x < 0, \\ \mathrm{g} &: x \mapsto \sin x, \quad 0 \le x \le 2\pi. \end{aligned}$$

1. Find the first derivative of $\mathrm{f}(x)$.

$\mathrm{f}(x) = x - x^{-1}$

$\mathrm{f}'(x) = 1 - (-1)x^{-2} = 1 + \frac{1}{x^2}$

2. Use the sign of the derivative to conclude the existence of the inverse function.

For $x < 0$, $x^2 > 0 \implies \frac{1}{x^2} > 0$.

Thus, $\mathrm{f}'(x) > 0$ for all $x$ in the domain.
Since its derivative is always positive, $\mathrm{f}$ is a strictly increasing function.
This implies that $\mathrm{f}$ is a one-to-one function, hence $\mathrm{f}^{-1}$ exists.

1. Set up an equation equating $y$ to $\mathrm{f}(x)$ and rearrange it into a quadratic equation in terms of $x$.

Let $y = x - \frac{1}{x}$.
Multiplying throughout by $x$ (since $x \ne 0$) and rearranging gives:
$xy = x^2 - 1$
$x^2 - yx - 1 = 0$

2. Apply the quadratic formula to solve for $x$ in terms of $y$.

Using the quadratic formula for $ax^2 + bx + c = 0$ with $a=1$, $b=-y$, and $c=-1$:
$x = \frac{-(-y) \pm \sqrt{(-y)^2 - 4(1)(-1)}}{2(1)}$
$x = \frac{y \pm \sqrt{y^2 + 4}}{2}$

3. Choose the correct sign based on the domain of $\mathrm{f}$.

The domain of $\mathrm{f}$ is $x < 0$.
Notice that $\sqrt{y^2 + 4} > \sqrt{y^2} = |y| \ge y$.
This means $y + \sqrt{y^2 + 4}$ is always positive, and $y - \sqrt{y^2 + 4}$ is always negative.
Therefore, we must reject the positive root to satisfy $x < 0$:
$x = \frac{y - \sqrt{y^2 + 4}}{2}$

4. Determine the domain of the inverse function and express it in mapping notation.

The domain of $\mathrm{f}^{-1}$ is the range of $\mathrm{f}$.
As $x \to -\infty$, $\mathrm{f}(x) \to -\infty$.
As $x \to 0^-$, $\mathrm{f}(x) \to \infty$.
Since $\mathrm{f}$ is continuous for $x < 0$, its range is the set of all real numbers ($\mathbb{R}$).
Expressing the inverse function in the requested similar form:
$\mathrm{f}^{-1} : x \mapsto \frac{x - \sqrt{x^2 + 4}}{2}, \quad x \in \mathbb{R}$

1. Find the axial intercepts and asymptotes of $y = \mathrm{f}(x)$ for its sketch.

For $y = x - \frac{1}{x}$ with $x < 0$:
When $y = 0$, we have $x - \frac{1}{x} = 0 \implies x^2 = 1$.
Since $x < 0$, $x = -1$.
The curve crosses the $x$-axis at $(-1, 0)$. It does not cross the $y$-axis since $x < 0$.
As $x \to 0^-$, $-\frac{1}{x} \to \infty \implies y \to \infty$, so $x = 0$ is a vertical asymptote.
As $x \to -\infty$, $-\frac{1}{x} \to 0$, so $y \to x$, meaning $y = x$ is an oblique asymptote.

2. Use the relationship between a function and its inverse to determine the features of $y = \mathrm{f}^{-1}(x)$.

The graph of $y = \mathrm{f}^{-1}(x)$ is the reflection of $y = \mathrm{f}(x)$ in the line $y = x$.
Reflecting the intercept $(-1, 0)$ gives a $y$-axis crossing at $(0, -1)$. It does not cross the $x$-axis.
Reflecting the vertical asymptote $x = 0$ gives a horizontal asymptote $y = 0$.
The oblique asymptote $y = x$ reflects onto itself.

3. Determine the equation and domain of $y = \mathrm{f}^{-1}\mathrm{f}(x)$ to sketch it and conclude the axes crossings.

By definition of inverse functions, $\mathrm{f}^{-1}\mathrm{f}(x) = x$ for all $x$ in the domain of $\mathrm{f}$.
Since the domain of $\mathrm{f}$ is $x < 0$, the graph of $y = \mathrm{f}^{-1}\mathrm{f}(x)$ is the straight line $y = x$ restricted to $x < 0$.
This forms a ray starting from, but not including, the origin $(0,0)$.
It does not cross either axis.

Summary of crossings:

• $y = \mathrm{f}(x)$ crosses at $(-1, 0)$.
• $y = \mathrm{f}^{-1}(x)$ crosses at $(0, -1)$.
• $y = \mathrm{f}^{-1}\mathrm{f}(x)$ does not cross any axis.

1. State the domain of $\mathrm{f}$ and find the range of $\mathrm{g}$.

From the definition of $\mathrm{f}$, the domain is $D_{\mathrm{f}} = (-\infty, 0)$.
For the function $\mathrm{g}(x) = \sin x$ over the domain $0 \le x \le 2\pi$, the range is $R_{\mathrm{g}} = [-1, 1]$.

2. Apply the condition for the existence of composite functions.

For the composite function $\mathrm{fg}$ to exist, the range of the inner function $\mathrm{g}$ must be a subset of the domain of the outer function $\mathrm{f}$, i.e., $R_{\mathrm{g}} \subseteq D_{\mathrm{f}}$.
Since $[-1, 1]$ is not a subset of $(-\infty, 0)$, we have $R_{\mathrm{g}} \not\subseteq D_{\mathrm{f}}$.
Therefore, the composite function $\mathrm{fg}$ does not exist.

1. Define the composite function $\mathrm{fh}$ and state its domain.

$\mathrm{fh}(x) = \mathrm{f}(\mathrm{h}(x)) = \mathrm{f}(\sin x) = \sin x - \frac{1}{\sin x}$
The domain of $\mathrm{fh}$ is the domain of the inner function $\mathrm{h}$.
Thus, $\mathrm{fh}: x \mapsto \sin x - \frac{1}{\sin x}, \quad \pi < x < 2\pi$.

2. Determine the range of the inner function $\mathrm{h}$.

For $\pi < x < 2\pi$, the graph of $y = \sin x$ reaches its minimum of $-1$ at $x = \frac{3\pi}{2}$ and approaches $0$ (but never reaches it) as $x \to \pi$ and $x \to 2\pi$.
Thus, the range of $\mathrm{h}$ is $[-1, 0)$.

3. Determine the range of $\mathrm{fh}$ by applying the outer function $\mathrm{f}$ to the range of $\mathrm{h}$.

Let $u = \mathrm{h}(x)$. We need the range of $\mathrm{f}(u) = u - \frac{1}{u}$ for $u \in [-1, 0)$.
Since $\mathrm{f}'(u) = 1 + \frac{1}{u^2} > 0$ for all $u \neq 0$, the function $\mathrm{f}(u)$ is strictly increasing on $[-1, 0)$.
When $u = -1$, $\mathrm{f}(-1) = -1 - (-1) = 0$.
As $u \to 0^-$, $\frac{1}{u} \to -\infty$, so $\mathrm{f}(u) \to 0 - (-\infty) = +\infty$.
Therefore, the range of $\mathrm{fh}$ is $[0, \infty)$.