a) Step 1: Find the inverse function $f^{-1}(x)$. Given $f(x) = \frac{x}{x-3}$. Let $y = f(x)$, so $y = \frac{x}{x-3}$. To find the inverse, swap $x$ and $y$: $$ x = \frac{y}{y-3} $$ Now, solve for $y$: $$ x(y-3) = y $$ $$ xy - 3x = y $$ $$ xy - y = 3x $$ $$ y(x-1) = 3x $$ $$ y = \frac{3x}{x-1} $$ So, $f^{-1}(x) = \frac{3x}{x-1}$. Step 2: State the domain of $f^{-1}(x)$. The domain of $f^{-1}(x)$ is restricted when the denominator is zero. $$ x-1 \neq 0 \implies x \neq 1 $$ The domain of $f^{-1}(x)$ is $\{x \in \mathbb{R} \mid x \neq 1\}$. The inverse function is $\boxed{f^{-1}(x) = \frac{3x}{x-1}}$ with domain $\boxed{\{x \in \mathbb{R} \mid x \neq 1\}}$. b) Step 1: Determine the composite function $fg(x)$. Given $f(x) = \frac{x}{x-3}$ and $g(x) = \sqrt{x+2}$. The composite function $fg(x)$ is $f(g(x))$: $$ fg(x) = f(\sqrt{x+2}) = \frac{\sqrt{x+2}}{\sqrt{x+2}-3} $$ Step 2: Find the domain of $fg(x)$. For $g(x) = \sqrt{x+2}$ to be defined, $x+2 \ge 0 \implies x \ge -2$. For $f(g(x))$ to be defined, the denominator $\sqrt{x+2}-3$ cannot be zero. $$ \sqrt{x+2}-3 \neq 0 $$ $$ \sqrt{x+2} \neq 3 $$ $$ x+2 \neq 3^2 $$ $$ x+2 \neq 9 $$ $$ x \neq 7 $$ Combining these conditions, the domain of $fg(x)$ is $\{x \in \mathbb{R} \mid x \ge -2 \text{ and } x \neq 7\}$. Step 3: Find the range of $fg(x)$. Let $u = g(x) = \sqrt{x+2}$. From the domain of $fg(x)$, we know $x \ge -2$ and $x \neq 7$. This implies $u = \sqrt{x+2} \ge \sqrt{-2+2} = 0$. Also, $u = \sqrt{x+2} \neq \sqrt{7+2} = \sqrt{9} = 3$. So, the range of $u$ is $[0, \infty) \setminus \{3\}$. Now we need to find the range of $f(u) = \frac{u}{u-3}$ for $u \in [0, \infty) \setminus \{3\}$. Let $y = \frac{u}{u-3}$. To find the range, we can express $u$ in terms of $y$: $$ y(u-3) = u $$ $$ yu - 3y = u $$ $$ yu - u = 3y $$ $$ u(y-1) = 3y $$