Let's go through the steps of solving the problem from the image.

1. Functions and their compositions:

• $f(x) = 9x^2 + 3$, with domain $M_f = (-\infty, -1]$
• $g(x) = -\frac{1}{3}\sqrt{x-3}$, with domain $M_g = [4, \infty)$

2. Compositions of functions $f \circ g$ and $g \circ f$:

• To find $f(g(x))$, substitute $g(x)$ into $f(x)$: $f(g(x)) = 9 \left( -\frac{1}{3}\sqrt{x-3} \right)^2 + 3 = 9 \cdot \frac{1}{9}(x - 3) + 3 = x - 3 + 3 = x$ Thus, $f(g(x)) = x$.

• Now for the domain of $f \circ g$ (denoted as $M_{f \circ g}$), we need $g(x)$ to produce values in the domain of $f(x)$, i.e., $M_f = (-\infty, -1]$. This happens when $g(x) \leq -1$.

  We know $g(x) = -\frac{1}{3}\sqrt{x-3}$, and for $g(x) \leq -1$, we solve: $-\frac{1}{3}\sqrt{x-3} \leq -1 \implies \sqrt{x-3} \geq 3 \implies x - 3 \geq 9 \implies x \geq 12$ Therefore, the domain of $f \circ g$ is $M_{f \circ g} = [12, \infty)$.

• To find $g(f(x))$, substitute $f(x)$ into $g(x)$: $g(f(x)) = -\frac{1}{3}\sqrt{9x^2 + 3 - 3} = -\frac{1}{3}\sqrt{9x^2} = -x$ Thus, $g(f(x)) = -x$.

• For the domain of $g \circ f$ (denoted as $M_{g \circ f}$), we need $f(x) \geq 4$ because the domain of $g(x)$ is $M_g = [4, \infty)$. So: $9x^2 + 3 \geq 4 \implies 9x^2 \geq 1 \implies x^2 \geq \frac{1}{9} \implies x \leq -\frac{1}{3}$ Since $M_f = (-\infty, -1]$, the domain of $g \circ f$ is $M_{g \circ f} = (-\infty, -\frac{1}{3}]$.

3. Inverse of $h(x)$:

Given $h(x) = \sqrt{x^2 + 6} - 2$, with domain $x \in [-2, 0]$, we solve for $h^{-1}(x)$.

To find the inverse, solve $y = \sqrt{x^2 + 6} - 2$: $y + 2 = \sqrt{x^2 + 6} \implies (y + 2)^2 = x^2 + 6 \implies x^2 = (y + 2)^2 - 6$ Taking the square root: $x = \pm\sqrt{(y + 2)^2 - 6}$ Since the domain of $h(x)$ is $[-2, 0]$, we take the negative root: $h^{-1}(x) = -\sqrt{(x + 2)^2 - 6}$

The domain $M_{h^{-1}}$ corresponds to the range of $h(x)$. For $x \in [-2, 0]$, the range of $h(x)$ is $[\sqrt{10} - 2, \sqrt{6} - 2]$, so: $M_{h^{-1}} = [\sqrt{6} - 2, \sqrt{10} - 2]$

Final Answers:

$f \circ g = x, \quad M_{f \circ g} = [12, \infty)$ $g \circ f = -x, \quad M_{g \circ f} = (-\infty, -\frac{1}{3}]$ $h^{-1}(x) = -\sqrt{(x + 2)^2 - 6}, \quad M_{h^{-1}} = [\sqrt{6} - 2, \sqrt{10} - 2]$

Would you like further details on any specific part? Here are some related questions:

1. How do we determine the domain of a composite function?
2. What is the process of finding the inverse of a function?
3. Why do we take only the negative root in $h^{-1}(x)$?
4. How does the domain of a square root function affect the solution?
5. What are the implications of restricting a domain to certain intervals?

Tip: When composing functions, always make sure the range of the inner function fits the domain of the outer function.