https://file.aiquickdraw.com/mathbot/file/1727186016550ncdqb12v.jpg

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

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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.

Olkoon f(x) = 9x^2 + 3 funktio, jonka määritysjoukko Mf =] -∞, -1]. Olkoon edelleen g(x) = -1/3√(x−3) funktio, jonka määritysjoukko Mg = [4, ∞[. Määrää yhdistetyt funktiot f∘g ja g∘f määritysjoukkoineen Mf∘g, Mg∘f. Sievennä vastauksesi. Olkoon h(x) = √(x^2+6)−2, x∈[−2, 0]. Määrää funktion h käänteisfunktio h⁻¹(x) määritysjoukkoineen Mh⁻¹. Anna määritysjoukot muodossa kaarisuluissa {a,b}. Ääretön kirjoitetaan inf.

Learn how to compute composite functions f∘g and g∘f, as well as the inverse function h⁻¹(x). This problem involves determining the domains of composite functions and finding the inverse of a square root function. Suitable for advanced high school students in Grades 11-12.

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