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I see that the image contains a mathematical problem regarding two functions \( f \) and \( g \). Here's a breakdown of the details from the image:

The functions \( f \) and \( g \) are defined as:
\[
f(x) = \sqrt{3 - x}, \quad x \leq 3
\]
\[
g(x) = x^2 - 1, \quad x \in A
\]

The following tasks are required:

1. **Find \( f^{-1} \)** (the inverse of \( f \)) and determine its domain and range.
2. **Determine the interval of \( A \) such that \( R_g \subseteq D_f \)** (where \( R_g \) is the range of \( g \), and \( D_f \) is the domain of \( f \)).

### Solution:

#### 1. Finding \( f^{-1}(x) \) and its domain and range:

Given:
\[
f(x) = \sqrt{3 - x}, \quad x \leq 3
\]

To find \( f^{-1} \), we reverse the function's operations:

- Start with \( y = f(x) = \sqrt{3 - x} \)
- Square both sides to remove the square root:
  \[
  y^2 = 3 - x
  \]
- Solve for \( x \):
  \[
  x = 3 - y^2
  \]

Thus, the inverse function is:
\[
f^{-1}(x) = 3 - x^2
\]

Now, we determine the domain and range of the inverse function:

- **Domain**: Since the original function \( f(x) \) is defined for \( x \leq 3 \), the range of \( f(x) \) (and thus the domain of \( f^{-1} \)) is \( [0, \infty) \) because \( \sqrt{3 - x} \geq 0 \).
  
- **Range**: The inverse function \( f^{-1}(x) = 3 - x^2 \) is defined for \( x \in [0, \infty) \), and the range will be \( (-\infty, 3] \), as squaring \( x \) produces values \( 0 \leq x^2 \leq 3 \).

#### 2. Finding the interval of \( A \) such that \( R_g \subseteq D_f \):

- **Domain of \( f \), \( D_f \)**, is \( (-\infty, 3] \).
- The function \( g(x) = x^2 - 1 \) needs to map its range into \( D_f \). We first find the range of \( g(x) \):

  \[
  R_g = [g(\text{min value of } x), g(\text{max value of } x)]
  \]

  Since \( g(x) = x^2 - 1 \), the minimum value occurs at \( x = 0 \), yielding \( g(0) = -1 \). As \( x^2 \geq 0 \), the maximum value approaches infinity as \( |x| \to \infty \). So the range of \( g(x) \) is \( [-1, \infty) \).

For \( R_g \subseteq D_f \), we need \( [-1, \infty) \subseteq (-\infty, 3] \). Therefore, we restrict \( A \) such that the range of \( g(x) \) stays within \( [-1, 3] \). To find the values of \( x \) for which this holds:

\[
-1 \leq x^2 - 1 \leq 3
\]
Solving this inequality:
\[
0 \leq x^2 \leq 4
\]
Thus, \( -2 \leq x \leq 2 \). Therefore, \( A = [-2, 2] \).

### Summary:
1. The inverse of \( f(x) \) is \( f^{-1}(x) = 3 - x^2 \), with a domain of \( [0, \infty) \) and a range of \( (-\infty, 3] \).
2. The interval of \( A \) is \( [-2, 2] \) such that \( R_g \subseteq D_f \).

---

Let me know if you need further details or clarification! Here are some related questions:

1. How do you find the inverse of a function?
2. What are the steps to determine the range of a quadratic function?
3. How do inequalities affect the solution when finding domains or ranges?
4. What is the significance of restricting domains for inverse functions?
5. How would you find the domain of a composite function like \( f(g(x)) \)?

**Tip**: When working with inverse functions, always ensure that the domain and range are correctly identified, as they switch roles between the function and its inverse.

The functions f and g are defined as f(x) = sqrt(3 - x), x <= 3, and g(x) = x^2 - 1, x ∈ A. Find f⁻¹ and hence, determine its domain and range. Determine the interval of A such that R_g ⊆ D_f.

Learn how to find the inverse of f(x) = sqrt(3 - x), including determining its domain and range. This problem also covers how to find the interval for A such that R_g ⊆ D_f for g(x) = x^2 - 1. Suitable for Grades 11-12 students studying inverse functions and quadratic functions.

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