Answer:
Explanation:
Let's analyze the given function \( f(x) = \sqrt{\sqrt{x + 2} - 1} \) and determine its domain and range, as well as the domain and range of its inverse function \( f^{-1}(x) \).
1. Domain of \( f(x) \):
- The expression inside the square root, \( \sqrt{x + 2} - 1 \), must be greater than or equal to 0 to ensure real values under the square root.
- So, \( x + 2 - 1 \geq 0 \).
- Solve for \( x \):
\( x + 1 \geq 0 \)
\( x \geq -1 \)
Therefore, the domain of \( f(x) \) is \( x \geq -1 \).
2. Range of \( f(x) \):
- To find the range, we need to consider the possible output values of \( f(x) \).
- Since the square root of a non-negative number is always non-negative, \( \sqrt{\text{anything}} \) is always non-negative.
- Subtracting 1 from a non-negative number keeps it non-negative.
So, the range of \( f(x) \) is \( y \geq 0 \).
3. Domain of \( f^{-1}(x) \):
- To find the domain of the inverse function, we need to consider the possible input values.
- The range of \( f(x) \) is \( y \geq 0 \), which means the input values for \( f^{-1}(x) \) should be \( x \geq 0 \).
Therefore, the domain of \( f^{-1}(x) \) is \( x \geq 0 \).
4. Range of \( f^{-1}(x) \):
- The range of the inverse function \( f^{-1}(x) \) corresponds to the domain of the original function \( f(x) \).
- So, the range of \( f^{-1}(x) \) is \( y \geq -1 \).
Now, let's fill in the statements:
- The domain of \( f(x) \) is \( x \geq -1 \).
- The range of \( f(x) \) is \( y \geq 0 \).
- The range of \( f^{-1}(x) \) is \( y \geq -1 \).
- The domain of \( f^{-1}(x) \) is \( x \geq 0 \).