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To determine the domain of the inverse function \( f^{-1}(x) \) for the function \( f(x) = x^2 + 2 \) with \( f : \mathbb{R}^+ \rightarrow \mathbb{R} \), let’s go through the steps:

1. **Understand the Range of \( f(x) \):**
   - Given \( f(x) = x^2 + 2 \), and since \( x \in \mathbb{R}^+ \), we know \( x \geq 0 \).
   - For \( x = 0 \), \( f(0) = 0^2 + 2 = 2 \).
   - As \( x \) increases, \( f(x) \) increases because \( x^2 + 2 \) is a continuous and increasing function.
   - Therefore, the range of \( f(x) \) is \( [2, \infty) \).

2. **Domain of \( f^{-1}(x) \):**
   - The domain of \( f^{-1}(x) \) is the range of \( f(x) \).
   - Since the range of \( f(x) \) is \( [2, \infty) \), the domain of \( f^{-1}(x) \) must also be \( [2, \infty) \).

3. **Conclusion:**
   - The correct answer is **C.** \( [2, \infty[ \).

Would you like further details on this solution or have any questions?

Here are five related questions:
1. How do we find the range of a function in general?
2. What does the notation \( \mathbb{R}^+ \) represent in domain constraints?
3. Why is it essential for a function to be one-to-one to have an inverse?
4. What happens if the function \( f(x) \) is defined for all real numbers instead of \( \mathbb{R}^+ \)?
5. Can we find the explicit form of \( f^{-1}(x) \) for \( f(x) = x^2 + 2 \)?

**Tip:** The domain of an inverse function is the range of the original function. Always verify the range first to determine the inverse’s domain.

If f: R⁺ → R, f(x) = x² + 2, then the domain of f⁻¹(x) is:

Explore how to determine the domain of the inverse function f⁻¹(x) for the function f(x) = x² + 2, where f: R⁺ → R. Understand key concepts about functions, inverse functions, and domain and range, suitable for high school math.

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