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.