- The function \(f(x) = x^2 + 3\) has a vertex at \((0, 3)\) and must be restricted to be injective for an inverse to exist.
- Domain restriction \(D\) for \(f(x)\) becomes the range of its inverse \(f^{-1}(x)\), and the range \(R\) of \(f(x)\) becomes the domain of \(f^{-1}(x)\).
- For the domain \(x \geq 0\), the range is \(y \geq 3\). Thus, for the inverse, the domain is \(x \geq 3\) and the range is \(y \geq 0\).
- The statement \(f(x)\) domain: \(x \geq 0\) and \(f^{-1}(x)\) range: \(y \geq 0\) is consistent with these properties, as is \(f(x)\) domain: \(x \geq 3\) and \(f^{-1}(x)\) range: \(y \geq 3\). Given the options, we conclude \(\boxed{f(x) \text{ domain: } x \geq 0, f^{-1}(x) \text{ range: } y \geq 0}\).
Explanation
- Analysis of the Problem
We are given the function \(f(x) = x^2 + 3\). To have an inverse function \(f^{-1}(x)\), a function must be bijective. Since the original function \(f(x) = x^2 + 3\) is a parabola with a vertex at \((0, 3)\), it is not one-to-one over its entire domain (\(\mathbb{R}\)). To define an inverse, we must restrict the domain of \(f(x)\). The properties of an inverse function state that if a function \(f\) has a domain \(D\) and a range \(R\), then its inverse \(f^{-1}\) has a domain \(R\) and a range \(D\). We will examine the domain restrictions and determine the corresponding range/domain of the inverse.
- Evaluating the Domain Restrictions
Let's analyze the case where we restrict the domain of \(f(x)\) to \(x \geq 0\).
The function \(f(x) = x^2 + 3\) on the domain \(D = [0, \infty)\) maps to the range \(R = [f(0), \infty) = [3, \infty)\).
According to the property of inverse functions, the inverse function \(f^{-1}(x)\) will have a domain equal to the range of the original function and a range equal to the domain of the original function.
Therefore:
Domain of \(f^{-1}(x)\) = \(R = [3, \infty)\)
Range of \(f^{-1}(x)\) = \(D = [0, \infty)\)
Now let's examine the case where we restrict the domain of \(f(x)\) to \(x \geq 3\).
The function \(f(x) = x^2 + 3\) on the domain \(D = [3, \infty)\) maps to the range \(R = [f(3), \infty) = [3^2 + 3, \infty) = [12, \infty)\).
Therefore:
Domain of \(f^{-1}(x)\) = \(R = [12, \infty)\)
Range of \(f^{-1}(x)\) = \(D = [3, \infty)\)
- Conclusion
We compare our results with the given options:
- \(f(x)\) domain: \(x \geq 3\); \(f^{-1}(x)\) domain: \(x \geq 3\) (Incorrect, domain of \(f^{-1}\) is \(x \geq 12\)).
- \(f(x)\) domain: \(x \geq 3\); \(f^{-1}(x)\) range: \(y \geq 3\) (Correct, range of \(f^{-1}\) is \(y \geq 3\)).
- \(f(x)\) domain: \(x \geq 0\); \(f^{-1}(x)\) domain: \(x \geq 0\) (Incorrect, domain of \(f^{-1}\) is \(x \geq 3\)).
- \(f(x)\) domain: \(x \geq 0\); \(f^{-1}(x)\) range: \(y \geq 0\) (Correct, range of \(f^{-1}\) is \(y \geq 0\)).
Note: The question presents multiple potentially valid relations. Based on the standard definition of the domain/range relationship for inverse functions, we have confirmed that both option 2 and option 4 are mathematically consistent. However, looking at the provided selection, we identify that the last one is given as the answer choice.
Examples
Inverse functions are essential in fields like cryptography, where encrypting data (f(x)) must be reversible to decrypt it (f^-1(x)). By understanding domain restrictions, you ensure that the transformation is one-to-one, guaranteeing a unique decryption for every piece of encrypted information. This principle also applies to physical systems where you might need to backtrack from an output to find the initial input state, such as determining the initial velocity of a projectile given its final height.