To find the range of the inverse function of \( f(x) = \sqrt{x} \), we first need to determine the domain of the original function \( f(x) \). The function \( f(x) = \sqrt{x} \) is defined for \( x \geq 0 \). The inverse function \( f^{-1}(x) \) will have a domain that is the range of \( f(x) \), which is \( [0, \infty) \). Therefore, the range of \( f^{-1}(x) \) will be the domain of \( f(x) \), which is also \( [0, \infty) \).
The function \( f(x) = \sqrt{x} \) is defined for \( x \geq 0 \). This is because the square root function is only defined for non-negative values of \( x \).
The range of \( f(x) = \sqrt{x} \) is \( [0, \infty) \). As \( x \) increases from 0 to infinity, \( \sqrt{x} \) also increases from 0 to infinity.
The domain of the inverse function \( f^{-1}(x) \) is the range of the original function \( f(x) \). Therefore, the domain of \( f^{-1}(x) \) is \( [0, \infty) \).
The range of the inverse function \( f^{-1}(x) \) is the domain of the original function \( f(x) \). Therefore, the range of \( f^{-1}(x) \) is \( [0, \infty) \).
Answered
