Given $f(x) = 1 + \\sqrt{x}$, what are the domain and range of $f^{-1}(x)$? A - HSC - SSCE Mathematics Extension 1 - Question 2 - 2020 - Paper 1

The expression $(x - 1)^2$ is defined for all real numbers. However, we also have to consider the original function $f(x)$. Since $f(x) = 1 + \sqrt{x}$ results in $y \geq 1$, we find that for $f^{-1}(x)$, the input $x$ must be such that:

$x - 1 \geq 0 \\ \Rightarrow x \geq 1.$

Hence, the domain of $f^{-1}(x)$ is $x \geq 1$.

Since the expression $y = (x - 1)^2$ can take any non-negative value as $x$ varies from 1 to infinity:

$y \geq 0.$

Thus, the range of $f^{-1}(x)$ is $y \geq 0$.