Given the function f(x) defined as:
[tex]f(x)=\sqrt[]{x-12}[/tex](a)
To find the inverse of f(x), we express the function as:
[tex]y=\sqrt[]{x-2}[/tex]Now, we take the square on both sides:
[tex]\begin{gathered} y^2=x-12 \\ \Rightarrow x=y^2+12 \end{gathered}[/tex]We change the notation:
[tex]\begin{gathered} x\rightarrow f^{-1}(x) \\ y\rightarrow x \end{gathered}[/tex]Then, the inverse function is:
[tex]f^{-1}(x)=x^2+12[/tex]For x ≥ 0
(c)
The domain of f(x) are those values such that:
[tex]\begin{gathered} x-12\ge0\Rightarrow x\ge12 \\ \text{Dom}_f=\lbrack12,\infty) \end{gathered}[/tex]And the range is the set of all positive numbers (including 0):
[tex]\text{Ran}_f=\lbrack0,\infty)[/tex]For the inverse, the domain of f(x) is its range, and the range of f(x) is its domain:
[tex]\begin{gathered} \text{Dom}_{f^{-1}}=\lbrack0,\infty) \\ \text{Ran}_{f^{-1}}=\lbrack12,\infty) \end{gathered}[/tex]