Respuesta :

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]