The function given is,
[tex]f(x)=(x-1)^2-4[/tex]
Given that
[tex]y=f(x)[/tex]
Therefore,
[tex]y=(x-1)^2-4[/tex]
Replace x with y
[tex]x=(y-1)^2-4[/tex]
Solve for y
[tex]\begin{gathered} x+4=(y-1_{})^2 \\ \end{gathered}[/tex]
Square-rooting both sides
[tex]\begin{gathered} \pm\sqrt[]{x+4}=\sqrt[]{(y-1)^2} \\ \pm\sqrt[]{x+4}=y-1 \end{gathered}[/tex]
Add 1 to both sides
[tex]\begin{gathered} \pm\sqrt[]{x+4}+1=y-1+1 \\ \pm\sqrt[]{x+4}+1=y \\ \Rightarrow y=\pm\sqrt[]{x+4}+1 \end{gathered}[/tex]
Hence,
[tex]f^{-1}(x)=\pm\sqrt[]{x+4}+1[/tex]
The domain of the inverse function will be,
[tex]x\ge-4[/tex]
The correct answer is Option B.
