Answer:
[tex]\huge\boxed{D) \ f^{-1}(x) = (x-3)^2+2}[/tex]
Step-by-step explanation:
[tex]f(x) = \sqrt{x-2} + 3[/tex]
Replace f(x) by y
[tex]y = \sqrt{x-2} + 3[/tex]
Interchange x and y
[tex]x = \sqrt{y-2} + 3[/tex]
Solve for y
[tex]x = \sqrt{y-2} + 3[/tex]
Subtracting 3 to both sides
[tex]x - 3 = \sqrt{y-2}[/tex]
Taking square on both sides
[tex](x-3)^2 = y-2[/tex]
Adding 2 to both sides
[tex]y = (x-3)^2 + 2[/tex]
Replace [tex]y = f^{-1}(x)[/tex]
[tex]f^{-1}(x) = (x-3)^2+2[/tex]
