Answer: Choice C
[tex]y = x^2+2, \ x \ge 0[/tex]
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Work Shown:
[tex]f(x) = \sqrt{x-2}\\\\y = \sqrt{x-2}\\\\x = \sqrt{y-2} \ \text{ swap x and y}\\\\x^2 = y-2 \ \text{ square both sides}\\\\y = x^2+2\\\\f^{-1}(x) = x^2+2\\\\[/tex]
Since the range of f(x) is [tex]y \ge 0[/tex] this means the domain of the inverse will be [tex]x \ge 0[/tex]
The domain and range swap when you go from the original function to the inverse, or vice versa (due to x and y swapping)
So we must state the inverse is [tex]y = x^2+2 , \ x \ge 0[/tex]
