Answer: Choice D
[tex]f^{-1}(x) = (x-3)^2+2\\\\[/tex]
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Explanation:
First we replace f(x) with y. This is because both y and f(x) are outputs of a function.
To find the inverse, we swap x and y and solve for y like so
[tex]y = \sqrt{x-2}+3\\\\x = \sqrt{y-2}+3 \ \text{ .... swap x and y; isolate y}\\\\x-3 = \sqrt{y-2}\\\\(x-3)^2 = y-2 \ \text{ ... square both sides}\\\\(x-3)^2+2 = y\\\\y = (x-3)^2+2\\\\f^{-1}(x) = (x-3)^2+2\\\\[/tex]
Note: because the range of the original function is [tex]y \ge 3[/tex], this means the domain of the inverse is [tex]x \ge 3[/tex]. The domain and range swap roles because of the swap of x and y.
As the graph shows below, the original and its inverse are symmetrical about the mirror line y = x. One curve is the mirror image of the other over this dashed line.
