Respuesta :
We have been given a quadratic function [tex]f(x)=(x-5)^{2} +2[/tex] and we need to restrict the domain such that it becomes a one to one function.
We know that vertex of this quadratic function occurs at (5,2).
Further, we know that range of this function is [tex][2,\infty)[/tex].
If we restrict the domain of this function to either [tex](-\infty,5][/tex] or [tex][5,\infty)[/tex], it will become one to one function.
Let us know find its inverse.
[tex]y=(x-5)^{2}+2[/tex]
Upon interchanging x and y, we get:
[tex]x=(y-5)^{2}+2[/tex]
Let us now solve this function for y.
[tex](y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\[/tex]
Hence, the inverse function would be [tex]f^{-1}(x)=5+\sqrt{x-2}[/tex] if we restrict the domain of original function to [tex][5,\infty)[/tex] and the inverse function would be [tex]f^{-1}(x)=5-\sqrt{x-2}[/tex] if we restrict the domain to [tex](-\infty,5][/tex].
