Question 169150
Remember, the domain of the inverse function is simply the range of the original function. Since the range of the original function is {{{y>=0}}}, the domain of the inverse function is {{{x>=0}}}



Likewise, the range of the inverse function is the domain of the original function. Since the domain of the original function is {{{x>=3}}}, this means that the range of the inverse function is {{{y>=3}}}



If those two previous statements made no sense whatsoever, then just keep in mind that the domain and range switch when it comes to the inverse function.




Finding the inverse:



{{{f(x)=sqrt(x-3)}}} Start with the given function



{{{y=sqrt(x-3)}}} Replace f(x) with y



{{{x=sqrt(y-3)}}} Switch x and y. The goal now is to solve for y



{{{x^2=y-3}}} Square both sides



{{{x^2+3=y}}} Add 3 to both sides.



So the answer is {{{y=x^2+3}}} which means that the inverse function is *[Tex \LARGE f^{-1}(x)=x^{2}+3] with the restriction {{{x>=0}}}







"Show that f of f^-1(x)=x"



{{{f(x)=sqrt(x-3)}}} Start with the given function



{{{f(f^(-1)(x))=sqrt(x^2+3-3)}}} Plug in *[Tex \LARGE f^{-1}(x)=x^{2}+3]. In other words, replace each "x" in f(x) with {{{x^2+3}}}



{{{f(f^(-1)(x))=sqrt(x^2)}}} Combine like terms.



{{{f(f^(-1)(x))=x}}} Take the square root of {{{x^2}}} to get x



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"Show that f^-1 of f(x)=x"



*[Tex \LARGE f^{-1}(x)=x^{2}+3] Start with the inverse function



*[Tex \LARGE f^{-1}(f(x))=\left(\sqrt{x-3}\right)^{2}+3] Plug in {{{f(x)=sqrt(x-3)}}}



*[Tex \LARGE f^{-1}(f(x))=x-3+3] Square {{{sqrt(x-3)}}} to get {{{x-3}}}



*[Tex \LARGE f^{-1}(f(x))=x] Combine like terms.