Answer:
[tex]\boxed{\sf \ \ \ f^{-1}(x)=1+\sqrt{x} \ \ \ }[/tex]
Step-by-step explanation:
the minimum of f is for x = 1 and f(1)=0
f is decreasing for x<=1 and increasing for x>=1
so the domain is [tex][1;+\infty[[/tex]
[tex]f^{-1}(f(x))=x =f^{-1}((x-1)^2)\\\\f(f^{-1}(x))=x=(f^{-1}(x)-1)^2[/tex]
x>=1 so x is positive and we can take the square root
so [tex]f^{-1}(x)=1+\sqrt{x}[/tex]
