Question

Please help. I’ll mark you as brainliest if correct!

Answer 1

Answer:

$\boxed{\sf \ \ \ f^{-1}(x)=1+\sqrt{x} \ \ \ }$

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 $[1;+\infty[$

$f^{-1}(f(x))=x =f^{-1}((x-1)^2)\\\\f(f^{-1}(x))=x=(f^{-1}(x)-1)^2$

x>=1 so x is positive and we can take the square root

so $f^{-1}(x)=1+\sqrt{x}$