Finding the inverse function

Question

The question is to find the inverse function of

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

I first found that the domain of definition is $\,x\ge 0$

Then studied the variation of the function and it is decreasing between $0$ and $1$ and increasing otherwise. Thus there are $2$ inverse functions to be found.
How can I find them. Any hints?

Answer 1

So $y = x-2\sqrt{x}+1 = (\sqrt{x}-1)^2$

$x = (1\pm\sqrt{y})^2=1 \pm 2\sqrt{y} + y$

So $f^{-1}(x)=x\pm2\sqrt{x}+1$

Specifically $f^{-1}(x) = x+2\sqrt{x}+1$ when $x\ge1$ and $f^{-1}(x) = x-2\sqrt{x}+1$ when $0\le x\le1$.