The question is to find the inverse function of


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?

  • 2
    $\begingroup$ how do you usualy find inverse??? $y=x-2\sqrt{x}+1$ and then solve for $x$.. Have you done that? $\endgroup$ – user87543 Nov 3 '13 at 9:29
  • $\begingroup$ I did but the problem is with the sqrt(x) $\endgroup$ – noname Nov 3 '13 at 9:40
  • $\begingroup$ How to get rid of it $\endgroup$ – noname Nov 3 '13 at 9:40
  • $\begingroup$ substitute $t=\sqrt{x}$ $\endgroup$ – Ömer Nov 3 '13 at 9:42
  • $\begingroup$ if you want to get rid of "root" in "square root" you have to "Square" it..... $\endgroup$ – user87543 Nov 3 '13 at 9:43

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 $0\le x\le1$.


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