1

I can't figure out how to find the inverse of $\frac{\sqrt{x}}{\sqrt{x+1}}$

Here's my process:

$y = \frac{\sqrt{x}}{\sqrt{x+1}}\\ y\sqrt{x+1} = \sqrt{x}\\y²(x+1)=x\\xy²+y²=x\\xy²-x=-y²\\x(y²-1)=-y²\\x=\frac{-y²}{y²-1}$

But this is incorrect for some reason, I just don't see anything wrong with it.

Could you help me get past this?

Thanks.

questionnaire
  • 11
  • 2
  • 1
    Looks fine to me. Why do you think your answer is incorrect? – Brevan Ellefsen Nov 21 '22 at 16:37
  • 1
    You should first check that the original function $f$ is injective. You should also state the domain and range of the original function $f$ and the inverse function $f^{-1}.$ That should clarify some things. – Adam Rubinson Nov 21 '22 at 16:40
  • You need the inverse of $\sqrt\frac x{x+1}=\sqrt{1-\frac1{x+1}}$ only – Тyma Gaidash Nov 21 '22 at 16:49
  • 1
    "But this is incorrect for some reason" what reason? Why do you think it is incorrect. One thing important to note is the domain of $\frac {\sqrt x}{\sqrt {x+1}}$ is $x\in [0,\infty)$ and the co-domain/range is $\frac {\sqrt x}{\sqrt{x+1}} \in [0, 1)$. It is injective so inverse function exists. Perhaps on first blush it looks like $\frac {-y^2}{y^2-1}$ is frequently negative and is undefined if $y=\pm 1$ but... that's okay. As the domain of the inverse is the codomain of the original $y\in[0,1)$ and so $\frac{-y^2}{y^2-1}=\frac{y^2}{1-y^2}>0$ and $\pm 1$ are not in the domain. – fleablood Nov 21 '22 at 17:14
  • If you are trying to enter it into some automated system that is rejecting it, one possible explanation is that it expects $\frac{y^2}{1-y^2}$ instead of the form you have. – Paul Sinclair Nov 22 '22 at 18:14

0 Answers0