Let $f:[-1,1]\rightarrow \mathbb R$ and $f(x)=\frac {x} {1+\sqrt{1-x^2}}$. Show that $f$ is continuous, $f([-1,1]) = [-1,1]$ and $f$ has an inverse function $f^{-1}: [-1,1]\rightarrow[-1,1]$.
I think it all comes down to showing that it's monotonously increasing. If I proove that, then I can show that $f([-1,1]) = [-1,1]$ and that it has an inverse function follows from it being strictly monotonously increasing. (please correct me if I said something false!)
Do I have to show something special to prove that it's continuous? Clearly $x$ is continuous and $1+\sqrt{1-x^2}$ also has no point that could present a problem in this interval. Can I just write this?
For the monotony proof I'm a little stuck. I have the following:
$\frac {x} {1+\sqrt{1-x^2}} < \frac {x+1} {1+\sqrt{1-(x+1)^2}}$
$x\sqrt{1-(x+1)^2} < x(1+\sqrt{1-x^2})$
$x\sqrt{-x^2+2x+2} < x\sqrt{1-x^2}+1+\sqrt{1-x^2}$
I can not reduce this anymore and from here it's not clear that it's monotone. Any tips? Many thanks guys!
PS: we are not allowed to use the derivative here.