determmine image of $f$ Let $f(x)=\dfrac{1}{x+1}\sqrt{x^2-1}$
$D_f={(-\infty,-1)}\cup {[1, +\infty)}$
How can i show algebraically that $$f\left( (1,+\infty) \right)=(0,1)$$
Indeed,
let $y\in \mathbb{R} $
$$
\begin{align}
 y\in f\left( ]1,+\infty[ \right) &\iff f(x)=y\\
 &\iff \dfrac{1}{x+1}\sqrt{x^2-1}=y\\
 &\iff  \sqrt{x^2-1}=yx+y\\
&\iff \begin{cases}
 |x^{2}-1|=y^{2}(x+1)^{2}&  \\
yx+y>0& \\
x> 1 &\\
\end{cases} \\
 &\iff \begin{cases}
 x-1=y^{2}(x+1)&  \\
y(x+1)>0& \\
x> 1 &
\end{cases}\\
 &\iff \begin{cases}
 x-y^{2}x=y^{2}+1&  \\
y(x+1)>0& \\
x> 1 &
\end{cases}\\
 &\iff \begin{cases}
 x=\dfrac{y^{2}+1}{1-y^{2}}&  \\
y(x+1)>0& \\
x> 1 &
\end{cases}\\
&\iff \dfrac{y^{2}+1}{1-y^{2}}>1
\end{align}$$
i'm trying to show that $y\in (0,1)$
I'm stuck here 

for $x>1$
$$f(x)=\sqrt{1-\frac{2}{x+1}}$$
we see that $f$ is continuous and strictly increasing at $(1,+\infty)$.
but
$$\lim_{x\to 1^+}f(x)=0$$
and
$$\lim_{x\to+\infty}f(x)=1$$
thus
$$f((1,+\infty))=(0,1)$$

algebraically

let $x>1$
$$f(x)=y\implies 1-\frac{2}{1+x}=y^2$$
$$\implies x=\frac{2}{1-y^2}-1$$


*

*$$x>1\implies 0 <\frac{2}{1+x}<1\implies 0<y<1$$

*$$0<y<1\iff 0<1-y^2<1\implies \frac{1}{1-y^2}>1\implies x>1$$
