Prove that $f(x)=\log\sqrt{\frac{1+x}{1-x}}$ is surjective from $(-1,1)$ to $\mathbb{R}$. I have to prove that the function $\;f:(-1,1)\to \mathbb{R}\;$ defined by $f(x)=\log\sqrt{\frac{1+x}{1-x}}\;$ is bijective.
I have already proved that it is injective:
$$f(x)=f(y)$$
$$\log\sqrt{\frac{1+x}{1-x}}=\log\sqrt{\frac{1+y}{1-y}}$$
$$\log\sqrt{\frac{(1+x)(1-y)}{(1-x)(1+y)}}=0$$
$$\frac{1+x-y-xy}{1-x+y-xy}=1$$
$$x=y$$
But now, how can I prove that the function is surjective?

We have that $f(x)$ is defined in $(-1,1)$ and
$$f(x)=\log\sqrt{\frac{1+x}{1-x}}\implies f'(x)=\frac1{1-x^2}>0$$
then $f(x)$ is injective, moreover
$$\lim_{x\to 1^-} f(x)=\infty \quad \lim_{x\to -1^+} f(x)=-\infty$$
and since $f(x)$ is continuous by IVT it is surjective.
