Problem source

In this question, the  function $\sin^{-1}$ is defined to have domain $ -1\le x \le 1$ 
and range \linebreak $ - \frac{1}{2}\pi \le x \le \frac{1}{2}\pi$ and the function $\tan^{-1}$ is defined to have the real numbers as its
domain and range $ - \frac{1}{2}\pi < x < \frac{1}{2}\pi$.
\begin{questionparts}
\item
Let 
$$
\g(x) = \displaystyle {2x \over 1 + x^2}\;,  \ \ \ \ \ \ \ \ \ \ -\infty <x<\infty\;.
$$
Sketch the graph of $\g(x)$ and state the range of $\g$.
\item
Let 
\[
\displaystyle \f \l x \r = \sin^{-1} \l {2x \over 1 + x^2} \r \;,\ 
\ \ \ \ \ \ \ \ -\infty < x < \infty\;.
\] 
Show that
$
\f(x ) =  
2 \tan^{-1} x$   for $ -1 \le x \le 1\,$ and $\f(x) =
\pi - 2 \tan^{-1} x $ for $x\ge1\,$. 
Sketch the graph of $\f(x)$.
\end{questionparts}