Answer
For the function $sec^{-1}~x$:
The domain is $(-\infty, -1]\cup[1,\infty)$
The range is $[0,\frac{\pi}{2})\cup (\frac{\pi}{2}, \pi]$
We can see a sketch of the graph of $~~sec^{-1}~x~~$ below.
Note there is a horizontal asymptote at $y = \frac{\pi}{2}$
Work Step by Step
Consider the function $sec~x$:
The domain is all real numbers except $\frac{\pi}{2}+\pi~n$, where $n$ is an integer
The range is $(-\infty, -1]\cup[1,\infty)$
We can consider the function $~~sec^{-1}~x~~$ as the inverse function of $~~sec~x~~$ by considering the domain of $~~sec~x~~$ restricted to $[0,\frac{\pi}{2})\cup (\frac{\pi}{2}, \pi]$
Then for the function $sec^{-1}~x$:
The domain is $(-\infty, -1]\cup[1,\infty)$
The range is $[0,\frac{\pi}{2})\cup (\frac{\pi}{2}, \pi]$
We can see a sketch of the graph of $~~sec^{-1}~x~~$ below.
Note there is a horizontal asymptote at $y = \frac{\pi}{2}$
