Answer:
Domain is [tex]D=\bold R-(n+1)\dfrac{\pi}{2}[/tex]
Range is [tex]R=\bold R-(-1,1)[/tex]
Other details are included in the explanation part.
Step-by-step explanation:
The given function is [tex]f(x) = \rm {sec }\;2\mathit x[/tex].
The function is a secant function which is the reciprocal of cosine function.
The graph of the function is is attached as an image.
The domain of the function is the range of x where it is defined. The domain of the function will be,
[tex]D=\bold R-(n+1)\dfrac{\pi}{2}[/tex] here, R is the real number.
The range of the function is the range of value of function for every value of x which is,
[tex]R=\bold R-(-1,1)[/tex]
it means that the range is real numbers excluding the range of (-1,+1).
The vertical asymptotes will form where the curve tends to parallel to the y-axis. So, the vertical asymptotes are at [tex](n+1)\dfrac{\pi}{2}[/tex].
The period of the function is [tex]2\pi[/tex] as the function repeats itself after that period.
The function goes to infinity and hence, it doesn't has any amplitude.
For more details, refer the link:
https://brainly.com/question/12017601?referrer=searchResults
