How do you graph, identify the domain, range, and asymptotes for #y=sec(1/2)x#?

1 Answer
Jul 20, 2018

Answer:

Domain; #x in ( - oo, oo )#. Range of.
# y = sec (x/2 ) in ( - oo, 1] U [ 1, oo ) rArr y notin ( - 1, 1 )#.
Asymptotes: x = ( 2k + 1 )pi, k = 0, +-1, +-2, +-3, ...#

Explanation:

The period of # y = sec (x/2 )# is #(2pi)/(1/2) = 4pi#.

Cosine range is #[ - 1, 1 ]# Its reciprocal

secant range is #( - oo, 1] U [ 1, oo )#. So,

# y = sec (x/2 ) in ( - oo, 1] U [ 1, oo ) rArr y notin ( - 1, 1 )#.

Domain; #x in ( - oo, oo )#

Asymptotes: x = ( 2k + 1 )pi, k = 0, +-1, +-2, +-3, ...#,

by solving the denominator #cos ( x/2 ) = 0#.

See illustrative graph, with the markings os asymptotes near O and

the out-of- bounds range #( - 1, 1 ):.
graph{(y cos ( x/2 ) - 1)(y^2-1)(x^2-(pi)^2) = 0[-10 10 -5 5]}

Slide the graph #rarr uarr larr darr#, to view the extended graph.