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

1 Answer
A. S. Adikesavan
Jul 20, 2018

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.

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