How do you graph #y = arcsec(x)#?

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Answer 1 · 2018-07-17T06:54:27

See graph and my idiosyncratic explanation.

#y = arcsec x = arc cos(1/x) #,

Conventionally limited ( for trigonometric arccos ) #y in [ 0, pi ]#.

See the truncated-graph, with asymptote #y = pi/2# and

the limiting lines #y = 0 and y = pi#.
graph{(y-arccos(1/x))(y-pi/2)(y-pi)(y)=0}

See the wholesome-inverse graph for #y = (sec)^(-1)x#,

using the inverse #x = sec y#

graph{(x cos y - 1)(x^2-0.25) = 0[-50 50 -25 25]}

You can see the effect of the small operator #sec^(-1# becoming

great #(sec)^(-1)#, like Universe before the Earth.

Observe in both the graphs that #y notin ( - 1, 1 )#

Please note, that piecewise/wholesome,

the graphs of #y = f( x ) and x = f^( - 1 ) ( y )# are one and the same.