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.

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