How do you find the domain and range of # arctan(x^2)#?

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Answer 1 · 2018-07-22T12:07:05

Range: #y = arctan (x^2) in [0, pi/2 )#,
sans the asymptotic #y = pi/2#. .
Domain: #x in ( - oo, oo )#.

#y = arctan x^2 rArr 0#, as #x^2 to 0 rArr x to 0#.

By convention, arctan values are confined to #( -pi/2, pi/2 )#.

Inversely, #x = +- sqrt( tan y), tan y >=0 rArr y in [0, pi/2)#

Here, it is halved, as #x^2 >= 0#. See illustrative graph.

graph{(y-arctan(x^2))(y-pi/2)=0}.

For the interested readers, some related information;

Using the piecewise-wholesome inverse operator (tan)^(-1),

instead of #tan^(-1)#,

#y = (tan)^(-1)(x^2)#

and using its inverse #x^2 = tan y#

the graph that is same for both is created.

graph{x^2- tan y= 0}

The y-negative graphs are constituents of

#y = (tan)^(-1)( x^2 )= kpi + arctan x^2, k = 0, +-1, +-2, +-3, .#