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

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

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