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

Jul 22, 2018

Range: $y = \arctan \left({x}^{2}\right) \in \left[0 , \frac{\pi}{2}\right)$,
sans the asymptotic $y = \frac{\pi}{2}$. .
Domain: $x \in \left(- \infty , \infty\right)$.

#### Explanation:

$y = \arctan {x}^{2} \Rightarrow 0$, as ${x}^{2} \to 0 \Rightarrow x \to 0$.

By convention, arctan values are confined to $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$.

Inversely, $x = \pm \sqrt{\tan y} , \tan y \ge 0 \Rightarrow y \in \left[0 , \frac{\pi}{2}\right)$

Here, it is halved, as ${x}^{2} \ge 0$. 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 ${\tan}^{- 1}$,

$y = {\left(\tan\right)}^{- 1} \left({x}^{2}\right)$

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

$y = {\left(\tan\right)}^{- 1} \left({x}^{2}\right) = k \pi + \arctan {x}^{2} , k = 0 , \pm 1 , \pm 2 , \pm 3 , .$