# Calculate the inverse for $\arctan(x^2+1), x≥0$

I have no idea how to solve this problem.

Calculate the inverse for the function:

$$f(x) = \arctan(x^2+1),\quad x≥0 .$$

Also specify $D_{f^{-1}}$ and $V_{f^{-1}}$.

I would really appreciate your help

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write $x = \arctan (y^2 + 1)$ and solve for $y$ –  Will Jagy Oct 1 '13 at 14:17
I might guess $D$ means "domain". But $V$? –  Ron Gordon Oct 1 '13 at 14:47

$$f(x)=y = \arctan(x^2+1), x≥0$$
$$\tan y =\tan( \arctan(x^2+1))=x^2+1$$
$$x^2=\tan y -1$$
$$x=\sqrt{\tan y-1}, \tan y\geq 1$$