# Inverse Trig. (1 Viewer)

#### fancyyyyyy <3

##### New Member
Hi how would i find the domain and range of x sin^-1(x^2)

#### CM_Tutor

##### Moderator
Moderator
The inverse sine function $\bg_white \sin^{-1}{x}$ can only operate on numbers in the domain of $\bg_white -1 \leqslant x \leqslant 1$.

For the given function $\bg_white x\sin^{-1}{x^2}$, the only restriction is that $\bg_white -1 \leqslant x^2 \leqslant 1$, which you must solve to find that the domain is actually $\bg_white \left\{x\in\mathbb{R}: -1 \leqslant x \leqslant 1\right\}$ or, in interval notation, $\bg_white x\in[-1, 1]$.

Had the question asked for the domain of $\bg_white x\sin^{-1}{\left(x^2-4\right)}$, you would have needed to solve $\bg_white -1 \leqslant x^2 - 4 \leqslant 1$ to get a domain of $\bg_white x\in\left[-\sqrt{5}, -\sqrt{3}\right]\cup\left[\sqrt{3}, \sqrt{5}\right]$.

Finally, had the question asked for the domain of $\bg_white \frac{x\sin^{-1}{\left(x^2-4\right)}}{x^2-4}$, you would still have needed to solve $\bg_white -1 \leqslant x^2 - 4 \leqslant 1$ but subject to the additional constraint that $\bg_white x^2-4 \neq 0$, producing a domain of $\bg_white x\in\left[-\sqrt{5}, -2\right)\cup\left(-2, -\sqrt{3}\right]\cup\left[\sqrt{3}, 2\right)\cup\left(2, \sqrt{5}\right]$.