Inverse Trig. (1 Viewer)

fancyyyyyy <3 · Apr 7, 2021

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

CM_Tutor · Apr 7, 2021

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

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

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

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

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Inverse Trig. (1 Viewer)

fancyyyyyy <3

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