Question

Find the domain of inverse sine of (x^2-1)

Answer 1

is that the answer?

Answer 2

close. x can be negative

Answer 3

Since the domain of \(\sin^{-1}(y)\) would be \(-1\leq y\leq 1\) then we would say \[
-1\leq x^2 - 1\leq 1
\]And solve for \(x\).

Answer 4

so what would be the result?

Answer 5

\[
0\leq x^2\leq 2
\]\[
0\leq |x| \leq \sqrt 2
\]Finally \[
-\sqrt 2\leq x \leq \sqrt 2
\]

Answer 6

why is it -root 2

Answer 7

\(\sqrt{x^2} = |x|\)
and
\(|x|\leq a \implies -a\leq x \leq a\)

Answer 8

ok, thanks!