1000 Problems in Calculus (Problem 2)

Problem 2:
Find the domain of $\dfrac{1}{\sqrt{x^2-1}}$ .

Since we have $x^2-1$ under the square root and on denominator, $x^2-1> 0$ , it means that $x> 1$ or $x< -1$ , so the domain is $(-\infty,-1) \cup (1, \infty)$ .

See problem 1 and Problem 3.