Find the Domain of \(f(x)=\frac1{\sqrt{x^2-1}}\)

Solution

Given: $$ f(x)=\frac1{\sqrt{x^2-1}} $$

Since the square root is in the denominator:

(i) The expression inside the root must be non-negative.
(ii) The denominator cannot be zero.

Therefore, $$ x^2-1>0 $$

$$ (x-1)(x+1)>0 $$

This is true when $$ x<-1 \quad \text{or} \quad x>1 $$

Hence, the domain is: $$ (-\infty,-1)\cup(1,\infty) $$