Domain of f(x) = (x^2 - 1)^{-1/2} is | vedclass

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Question

(A) The function is defined as $f(x) = \frac{1}{\sqrt{x^2 - 1}}$.For $f(x)$ to be defined,the expression inside the square root must be strictly great

Vedclass · Accepted Answer

$( - \infty, -1) \cup (1, \infty)$

Vedclass · Answer

(A) The function is defined as $f(x) = \frac{1}{\sqrt{x^2 - 1}}$.For $f(x)$ to be defined,the expression inside the square root must be strictly greater than zero because it is in the denominator.Therefore,$x^2 - 1 > 0$.This implies $x^2 > 1$.Solving the inequality $x^2 > 1$,we get $|x| > 1$.This means $x > 1$ or $x In interval notation,this is expressed as $x \in ( - \infty, -1) \cup (1, \infty)$.

Domain of $f(x) = (x^2 - 1)^{-1/2}$ is

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