Find the domain and the range of each of the following real

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

Need to find: Where the functions are defined.

The condition for the function to be defined,

$x^{2}-1>0$

$\Rightarrow x^{2}>1$

$\Rightarrow x>1$

So, the domain of the function is the set of all the real numbers greater than 1 .

The domain of the function, $\mathrm{D}_{\mathrm{f}(\mathrm{x})}=(1, \infty)$.

Now putting any value of $x$ within the domain set we get the value of the function always a fraction whose denominator is not equals to 0 .

The range of the function, $\operatorname{Rf}(x)=(0,1)$