Find the domain and the range of the real function: f(x)=1 x^2-1 - Vidyadip

Question

Find the domain and the range of the real function: $f(x)=\frac{1}{\sqrt{x^2-1}}$

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Answer

Here we have, $f(x)=\frac{1}{\sqrt{x^2-1}}$
we need to find where the function is defined
The condition for the function to be defined
$\begin{array}{l}x^2-1>0 \\ \Rightarrow x^2>1 \\ \Rightarrow x>1\end{array}$
So, the domain of the function is the set of all the real numbers greater than 1
The domain of the function, $D _{\{ f ( x )\}}=(1, \infty)$
Now put any value of x within the domain set we get the value of the function always a fraction whose denominator is not equalled to 0
The range of the function, $R _{ f ( x )}=(0,1)$.

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