What is the range of $ f(x)=\sqrt \frac{x^2}{x^2-1}$ $$\sqrt{\frac{x^2}{x^2-1}}$$
I am stuck at finding the range of the given function mathematically. I could deduce from  its graph that the range is $[1,\infty)$.
But that is incorrect. Could anyone help me? It would help improve my concept
 A: Usually, it is a good idea to use theory of equations to solve range of expressions like these. Assume $y = \sqrt{\dfrac{x^2}{x^2 - 1}}$. Then
\begin{align}
y^2 = \dfrac{x^2}{x^2 - 1} \\
x^2(y^2 - 1) - y^2 = 0
\end{align}
Since $x \in \mathbb{R}$, the $D \ge 0$. Hence, $0 - 4(-y^2)(y^2 - 1)\ge 0$. Solve the inequality.
A: The domain is $(-\infty,-1)\cup\{0\}\cup(1,\infty)$ so that the rang is $(1,+\infty)\cup\{0\}$. It is easy to verify that as $x\rightarrow 1^\pm$, $y$ tends to $\infty$; and as $x\rightarrow\pm\infty$, $y$ tends to unity ($1$). You can also see that, in the  subdomain of $\{0\}\cup(1,\infty)$, the inverse of the function is itself! alike the functions $f(x)=x$ and $f(x)=\frac{1}{x}$.
A: Note that the domain is $|x|>1$. Then, the range is
$$f(x)=\sqrt \frac{x^2}{x^2-1} = \sqrt{ 1+  \frac{1}{x^2-1}}>1$$
A: The range here could be find out by getting an Idea about the domain. Here assuming it is defined only on real numbers then the domain will be ${0}\cup(-\infty,-1)\cup(1,\infty)$ and hence the range of this function will be ${0}\cup(1,\infty)$
A: $$
f(x)=\sqrt{\frac{x^{2}}{x^{2}-1}}
$$
if u express it in term of  $y$  will get
$$
x=\sqrt{\frac{y^{2}}{y^{2}-1}}
$$
Now find domain of $f(y)$
this will de your range,
$y \geq 0$...(1)
$$
\begin{array}{l}
\left(y^{2}-1\right) > 0:......(2) \\
(y+1)(y-1) > 0
\end{array}
$$
By wavy curve method
$$
(-\infty,-1) \cup(1, \infty)
$$
Combining 1 and 2, Domain of $f(y)$
$$
(1,\infty )
$$
