Finding the range of the function $f(x)=\frac{\sqrt{x^{2}-4}}{\sqrt{x-2}}$ My question regards finding the range of the following function:
$$f(x)=\frac{\sqrt{x^{2}-4}}{\sqrt{x-2}}.$$
I have found the domain of the function to be $[-2, 2)$ $\cup$ $(2, \infty)$. According to my teacher's notes, this is correct. However, I have almost no clue how to go about finding the range without simply plugging in values of $x$ from the domain and seeing a general pattern of values of $f(x)$ from which to educatedly estimate the range. Could someone please help me find the range of this function?
 A: We have $x \ne 2 \implies f(x) = \sqrt{x+2}\ne 2$ also. Thus $\text{Range}(f)= \{y: y > 2\}$. Note that $x > 2\implies y = \sqrt{x+2} > \sqrt{2+2} = 2$, hence the answer.
A: The domain of your function is $x>2$. The function mentioned in the comments which is "almost" like yours, $g(x) = \sqrt{x+2}$ can therefore get nowhere near $0$ as an output. Since $g(x)$ is increasing, $g(2)=2$ is the minimum it attains on $[2,\infty)$. But $f$ is undefined at $2$ so your range is $(2,\infty)$. 
It is not possible for the function to output $1$, for example.
A: first of all $[-2, 2)$ is not in the domain since in the denominator is ${\sqrt{x-2}}.$
the function isn't defined for $x\le 2$ making the domain $(2,\infty)$ (i.e. $x>2$)
knowing that let's simplify the function:
$$f(x)=\frac{\sqrt{x^{2}-4}}{\sqrt{x-2}} = \frac{\sqrt{(x-2)(x+2)}}{\sqrt{x-2}}=\frac{\sqrt{x-2} \cdot \sqrt{x+2}}{\sqrt{x-2}}=\sqrt{x+2}$$
since the domain is $(2,\infty)$ we plug both end values into the new $f(x)$, and we get the range
$$f(2)=\sqrt{2+2}=\sqrt{4}=2 ,f(\infty)=\sqrt{\infty+2}=\sqrt{\infty}=\infty$$
thus the range is $(2,\infty)$
NOTE: it's a coincidence that it's the same as the domain due to the choice of numbers
