How to find the domain and range of $f(x) = \sqrt{x^2-2x+5}$? This is the function:
$$f(x) = \sqrt{x^2-2x+5}$$
Edit: normally what I would do is this: Since it's a square root function, the thing inside the root has to be $\ge 0$. So, $(x^2 - 2x+5)\ge 0$. Then I would factor the stuff in the brackets so that I get () (__). But since this has complex roots, I don't know what to do 
Edit: Thanks for the help with the domain! For the range, I found the inverse of the function and did this:
$$x = \sqrt{(y-1)^2 +4}$$ 
$$x^2 -4 = (y-1)^2 $$
$$\sqrt{x^2 -4}+1 = y$$
And then proceeded to find the domain of the inverse:
$$x^2 -4 \ge 0$$
$$x^2 \ge 4$$
$$x \ge  +-2$$
Is this correct? How do I know if it is correct without graphing it out? 
 A: Hint:$$x^2-2x+5=(x-1)^2+4\geq0$$ for all$x\in(-\infty,+\infty)=\mathbb R$
A: 
I found the inverse of the function ... And then proceeded to find the domain of the inverse 

I understand the logic of doing this when finding the range, since range is kind of the  domain of inverse function... provided there is an inverse function. Not every function has an inverse: if several values of $x$ correspond to the same $y$, we cannot invert this relationship. 
Since the range concerns the values of dependent variable ($y$ in your case), it helps to look at a formula where $y$ is isolated -- which it was from the beginning, 
$y = \sqrt{x^2-2x+5}$. We want to know what kind of values this square root can produce. That depends on what goes under the square root, of course. Under the square root we have $x^2-4x+5 = (x-1)^2+4$. So, this is something that takes on all values from $4$ upward, and none below $4$. So you consider what sort of square root comes out of this, and arrive at the answer for the range.
