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?
Best Answer
Hint:$$x^2-2x+5=(x-1)^2+4\geq0$$ for all$x\in(-\infty,+\infty)=\mathbb R$