I need to find the domain of this function:
$$
f(x) = \sqrt{\frac{x-3}{x^2-3x+2}}
$$
I understand that initially the denominator cannot be zero, and since it's just a formatted equation with roots 1 and 2, I know that 1 and 2 cannot be in the domain. However I can't get along when checking for the whole fraction to not be negative. I'm working with this:
$$
\sqrt{\frac{x-3}{x^2-3x+2}} \geq 0 \quad\rightarrow\quad x-3 \geq x^2-3x+2
$$
How shall I proceed here? Factored version of the equation $(x-2)(x-1)$ does not appear to be helpful.
Best Answer
$$ f(x) = \sqrt{\frac{x-3}{x^2-3x+2}}= \sqrt{\frac{x-3}{(x-1)(x-2)}} $$
So the domain is when $$\frac{x-3}{(x-1)(x-2)}\ge0$$ and $$x\neq1$$ and $$x\neq2$$
$$\mathrm{Domain}=(1, 2)\cup[3, \infty)$$