# Finding Domain of a Function with a Fraction Inside a Square Root

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.

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You need to solve the inequality $\frac{x-3}{x^2-3x+2}\geq0$.

The algorithm to do so it to factor numerator and denominator.

Then divide the real line in intervals bounded by the roots of the numerator and denominator.

Finally determine the sign of the fraction on each of these intervals by computing the sign of each of the factor on them.

In your case we get $\frac{x-3}{(x-1)(x-2)}$. We get the intervals $(-\infty,1]$, $(1,2]$, $(2,3]$, and $(3,\infty)$.

The sign of the fraction on them turns out to be $-,+,-,+$, respectively.

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Thank you very much. I didn't remember about this procedure but now makes all sense. Much appreciated –  sidyll Jun 3 '14 at 1:48

$$f(x) = \sqrt{\frac{x-3}{x^2-3x+2}}= \sqrt{\frac{x-3}{(x-1)(x-2)}}$$

So the domain is when $$\frac{x-3}{(x-1)(x-2)}\ge0$$ and $$x\neq1$$ and $$x\neq2$$

$$\mathrm{Domain}=(1, 2)\cup[3, \infty]$$

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Thank you for your answer ! I didn't think about approaching it at once like here, but I don't get the part of finding the domain so quickly, maybe a trick? Sorry I'm unable to see the "middle path" of it! –  sidyll Jun 3 '14 at 1:49