# Finding the Domain of Square Root

• October 4th 2011, 02:22 PM
Gfan2010
Finding the Domain of Square Root
$f(x) = \sqrt{x^2-5x+6}$

$f(x) = \sqrt{(x-2)(x-3)}$

$(x-2)(x-3) \geq 0$

$x-2 \geq 0 \Longrightarrow x \geq 2$

$x-3 \geq 0 \Longrightarrow x \geq 3$

I know from the graph that it should be $x \leq 2$

So what am I doing wrong?
• October 4th 2011, 02:29 PM
pickslides
Re: Finding the Domain of Square Root
Your workng looks good to give you some bounds on the solution

Consider $\displaystyle (-\infty ,2]\cap[3,\infty)$
• October 4th 2011, 02:52 PM
mr fantastic
Re: Finding the Domain of Square Root
Quote:

Originally Posted by Gfan2010
$f(x) = \sqrt{x^2-5x+6}$

$f(x) = \sqrt{(x-2)(x-3)}$

$(x-2)(x-3) \geq 0$

$x-2 \geq 0 \Longrightarrow x \geq 2$

$x-3 \geq 0 \Longrightarrow x \geq 3$

I know from the graph that it should be $x \leq 2$

So what am I doing wrong?

Case 1: x - 2 >= 0 AND x - 3 >= 0 => x >= 2 AND x > 3 => x >= 3.

Case 2: x - 2 =< 0 AND x - 3 =< 0 => .......

Personally, I think it's much better to draw a graph of y = (x - 2)(x - 3) and then use it to see the values of x for which y >= 0.