# Finding the Domain of Square Root

• Oct 4th 2011, 01:22 PM
Gfan2010
Finding the Domain of Square Root
$\displaystyle f(x) = \sqrt{x^2-5x+6}$

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

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

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

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

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

So what am I doing wrong?
• Oct 4th 2011, 01:29 PM
pickslides
Re: Finding the Domain of Square Root
Your workng looks good to give you some bounds on the solution

Consider $\displaystyle \displaystyle (-\infty ,2]\cap[3,\infty)$
• Oct 4th 2011, 01:52 PM
mr fantastic
Re: Finding the Domain of Square Root
Quote:

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

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

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

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

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

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

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.