Determine the domain of this function

f(x)= square root (x + 6)/ x^2-4x+3

please explain how to do this and please put answer in interval form.

ok so i factor out the denominator (x-3)=0 (x-1)=0

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- Nov 11th 2009, 09:12 PMrefreshDetermine the domain of this function
Determine the domain of this function

f(x)= square root (x + 6)/ x^2-4x+3

please explain how to do this and please put answer in interval form.

ok so i factor out the denominator (x-3)=0 (x-1)=0 - Nov 11th 2009, 09:16 PMredsoxfan325
$\displaystyle f(x)=\sqrt{x}+\frac{6}{x^2-4x+3}$

To find the domain of a function, you start with a "default" domain of $\displaystyle \mathbb{R}$ and remove all points where the function is not defined. $\displaystyle \sqrt{x}$ can't take any $\displaystyle x<0$, and $\displaystyle \frac{6}{x^2-4x+3}$ is undefined where $\displaystyle x^2-4x+3=0$. Every other input value is fine. - Nov 11th 2009, 09:24 PMrefresh
- Nov 11th 2009, 09:26 PMredsoxfan325
- Nov 11th 2009, 09:30 PMrefresh
- Nov 11th 2009, 09:31 PMredsoxfan325
What does the book say?

Edit 1: Just noticed that you said $\displaystyle \sqrt{\frac{x+6}{(x-1)(x-3)}}$. In this case you need to find the places where $\displaystyle \frac{x+6}{(x-1)(x-3)}<0$.

On $\displaystyle (-\infty,-6)$, $\displaystyle (x-1)(x-3)>0$ and $\displaystyle x+6<0$, so $\displaystyle \frac{x+6}{(x-1)(x-3)}<0$.

On $\displaystyle [-6,1)$, $\displaystyle (x-1)(x-3)>0$ and $\displaystyle x+6>0$, so $\displaystyle \frac{x+6}{(x-1)(x-3)}>0$.

On $\displaystyle [1,3]$, $\displaystyle (x-1)(x-3)\leq0$ and $\displaystyle x+6>0$, so $\displaystyle \frac{x+6}{(x-1)(x-3)}<0$. (Note that this interval includes the two points where the denominator is zero.)

On $\displaystyle (3,\infty)$, $\displaystyle (x-1)(x-3)>0$ and $\displaystyle x+6>0$, so $\displaystyle \frac{x+6}{(x-1)(x-3)}>0$.

So the domain is $\displaystyle [-6,1)\cup(3,\infty)$.

Edit 2: It's definitely not all real numbers. Either the book is wrong or you're reading the answer to the wrong problem. - Nov 11th 2009, 09:36 PMrefresh
- Nov 11th 2009, 10:34 PMrefresh
- Nov 11th 2009, 10:47 PMredsoxfan325
That makes it easier then. $\displaystyle (-\infty, 6)$ is still off limits, but now only the two points 1 and 3 (as opposed to the interval $\displaystyle [1,3]$) are problematic, so the answer is what the books says:

$\displaystyle [-6,1)\cup(1,3)\cup(3,\infty)$ - Nov 11th 2009, 11:05 PMrefresh
- Nov 12th 2009, 08:35 AMredsoxfan325
You would be donating money to the website, not me.