# Determine the domain of this function

• Nov 11th 2009, 09:12 PM
refresh
Determine the domain of this function
Determine the domain of this function

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

ok so i factor out the denominator (x-3)=0 (x-1)=0
• Nov 11th 2009, 09:16 PM
redsoxfan325
Quote:

Originally Posted by refresh
Determine the domain of this function

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

$\displaystyle f(x)=\sqrt{x}+\frac{6}{x^2-4x+3}$

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 PM
refresh
Quote:

Originally Posted by redsoxfan325
$\displaystyle f(x)=\sqrt{x}+\frac{6}{x^2-4x+3}$

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.

i got x=3 and x=1 how do i put this in interval form???
and the question was f(x)= square root (x+6)/x^2-4x+3
• Nov 11th 2009, 09:26 PM
redsoxfan325
Quote:

Originally Posted by refresh
i got x=3 and x=1 how do i put this in interval form???

Since $\displaystyle x\geq0$ and $\displaystyle x\neq 1$ and $\displaystyle x\neq3$, the interval form is:

$\displaystyle \text{dom}(f)=[0,1)\cup(1,3)\cup(3,\infty)$
• Nov 11th 2009, 09:30 PM
refresh
Quote:

Originally Posted by redsoxfan325
Since $\displaystyle x\geq0$ and $\displaystyle x\neq 1$ and $\displaystyle x\neq3$, the interval form is:

$\displaystyle \text{dom}(f)=[0,1)\cup(1,3)\cup(3,\infty)$

hmmm thats odd the back of the bok has a different answer??? maybe the book is wrong.
• Nov 11th 2009, 09:31 PM
redsoxfan325
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 PM
refresh
Quote:

Originally Posted by redsoxfan325
What does the book say?

all real numbers x
• Nov 11th 2009, 10:34 PM
refresh
Quote:

Originally Posted by redsoxfan325
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.

im sorry i read the wrong answer ok this is what the book said.

[-6,1) U (1,3) U (3,infinity)

by the way the only the (x+6) is in the squareroot.
• Nov 11th 2009, 10:47 PM
redsoxfan325
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 PM
refresh
Quote:

Originally Posted by redsoxfan325
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:

$\displaystyle [-6,1)\cup(1,3)\cup(3,\infty)$

hey thanks for all your help. im new to this site so i don't really know how everything works yet so i have a quick question. there is a donate button on the bottom, if i donate will you get the money or what?
• Nov 12th 2009, 08:35 AM
redsoxfan325
You would be donating money to the website, not me.