Determine the domain of this function

**refresh** · November 11th 2009, 09:12 PM

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

**redsoxfan325** · November 11th 2009, 09:16 PM

Originally Posted by refresh

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.

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

To find the domain of a function, you start with a "default" domain of $\mathbb{R}$ and remove all points where the function is not defined. $\sqrt{x}$ can't take any $x<0$ , and $\frac{6}{x^2-4x+3}$ is undefined where $x^2-4x+3=0$ . Every other input value is fine.

**refresh** · November 11th 2009, 09:24 PM

Originally Posted by redsoxfan325

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

To find the domain of a function, you start with a "default" domain of $\mathbb{R}$ and remove all points where the function is not defined. $\sqrt{x}$ can't take any $x<0$ , and $\frac{6}{x^2-4x+3}$ is undefined where $x^2-4x+3=0$ . 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

**redsoxfan325** · November 11th 2009, 09:26 PM

Originally Posted by refresh

i got x=3 and x=1 how do i put this in interval form???

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

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

**refresh** · November 11th 2009, 09:30 PM

Originally Posted by redsoxfan325

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

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

hmmm thats odd the back of the bok has a different answer??? maybe the book is wrong.

**redsoxfan325** · November 11th 2009, 09:31 PM

What does the book say?

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

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

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

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

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

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

Edit 2: It's definitely not all real numbers. Either the book is wrong or you're reading the answer to the wrong problem.

**refresh** · November 11th 2009, 09:36 PM

Originally Posted by redsoxfan325

What does the book say?

all real numbers x

**refresh** · November 11th 2009, 10:34 PM

Originally Posted by redsoxfan325

What does the book say?

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

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

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

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

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

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

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.

**redsoxfan325** · November 11th 2009, 10:47 PM

That makes it easier then. $(-\infty, 6)$ is still off limits, but now only the two points 1 and 3 (as opposed to the interval $[1,3]$ ) are problematic, so the answer is what the books says:

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

**refresh** · November 11th 2009, 11:05 PM

Originally Posted by redsoxfan325

That makes it easier then. $(-\infty, 6)$ is still off limits, but now only the two points 1 and 3 (as opposed to the interval $[1,3]$ ) are problematic, so the answer is what the books says:

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

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?

**redsoxfan325** · November 12th 2009, 08:35 AM

You would be donating money to the website, not me.

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