# Thread: Determine the domain of this function

1. ## 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

2. Originally Posted by refresh
Determine the domain of this function

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

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

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.

3. 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

4. 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:

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

5. 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.

6. 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.

7. Originally Posted by redsoxfan325
What does the book say?
all real numbers x

8. 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.

9. 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:

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

10. 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?

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