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
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
$\displaystyle f(x)=\sqrt{x}+\frac{6}{x^2-4x+3}$
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.
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.
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:
$\displaystyle [-6,1)\cup(1,3)\cup(3,\infty)$