### Instruction: What is the domain of the function $f(x)=\frac{x+6}{\sqrt{x^2-3x-4}}$?### Response: To find the domain of a function, I need to identify the values of x that make the function undefined or invalid.
One thing that can make a function undefined is dividing by zero, so I need to check when the denominator of the fraction is zero.
To do that, I can factor the quadratic expression under the square root sign: $x^2-3x-4=(x-4)(x+1)$.
Then, the denominator is zero when $x-4=0$ or $x+1=0$, which means when $x=4$ or $x=-1$.
So, I need to exclude those two values from the domain.
Another thing that can make a function invalid is taking the square root of a negative number, so I need to check when the expression under the square root sign is negative.
To do that, I can use a sign chart to analyze the quadratic expression.
I know that the expression is zero when $x=4$ or $x=-1$, so those are the critical points that divide the x-axis into three intervals.
I can pick a test value in each interval and plug it into the expression to see the sign of the result.
For example, if I pick $x=-2$, then $x^2-3x-4=(-2)^2-3(-2)-4=4+6-4=6$, which is positive.
This means that the expression is positive in the interval $(-\infty,-1)$.
Similarly, if I pick $x=0$, then $x^2-3x-4=0^2-3(0)-4=-4$, which is negative.
This means that the expression is negative in the interval $(-1,4)$.
Finally, if I pick $x=5$, then $x^2-3x-4=(5)^2-3(5)-4=25-15-4=6$, which is positive.
This means that the expression is positive in the interval $(4,\infty)$.
So, I need to exclude the interval $(-1,4)$ from the domain, since the expression under the square root sign is negative there.
The remaining intervals are $(-\infty,-1)$ and $(4,\infty)$, which are the domain of the function.
I can write the domain using interval notation or set-builder notation.