To determine the domain of the function, we must consider where x cannot exist. The only limitation on the function is the denominator, which cannot equal zero. 

To find the x-values where this occurs, we must set the denominator equal to zero and solve for x:

$\displaystyle (x^3-27)(x-2)=0$

$\displaystyle x=3, 2$

These are the only two limitations on the domain of the function, so the domain of the function is

$\displaystyle (-\infty, 2)\cup(2, 3)\cup(3, \infty)$

Note that round brackets were used for all of the intervals, because none of the bounds of the intervals are included in the domain.

To determine the domain of the function, we must find the x-values that would give us an undefined result when we plug them into the function. On the numerator, we know that the natural log function can never equal zero or be negative. In the denominator, we can never have a zero. With these conditions in mind, we must now find the x values that makes these undefined situations occur.

For the numerator:

$\displaystyle x^2-9\leq 0$

$\displaystyle x^2\leq 9$

$\displaystyle \left |x \right |\leq 3$

$\displaystyle x\leq3$ and $\displaystyle x\geq -3$

For the denominator:

$\displaystyle x+3=0$

$\displaystyle x=-3$

Now that we know where x cannot be, we can now write the domain, making sure to use round brackets for the endpoints of the intervals:

$\displaystyle (-\infty, -3)\cup(3, \infty)$

To find the domain of the function, we must ask ourselves where x cannot be if we want the function to be defined. 

The only limitation on the function is that we can't have zero in the denominator (this is undefined). To find the x value(s) where this occurs, we set the denominator equal to zero and solve for x:

$\displaystyle x+3=0$

$\displaystyle x=-3$

This is the only x value that we have an undefined result for the function, so our domain is

$\displaystyle (-\infty, -3)\cup(-3, \infty)$.

Note that we use no square brackets, because we are not including the endpoints of the intervals in the domain.

Range asked for the values of y expressed by the function. The absolute value function is always positive with respect to y. The point 0 is included in the range requiring a square bracket. Infinity is not a numerical value so it is bound by a parentheses. $\displaystyle [0,\infty)$

Domain is finding the acceptable $\displaystyle x$ values that will make the function generate real $\displaystyle y$values. $\displaystyle y=x+4$ is a linear function therefore any $\displaystyle x$ values will always generate real $\displaystyle y$values. Answer is all real numbers.

Domain is finding the acceptable $\displaystyle x$ values that will make the function generate real $\displaystyle y$values. $\displaystyle y=x^2+5x+6$ is a quadratic function therefore any $\displaystyle x$ values will always generate real $\displaystyle y$values. Answer is all real numbers.

Domain is finding the acceptable $\displaystyle x$ values that will make the function generate real $\displaystyle y$values. $\displaystyle y=\left | x+5\right |$ is an absolute value function therefore any $\displaystyle x$ values will always generate real $\displaystyle y$values. Answer is all real numbers.

Domain is finding the acceptable $\displaystyle x$ values that will make the function generate real $\displaystyle y$values. Because this function is a fractional expression, we need to check the denominator. Remember, the denominator must not be zero. This would make the function undefined. 

$\displaystyle f(x)=\frac{2}{x+3}$ We only need to look at $\displaystyle x+3$. Since we know it can't equal zero, we set that expression to zero to determine the $\displaystyle x$-value that makes this function undefined.

$\displaystyle x+3=0$ Subtract $\displaystyle 3$ on both sides.

$\displaystyle x=-3$

Domain is finding the acceptable $\displaystyle x$ values that will make the function generate real $\displaystyle y$values. Because we have a radical, we have to remember the smallest possible value inside the radical is zero. Anything less means we will be dealing with imaginary numbers.

$\displaystyle y=\sqrt{x+9}$ So let's set $\displaystyle x+9=0$ to ensure the smallest value inside is zero and makes all possible $\displaystyle y$ values real.

$\displaystyle x+9=0$ Subtract $\displaystyle 9$ on both sides.

$\displaystyle x=-9$ This means that all values greater than or equal to $\displaystyle -9$ will satisfy the equation $\displaystyle (-9\leq x)$.

Because the domain is specifying certain numbers not being acceptable in the domain, we can eliminate the radical choice. The cubic equation is also wrong as it's not a fractional function. Let's look at the remaining choices since they are fractional expressions. Our main focus will be setting that denominator to zero as it makes the function undefined.

$\displaystyle y=\frac{4(x-2)(x-4)}{(x+1)}$ Therefore $\displaystyle x+1=0$ or $\displaystyle x=-1$. This doesn't satisfy it as we are looking for values of $\displaystyle 2,4$ to also not work. This answer is wrong.

$\displaystyle y=\frac{9}{(x^3-8)}$ Therefore $\displaystyle x^3-8=0$, $\displaystyle x^3=8$, $\displaystyle x=\pm2$. This doesn't satisfy it as we are looking for values of $\displaystyle -1, 4$ to also not work. Also $\displaystyle -2$ is acceptable according to the domain in the question. This answer is wrong.

$\displaystyle y=\frac{7}{(x-2)(x-4)(x+1)}$ Therefore $\displaystyle x-2=0$, $\displaystyle x=2$. Also $\displaystyle x-4=0$, $\displaystyle x=4$. Finally $\displaystyle x+1=0$, $\displaystyle x=-1$. This is the correct answer.