The numerator, being a polynomial, is not restricting our domain. The domain is, however, restricted by the polynomial in the denominator, which must be nonzero. Therefore, we set the denominator equal to zero to determine the excluded values:

Therefore, the domain of  is the set of all real numbers except  - that is, 

The domain of  is restricted by two different denominators, neither of which can be equal to 0, so the excluded values are:

The correct response is therefore .

The numerator, being a polynomial, does not restrict our domain. The denominator, however, does restrict it to the values for which it is not equal to 0. We set the denominator equal to 0 to find the excluded values:

The domain, in interval notation, is therefore

.

The domain of  is restricted by two things.

First, the expression within the radical in the numerator must be nonnegative. We therefore solve for  in the inequality

,

or, in interval notation, 

Second, the expression in the denominator must be nonzero.  Therefore, we set the denominator equal to zero to determine the excluded value(s):

We exclude 4 from , so the correct response is 

There are two things restricting the domain of .

One is the radical symbol in the numerator. The expression inside the radical must be nonnegative, so solve the inequality:

,

or, in interval notation, 

The other is the denominator, which must be equal to 0, so set, and solve for  in, the equation:

 is already excluded from the domain; we exclude 4, so the domain is

.

The numerator, being a polynomial, is not restricting our domain. The domain is, however, restricted by the expression in the denominator, which must be nonzero. Furthermore, the radicand must be nonnegative. Combined, these facts mean that the radicand must be positive, and that the following inequality be solved:

 or, equivalently, .

In interval form, this is 

The domain of the product of two functions is the intersection of the domains of the individual functions.

The domain of  is restricted to all values of  that yield a nonzero denominator. Since this means that 

,

then, subsequently,

,

so the domain is the set of all real numbers except 4.

Similarly, the domain of  is restricted to all values of  that yield a nonzero denominator. This set is found to be the set of all real numbers except 7.

The intersection of these sets is the set of all real numbers except 4 and 7, or

.

The definition of  has two denominators of two fractions (one within the other), so we must exclude the values of  that make either denominator equal to zero.

One denominator is . Since it cannot be zero, we have 

,

and, subsequently,

.

The value 2 is excluded from the domain.

The other denominator is . If it is equal to zero, then

Therefore, this value is also excluded from the domain.

The correct domain is the set .

The numerator , being a polynomial, does not restrict the domain.

 appears as a radicand of a square root in the definition of , so  must be restricted to nonnegative numbers. Also,  is a denominator, which means 0 must also be excluded. Therefore, we are restricted so far to positive numbers.

There is one more denominator, which is ; it must be nonzero. We set this equal to zero to determine any additional value(s) that must be excluded:

Therefore, the domain is the set of all positive numbers except 9 - or

.

Given a fractional algebraic equation with variables in the numerator and denominator of one side and the other side equal to zero, we rely on a simple concept.  Zero divided by anything equals zero. That means we can focus in on what values make the numerator (the top part of the fraction) zero, or in other words,

The expression  is a difference of squares that can be factored as 

Solving this for  gives either  or .  That means either of these values will make our numerator equal zero.  We might be tempted to conclude that both are valid answers.  However, our statement earlier that zero divided by anything is zero has one caveat. We can never divide by zero itself.  That means that any values that make our denominator zero must be rejected.  Therefore we must also look at the denominator.

The left side factors as follows

This means that if  is  or , we end up dividing by zero.  That means that  cannot be a valid solution, leaving  as the only valid answer.  Therefore only #3 is correct.