We need to be careful here not to confuse the domain and range of a function. The problem specifically concerns the range of the function, which is the set of possible numbers of . It can be helpful to think of the range as all the possible y-values we could have on the points on the graph of .

Notice that  has  in its equation. Whenever we have an absolute value of some quantity, the result will always be equal to or greater than zero. In other words, |4-x| 0. We are asked to find the smallest value in the range of , so let's consider the smallest value of , which would have to be zero. Let's see what would happen to  if .

This means that when , . Let's see what happens when  gets larger. For example, let's let .

As we can see, as  gets larger, so does . We want  to be as small as possible, so we are going to want  to be equal to zero. And, as we already determiend,  equals  when .

The answer is .

 is the same as . 

To find the inverse simply exchange  and  and solve for . 

So we get  which leads to .

The domain of the inverse of a relation is the same as the range of the original relation. In other words, the y-values of the relation are the x-values of the inverse.

For the original relation, the range is: .

Thus, the domain for the inverse relation will also be .

The domain means what real number can you plug in that would still make the function work. For this case, we have to worry about the denominator so that it does not equal 0, so we solve the following. 7x – 1 = 0, 7x = 1, x = 1/7, so when x ≠ 1/7 the function will work.

The domain of a function or relation is the set of all possible x-values. Here that is simply a list of the x-coordinates of all of the coordinate pairs. So the domain is {–2, –3, 2, 4}.

Because x and y must be negative integers greater than negative five, then x and y can only be equal to the following values:

x can equal -4, -3, -2, or -1

y can equal -4, -3, -2, or -1

Now we can try all of the combinations of x and y, and see what x # y would equal. It is helpful to note that x # y is the same as y # x because 2yx + y + x = 2xy + x  + y. This means that the order of x and y doesn't matter. 

-4 # -4 = 2(-4)(-4) + -4 + -4 = 24

-4 # -3 = 2(-4)(-3) + -4 + -3 = 17

-4 # -2 = 10

-4 # -1 = 3

-3 # -3 = 12

-3 # -2 = 7

-3 # -1 = 2

-2 # -2 = 4

-2 # -1 = 1

-1 # -1 = 0

We don't need to find -3 # -4, -2 # -4, etc, because x # y = y # x . 

The smallest value of x # y must be 0.

The domain of a function includes all of the values of x for which that function is defined. In other words, the domain is all of the real values of x that will produce a real number. Let's look at the domains of each function one at a time.

First, let's examine

In general, when we are examining the domain of a function, we want to find places where we end up with zeros in the denominators or square-roots of negative numbers. For example, in the function , if we let x = 4, then we would be forced to evaluate 1/0, which isn't possible. We can never divide by zero. Thus, this function is not defined over all real values of x. We can eliminate it from the answer choices.

Next, let's look at . Let's set the denominator equal to zero to see if there are any values of x which might give us a zero in the denominator.

Subtract one from both sides.

Take the cube root of both sides.

Thus, if x = –1, then f(x) will be equal to 1/0, which isn't defined, because we can't divide by zero. Therefore, we can eliminate this answer choice.

Now, let's analyze .

We can never take the square root of a negative number. Thus, if , then f(x) won't be defined. For example, if x = 4, then , which would produce an imaginary number. Therefore, this function can't be the correct answer.

Next, let's look at . It will help us to rewrite f(x) in a form using square roots. In general, . As a result, we can rewrite f(x) as follows:

. In this form, we can see that if is negative, then f(x) won't be defined. Thus, if x = –2, for example, we would be forced to find the square root of –8, which produces an imaginary result. So, this function isn't the correct answer either.

By process of elimination the answer must be . The reasons that this function is defined for all values of x is because the denominator can never be zero. Thus, we can pick any value of x from negative to positive infinity, and we will get a real value for f(x).

The answer is

The domain of f(x) is defined as all of the values of x for which f(x) is defined.

The first value we have to consider is 1. Let's find f(1):

f(1) = (1³ – 3(1)² +2(1))^–1

= (1 – 3(1) + 2)^–1

= (1 – 3 + 2)^–1

= (0)^–1.

Remember that, in general, a^–1 = 1/a. Thus, 0^–1 = 1/0. However, it is impossible to have 0 in the denominator of a fraction, because it is impossible to divide anything by zero. Thus, 0^–1 is undefined. Since f(1) is undefined, 1 cannot belong to the domain of f(x).

Now let's find f(2):

f(2) = (2³ – 3(2)² +2(2))^–1

= (8 – 3(4) + 4)^–1

=(8 – 12 + 4)^–1

= 0^–1

Because f(2) is undefined, 2 is not in the domain of f(x).

Finally, we can look at f(–1):

f(–1) = ((–1)³ – 3(–1)² +2(–1))^–1

= (–1 – 3(1) – 2)^–1

= (–1 – 3 – 2)^–1 = (–6)^–1 = –1/6.

f(–1) is defined, so –1 belongs to the domain of f(x).

Therefore only III is in the domain of f(x).