The range could be anything. Without parameters specified, the domain of {1,2,3,4} could have any range.
This problem is unsolvable.
You can define the domain as anything you like and that will determine the range. Or, you can define the range as anything you like and that will determine the domain. For example: domain = {1, 2, 3, 4, ... } then range = {-3, 0, 5, 12, ... } or range = {1, 2, 3, 4, ... } then domain = {sqrt(5), sqrt(6), sqrt(7), sqrt(8), ...}. There is, of course, no need to restrict either set to integers but then it was easier to work out one set from the other.
If this is the whole of the function, then the domain is {2, 1, -3, -1}. That set can be put in increasing order if you wish but that is not necessary.
Find all possible "x" and "y" values for domain and range. Then put it in inequality form. For example the domain and range for the equation 2x-3/x-5 would be: Domain: All Reals; x>5 Range: All Reals
No it is not. The number 3, in the domain, gets mapped to more than one number in the range.
The domain consists of the set {3, 4, 5, 8}
That depends on the specific function.
x y -3 2 -1 6 1 -2 3 5
The range is {-5, -2, 1, 4}
The domain and the range depends on the context. For example, the domain and the range can be the whole of the complex field. Or I could define the domain as {-2, 1, 5} and then the range would be {0, 3, -21}. When either one of the range and domain is defined, the other is implied.
You can define the domain as anything you like and that will determine the range. Or, you can define the range as anything you like and that will determine the domain. For example: domain = {1, 2, 3, 4, ... } then range = {-3, 0, 5, 12, ... } or range = {1, 2, 3, 4, ... } then domain = {sqrt(5), sqrt(6), sqrt(7), sqrt(8), ...}. There is, of course, no need to restrict either set to integers but then it was easier to work out one set from the other.
The domain is (-infinity, infinity) The range is (-3, infinity) and the asymptote is y = -3
Yes. Typical example: y = x2. To avoid comparing infinite sets, restrict the function to integers between -3 and +3. Domain = -3, -2 , ... , 2 , 3. So |Domain| = 7 Range = 0, 1, 4, 9 so |Range| = 4 You have a function that is many-to-one. One consequence is that, without redefining its domain, the function cannot have an inverse.
If this is the whole of the function, then the domain is {2, 1, -3, -1}. That set can be put in increasing order if you wish but that is not necessary.
The domain is {-1, 0, 1, 3}.
The domain is {-1, 0, 2, 4}.
the range for 5 2 1 and 3 is 4.
11