That depends on the specific function.
x y -3 2 -1 6 1 -2 3 5
The domain is {-1, 0, 1, 3}.
The domain is {-1, 0, 2, 4}.
the range for 5 2 1 and 3 is 4.
There are 24 possible functions: One of these is f(0) = 2 f(0.5) = 4.5 f(2) = 0.5 f(3) = 0 The four numbers in the range can be placed opposite the domain in any order.
x y -3 2 -1 6 1 -2 3 5
The range could be anything. Without parameters specified, the domain of {1,2,3,4} could have any range. This problem is unsolvable.
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.
The domain includes the numbers that serve as the input to a function. In contrast the range are the outputs which corresponds to the domain. for example in the equation f(x)=(2-x)^2 the domain is given by the numbers you choose to substitute in for x and the range is given by the values of "f" that correspond to those substitutions. when you graph the function "over all real numbers" you are choosing the domain to go from negative infinity to positive infinity. If i choose the domain to be all integers from 0 to 4 then we would have the following: f(0)=(2-0)^2=4 f(1)=(2-1)^2=1 f(2)=(2-2)^2=0 f(3)=(2-3)^2=1 f(4)=(2-4)^2=4 so we chose the domain to be (0,1,2,3,4) and that gave us the following range (4,1,0) If our function was instead something like this: f(x,y)=x+y then we have two input variables, x and y. so this means the domain must be specified for each variable just like it was specified for the single variable in the previous example. here is an example: lets choose our range for this same function, f(x,y)=x+y, to be x: {0,1,2} and y: {0,1,2}. then we would have the following f(0,0)=0+0=0 f(0,1)=0+1=1 f(0,2)=0+2=2 f(1,0)=1+0=1 f(1,1)=1+1=2 f(1,2)=1+2=3 f(2,0)=2+0=2 f(2,1)=2+1=3 f(2,2)=2+2=4 so the corresponding range for that domain would be (0,1,2,3,1) You can see how complex things get with multiple variables as inputs to a function. But what happens when we assign an interval to each x and y domain? In this case since the domain is more complex it is not given by a single interval (e.g x goes from 0 to infinity) but rather by two intervals which comprise a region. As an example of this lets choose our domain to be all of the points (x,y) that would make up the region inside of a circle of radius 1 around the origin of a Cartesian coordinate system. An easier way to specify this domain is by saying: "the domain is given by all of the points where x^2+y^2<1" this is because that is the equation for a disc centered at the origin. so how do we find the range for this domain? It would take an infinite amount of time for me to substitute random values in for x and y as i have done before. Another more practical way to visualize the range is to use a coordinate system with 3 axes, an x-axis, a y-axis, and a range-axis. the range can be determined by graphing the function, f(x,y)=x+y, on this system of coordinates but only including those values that lie over the circular region (our chosen domain) on the xy-plane of the coordinate system. What happens if we extend this domain and range analysis to functions of not one, not two, but three variables. if our function is something like, f(x,y,z)=x+y+z then we would not be able to graph the range on a three dimensional, x, y, z coordinate system. If time is one of our domain variables in a 3 dimensional function, f(x,y,t) then we can graph the range on a coordinate system with an x-axis, y-axis, and Range-axis. the time time domain would not be given by a spacial interval but a temporal interval. so the graph would begin building itself at time t=0 and the range would be swept out as its shadow hovers over the x,y domain.