Question 1022162: Which is not a function?
(x,1)(z,w)(w,z)
(x,1)(y,1)(w,1)
(x,y)(y,y)(w,y)
All
None
plz explain too
Found 2 solutions by Theo, Edwin McCravy:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the rule for whether a relation is a function is:
you can have one and only distinct value of the dependent variable for each and every distinct value of the independent variable.
the point pairs (1,2) and (1,3) do not represent a function because you have more than 1 distinct value of the dependent variable (2 and 3), for each and every distinct value of the independent variable (1).
the point pairs (2,1) and (3,1) do represent a function because you have one and only one distinct value of the dependent variable (1) for each and every distinct value of the independent variable (2 and 3).
here they're using variable names instead of numbers.
the thing to remember here is that the same variable name points to the same value while a different variable name points to a different value.
therefore (x,y) and (x,z) would not be a function, while (y,x) and z,x) would be a function.
(x,y) and (x,z) would not be a function because multiple values of the dependent variable (represented by y and z) exist for the same value of the independent variable (represented by x).
(y,x)and (z,x) are functions because, while you have the same value for the dependent variable (represented by x), that same value of x does not correspond to the same value of the independent variable (represented by y and z).
in other words:
you can have y point to x and z point to x and you're ok - it's still a function, but you can't have x point to y and x also point so z - it's not a function in that case.
all of your sets of points are functions because it cannot be proven that they're not, since none of them has multiple values of the dependent variable points to by the same value of the independent variable.
let's look at them in turn.
(x,1)(z,w)(w,z)
x points to 1
z points to w
w points to z.
since the value of each of the independent variables are different (x,z,w), there are no instances of the same value of the independent variable pointing to multiple values of the dependent variable.
(x,1)(y,1)(w,1)
same deal.
x,y,w all point to 1.
since the value of each of the independent variables are different (x,y,w), there are no instances of the same value of the independent variable pointing to multiple values of the dependent variable.
(x,y)(y,y)(w,y)
sane deal again.
since the value of each of the independent variables are different (x,y,w), there are no instances of the same value of the independent variable pointing to multiple values of the dependent variable.
all of the sets of points pass the function so they're all functions.
your answer should be none.
here's some examples of where the test is not passed.
(x,1)(x,w)(w,z)
(x,1)(y,1)(y,3)
(w,y)(y,y)(w,z)
in the first case, x points to 1 and w.
in the second case y points to 1 and 3.
in the third case w points to y and z.
the independent variable is the first element in the point pair.
the dependent variable is the second element in the point pair.
my convention is that the value of the independent variable points to the value of the dependent variable. this convention is followed in all of the above.
Answer by Edwin McCravy(20054)  (Show Source):
You can put this solution on YOUR website!
To put it is more concisely:
To have a function,
The FIRST coordinates must ALL be DIFFERENT!
(It doesn't matter if some of the SECOND coordinates
are the same, or if they are the same as some of the
FIRST coordinates. We go only by the FIRST coordinates)
------------------------------------------
{(x,1)(z,w)(w,z)} IS a function because the 3 FIRST coordinates
x, z and w are all different.
{(x,1)(y,1)(w,1)} IS a function because the 3 FIRST coordinates
x, y and w are all different. It doesn't matter that the SECOND
coordinates are the same.
{(x,y)(y,y)(w,y)} IS a function because the 3 FIRST coordinates
x, y and w are all different. Again, it doesn't matter that the
SECOND coordinates are the same.
Edwin
|
|
|
