Question

Determine which relation is a function.

A.
{(–4, 3), (–2, 3), (–1, 2), (2, 5), (3, 2)}

B.
{(–4, 1), (–2, 3), (–2, 1), (–1, 5), (3, 2)}

C.
{(–4, 1), (–2, 3), (–1, 2), (3, 5), (3, 2)}

D.
{(–4, 1), (–2, 3), (–1, 1), (–1, 5), (3, 2)}

Answer 1

I'm thinking its C? but I'm not 100% sure

Answer 2

To answer this you need to remember that by definition, a function must have exactly one output value for every input value. So which set of points does not have multiple outputs for the same input?

Answer 3

Problem with C is (3,5) and (3,2) will make it NOT a function.
A function must have only one image (y) for each value of the domain (x).
If I want to know f(3), I would not know f(3)=5 or f(3)=2 for the relation (C).

Answer 4

Oh ok @mathmate @blacksteel lemme work this out and I'll see what I come up with.

Answer 5

@mathmate @Blacksteel Is the right answer D?

Answer 6

If D were a function, what would be the value of f(-1)?

Answer 7

It's A because the domain doesn't repeat.

Answer 8

The domain could repeat, as long as it is mapped to the same image, then it is still a function. It's just that there is a repeated ordered pair.
Yes, A is the correct answer. Well done! :)

Answer 9

I hate to prove you wrong, but:

You're wrong XD, sorry.

Answer 10

This is not a proof, just someone's simplified opinion.
What's important is not what appears in the domain, but what's in the image.
Nothing says you cannot repeat an ordered pair. (we cannot repeat in a set).
(5,1),(6,1),(7,1),(5,1) is a perfectly valid function.
If you put them in a set, as in the case above, then it cannot repeat, for example,
{(5,1),(6,1),(7,1),(5,1)} is not a valid set, so yes, a valid set won't repeat.

Answer 11

@Maria195 by the way, it is a very good idea to challenge answers, this is what make mathematics skills grow by having a thorough understanding.

Answer 12

Yeah, but I just learned about this in math class a month ago. I was pretty sure the domain can't repeat, but the range can.

And you're right about that :)

Answer 13

@maria195
Instead of memorizing rules about domain and range (which can be confusing under different contexts), you just have to understand that a function must give a unique image for any given value of the domain.
(5,3),(6,4) is a function because f(5)=3, f(6)=4, no ambiguities.
(5,3),(6,3),(6,4) is not a function, because f(5)=3, f(6)=3 or f(6)=4? the value is not unique, therefore it is not a function.
On the other hand, if we have
(5,3),(5,3),(6,4),(7,5) we have f(5)=3,f(6)=4,f(7)=5 without ambiguity, it is still a function.
If the ordered pairs are put in a set, then you will not see (5,3) repeating, because sets cannot have repeating elements. (a list of unique elements enclosed in braces {} is a set.)
{(5,3),(5,3),(6,4),(7,5)} is an invalid set because of the repeated ordered pair, and it should not appear in a question.
IF ordered pairs are enclosed in braces (thus forming a set), then whenever the domain repeats, it is not a function, because the images are necessarily different.
But if the ordered pairs are not enclosed in braces, or if they form union of two sets, you can see repeating ordered pairs (thus "repeated domain", but with identical image), it can still represent a function. For example:
\(\{(5,3),(6,4)\}\cup \{(7,5),(5,3)\}\) is a function, but
\(\{(5,3),(6,4)\}\cup \{(7,5),(5,4)\}\) is not, because we don't know what f(5) is.
Or
(5,4),(6,5),(7,6),(5,4) can be used to form a function, because there is no ambiguity.

In short, do not depend solely on the repetition of the domain to decide if the relation is a function.
Figure out if the relation gives a \(unique\) value of image for \(every\) value of the domain.