Question

Is the following relation a function?

Answer 1

if you can draw a vertical line (up and down) that passes through the graph of the function more than once, then it is NOT a function

Answer 2

can you draw such a line?

Answer 3

Or you may notice an \(x\) value that goes with two \(y\) values. Then we again have that it is NOT a function.

Answer 4

yes you can

Answer 5

then it is NOT a function

Answer 6

thank you so much! I appreciate it :)

Answer 7

I have a question here.

Answer 8

np

Answer 9

sup

Answer 10

Also notice @jgirl128 that the two methods a mentioned are exactly the same. Do you see why?

Answer 11

You can ignore this part jgirl.
That technically is a function. I mean it's a function of y, but it's definitely a function. Why do text books call this "not a function" when our calculus books just teach us otherwise?

Answer 12

it is NOT a function

Answer 13

A function of y. I can integrate on it even... Lol

Answer 14

its not a function, you may integrate over non functional relations

Answer 15

go look at the definition of a function

Answer 16

you MAY NOT have an x values that maps to two different y values.

Answer 17

I submitted my work and zzrockers right, it's not a function

Answer 18

here we have x=5 obviously maps to two different y values

Answer 19

I don't think you're hearing my point. It's not a function on the x-axis. But it is on the y-axis.

Answer 20

correct

Answer 21

x is a function of y

Answer 22

it is standard to ask if the function is a function of x

Answer 23

So it is a function y. Instead of your f(x) you got f(y).

Answer 24

yes as long as you designate that the set off y values is the domain and not the codomain

Answer 25

I think that answered my question.

Answer 26

good question:)

Answer 27

\(y=f(x)\) is not a function. \(x=g(y)\) is.

Answer 28

They're setting x as sort of the default domain. Got it.

Answer 29

Can f(y) = x be a function though?

Answer 30

I mean I'm pretty sure that I've seen Stewart do f(y) = x before.

Answer 31

yeah technically the function is the set of ordered pairs (or more technically the set of ordered pairs of representative from an equivalence class) {(a,b),(c,d)....} and when we show the graph we assume the first element in the ordered pair to be the domain

Answer 32

:)

Answer 33

because our notation is \(f:A\rightarrow B\)
so
\((a,b)\) makes sense as standard

Answer 34

sorry, I dont know why it sent that again

Answer 35

the problem with just using f(y) is that f is the rule that gives our relation, and that relation is based on the fact that our domain has been picked

Answer 36

\[
f(y) =x
\]is misleading because typically you expect a function definition. However it is just transitive property of equality.

Answer 37

f(y) actually makes no sense at all if y is in the codomain

Answer 38

Well when you do shells, you typically integrate on the x-axis. The functions that you are working with are COMMONLY placed on the y-axis. Yeah, it sucks. But sometimes it's better to put a function on the y-axis rather than make the integration a living hell with washers. I'm not arguing that f(y) = x is a good form, but a necessary form. Dimensions change, and we have to accommodate. So be it.

Answer 39

again f(y) makes no sense, if you are thinking about "moving the function" you are really just coming up with a new function. they are not the same functions

Answer 40

No, \(x=f(y)\) makes plenty of sense.

Answer 41

a function is defined on its domain which we call x, and its range which we call y is a subset of its codomain (we dont really talk about this untill upper math) So taking something from the range and putting it in the function makes absolutely no sense at all

Answer 42

you cant put elements from the codomain of f into f.

Answer 43

it is not defined on the codomain.....

Answer 44

this is all by definition of a function

Answer 45

The codomain and the domain in this case is all real numbers...

Answer 46

yes in this special case we have y = x for some x

Answer 47

I was talking more general, and why it is dangerous to play that game:)

Answer 48

this would of course not be the case for \(f:\mathbb{R}\rightarrow \mathbb{R}^2, f(x) = (x,x)\)

Answer 49

but again f here is not a function, so I maintain the f(y) makes no sense

Answer 50

im sticking to it:)

Answer 51

First of all, even if we defined \(y=f(x)\), the codomain is still all real numbers. The image is the only thing that isn't all real number.

Second of all, the statement \(x=f(y)\) would be false, because it is asserting an equivalence which is not necessary true. However when someone says \(x=f(y)\) it is likely not in the context of \(y=f(x)\).

Answer 52

also we are assuming the domain is all real numbers when it need not be...again though I was speaking more in general. I normally don't work with R.

Answer 53

In the graph, we see a continuous function whose axis are labeled with real numbers.

Answer 54

lol you made your point, but I still claim, and again will stick to it, that f is NOT a function so the notation f(anything) does not make sense as this is function notation and f is NOT a function

Answer 55

you may define things to work out, but this was not the question...

Answer 56

and of course this graph does not say real numbers anywhere on it.

Answer 57

:)

Answer 58

could be the set of continuous linear functionals.... I dont assume.