One-to-one / onto funtions

Iancar29

Hi guys, me again.

How do i go about checking if the following functions are one-to-one and onto?


f(x) = 5x/(x-4)

of what i know one to one means f(a) must = f(b)?

So filling in then to this eqn i get...

5a/(a-4) = 5b/(b-4)

5ab -20 = 5ab -20

0 = 0 ? ... not 1-1 then?

For onto functions all i know that if it graphs to be a straight line it is an onto function?
BUt how do i figure it out with just the equation.

f(x) = 5x/(x-4)

ANy help much appreciated guys

Michael Collins

Iancar29 wrote: »

Hi guys, me again.

How do i go about checking if the following functions are one-to-one and onto?


f(x) = 5x/(x-4)

of what i know one to one means f(a) must = f(b)?

So filling in then to this eqn i get...

5a/(a-4) = 5b/(b-4)

5ab -20 = 5ab -20

0 = 0 ? ... not 1-1 then?

You're on the right track. What you want to test is: if two different elements in your domain 'a' and 'b', map to the same element in your co-domain i.e. f(a) = f(b), then for the function to be one-to-tone, this must mean a=b, i.e. they were really the same element.


For example, the function above is one-to-one, because each element in the codomain (Y) is mapped to by, at most, one element in the domain (X). (The codomain element, C, isn't mapped to by anything, but that's fine).



This function is not one-to-one, because the codomain element, C, is mapped to by two separate domain elements, 3 and 4.

So, you have started correctly by setting f(a)=f(b) -- now see if you can deduce that this implies a=b.

Important note: to actually answer this equation you need to know what your domain and codomain actually are! If it is not stated in your question you can probably just assume both domains are the set of all real numbers - but really it should be given in the question.

For onto functions all i know that if it graphs to be a straight line it is an onto function?
BUt how do i figure it out with just the equation.

f(x) = 5x/(x-4)

ANy help much appreciated guys

For onto, you need to make sure every element in your codomain is mapped to by at least one element of your domain. Again we MUST know what each domain is, or we can't answer the question.

This can be a little tricky, (not too hard really, and I'll show you if you want), but we can use a handy method here. Since, you will find that the function is actually one-to-one, then if it is also onto, it must be invertible i.e. you can come up with a function that takes each element in the codomain Y back to its original domain element X. (This is not possible in the first diagram above because the codomain element, C, is not mapped to by a domain element (this function is not onto). Nor is it possible in the second diagram, as the codomain element, C, has two domain elements mapping to it - how would we pick an element to map back to?)

Now our function is

f(x) = 5x/(x-4),

so the first step is to set y = f(x). Now, see if you can get x in terms of y.

If we assume the domain and codomain are [latex] \in {\bf R} [/latex] i.e. the reals, then can you see any real number, y, that doesn't give you a corresponding real x? Hint: there is one!

Iancar29

Cheers MC for the great detail. Yes, btw it is of the set of real numbers.

So for the One-to-one relationship i got

5ab - 20 = 5ab -20
5ab = 5ab
ab = ab
a = b therefore one-to-one


Then for onto

y= 5x/x-4

yx -4y = 5x

-4y = 5x -yx

-4y = x(5-y)

-4y/5-y = x

Kinda stuck from there ... ?

@ y = 5

20/0 = x ? .... hope i didnt just blow up the maths forum by trying to divide by 0 ! lol

TheBody

Iancar29 wrote: »

Cheers MC for the great detail. Yes, btw it is of the set of real numbers.

So for the One-to-one relationship i got

5ab - 20 = 5ab -20
5ab = 5ab
ab = ab
a = b therefore one-to-one


Then for onto

y= 5x/x-4

yx -4y = 5x

-4y = 5x -yx

-4y = x(5-y)

-4y/5-y = x

Kinda stuck from there ... ?

@ y = 5

20/0 = x ? .... hope i didnt just blow up the maths forum by trying to divide by 0 ! lol

Yes, 5 is a value not in the range of f(x). So f(x) is not onto because you can't get the value of 5 from the function. It might be worth your while to plot the function to see the asymptotes. Just lash it into some graphing sortware like GeoGebra to see what it looks like.

Informally, a function from X to Y is one to one if to each y value in the range there corresponds exactly one x value in the domain.
A function from X to Y is onto if its range consists of all of Y.

Michael Collins

TheBody wrote: »

...Informally, a function from X to Y is one to one if to each y value in the range there corresponds exactly one x value in the domain...

Just a slight issue here. This description perhaps implies that the function given in the first diagram in my post above would not be one-to-one since C has no element of the domain mapping to it. (It's actually the description of a bijective function i.e. both one-to-one (injective) and onto (surjective), so it's technically not wrong I suppose!) So I'd modify your sentance slightly:

Informally, a function from X to Y is one to one if to each y value in the range there corresponds at most one x value in the domain

Iancar29

Cheers guys , much appreciated! ... Its hard getting to grips with it all when the question can be asked either in function, diagram or graph form.

MathsManiac

I think the point MC first made needs to be made even more strongly: you cannot answer the question without being told the domain and codomain.

You cannot assume the domain is all of R, because it can't be all of R. Since the given rule does not assign any value to 4, it is clear that 4 is not in the domain of the function. You could assume the domain is R\{4}. But if you're going to assume that, without being told, why can you not assume that the codomain is R\{5}? In that case, the function is surjective (onto).

If the domain is not going to be given to you in the question, then you need to have some clear rule stated in advance for what domain you are supposed to assume. For example, you might be told: "In all of the following, assume the domain is the set of all real numbers for which the given rule gives a well-defined answer."

Michael Collins

MathsManiac wrote: »

...You cannot assume the domain is all of R, because it can't be all of R. Since the given rule does not assign any value to 4, it is clear that 4 is not in the domain of the function. You could assume the domain is R\{4}. But if you're going to assume that, without being told, why can you not assume that the codomain is R\{5}? In that case, the function is surjective (onto).
...

Yes indeed! Given that these concepts are quite technical you'd hope that the question is stated equally rigorously to avoid such confusion.

Iancar29 wrote: »

Cheers MC for the great detail. Yes, btw it is of the set of real numbers.

So for the One-to-one relationship i got

5ab - 20 = 5ab -20
5ab = 5ab
ab = ab
a = b* therefore one-to-one
...

I missed an error in your deduction here. The first equation should be

5ab - 20b = 5ab - 20a

so you get a = b.

The * equation you have above does not follow from your previous line!