Chapter 1 - 5 HELP!

[Rosie]

Can someone please solve this question for me.
If ax³ + x² + 6 is exactly divisible by x + 1, the value of a is:

[ed_saifa]

Since you know that is a factor of then we know that
Therefore

[Rosie]

Does anyone know how to change your topic name. Can you do that?

[Mao]

go to the first post (by you) and click "modify"

[Rosie]

Passage in my textbook:
The definition of a function tells us that for each x in the domain of f there is a unique element,y, in the range such that (x,y) f. The element y is called the image of x under f or the value of f at x and is denoted by f(x). If (x,y) f, then x is called a pre-image of y.      

I'm preparing my reference book for methods and this paragraph makes no sense to me. Can anyone explain it to me?

What does image and pre-image mean. (is image the x values and pre-image the y values)

I also wanted to ask what the codomain is?

[Mao]

mmm nice question:

1. i have never heard of "image" or "pre-image" used in the sense of the cartesian plane.

2. don't worry about the codomain. For the purpose of methods, just accept that for the functions you will be dealing with, the codomain will be R.
but if you do want a definition, it is the plane at which the values of the function exist on.

[kj_]

the words pre- image and image are just fancy terms for subbing a value in as f(x) and subbing a value in as x. i can't remember which one is which, though - but i've never seen them in any question for 3/4 methods, so unless you were referring to 1/2 methods, don't worry about it at all

edit: just went through my 1/2 book, and to clarify,

preimage, say, f(x) = x +1, the preimage of 7, let 7 = f(x) and solve for x
image, say, f(x) = x + 1, the image of 1, let 1 = x, and solve for f(x)

you should be able to try and put a literal meaning to the words "pre-image" and "image" now

[/0]

Put simply:



E.g Consider

The bit will be the domain, and will be the codomain. As you can see from observing the graph, the range is actually

[Rosie]

I understand now but it is in my 3/4 methods book. There's actually one or two questions about finding the pre-image. Thanks for that

Q. For {(2,y):y Z} Is this a function and state the domain and range.
Q. For y 3x+2 Is this a function and state the domain and range.
How would you go about this. Thanks

[/0]

Q.1

You could rephrase it as: the points (2,y) such that y is an integer, ...(2, -1), (2, 0), (2, 1), (2, 2)...

So the domain will be {2} and the range will be Z.

Q.2

isn't a function because for each x coordinate there is more than one y coordinate. The domain and range will both be R because if u draw the graph the required region will go from to in both the x and y directions.