Domain and codomain

Loki123 said:
[0,3/2] i assume...
can you just tell me when would it be (0,3/2), if the interval of x values was (a,b)??
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Yes, since your function is not defined at a and b in that case. But in the actual problem, you made sure that the function is defined by setting [imath]-x^2+x+2\ge0[/imath]
 
Dr.Peterson said:
As I understand it, what you've found is the range, not (necessarily) the codomain, which is commonly a superset of the range..

What definition have you been given for "codomain"? Your definition may differ from mine.
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this, but I don't really get it.
In mathematics, the codomain or domain of the value of the function f: X → Y is the set Y. The domain of the function f is the set X. The image of the function f is the set f (X) defined by {f (x): x ∈ X}. It follows from these definitions that the image of the function f is always a subset of the codomains of f.
 
Loki123 said:
this, but I don't really get it.
In mathematics, the codomain or domain of the value of the function f: X → Y is the set Y. The domain of the function f is the set X. The image of the function f is the set f (X) defined by {f (x): x ∈ X}. It follows from these definitions that the image of the function f is always a subset of the codomains of f.
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That is the definition I know. But it implies that you need to be told what the codomain is; it is the set from which values of f(x) may be taken, but not necessarily only the actual values that f(x) takes (which is the range, or image).

For that reason, I don't think the question should even be asked!

If it is, then it is reasonable to give either the range, or any set containing it, as the codomain. So your new answer, [imath][o,\frac{3}{2})[/imath] is a possible one, but so is your original answer, [imath][0,+\infty)[/imath], because that is the set from which potential values of a square root are taken. It is possible that you have been taught to do that.
 
Based on your definition above:

You seem to know what a domain is. Great!

Let f:x->y. The set y is always the codomain by definition. You can't argue or not understand this, as it is simply by definition.

Now the range will only include values in the codomain--maybe no values, maybe some and maybe all.

For example, suppose f: R->R and f(x)=x^2

Since we can square any number in R, the domain is R.
The codomain is GIVEN as R.
The range however is [0, oo)

Is this now clear?

For the record, in your stated problem there was no codomain listed!
 
Steven G said:
Based on your definition above:

You seem to know what a domain is. Great!

Let f:x->y. The set y is always the codomain by definition. You can't argue or not understand this, as it is simply by definition.

Now the range will only include values in the codomain--maybe no values, maybe some and maybe all.

For example, suppose f: R->R and f(x)=x^2

Since we can square any number in R, the domain is R.
The codomain is GIVEN as R.
The range however is [0, oo)

Is this now clear?

For the record, in your stated problem there was no codomain listed!
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how can we have two Rs
 
Loki123 said:
how can we have two Rs
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Why not? Suppose f(x) = x. Can't the domain = R while the codomain also equals R.

The domain, as you know, is the set of x values that give back a correspondent f(x) value.
Many functions, just think polynomials, have R as its domain. The codomain can be any set, including R, as long as this set contains the range.

Remember that a function does not need to be onto. That is given a codomain of Y, there does not have to be any x value such that f of this value equals y in Y. That is while the codomain includes the range = Y, it can be larger than Y
 
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Steven G said:
Why not? Suppose f(x) = x. Can't the domain = R while the codomain also equals R.

The domain, as you know, is the set of x values that give back a correspondent f(x) value.
Many functions, just think polynomials, have R as its domain. The codomain can be any set, including R, as long as this set contains the range.

Remember that a function does not need to be onto. That is given a codomain of Y, there does not have to be any x value such that f of this value equals y in Y. That is while the codomain includes the range = Y, it can be larger than Y
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okay, then what
 
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