I am having trouble with codomain.
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i forgot it's -x^2.... what about codomain ?Your domain isn't correct. Think again. Try x=-2 or 3. Does it work?
What is the answer for the domain first?i forgot it's -x^2.... what about codomain ?
[-1,2]What is the answer for the domain first?
Correct. You've learned about finding the max/min of a function using derivative.[-1,2]
3/2Correct. You've learned about finding the max/min of a function using derivative.
Find max/min of [imath]y=\sqrt{-x^2+x+2}, x\in[-1,2][/imath]
What is 3/2?
oh right, max is 3/2, min is 0What is 3/2?
When finding relative max/min of bounded functions, you'd also need to consider their endpoints i.e. x=-1 and x=2
So the co-domain is.....?oh right, max is 3/2, min is 0
[0,3/2] i assume...So the co-domain is.....?
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][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)??
okay! thanks!!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]
I am having trouble with codomain.
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[0,3/2] i assume...
this, but I don't really get it.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.
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).this, but I don't really get it.
In mathematics, the codomainor domain of the valueof 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.
how can we have two RsBased 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!
Why not? Suppose f(x) = x. Can't the domain = R while the codomain also equals R.how can we have two Rs
okay, then whatWhy 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