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# C3 functions (domain) Watch

1. I need to find the domain of : x^2 - 5 /2

Can't x be anything?
The exam has got the answer as x>(equal to of greater then) 0
2. It'll probably be that because of the context of the question as they'll probably want you to find the inverse function in a later part of the question and you can't inverse a many to one function
3. Well f^-1 (x) = (x^2 - 5)/2

The domain should be any real number right but they have it as equal to or greater then 0.

http://store.aqa.org.uk/qual/gceasa/...W-QP-JUN09.PDF
http://store.aqa.org.uk/qual/gceasa/...W-MS-JUN09.PDF
2bii
4. Is the function supposed to be one-one?
5. Did you mean q 2b ii?

the domain of f^-1(x) is greater than or equal to 0 because it is the same as the range of f(x)
6. Think about it, the domain of will be the same as the range of
7. (Original post by uxa595)
Well f^-1 (x) = (x^2 - 5)/2

The domain should be any real number right but they have it as equal to or greater then 0.

http://store.aqa.org.uk/qual/gceasa/...W-QP-JUN09.PDF
http://store.aqa.org.uk/qual/gceasa/...W-MS-JUN09.PDF
2cii
If your function were just then your domain would just be , but because we're looking at this as the inverse of another function, this won't necessarily be the case. Here, , which has range .
8. (Original post by nuodai)
If your function were just then your domain would just be , but because we're looking at this as the inverse of another function, this won't necessarily be the case. Here, , which has range .

Yes, thats what i initally got, as the range of f(x) which i just got has to be the domain of the inverse.

But i found it strange as you could have x to be anything if it were on its own.

so, all your doing is reflecting f in y=x, and looking at that part of the curve?
So your only looking at one side of the curve which is a flection of f(x)?
9. (Original post by uxa595)
Yes, thats what i initally got, as the range of f(x) which i just got has to be the domain of the inverse.

But i found it strange as you could have x to be anything if it were on its own.

so, all your doing is reflecting f in y=x, and looking at that part of the curve?
So your only looking at one side of the curve which is a flection of f(x)?
Yup. The idea is that we could define:

These are different functions, despite being "the same", because they are defined on different domains. Because we're looking at the function that "undoes" , and because this only takes non-negative values, we must have that the inverse of is , and not .

The way functions are taught at C3 is a bit weird. When it asks you for the domain of a function, it usually means "the largest possible set of real numbers that can be a domain of" the function. For example, for the function , the domain could be (which is the largest possible real domain), in which case the range would also be ; or it could be , in which case the range would be . You could even define the domain to be , in which case the range is ... but when you're asked the question in C3, it would expect as an answer.

So in this case, although the function can be defined on all the real numbers, because we're looking at it as the inverse of , it doesn't make sense for us to allow negative values of .

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