A tuff question...
Author
Hi,
I m reading a text-book on differential calculus and i have a doubt regarding a solved example given in the book.
The question given is:
A function R->R (from R to R) where R is the set of all real numbers is defined by:
f(x) = {A(x^2) + 6x - 8} / {A + 6x - 8(x^2)}
Find integral values of ''A'' for which the function ''f'' is onto.
The answer for A turns out to be:
A >= 2 and A <= 14
My doubt:
If A is substituted in f(x) as 2, then f(x) is not defined for x = 1 and x = -1/4, in which case f(x) ceases to be a function (because f(x) is not defined for all the elements of the domain viz. set of all real numbers).
Can anybody explain this? Would typing in the solution given in the book help?
Thank you!
Obviously f(x) isn''t defined for all x in R, but it is still ''onto'' since for any y in R there exist some x such that y=f(x).
Author
Well, the book defines a function as "a relation from one set (called domain) to another (called co-domain) such that EACH element of the domain is related to exactly one element of the co-domain".
Any ideas?
Any ideas?
The only case I can see where f(x) is not a function from R to R is when the denominator has one or more roots. Thus find the A''s so that no real root exists (the discriminant is less than 0). But I can''t get the values you gave us.
quote:Original post by pacman2003
Well, the book defines a function as "a relation from one set (called domain) to another (called co-domain) such that EACH element of the domain is related to exactly one element of the co-domain".
True, but the domain need not be the entire set of real numbers. It could be any subset, like in this case. The same is true for the co-domain (or "range" as I thought it was called in english).
This is dangerously close to a homework question, may actually be a homework question. The discussion so far is okay, but please keep this at a general discussion level and do not provide a derived answer.
Review the Forum FAQ for our policy on homework!
Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.
[edited by - grhodes_at_work on January 25, 2004 11:18:38 AM]
Review the Forum FAQ for our policy on homework!
Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.
[edited by - grhodes_at_work on January 25, 2004 11:18:38 AM]
pacman
erhaps ud be better off asking ur question on a forum specially dedicated to math. this one is for math and physics related only to game development
erhaps ud be better off asking ur question on a forum specially dedicated to math. this one is for math and physics related only to game development
quote:Original post by Anonymous Poster
True, but the domain need not be the entire set of real numbers. It could be any subset, like in this case. The same is true for the co-domain (or "range" as I thought it was called in english).
The co-domain and range are not the same thing. The codomain is the set onto which you are mapping, and the range is the actual set of points which are mapped onto, e.g. for f(x) = x*x, the domain could be all the real numbers and the codomain all the real numbers, but the range would only be all the positive real numbers and zero.
Author
(No grhodes, this is not a homework question. I do not need the derived answer because its already given in the book - this is a SOLVED example.)
If A = 2 is substituted in F[x], we have
f[x] = (2x^2 + 6x - 8)/(-8x^2 + 6x + 2)
= (x+4)(x-1)/((4x+1)(x-1))
and it is actually improper to cancel the factor of (x-1) from the numerator and denominator, because for x = 1, we have F[1] = 0/0 which is undefined. Thus there is a gap in F at x = 1, and F is not onto, since there is no value of x such that F[x] = -1.
There is a similar gap at x = -1 when A = 14.
So I think that the given answer IS wrong in this respect: it should be 2 < A < 14 and NOT 2 <= A <= 14.
If A = 2 is substituted in F[x], we have
f[x] = (2x^2 + 6x - 8)/(-8x^2 + 6x + 2)
= (x+4)(x-1)/((4x+1)(x-1))
and it is actually improper to cancel the factor of (x-1) from the numerator and denominator, because for x = 1, we have F[1] = 0/0 which is undefined. Thus there is a gap in F at x = 1, and F is not onto, since there is no value of x such that F[x] = -1.
There is a similar gap at x = -1 when A = 14.
So I think that the given answer IS wrong in this respect: it should be 2 < A < 14 and NOT 2 <= A <= 14.
I have to agree with the original poster that a function that is not defined for some real number is not a function R->R. You can call it a function from a subset of R to R. It is a common abuse of language to call those things "functions R->R", when technically they are not.
[edited by - Alvaro on January 27, 2004 3:15:07 PM]
[edited by - Alvaro on January 27, 2004 3:15:07 PM]
This topic is closed to new replies.
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