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Homework Help: Function and domain

  1. Mar 17, 2010 #1
    hi ,,,,

    1. The problem statement, all variables and given/known data

    these are the function

    1) f(x)=A(x)/B(x)

    2) f(x)=√(A(x))

    3) f(x)=√(A(x))/(B(x))

    4) f(x)=(A(x))/√(B(x))

    5) f(x)=√(A(x))/√(B(x))

    6) f(x)=√((A(x))/(B(x)))

    3. The attempt at a solution

    the domain are they right

    1) { x∈ R : B(x) ≠ 0 }

    2) { x∈ R : A(x) ≥ 0 }

    well the other I don't know how to makes them ,,, can you give me some infos ,, i dont want the domains because i want to make them to learn and don't forget them later ,,, :rolleyes:

    i dont know how to have the domains because in the lecture the doctor give a example on potato so i got confused :confused::confused:
  2. jcsd
  3. Mar 17, 2010 #2


    User Avatar
    Homework Helper

    How can you use 2) to find the domain on 3)?
  4. Mar 17, 2010 #3
    The one point about this problem is to realize that in order for x to be an element of the real numbers, the radicand (or the stuff under the square root) cannot be negative, so your objective is to find a domain where there are no imaginary numbers and one would not have to divide by zero, which would make the function undefined...for the first one, x [tex]\in[/tex] [tex]\Re[/tex] means that x is an element of the real number set, and B(x) cannot equal zero because this would make the denominator equal to zero and therefore make the function undefined (that one is easy though). The main problems are going to occur when you have fractions or square roots (this is true for every domain, not just the ones shown).

    So as another example...for 4), we have two problems with B(x) and no problems with A(x), so first of all, B(x) cannot be negative because it is within a square root (in order for x to be an element of the real numbers), so we now have B(x) [tex]\geq[/tex] 0, but it is also in the denominator, and the square root of zero equals zero, so B(x) cannot equal zero...so therefore in the "greater than or equal to" sign, we can eliminate the "or equal to" and simply have B(x) > 0...so our final domain is {x [tex]\in[/tex] [tex]\Re[/tex]: B(x) > 0} which in English means that, for x to be an element of the real numbers, B(x) must be greater than 0, or in other words positive.
  5. Mar 17, 2010 #4
    3) {x∈ R : A(x) ≥ 0 and B(x) ≠ 0 }
    4) {x∈ R : A(x) ≠ 0 and B(x) > 0 }
    5) {x∈ R : A(x) ≥ 0 and B(x) > 0 }
    6) maybe like the " 5 " but am not sure ,,,

    are they right ? or need some changes ?
  6. Mar 17, 2010 #5
    number four is incorrect because A(x) can equal zero...I don't know if I mentioned that, but there are never problem spots in the numerator unless there is a radical in the numerator...in this case, f(x) = A(x)/sqrt(B(x)) so there is no problems in the numerator...the only problem is in the denominator because B(x) is both in a radical and on the bottom of the fraction, so it cannot equal zero and must be positive, or B(x) > 0, which you had...you just have to get rid of A(x) ≠ 0... so the correct domain would be {x∈ R : B(x) > 0}
    Number six is exactly like five, in fact it is a duplicate...we are simply combining the numerator and denominator into one radical, which has the exact same domain, since the functions are equivalent.
  7. Mar 18, 2010 #6
    6) {x∈ R : A(x) ≥ 0 and B(x) > 0 } ?

    but our doctor said it's not the same,,,,
  8. Mar 18, 2010 #7
    Well, essentially what he wants you to say, is that the ratio of A(x) to B(x) must be greater than zero, rather than each individual subfunction...so basically, you could rewrite the domain as {x∈ R : A(x)/B(x) ≥ 0 and B(x) > 0 }, but technically, the other domain that we already wrote is correct because the only way for the ratio of A(x) to B(x) to be greater than or equal to zero is dependent on the numerator, which is A(x)...so A(x) ≥ 0 is really true...I don't know how else one might write it, unless I misunderstood the function notation, is it the fraction of A(x) over B(x) all within the radical?...if it is, then what I said should be true, unless your professor has some trick he wants you to pick up on...if its not, then let me know what you meant...
  9. Mar 18, 2010 #8
    Is it... [tex]\sqrt{\frac{A(x)}{B(x)}}[/tex] ?????
  10. Mar 20, 2010 #9
    ya it's ,,,,,

    both of them in one root ,,,,

    tomorrow I have class ,,, I will ask him and will see the answer ,,,,,
  11. Mar 21, 2010 #10
    well he told me it's {x∈ R : √((A(x))/(B(x))) ≥ 0 }

    i dont know if it's right or no ,,,, can someone check it for me ,,,,
  12. Mar 21, 2010 #11


    Staff: Mentor

    Yes, but that's not where you should quit. For the radical to be nonnegative, B(x) != 0, and A(x)/B(x) > 0. The latter condition means that A(x) and B(x) are the same sign: both positive or both negative.
  13. Mar 21, 2010 #12
    {x∈ R : A(x)/B(x) ≥ 0 }

    no need for the root ,,, now right ..? that's what he say ,,,
    sorry about the first one i wrote it fast so, i wasn't see what i was typing
  14. Mar 21, 2010 #13
    B(x) cannot equal zero, it is in the denominator, so that is definitely not right...and I already told you that A(x)/B(x) ≥ 0...it can equal zero as well...but if you multiply both sides of the inequality by B(x) (which you can do), you find that A(x) ≥ 0 which is the correct domain because A(x) is the only thing that makes the function zero, but B(x) must also be positive (NOT EQUAL TO ZERO) in order to make the ratio positive, so B(x) > 0.
  15. Mar 21, 2010 #14
    The main point is that there are actually multiple ways to write the domain, as long as they are consistent with the fact that everything under the root must be nonnegative and the denominator of the fraction cannot ever be zero...in order to ensure that, your professor has given you an accurate domain of A(x)/B(x) ≥ 0 ....that is definitely correct because what he is saying is that the ratio CAN EQUAL ZERO, which is true, and the ratio MUST BE NONNEGATIVE in order for x to be in the set of real numbers...
  16. Mar 21, 2010 #15


    Staff: Mentor

    Your logic is faulty. If A(x) < 0 and B(x) < 0, then A(x)/B(x) > 0.

    If A(x)/B(x) >= 0, then multiplying by B(x) does not necessarily imply that A(x) >= 0. Remember that multiplying an inequality by a negative number reverses the direction of the inequality.
  17. Mar 22, 2010 #16
    aha ,,, but yet I didn't understood you mark ,,,

    you want me to find new domain ?
  18. Mar 22, 2010 #17


    Staff: Mentor

    No, what I'm saying is that there are a couple of ways of describing the domain of f(x) = sqrt(A(x)/B(x)). Your teacher gave you a starting point -- {x | sqrt(A(x)/B(x)) >= 0}. That describes the domain, but doesn't give you any explicit conditions on the functions inside the radical. Your teacher probably wanted you to start with this inequality, and give those conditions on A(x) and B(x).

    The conditions I gave earlier are:
    1. B(x) != 0
    2. A(x) and B(x) both positive or A(x) and B(x) both negative.

    This is the principal difference between the function of this problem and f(x) = sqrt(A(x))/sqrt(B(x)). For that function, you must have B(x) > 0 and A(x) >= 0.
  19. Mar 22, 2010 #18
    aha ,,, now I see what u are meaning
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