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Homework Help: Are the following functions? Give domain and range

  1. Mar 21, 2008 #1
    1. The problem statement, all variables and given/known data

    State if the following is a function. If yes, write domain and range.

    A. {(x,y): y=(x^1/2) ; x,y are real numbers and x>0}
    B. {(x,y): y^2=x ; x is real numbers and x>0}

    2. Relevant equations

    A relation is called function if each element of the domain necessarily has a unique (one and only one) image in th range.

    3. The attempt at a solution

    My book says, A is a function.
    Domain= Set of positive real numbers
    Range= R+ - {0}

    But I think that A is not a function as for say x=25, y=+5 or -5. 2 images, isnt it??? Then how can it be a function.

    As about B my book and me both have the same opinion. B is not a function.

    Pleas help me to figure out if I am right!
  2. jcsd
  3. Mar 21, 2008 #2
    Disagree. A is a function. 25^(1/2) is 5, if indeed a really is y=x^(1/2).

    Obviously, y^2=x is not a function, because x will have two values in that case (graph y = sqrt(x) and y1=-(sqrt(x)).
  4. Mar 21, 2008 #3
    You're thinking of the other way around. As Orbital Power said, the + or - square root idea applies to when you have one term squared and then you take the square root of the other side. It doesn't necessarily go the other way in this case, though. What you are given is ONLY the positive square root. Every positive number you put into a positive square root is going to be just that. Whereas if you start by taking the square of each side, you have to do the plus or minus square root idea. Why? Well, think about part B. Using your example, both y=-5 and y=5 will result in the same x value. But when you take the positive square root of a number, can you truly come up with a negative? There is no negative outside of the root to change the value and having a negative under a square root results in... funky things that aren't simply negative numbers...

    If it helps, graph it on your calculator. You'll see that for all positive values of x, a positive value of y corresponds to it.
    Now, for the second part, it goes back to the idea of starting with y^2 and getting the graph of sqrt(x) and -sqrt(x) together. You already know that, though =) Just remember it doesn't go both ways, so to say.
    Last edited: Mar 21, 2008
  5. Mar 24, 2008 #4
    Ritwik06 makes a valid point. By convention,when the square root of a variable/number is asked, we take the positive value. But the root of 25 IS both 5 and -5. I would say that, ignoring convention, y=x^(1/2) is not a function. But since the textbook almost certainly means (by convention) the positive square root function, y=sqrt x is a fcn.

    Go with what Seichan says, but now you know the deal behind it all... ;)
  6. Mar 24, 2008 #5
    Now I really wonder if 'B' really is a function or not.... It doesn't seems to be so??!!

    I think Ritwick06 has a right problem!!!

    Quote from Seichan:
    If it helps, graph it on your calculator. You'll see that for all positive values of x, a positive value of y corresponds to it.

    Why should it be so?
  7. Mar 25, 2008 #6


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    The problem should have been phrased "is y a function of x". In B, y is not a function of x but x is a function of y.
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