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Graph Question

  1. Jul 26, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]y= \frac{x + \frac{1}{x}+1}{x+\frac{1}{x}+2}[/tex]

    That is equal to [tex]y = \frac{x^2+x+1}{(x+1)^2}[/tex]

    Whenever I graph either one of those they are the same graph (as well they should be) and they have a value for x=0. But why should x = 0 be in the domain? And also if you go by the second equation then x = 0 appears that it should be in the domain even though it shouldn't?

    If it shouldn't be in the domain, then whenever you see an equation for a graph, how do you know whether or not it stemmed from another equation where 0 shouldn't be in the domain?
  2. jcsd
  3. Jul 26, 2008 #2


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    Staff Emeritus
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    They are the same for all x except x= 0. The first is not defined at x= 0: x= 0 is not in its domain, as you say. x= 0 is in the domain of the second function.

    When you graph them you should NOT get the same graph- but they differ ONLY at x= 0 so your caculator (I imagine that is how you are graphing) can't tell the difference- it's "skipping over" x= 0 when it is doing the calculation. If you graphed them by hand, the graph of the first should show a "hole" one point wide at x= 0 which the other does not.

    The second function is continuous at x= 0, the first has a "removable" discontinuity at x= 0. You would find that the limit, as x goes to 0, of the first is exactly the same as the limit of the first, 1, but it is not continuous because the function itself is not defined there. Of course, you cannot "see" a single point but it is a good idea to draw a small open circle at a missing point. As for telling what the function was from the graph- you can't in general. Even if there were no missing points, you are only seeing a small portion of the graph. You have to have some other information (or assumptions) to be able to deduce the function from the graph.

    That is something that probably is not emphasized as much as it should be:
    (x2-1)/(x-1)= (x-1)(x+1)/(x-1) is NOT exactly the same as x-1: they are the same for all x except x= 1. When x= 1, x-1= 0 and you cannot cancel the two "x- 1"s because 0/0 is NOT equal to 1.
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