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Functions problem help needed

  1. Sep 12, 2013 #1
    1. The problem statement, all variables and given/known data
    Two students are arguing over the following equations. One student claims that the graphs are the same, the other student claims they are different. The equations are as follows:


    2. Relevant equations
    y = x - 1 and y = x^2 - 1 / x + 1 (x + 1 is the divisor)

    3. The attempt at a solution
    I believe the graphs are different, but an algebraic solution must be included. Please help solving this problem algebraically. Any help is greatly appreciated. Thank you.
     
  2. jcsd
  3. Sep 12, 2013 #2

    eumyang

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    Well, they are different. You'll have to simplify the 2nd function, but in doing so, an assumption has to be made. What is that assumption?
     
  4. Sep 12, 2013 #3
    Well, when the second equation is simplified: x - 1 is left over, correct? I'm not sure what exact assumption you are referring to.
     
  5. Sep 12, 2013 #4
    Don't simplify anything. Does a value for the second function exist if X = -1?

    Simplifying a multidimensional equation to make it "easier" to solve as a linear one is not correct. However many dimensions an equation has, that many solutions you must also have. If you "simplify" to a lower dimension you will lose solutions and your entire solution would be incorrect.
     
  6. Sep 12, 2013 #5

    Ray Vickson

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    Never, never write what you wrote, which was
    [tex] y = x^2 - \frac{1}{x} + 1[/tex]
    when you mean
    [tex] y = \frac{x^2 - 1}{x+1}.[/tex]
    Use parentheses, like this: y = (x^2-1)/(x+1). It only takes an extra 1--2 seconds to type, and it makes everything clear and unambiguous.

    Anyway, ask yourself: does the first form make sense for alll x? Does the second form make sense for all x? If not, what goes wront?
     
  7. Sep 12, 2013 #6

    epenguin

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    Well you make me unsure what official doctrine is about whether f(x)g(x)/g(x) exists where g(x) = 0 .

    But let me point out that that was not the question, the question was about graphs. Assuming every point satisfying y = the function has to be on a black line, are you able to show me graphs for these two functions, if there are two, which are different? :tongue2:
     
  8. Sep 13, 2013 #7
    I apologize for not using parentheses. It's my first time using this forum. I have solved the problem if I simplify the multidimensional equation, but I'm not sure how that is not correct. Thank you, everyone.
     
  9. Sep 13, 2013 #8
    I made a post earlier, but I didn't realize that I solved it for you. Be careful when you simplify, though. Revise the material that addresses continous functions and the conditions. You said you solved the problem if you simplify, well are you sure you didn't divide by 0?
     
  10. Sep 13, 2013 #9

    epenguin

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    What is your, or anyone's, answer?
     
  11. Sep 13, 2013 #10
    I guess the graphs would be different. The second function is not defined at x=-1 so I think the graph would be a straight line (it reduces to linear after factorisation) with a discontinuity at x=-1 (shown by a small circle?).
     
  12. Sep 13, 2013 #11

    epenguin

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    I disagree, see #7.

    But I indicate there how anyone can prove me wrong. :smile:
    Graphically.
     
  13. Sep 13, 2013 #12
    See attachment. That's what I think the graph for second function would be but I would rather wait for someone else to confirm.
     

    Attached Files:

    Last edited: Sep 13, 2013
  14. Sep 13, 2013 #13
    This is a post that got deleted since it was considered as solving the problem. I hope it is alright to repost it now since people demand blood:

    #13 Your graph looks like F(x) = 2x to me
     
    Last edited: Sep 13, 2013
  15. Sep 13, 2013 #14

    epenguin

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    Elementary mistake there I think. Using the formula as written in #6 I think the slope should be 1 not 2. And also we should have the graph of the first condition for comparison.
     
    Last edited: Sep 13, 2013
  16. Sep 13, 2013 #15
    Woops. Fixed it.
     
  17. Sep 13, 2013 #16

    epenguin

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    Does the graph of y = (x - 1) look different?
     
  18. Sep 13, 2013 #17
    Yes, it does not contain a discontinuity at x = -1.
     
  19. Sep 13, 2013 #18

    epenguin

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    And your graph does? I can't see a discontinuity on your graph either.
     
  20. Sep 13, 2013 #19
  21. Sep 13, 2013 #20
    After thinking about Riemann's theorem on removable singularities I get the feeling it's correc to state that they are the same functions or different, since Riemann doesn't say it's incorrect to imply a discontinuity, even though holomorphically extendable, makes the difference between our 2 functions. From hereon, I am a neutral party :D
     
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