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I Graphing a strange equation

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  1. Apr 23, 2017 #1
    Hello Forum,

    I've gone about graphing the below equation by inserting values in for y and then solving for x:

    y(x-1)=x2-1

    For instance, I say y=3 and then solve:

    3(x-1)=x2-1

    3x-3=x2-1

    3x=x2-1+3

    3x=x2+2

    0=x2-3x+2

    0=(x-1)(x-2)

    Thus x = 1 and 2 and so I plot co-ordinates (1,3) and (2,3).

    Now I repeated this quite a few times with different values and it appears that there is a vertical line in the graph where x=1.

    Can anyone explain why this vertical line feature appears in the graph?
     
  2. jcsd
  3. Apr 23, 2017 #2

    FactChecker

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    At x=1, any value of y will satisfy the original equation. The original equation at x=1 is y×0=0. So any value of y will work.
     
  4. Apr 23, 2017 #3

    Mark44

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    You've gone to a lot of effort below that is unnecessary. Divide both sides by x - 1 to get
    ##y = \frac{x^2 - 1}{x - 1} = \frac{(x - 1)(x + 1)}{x - 1} = x + 1##, if ##x \ne 1##.
    Your first point is incorrect. If x = 1, the equation you started with becomes y * 0 = 0, which, as already stated, can't be solved for y.
    There is no vertical line. What you have in your equation is a removable discontinuity, a single point at which the equation is not defined -- at (1, 2). For any other value of x, a y value can be found.
     
  5. Apr 23, 2017 #4

    mfb

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    OP doesn't solve for y. Finding possible x-values for a given value of y is perfectly possible. It is not the most efficient way, but it is not wrong.
    Of course there is. x=1 is a solution for all y. The equation is perfectly defined at x=1. It just reads 0=0, which is true. No one said you have to be able to solve for y.
     
  6. Apr 23, 2017 #5

    Mark44

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    Finding possible x-values is perfectly possible, but I can't help wondering if we're seeing later work in the problem, and not the original problem, such as if it were "sketch a graph of ##y = f(x) = \frac{x^2 - 1}{x - 1}##."


    Right, but again, this might be the answer to the question as posed in this thread, but not the one really intended.

    To solve for x in the equation ##y(x - 1) = x^2 - 1## you would need to get all the terms involving x on the same side, by dividing both sides by x - 1. One would have to stipulate that x could not be 1, for obvious reasons. That would eliminate the possibility of an infinite number of y values when x = 1.
     
  7. Apr 23, 2017 #6

    mfb

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    To solve for x, the correct answer is "x=1 or x=y-1".

    While we can speculate if OP's question was the original problem statement, if you answer a different question than the question that got asked, I think you should at least point out that you are not answering the question that was asked.
     
  8. Apr 23, 2017 #7

    Mark44

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    I agree. However, we get many questions that stem from incorrect work or are poorly posed, so a bit of interpretation is called for.

    @Ned Eterminita, was the question you posted the original problem statement?
     
  9. Apr 23, 2017 #8
    This is how the question was posed:

    Yo Cuz,

    You want to make $20???

    Graph this equation [ y(x-1)=x^2-1] and then explain its unusual feature. If you do it correctly I'll give you $20.

    Look forward to seeing you and your graph at Tony's wedding next Saturday.

    Ciao

    So yea, I've got a very strange cousin. He's always been into riddles, paradoxes and generally thinking about things too deeply. While I'm not someone to do math in my free time, I will do it for money.

    So at first, I went about moving (x-1) to the right hand side and then simplifying to y=x+1. This produces no unusual features. I thought about it for a while and realised that I made the mistake of assuming that 0/0 = 1.

    On second approach I've gone about inserting values for y and then solving for x, this way I avoid division by zero. This time I got the vertical line, which makes sense considering that y can be anything for y*0= 0.

    So what do you recon guys, will I be getting that $20?

    P.S. Thanks for the help
     
  10. Apr 24, 2017 #9
    Hint:

    Think about f(x) = x^2 and g(x) = x-1. What happens when you compose these functions? Try f(g(x)) and g(f(x)). The order does matter.

    I mention composition because learning how to compose and decompose functions is a very important skill.
     
    Last edited: Apr 24, 2017
  11. Apr 24, 2017 #10
    Hi Aufbauwerk, I can't make sense of your approach to the question.
     
  12. Apr 24, 2017 #11
    Sorry, it was not a good hint because I was tired.

    I will make up for it by giving the correct explanation for you.

    We have f(x) = y = (x^2 - 1)/(x-1).

    Looks like it has a discontinuity at x = 1, because the denominator = 0.

    But the problem goes away when you factor the numerator, so y = f(x) = (x+1)(x-1)/(x-1) = x + 1.

    In fact it's just a straight line. f(1) = 2.

    [EDIT Sorry, I was a bit careless. I should say lim f(x) as x -> 1 = 2. I've only seen this problem in the context of doing the limit.]
     
    Last edited: Apr 24, 2017
  13. Apr 24, 2017 #12

    FactChecker

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    The fraction is undefined at x=1. The original statement of the OP is fine as long as we realize that any value of y satisfies y⋅0 = 0. That is why the OP finds a vertical line at x=1.
     
  14. Apr 24, 2017 #13
    Yes I got sloppy. Sorry. I'm thinking of the limit of f(x) as x goes to 1. I thought that was the question.

    But if you want to graph the original function just put a little circle at f(1) to indicate a discontinuity. I'm not sure what's the point, if we are not talking about the limit.
     
    Last edited: Apr 24, 2017
  15. Apr 24, 2017 #14
    P.S. drawing the straight line x=1 to indicate the discontinuity (due to the function f(x) being undefined for x=1) is not correct. We just draw a little circle to indicate that at one point only the function is undefined. At least that's the way it's done in the textbooks.
     
  16. Apr 24, 2017 #15
    This actually raises an interesting, or maybe a silly, question. It seems very basic.

    Suppose we start with the function f(x) = x + 1. This is defined for all x and is a straight line.

    Now we multiply both numerator and denominator by x - 1. So we get the function of the problem, f(x) = (x+1)(x-1)/(x-1) which is undefined for x = 1.

    This seems nonsensical. Is it even valid to multiply by (x-1)/(x-1)? If it is, why isn't it equally valid to cancel the x-1 terms?

    In other words, it seems strange that we can use algebra to transform a function that is defined everywhere, to one that is undefined at one point, but we can't do the reverse operation and get back to the original function.
     
  17. Apr 24, 2017 #16

    FactChecker

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    Just answering this from the original post:
    Every point on the vertical line is a valid solution of y(x-1)=x2-1. There is no division by 0.
     
  18. Apr 24, 2017 #17
    I gave my cousin a call, because it seemed strange that he would offer me $20 with no down side.

    As it turns out, FactChecker was right and I'm getting that $20 (Thanks FactChecker) but yes he had an ulterior motive. He was using it to butter me up for some crank paper hes drafted called "Indeterminate is not synonymous with undefined" (You'll find it if you google it supposedly) that he wants me to show to my maths lecturer.

    Basically, he thinks that 0/0 should be depicted with a vertical line rather than a hole because its indeterminate (aka could be anything) and not undefined like 1/0. It seems that he's reached that conclusion following Aufbauwerk's reasoning. Though maybe entropy is a thing in algebra too?
     
  19. Apr 24, 2017 #18

    FactChecker

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    You should tell your friend that y(x-1)=x2-1 and y=(x2-1)/(x-1) are not the same thing. They do not have the same set of solutions. He will get nowhere with mathematicians.
     
  20. Apr 24, 2017 #19
    This thread reminds me of a bad dream. I feel like I've entered some alternative reality. If f(x) is a function, then there is only one value of f(x) for each value of x that is in the domain of f(x). Vertical lines do not enter into it at all.

    Sorry, but all I can think of now is to press the panic button and respond with a video. It's a good video for mental clarity. Good night and goodbye to this thread. I hope someone learned something.

    I wonder if they need AI programmers in Belarus.

     
  21. Apr 25, 2017 #20

    Mark44

    Staff: Mentor

    Your cousin is correct in saying that the two are not synonymous.

    ##[\frac 0 0]## is just one of a number of indeterminate forms, which also include ##[\frac {\pm \infty} \infty]##, ##[\infty - \infty ]##, ##[1^\infty]##, and one or two more. They are usually presented in brackets to emphasize that we're not talking about a specific thing, but rather a form. These indeterminate forms usually show up in the context of taking a limit.
    The reason they have "indeterminate" in their descriptions is that you can't determine in advance how the limit will turn out. For example, these three limits are all indeterminate forms:
    ##\lim_{x \to 0} \frac x x##
    ##\lim_{x \to 0} \frac {x^2} x##
    ##\lim_{x \to 0} \frac x {x^2}##
    Although all are of the (##[\frac 0 0]##) form, the first limit is 1, the second limit is ##\infty##, and the third limit is 0.
    ##\frac 0 0## on its own is undefined, for the simple reason that division by zero is undefined. If we're working with the limit of a fraction, and the denominator is approaching zero, the limit can exist, even if the numerator is also approaching zero, with the difference being how quickly each of them is approaching zero.

    Your cousin's idea that 0/0 should be depicted, let alone as a vertical line -- no mathematician is going to pay any attention to that.

    Finally, entropy doesn't enter into the picture at all.
     
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