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Open interval (graph)

  1. Jun 17, 2010 #1
    Im stuck in the 4 red boxes ive highlighted.

    1st box: what is the method/quickest way to draw the graph?
    2nd box: what does it mean by open interval?
    3rd box: how would i sketch the preimage? what does preimage really mean?
    4th box: how would i prove it for open interval?

    [PLAIN]http://img189.imageshack.us/img189/526/untitlex312x31d.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 17, 2010 #2

    Mark44

    Staff: Mentor

    You could plot points. That's not what I would do, though.
    I would draw a graph of y = |x - 1| (has a vee shape, with the low point at (1, 0)). Then I would draw a graph of y = |x - 1| - 2, which is a translation downward of the previous graph. Finally, I would draw a graph of y = ||x - 1| - 2|. Any parts of the previous graph that are below the x-axis should be reflected across the x-axis. Any parts of the previous graph that are above the x-axis should be left alone.
    An open interval is an interval that does not include the endpoints. There are two ways to denote intervals: by inequalities such as 2 < x < 4 and 0 <= x <= 5; by interval notation such as (2, 4) and [0, 5]. The first inequality and its equivalent interval notation represent an open interval. The second inequality and its equivalent interval notation represent a closed interval. An interval can also be half-open and half-closed if one endpoint is included and the other not included in the interval.
    The preimage of a function is the set of all numbers or points that are valid inputs for the function. What they seem to be asking for in this problem is the set of all points in the plane such that f(x, y) = 4. Note that for this function there are a lot of points that satisfy the equation f(x, y) = 4.
    Before tackling how you would prove it, what do you think |A| means in this problem?
     
    Last edited by a moderator: May 4, 2017
  4. Jun 17, 2010 #3
    thanks, so:

    erm, does it mean the size of set A? so it means that when |A| = |B|, the size are the same?
     
  5. Jun 17, 2010 #4

    Mark44

    Staff: Mentor

    Can you be more specific about what you mean by size? For example, the interval (-2, 2) is twice as long as the interval (-1, 1), and you're supposed to show that these intervals are the same "size", so size here can't mean the length of the interval.
     
  6. Jun 17, 2010 #5
    Hmm sorry, im really stuck. But I guess I was meaning cardinality?
     
  7. Jun 17, 2010 #6

    Mark44

    Staff: Mentor

    Right. I just wanted to make sure you understood what was meant here. To show that the set A = {x | -1 < x < 1} has the same cardinality as set B = {x | -2 < x < 2}, you need to find a one-to-one function that maps each element of A to an element of B. There is a very simple function that does that.
     
  8. Jun 17, 2010 #7
    ps
    Mark is the man.
     
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