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Domain In Interval Notation?

  1. Sep 17, 2007 #1
    1. Find the domain of the function 5x/8x^2+9 in interval notation.



    2.



    3. I believe the domain is all real numbers, but do not know what that would look like in interval notation. Could someone show me?
     
  2. jcsd
  3. Sep 17, 2007 #2

    dynamicsolo

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    You would be right: since the denominator is always positive (in fact, never less than 9), the ratio is always defined. In interval notation, the set of real numbers looks like
    ( -infinity , +infinity) .

    [Sorry, I haven't worked out yet how to get the symbols on this browser; I can see I'll need to review TeX...]
     
  4. Sep 17, 2007 #3
    Thank you very much.
     
  5. Sep 17, 2007 #4
    Would the domain of a graph that has more than one integer on the same Y coordinate, but none repeating in the X be considered all real numbers?
     
  6. Sep 17, 2007 #5

    HallsofIvy

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    It would be [ tex ][-\infty, \infty [ /tex ]
    [tex][-\infty, \infty][/tex]
     
  7. Sep 17, 2007 #6
    Thank you.
     
  8. Sep 17, 2007 #7

    dynamicsolo

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    One would want parentheses rather than brackets, no? [tex](-\infty, \infty)[/tex]
     
  9. Sep 17, 2007 #8

    dynamicsolo

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    I think HallsofIvy was answering my TeX question. If I understand your question, I think the condition doesn't tell you much about the domain. Since the domain of a (numerical) function is the set of numbers for which the calculations of the function are defined, we'd have to know what is being used in the graph. A semicircle above the X-axis ( X^2 + Y^2 = R^2, Y>=0 ) would have more than one occurrence of Y-coordinates and no repetitions of X-coordinates, but the domain of the graph would be
    [-R,R]. We'd need more information about the graph to resolve this question.
     
  10. Sep 17, 2007 #9
    Write the function whose graph is the graph of y=x^3, but is stretched horizontally by a factor of 3.

    I think it should look like y=1/3x^3 but am unsure. Does anyone know if this is close?
     
  11. Sep 17, 2007 #10

    HallsofIvy

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    You are right, of course, the proper answer is [itex](-\infty, \infty)[/itex] I was focusing on how to show that in LaTex and didn't get it right.

    By the way, you can see the code on any LaTex by just clicking on it. There is also a "LaTex thread" under "tutorials".
     
  12. Sep 17, 2007 #11

    dynamicsolo

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    Thanks! I won't say how long it's been since I had to use LaTeX for anything; goodness knows where the "manual" I had has gotten to. I found out about viewing the code already: I've been nicking people's equations to use or modify from time to time so far. I'll have a look at the tutorial here.
     
  13. Sep 17, 2007 #12

    dynamicsolo

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    To make a horizontal scale change of the function f(x) by a factor k>0, you would use f(kx) . If k>1, you get a compression or "squash"; if 0 < k < 1, you get a "stretch". You can think of it as if you were running along the graph k times faster, plotting points in the interval [0, kx] on the interval [0, x].

    So you're thinking in the right direction, but the function would need to be
    y = [(1/3)x]^3 or (x/3)^3 . Notice that, since this gives you y = (1/27)(x^3), stretching y = x^3 horizontally by a factor of 3 gives you the same graph as if you had squashed the original function *vertically* by a factor of 27 (that is, as if you'd multiplied the original function's values by (1/27) ).
     
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