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What does it mean when a function is undefined?

  1. Mar 9, 2012 #1
    If a function is undefined at (0,0), does that mean the function does not pass through the origin? in general, if the function is undefined at a certain point, is that because it does not pass through that specific point?
  2. jcsd
  3. Mar 9, 2012 #2


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    hi lonewolf219! :smile:
    i don't know what you mean by "pass through" :confused:

    functions don't move, they just are

    the function is defined on the subspace consisting of the whole space except that point
  4. Mar 9, 2012 #3
    Thanks Tiny Tim for answering..


    The graph of this function doesn't "pass through" the point (20,0). Am I correct to say the function y=x^2 is undefined at the point (20, 0) because this point does not lay on the curve of the function? Or would you say the function does not exist at that point...

    If I understood what you meant, a function is undefined if it is not continuous at that point (like a cusp, or a jump)?
  5. Mar 9, 2012 #4


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    hi lonewolf219! :smile:

    (try using the X2 button just above the Reply box :wink:)
    ah, i see what you mean

    no, the function y = x2 isn't defined on ℝ2, it's only defined on ℝ

    y is defined at all values of x, not of (x,y)

    (eg y is defined at 0,and the graph goes through (0,0))
    if a function z is undefined at (0,0), that means the graph (a surface) has a hole in above the origin :wink:
  6. Mar 9, 2012 #5


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    You need to be careful about exactly what a function is. The function y= f(x) maps real numbers to real numbers. It is, or is not, defined for individual real numbers, not pairs of numbers. [itex]f(x)= x^2[/itex] is defined for all real numbers. The function [itex]f(x)= 1/x^2[/itex] is defined for all x except x= 0 where it is "undefined".

    Similarly, [itex]f(x,y)= 1/(x^2+ y^2)[/itex] is a function that maps (x, y), a point in R2, to a real number. It is undefined for (0, 0) and defined for all other (x, y).
  7. Mar 9, 2012 #6
    I hope I know what you guys know one day...

    you guys are brilliant!
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