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Same expressions, but different functions

  1. Jan 31, 2010 #1
    f(x) = x/x

    g(x) = 1

    so... those are different functions? i dont know how to treat expressions like that.


    When expression is not defined for some x=a, but after some algebraic transformations i get expresion defined for x=a .... what happends?

    Can i make algebraic transformations on expressions with an unknown number without knowing it's value?
    I know can.
    But can i do this with epression that is part of a function?
     
  2. jcsd
  3. Jan 31, 2010 #2

    TD

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    Homework Helper

    You talk about "functions", but then it becomes very relevant to mention the domain. Assuming you mean the "maximal domain", i.e. all real numbers for which the expression is meaningful, then both functions are indeed different. The first expression is not defined for x = 0, since division by zero is undefined. The second expression of course also exists for x = 0; so yes: they are different.
    The graph of f would be exactly the same as the graph of g, except for a "small hole" (or "perforation") at (0,1). The algebraic manipulation you would perform to go from x/x to 1, is only allowed for x nonzero so if you start with x/x, you can only simplify it to "1" if you keep track of the condition that this only holds for all nonzero x.
     
  4. Jan 31, 2010 #3
    And even if you wrote f(x) = g(x) = x, they might still have different codomains and so be different functions.
     
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