Domain of x^2-6x+9 / x^2

  1. Apr 29, 2006 #1
    Here is the problem again
    x^2-6x+9 / x^2

    I think the answer is "all real numbers", but I don't know. I'm not used to seeing only x^2. Most of the ones I have done are x^2 - 4 or something like that.
     
  2. jcsd
  3. Apr 30, 2006 #2

    Curious3141

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    The question as it is written makes little sense. The domain has to be defined in the first place for a function to mean anything. So the domain can be a subset of the reals, or complex numbers, or even integers. Given a particular domain, it is a perfectly valid question to determine the range of the function.

    But there is one real value for x where the function ceases to be well-defined, and I think the question is asking you to find this. What happens when x = 0 ?
     
  4. Apr 30, 2006 #3

    HallsofIvy

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    As Curious3141 said, strictly speaking, the domain has to be "given" along with the formula describing a function. A lot of the time, however, it is understood that the domain is "all values of x for which the formula gives a valid result". One of the first things you should have learned about "domain" is "you can't divide by 0". Thus Curious3141's question "what happens when x= 0?"
     
  5. Apr 30, 2006 #4
    Undefined!
     
  6. Apr 30, 2006 #5

    HallsofIvy

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    And therefore, the domain of (x^2-6x+9 )/ x^2 is?
     
  7. Apr 30, 2006 #6
    Any value of x for which you can evaluate the term.
     
  8. Apr 30, 2006 #7

    Curious3141

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    A nice way of representing the domain is R\{0} which means all the reals except zero. Another way is to state the domain is [tex](-\infty,0) \cup (0,\infty)[/tex] because the open interval excludes the point at zero.

    If you're working in a system other than the reals, amend accordingly.
     
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