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Domain of x^2-6x+9 / x^2 |
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| Apr29-06, 11:56 PM | #1 |
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Domain of x^2-6x+9 / x^2
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. |
| Apr30-06, 12:12 AM | #2 |
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Recognitions:
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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 ? |
| Apr30-06, 10:33 AM | #3 |
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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?"
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| Apr30-06, 03:56 PM | #4 |
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Domain of x^2-6x+9 / x^2
Undefined!
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| Apr30-06, 06:07 PM | #5 |
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And therefore, the domain of (x^2-6x+9 )/ x^2 is?
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| Apr30-06, 06:18 PM | #6 |
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Any value of x for which you can evaluate the term.
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| Apr30-06, 07:06 PM | #7 |
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Recognitions:
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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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