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Finding the domain without a calculator

  1. Sep 14, 2004 #1
    Finding the domain is easy with a graphing calculator but what if I was to do it without one. Is there a way to do it?

    How would I find the domain of f(x)=(x-1)/(x^2+1)
     
  2. jcsd
  3. Sep 14, 2004 #2

    Pyrrhus

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    Well in a rational expression, you have to look at the denominator, because if a value is put in the denominator and it makes it 0 then it will go to infinite, so you have to solve the denominator and find the values that makes it 0, and in the domain you have to specify any value of the Reals except those values that makes it 0, of course in the proper notation.

    In your sample, you should do x^2 + 1 =0 so, x^2 = -1 and x = sqrt(-1), so that will be x = i, so an imaginary number will make it 0, that's great so this means the domain will be all the real numbers, because no real number can make it 0.
     
  4. Sep 14, 2004 #3
    what if the denominator had a sqrt such as sqrt(x-2)/(x^2-x)
     
  5. Sep 14, 2004 #4

    Pyrrhus

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    I will assume the following:

    sqrt(x-2)/sqrt(x^2-x) well factoring the deniminator

    sqrt(x-2)/sqrt(x(x-1)), read above what i said about the denominator.

    Now this is a little addon:
    take for example the numerator sqrt(x-2), if you had a value like x=2, it will make it 0 right? what about if you had a value like x=1 wouldn't that make it sqrt(-1), whoa, so now it has gone out of the real numbers into the imaginary numbers... and a value below 2 will put it into imaginary numbers..., so what should the restriction be? obviously x>=2.

    Now try to find the domain for this.
     
  6. Sep 14, 2004 #5
    Same idea; solve the denominator for 0. The domain is all those values that don't make it 0
     
  7. Sep 14, 2004 #6
    so when there is a sqrt in the numerator, I must take that into account when finding the domain right? so it's different than just solving for the denominator is that correct?
     
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