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How do you find the range of 1/{(x-1)(x+2)} for [0, 6]?

  1. Feb 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Need to find the range of 1/{(x-1)(x+2)} for [0, 6]

    2. Relevant equations
    3. The attempt at a solution

    The domain is [0, 1) AND (1, 6] because 1/0 is an invalid operation. So I tried and this is what I got: [-0.5, ∞) and [0.025, ∞). Wrong, of course :)

    How do I go about arriving to the right solution?

    Thanks.
     
  2. jcsd
  3. Feb 12, 2012 #2

    SammyS

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    Can you find the range of (x-1)(x+2) for [0, 1) ∪ (1, 6] ?
     
  4. Feb 12, 2012 #3
    Hopefully I am right: [-2,0) and (0,40]
     
  5. Feb 12, 2012 #4

    SammyS

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    Yes, that's correct.

    Do you see how that's related to the range of 1/((x-1)(x+2)) ?
     
  6. Feb 12, 2012 #5
    Honestly not. Can I get a hint or something :)
     
  7. Feb 12, 2012 #6

    Mentallic

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    If the values of a variable k range between 2 to 5, then what are the possible values of 1/k?
     
  8. Feb 12, 2012 #7
    Re: How do you http://www.physicsforums.cfind [Broken] the range of 1/{(x-1)(x+2)} for [0, 6]?

    [-0.5, infinity) and [0.025, infinity) ?

    But it's supposed to be (-infinity, -0.5] instead of [-0.5, infinity) so I was wondering how 1/0 gives negative infinity. This is what originally confused me.
     
    Last edited by a moderator: May 5, 2017
  9. Feb 12, 2012 #8

    Mentallic

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    Well no :tongue: but I see we're just going to disregard my example question since you already have the hang of it.

    Well for the quadratic (x-1)(x+2) with [itex]x\in [0,1)[/itex], the range is negative. So for the reciprocal function 1/((x-1)(x+2)) it's still going to be negative, thus this is how you should quickly realize that as x approaches 1 from the left side (0.99... etc.), it should be approaching [itex]-\infty[/itex] rather than [itex]+\infty[/itex]
     
  10. Feb 12, 2012 #9
    Whoa, this is WAY above my head. [itex]x\in [0,1)[/itex] for one. Have no idea what that's supposed to mean. Can you, please, tell me what that is? Thank You.
     
  11. Feb 12, 2012 #10

    Mentallic

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    It's nothing special, it just means the values of x in that set (0 to 1, but not including 1). It means the same thing as what Sammy posted in post #2.
     
  12. Feb 12, 2012 #11
    Thank You very much, Mentallic. Will look into it a bit closer.
     
  13. Feb 12, 2012 #12

    SammyS

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    Re: How do you http://www.physicsforums.cfind [Broken] the range of 1/{(x-1)(x+2)} for [0, 6]?

    For number, z, where z is between -2 and 0, what is 1/z? For one thing, isn't 1/z negative if z is negative.

    ... And if z is a negative number very close to zero, such as z=-0.001, isn't 1/z a very "large" negative number, such as -1000 ?
     
    Last edited by a moderator: May 5, 2017
  14. Feb 13, 2012 #13
    Re: How do you http://www.physicsforums.cfind [Broken] the range of 1/{(x-1)(x+2)} for [0, 6]?

    Oh, now I understand the flaw in my thinking- I kept imagining a positive 1 between -2 and 0, for some reason. Thank You, SammyS.
     
    Last edited by a moderator: May 5, 2017
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