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Math Help - last of inequalities

  1. #1
    Senior Member furor celtica's Avatar
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    last of inequalities

    provided that x is real, prove that the function (2(3x-1))/(3(x^2-9) can take all real values. i believe this can be done by proving that there are no real roots to the quadratic equation obtained by introducing y=f(x). but why?
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  2. #2
    Junior Member Rachel.F's Avatar
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    To prove that you have to find the domain:

    3(x^2-9) \ne 0 <=> x\ne -3 \wedge x\ne 3

    I think it can't take all real values it can't take the values -3 and 3
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  3. #3
    Lord of certain Rings
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    Quote Originally Posted by furor celtica View Post
    provided that x is real, prove that the function (2(3x-1))/(3(x^2-9) can take all real values. i believe this can be done by proving that there are no real roots to the quadratic equation obtained by introducing y=f(x). but why?
    Hmm ... not really...Its some other quadratic equation....

    Let y = \frac{6x-2}{3x^2-27}. Rewriting will give you a quadratic equation, with y in co-efficients. Show that for every real y, the quadratic equation has real roots
    y = \frac{6x-2}{3x^2-27} \implies 3yx^2 - 6x + 2 - 27y = 0. By the quadratic formula, we have x = \frac{6 \pm \sqrt{54y^2 - 4y + 36}}{6y}. Now for x to be real always, 54y^2 - 4y + 36 must be non-negative for all values of y. This quadratic expression is clearly positive since the discriminant is negative.
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  4. #4
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    Hello, furor celtica!

    This is probably not considered a proof.


    Provided that x is real, prove that: . f(x) \:=\: \frac{2(3x-1)}{3(x^2-9)} can take all real values.

    We note that: . \lim_{x\to\infty}f(x) \:=\:0 .and there are vertical asymptotes at x = \pm3

    Part of the graph looks like this:
    Code:
                  :     |     :
                  :     |     :*
                  :     |     :
                  :     |     : *
                  :     |     :   *
                  :     |     :      *
                  :     |     :           *
    - - - - - - - + - - | - - + - - - - - - -
      *          3:     :     :3
           *      :     |     :
              *   :     |     :
                * :     |     :
                  :     |     :
                 *:     |     :
                  :     |

    We see that f(x) takes all real values except 0.

    And this provided by f\left(\tfrac{1}{3}\right) \:=\:0.

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  5. #5
    Senior Member furor celtica's Avatar
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    sorry i made a little mistake in the problem, so im rewriting it and giving my proof, i'd like to know if it makes sense.
    f(x)=y=2(3x+1) / 3(x^2 - 9)
    => 3x^2y - 6x - 27y -2
    => 36 - 12y(-27y -2) >= 0
    => 27y^2 + 2y + 3 >= 0
    => 27y^2 + 2y + 3 has no real roots
    f(x) can take on all real values


    is this proof conclusive?
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  6. #6
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    Quote Originally Posted by furor celtica and edited by Mr F View Post
    sorry i made a little mistake in the problem, so im rewriting it and giving my proof, i'd like to know if it makes sense.
    f(x)=y=2(3x+1) / 3(x^2 - 9)

    => 3x^2y - 6x - 27y -2 = 0.

    Discriminant = 36 - 12y(-27y -2)

    = 12(27y^2 + 2y + 3).

    But 27y^2 + 2y + 3 > 0 for all values of y (why?)

    Therefore f(x) can take on all real values (why?) except possibly for y = 0 (why?).

    [snip]
    See above for my non-definitive edit of your solution. You still need to:

    1. Explain where each step comes from.

    2. Clearly demonstrate that discriminant > 0.

    3. Clearly explain why it follows from discriminant > 0 that y can take on on all real values except possibly for y = 0.

    4. Explain why y = 0 is possible.

    Quote Originally Posted by Isomorphism View Post
    Hmm ... not really...Its some other quadratic equation....

    Let y = \frac{6x-2}{3x^2-27}. Rewriting will give you a quadratic equation, with y in co-efficients. Show that for every real y, the quadratic equation has real roots
    y = \frac{6x-2}{3x^2-27} \implies 3yx^2 - 6x + 2 - 27y = 0. By the quadratic formula, we have x = \frac{6 \pm \sqrt{54y^2 - 4y + 36}}{6y}. Now for x to be real always, 54y^2 - 4y + 36 must be non-negative for all values of y. This quadratic expression is clearly positive since the discriminant is negative.
    All values of y include y = 0 and there is a difficulty that needs resolving if y = 0 because y = 0 \Rightarrow x = \frac{6 \pm 6}{0} ....
    Last edited by mr fantastic; June 11th 2009 at 05:16 AM. Reason: Merged posts
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  7. #7
    Lord of certain Rings
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    Quote Originally Posted by mr fantastic View Post
    All values of y include y = 0 and there is a difficulty that needs resolving if y = 0 because y = 0 \Rightarrow x = \frac{6 \pm 6}{0} ....
    Yes that is true. When I divided something by 'y' in the quadratic formula, I should have been careful.
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  8. #8
    Senior Member furor celtica's Avatar
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    but is the conclusion correct for all that? and i am not too sure myself of the connection between the 'no real roots' and the conclusion that 'all values can be taken', its just that way in my textbook and i dont have a teacher. can someone suggest a link?
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