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Math Help - Number of solutions

  1. #1
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    Number of solutions

    Prove that f(x) = 4x^4+2x^2-6x+2=0 has exactly two solutions.

    Proof so far.

    Now f'(x)=4x^3+4x-6, set it equals to zero, we have  2x^3+2x-3=0, so this equation have 3 solutions, meaning we have 3 extremas, meaning the equation can cross the x-axis twice.

    Would that be enough to prove it?
    Last edited by tttcomrader; November 13th 2008 at 07:35 AM.
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  2. #2
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    f'(x)=[4x^3+4x-6]'=12x^2+4
    \forall \mbox{ }12x^2+4>0
    The derivative is always positive, so value of this finction constantly rises and there's just one root (solution).
    Is that what you meant?
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  3. #3
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    Quote Originally Posted by tttcomrader View Post
    Prove that f(x) = 4x^3+4x-6=0 has exactly two solutions.

    Proof so far.

    Now f'(x)=4x^3+4x-6
    No, that is NOT f', it is just f itelf!

    , set it equals to zero, we have  2x^3+2x-3=0, so this equation have 3 solutions, meaning we have 3 extremas, meaning the equation can cross the x-axis twice.

    Would that be enough to prove it?
    No, [tex]2x^3+ 2x- 3= 0[/itex] is just your original equation divided by 2. IF it has 3 solutions, then the original certainly does NOT have exactly two solutions! The equation you start with has only one real solution and two complex solutions. Did you miswrite the function?
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  4. #4
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    Oh, it should have been x^4+2x^2-6x+2, I accidentally put in the derivative instead of the original function.
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