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Math Help - How do you show that x^4-4x+1 has exactly 2 real roots?

  1. #1
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    How do you show that x^4-4x+1 has exactly 2 real roots?

    Using Intermediate Value Theorem and Rolle's Theorem, is there a way to show that x^4-4x+1 has exactly 2 real roots?

    I have discovered that a root exists in the interval (0,1) and (1,2). So there must be at least two real roots to the equation. However, I am having difficulty continuing the proof to show that these are the only 2 real roots.

    Any help would be appreciated.

    Thanks in advance.
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  2. #2
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    f'(x) = 4x^3 - 4

    Let's find all local maxima and minima by setting f'(x) = 0:
    \begin{array}{rcl} 0 & = & 4x^3 - 4  \\ 0 & = & x^3 - 1 \\ 0 & = & (x - 1)(x^2+x+1) \end{array}

    So: x - 1 = 0 \ \Rightarrow \ x = 1 and that is all since x^2 + x + 1 = 0 has no real roots. So we know there is only one value such that f'(x) = 0

    Now for the proof. Assume that there are 3 real roots a_{1}< a_{2}< a_{3} so f(a_{1}) = f(a_{2}) = f(a_{3}).

    Now see if you can use Rolle's Theorem to finish off the proof, keeping in mind f(x) has only 1 value of x such that f'(x) = 0.
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  3. #3
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    nice proof by contradiction.

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  4. #4
    Moo
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    If you see the sign of the derivative, you'll see that f is strictly decreasing until 1, then strictly increasing.

    The minimum is f(1)=1-4+4=-4 < 0

    Therefore there is one and only one value from - infinity to 1 such that f(x)=0 and one and only one value from 1 to + infinity such that f(x)=0

    Rolle or IVT can be applied to prove it.
    Last edited by Moo; September 14th 2008 at 12:07 AM.
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    Quote Originally Posted by Moo View Post
    Rolle or IVT can be applied to prove it.
    This is a special case of Rolle's theorem, but I like to use the generalized result:

    For any integer n > 0, if f is continuous on the interval [a,b], f is differentiable on the interval (a, b), and f(x)=0 has n solutions in [a, b], then f'(x) = 0 has at least (n-1) solutions in (a, b).

    If you were to show n-1=1, then n=2. Therefore, f(x) has 2 solutions in which f(x) = 0.
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  6. #6
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    So we know there is only one value such that
    Quote Originally Posted by Moo View Post
    In fact, there are two.
    you lost me with that statement.
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  7. #7
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    I'm confused too .. There's only one value of x such that f'(x) = 0

    To add to my earlier post, note that showing that the quartic has at most 2 real roots doesn't prove that it actually does. For example, you can use the same reasoning to g(x) = x^4 - 4x + 5 as I did earlier but this function doesn't have any real roots.

    That's why noting f(1) < 0 is crucial and this is where IVT comes in.

    Aaand .. Moo's method was a lot faster and simpler.
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  8. #8
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    Much appreciated guys. Thanks.
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  9. #9
    Moo
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    Quote Originally Posted by skeeter View Post
    you lost me with that statement.
    Huh... I misread !
    Pardon me
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