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Proving roots using mean value theorem

  1. Mar 21, 2006 #1
    Prove [tex] x^4 + 4x + c = 0 [/tex] has at most two real roots

    My thinking is that to prove this I would assume that it has three real roots and look for a contradiction.

    So I set [tex] f(x) = x^4 + 4x + c [/tex] and assume three real roots [tex] x_1, x_2, x_3 [/tex] such that [tex] f(x_1) = f(x_2) = f(x_3) = 0 [/tex]

    By MVT I know that there must exist [tex] c_1 [/tex] on the interval [tex] (x_1, x_2) [/tex] such that [tex] f'(c_1) = 0 [/tex] and [tex] c_2 [/tex] on the interval [tex] (x_2, x_3) [/tex] such that [tex] f'(c_2) = 0 [/tex] and [tex] c_3 [/tex] on the interval [tex] (x_1, x_3) [/tex] such that [tex] f'(c_3) = 0 [/tex]

    Now I'm a little confused. Do I try to find three values of c that will satisfy [tex] f'(x) = 4x^3 + 4 = 0 [/tex]? Only the value x = -1 would make this true, so is that my contradiction? That MVT predicts three values and only one exists?

    Thanks
    Jeff
     
    Last edited: Mar 21, 2006
  2. jcsd
  3. Mar 21, 2006 #2

    arildno

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    Dearly Missed

    Hint:
    How many STATIONARY points does the function have?
     
  4. Mar 21, 2006 #3
    Only one, at (-1, 5+c). Since the derivative is negative to the left of that point and positive to the right, it must be a minimum of the function.
     
    Last edited: Mar 21, 2006
  5. Mar 21, 2006 #4

    Galileo

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    I assume you orden x1 < x2 < x3. It could be the case that f'(c3) equals f'(c1) or f'(c2), so according to the MVT you'll get at least 2 points where f' is zero. You've shown there's only one, so the assumption of three distinct roots is false. Not sure why you want to find particular values of c to make f' zero. The value of c is irrelevant to the problem.
     
  6. Mar 21, 2006 #5
    Thanks for the clarification
     
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