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The Mean Value Theorem

  1. Nov 15, 2006 #1
    This theorem is confusing me even though it is sittin right in front of me.. I am given an equation x^4 + 4x + c = 0 and asked to find at most two real roots??

    I know we need to take the derivative, but from there I am lost.
     
  2. jcsd
  3. Nov 15, 2006 #2
    If [tex] f(x) [/tex] is differentiable in the open interval [tex] (a,b) [/tex] and continuous on the closed interval [tex] [a,b] [/tex], then there is at least one point [tex] c [/tex] in [tex] (a,b) [/tex] such that:

    [tex] f'(c) = \frac{f(b)-f(a)}{b-a} [/tex]

    Assume that there are two real roots [tex] c_{1} [/tex] and [tex] c_{2} [/tex] where [tex] c_{1} < c_{2} [/tex].


    Then [tex] f(c_{1}) = 0 = f(c_{2}) [/tex].

    Thus [tex] 4x^{3} + 4 = 0 [/tex]
     
    Last edited: Nov 15, 2006
  4. Nov 15, 2006 #3

    mathwonk

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    basic principle of garphing: a graph can only change direction at a critical pont, and not always then.
     
  5. Nov 19, 2006 #4
    x^(4) + 4x + c = 0
    The function is a polynomial and is differentiable and continuous. Suppose a and b are distinct roots. There exists a c in which a<c<b such that 0 = f(b) - f(a). Since f'(x)= 4x^(2) + 4>0, f(a) != f(b). This is a contradiction; hence, a and b cannot both be roots.
     
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