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Given function f defined by f(x) = ( 1- x)³. What are all values of c, in the closed interval [0,3], that satisfy the conditions of the Mean Value Theorem?

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Erica

  1. The mean value theorem states that a continuous function between x=a and x=b will have at least one tangent parallel to the chord AB.
    For f(x)=(1-x)³, the chord between 0 and 3 has a slope of
    s=(f(3)-f(0))/(3-0)=-3

    The value(s) of c required must satisfy
    f'(c)=-3
    So, differentiate with respect to x:
    f'(x)=-3(1-x)²
    and solve for
    f'(x)=-3
    to get
    x=0 and x=2

    Reject any solution that is not on the interval [0,3].

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    MathMate

  2. mean value theorem of x^3+x-6 with range of [0,2]

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    Noran

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