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Mean Value Theorem, Intermediate Value Theorem

  1. Sep 24, 2009 #1
    1. The problem statement, all variables and given/known data
    Let [tex]h[/tex] be a differentiable function defined on the interval [tex][0,3][/tex], and assume that [tex]h(0) = 1, h(1) = 2[/tex] and [tex]h(3) = 2[/tex].

    (c) Argue that [tex]h'(x) = 1/4[/tex] at some point in the domain.


    2. Relevant equations
    (a) Argue that there exists a point [tex]d \in [0,3][/tex] where [tex]h(d) = d[/tex].
    (b) Argue that at some point [tex]c[/tex], we have [tex]h'(c) = 1/3[/tex]


    3. The attempt at a solution
    This question comes in 3 parts and I've already solved parts (a) and (b). Part (a) is simply by the intermediate value theorem, using the fact that differentiation implies continuity and that continuous functions have the intermediate value property. Part (b) is by the mean value theorem. So those two parts I'm fine, but I'm kind of stuck on part (c).

    I'm convinced that part (c) has something to do with the mean value theorem. And tips would be appreciated here! Thanks!
     
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  3. Sep 24, 2009 #2

    LCKurtz

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    Hint: Do you know that the derivative f'of a differentiable function f satisfies the intermediate value theorem even though f' may not be continuous? That plus what you know...
     
  4. Sep 24, 2009 #3
    Ah.. yes, I know that --- but somehow it totally escaped me when I was looking at this problem. Then this is easy!

    By part (b), [tex]\exists c_1[/tex] s.t. [tex]h'(c_1) = 1/3[/tex]. Applying the mean value theorem again on [tex][1,3][/tex], [tex]\exists c_2 \in (1,3)[/tex] such that [tex]h'(c_2) = \frac{h(3) - h(1)}{3 - 1} = \frac{2 - 2}{3 - 1} = 0[/tex]. And since we have [tex]h'(c_2) = 0 < 1/4 < 1/3 = h'(c_1)[/tex], then by the intermediate value theorem for derivatives, [tex]\exists x \in (c_1, c_2) \subseteq [0,3][/tex] s.t. [tex]h'(x) = 1/4[/tex].

    Thanks again :)
     
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