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(Sort of) Intermediate Value Theorem

  1. Sep 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Suppose that f is continuous on [a,b], its right-hand derivative [itex]f'_{+}(a)[/itex] exists, and [itex]\mu[/itex] is between [itex]f'_{+}(a)[/itex] and [itex]\frac{f(b)-f(a)}{b-a}[/itex]. Show that [itex]f(c) - f(a) = \mu (c-a)[/itex] for some [itex]c \in (a,b)[/itex].



    2. Relevant equations
    [itex] f'_{+}(a) = \lim_{x \rightarrow a^+}\frac{f(x)-f(a)}{x-a} [/itex]


    3. The attempt at a solution
    I spent a little while trying to figure out exactly what I am trying to prove and it definitely makes sense now. Intuitively I can see that it is true. My first instinct was to use the IVT for derivatives or the MVT, but those both require that f be differentiable, which I don't know to be the case here.
    So I looked at using the regular IVT, since it only requires that f be continuous on [a,b], but also needs f(a) not equal to f(b). I was going to for now ignore the case where f(a) = f(b), but I was having trouble showing that [itex]\mu [/itex] is between the required bounds.
    Am I even on the right track or is there another approach that makes more sense?


    Thanks.
     
  2. jcsd
  3. Sep 8, 2013 #2

    Office_Shredder

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    It might help to know that the function
    [tex]\frac{ f(x) - f(a)}{x-a}[/tex]
    is a continuous function on (a,b)
     
  4. Sep 8, 2013 #3

    pasmith

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    The question suggests that it would be worth considering
    [tex]g(x) = \frac{f(x) - f(a)}{x - a}[/tex]
    which is certainly defined and continuous for [itex]a < x \leq b[/itex]. Since [itex]f'_{+}(a)[/itex] exists we can make [itex]g[/itex] continuous at [itex]x = a[/itex] by setting [itex]g(a) = f'_{+}(a)[/itex].

    Now [itex]g[/itex] is continuous on [itex][a,b][/itex] and you can apply the IVT.
     
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