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Differentiable functions proof

  1. Nov 27, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider a function g : (a, b)-->R. Assume that g is differentiable at some point c in
    (a,b) and that g'(c) is not = 0. Show that there is a delta > 0 so that g(x) is unequal to g(c) for all x in V_delta(c)\{c}intersect(a,b)


    2. Relevant equations



    3. The attempt at a solution

    Alright, this is what I tried so far:

    Since g is differentiable at c that means that

    g'(c)=lim_x-->c (g(x)-g(c)/x-c)

    since we are assuming that g'(c) is unequal to 0 that means that
    lim_x-->c (g(x)-g(c)/x-c) is unequal to zero and thus g(x)-g(c) is unequal to zero
    and therefore g(x) is unequal to g(c).

    I don't really know where the delta comes in though, am I supposed to use the definition of a limit to show that there exists an epsilon greater than zero so that for |g'(c)-L|<epsilon?

    Any help would be great!!
     
  2. jcsd
  3. Nov 27, 2009 #2

    HallsofIvy

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    The fact that [itex]\lim_{x\to c} (g(x)-g(c))/(x-c)= g'(c)[/itex] is not 0 does NOT imply that (g(x)- g(c))/9x-c) is not 0! It does mean that you can take [itex]\epsilon[/itex] to be |g'(c)/2|[/itex] and then, for some [itex]\delta> 0[/itex] if [itex]|x-c|< \delta[/itex], [itex]||g(x)- g(c))/(x-c)- g'(c)|< |g'(c)|/2[/itex]. Write that as [itex]-|g'(c)|/2< (g(x)- g(c))/(x-c)- g'(c)< |g'(c)|/2[/itex] and see where that leads.
     
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