1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Differentiable functions proof
  1. Differentiation Proof (Replies: 1)