Results 1 to 6 of 6

Math Help - Rolle/MVT Problem

  1. #1
    Newbie
    Joined
    Feb 2009
    From
    Phoenix
    Posts
    6

    Rolle/MVT Problem

    I've been stating at this problem for a while and have no idea where to even start. Not looking for an answer, but really just a nudge in the right direction.

    The homework is in a section covering Rolle's Theorem and the Mean Value Theorem.

    Suppose f is differentiable on (a,b) except possibly at x_o \in (a,b) and is continuous on  [a,b] ; assume lim(x \rightarrow x_o) f'(x) exists. Prove that f is differentiable at x_o and f' is continuous at x_o.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Junesong View Post
    I've been stating at this problem for a while and have no idea where to even start. Not looking for an answer, but really just a nudge in the right direction.

    The homework is in a section covering Rolle's Theorem and the Mean Value Theorem.

    Suppose f is differentiable on (a,b) except possibly at x_o \in (a,b) and is continuous on  [a,b] ; assume lim(x \rightarrow x_o) f'(x) exists. Prove that f is differentiable at x_o and f' is continuous at x_o.
    maybe it is me, but there seems to be something off about this question. first it says we are not (or might not be) differentiable at a point, and then it asks us to prove that we are? f' continuous at x_0 means that \lim_{x \to x_0}f'(x) = f'(x_0). the right side of this equation gives us problems when we are in the case where f is not differentiable at x_0
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    Since \lim_{x \rightarrow x_0}f'(x) exists the second part of the question follows immediately. This shows f'(x) is sequentially continuous and is therefore continuous.

    For the first part, since f(x) is continuous on [a,b] and differentiable on (a,b) then, by MVT, \exists \ x_0 \in (a,b) \ s.t \ f'(x_0)=\frac{f(b)-f(a)}{b-a}.

    We already know that f'(x_0)=\lim_{x \rightarrow x_0} f'(x) from the question. Therefore we have:

    f'(x_0)=\lim_{x \rightarrow x_0}f'(x)=\frac{f(b)-f(a)}{b-a}.

    From the question f(b),f(a) and b-a are defined ( f(b) and f(a) are defined since f(x) is continuous and x_0 \neq a,b from how we defined x_0. Also, b-a \neq 0 from how we defined a and b.) This gives the conclusion that f'(x_0) exists so f(x) is differentiable at x_0 as required.

    It is an oddly phrased question though
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Feb 2009
    From
    Phoenix
    Posts
    6
    Quote Originally Posted by Showcase_22 View Post
    For the first part, since f(x) is continuous on [a,b] and differentiable on (a,b) then, by MVT, \exists \ x_0 \in (a,b) \ s.t \ f'(x_0)=\frac{f(b)-f(a)}{b-a}.
    How do we know that the x_o that you used in the MVT is the same x_o mentioned in the problem? As Jhevon stated, the problem is worded very oddly and that may be a lot of the problem in me grasping it.

    MVT states that \exists c \in (a,b) such that f'(c)=\frac{f(b)-f(a)}{b-a}, but if x_o \in c defines a single point, then we can clearly have that c \neq x_o.

    I was thinking, and I don't know if this also holds, but we know by MVT that f'(c)=\frac{f(b)-f(a)}{b-a}. For any other point x \in (a,b) such that x \neq c, is there guaranteed a subinterval x \in (m,n) \subset (a,b) such that f'(x)=\frac{f(n)-f(m)}{n-m}?

    Sorry if this seems jumbled a bit, but I greatly appreciate all the help.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    oh, I thought x_0 was a generic point on (a,b).

    What I wrote still holds, turn the x_0 in my proof to a c and show that it's continuous and differentiable for c. Therefore x_0 can equal c so f(x) must be differentiable at x_0.

    (If every f(t) is continuous and differentiable where t \in (a,b) then if x_0 \in (a,b) then f(x_0) must be must continuous and f must be differentiable at x_0).

    x_0 \in c
    This is incorrect. \in can only be used for sets but c is a point.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Feb 2009
    From
    Phoenix
    Posts
    6
    Quote Originally Posted by Showcase_22 View Post
    This is incorrect. \in can only be used for sets but c is a point.
    I definitely meant x_o \in (a,b), not x_o \in c. That's what I get for working at 7am...

    Thanks for all the help, it's much appreciated.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Rolle's Theorm example
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: September 22nd 2011, 09:45 PM
  2. Rolle's Theorem Problem
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 30th 2010, 10:29 AM
  3. Rolle's Theorem
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: February 27th 2010, 02:59 PM
  4. Rolle's Theorem Problem:
    Posted in the Calculus Forum
    Replies: 6
    Last Post: October 29th 2009, 05:11 PM
  5. Rolle's Thm and the mean value thm
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 29th 2008, 04:27 PM

Search Tags


/mathhelpforum @mathhelpforum