Results 1 to 7 of 7

Math Help - Please Help with Mean Value Theorem Proof

  1. #1
    Newbie
    Joined
    Nov 2006
    Posts
    22

    Smile Please Help with Mean Value Theorem Proof

    Use the Mean Value Theorem to show that:

    a)Suppose f is a diferentiable function on the interval a < b, and suppose f '(x) is not equal to 0 for all x Element Symbol (a,b). Show that f is one-to-one on the interval (a,b).

    b) Assume that |f ' (x)| < or equal to C < 1 for all x. Show that f (x) = x has at most one solution.


    Please someone help me with this Mean Value Theorem Proof. Thanks
    Last edited by CaptainBlack; December 19th 2006 at 02:17 AM. Reason: accidental
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Use the Mean Value Theorem to show that:

    a)Suppose f is a diferentiable function on the interval a < b, and suppose f '(x) is not equal to 0 for all x Element Symbol (a,b). Show that f is one-to-one on the interval (a,b).
    Suppose f is not one-one on the interval then there exists u, v in (a,b) such that f(u)=f(v).
    Then by the mean value theorem there exists a point w in (u,v) such that f'(w)=0, a contradiction.

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2006
    Posts
    22
    I have a test in a couple of hours. I can't state part a with a contradiction and I need a full blown proof if someone can help me with that.

    Also for part b, the assumption that f is differentiable on
    (a,b) should have been explicitly stated. If someone can help me finish these 2 parts off into one giant proof , it would be great appreciated. Thank you
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2006
    Posts
    22
    If anyone can still help me with this, it would be greatly appreciated. Thanks
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,966
    Thanks
    1785
    Awards
    1
    Surely you have been helped with this.
    The above posting is definitive in this question.
    What more do you want?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Swamifez View Post
    Use the Mean Value Theorem to show that:

    a)Suppose f is a diferentiable function on the interval a < b, and suppose f '(x) is not equal to 0 for all x Element Symbol (a,b). Show that f is one-to-one on the interval (a,b).

    b) Assume that |f ' (x)| < or equal to C < 1 for all x. Show that f (x) = x has at most one solution.

    b) consider g(x)=f(x)-x, then g'(x) != 0 on any interval [-u,u], since:

    |g'(x)|=|f(x)-1|>=||f(x)|-1| != 0.


    Hence by a) g(x) is one-one on [-u,u], so g(x)=0 has at most one root on
    such an interval, and as u is arbitary at most one root in R.

    RonL
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,966
    Thanks
    1785
    Awards
    1
    First, thank you for explaining what you are doing with this sequence of questions. It would have been helpful to know that up front.

    Now for another way to do part (b) which can be done as quite separate from part (a). Assume that there are points c & d in [a,b] such that f(c)=c & f(d)=d and c is not d. Using the mean value theorem \exists e \in [a,b]\left[ {f'(e) = \frac{{f(d) - f(c)}}{{d - c}} = 1} \right].
    But \left| {f'(e)} \right| \le C < 1 which is a contradiction.
    Thus there is a most one x in [a,b] such that f(x)=x.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. who can proof the lie's theorem?
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: December 28th 2011, 08:26 AM
  2. Mean Value Theorem Proof
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: March 26th 2010, 12:08 PM
  3. Proof of theorem...
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: January 31st 2010, 03:40 PM
  4. proof using mean value theorem
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: November 16th 2009, 05:50 PM
  5. proof of a theorem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 4th 2008, 07:36 PM

Search Tags


/mathhelpforum @mathhelpforum