Results 1 to 6 of 6

Math Help - Intermediate value theorem

  1. #1
    Newbie
    Joined
    Feb 2008
    Posts
    12

    Intermediate value theorem

    Can anyone help me with this?

    Let f be a continuous function on [0,4]. Show that there must be values c and d in [0,4] such that:

    c - d = 2 and f(c) - f(d)= [f(4) - f(0)]/2


    I have been struggling with this one for a while now. The best thing I can think of is to split [0,4] into two intervals and use the IVT to show that f(c) and f(d) exist, but I don't know how the fact that their difference is the average of the interval affects things. Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    weird.. it must have been easier if it was f'(c) - f'(d)..
    hahaha.. i am looking forward for the solution..
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2008
    Posts
    12

    Further workings.

    I think the problem is a special case of the general rule (at least it only makes sense):



    Let f be a continuous function on [beg,end]. Show that there must be values c and d in [beg,end] such that:

    c - d = (beg+end)/2 and f(c) - f(d)= [f(end) - f(beg)]/2


    But I still don't know how to show this is true. I have checked with all kinds of functions for all kinds of intervals, and it appears true.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by spammanon View Post
    I think the problem is a special case of the general rule (at least it only makes sense):



    Let f be a continuous function on [beg,end]. Show that there must be values c and d in [beg,end] such that:

    c - d = (beg+end)/2 and f(c) - f(d)= [f(end) - f(beg)]/2


    But I still don't know how to show this is true. I have checked with all kinds of functions for all kinds of intervals, and it appears true.
    I haven't had a really hard look, but the Mean Value Theorem might come into it.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member JaneBennet's Avatar
    Joined
    Dec 2007
    Posts
    293
    You can NOT use the mean-value theorem. The MVT is only applicable to differentiable functions but nothing is stated in the question about f being differentiable.

    This is my partial proof, which works only if \mathrm{f}(2)\ne\frac{\mathrm{f}(4)+\mathrm{f}(0)}  {2}.

    Define \mathrm{g}:[0,2]\to\mathbb{R} by \mathrm{g}(x)=\mathrm{f}(x+2)-\mathrm{f}(x)-\frac{\mathrm{f}(4)-\mathrm{f}(0)}{2}.

    Thus:

    {g}(0)=\mathrm{f}(2)-\frac{\mathrm{f}(4)}{2}-\frac{\mathrm{f}(0)}{2}

    {g}(2)=-\mathrm{f}(2)+\frac{\mathrm{f}(4)}{2}+\frac{\mathr  m{f}(0)}{2}

    Hence if \mathrm{f}(2)\ne\frac{\mathrm{f}(4)+\mathrm{f}(0)}  {2}, {g}(0) and {g}(2) are nonzero and opposite in sign; hence by the IVT, \exists\,d\in[0,2] such that {g}(d)=0.

    So it remains to prove the result for the case \mathrm{f}(2)=\frac{\mathrm{f}(4)+\mathrm{f}(0)}{2  },
    Last edited by JaneBennet; March 1st 2008 at 03:49 AM. Reason: Got a sign wrong
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Feb 2008
    Posts
    12

    Looks like you have almost got it.

    Thanks for your work so far! I will keep going and see if I can figure out the rest from what you posted. I would appreciate it if you found the complete proof too!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Intermediate Value Theorem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 26th 2011, 02:43 PM
  2. Intermediate Value Theorem
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: April 19th 2009, 12:55 AM
  3. Intermediate Value Theorem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 6th 2009, 09:49 AM
  4. Intermediate Value Theorem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 22nd 2008, 01:33 PM
  5. intermediate value theorem/rolle's theorem
    Posted in the Calculus Forum
    Replies: 6
    Last Post: December 8th 2007, 02:55 PM

Search Tags


/mathhelpforum @mathhelpforum