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Homework Help: How to finish this proof

  1. Mar 13, 2009 #1
    f is differentiable continuously(which means that f(x) and f'(x) are continues and differentiable) on [a,b]
    suppose that |f'(x)|<1 for x in [a,b]

    prove that there is 0<=k<1 so there is x1,x2 in [a,b]

    i started from the data that i was given and i know that f(x) is differentiable
    f'(x)=\lim _{h->0}\frac{f(x+h)-f(h))}{h}
    i use

    |f'(x)|=\lim _{h->0}\frac{|f(x+h)-f(h))|}{|h|}<1

    so f(x) is continues and differentiable
    so i use mvt on x1 and x2
    and i combine that with

    |f(x1)-f(x2)| <|x1-x2|

    what now?
    Last edited: Mar 13, 2009
  2. jcsd
  3. Mar 13, 2009 #2
    Show that [tex]|f'(x)|\le k[/tex] for all x in [a,b] and for some 0<k<1. Hint: A continuous function defined on a compact interval has a maximum.
  4. Mar 13, 2009 #3
    they sign L as the minimum of f'(x)
    and they sign l as the maximum of f'(x)

    then they say that k=max{|f(L)|,|f(l)|}

    but its not true
    i am given that 0=<k<1
    and i dont know if i get the same values

    k=max{|f(L)|,|f(l)|} equals 0=<k<1
    Last edited: Mar 13, 2009
  5. Mar 13, 2009 #4
    I was refering to the http://en.wikipedia.org/wiki/Extreme_value_theorem" [Broken].

    What does it tell you about the function |f'(x)| on the interval [a,b]?
    You will find that is has a maximum M. Now use |f'(x)|<1 to show that M<1.
    Last edited by a moderator: May 4, 2017
  6. Mar 13, 2009 #5
    why this maximum is K ??
    and not 1 ??
  7. Mar 13, 2009 #6
    Use the following: If M is the maximum of |f'(x)| then there exists a point a with |f'(a)|=M (by definition), but |f'(x)|<1 for all x (including a), so what does that tell you about M?
  8. Mar 13, 2009 #7
  9. Mar 13, 2009 #8
    i have been told that l,L are the maximum and minimum points on [a,b] interval so

    what is the meaning of
    why not just say that
    we see that f'(L) is the maximum in the innequality
  10. Mar 13, 2009 #9
    ok here is the complete proof as i see it

    if we take K as maximum of the sequence
    we get

    but how to show that k
    is 0<=k<1
  11. Mar 13, 2009 #10
    Look at posts #6 and #7.
  12. Mar 13, 2009 #11
    ok i agree that k<1

    but its -1<k<1 because of |f'(x)|<1

  13. Mar 13, 2009 #12
    If k=max |f'(x)| then k>=0 because [tex]|f'(x)|\ge 0[/tex] for all x.
    Modify your original proof where you used |f'(x)|<1 and use [tex]|f'(x)|\le k[/tex] instead to get a better estimate.
  14. Mar 13, 2009 #13
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