1. Not finding help here? Sign up for a free 30min 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!

Application of the mean value theorem.

  1. Nov 21, 2012 #1
    1. The problem statement, all variables and given/known data

    Let f and g be two continuous functions on [a,b] and differentiable on ]a,b[ such that for every x in ]a,b[ : f'(x)<g'(x)

    2. Relevant equations

    Show that f(b)-f(a)<g(b)-g(a)

    3. The attempt at a solution

    So I said that there exists a c in ]a,b[ such that f'(c)=(f(b)-f(a))/(b-a) and g'(c)=(g(b)-g(a))/(b-a) and since f'(x)<g'(x) then f'(c)<g'(c) and that implies that f'(c)/g'(c)<1 and when substituting in what we found we get that (f(b)-f(a))/(g(b)-g(a))<1 and that implies that f(b)-f(a)<g(b)-g(a). Is this correct??
     
    Last edited: Nov 21, 2012
  2. jcsd
  3. Nov 21, 2012 #2
    Yes, that is correct, but I don't understand why you did the division (also this is dangerous- what if [itex]g'(c) =0[/itex]). Do you see how you can do it with out the division?
     
  4. Nov 21, 2012 #3
    No I can't see how, but i can suppose that g'(c)≠0 right?
     
  5. Nov 21, 2012 #4
    I don't see how you can suppose that (I might be missing something, but if you can, you will have to prove why it is true, which isn't necessary.) For example, if [itex]f(x)=-5x[/itex] and [itex]g(x)=-x^2+1[/itex] and [itex]a=-1[/itex] and [itex]b=1[/itex] the smallest that [itex]g'[/itex] gets is -2. But, [itex]f'(x)=-5 < -2 < g'(x)[/itex]. Right?

    So, you know that:
    [itex]f'(c) < g'(c)[/itex] and
    [itex]f'(c)=\frac{f(b)-f(a)}{b-a}[/itex] and [itex]g'(c)=\frac{g(b)-g(a)}{b-a}[/itex].

    Now do you see?
     
  6. Nov 21, 2012 #5
    So all you are going to do now is say that since f'(c)<g'(c) and when we cancel the b-a we will get f(b)-f(a)<g(b)-g(a) right??
     
  7. Nov 21, 2012 #6
  8. Nov 21, 2012 #7
    Alright thanx!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Application of the mean value theorem.
  1. Mean Value Theorem (Replies: 2)

Loading...