1. Limited time only! Sign up for a free 30min personal 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!

Rolle’s Theorem

  1. Feb 27, 2010 #1
    1. The problem statement, all variables and given/known data

    Let the function:


    f : I→ I be continuous on I and differentiable on the open set I
    for I := [0,1]


    Now I need to use Rolle’s Theorem to show that if f'(x) is not equal to 1 in (0, 1), then there is exactly one such point t



    2. Relevant equations

    I know that theres at least 1 point t ∈ [0, 1] such that f(t) = t.

    3. The attempt at a solution

    I've tried 3 different proofs for this, but none of them are giving me uniqueness of t. Please help!
     
  2. jcsd
  3. Feb 27, 2010 #2
    Did you mean that there is exactly one t for which f(t) = t in your problem statement? As it stands, it's reading as if f'(t) does not equal 1 in (0,1), then there is exactly one point for which it does not equal 1, which makes no sense.

    Assuming you meant the former statement. You're right in that there is at least one, which is due to the simplest of fixed point theorems. Now if there were two such points, say x_1 and x_2 consider g(x) = f(x) - x and apply Rolle's theorem to get a contradiction.
     
  4. Feb 27, 2010 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Suppose there are two points t1 and t2 such that f(t1)=t1 and f(t2)=t2. Apply Rolle's theorem to the function g(t)=f(t)-t on the interval [t1,t2].
     
  5. Feb 27, 2010 #4
    So this is how I've been starting, as you said, and I think I'm confusing myself... can't I only apply Rolle's Theorem if f(x1) = f(x2)?

    I have:
    suppose f'(x) not= 1 on [0,1] and suppose there exists 2 pts, x1 and x2 in [0,1] such that
    f(x1) = x1
    f(x2) = x2
    and let some function g be defined as g(x):=f(x)-x for x in [x1,x2]

    To apply Rolle's Theorem, don't I need g(x1) = g(x2)?

    I appreciate your help, I'm just still confused as to how I can actually apply Rolle's :(
     
  6. Feb 27, 2010 #5
    But what is g(x1)? It is f(x1) - x1 = 0, since x1 is a fixed point of f. Similarly for g(x2). So Rolle's Theorem tells you something about g.
     
  7. Feb 28, 2010 #6
    Which is clearly a contradiction!!! I realized my mistake: at the end I kept using Rolle's to say there exists a point c where f'(c)=0 instead of looking at the new function. It seems so easy now... I'm sorry! Thank you for your time, I really appreciate it!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Rolle’s Theorem
  1. Gauss' Theorem (Replies: 12)

  2. Liouville's Theorem (Replies: 8)

  3. Green's Theorem (Replies: 2)

Loading...