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!

MVT question + fixed points

  1. May 2, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider the function g: [0,∞) -> R defined by G(x)=x+e-2x

    A. Use the mean value theorem to prove that |g(x2)-g(x1)|<|x2-x1| for all x1,x2 E [0,∞) with x1≠x2.

    B. Find all fixed points of g on [0,∞).

    2. Relevant equations

    MVT: f'(c) = f(b)-f(a)/b-a.

    3. The attempt at a solution

    It seems we are being asked to prove in part A. that |g(x2)-g(x1)|/|x2-x1|<1.

    This should be proved if we show that g'(x)<1 for x E [0,∞).

    g'(x) = 1-2e-2x. By inspection g'(x) is clearly less than 1 for x values between 0 and ∞.

    Is this sufficient an explanation? M.V.T. states that the gradient of a function must, at at least one point (c,f(c)) between the points (a,f(a)) and (b,f(b)), equal the slope over the whole region [a,b] (as the equation shows). My method is essentially the same but replaces the 'rise over run' bit of the MVT simply with f'(x).

    Am I on the right track, is there a better way to explain this, perhaps more directly involving MVT...? Thanks

    have not yet tackled question B but any hints would be appreciated
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?



Similar Discussions: MVT question + fixed points
Loading...