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!

Homework Help: Mean Value theorem

  1. Jul 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Let a>b be Real numbers and
    f, g: [a,b] --> R be continuous and differentiable on (a,b)

    Show g is injective on [a,b] if g'(x) != 0 for all x in (a,b)

    2. Relevant equations
    Rolle's theorem: Continuity and differentiability (in the conditions above) imply that
    f(a) = f(b) and there exists c in (a,b) such that f'(c) = 0

    3. The attempt at a solution
    Well first I don't know exactly what injective means (what is "distinctness"). What I do understand is Rolle's theorem: that there is a turning point or point of zero gradient between any two points that have the same y-value (if that's right). So in this question there is no turning point or zero gradient in the interval [a,b] - but I don't know what the function is restricted to look like. I'm thinking it could be a horizontal straight line, a parabola, or a squiggly thing that starts and ends between two horizontal points. I'm really quite clueless how to prove something for all situations

    If you could just give me a starting point or outline,
    Thanks
     
  2. jcsd
  3. Jul 18, 2010 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You may have heard the term 1-1 before.

    The definition for injective (1-1 is the same thing) is: if g(x)=g(y) then x=y. For example, the function y=x2 is not injective because (-1)2=12, but the function y=x3 is injective because if a3=b3 this forces a=b

    In fact in your post you almost stated a solution to the problem: a re-wording of this definition is that g(x) is not injective if there are two different values of x for which g(x) is the same
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook