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If f(a)=g(a) and f'(x)>g'(x) for all x, use MVT to prove f(x)>g(x)

  1. Dec 8, 2011 #1
    1. The problem statement, all variables and given/known data
    Give a graphical argument that if f(a)=g(a) and f'(x)>g'(x) for all x>a, then f(x)>g(x) for all x>a. Use the Mean Value Theorem to prove it.


    2. Relevant equations



    3. The attempt at a solution
    I have sketched a graphical argument to show that f(x)>g(x). This is what I got:

    Let h(x) = f(x) - g(x).
    Then h'(x) = f'(x) - g'(x) > 0 for any x > a.
    Now apply the MVT on the interval [a, x].

    So.. h'(c) = (h(x) - h(a))/(x -a)

    And then
    [tex]h'(c) =\frac{f(x) - f(a) - g(x) + g(a)}{x - a} > 0[/tex]
    [tex]h'(c) =f(x) - f(a) - g(x) + g(a) > 0[/tex]
    [tex]h'(c) =f(x) - f(a) > g(x) - g(a)[/tex]
    [tex]h'(c) =\frac{f(x) - f(a)}{x-a} > \frac{g(x) - g(a)}{x-a}[/tex]
    ==> h'(c) = f(x) > g(x)
    ==> f(x) > f(x) for all x>a

    But I feel like this is wrong?!
     
  2. jcsd
  3. Dec 8, 2011 #2

    Fredrik

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    The equalities that all begin with h'(c)= look very wrong. They're saying that a single number h'(c) is equal to several different numbers. That clearly can't be true. I also don't see why you conclude that h'(c)>0.
     
  4. Dec 8, 2011 #3
    Well, I know to set h(x) = f(x) - g(x) but I don't know where to go from there. lol
     
  5. Dec 8, 2011 #4

    Fredrik

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    The good news is that that's the right way to start. :smile:

    Can you tell me exactly what the MVT says about h on the interval [a,x]?
     
  6. Dec 8, 2011 #5

    Mark44

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