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Proof f(x)>g(x) in an interval

  1. Dec 20, 2016 #1
    1. The problem statement, all variables and given/known data
    let fx, gx be continuous in [a,b] and differentiable in (a,b). at the end of the interval f(a) >= g(a).

    and f'(x) >g'(x) for a<x<b.

    proof f(x) > g(x) for a<x<=b

    Attempt:
    There is a statement says that if the f'x = g'x for x in [a,b] , then there exists k such that f'x - g'x = k for any x in [a,b]

    but f'(x) = g'(x) + t(x), where t(x) isn't have to be a line. and i cannot use the statement
     
  2. jcsd
  3. Dec 20, 2016 #2

    BvU

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    Shouldn't that be a < sign ?
     
  4. Dec 20, 2016 #3
    nope.
     
  5. Dec 20, 2016 #4

    BvU

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    Might be hard to prove, then:
    a=0, b =2, f(x) = x, g(x) = 1 satisifies all criteria, yet f < g on half the interval.

    [edit] Ah, the end of the interval is at the beginning. My bad o:) :rolleyes:
     
  6. Dec 20, 2016 #5
    but f(a) is not equal or or bigger thatn g(a).
     
  7. Dec 20, 2016 #6

    BvU

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    Yeeah, I get it. Was wrongfooted by the 'end' term.

    With f' - g' > 0 and the definition of derivative you should be able to go through the interval without f-g diminishing from its starting value f(a)-g(a).
     
  8. Dec 20, 2016 #7
    f'(x) - g'(x) = lim(f(x+h) -f(x) -g(x+h) -g(x))/h > 0

    like that? how can i take out f(x) - g(x)
     
  9. Dec 20, 2016 #8

    BvU

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    Ergo there is a h> 0 such that f(a+h)-g(a+h) > f(a)-g(h)
     
  10. Dec 20, 2016 #9
    wait but we need x no?, a is the start of the interval
     
  11. Dec 20, 2016 #10

    BvU

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    So a+h is the new 'beginning' for the next step. Crux is if we can prove this can be continued all through the a,b interval in a finite number of steps.
     
  12. Dec 20, 2016 #11
    but why doesn't the limit and the h at the denominator making problems here?
    what i've got is
    lim (f(x+h) - g(x+h))/h > lim (f(x) + g(x))/h
     
  13. Dec 20, 2016 #12
    ok i think i got it, the limit doesn't change it much
     
  14. Dec 20, 2016 #13

    PeroK

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    I'm not sure looking at the definition of the derivative is the best way forward. In any case, it's best to introduce a new function, let's call it ##k## such that:

    ##k(x) = f(x) - g(x)##

    What can you say about ##k##? And what theorems do you know?
     
  15. Dec 20, 2016 #14
    it is continuous and it is differentiable.
    However there is a theorem i know regarding k(x) being constant, but i cannot assume it here.
     
  16. Dec 20, 2016 #15

    PeroK

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    Do you know the Mean Value Theorem?
     
  17. Dec 20, 2016 #16
  18. Dec 20, 2016 #17

    PeroK

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    It might be useful here!
     
  19. Dec 20, 2016 #18
    So;
    (f(b) - f(a))/b-a > (g(b)-g(a))/b-a

    and i can pick any b-a such that the diffrence is very small, so i can use x instead of it ?? so f(b) - f(a) = f(b)
     
  20. Dec 20, 2016 #19

    PeroK

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    You need to switch your attention to ##k(x)##. Also, I suggest a proof by contradiction using the MVT.
     
  21. Dec 20, 2016 #20
    ok K'(x) >0 for any x in the interval.
     
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