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Positive derivative and strictly increasing

  1. Aug 30, 2010 #1
    Isn't it true that if a function [itex]f(t)[/itex] satisfies [itex]f'(t) > 0[/itex] on some interval, then [itex]f[/itex] is STRICTLY increasing on that interval; i.e., that [itex]s < t[/itex] implies that [itex]f(s) < f(t)[/itex]?
     
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  3. Aug 30, 2010 #2

    Office_Shredder

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    You can prove this using the mean value theorem, by assuming there are points s and t with [tex]s<t[/tex] and [tex]f(s)\geq f(t)[/tex] and developing a contradiction
     
  4. Aug 30, 2010 #3
    Yes.
    In fact:
    if x2>x1 implies that
    f(x2)>f(x1)


    the derivative is:
    [f(x2)-f(x1)]/(x2-x1)
    (as x2 approach x1)

    The numerator is positive and also the denominator (by hypothesis ) implies that:
    f'(x)>0
     
  5. Aug 30, 2010 #4
    Thanks, guys. I was pretty sure this was a MVT exercise, but I was so tired last night I couldn't force myself to get out of bed and look for paper and pencil to run through it :) I'm amazed I was actually able to phrase the question correctly.
     
  6. Jan 16, 2011 #5
    Related to the topic at hand, if, on the other hand, f'(a) > 0, is there necessarily an interval of positive length that is centered at c, on which f is strictly increasing? I've though about this for a while now, and it seems to me the answer is no, but I just can't come up with a counterexample. I can imagine an ever oscillating function that never reaches f(a) when x > a, but since it constantly oscillates, you also can't pinpoint the endpoint of an interval, on which f would be strictly increasing.

    Any thoughts on this, is this a good counterexample? And does anyone perhaps have a specific function in mind, one that isn't just vaguely described such as above in my previous paragraph?
     
    Last edited: Jan 16, 2011
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