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Proving a function is increasing

  1. Jun 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Suppose f is differentiable on [0,1] and that f(0)=0. Prove that if f' is increasing on (0,1) then f(x)/x is increasing on (0,1)

    2. Relevant equations

    3. The attempt at a solution
    I've tried several things. I tried applying the definition of increasing on f', but I don't really get anywhere with that. I've used the Mean Value Theorem, but I'm not sure what to apply it to. I figured the interval (0,1) and I get there is a c in (0,1) such that f'(c)=f(1). I applied it again on the interval (0,c) to get another d such that f'(d)=f(c)/c. Since f' is increasing, then f(c)/c[tex]\leq[/tex]f(1)/1. I could apply it again and again for smaller and smaller intervals closer to 0, but I don't know how to prove it's increasing on the whole interval.
     
  2. jcsd
  3. Jun 25, 2009 #2
    Remember that f(x) = f(x) - f(0) and x = x - 0. Form the difference quotient and use the MVT.
     
    Last edited: Jun 25, 2009
  4. Jun 25, 2009 #3

    Dick

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    f(x)/x increasing means (f(x)/x)'>=0. Suppose (f(x)/x)'<0 at some point x=c. Can you show that would lead to a contradiction using the MVT?
     
  5. Jun 25, 2009 #4
    Then [tex](f(x)/x)' = \frac{xf'(x)-f(x)}{x^{2}}<0[/tex] for come c in (0,1) implies f'(c)<f(c)/c. But, by the MVT, on (0,c) there is a d such that f'(d)=f(c)/c. Since f' is increasing, [tex]f'(d)\leq f'(c) \Rightarrow f(c)/c<f(c)/c[/tex]. I hope this is the right contradiction. I was unsure whether the question meant strictly increasing or not, so I guess this answers it
    I don't know why it never occured to me to use the derivative of f(x)/x...
    Thanks for the help!
     
  6. Jun 25, 2009 #5

    Dick

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    Sure, that's the contradiction. You've got d<c, f(c)/c=f'(d) but f'(c)<f(c)/c. So f'(c)<f'(d). That's not increasing.
     
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