Proving a function is increasing

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Homework Statement


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)

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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\leqf(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.
 
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Remember that f(x) = f(x) - f(0) and x = x - 0. Form the difference quotient and use the MVT.
 
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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?
 
Then (f(x)/x)&#039; = \frac{xf&#039;(x)-f(x)}{x^{2}}&lt;0 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, f&#039;(d)\leq f&#039;(c) \Rightarrow f(c)/c&lt;f(c)/c. 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 occurred to me to use the derivative of f(x)/x...
Thanks for the help!
 
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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