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Consider a function f:[0,1] -> R with f(0)=0
having a finite derivative at each x in (0,1)
Prove that if f' is an increasing function (at least on the interval (0,1))
then h(x)=f(x)/x is also increasing

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    I think this is a poorly chosen title: an increasing derivative *does not* imply an increasing function, of course. Neither is this the actual question.2010-11-23
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    f(x) = -sin(x) is a counter example.2010-11-23

2 Answers 2

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Hint: show that $h'(x)$ is positive on $(0,1)$, using the mean value theorem applied to $f(x)-f(0)$.

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    -1 for doing somebody else's homework for him, instead of prodding him towards a solution. This wasn't a hint, but the solution.2010-11-23
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    I'm trying to apply MVT and I'm not getting an obvious solution being h'(x) is positive on (0,1) since h(x) isn't cts on [0,1]2010-11-23
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    Maybe I don't understand your comment. Can you please formulate it differently? ($h'(x) > 0$ for $x \in (0,1)$ for sure). Also note that MVT should be applied to $f$, not $h$.2010-11-23
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This seems to be a straightforward application of Cauchy's Mean value theorem.

http://en.wikipedia.org/wiki/Cauchy%27s_mean_value_theorem#Cauchy.27s_mean_value_theorem