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We know that a function $f: [a,b] \to \mathbb{R}$ continuous on $[a,b]$ and differentiable on $(a,b)$, and if $f'>0 \mbox{ on} (a,b)$ , f is strictly increasing on $[a,b]$. Is there any counterexample that shows the converse fails?

I have been trying to come up with simple examples but they all involve functions that are discontinuous or has derivative $f'=0$ which does not agree with the hypothesis hmmm

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1 Answer 1

up vote 9 down vote accepted

Consider $f(x)=x^3$ on $[-1,1]$. It is strictly increasing, but has zero derivative at $0$.

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you are absolutely right –  Daniel Jul 10 '12 at 23:02
2  
@jsk Note that you can come up with counterexamples like this by taking a nonnegative continuous function with a $0$ (in this case $x^2/3$) and integrating. –  Alex Becker Jul 10 '12 at 23:49
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@AlexBecker You surely meant $3x^2$. –  Pedro Tamaroff Jul 11 '12 at 1:29
    
@PeterTamaroff Indeed. Slip of the fingers. –  Alex Becker Jul 11 '12 at 2:26

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