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If a function $f(x)$ is continuous on the interval $(a, b)$ and has a local maximum at the point $c \in (a, b)$, then in a sufficiently neighborhood of the point $x = c$, the function is increasing for all $x < c$ and decreasing for all $x > c$.

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c element of (a,b) not c 2 (a,b) –  user57175 Jan 13 '13 at 22:32
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Surely you meant to write that you have to provide counterexamples. In your questions there's lacking some self work, self ideas and effort, and even some "please". –  DonAntonio Jan 13 '13 at 22:36
    
very sory,, i only copy and paste.. very very sorry because i didn`t know how to answerd it –  user57175 Jan 13 '13 at 22:39
    
can someone please help me –  user57175 Jan 13 '13 at 22:40
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i only copy and paste... This might be part of the problem. –  Did Jan 14 '13 at 13:36
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1 Answer

The function $$f(x)=\begin{cases} -\left\lvert x\sin\left(\dfrac{1}{x}\right)\right\rvert, \ x\neq0 \\ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ \,, \ x=0 \end{cases}$$ is continuous with local max at $x=0$. Is not increasing on $(-\epsilon,0)$ or decreasing on $(0,\epsilon)$ for any $\epsilon>0$.

See this and this.

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