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Let $f(x) = \frac{x}{2} + x^2\sin(\frac{1}{x})$.

I need to prove that there is a neighborhood of 0 that f is monotone there.

I tried to check the sign of the first derivative and then conclude something about monotonicity (using Lagrange mean value theorem), but couldn't really make a progress.

I'll be happy for help from you guys.

Thanks.

1 Answers 1

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It's not true:

First, we need to define $f(0)=0$. Then we have:

\tag{1}f'(x)={1\over 2} +2x\sin(1/x)+ \cos(1/x)\quad\text{for}\quad x\ne 0.

and

f'(0) =\lim\limits_{h\rightarrow 0} { (h/2)+h^2\sin(1/ h)\over h } =\lim\limits_{h\rightarrow 0}\bigl[\, { {\textstyle{1\over2}}+h\sin(1/ h) }\,\Bigr]={\textstyle{1\over 2}}. So, $f$ is differentiable everywhere.

To show that $f$ is monotone on no interval, use the

Hint: For $x$ near $0$, what can you say about the sign of f'?



Solution:

In any neighborhood $O=(-\delta,\delta)$ of $0$:

For any $x\in O$ the term $2x\sin(1/x)$ of $(1)$ is small. In fact, we have $|2x\sin(1/x)|\le 2|x|$. But we can find values $x_1$ and $x_2$ in $O$ as small as we like such that $\cos(1/x_1)=-1$ and $\cos(1/x_2)= 1$. Keeping these thoughts in mind and looking at the form of f' given by $(1)$, it follows that f' takes both positive and negative values in $O$.

Thus, since a differentiable function is monotonic on an interval if and only if its derivative does not change sign on the interval, $f$ is monotonic in no neighborhood of 0.

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    @AmihaiZivan Yes, that would be even better.2012-04-01