3
$\begingroup$

Is the function $y=x/2 + x^{2}\sin(1/x)$ monotonic near $0$?

The derivative $f'$ obviously goes positive and negative near $0$, because $$f'(x)= \frac12 + 2x\sin(1/x) - \cos(1/x))$$ Does that mean that $f$ is not monotonic near $0$?

  • 2
    If $f'$ takes both positive and negative values in every neighborhood of $0$, then $f$ cannot be monotonic in any neighborhood of $0$.2012-12-18
  • 0
    Think of it this way: if the derivative takes on both positive and negative values near 0, then the function both increases and decreases near 0. So, it isn't either just increasing or just decreasing no matter how close to 0 we get, meaning it can't be monotone at 0.2012-12-18
  • 0
    Thank you guys... i wasn't sure about f but for small x 2xsin in f ' is ~0 so 1/2 - cos(1/x) i think can be both + and - (depending on cos)...2012-12-18
  • 0
    Nevertheless, you of course do have $f(x)>f(0)$ for $x>0$ and $f(x) for $x<0$.2012-12-18
  • 0
    Show (by continuity) that for any $\epsilon$, there is an $a$ with $|a|\lt \epsilon$, and an *interval* about $a$, such that $f'(x)\lt 0$ in that interval. So by Mean Value Theorem, $f$ is decreaing in that interval. Repeat with $f'(x)\gt 0$.2012-12-18

1 Answers 1