The function is
$f(x) = \begin{cases} x^2\sin(1/x) & \text{if $x \neq 0$} \\ 0 & \text{if $x = 0$} \end{cases} $
I have proved that $f$ is differentiable at $0$, and $f^{'}(0) = 0$.
Now I have to show that $f$ has neither a local max nor local min at $0$. I know that by showing that if there is no sign change around $0$, then $f$ has no local max or min.
However, I are there other ways to prove this? I was thinking about how if proving if $f$ is either increasing, decreasing, or constant around $0$, then $f$ does not have any local max or min at $0$ .
2 Answers
You can prove this simple by the definition of the local extrema. Function does have a local extrema at the point $x \in X$ if there is a neighbourhood $V$, such that $f(x)$ is either the minimum of $f(V)$ or the maximum of $f(V)$.
Let $S_\delta(x) = \{ y \mid |y_i - x_i| \leq \delta \}$.
You can show, that for any $\delta > 0, \ f(S_\delta(x))$ will contain both strictly positive and strictly negative values?
Consider $f(\frac 2 {(2n+1) \pi})=(\frac 2 {(2n+1) \pi} )^{2} (-1)^{n}$. You can see that every interval around $0$ has points where the value of $f$ is greater than $f(0)$ and points where the value of $f$ is less than $f(0)$.
