Can critical point that $f''$ has different sign in its every neighborhood be a local extreme point? Suppose that $f$ is a second order derivable function on $[0,1)$, and $f'(0)=0$. It is true that:


*

*If there exits $\delta>0$ such that $f''(x)\geq0$ for all $x\in[0,\delta)$, then $0$ is a local minimum point.

*If there exits $\delta>0$ such that $f''(x)\leq0$ for all $x\in[0,\delta)$, then $0$ is a local maximum point.


Question:
If  $f''(0)=0$ and for any $\delta>0$ there exits $x_1,x_2\in [0,\delta)$ such that $f''(x_1)>0$ and $f''(x_2)<0$, 
is it possible that $0$ is a local extreme point of $f$ ?
 A: Take $f(x)=1.1x^6+x^6\sin(1/x)=x^6\left(1.1+\sin(1/x)\right)$, with $f(0)$ defined to be $0$. (I suppose you may want to prove that this makes a continuous function.) Here is a picture of $y=f(x)$.

The factorization reveals that for $x\neq0$ the outputs are positive. So this function has a minimum at $x=0$. 
$$f'(x)=6.6x^5+6x^5\sin(1/x)-x^4\cos(1/x)$$
$$f''(x)=33x^4+30x^4\sin(1/x)-4x^3\cos(1/x)-x^2\sin(1/x)$$
for $x>0$. Additionally, it can directly be shown using the definition of the derivative that $f'(0)=0$ and $f''(0)=0$. And you can show that theses derivatives are continuous.
I have not pursued a formal proof, but it appears to me that $f''(x)$ takes both negative and positive values on any $[0,\delta)$. Basically, the first three terms of $f''(x)$ are negligible near $0$ compared to the last term, which oscillates in a well-understood way between $y=x^2$ and $y=-x^2$. Here is a picture of $y=f''(x)$:

So I believe this function is an example that answers your question: yes, it is possible.

I found this function by asking what kind of $$g(x)+h(x)\sin(1/x)$$ would meet your criteria.
