The derivative of a point is greater than 0, and the non-monotonic increment function is included in the open interval.

As we have discussed in the post, a continuous function can lead to an example of a discontinuity in the derivative function:

Http://www.cnblogs.com/zhangwenbiao/p/5426699.html

This function is $g (x) =x^{2}\sin \left (\frac{1}{x}\right) $, supplementing the definition $g (0) =0$. can be calculated $g ' (0) =0$

We define the function

$ $f (x) =\epsilon x+x^{2}\sin \left (\frac{1}{x}\right) $$

Supplemental definition $f (0) =0$.

$ $f ' (0) =\epsilon$$

Take $1>\epsilon>0$, when $x\neq 0$,

$ $f ' (x) =\epsilon+2x\sin \left (\frac{1}{x}\right)-\cos \left (\frac{1}{x}\right) $$

Take $x_{k}=\frac{1}{k\pi}$,

$ $f ' (X_{k}) =\epsilon-( -1) ^{k}, k=1,2,3,\cdots$$

Therefore, the interval derivative symbol containing 0 cannot obtain a consistent, non-monotonic function.

Question: What is the extremum point of this function?

The derivative of a point is greater than 0, and the non-monotonic increment function is included in the open interval.