I have a conjecture on local max/min , can any of you propose a contradiction?

Question

If $f$ is a non-piecewise function defined continuous on an interval $I$, and within that interval $I$, there exists a value $x$, such that $f`(x)$ (derivative of $f$) does not exist , then at that value $x$, is a local $\max/\min$ value.

Answer 1

I'm surprised no one has said $$f(x)=\sqrt[3] x$$ The derivative at $x=0$ is undefined (or $\infty$ if you like) but that is not an extremum of the function.

Answer 2

This is not true.

Consider $$\begin{cases} f(x) = 1-x & \text{if } 0\le x< 1\\ f(1) = 0& \\ f(x) = \frac 12(1-x) & \text{if } 1< x\le 2\end{cases} $$

Answer 3

False. Consider, for example,

$$f(x) = \begin{cases} x^3 & \text{ if } x<0 \\ x & \text{ if } x \ge 0\end{cases}.$$

Then $f$ is not differentiable at $0$, but $f$ is strictly increasing around $0$.

Answer 4

Very not true. For example:

$$f(x) = \begin{cases}x&\text{ if } x\leq 0\\ 2x&\text{ if } x\geq0\end{cases}$$ on $I=[-1,1]$.