Example where $g'(x_*) = 0$ and $g''(x_*) \geq 0$ but $x_*$ is not a local minimum for g.
It is true if $g'(x_*) = 0$ and $g''(x_*) > 0$, then $x_*$ is a local minimum for $g$.
So I guess the example is when $g'(x_*) = 0$ and $g''(x_*) = 0$.
I thought of constant function, but it is not a counter example.
What could an example for this?
Thank you!
I just started by trying to draw a line which passes the vertical line test on a piece of paper meeting the conditions prescribed above (which is my approach to almost any problem in real analysis. That is, translating the problem into analytic geometry):
It wasn't hard to immediately figure out that
G(x)=$x^3$ on [-$\delta$,$\delta$] for any $\delta$ >0, meets the conditions.
Look at its graph, (or do the computation). You'll find that
$g'(0)=0, g''(0)=0 \geq 0$.
Adam V. Nease