Are local min/max of a cubic function determined by the zeros alone?

[karush]

Just curious are cubic functions dirvel from just having the zeros, does that always determine where the local min/max is. I notice many cubic graphs given on homework show where the zeros are but the local min/max is not given.

For example
$$y=\left(x-4\right)\left(x+1\right)(x+2)={x}^{3}-{x}^{2 }-10x-8$$
$$y'=3{x}^{2 }-2x-10$$

$y'=0$ is $ - 1.5226,2.1893$ and min=-24.1926 max=1.3778

So I presume the local min/max are fixed values given the zeros

 

[MarkFL]

If a cubic function has relative extrema, then they will occur at places where the first derivative has roots. But, at the roots of the first derivative, you won't always find an extremum...consider $y=x^3$.

 

[karush]

OK I thought the humps could be moved despite the zeros but doesn't look like it.
So the only to find the extreme is by the derivative

 

[MarkFL]

OK I thought the humps could be moved despite the zeros but doesn't look like it.
So the only to find the extreme is by the derivative

Yes, the first derivative will tell you where the function itself is increasing/decreasing/turning. If the first derivative has differing signs on either side of a root, then you know that root corresponds with an extremum for the function. This will happen for all roots as long as they are all of odd multiplicity.

Observe that in the example I gave of $y=x^3$, the first derivative has a root of even multiplicity, and so its sign does not change as it crosses this first derivative root.