Suppose $p=x^3+ax^2+bx+c$ and it has both a local max and min, why is the midpoint of the line segment connecting local values a point of inflection? How can you do a $\Rightarrow$ proof?
So I get that $p^{\prime}$ has two local extreme values if the discriminant $a^2-3b\gt 0 \iff b\lt \frac {a^2}{3}$. And calculating its POI I get $x=-\frac{a}{3}$. How do you make a connection to the line segment?
 A: Hint: Make the change of variable $x=t-\frac{a}{3}$. That is just a shift, it does not change the geometry. 
Now our cubic has equation of the shape $y=t^3-dt+e$, and computations are easy. To make things even simpler looking, note that we can assume that $e=0$ and that $d=3k^2$ for some positive constant $k$. Then computation becomes unnecessary, since symmetry takes care of things.
A: Another way is to look at the first derivative of the cubic curve - a parabola.  But the parabola is symmetric about the point where its derivative vanishes, its vertex;
In particular, the two points where the parabola intersects the horizontal line (parabola's two solutions which are simultaneously and at the same time the local min/max of the original cubic) are going to be symmetrically located with respect to the vertex of the parabola -
which corresponds to the point of inflexion of the cubic curve, on account of being the point where the derivative of the derivative vanishes.
A: (Assuming in advance that the extreme points are different.)The points discussed are local. Therefore $x_{1,2}={-2a\pm 2\sqrt{a^2-3b}\over 6}$ exist both. The midpoint of the aforementioned line segment has $x$ coordinate $c={-2\over 3}a$. It is easy to see that $p''(c)=0$ which means $c$ is potentially an exrema of $p'$. $P'''(c)=p'''=6>0$ which means $p'(c)$ is a local minimum. Therefore $p''$ goes from negative values to positive values at $c$, meaning $p$ is goes from being concave down to being concave up which means this is an inflation point.
