Proving that extrema of cubic with 3 distinct roots always happen to fall between the roots By Rolle's Theorem, it is possible to prove that between points $a$ and $b$ there is a point $c$ at which the value of $f'(c)=0$. 
Now, consider a cubic polynomial function with 3 distinct real roots,
$f(x)=A(x-a)(x-b)(x-c)$
It is now necessary to prove that $x$-coordinates of extrema of $f(x)$ fall between $a, b$ and $c$. Or otherwise stated,
$a<x_1<b<x_2<c$, where $x_1$ and $x_2$ are $x$-coordinates of extrema.
The problem is, however, in the fact that this should be proven without the usage of Rolle's Theorem. I can prove this fact for quadratic, I get $x=(a+b)/2$, which suggests that extremum is halfway between the roots. With cubic, however, it is much more difficult. Any ideas?

Translating and scaling, you can assume that $a=0$ and $b=1$. Then
$$f(x)=x^3-(c+1)x^2+cx$$
$$f'(x)=3x^2-2(c+1)x+c$$
The roots of $f'$ are
$$\frac{2c+2\pm\sqrt{4c^2-4c+4}}6=\frac{c+1\pm\sqrt{c^2-c+1}}3$$
Note that the discriminant is increasing for $c>1/2$. Then, since $c>b=1$, we have
$$\frac{c+1+\sqrt{c^2-c+1}}3>\frac{1+1+1}3=1$$
and
$$\frac{c+1-\sqrt{c^2-c+1}}3=\frac{c+1-\sqrt{(c-\frac12)^2+\frac34}}3<\frac{c+1-(c-\frac12)}3=\frac12<1$$
