How come I don't get a contradiction by Rolle's theorem here? We have $f(x)=x^4 -2x^3 -12x^2 -6x +11$, $x\in\mathbb{R}$
I need to prove that $f$ has exactly three extrema $x_1$,$x_2$,$x_3$ with: $-2<x_1<-1<x_2<0<x_3$
The functions possible extrema points since it is differentiable on $\mathbb{R}$ are where $f'(x)=0$.
The derivatives for this function are: $f'(x)=2(x^3-3x^2-12x-3)$
$f''(x)=12(x^2-x-2)$
$f'''(x)=12(2x-1)$
I calculated $f'(-2)=-14$, $f'(-1)=8$, $f'(0)=-6$ and $\lim_{x\to\infty}f'(x)=\infty$ so $\exists \kappa>0$ s.t $f'(\kappa)=0$
By applying the $\textit{Bolzano-Weierstrass Theorem}$ on $[-2,-1]$ ,$[-1,0]$ ,$[0,\kappa]$
We get $f'(x_1)=f'(x_2)=f'(x_3)=0$
I know realize that proving exactly 3 solutions can be done simply by saying that $f'$ is a 3rd degree polynomial but I'm still curious as to why my other method didn't work.
Let $f'$ have 4 roots $r_1$,$r_2$,$r_3$,$r_4\in \mathbb{R}$ with $r_1<r_2<r_3<r_4$
By applying $\textit{Rolle's Theorem} $ for $f'$ on $[r_1,r_2]$, $[r_3,r_4]$
$\exists \xi_1 \in(r_1,r_2):f''(\xi_1)=0$ and $\exists \xi_2 \in(r_3,r_4):f''(\xi_2)=0$
By applying $\textit{Rolle's Theorem}$ for $f''$on $[\xi_1,\xi_2]$ then $\exists \xi \in (\xi_1,\xi_2):f'''(\xi)=0 \Leftrightarrow 12(2\xi-1)=0 \Leftrightarrow \xi=\frac{1}{2} $
Where is the contradiction? Why didn't it come up?

Didn't apply $\textit{Rolle's Theorem}$ for $[r_2,r_3]$ which you in the end give us a $f^{(4)}(x_0)=0$ which gives us a contradiction where $24=0$.
