Let $f\left(x\right)=(x-4)(x-5)(x-6)(x-7)$ then

Solution:

We have,

$f\left(x\right)=(x-4)(x-5)(x-6)(x-7).$

Clearly, $f\left(4\right)=f\left(5\right)=f\left(6\right)=f\left(7\right)=0$

By Rolle's theorem, there exist ${\alpha }_1\in (4,5),{\alpha }_2\in (5,6),$

${\alpha }_2\in \left(6.7\right)$ such that $f^{\mathrm{\prime }}\left({\alpha }_i\right)=0,i=1,2,3.$

Since $f\text{'} \left(x\right)$  is a cubic polynomial. Therefore, ${\alpha }_1,{\alpha }_2,{\alpha }_3$ are the only roots of $f^{\mathrm{\prime }}\left(x\right)=0.$