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

the equation f’(x)=0 has only one root  

f’(x)=0 has four roots

three roots of  f’(x)=0 lie in $\left(4,5\right)\cup \left(5,6\right)\cup \left(6,7\right)$

three roots of  f’(x)=0 lie in $\left(3,4\right)\cup \left(4,5\right)\cup \left(5,6\right)$