If $f\left(x\right)=x(x-2)(x-4),1\le x\le 4$ then a number satisfy ing the conclusion of the mean value theorem is

$f^{\text{'}}\left(x\right)=(x-2)(x-4)+x(x-4)+x(x-2)=$

$3x^2-12x+8.\text{ Also }f\left(4\right)=0\text{ and }f\left(1\right)=3$ Thus $\frac{f\left(4\right)-f\left(1\right)}{4-1}$

$=-1$ We must have $-1=f^{\text{'}}\left(x\right)$

$\Rightarrow 3x^2-12x+9=0$

$\Rightarrow x^2-4x+3=0$

$\Rightarrow x=1 \text{ or } x=3$