If Mean value theorem holds for the function \(\mathrm{f}(x)=(x-1)(x-2)(x-3), x \in[0,4] \quad\) then the values of c as per the theorem are

\(\begin{aligned} & \mathrm{f}(x)=(x-1)(x-2)(x-3) \\ & \mathrm{f}(4)=(4-1)(4-2)(4-3) \\ & \mathrm{f}(4)=6 \\ & \mathrm{f}(0)=(0-1)(0-2)(0-3)=-6\end{aligned}\)
\(\therefore \quad\) Using LMVT
\(f^{\prime}(c)=\frac{f(4)-f(0)}{4-0}=\frac{6-(-6)}{4}=3\)
\(\begin{aligned}
& \therefore \quad \mathrm{f}^{\prime}(\mathrm{c})=3 ...(i)\\
& \text { Now, } \mathrm{f}(x)=(x-1)(x-2)(x-3) \\
&=x^3-6 x^2+11 x-6
\end{aligned}\)
\(\begin{aligned}
\therefore \quad & f^{\prime}(x)=3 x^2-12 x+11 \\
& f^{\prime}(\mathrm{c})=3 \\
\Rightarrow & 3 \mathrm{c}^2-12 \mathrm{c}+11=3 \\
\Rightarrow & 3 \mathrm{c}^2-12 \mathrm{c}+8=0 \\
\Rightarrow & \mathrm{c}=\frac{12 \pm \sqrt{48}}{6}=2 \pm \frac{2 \sqrt{3}}{3}=2 \pm \frac{2}{\sqrt{3}}
\end{aligned}\)
...[From (i)]