Let \(\mathrm{f}(x)=(x-1)(x-2)(x-3), x \in[0,4]\), Values of C will be _______ [if L.M.V.T. (Lagrange's Mean Value Theorem) can be applied].

\(\begin{aligned}
& \text {Let } y=(x-1)(x-2)(x-3) \\
& \log y=\log (x-1)+\log (x-2)+\log (x-3)
\end{aligned}\)
Differentiating w.r.t. \(x\), we get
\(\frac{1}{y} \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{1}{x-1}+\frac{1}{x-2}+\frac{1}{x-3}\)
\(\therefore \mathrm{f}^{\prime}(x)=\frac{\mathrm{d} y}{\mathrm{~d} x}=(x-2)(x-3)+(x-1)\)\((x-3) +(x-1)(x-2) \)
\(\therefore \mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(4)-\mathrm{f}(0)}{4-0}=\frac{6-(-6)}{4}=3, \text { for } \mathrm{c} \in[0,4] \)
\( \therefore (\mathrm{c}-2)(\mathrm{c}-3)+(\mathrm{c}-1)(\mathrm{c}-3)+(\mathrm{c}-1)\)\((\mathrm{c}-2)=3 \)
\( \therefore 3 \mathrm{c}^2-12 \mathrm{c}+11=3 \)
\( \therefore 3 \mathrm{c}^2-12 \mathrm{c}+8=0 \)
\( \therefore \mathrm{c}=\frac{12 \pm \sqrt{144-96}}{6}=\frac{6 \pm 2 \sqrt{3}}{3}\)