For the function $\text{⨍}\left(\mathrm{x}\right)\text{=}\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right)$ defined on $\left[0,\frac{1}{2}\right],$ the value of 'c' satisfying Lagrange's mean value theorem, is 
 

$$\begin{gathered} {\mathrm{f}}^1\left(\mathrm{x}\right)=(\mathrm{x}-1) \left(1\right)+(\mathrm{x}-2) \left(1\right) \\ =2\mathrm{x}-3 \\ {\mathrm{f}}^1\left(\mathrm{c}\right)=\frac{\mathrm{f}\left({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)-\mathrm{f}\left(0\right)}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-0} \Rightarrow 2\mathrm{c}-3=\frac{\left({\displaystyle \frac{-1}{2}}\right) \left({\displaystyle \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)-\left(2\right)}{{\displaystyle \frac{1}{2}}} \\ 2\mathrm{c}=3=\frac{{\displaystyle \frac{3}{4}}-2}{{\displaystyle \frac{1}{2}}}=\frac{-5}{2} \\ 2\mathrm{c}=3-\frac{5}{2}=\frac{1}{2} \Rightarrow c=\frac{1}{4} \end{gathered}$$