The minimum value of (x–a) (x–b) is

$\mathrm{f}\left(\mathrm{x}\right) = \left(\mathrm{x}-\mathrm{a}\right)\left(\mathrm{x}-\mathrm{b}\right)$

$\begin{array}{rr}{\mathrm{f}}^{\mathrm{\text{'}}}\left(\mathrm{x}\right)=(\mathrm{x}-\mathrm{a})\left(1\right)+(\mathrm{x}-\mathrm{b})\left(1\right)& \end{array}$

$\begin{array}{rr} =2x-(a+b)& \end{array}$

$\begin{array}{rr}\mathrm{f}\left(\mathrm{x}\right)\text{ is minimum (or) maximum }& \end{array}$

$\begin{array}{rr}\Rightarrow {\mathrm{f}}^{\mathrm{\text{'}}}\left(\mathrm{x}\right)=0\Rightarrow x=\frac{\mathrm{a}+\mathrm{b}}{2}& \mathrm{ }\end{array}$

$\begin{array}{rr}\text{ minimum value in }& \end{array}$

$\begin{array}{rr}\mathrm{f}\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)=\left(\frac{\mathrm{a}+\mathrm{b}}{2}-\mathrm{a}\right)\left(\frac{\mathrm{a}+\mathrm{b}}{2}-\mathrm{b}\right)& \end{array}$

$\begin{array}{rr}=\left(\frac{b-a}{2}\right)\left(\frac{a-b}{2}\right)& \end{array}$

$\begin{array}{rr}=\frac{-(\mathrm{a}-\mathrm{b}{)}^2}{4}& \mathrm{ }\end{array}$