The function $f\left(x\right) = {(x – 1)}^2+{(x – 2)}^2+{(x – 3)}^2+....+ {(x–9)}^2$ has minimum value at x=

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)={\left(\mathrm{x}-1\right)}^2+{\left(\mathrm{x}-2\right)}^2+----+{\left(\mathrm{x}-9\right)}^2 \\ {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=2\left(\mathrm{x}-1\right)+2\left(\mathrm{x}-2\right)+----+2\left(\mathrm{x}-9\right) \end{gathered}$$

$\mathrm{f}\left(\mathrm{x}\right)$ is minimum or maximum

$$\begin{gathered} \Rightarrow {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)=0 \\ 2\left(\mathrm{x}-1\right)+2\left(\mathrm{x}-2\right)+---+2\left(\mathrm{x}-9\right)=0 \\ 9\mathrm{x}-\left(1+2+---+9\right)=0 \\ \Rightarrow 9\mathrm{x}=\frac{9\left(10\right)}{2} \\ \Rightarrow \mathrm{x}=5 \end{gathered}$$