Approximate value of $\sqrt{4.08}$ is

$$\begin{gathered} \mathrm{x}=4 \delta x=0.08 \mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}} \\ \mathrm{f}(\mathrm{x}+\delta x)={\mathrm{f}}^1\left(\mathrm{x}\right) \delta x+\mathrm{f}\left(\mathrm{x}\right) \\ =\frac{1}{2\sqrt{\mathrm{x}}} \delta x+\mathrm{f}\left(\mathrm{x}\right) \\ =\frac{0.08}{4}+2 \\ =0.02+2 \\ =2.02 \end{gathered}$$