The minimum value of $f\left(x\right)=x+\frac{4}{x+2}$  is

$$\begin{gathered} f\left(x\right)=x+\frac{4}{x+2} \\ f\text{'}\left(x\right)=1-\frac{4}{{(x+2)}^2} \\ f\text{'}\left(x\right)=0 \\ 1=\frac{4}{{(x+2)}^2}\Rightarrow x+2=\pm 2 \\ x=0, x=-4 \end{gathered}$$
$$\begin{gathered} \Rightarrow f^{\prime \prime }\left(x\right)=-\frac{8}{{(x+2)}^3} \\ at x=0 f"\left(0\right)=-1<0 \end{gathered}$$

$\therefore$ The minimum value = $f\left(0\right)=2$