The minimum value of f x = x + 4 x + 2 is

$f\left(x\right)=x+\frac{4}{x+2}$

For minima, $f^{\text{'}}\left(x\right)=0$

$1-\frac{4}{{\left(x+2\right)}^2}=0$

${\left(x+2\right)}^2=4\Rightarrow x+2=\pm 2\Rightarrow x=0,-4$

$f^{"}\left(x\right)=8{\left(x+2\right)}^{-3}$

At $x=0, f\left(x\right)$ will have minima.

i.e. the minimum value $f\left(0\right)=2$