Step 1: Shift to a friendlier variable.
Let $t=x+2$, so $x=t-2$ and $f=x+\dfrac{4}{x+2}=(t-2)+\dfrac{4}{t}=t+\dfrac{4}{t}-2$.
Step 2: Focus on the core $t+\dfrac4t$.
The minimum of $f$ over the branch where $t>0$ depends on minimizing $t+\dfrac{4}{t}$.
Step 3: Apply AM to GM.
For $t>0$, $t+\dfrac{4}{t}\ge 2\sqrt{t\cdot\dfrac4t}=2\cdot2=4$, with equality when $t=\dfrac4t$, i.e. $t=2$.
Step 4: Find the minimum value.
So the least value of $f$ on $t>0$ is $4-2=2$, attained at $t=2$, i.e. $x=0$.
Step 5: Confirm with the second derivative.
$f'(x)=1-\dfrac{4}{(x+2)^2}$ vanishes at $x=0$, and $f''(x)=\dfrac{8}{(x+2)^3}>0$ there, confirming a minimum.
Step 6: Read off the answer.
$f(0)=0+\dfrac{4}{2}=2$.
\[ \boxed{2} \]