If a function \(f(a) = max (a, 0)\) then the smallest integer value of \(x\) for which the equation \(f(x - 3) + 2f(x + 1) = 8\) holds true is

The smallest integer solution is $3$. At $x=3$, $f(x-3)=0$ and $2f(x+1)=2(4)=8$, while every smaller integer has $f(x-3)=0$ and $f(x+1)\leq3$. The value $4$ comes from dropping the $+1$ in $2(x+1)$.