Smallest Possible Value of b

$g(x) = f(x+1) - f(x-1) = [(x+1)^2 + a(x+1) + b] - [(x-1)^2 + a(x-1) + b] = 4x + 2a$. Given $g(20) = 72 \implies 4(20) + 2a = 72 \implies 80 + 2a = 72 \implies 2a = -8 \implies a = -4$. Since $f(x) = x^2 - 4x + b \ge 0$ for all real $x$, the discriminant must be $\le 0$: $a^2 - 4b \le 0 \implies (-4)^2 - 4b \le 0 \implies 16 \le 4b \implies b \ge 4$ So the smallest possible value of $b$ is 4.