Solution: Given that for any polynomial of degree at most $2$, we have \[ \frac{\mathrm{d}f}{\mathrm{d}x} = a f(x) + 2 f(x + 1) + b f(x + 2),~ \text{for all } x \in \mathbb{R}. \] As the above equation is true for any polynomial of degree at most $2$, so we take $f(x) = x$. Then we have \begin{align*} \frac{\mathrm{d}f}{\mathrm{d}x} = 1 & \implies a x + 2(x+1) + b (x + 2) = 1\\ & \implies (a+b+2)x + 2b + 2 = 1 \\ & \implies \begin{dcases} a + b + 2 = 0 \\ 2b + 2 = 1. \end{dcases} \\ & \implies a = -\frac{3}{2} \text{ and } b = -\frac{1}{2}. \end{align*} Therefore, \[ 4a + 3b = -6 - 1.5 = -7.5. \]