Suppose that $f\colon \mathbb{R} \to \mathbb{R}$ is a continuous function on the interval $[-3, 3]$ and a differentiable function in the interval $(-3, 3)$ such that for every $x$ in the interval, $f^{\prime}(x) \le 2$. If $f(-3) = 7$, then $f(3)$ is at most _______.

By Lagrange's Mean Value Theorem (LMVT), since $f$ is continuous on $[-3, 3]$ and differentiable on $(-3, 3)$, there exists some $c \in (-3, 3)$ such that:

Rewriting this expression:

Given that $f^{\prime}(x) \le 2$ for all $x \in (-3, 3)$, it follows that $f^{\prime}(c) \le 2$. Thus:

Substitute the given value $f(-3) = 7$:

Hence, the maximum possible value of $f(3)$ is 19.