Suppose that $ f : \mathbb{R} \rightarrow \mathbb{R} $ is a continuous function on the interval $ [-3,3] $ and differentiable on $ (-3,3) $ such that for every $ x $ in the interval, $ f'(x) \le 2 $. If $ f(-3) = 7 $, then $ f(3) $ is at most $\underline{\hspace{2cm}}$.

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Step 1: Apply the Mean Value Theorem. Since $ f $ is continuous on $ [-3,3] $ and differentiable on $ (-3,3) $, the Mean Value Theorem applies. Thus, there exists some $ c \in (-3,3) $ such that: $$ f'(c) = \frac{f(3) - f(-3)}{3 - (-3)} = \frac{f(3) - 7}{6} $$ Step 2: Use the given bound on the derivative. It is given that $ f'(x) \le 2 $ for all $ x $. Hence: $$ \frac{f(3) - 7}{6} \le 2 $$ Step 3: Solve for $ f(3) $. $$ f(3) - 7 \le 12 $$ $$ f(3) \le 19 $$ % Final Answer Final Answer: $$ \boxed{19} $$

Suppose that \( f : \mathbb{R} \rightarrow \mathbb{R} \) is a continuous function on the interval \( [-3,3] \) and differentiable on \( (-3,3) \) such that for every \( x \) in the interval, \( f'(x) \le 2 \). If \( f(-3) = 7 \), then \( f(3) \) is at most \(\underline{\hspace{2cm}}\).

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Correct Answer: 19

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