Question

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 \(\underline{\hspace{2cm}}\).

Hint:

Bounds on derivatives give bounds on function values via the Mean Value Theorem.

Answer 1

Correct Answer: 19

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: \[ \boxed{19} \]