Question:

Step 1: Identify the interval and apply the Mean Value Theorem. We are given that $f(x)$ is continuous and derivable on $[-2, 5]$. The length of this interval is $5 - (-2) = 7$. By the Mean Value Theorem, there exists some $c \in (-2, 5)$ such that: $$f(5) - f(-2) = f'(c) \cdot (5 - (-2)) = 7 \cdot f'(c)$$ Step 2: Express $f(5)$ in terms of $f(-2)$ and $f'(c)$. Rearranging the equation from Step 1: $$f(5) = f(-2) + 7f'(c)$$ Step 3: Use the constraint on the derivative to find bounds on $f(5)$. We are given that $-4 \leq f'(x) \leq 3$ for all $x \in (-2, 5)$. This means $-4 \leq f'(c) \leq 3$. Multiplying the inequality by 7: $$-28 \leq 7f'(c) \leq 21$$ Adding $f(-2)$ to all parts: $$f(-2) - 28 \leq f(-2) + 7f'(c) \leq f(-2) + 21$$ Therefore: $$f(-2) - 28 \leq f(5) \leq f(-2) + 21$$ Step 4: Calculate the difference between maximum and minimum values of $f(5)$. From Step 3, we have: - Maximum value of $f(5)$ is $f(-2) + 21$ - Minimum value of $f(5)$ is $f(-2) - 28$ The difference is: $$\text{Maximum} - \text{Minimum} = [f(-2) + 21] - [f(-2) - 28] = 21 + 28 = 49$$ **Final Answer:** The difference between the maximum and minimum values of $f(5)$ is $\boxed{49}$. This corresponds to **Option 4**.