To solve this problem, we will apply the Mean Value Theorem (MVT).

The Mean Value Theorem states that if $f(x)$ is continuous on the closed interval $[3,7]$ and differentiable on the open interval $(3,7)$, then there exists some $c \in (3,7)$ such that:

$f'(c) = \frac{f(7) - f(3)}{7 - 3}$

This simplifies to:

$f'(c) = \frac{f(7) - f(3)}{4}$

From the problem, we are given that:

$-2 \leq f'(x) \leq 3 \quad \text{for all} \quad x \in (3,7)$

Thus, applying this to the Mean Value Theorem:

$-2 \leq \frac{f(7) - f(3)}{4} \leq 3$

Now, multiply the entire inequality by 4 to estimate $f(7) - f(3)$:

$-8 \leq f(7) - f(3) \leq 12$

Final Answer:

$-8 \leq f(7) - f(3) \leq 12$

Would you like more details or have any further questions?

Here are 5 related questions you might find helpful:

1. How does the Mean Value Theorem apply to non-linear functions?
2. Can the bounds on $f'(x)$ ever affect the continuity of $f(x)$?
3. What other theorems are similar to the Mean Value Theorem?
4. How can we generalize this result for any interval $[a,b]$?
5. How does the MVT help in estimating the maximum and minimum values of a function?

Tip: The Mean Value Theorem is a powerful tool to connect the values of a function to its derivative and is often used in estimating bounds on function values.