To find the maximum possible value of $g(0)$ given that $g(x)$ is continuous and differentiable on $[-7, 0]$ with the information provided, we can use the Mean Value Theorem (MVT).

Step-by-Step Solution:

1. Mean Value Theorem (MVT): Since $g(x)$ is continuous on $[-7, 0]$ and differentiable on $(-7, 0)$, the Mean Value Theorem applies. According to the MVT, there exists a point $c \in (-7, 0)$ such that: $g'(c) = \frac{g(0) - g(-7)}{0 - (-7)} = \frac{g(0) + 3}{7}.$

2. Applying the Derivative Condition: We are given that $g'(x) \leq 2$ for all $x \in [-7, 0]$. Therefore, $\frac{g(0) + 3}{7} \leq 2.$

3. Solving the Inequality: Multiply both sides by 7 to solve for $g(0)$: $g(0) + 3 \leq 14.$ $g(0) \leq 11.$

Conclusion:

The maximum possible value of $g(0)$ is $\boxed{11}$.

Would you like further details or have any questions about this problem?

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Here are some related questions to deepen your understanding:

1. How would the solution change if $g' (x) \leq 3$ instead of 2?
2. Why is the Mean Value Theorem applicable to this problem?
3. What if $g(-7) = -5$ instead of $g(-7) = -3$?
4. How does continuity and differentiability of $g(x)$ affect the application of MVT here?
5. Could $g(0)$ exceed 11 if $g(x)$ had discontinuities?

Tip: In problems involving maximum values and derivatives, always consider if the Mean Value Theorem or other derivative-based inequalities apply.