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The global maximum of the function \( y = x^3 - 3x^2 - 3x + 6 \) on the interval \([-2, 4]\) occurs at \(x = 4\), and the maximum value is \(10.00\) (to two decimal places).

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Here are 5 related questions for deeper exploration:
1. How do we classify critical points of a function?
2. What is the second derivative test for identifying maxima and minima?
3. How do boundary points affect the global extremum of a function?
4. Why is finding the derivative crucial for optimization problems?
5. How would the graph of this cubic function look in the given interval?

**Tip**: When solving for maximum and minimum values on a closed interval, always check both critical points and endpoints!

The global maximum over the indicated interval for y = x^3 - 3x^2 - 3x + 6, -2 ≤ x ≤ 4 is ____. Type your answer to 2 decimal places.

This problem demonstrates how to find the global maximum of the cubic function y = x^3 - 3x^2 - 3x + 6 on the interval [-2, 4] using calculus techniques such as derivatives and critical points. Suitable for high school calculus students.

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