To find the maximum and minimum values of the function $y = \frac{x^3}{3} - \frac{x^2}{2} - 6x + \frac{5}{3}$, we need to follow these steps:

1. Find the first derivative: $y' = x^2 - x - 6$

2. Set the first derivative to zero to find critical points: $x^2 - x - 6 = 0$ Factoring gives: $(x - 3)(x + 2) = 0$ So, $x = 3$ and $x = -2$.

3. Determine whether these points are maximum or minimum using the second derivative: $y'' = 2x - 1$ Evaluate the second derivative at the critical points:

   • For $x = 3$: $y''(3) = 2(3) - 1 = 5 \quad (\text{positive, so } x = 3 \text{ is a local minimum})$
   • For $x = -2$: $y''(-2) = 2(-2) - 1 = -5 \quad (\text{negative, so } x = -2 \text{ is a local maximum})$

4. Calculate the corresponding $y$-values:

   • For $x = 3$: $y(3) = \frac{27}{3} - \frac{9}{2} - 18 + \frac{5}{3} = 9 - 4.5 - 18 + 1.67 \approx -11.83$
   • For $x = -2$: $y(-2) = \frac{-8}{3} - 2 + 12 + \frac{5}{3} = \frac{-8 + 5}{3} + 10 = \frac{-3}{3} + 10 = 9$

Summary of Results:

• Local Maximum: At $(-2, 9)$
• Local Minimum: At $(3, -11.83)$

If you need any further details or have questions about the process, let me know!

Further Questions:

1. How do you find the second derivative test for other functions?
2. What is the significance of critical points in calculus?
3. Can this function have any global extrema?
4. How can we visualize this function graphically?
5. What other methods can be used to find maxima and minima?

Tip: Always check the endpoints of the interval if you're looking for global extrema on a closed interval.