To solve this, we will analyze the given function $y = \frac{1}{3}x^3 + \frac{1}{2}x^2 - 6x + 8$.

(a) Finding the critical points

Critical points occur where the first derivative $y'$ is zero or undefined.

1. Find the first derivative: $y' = \frac{d}{dx} \left( \frac{1}{3}x^3 + \frac{1}{2}x^2 - 6x + 8 \right)$ $y' = x^2 + x - 6$

2. Set the derivative equal to zero: $x^2 + x - 6 = 0$

3. Solve the quadratic equation: Factorize: $x^2 + x - 6 = (x + 3)(x - 2) = 0$ So, the critical points are: $x = -3 \quad \text{and} \quad x = 2$

(b) Maximum and minimum values using the First Derivative Test

To classify the critical points, analyze the sign of $y'$ around $x = -3$ and $x = 2$.

1. Intervals to test:

   • $(-\infty, -3)$
   • $(-3, 2)$
   • $(2, \infty)$

2. Test a point in each interval:

   • For $x = -4$ (in $(-\infty, -3)$): $y'(-4) = (-4)^2 + (-4) - 6 = 16 - 4 - 6 = 6 > 0 \quad (\text{Positive})$
   • For $x = 0$ (in $(-3, 2)$): $y'(0) = (0)^2 + 0 - 6 = -6 < 0 \quad (\text{Negative})$
   • For $x = 3$ (in $(2, \infty)$): $y'(3) = (3)^2 + 3 - 6 = 9 + 3 - 6 = 6 > 0 \quad (\text{Positive})$

3. Interpret the results:

   • At $x = -3$: $y'$ changes from positive to negative, so $x = -3$ is a local maximum.
   • At $x = 2$: $y'$ changes from negative to positive, so $x = 2$ is a local minimum.

Maximum and Minimum Values:

1. Find $y$ at $x = -3$ and $x = 2$: Substitute into $y = \frac{1}{3}x^3 + \frac{1}{2}x^2 - 6x + 8$.

   • For $x = -3$: $y(-3) = \frac{1}{3}(-3)^3 + \frac{1}{2}(-3)^2 - 6(-3) + 8$ $y(-3) = \frac{1}{3}(-27) + \frac{1}{2}(9) + 18 + 8$ $y(-3) = -9 + 4.5 + 18 + 8 = 21.5$

   • For $x = 2$: $y(2) = \frac{1}{3}(2)^3 + \frac{1}{2}(2)^2 - 6(2) + 8$ $y(2) = \frac{1}{3}(8) + \frac{1}{2}(4) - 12 + 8$ $y(2) = \frac{8}{3} + 2 - 12 + 8 = \frac{8}{3} - 2 = -\frac{10}{3}$

Final Answer:

• Critical points: $x = -3$ and $x = 2$
• Local maximum: $y = 21.5$ at $x = -3$
• Local minimum: $y = -\frac{10}{3}$ at $x = 2$

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Would you like further details or help with a visualization?

Related Questions:

1. What is the concavity of $y$ at the critical points?
2. How would the graph of $y$ look around $x = -3$ and $x = 2$?
3. What is the second derivative of $y$, and what does it tell us?
4. Are there any points of inflection for $y$?
5. What is the behavior of $y$ as $x \to \pm \infty$?

Tip: The First Derivative Test is a simple way to classify critical points, but the Second Derivative Test can often confirm your findings with less work!